**An Alternative Derivation of the Static Dry Adiabatic **

**Temperature Lapse Rate**

**Hans Jelbring **

**BSc, meteorologist, Stockholm University, Civil engineer, electronics, Royal Institute of Technology, Stockholm, PhD, institution of Paleogeophysics & geodynamics, Stockholm University**

**Abstract**

The ”static” dry adiabatic temperature lapse rate is derived for a hypothetical energetically isolated model atmosphere lacking advection and convection. The method of derivation is to investigate the energetic situation for two small equal atmospheric air masses at different altitudes in a vertical column of air. The difference of total energy between these masses is calculated. The ideal gas law is assumed to be valid. This derivation is just one version of several others but might be easier to understand for laymen. The adequate theory needed should have been learnt at high school in natural sciences.

**Background**

The “dynamic” dry adiabatic temperature lapse rate can be and has been derived in different ways and it turns out to be dT/dz = –g/Cp* where g is the gravity constant and Cp* is the heat capacity of air at constant pressure. By “dynamic” it is meant that the derivation is based on following an energetically isolated “air parcel” of constant mass when moving vertically which is the reason for the epithet “dynamic”. One method to derive the result is to use the equation of state P = ρRT, take its total derivative and do a lot of calculus ( ref 1).

It has been shown by Jelbring 2003 (ref 2) that the an energetically closed planetary atmosphere under the impact of gravity which is allowed to come to rest for a long time (no winds and no temperature inversions) has to develop a “static” dry adiabatic temperature lapse rate that is equal to the “dynamic” one mentioned above. That derivation rests on the use of first principle physics (1:st and 2:nd law of thermodynamics).

This derivation considers two air parcels of equal and suitable mass (a billion molecules) which have to carry an equal amount of total energy regardless of their altitude if an adiabatic condition is assumed. It seems that the proof delivered in reference 2 has been hard to understand both by professional scientists and by laymen. Hence, the major reason for writing this article using an alternative method for the derivation. This is to make it easier for anybody to grasp the physics behind the observed temperature (energy) structure that can be observed in our atmosphere and in an even more illuminating way in the Venusian atmosphere.

It is conceptually important to accept that the “dynamic” and “static” dry adiabatic temperature lapse rates are approximately identical. The kinetic (vertical) energy in the first one is very small compared to other energies involved and can be omitted. The existence of the “static” adiabatic temperature lapse rate directly implies that there has to be a substantial “Greenhouse Effect” (GE) on any real planetary atmosphere as long as there exists agents in the atmosphere that are able to emit IR radiation to space from altitudes above the surface of the planet. Observational evidence shows this to be the case in all known planetary atmospheres. These agents can be any solid or liquid matter suspended in the atmosphere (dust, clouds and salt particles) and also so called “greenhouse gases”.

**Methodology **

The energy states of two air parcels in a vertical air column with identical masses m1 = m0 and m2 = m0 at two locations (L1 and L2) are investigated. The altitudes of L1 and L2 is z1 and z2 where z2 > z1.m1 and m2 are carrying the total energies E1 and E2. ∆E = E1 – E2 is investigated and described with the help of mathematical formulae. The adiabatic relationship is found by setting ∆E = 0.

The following assumptions are made:

- The atmosphere is energetically insulated.
- There are no winds in the enclosed atmosphere.
- The atmosphere consists of a mixture of ideal gases and the Ideal Gas Law is valid.
- Gravity (g) is considered constant at L1 and L2.
- Cp* = Cv* + R* The star indicates that the dimension is (Joule/(kg K)).

Cp* = 1004, R* = 287 and Cv* = 717 Joule/(kg K) in air.

**Statement: **

The static dry adiabatic temperature lapse rate dT/dz = – g/Cp* will develop in any model planetary atmosphere that is insulated from energy input and output at the surface and through a concentric spherical shell that surrounds the troposphere.

**Proof:**

The following equations describe the energy situation for m1 and m2 both equal to m0.

E2 = E1 + m0 g ∆z + m0 Cv* ∆T + (P1 V1 – P2 V2) ( 1)

∆E = m0 g ∆z + m0 Cv* ∆T + (P1 V1 – P2 V2) (2)

The first term at the right hand side is gravitational potential energy difference. The second term is the increase (actually decrease since ∆T is negative) in molecular kinetic energy and the third term is the change in work done on the atmosphere by m1 and m2 at the two static locations under consideration.

It follows from P1 V1 = m0 R* T1 and P2 V2 = m0 R* T2 that

(P1 V1 – P2 V2) = m0 R* (T1-T2) where ∆T = T1-T2. Enter this into equation (2) and we get

∆E = m0 g ∆z + m0 Cv* ∆T + m0 R* ∆T (3)

R* = Cp* – Cv* gives

∆E = m0 g ∆z + m0 Cv* ∆T + m0 Cp* ∆T – m0 Cv * ∆T or

∆E = m0 g ∆z + m0 Cp* ∆T (4)

The definition of an adiabatic energy situation is given by ∆E = 0 which leads to

g ∆z + Cp* ∆T = 0 or (5)

∆T/ ∆z = -g/Cp*

When ∆z goes to zero, ∆T/ ∆z goes to dT/dt. Therefore we get:

dT/dz = -g/Cp*; dT/dz = – 9.81/1004; dT/dz = -0.00978 K/m or 9.78 K/km

**Some comments**

It is obvious that the temperature profile at a specific location depends on latitude (solar insolation), orography, water vapour, tides and whether the surface happens to be land or ocean. In other words, the temperature profile at a specific location depends on several physical factors but the most important one is very probably the tendency for Earth’s (and the atmosphere of any planet) to seek an energetic equilibrium. This is what the existence of the static dry adiabatic temperature lapse rate shows and that is why it is important. On earth the dry adiabatic temperature lapse rate is best verified by observations during afternoons after a sunny day or in Antarctica where very strong katabatic winds prevail. In these situations dT/dz is close to -9.8 K/km. On the other hand the US 1976 Standard Atmosphere has a temperature lapse rate of -6.5 K/km under 10000 m altitude and is isothermal at the tropopause. Hence, the static dry adiabatic temperature lapse rate is directly confirmed on earth during specific physical conditions that don´t always prevail. But it does evolve close to Earth’s surface every sunny day over land.

The absolute best observational evidence of the impact of “energy dissipation” according to the second law of thermodynamics is found in the Venusian atmosphere where the lapse rate is close to the theoretical adiabatic one from the surface to about 40 km altitude. The reason is simply its thickness which facilitates an approximate even total energy distribution per mass unit in its atmosphere, from the surface to its upper troposphere.

It has been known for a 100 years that the surface of Earth is warmer than it should have been if radiating as a “black body” into space. This is the reason why a “Greenhouse Effect” (GE) has been suggested. The accepted value of GE is 33 K which is used by NASA. The value can be challenged. There is no doubt that most of this effect can be traced to the physical process of energy/mass equalization in the atmosphere of earth. It follows from this insight that other physical processes affecting GE must have a relatively small impact. It is possible today to quantitatively examine how much of the GE is created by a number of physical processes but there is little initiative to do so since the cause of GE creation has been almost solely attributed to the impact of “greenhouse gases”. IPPC should be more than ashamed to translate a complex scientific climate problem into a reductionistic one variable political statement relating to the unproven danger of carbon dioxide increase in our atmosphere.

**References:**

- Holton J.R., An Introduction to dynamic Meteorology, second edition, Academic Press, 1979, pp 47-49
- Jelbring H. “The Greenhouse Effect as a Function of Atmospheric Mass.”, Energy & Environment, Vol 14, 2&3, 2003, 351-356

IPCC’s argument-from-ignorance is all it has.

I make the Venusian lapse rate about 12 but it isn’t air. Figure comes from the computation of the line to the surface.

Figure 2

If your packets of air are in rigid, sealed and perfectly insulated capsules, you can raise them or lower them in a gravitational field as much as you like and they will undergo no change in temperature, even though you have changed their gravitational potential energy. We can see, therefore, that the change in gravitational potential energy of air with altitude does not explain the adiabatic change in air temperature with altitude.

[Reply] Watching Robert Brown waving his arms around catching butterflies in jam jars is amusing, but unedifying. What we have here with Hans’ paper is a properly presented scientific proof. Please address it. – RogCanspeccy, here’s the full explanation of Robert Brown’s jam jar fallacy from Trick on the WUWT ‘perpetuum mobile’ thread:

Trick says:

January 22, 2012 at 2:14 pm

Robert Brown says at 1/22 8:04am:

“Therefore consider the “jar” argument once again.”

Ok. This may be long, fair warning. Pour a libation of choice, may only interest Robert.

“Grab a jar of air at the bottom of your “equilibrium” room with a DALR. Its pressure is a bit higher and temperature is a bit higher than air at the top.”

Ok. Willis’ cv equilibrium not mine; I haven’t added anything to Willis’ adiabatic GHG-free equilibrated tall air column under presence of inexplicable gravity field. I add for clarification assuming your permission: an open Bjar, b for bottom, and its separate lid actually. Assume here both jar & lid were inside Willis 1 cv and in energy equilibrium (i.e. no “hot” jars or “cold” lids suddenly cross cv.)

“Grab a second jar of air at the top, where pressure and temperature are both a bit lower.”

Ok. Tjar, t for top. With same clarification as Bjar above.

“So far, we know nothing about the density of said air.”

In my view this is incorrect Robert. Within Willis cv in gravity field, can define h as 0 at bottom of cv. We know PV=nRT (for all of h) and thus rearranging CAN find density (of h) = n/V = P/RT (all of h).

“Nothing in fluid dynamics requires a fluid to be compressible….I don’t know how you want to idealize your “air”…. Gravity, however, is the same at the top and at the bottom.”

Ok. Lotsa’ thought experiments exist using incompressible fluids. Not my air – I’ll use Willis’ GHG-free atmospheric non-idealized air.

“Moving the jar at the top to the bottom… gently enough that one doesn’t slam the fluid molecules…”

Ok. Before Tjar is moved down to a decreased h, I’ll assume with Robert’s permission, the lid has been closed completely and we have a second body (Tjar + gas) within the cv, nothing across Willis’ 1 cv.

“The fluid in the two jars is not in thermal equilibrium. Surely you agree?”

YES for Tjar moved down (deliberately –w/no time to re-establish equilibrium) and “gently” thus heat flows before re-establishing cv equilibrium since Tjar is touching Willis air.

NO for 3rd body closed Bjar still presumably at the bottom, Robert hasn’t moved it yet. BJar is still in cv energy equilibrium near h=0.

“…the two jars have any sort of thermal pathway” opened between them, they will come to thermal equilibrium at a temperature in between T_b and T_t, one we can actually compute as it will depend on their heat capacities at constant volume which depend on N only.”

Robert didn’t write it, but I presume just a trivial oversight, that closed completely Bjar was then raised, deliberately and gently as Tjar was lowered. So YES, the thermal path way being established between T&Bjars at some h >0, establishes the jars new equilibrium (Caballero’s mean KE) temperature (of h) thru heat flow that ceases eventually.

“Note well — extremely well, if you please — that the heat capacity of the gas gets no contribution from gravity.”

Ok. Noted extremely well.

“From this alone you should be able to see that gravity is decoupled from the problem, because your belief that dT/dh (where h is the height of a jar) gets a contribution from gravity at equilibrium is incorrect.”

No. Robert & I part views here. Total Tjar energy = its KE+PE & is not ever changed; it is always conserved & constant. No total energy can be harmed or created by moving the jars. Robert is incorrect to write gravity field can be decoupled (i.e. ngh term ignored). Accepting that means no conservation of total energy & then a thermo law is broken.

Since gravity potential energy cannot be decoupled & conserve TE, the PE of Tjar is reduced (of h), the PE of Bjar is increased (of h). When the established thermal pathway allows Tjar and Bjar to reach energy equilibrium, the KE of the molecules inside each will be the same as the KE of the molecules in Willis’ cv air at h. (After Caballero, we are talking KE mean here, actually each jar has a parcel of molecules.)

After Caballero, each of those jar KE means are the respective jar equilibrium temperatures (of h). Tjar (KE +PE) = constant = Bjar (KE + PE) = Willis’ air (KE + PE) (all of h). In the Tjar case, mean KE reduces as PE increases. In the Bjar case, mean KE decreases as PE increases.

“…your belief that dT/dh (where h is the height of a jar) gets a contribution from gravity at equilibrium is incorrect. dT/dh = 0 (where the derivative should be a partial derivative btw). Adiabatic lapse is a non-equilibrium state.”

In my view, my belief is correct since no thermo law is broken. My view/belief invokes immutable conservation of energy in the 1 cv (as others have stated in this thread). dT/dh is not = 0 which it cannot be and keep total energy = constant. (as an aside, not a partial derivative I think, I could be wrong – trivial matter). Here in presence of inexplicable gravity field, adiabatic lapse IS an equilibrium state. Presence of this g field results in the ngh = PE term necessarily and always materializing. Curiously, this stratification also enables buoyancy – a matter for another day.

“Even if we leave them where they were in the first place and simply run a heat-superconducting wire between them, we expect heat to flow in the wire because they are not in thermal equilibrium and thermal equilibrium does not depend on where you are!”

This gets a little more interesting. Note extremely well here, if Robert runs his heat-superconducting wire between T&Bjars in their original positions (at hugely different h), in my view nothing of interest happens. (NB: don’t need exotic super-conductor wire here, an ordinary works, but maybe SC enables an easier-to-see solution). This (now 4th body) wire is in thermal contact with Willis’ cv gas all the way from top to bottom. At system equilibrium in the cv, that wire will be in thermal contact with the KE of Willis gas at each continuous h so the T of that SC wire will necessarily vary at each h. By immutable thermo law!

The bottom terminal end of the superconducting wire (ending in Bjar) will be at thermal equilibrium of Bjar KE and vice versa all along its length (of h) in Willis air KE to Tjar KE which is also in energy equilibrium. There is never any 1 body in the system cv that is not in equilibrium thus no energy can flow between any of the4 bodies forever – esp. with conservation of energy invoked.

“Once again, you are stuck.”

No, conservation of total energy rescued me. As it has for many of my exploits.

“You have a closed system that — you claim — is in thermal equilibrium with two different temperatures, one at the top and one at the bottom. Yet that means that an ordinary thermometer carried from the bottom to the top would read two values, and, just as would be the case for any two systems whose temperature is measured, means that heat would flow between them if it could.”

Yes, two thermometers will read two different temperatures (KE) even in the equilibrated CV because of conservation of energy in a gravity field. Heat will now NOT flow in Willis’ cv, since all 4 bodies is – at every continuous h in stratified T- in energy equilibrium. No irreversibility. Entropy is constant. No heat flow.

“Air is a conductor of heat and heat will flow from the bottom to the top until the system is in equilibrium.”

Yes initially & stop flowing heat (no energy flow forever here) when balanced – just don’t go across Willis cv, Robert. Same system, different h’s. There are NOT two separate cv.s, here one hot and one cold. Only Willis’ original 1 cv, which in energy equilibrium & touching everywhere.

Yes GHG-free air conducts KE (i.e. T) to the wire from each h at molecular mean KE (Caballero’s temperature defn.), thanks for pointing out. Willis cv can achieve stratified T equilibrium, all 4 bodies touch in energy equilibrium at continuous h’s thru the wire, no Perpetuum Mobile. Nothing crosses Willis’ cv. A drinking bird (got a laugh out of me) could be added but it stops drinking energy at equilibrium.

“”Watching Robert Brown waving his arms around catching butterflies in jam jars is amusing, but unedifying””

Robert just said this over at WUWT

Personally, I think the DALR is caused by the greenhouse effect and gravity, working together to maintain the heat differentials that drive the troposphere. Heresy, I’m sure, on this blog, but there it is.

rgb

Defiantly confused, and definitely coming across.

Can you see the problem Canspeccy? Robert doesn’t understand the changing density of a compressible gas. In fact, he then asked trick:

Robert Brown says:

January 22, 2012 at 3:02 pm

In my view this is incorrect Robert. Within Willis cv in gravity field, can define h as 0 at bottom of cv. We know PV=nRT (for all of h) and thus rearranging CAN find density (of h) = n/V = P/RT (all of h).

I don’t know whether to do an itemized reply all at once or reply as you make mistakes. I’m also not sure what “cv in gravity field” stands for.

What is cv?Also, if you look at the explicit context of my remark, it was fluids, and I later said that if one presumes an ideal gas… The point being that real gases have compressibilities that are not idea[l].Trick replies:

Trick says:

January 22, 2012 at 3:15 pm

Robert Brown says at 1/22 3:03pm:

“What is cv?”

Control volume.

—————————————————————

Physicists farting around with conceptual ideal gases in thought experiments is one thing. When they start telling us that that is how the real atmosphere is, they are bullshitting. Meteorologists and engineers deal with real gases in real gravitational fields.

“the tendency for Earth’s (and the atmosphere of any planet) to seek an energetic equilibrium. This is what the existence of the static dry adiabatic temperature lapse rate shows and that is why it is important”

Precisely.

A negative system response occurs, acting against ANYTHING that tries to shift the atmosphere away from the sun and pressure induced dry adiabatic lapse rate.

But one does have to accept SOME climate consequence if the lapse rate isn’t allowed to change much if at all.

Hence my long standing proposal concerning surface pressure redistribution and latitudinally shifting climate zones.

Then the only question is as to how far a bit more CO2 can shift them as compared to natural solar and oceanic influences. I’d say miniscule compared to the changes observed from MWP to LIA to date.

It is all coming together nicely now that we have support for the ‘old’ pre GHG theories about planetary surface temperatures from Jelbring, Nikolov and Zeller and apparently a few others too whose names have not yet come to the fore.

The improved planetary surface measurements and the corrected S – B equations provided by N & Z are the catalyst for the revived interest in that old understanding and the creation of NTE/ATE is a very useful step forward.

I was lucky not to have absorbed the radiative nonsense. All my work has been based on the earlier theories involving the Gas Laws.

Take a column of air of height h in thermal equilibrium with a total energy content of e that is supplied through conduction/convection/radiation from the ground.

1kg of air at the bottom of the column will have the same total energy as 1kg at the top of the column. 1kg of air at the bottom of the column will have a volume v that will be less than the volume of 1kg of air at the top of the column. 1kg of air at the bottom of the column will have a temperature t that is greater than the temperature of 1kg or air at the top of the column.

If we magically increase gravity, the total energy in the column will remain the same. 1kg of air at the bottom of the column will still hold the same energy as 1kg at the top of the column. However, 1kg of air at the bottom of the column will now occupy a smaller volume than before we increased gravity as will 1kg of air at the top of the column. As they both hold the same amount of energy in a smaller volume, the temperature of both will have risen.

“The definition of an adiabatic energy situation is given by ∆E = 0 “How so? Adiabatic simply means that the region does not gain or lose heat (assumption 1). It doesn’t say that two arbitrary parcels of fluid must have the same E. I see nothing in your assumptions to justify this.

May I remind of the following discussion:

https://tallbloke.wordpress.com/2012/01/01/hans-jelbring-the-greenhouse-effect-as-a-function-of-atmospheric-mass/#comment-13413

and a few subsequent posts.

Seems the oldfashioned DALR and WALR have been rebranded in Dynamic Adiabatic Lapse Rate

or more specific DDALR and DWALR.

The Environmental lapse rate of the static atmosphere with a temp. gradient of 9,8K/km

(indifferent Environmental lapse rate) has been rebranded Static Adiabatic Lapse Rate.

No new names yet for an unstable or stable Environmental lapse rate.

How an atmosphere that receives a lot of energy from the sun and loses about the same amount at the top can have a lapse rate that is called ADIABATIC escapes me at the moment.

Nick Stokes says:

January 25, 2012 at 10:23 am

“The definition of an adiabatic energy situation is given by ∆E = 0 “ (stated by Jelbring)

“How so? Adiabatic simply means that the region does not gain or lose heat (assumption 1). It doesn’t say that two arbitrary parcels of fluid must have the same E. I see nothing in your assumptions to justify this.”

Thanks for your interest and your straightforward statement. Let me try to explain.

A. Your first sentence is correct of you exchange the word heat to energy. It is energy that cannot be destroyed or created from nothing according to the 1:s law of thermodynamics.

B. The second law of thermodynamics assure that the energy content per mass unit in any part of the isolated atmospheric mass by time gets equal. Just wait long enough. Friction energy is levelled our [off?] after about two weeks (no winds). Conduction processes are the slowest one when winds have stopped blowing. Any temperature inversion will then disappear gradually until the static adiabatic temperature lapse rate is reached in the whole atmosphere.

C. In this article I have chosen two masses with equal mass = m0 for simplicity. Since they both have an equal mass they have to posses equal total energy when the energetic equilibrium is reached. That is when ∆E = 0. Hope this explanation is to some help.

Tides are assumed not to affect the inclosed insulated atmospheric mass.

tallbloke says:

January 25, 2012 at 8:35 am

This kind of discussions become much simpler if you imagine a parcel of air contained in a superlatex balloon. This latex can let the air inside expand and shrink without any restrictions

(no resistance when expanding, no work done when shrinking)

If you let the balloon rise, it will expand, doing work against the surrounding air and cooling with the DALR. Total energy remains the same, because the potential energy increases.

At eg. 2km altitude the temp has decreased 19,6K.

Now bring it back to the surface, balloon shrinks, temp. increases.

Back at the surface the temp and the volume are the same again as when the trip started.

This is because this process is supposed to be adiabatic.

If there is energy exchange with the surrounding air the picture changes.

BenAW says:

January 25, 2012 at 11:33 am

“How an atmosphere that receives a lot of energy from the sun and loses about the same amount at the top can have a lapse rate that is called ADIABATIC escapes me at the moment.”

You are correct that it cannot. One reason to write about static dry adiabatic temperature lapse rate is to explicitly identify a very strong physical process (energy dissipation) that haven´t been properly understood within meteorology and climatology.

Why the energy dissipation process (an array of physical processes) has to be important when treating earth´s climate system is a topic which awaits to be discussed seriously. Just notice that earths atmosphere can be seen as an approximate steady state system in regard to long term energy variations (longer than a year).

(steady state: energy input = energy output). As a matter of fact Willis´ suggested model (with 1000 minor suns) is a steady state system which results are useless for applications in real atmospheres. It was basically constructed to spread confusion instead of scientific enlightment and he really succeeded with that plus delivering “en masse” ad hominem at the same time. His claim as being a defender of science seems as a joke to me.

Hans Jelbring

Hans,

“The second law of thermodynamics assure that the energy content per mass unit in any part of the isolated atmospheric mass by time gets equal. Just wait long enough.”Well, first you said it followed from the definition of adiabatic. Now you say that it is a consequence of the second law, and is only true after some time. I think if you want to write this down as a line of a proof, it needs proof itself, starting with a proper explanation.

[Reply] Nick, Read the references. See Jelbring 2003.Some help with the laws of thermodynamics for those obviously in need of it:

http://www.emc.maricopa.edu/faculty/farabee/biobk/biobookener1.html

Laws of Thermodynamics

Energyexists in many forms, such as heat, light, chemical energy, and electrical energy.Energyis the ability to bring about change or to do work.Thermodynamics is the study of energy.First Law of Thermodynamics:

Energycan be changed from one form to another, but it cannot be created or destroyed. The total amount ofenergyand matter in the Universe remains constant, merely changing from one form to another. The First Law of Thermodynamics (Conservation) states thatenergyis always conserved, it cannot be created or destroyed. In essence,energycan be converted from one form into another. Click here for another page (developed by Dr. John Pratte, Clayton State Univ., GA) covering thermodynamics.The Second Law of Thermodynamics states that “in all

energyexchanges, if noenergyenters or leaves the system, the potentialenergyof the state will always be less than that of the initial state.” This is also commonly referred to as entropy. A watchspring-driven watch will run until the potentialenergyin the spring is converted, and not again untilenergyis reapplied to the spring to rewind it.———————-

Nick, I’m not seeing the words

‘heat’or‘temperature’in these definitions, so please could you clarify your point. Thanks.markus says:January 25, 2012 at 8:36 am

“”Watching Robert Brown waving his arms around catching butterflies in jam jars is amusing, but unedifying””

Robert just said this over at WUWT

Personally, I think the DALR is caused by the greenhouse effect and gravity, working together to maintain the heat differentials that drive the troposphere. Heresy, I’m sure, on this blog, but there it is.

rgb

Defiantly confused, and definitely coming across.Well spotted Markus!

Maybe soon he’ll realise that GHG’s predominantly cool the upper atmosphere , and pressure+sunshine predominantly warms the near surface atmosphere.

TB,

My point is that there is nothing in the statement of the second law that says that the energy E, as written by Hans, has to be the same for two different masses in two different places. He claims it can be deduced – well, lets see it done.

If you want a demo of why I think it is wrong, consider Eq 2 applied to water. It’s essentially incompressible, so you’d leave out the PV terms. If E1=E2 is a consequence of the second law, it will apply to liquids as well as gases. That means that a column of water would have to conform to a lapse rate.

Nick, it will apply to a column of seawater, because it contains oxygen, or all the fish would die.

Intuitively you would think that it would be a smaller lapse rate. Interestingly, Graeff’s experiments found a larger lapse rate for water than air. Go figure.

Water is more efficient at convecting however, so the sea bottom is full of colder and therefore denser water. I think the lapse rate will likely be found to be responsible for entraining solar energy into the depths to some as yet unknown extent.

Nick Stokes says:January 25, 2012 at 12:13 pm

TB,My point is that there is nothing in the statement of the second law that says that the energy E, as written by Hans, has to be the same for two different masses in two different places. He claims it can be deduced – well, lets see it done.

The entropy embodied in the second law ensures equipartition of energy after sufficient time has elapsed to establish equilibrium..

The Robert Brown comment in full:

Robert Brown says:

January 24, 2012 at 11:09 pm

What maintains the lapse rate temperature difference?”

Mostly gravity.

Mostly gravity plus the differential heating and cooling. Move your house to Antarctica and look up mid-July. See all that sky that is warmer than you are?

But generally, I agree with your reply. As I stated, my objection is specific to EEJ

[Jelbring 2003 E&E)]— the DALR is not a stable thermal equilibrium, which is precisely what EEJ asserts. I’m not suggesting that there is no ALR, as a general rule, only that a) it isn’t precise, constant, ubiquitous; b) that it depends on differential heating and cooling and active transport in the atmosphere, and goes away when you stop heating the ground underneath it. The layer where the DALR approximately holds is the troposphere, the layer with vertical convective mixing, and it goes away as the ground temperature drops — making it look a whole lot more like an effect, rather than a cause, of warmer ground temperatures.Personally,

I think the DALR is caused by the greenhouse effect and gravity, working togetherto maintain the heat differentials that drive the troposphere. Heresy, I’m sure, on this blog, but there it is.rgb

[Note]My expansion of the reference andbold. – TB.“The entropy embodied in the second law ensures equipartition of energy after sufficient time has elapsed to establish equilibrium..”Evidence? Links? Anything?

Nick, try this, I haven’t had time to study it, and have a lot to do, but this should keep you busy until I get back.

http://www.math.uconn.edu/~kconrad/blurbs/analysis/entropypost.pdf

TB,

No, it didn’t take long. There’s nothing in that paper to justify your claim.

Nick, I’ll believe you are serious about needing entropy proved when you start levelling the same complaint at Robert Brown and Willis on their threads which accuse N&Z and Jelbring of breaking the second law and wrongly asserting isothermality. Until then, I’m afraid I’m taking the view that you are trolling and will no longer respond.

Cheers

Rog

Well. TB, I think Willis and Robert Brown had it right there. But here we have a purported proof with a totally unexplained step. And it’s the line that is vital to the result.

Hans says: “Your first sentence is correct of you exchange the word heat to energy. ”

No, his first sentence is correct as it was stated. Adiabatic refers to heat only. It is common to talk about “adiabatic compression” where no heat is exchanged, but the energy of the system changes because of work done.

In particular, take an ideal gas and do an adiabatic compression. Then do an adiabatic free expansion back to the original volume. This completely adiabatic process will raise the temperature and the energy of the system.

Nick, whoever you think is right or wrong, both parties rely on entropy for their view. So where is the equality of treatment by you?

As requested earlier, please explicitly state the laws of thermodynamics you are working from.

Definitions please, no more thread bombing until they are forthcoming..

Nick, I tend to keep quiet for various reasons but broadly I don’t know the answer so I am waiting.

Both Brown and Willis produced fantasy limit case worlds dealing in singularities so that zeros and infinity are present. All manner of maths weirdness is likely.

Taking the Brown case, sanely it is a system in transient, will be radiating, now figure that out.

In addition yes, the bar can conduct and new heat will appear *without* breaking any laws, pea and thimble problem which Brown didn’t notice. The point is it is a closed system which does no work (he put it in impossible isolation).

This is identical to the “work” done by atoms wobbling around, exactly nothing happens externally.

If you could harness an atom, pull the energy out of it, the energy is transferred to elsewhere and rather likely will end up leaking back, if not you just cooled it to 0K, moved the energy out. Now was that to outside or inside?

I remain unconvinced a gravity induced gradient is significant even though a theoretic gradient seems possible.

Well Doc I see you have invented a universe where convection doesn’t take place.

“2. 2.There are no winds in the enclosed atmosphere.”

Good thing too in my judgement because convection needs work to take place doesn’t it!

Really, anyone can invent there own gedanken universe to demonstrate something.

E.g.

I could invent a closed system that was initially composed of a diffuse cloud of particles in

thermodynamic equilibrium.

Now that system would be near maximum entropy. However the addition of gravity starts to cause the particles to compress and voila I now have a system like the solar system or a galaxy or the universe even and I have reduced the entropy of the system without the addition of energy or increasing the entropy of another system.

Perhaps someone could say that this proves gravity can reverse the flow of entropy in a closed system and the Thermodynamic laws need revising.

However, in the real universe we inhabit we cannot create such a system in such an inital state. Perhaps a supreme, all powerful being could but I am not holding my breath! The existence of such a being would invalidate all known physical laws in any event.

So we have to be careful with gedanken experiments. I tend to the view that if such a system is proposed, that is in breach of thermodynamic laws, that it is either in error or could not ever be created in the universe we inhabit.

So in your universe convection (and therefore work) doesn’t take place getting round the 2nd Law and in mine gravity can reverse the flow of entropy.

Why didn’t you allow convection and just say the 2nd Law is not appicable in my gedanken universe, it would be a lot simpler?

So we have created two universes ‘proving’ different things but what does this really have to say about the real universe?

Alan

Well, Rog, again I’m just asking for the justification of a step in the proof – I’m not “working from” a particular version of the laws of thermo.

But the second law has been quoted here – I’m happy with the Clausius version:

“No process is possible whose sole result is the transfer of heat from a cooler to a hotter body.”(tho I’d settle for any of the classic versions on that page).

tchannon says: January 25, 2012 at 2:34 pm“Nick, I tend to keep quiet for various reasons but broadly I don’t know the answer so I am waiting.”

I wrote a blog post on that. The simplest proof in my view is that every temperature gradient is constantly generating entropy according to the formula

dE_v/dt = k/T^2 (∇T)^2

where E is entropy per unit volume and k is the conductivity, which here reduces to molecular conductivity. So a steady system with a temperature gradient is constantly creating entropy. Gravity etc do not enter the argument.

So you can’t have a steady state with a temp gradient, unless k=0. Of course in the real atmosphere there is a lot else going on, and in fact turbulence (for lapse < DALR) acts as a heat pump, countering the entropy gain at an energy cost.

“So a steady system with a temperature gradient is constantly creating entropy. Gravity etc do not enter the argument.”

Well that’s where you are going wrong, you won’t get a temperature gradient in a system at equilibrium unless it is in a gravitational field.

Anyway, thanks for the thermo 2nd law definition you’re using. I’ll add it to the collection. I’ll just note that since Clausius in the 1860’s scientists have been formulating the laws of thermodynamics in energy terms not heat terms as I quoted above. Seems the news didn’t reach Duke University or your corner of Australia yet…

Alan Millar says:

“Good thing too in my judgement because convection needs work to take place doesn’t it!”

Well it depends on the definition or type of convection.

This mainstream Climate Science publication defines convection as the turbulent movement of the air parcel when the buoyancy force is greater than the gravitational force.

In this casework is done.

(See bottom of page 13)

If the air parcel remains at rest or moves with constant speed ( same thing according to Newton) then no work is done.

This seems a puzzle to most!

Surely they say, if the parcel moves up against the gravitational field it gains potential energy.

What they don’t see is the surrounding air of the same density moving down to fill the gap left.

The two processes cancel.

Both types of air movement conserve energy which originally comes from solar radiation.

This pumping up the level of the troposphere gives rise to the diurnal bulge.

When nighttime comes the troposphere level shrinks but there is more than enough energy stored to buffer the climate temperature till the sun driven phase restarts.

This effect Academician Oleg Sorokhtin refers to as ” The adiabatic theory of greenhouse effect”

One gripe I have is the tendency to keep calling it the ‘greenhouse effect’.

It has absolutely nothing in common with what most folk think of as the greenhouse effect and this only leads to confusion.

Click to access ch2_brasseurjacob_Jan11.pdf

Bryan: Yes, that;’s why N&Z have given it a theory neutral name : ATE: Atmospheric Thermal Enhancement

Tallbloke said: “Can you see the problem Canspeccy? Robert doesn’t understand the changing density of a compressible gas”

It isn’t a problem. I spoke of “packets of air in

rigid, sealed and perfectly insulated capsules” (the insulated bit was superfluous, since we are talking about adiabatic changes in temperature).As long as the volume of the

rigid, sealed capsule remains constant the volume of the air within will not change in volume either, and therefore, wherever the capsule is located in the gravitational field, i.e., whatever the gravitational energy potential of the air within, the temperature will remain constant.Can you not see the error in your reasoning now?

Heh. Canspeccy. Re-read Trick’s reply to Roberts butterfly jars armwaving and get a clue. I’m off duty for a bit.

Bryan says

January 25, 2012 at 4:07 pm

“If the air parcel remains at rest or moves with constant speed ( same thing according to Newton) then no work is done.”

What has that got to do with convection in a planetary atmosphere? It is just obfuscation.

Believe me if you think you can move a planet’s air without doing work then then you shouldn’t really be commenting on Thermodynamics.

Why do you think Jelbring put the second condition into his gedanken universe? If convection is entirely neutral as far as entropy goes why does he need to specify it?

Alan

I posted the following over at WUWT on Dr. Brown’s thread “Refutation of Stable Thermal Equilibrium Lapse Rates”. Thought it might apply here as well…

I have been following these threads, lurking from the sideline, for a while. Well, the time has come to add my $.02 worth. I have been studying meteorology for 40 some-odd years now & I shake my head in seeing some of the most common properties of the atmosphere being missed in these threads as it applies to these ‘thought experiments’.

1) if the several km-long tube is horizontal & the perfectly dry air is at a constant temperature throughout & is moved to the vertical, the dry adiabatic gradient will be produced (warm at the bottom, cool at the top w/ approx 8C/1000m gradient in between) due to the ‘work’ of gravity creating a pressure gradient to the compressible gas. Notice, no gradient will be produced if water is used instead of gas because water is non-compressible so no work will be done. If no heat is added or removed to the gas, the column will be in a neutral buoyant state (and will stay that way!!) – if a parcel of air is moved vertically by an outside force, it’s temperature will change to reflect the change in pressure but will still be the same temperature as it’s surroundings.

2) as to the experiment with the thermal conductive wire at the base & top of the tube, the author here is incorrect. If the wire moves heat from the bottom of the tube (the base cools) to the top of the tube ( the top heats), presuming, as the author, Dr. Brown, says, “…save to note that the internal conductivity of the ideal gas is completely neglected.”, the heat from the *local* area of the wire is all that will be moved from the bottom to the top ***and nothing else*** . Why, you ask?? In moving the heat from the bottom of the tube to the top is causing the lapse rate to become **more stable** – cool at the bottom with warm air above is an inversion which inhibits vertical mixing!! THAT is why the engine will not work as it is set up.

Just a few thoughts…

Jeff

Alan Millar

I said

“If the air parcel remains at rest or moves with constant speed ( same thing according to Newton) then no work is done.”

You replied

“What has that got to do with convection in a planetary atmosphere? It is just obfuscation”.

Well look up references to the Neutral Atmosphere

In the absence of TURBULENT convection (still dry air or low constant speed parcels of air) there is a still a lapse rate.

This is what is called the DALR = -9.8K/km

This is what meteorologists call the neutral atmosphere.

The neutral atmosphere can be quite stable particularly at night.

See page 31 and the residual layer.

Temperature is the integral of the number and energy of the particles encountered at the measuring surface over some time period.

Less particles arriving at the measuring surface during that time period means a lower temperature.

Less particles also means a lower pressure.

Thus temperature and pressure are directly linked.

Hence the dry lapse rate.

Gravity produces a pressure gadient.

Hence the dry lapse rate.

Seems like two side of the same coin to me.

Bryan says:

January 25, 2012 at 5:27 pm

“Well look up references to the Neutral Atmosphere

In the absence of TURBULENT convection (still dry air or low constant speed parcels of air) there is a still a lapse rate.

This is what is called the DALR = -9.8K/km

This is what meteorologists call the neutral atmosphere.”

Your understanding of the subject seems to be somewhat lacking here. We are not talking about equal lapse rates or pressures or temperature or anything else.

We are talking about the work performed to move the air in the first place or do you think you can move a particle up a gravity well without work?

Presuming you accept that and that work has been performed, then the 2nd Law says entropy has increased and can never be decreased in a closed system without the creation of energy.

Again why do you think Jelbring has specified his no convection rule in his gedanken universe?

Alan

Surely the “jam jar” experiment can be very easily tried by anyone who is intending to fly anywhere in the next day or two.

Take a simple electronic thermometer (like this and an empty thermos flask on your flight.

I’d suggest not assembling the equipment till you get thru security.

Next, place the temperature probe inside the room-temp flask, with nothing else present, and carefully screw on the lid. Take regular temp readings from this point till the plane lands. Do not leave the thermometer active between readings, in case the probe somehow alters things.

Are people here seriously suggesting that at 30,000 feet, where the gravity-induced temperature is about -30c or lower outside the plane, that the temperature readings inside the flask will have changed?

Really?

The second law of thermodynamics states that when two or more systems interact with each other they will after some period of time reach the same average energy per particle and, by definition, the same average temperature.

As gases are compressible, the gravity gradient causes more molecules on average to reside at the surface, where the field is strongest resulting in a greater density and a higher pressure. The greater density of particles results in a higher temperature at the bottom of a column compared to the top, while the total energy stays constant within the system.

In the real world, input of varying amounts of solar energy over each dt cause convective turbulence and, as a climate is a dynamic system, equilibrium is never achieved. It can be argued that as temperature is an macroscopic intensive variable without system equilibrium it’s use as a global climate metric is invalid.

To really comprehend what is going on we need to use different tools, such as are slowly being developed by the embryonic sciences of non-equilibrium thermodynamics and spatio-temporal chaos. In the meantime we are left with crude estimates and incorrect assumptions.

Alan Millar says:

“We are talking about the work performed to move the air in the first place or do you think you can move a particle up a gravity well without work?”

Alan have a look at a video of the Cartesian Diver – google it

You will be gobsmacked.

“Bryan says:

January 25, 2012 at 6:28 pm

Alan have a look at a video of the Cartesian Diver – google it”

Sorry Bryan i am not wasting my time studying anything which you claim proves that you can move a particle up a gravity well without performing work, or that you can perform work without increasing entropy. You are just doing the classic trick of obfuscation when you have adopted an indefensible position.

Mind you your example of the Cartesian Diver as relevant in any way is some sort of joke right?

If your position is that the Laws of Thermodynamics are faulty just say so.

Indeed, publish your proof and get a Nobel prize.

My statement is as simple as you can get. Perform work and entropy increases, which will lead to the ‘heat death’ of any closed system.

Again, for the umpteenth time, why do you think Jelbring specifically specified no convection could take place in his gedanken universe?

Alan

[Reply] Mind your manners. No work is done because it is cancelled out by descending air since pressure is always maintained by the enforced adaibatic exchange. Hans doesn’t ‘specify’ no convection. It is simply a consequence of the setup of the experiment. See his 2003 paper, and actually read it.@Tenuc:

non-equilibrium thermodynamics and spatio-temporal chaosMy guess is that in the future more simplicity will prevail, and thanks God no chaos at all, but humble common sense, as we revisit the simpler laws of nature, otherwise we will succumb under the rubbles of the falling edifice of of a confused occidental civilization.

Tallbloke says

“No work is done because it is cancelled out by descending air since pressure is always maintained by the enforced adaibatic exchange”

Work done is cancelled out by something you say.

What would that be? Some sort of anti work process like anti matter that cancels out the work done in raising a particle in a gravity well and fools the 2nd Law into thinking no work has taken place?

This Jelbring system lasts beyond the end of the non-Jelbring universe does it? The normal universe suffers ‘heat death’ but this system carries on into infinity moving particles up and down for ever without any energy being input!

If not what kills it?

Alan

[Reply] Is the penny dropping Alan? Are you seeing why in the gedanken experiment set up by Hans convection comes to a halt after a while? It is because energy spent moving air packets up the gravity well is given back to the system by the air packets which have to move down the gravity well to replace them to equalise pressure. Next step; what is the energy state at maximum entropy? and what is the resultant of the proportion of total energy content used by potential and kinetic energy in the air packets at low and high altitudes? Hint: this is where temperature comes into it.“The second law of thermodynamics states that when two or more systems interact with each other they will after some period of time reach the same average energy per particle and, by definition, the same average temperature.”

That is only true for equal pressure.

In any system that has a pressure gradient there will always be a equivalent temperature gradient assuming the energy profile of the system is static overall.

That is because all practical measures of temperature are integrators of both the number and the energy of particles seen at their measuring surfaces. The energy profile of the particles may be as the second law requires, equal throughout, but the temperature will vary with height because of the density changes.

Tallbloke says

” Is the penny dropping Alan? Are you seeing why in the gedanken experiment set up by Hans convection comes to a halt after a while? It is because energy spent moving air packets up the gravity well is given back to the system by the air packets which have to move down the gravity well to replace them to equalise pressure?”

I don’t care about the energy ‘spend’ as the energy in the system is unchanging and nothing is getting spent. I am concerned about the work that is being performed that always leads to a more disordered system.

No I don’t see why convection comes to a halt. That is unless you are saying the system suffers ‘heat death’ like the normal universe with temperatures equalising across the system which I would agree with.

However, I thought you and Jelbring are claiming that there will always be a temperature gradient into infinity.

As far as energy goes the system will have exactly the same amount of energy it had when it was created, just as the normal universe will when it dies.

So what are you saying. That there is a situation where we have a gravity well with hotter gases at the bottom but no convection can take place. What force is stopping these more energetic particles moving?

Alan

[Reply] You’re getting there. Read Jelbring’s two papers. Slowly and carefully.Is there anything wrong with my thought bubble of single (perfectly elastic) atom bouncing up and down in a perfectly insulated, frictionless tube?

At the top of the tube the atom comes to rest (momentarily) and has zero kinetic energy, zero temperature and a lot of potential energy. Gravity accelerates the atom down to the bottom of the tube where it has maximum kinetic energy and heat and zero potential energy. Since heat and temperature are a measure of kinetic energy alone, that establishes the lapse rate, zero temperature at the top, 50% kinetic energy halfway down and 100% kinetic and temperature at the bottom.

Gravity is constantly accelerating the atom and the total energy in the system never varies, it is simply oscillating between potential and kinetic energy. It can never reach kinetic (thermal) equilibrium unless both the kinetic and potential energy is taken from the system.

Stacking atoms in the tube doesn’t change the kinetic – potential equation at all.

Nice clear illustrative thinking and exposition Ghengis.

Genghis,

No, there’s nothing wrong. Indeed, you could take it further. A reasonable estimate of “temperature” is .5v^2/R. The grad of that is -g/R, and with the 1D constraint, you can identify R with cp. You have a DALR!

With more particles (but no collisions), same deal. That’s consistent with my entropy explanation. No collisions means zero conductivity (k), and you can maintain a temp grad indefinitely.

But once you allow collisions, KE will diffuse along its gradient, and that means upward. PE can’t diffuse – it’s fixed by height. So (as long as you have LTE) you get to uniform temperature.

Genghis Says

January 25, 2012 at 11:40 pm

“Gravity is constantly accelerating the atom and the total energy in the system never varies, it is simply oscillating between potential and kinetic energy.”

tallbloke says:

January 25, 2012 at 11:51 pm

“Nice clear illustrative thinking and exposition Gengis.”

Yes, nice clear explanation as to why the the Jelbring system is in violation of the Laws of Thermodynamics.

You boys do know that work is performed when energy changes its form, don’t you? That is at the heart of thermodynamics.

So we have a constant change in the form of energy, which causes constant work, which leads to maximum entropy.

Except of course it doesn’t in this system, because of what?

Ok, I get where you are coming from on this subject. I am just going to get some more BS or obfuscation. You are either clueless about thermodynamics or you are pushing some sort of agenda.

Perhaps you will be trying to sell people some sort of free energy machine in the future, who knows?

I wish you luck gentlemen, I am sure, when you look back on your contributions to this matter in the future it will make you proud.

For me I am done, I have better things to spend my time on, I have some paint I need to watch dry!

Alan

[Reply] You go and do that Alan. It’s a process you’re more likely to attain a strong grasp of.“I make the Venusian lapse rate about 12 but it isn’t air. Figure comes from the computation of the line to the surface.”

That’s strange, when I look at the Venus lapse up supplied, I get (740-275)/(0-60) or -7.75°C/km; maybe you could push it to (740-260)/(0-60) or -8.0°C/km. tchannon, how did you come up with twelve? Not that it matters much but I have always found Venus’s lapse to be very close to Earth’s even though Venus’s pressure is 92 times greater, it’s mass 94 times greater *and* it’s absorbed radiation is a merely a mean of 65-66W/m2. Your right, it is mostly carbon dioxide but that curious relationship just proves that a lapse rate has little relation, or no relation at all, with either the gas components *or* the radiation and where that radiation is being absorbed into the column or where radiation outward is being ejected.

I would just love for someone to put that in equations and somehow prove this wrong, but even Dr. Robert Brown of the Duke University physics department refused to directly take up that challenge.

http://wattsupwiththat.com/2012/01/24/refutation-of-stable-thermal-equilibrium-lapse-rates/

(note a small correction a few comments further down)

That is an empirical observation that says Jelbring, Nikolov and Zeller are all correct in this matter.

Here’s where the Earth-Venus comparisons get REAL interesting :

The height of the Earth’s tropopause, where the adiabatic lapse ceases to apply is 11km. Atmospheric pressure is 226hPA and the temperature is -56.5C (217K)

Apply the adiabatic lapse rate (6.49C/km) down to the surface, and the extrapolated temperature at the surface is 217+(11*6.49) = 288K. Exactly as is observed.

Confirmation can be found here :

http://en.wikipedia.org/wiki/International_Standard_Atmosphere

Now let’s assume that the tropopause on Venus, where the adiabatic lapse rate again ceases to apply, also occurs where the Atmospheric pressure is 226hPA.

From here http://www.datasync.com/~rsf1/vel/1918vpt.htm it appears to be close to 62km above the surface.

The excellent Harry Dale Huffman provides the temperature at the 200mb and 300mb levels in the Venusian atmosphere here :

http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html

From this we can calculate the temperature at the Venusian tropopause (226mb) as close to 251K

(T200mb + (0.26*(T300mb-T200mb))

Now let’s again apply the adiabatic lapse rate down to the surface, taking Wayne’s figure of 7.75C/km. The extrapolated surface temperature is 251+(62*7.75) = 731K. Again, very close to actual observations.

Greenhouse theorists would have you believe this is co-incidence, but personally I find the fact that the lapse rate ceases to apply at the same atmospheric pressure (226mb) in two atmospheres as different as those found on Venus and the Earth, very interesting indeed.

possible implications/

Anything is possible, you have it! That is VERY interesting.

Now just take your imagination one more step and say Venus’s albedo was 0.95 instead of 0.90 and the entire atmosphere was only receiving 33 W/m2 instead of 66 W/m2… would that make the surface temperature half… I hardly think so! It does make you think hard exactly what would change at all. The surface pressure would be the same but not the pressure at higher altitude levels. The density would change but it seems only at higher levels. The level of your 226 mb would change. The atmosphere would have collapsed to some degree. Wish I had better calculating tools for spot answers to such questions; those factors would be very interesting. I will eventually program or excel it and calculate those but it is so slow and these threads keep on truckin’.

We have two great laboratories about as diverse as possible at our disposal, the Earth and Venus, and after homing in on their properties, these relations seem to invalidate all that the consensus climate science is saying, especially in radiation and ghg aspects.

What irks me is that I myself have been letting them lead this ‘conversation’, time to stop. Looks like I have found a new home. I so like the slower pace, I’m no speed typist.

Ghengis, I don’t think your atom in a tube demonstrates anything bout the lapse rate. First, you’ve described this atom as if it must necessarily rise up the tube, but there is in fact no reason at all for it to rise up the tube, no matter how hot it is, because it is alone. It would spend most of its time close to the ground because it is, after all, attracted to the earth’s gravity, and it has just as much room to move there as it would have way up at the top of the tube. Energetic atoms rise when they are crowded by less energetic atoms, I think, sort of a spacially constrained diffusion.

Second, the laws of thermodynamics apply to bulk properties. They don’t say anything about individual atoms. They require a large enough number of molecules to provide a normal distribution of molecular states whose bulk/average effect is a given temperature, pressure etc.

Another interesting fact is that if we apply N & Z’s ATE ratio to temperatures at the 226mb levels, we get very similar values for both Venus (1.382) and Earth (1.403).

Hmmm……

Nick Stokes says:January 26, 2012 at 12:19 am

Genghis,

No, there’s nothing wrong. Indeed, you could take it further. A reasonable estimate of “temperature” is .5v^2/R. The grad of that is -g/R, and with the 1D constraint, you can identify R with cp. You have a DALR!

With more particles (but no collisions), same deal. That’s consistent with my entropy explanation. No collisions means zero conductivity (k), and you can maintain a temp grad indefinitely.

But once you allow collisions, KE will diffuse along its gradient, and that means upward. PE can’t diffuse – it’s fixed by height. So (as long as you have LTE) you get to uniform temperature.Wotcher Nick.

It’s an ingenious argument, but incorrect. Here’s why:

While the system isn’t at maximum entropy, the collisions will indeed thermalise energy, and this will lead to the adiabatic convection of the warmed packets of air. But because it’s adiabatically convecting, the warmer upward parcel is replaced by a cooler downward parcel which loses PE as it descends. Meanwhile the upward parcel expands and cools, losing KE and gaining PE. Eventually equilibrium is reached, and the kinetically stratified distribution means that at the horizontal plane, collisions don’t lead to net upward or downward motion, as the average KE for the z altitude is what it is. up and down collisions are random and cancel, with the suitable tradeoff of KE and PE occurring each time.

clazy8 says:January 26, 2012 at 6:23 am

Ghengis, I don’t think your atom in a tube demonstrates anything bout the lapse rate. First, you’ve described this atom as if it must necessarily rise up the tube, but there is in fact no reason at all for it to rise up the tube, no matter how hot it is, because it is alone. It would spend most of its time close to the ground because it is, after all, attracted to the earth’s gravity,Ghengis specified perfect elasticity. Think of one of those super bouncy balls, dropped from a height, which bounces back up to nearly the same height it started at. Ghengis’ atom will get all the way back up to the height it started from.

AiP: Yes, that’s what Huffman’s original paper pointed out. If the ‘effective height of emission’ coincides with that pressure, we’ve probably learned something important.

Wayne: Glad you’re finding that the need for me to run this site at the speed my crash-impaired brain works at to be of benefit. Less Joels more Jewels… or is that Joules 🙂

Ok another attempt Tallbloke to inject sanity here.

So you and Jelbring are saying that we can maintain a planetary atmosphere with a permanent temperarture gradient and at some point convection stops happening due to something that is unclear to me.

Anyway, if that is so, what is holding the atmosphere up in this system? It will collapse instantly you prevent the movement of air upwards.

Our atmosphere stays up because of the Pressure-Gradient Force which is constantly accelerating parcels of air from high pressure zones to low pressure zones and this balances out the gravitational pull. This constant acceleration of particles (that would be work!) apparently ceases in your system, presumably to protect the second Law but of course now you dont have any atmosphere anymore because it coiiapses instantly.

You see when Jelbring imposed his second condition, on the rules operating in his gedanken universe ,i.e. “2.There are no winds in the enclosed atmosphere”, presumably to try and protect the system from the second law then he did away with the Pressure-Gradient Force.because there will always be wind with that force present.

Therefore, no wind equals no Pressure-Gradient Force, which equals no atmosphere, which equals no Jelbringian system. QED!

Or there is a Pressure-Gradient Force, which equals wind, which equals work, which equals maximum entropy eventually, which equals no Jelbringian system. QED!

Or perhaps you can tell me different.

Alan

[reply] Pressure holds it up, because heat doesn’t escape from the system as defined in the papers. the is no wind because the planet doesn’t rotate as defined in the papers. Read the papers Alan, and stop trash talking Hans Jelbring or I’ll boot you off my site. First and last warning. If you come back with further inanity which shows you haven’t read the papers you are criticising, the comment will go in the bit bucket.

Hi Tallbloke, I guess I don’t understand what he’s getting at. What is he comparing this to?

Anything Is Possible states:

“…The height of the Earth’s tropopause, where the adiabatic lapse ceases to apply is 11km. Atmospheric pressure is 226hPA and the temperature is -56.5C (217K)”

One must remember the tropopuse is not held at a fixed height across the globe. It averages at around 11km…up around the polar regions *however* increases with distance away from the poles till it is around 17km near the Equator. What else is increasing as you get closer to the Equator, eh??? Surface temperature!! Purely a coincidence I’m sure 😉

Jeff

Clazy8: Ghengis is just showing in the simplest way posiible that when the bouncing atom has max potential energy it has minimum kinetic energy, and vise versa. The total energy is always the same, at whatever height the atom happens to be at the time. This relates to the temperature gradient which Hans Jelbring has proven will exist at energy equilibrium because fast moving (vibrating and colliding) atoms (molecules) are warm and vise versa.

It’s a little oversimplified but a good analogy.

JKrob says:January 26, 2012 at 2:49 pm (Edit)

Anything Is Possible states:

“…The height of the Earth’s tropopause, where the adiabatic lapse ceases to apply is 11km. Atmospheric pressure is 226hPA and the temperature is -56.5C (217K)”

One must remember the tropopuse is not held at a fixed height across the globe. It averages at around 11km…up around the polar regions *however* increases with distance away from the polesI wonder how much the polar tropopause height varies over the year, from endless day to endless night. That measurement plus the energy flow estimates would tell us a lot I think.

Alan Millar says:

January 26, 2012 at 1:05 am

tallbloke says:

January 25, 2012 at 11:51 pm

“Nice clear illustrative thinking and exposition Gengis.”

“Yes, nice clear explanation as to why the the Jelbring system is in violation of the Laws of Thermodynamics.

You boys do know that work is performed when energy changes its form, don’t you? That is at the heart of thermodynamics.”

The Jelbring system is not violating the Law of Thermodynamics. You have all information needed to specify your statment and you don´t do that for reasons I don´t know.

To repeat to you what is told in my article:

A system as discussed which is closed at an arbitrary tim T0 contain a specific constant amount of energy E0 at closure.

This energy can be arbitrarily distributed within limits given by some other gas laws such as the hydrostatic equation etc at closure.

The second law of thermodynamics guarantees that the energy E0, when enough time has passed, is distributed in such a way that any equal mass will contain an equal amount of energy. This rearrangment of energy will happen spontaneously wihtout any need how it is done (processes at work).

Your statement is wrong and you are violating the Law of thermodynamic in your statement since energy is just redistributed within the system. No work is done by the system on anything not belonging to the system. The total energy stays constant all the time during the spatial rearrangment of energy.

Hans Jelbring

Anything is possible says:

January 26, 2012 at 4:05 am

You are really catching on. Try checking the lapse rate on Titan. I predicted it 6 months before the sond hit its surface. Titan´s atmosphere is thicker than the one onearth.

Hans, is it on record anywhere what your prediction was? That could be important I think.

JKrob says …The height of the Earth’s tropopause……. increases with distance away from the poles till it is around 17km near the Equator. What else is increasing as you get closer to the Equator, eh??? Surface temperature!! Purely a coincidence I’m sure 😉 Jeff

It also has a day/night variation.

The height of the tropopause indicates the energy stored in the atmosphere both kinetic and gravitational potential.

This Sun driven gravitational thermodynamic reserve of energy is a major part of the climate conditions that keep the Earth surface well above its radiative bare rock temperature.

Tallbloke says:

January 26, 2012 at 7:41 am

“AiP: Yes, that’s what Huffman’s original paper pointed out. If the ‘effective height of emission’ coincides with that pressure, we’ve probably learned something important.”

The amazing fact I found out before writing my 2003 article was that the tropopause (coldest part of the atmosphere) started at a similar pressure in all atmosphere bearing planets except Mars and that pressure were between 0.1-0.2 bar which is quite amazing. This is why I am claiming that it is possible to calculate the surface pressure “backwards” from the known temperature the upper troposphere down to the surface using an approxiamte (dry) adiabatic temperature lapse rate = -g/Cp.

This observation does not work on Mars since its surface atmosphere is less than 0.1 bar.

tallbloke says:

January 26, 2012 at 3:53 pm

Hans, is it on record anywhere what your prediction was? That could be important I think.

Yes, it is somewhere among 15000 mails in my Climatesceptic files in an old computer that I want to stay away from. But some time I might find it.

Bryan says:

January 26, 2012 at 3:53 pm

“JKrob says …The height of the Earth’s tropopause……. increases with distance away from the poles till it is around 17km near the Equator. What else is increasing as you get closer to the Equator, eh??? Surface temperature!! Purely a coincidence I’m sure Jeff”

Solar irradiation promotes ascending air (especially over land) and the development of approximate adiabatic temperature profiles. Its better developed during day time and get partially destroyed during nights.

Subduction in polar areas creates inversions when air is descending. The tropopause starts seldom as high as 17000 m. It is more of a maximum value.

Hans says:

January 26, 2012 at 3:45 pm

Anything is possible says:

January 26, 2012 at 4:05 am

“You are really catching on. Try checking the lapse rate on Titan. I predicted it 6 months before the sond hit its surface. Titan´s atmosphere is thicker than the one onearth.”

Thanks.

I was trying to piece some figures together on Titan last night :

I got as far as a tropopause height of 32km, a tropopause temperature (based on solar radiation) of 68K, and a dry adiabatic lapse rate of 1.301C/Km.

Extrapolating a surface temperature from this gives a figure of 109K, somewhat higher than actually observed. BUT I also got to reading a paper that stated that methane is a condensing gas in Titan’s atmosphere, acting very much like water vapour on Earth. That implies that the actual environmental lapse rate may be somewhat lower than the number I have used. If you can lay your hands on that number, that would be very useful……

In reply to TB @ 7:41am :

It occurs to me that the tropopause represents the closest point to a planets surface where the temperature can be calculated by radiative theory alone – that seems to be the key. Below that, the dynamic processes in the atmosphere (ie : Meteorology) take over.

That also being the “effective height of emission” makes perfect sense to me because there is hardly anything above the tropopause (except the Ozone layer?) to interfere with the radiative exchange between Earth and outer space.

Tallbloke says

“reply] Pressure holds it up, because heat doesn’t escape from the system as defined in the papers. the is no wind because the planet doesn’t rotate as defined in the papers. Read the papers Alan, and stop trash talking Hans Jelbring or I’ll boot you off my site. First and last warning.”

That pressure is the Pressure-Gradient Force. That force cause winds, it will cause winds on a non-rotating planet just as on a rotating planet. A planet does not need to rotate to have winds. However. any planet with an atmosphere must have winds due to the Pressure-Gradient Force keeping its atmosphere up

If the Pressure-Gradient force is operating and it must if you don’t want your atmosphere to collapse then horizontal winds will occur. This Force accelerates parcels of air from high pressure zones to low pressure zones and it does it just as well horizontally as well as vertically.

Jelbring has specified that winds cannot occur in his system. He has done this for a reason. I am sure he realises he needs the Pressure-Gradient Force to operate in his system to keep his atmosphere up but he needs it to only accelerate parcels of air vertically and not horizontally. If parcels of air are being constantly accelerated horizontally then you cannot say that the air moving up and down is cancelling out the work being done and entropy will increase to maximum.

Now I am sure that you realise that it is impossible, in this universe, to make the Pressure-Gradient Force only act in a vertical direction.

Off course if you are making your own gedanken universe you could make it as one of the conditions as has been done here. However, it doesn’t say anything about this universe.

Alan

[Reply] I know my answer but I’ll wait to see if Hans wants to respond.Bryan says:

January 26, 2012 at 3:53 pm

“The height of the tropopause indicates the energy stored in the atmosphere both kinetic and gravitational potential.

This Sun driven gravitational thermodynamic reserve of energy is a major part of the climate conditions that keep the Earth surface well above its radiative bare rock temperature.”

What you say is true. Outgoing IR at the Southpole is about 120 W/m^2. In Arctic regions it is about 150 and in Australia during summer up to 350. The OLR from the equator (or rather the IPCZ) is lower depending on the cloud cover.

What you describe is a time dependant steady state climate system with seasonal variations.

The key issue is your words “stored energy”. This is what gravity does to an atmosphere. It forces the atmosphere to keep a storage of energy. If it doesn´t it has turned into a liquid or a solid and there wouldn´t be an atmosphere.

Alan Millar says:

January 26, 2012 at 5:55 pm

“Off course if you are making your own gedanken universe you could make it as one of the conditions as has been done here. However, it doesn’t say anything about this universe.”

Well, my model atmosphere is not treating the universe. It is restricted to planetary atmospheres and the properties of gases. Please, refer to what I have written and I will be grateful and answer. TB is welcome to answer the rest of your mail.

My reply to Alan Millar is this:

Either you haven’t read and understood the two papers, but think you have, and make categorical statements based on your belief in your own understanding, or you have read and understood the papers, but instead of addressing them on their own terms as an internally consistent and usefully applicable gedanken experiment, you are seeking to trash them by undermining their credibility as such.

Either way, good etiquette would be to get to the point and ask Hans to clarify whatever genuine question you may have, or simply state that your issue is the thermal gradient in the energy equilibrium which is the real bone of contention. It’s a one hundred and twenty year old unresolved paradox and I doubt you will be the one to convince anyone either way. My previous interaction with you on WUWT wasn’t a good start, and doesn’t bode well for any plea for you to be polite or conversational. People are rarely persuaded to a different point of view by someone who seems incapable of reasonable debate.

Anything is possible says:

January 26, 2012 at 4:59 pm

“Below 160 km, temperature data are direct measurements collected by the TEM sensor. The temperature profile in the upper atmosphere (thermosphere) is characterized by several temperature variations ”

“Thus the structure that we observe may vary with time. The horizontal lines mark the mesopause (152 K at 490 km), the stratopause (186 K at 250 km) and the tropopause (70.43 K at 44 km).”

See image at:

http://www.google.se/imgres?

This is the worst case of an URL I have ever seen try this one too).

Titan lapse rates Google Sweden image search

or

http://en.wikipedia.org/wiki/File:Titan_atmosphere_detail.svg

The chemical composition is 98.4% nitrogen. (Wikipedia)

g(titan) = 1.352 m/s (obscure source)

Assume 100% nitrogen in the atmosphere

Cp = 1039 (at 175 K) might be quite off since Cp should be at about 90K

Anyway dT/dz = 1,352/1059 dT/dz = 0.00128 k7m or 1.28 K/km

The upper tropopause is approximately at 25 km at 75K. The surface temperature should be at

75+1.28×25 = 110 K.

Wikipedia tells a surface temperature of 94 K. 1.28 K is a maximum value. Condensation processes exist as on earth which have a real dt/dz = -6.5 K/km instead of the theoretical -9.8 K/km

I did a better work last time and didn´t have my old calculations but this one seems quite close to yours. It is very hard to extract information from internet on this topic nowadays. All images seem to be protected.

Hope this was to some help.

Whilst searching for something else I came across this amusing web page.

http://www.theweatherprediction.com/basic/equations/

Includes this

“Interpretation: The dry adiabatic lapse rate is a direct function of gravity. Since gravity is basically a constant, the dry adiabatic lapse rate is basically a constant.” — Jeff Haby, Mississippi State

He also calculates for Venus, giving a slightly higher figure.

Thanks, Hans.

I don’t think we’re going to get any more precise than we already have on account of the tropopause being much more poorly defined on Titan than it is on Venus or on Earth.

Could that be on account of Titan’s low gravity?

tallbloke says:

January 26, 2012 at 8:34 pm

“Either way, good etiquette would be to get to the point and ask Hans to clarify whatever genuine question you may have,”

I have asked this question in various ways but I will ask of it you and Hans, again. It is a simple question.

The Pressure-Gradient Force is responsible for keeping the atmosphere up on all planets, including the Jelbring one, by accelerating parcels of air vertically, to lower pressure zones.

This Force has to also produce acceleration of parcels of air in a horizontal direction (winds). It is not a question of if, it has to.

What is stopping this Force producing horizontal movement (wind) on Jelbring’s planet?

Alan

Nice one Tim.

Alan, Thank you.

Wikipedia says this about the pressure gradient force:

“The pressure gradient force is not actually a ‘force’ but the acceleration of air due to pressure difference (a force per unit mass). It is usually responsible for accelerating a parcel of air from a high atmospheric pressure region to a low pressure region, resulting in wind.”

“In large-scale atmospheric flows, the coriolis force generally balances the pressure gradient force, producing winds blowing largely along the isobars;”

No spinning of planet Jelbring, no coriolis force…

No spinning of uniformly warm planet Jelbring, no day side-night side thermal imbalance…

No wind on planet Jelbring.

Jelbring did not stipulate “No wind”

He carefully designed planet Jelbring so that no wind would arise.

This is so that the important effect he seeks to demonstrate can be discerned without being lost and confused in a cauldron of weather.

I’ve trawled the private archives Hans is talking about. (and getting many security alerts on a blacklisted server address, something nearby is infected)

Hans was writing a lot about Titan various times 2003 onwards but I have failed to find a figure. Archive search is not wonderful.

Maybe Hans should put the old computer hard drive in a newer computer and mount it as a slave drive so he can trawl the mail archive.

Need a hand with that Hans?

tallbloke says:

January 26, 2012 at 9:46 pm

I am sorry Tallbloke but the coriolis force is not needed to create winds. It can have various effects on created winds such as cyclonic direction in the two hemispheres.

Winds happen on non-rotating planets, I don’t think there is any dispute about this in science.

The fact is, on Jelbring’s planet, parcels of air are being accelerated upwards and something is preventing parcels of air moving horizontally to fill the low pressure area created and that is not the lack of a coriolis effect.

So again what is it?

Alan

Alan Millar says:

January 26, 2012 at 9:16 pm

“What is stopping this Force producing horizontal movement (wind) on Jelbring’s planet?”

First there is not Jelbring planet. What I discuss is a insulated chamber inclosing the whole planet atmosphere or an insulated vertical column of air.

The answer to your question is: FRICTION

(but iot might take abour two weeks after the incluse before winds stopps.

A couple things which I’m not sure have been mentioned…

1. Any attempt to assume an ideal gas or use the ideal gas law in a gravity constrained container (atmosphere) is, I think, a huge error. Even if you had a perfectly elastic gas with no inter-molecular attractions, i.e. perfectly ideal by definition, it can’t behave ideally since gravity imposes the analog to inter-molecular attractions for any molecules with a vertical component to their motion. You can get away with it in small scales but on the scale of the atmosphere and tens of kilometers of air column, IMO, it is a mistake.

2. Over on WUWT there’s a thread that states that the energy content of a gas includes translational motion and potential energy of the parcel. I think this is also incorrect. I can seal up (really really tight) an energetic parcel of gas and raise it as high as I want and never impact it’s internal energy. I can also move it along as fast as I want and not impact it’s internal energy. That’s not to say that the PARCEL can’t take on PE and KE, of course it can, but these need to be considered separately from the internal energy as they don’t contribute to the temperature of the gas at all until you hit something. Throwing a baseball doesn’t make the inside hot. The internal energy of the gas is what determines its temperature isn’t it?

3. Lots of people are hung up on the (understandable) fallacy that hot air rises. It’s not hot air that rises it’s light air that rises, i.e. air that is lighter than the air above it. With gravity squeezing things it’s possible to have heavy, hot air underneath lighter cool air and having this condition violates no laws that I know of. This is why hot air balloons are big, else wise folks would balloon with scuba tanks filled with super heated nitrogen.

4. Lastly, here’s a question I’ve not seen addressed. If you draw an arbitrary horizontal boundary and consider the atmosphere on either side of it (under the influence of a pressure gradient), there’s a very interesting energy exchange. A molecule can cross the boundary going up as a function of its internal energy (from wizzing about) and it can cross it as a member of a parcel rising due to convection. In the former case the crossing has nothing to do with local temperature differences only relative energy levels. In the latter case the crossing is all about local density differences and only indirectly related to local temps. In both cases though, upward bound molecules are exchanging KE for PE but only the former effects the internal energy of the local parcel. The really interesting part is that for those molecules crossing because of their wizzing around activities there is a bias induced by the pressure gradient. A particle of energy X wizzing down has KE of X plus the KE due to gravity. The particle of energy X wizzing up KE of K minus the PE it is gaining. Doesn’t this mean there is a bias that develops where KE below the layer is enhanced by gravity while the KE above the layer is inhibited (transformed to PE)?

It seems to me that given the above there are way too many gross over simplifications flying around. Including, perhaps, my own.

Paul,

I like the self deprecating tail humour, something certain people need to cultivate.

You mention hot air rises because it is buoyant (no girlies in this case). I add that cold “air” can also rise, one of the really tricky problems on earth, the lowest density common gas which materialises by magic, h2o.

The stability problem, whether convection starts is I think known tricky.

Hans says:

January 26, 2012 at 11:07 pm

“What is stopping this Force producing horizontal movement (wind) on Jelbring’s planet?”

First there is not Jelbring planet. What I discuss is a insulated chamber inclosing the whole planet atmosphere or an insulated vertical column of air.

The answer to your question is: FRICTION

(but iot might take abour two weeks after the incluse before winds stopps.”

Thanks for replying Hans.

I agree that friction will definitely take place whilst parcels of air are moving, particularly against a solid surface.

So we agree, in your system of a solid chamber that initially at least some horizontal movement of air takes place until friction stops it but that vertical movement of parcels of air continues in order to keep your atmosphere up.

However, what is friction? It is the conversion of kinetic energy into heat.

So there are at least two very big problems I have with your system here.

1. Converting energy into heat is what entropy is all about. You say the horizontal movement of air will be stopped fairly quickly by friction. However, parcels of air are going to continue to rise indefinitely and therefore are also going to be losing kinetic energy to heat continuously by friction. Second Law says this cannot happen and the system will suffer ‘heat death’ as entropy reaches maximum.

So how do you turn this heat back into usable energy in your system? The 2nd Law says you cannot do this either, without the addition of further energy, which your system is not getting.

2 Assuming there is an answer to the first issue (which I think is the key one actually) the second question would be. If you agree that friction can stop the horizontal movement of air, what prevents friction stopping the vertical movement of air and collapsing your atmosphere instantly?

Alan

I remember flying one really nasty hot humid New England day with a friend in a little Cessna. He needed every bit of the runway even with his healthy plane. Not enough lift in that New England H2O. It was a real eye opener for me.

“It seems to me that given the above there are way too many gross over simplifications flying around. Including, perhaps, my own.”

(Re•duc•tion•ism

noun

1. The practice of analyzing and describing a complex phenomenon, esp. a mental, social, or biological phenomenon, in terms of phenomena that are held to represent a simpler or more fundamental level, esp. when this is said to provide a sufficient explanation

Reductionism can either mean (a) an approach to understanding the nature of complex things by reducing them to the interactions of their parts, or to simpler or more fundamental things or (b) a philosophical position that a complex system is nothing but the sum of its parts, and that an account of it can be reduced to accounts of individual constituents

[Wiki] )

That is the wiki on reducto, I don’t agree with you totally, some simplifications do lead to greater understanding, imagine trying to teach kids the mystics of Science of Atmospheric without relating it to their understandings.

I have previously posted a little ditty, hopefully something like that can be taught to kids, so they understand more. And frankly some of those lyrics that have been deposited by me, have also helped some of the greatest minds in the fields to understand the wrongness of the greenhouse paradigm.

The art. Rog posted on Baron Fourier nailed it for me. When I thought about the N&K principle, it clicked immediately. He did not distinguish the manner of mass, between its different composition, in the crust of the Earth.

Casting off greenhouse, I quickly saw the relationship of refrigeration when Rog posted that graph about the temperature through out the atmosphere as stratified, Then looking back Fouriers observations, it was obvious Co2 meant very little to heat distribution in a straight line within the Earth.

I then imagined the greenhouse, as the glass only, and hence a new perspective. But what mechanism drove our system. Clearly not a greenhouse, as its hotter at TOA. Then it struck, refrigeration heat pump. Thermostats, condensers and evaporators, when logically applied to the natural systems of atmosphere, it gelled, like a bolt from Heaven.

To to be sure, I’m not sure it wasn’t. Roger Tattersall, was the reason for my limited understandings.

Markus Fitzhenry.

I agree totally with the use of simplification as a learning tool. I’m sure at some point Albert Einstein said, “Hmmmm, I wonder what would happen to a man in a space ship moving at the speed of light?”

The good, bad, and ugly of blogs is that someone will make a simplifying assumption in a perfectly valid attempt to model some behavior for insight. Then someone will ‘show’ that it doesn’t work. Then someone jumps in to declare the author a dolt. Before long the entire thread is a mixture of pissing contests, credentialing, ego trips, and people who are still trying to learn and teach. This turns me off eventually and I stop reading the thread.

I love the blogging of science because, for me at least, it’s a great learning tool and it challenges me to think. Yet, it’s a bit like seeing sausage made, isn’t it? If people would see errors as teaching opportunities instead of platforms to show off, they would learn, and the mistakee would learn.

Win, win! It’s all about the tone.

Actually, I’m over here because the pee got too deep over at WUWT.

Alan Millar says:

January 27, 2012 at 12:56 am

Hans says:

January 26, 2012 at 11:07 pm

“What is stopping this Force producing horizontal movement (wind) on Jelbring’s planet?”

The answer to your question is: FRICTION/HJ

(but it might take abour two weeks after the incluse before winds stopps.)/HJ

“I agree that friction will definitely take place whilst parcels of air are moving, particularly against a solid surface.”

2. Assuming there is an answer to the first issue (which I think is the key one actually) the second question would be. If you agree that friction can stop the horizontal movement of air, what prevents friction stopping the vertical movement of air and collapsing your atmosphere instantly?

Alan you are absolutely amazing. You were asking the question above and I answered it. You were the one who wanted to know about horizontal motion. If you accept that FRICTION stopps horizontal motion it for sure has to stop vertical motion, too (both up and down). Vertical motion is also wind.

And please, point to my text why the described model atmosphere that I have assumed would collaps instantly if there were no vertical air motion?????

The title of my article is: “An Alternative Derivation of the Static Dry Adiabatic Temperature Lapse Rate”, with emphasise on STATIC meaning there are no vertical winds involved in the derivation.

There were such winds in the referenced derivation by Holton though. A major point with this paper is to prove that the dry adiabatic temperature lapse rate is -g/Cp also in a static atmosphere where no air parcel is mowing upwards.

Have you read the article above?

Paul Bahlin says:

January 27, 2012 at 12:32 am

“It seems to me that given the above there are way too many gross over simplifications flying around. Including, perhaps, my own.”

And

“1. Any attempt to assume an ideal gas or use the ideal gas law in a gravity constrained container (atmosphere) is, I think, a huge error.”

It is fine that you have an opinion but the aim of science is to separate what are opinions from what are facts. It is a very good approximation treating our atmosphere as a mixture of ideal gases.

On the contrary, there is none gross over simplification flying around. The model atmosphere can be physically constructed and hence it should be of value to discuss it as such. Only first principle physics is used in the derivation.

Just point to where the derivation is unclear and I will answer your questions after you have specified your statements about what I have written. I will be glad if you can point out where I have made an error but I refuse to discuss your unsupported opinion that I have made an error.

Hans:

Actually, I expressed no such opinion (about your paper). Your paper is, I think, a quite interesting and valid piece of simplification. I was referring to the folks who are beating you up in here and it’s my mistake not to have made that clearer.

See how easy that was?

And my argument about ideal gas was not aimed at you (again, my bad for not stating so) it was a general question and I raised it because I see it being made everywhere (not just this thread or blog) and I have raised this same question in other places, only to get peed on.

Let me try it another way… Let’s assume we have a perfectly ideal gas. Place it in a three sided container with the fourth side, gravity. Can this gas behave ideally, given that one of the sides of the container presents it with the exact same ‘wall bias’ as it would have should there be inter-molecular forces that you would find in a non-ideal gas.

BTW – I was peed on (by a scientist, yikes) for suggesting that an ideal gas is assumed to have (one of its properties) the property of no inter-molecular forces.

In your paper you reach your conclusion based on your (valid) simplifying assumption of an ideal gas and then issue a challenge to readers to come up with an explanation for the (unexpected) result. Perhaps this ideal gas, in an unideal container is part of the explanation.

Can anyone help with the following thought experiment:

I have a really really tall cylinder filled with n mols of an ideal gas and placed in a gravity field (a slice of an ideal gas atmosphere). There is a gravity imposed pressure gradient in the cylinder. I construct two cases:

1) Along with the pressure profile there is a temperature profile that dollows the DALR. The gas occpuies a volume. The cylinder contains an amount of energy equal to nU**, where U** is an appropriately average internal energy. The gas has a volume nV**, where V** is an appropriately average molar volume. These averages aren’t easy to calculate…

2) In this case there is a pressure profile but the system is isothermal. Let’s require that it have the same energy as case 1. It will also have a volume nV^^, where V^^ is a appropriately averaged volume for the isothermal case.

Now, because the system has the same energy the isothermal temperature will be somewhere between the high and low temperatures when the temperature profile follows the DALR. Changing the temperature profile to isothermal tends to increase the system entropy – warming the cool parts increases entropy more than cooling the warm parts.

But, let’s consider the volume. Does changing the temperature profile also change the volume? If so, it will also change the entropy. If the isothermal volume is less the entropy will decrease, and could partially or completely offset the entropy gained from changing the temperature profile.

Which case has the most entropy?

It seems to me that if you answer that question and you will know which is the thermodynamically stable condition.

Hans says:

January 27, 2012 at 1:02 pm

Hans

You have already agreed that there must be movement of parcels of air in your system when created, the Pressure-Gradient Force is present.

You have then said that friction will stop this movement eventually. Well I agree with you there.

How does friction stop movement?

It does this by converting kinetic energy into heat until there isn’t enough kinetic energy left in the system to initiate movement. I think that is also described as ‘heat death’ the whole basis of the 2nd Law of Thermodynamics.

No magic can reconvert that heat into kinetic energy in a closed system.

Your system will die Hans and it will have been killed because it was obeying the Laws of Thermodynamics. You cannot keep changing potential energy into kinetic energy to keep replacing the kinetic energy lost to friction and therefore heat at the bottom of the system, you just run out of energy.

Alan

[Reply] Kinetic energy is heat, heat is kinetic energy. Or it was when I learned thermodynamics anyway.

Hans says

“A major point with this paper is to prove that the dry adiabatic temperature lapse rate is -g/Cp also in a static atmosphere where no air parcel is mowing upwards. ”

This paper from Climate Science would agree with you

This is what meteorologists call the neutral atmosphere.

The neutral atmosphere can be quite stable.

See page 31 and the residual layer.

Click to access ch2_brasseurjacob_Jan11.pdf

[Reply] Kinetic energy is heat, heat is kinetic energy. Or it was when I learned thermodynamics anyway.

“heat is energy produced or transferred from one body, region, set of components, or thermodynamic system to another in any way other than as work”

“For a closed system (with no external transfer of matter), heat is defined as energy transferred to the system in any way other than as work. Heat transfer is an irreversible process”….Wiki

I think the problem here Tallbloke is that Hans and yourself are convinced that the Jelbring system is internally consistent and the math involved is correct.

Well that very well may be true.

However, whenever you come up with what you think is a totally new and never thought of idea in science then you need to be very careful, even if your math is sound you need to step back from the detail and consider the general position.

Earlier I gave you this example of what I shall call the Millar system.

“I could invent a closed system that was initially composed of a diffuse cloud of particles in

thermodynamic equilibrium.

Now that system would be near maximum entropy. However gravity starts to cause the particles to compress and voila I now have a system like the solar system or a galaxy even and I have reduced the entropy of the system without the addition of energy or increasing the entropy of another system.

Perhaps I could say that this proves gravity can reverse the flow of entropy in a closed system and the Thermodynamic laws need revising.

Now I could provide the math and it would be entirely consistent with the system I have created. Neither you or Hans or anyone could disprove it by any equations or Thermodynamic laws. We know for sure that this is what happens to clouds of particles.

However, in the real universe we inhabit we cannot create such a system in such an initial state. A supreme, all powerful being could but since I am not, the system can only ever exist in my mind.

Now this is also the case with the Jelbring system it cannot be created in such an initial system. If you could perhaps it all works.

However, in this universe it has to start in some other state and reach that state. However, it cannot, as the system starts out with lots of movement and conversion of kinetic energy to heat by friction. Hans has already agreed that would be the reality. The only way to stop that is to stop the movement but that doesn’t stop until the kinetic energy of the system approaches zero.

Alan

kdk33 says:

January 27, 2012 at 2:20 pm

“Which case has the most entropy?

It seems to me that if you answer that question and you will know which is the thermodynamically stable condition.”

This is a simple problem for meteorologists. The condition for an atmosphere with no winds has to fullfill the equation:

dp(z)/dz = – (density(z)) g for all z. (Hydrostatic balance). That atmosphere is stable AND static.

The dry adiabatic temperature lapse rate in a static atmosphere is the most stable atmosphere you can find. It will always develope if you inclose an atmosphere in an insulated column of air as time passes.

The statements above answers your question. Air with an dry adiabatic temperature lapse rate at the moment of enclosure can not rearrange its temperature profile. Any other initial temperature profile will become rearreanged to a dry adiabatic temperature lapse rate as time goes by.

In a real atmosphere the surface atmosphereic mass always has the minimum of total energy per mass unit in an air column. The reason is that if by chance (which happens in cumulus nimbus clouds) the air get very cold it sinks until it meets heavieer air or it hits the surface of earth. This process can actually ground air planes over the US great plains. There exists special restrictions for air craft due to this atmospheric process in US.

If you have two equal insulated columns of air with the same temperature at the surface the one

with the highest top temperature has most total energy when the inclusre happens. You can be sure that the latter one does not has a dry adiabatic temperature lapse rate and that the former has most entropy.

Bryan says:

January 27, 2012 at 3:42 pm

Hans says

“A major point with this paper is to prove that the dry adiabatic temperature lapse rate is -g/Cp also in a static atmosphere where no air parcel is mowing upwards. ”

This paper from Climate Science would agree with you

This is what meteorologists call the neutral atmosphere.

The neutral atmosphere can be quite stable.

See page 31 and the residual layer.

http://www-as.harvard.edu/education/brasseur_jacob/ch2_brasseurjacob_Jan11.pdf”

Thanks Bryan. It is nice toi see that Harvard seems to make progress in their education. MIT seems to be more confused.

Look at Figure 2.23. Theta is called Potential Temprature and is a misnomer since its actually means EQUAL TOTAL ENERGY PER MSS UNIT. Watch how often it is vertical. This indicates tha action of energy dissipation in our real atmosphre. Temperature is not found to be vertical in these graphs for the simple reason that there in nothing special to it in terms of total energy.

Alan Millar says:

January 27, 2012 at 2:21 pm

Alan, please stop acting as you are stupid when you aren´t. Tell me what agenda you have for your mails when you refuse to comment on the article I have written. It is very clear what assumptions I have made and how the derivations are made. Are you such a coward that you don´t dare to behave as a scientist? It is based on first principle physics. Just tell me where I am wrong IN MY ARTICLE or piss off.

Paul Bahlin says:

January 27, 2012 at 1:32 pm

“Hans:

Actually, I expressed no such opinion (about your paper). Your paper is, I think, a quite interesting and valid piece of simplification. I was referring to the folks who are beating you up in here and it’s my mistake not to have made that clearer.

And my argument about ideal gas was not aimed at you (again, my bad for not stating so) it was a general question and I raised it because I see it being made everywhere (not just this thread or blog) and I have raised this same question in other places…..

In your paper you reach your conclusion based on your (valid) simplifying assumption of an ideal gas and then issue a challenge to readers to come up with an explanation for the (unexpected) result. Perhaps this ideal gas, in an unideal container is part of the explanation.”

I accept your explanation and my misunderstanding. it is true that my paper is a simplifaction but it is in that way for a very impostant purpse, to show the importance of energy dissipation as an important physical process in any atmosphere bearing planet.

You can propose and treat the atmosphere as consisting of ideal gases (as a good approximation) in 99.9% of all time in any specific place. Lightning and extreme turbulence (tornados) will make the approximation less useful but such situations do not happen so often.

Hans says:

January 27, 2012 at 6:56 pm

Alan Millar says:

January 27, 2012 at 2:21 pm

Alan, please stop acting as you are stupid when you aren´t. Tell me what agenda you have for your mails when you refuse to comment on the article I have written. It is very clear what assumptions I have made and how the derivations are made.

It is impossible to bring your system into existence in this universe in that initial state and that is a fact and you know it.

[Reply] This is assertion and as such counts for nothing. Demonstrate it with reference to the papers.I assume, if I submitted an article with a bit of math in it detailing my ‘Millar’ system to you, that you gladly accept it as proof that the second Law is incomplete.

[Reply] By all means give it a try. I’m not publishing diatribe though. Nor am I going to continue as unpaid minder of your surly attitude. Go and read the house rules, and if you can’t abide by them, go and moan about being censored elsewhere.It all fits into your belief that E=E allows things to happen for ever.

[reply] Sigh. Read the definitions in Hans’ 2003 paper Alan. It specifies a perfectly insulated system which reaches maximum entropy after a suitable time has elapsed. In that state of max entropy, all circulation has ceased, and yet although energy is randomly spread such that work has ceased as the second law demands, there is still energy in the system because none has been lost. For this reason, ‘heat death’ does not occur.is it the house rules that allows Nickolov to insult and ad hom Willis Eschenbach?

Do you really need me to quote back to you the ad homs Willis hurled at Ned from his priviledged position with the loudhailer on WUWT before Ned finally snapped and responded in kind? I won’t allow this blog to descend into that kind of ruckus, which is why you won’t be commenting further here.Do you want a blog that is about true debate, or do want a talking shop for ‘people of the faith’?I run a blog where a requirement is that disputes over science are conducted courteously. You failed.

Goodbye Alan.

Hans,

It is not my intention to be disrespectful. I do mean to be clear. I am certain that the DALR satisfies the ISENTROPIC condition. I am also certain that adiabatic means no energy lost as HEAT – it says nothing about energy lost as work. You can confirm this via a Wiki search, which I provided, but has now been deleted. I am also certain that when an air particle expands it does work on it’s surroundings – it has to.

Whether the DALR is, for a dry ideal gas atmosphere, a steady state condition resulting from convection or an equilibrium condition is an interesting question. One cannot ignore the zeroth law of thermodynamics blithely.

Your exchange of potential energy enters my derivation of the DALR through the hydrostatic pressure profile: dp/dz = rho*g.

If you will look at the comments you will see I previously posed what i think is an interesting thought experiment. I did not find your answer satisfactory. To be honest, I don’t think you took it seriously. You should…

Please feel free to show me my mistake.

The dry adiabatic lapse rate is derived several ways, each of which assumes an ISENTROPIC NOT ADIABATIC atmosphere.

Example 1. A packet of dry air rises. Let’s first assert that it is adiabatic (hopefully not controversial), as it rises it expands and does work on it’s environemnt. But that work is reversible. Thus the process is adiabatic and reversible, hence isentropic. By definition.

Example 2. We know from calculus that dx/dy(w)*dy/dw(x)*dw/dx(y) = -1. Let us apply this identity to dT/dp(s), the temperature pressure relationship at constant entropy. We will find that dT/dp=-dS/dP * dT/ds. the second RHS term is Cp/T by the definitoin of entropy. The first RHS term is, by a Maxwell relationship, -dV/dT, which for an ideal gas is R/P (you can google Maxwell relationship). Making the substitutions: dT/dp = RT/PCp = V/Cp for an ideal gas. Lastly we recognize that the gravity imposed pressure gradient is dP/dz = rho*g, but rho is 1/V, so when we multiply: dT/dp * dp/dz = V/Cp * g/V = g/Cp.

Example 3. The state of an ideal gas is fully specified when two variables are specified. We shall choose T and p. The total entropy differential is: dS = dS/dT * dT + dS/dp * dp. at constant entropy, dS=0, then dT/dp = dS/dp * dT/dS and we proceed as above.

kdk33 says:

“Whether the DALR is, for a dry ideal gas atmosphere, a steady state condition resulting from convection or an equilibrium condition is an interesting question. One cannot ignore the zeroth law of thermodynamics blithely.”

Just in case I’m not understanding this passage could you confirm

“DALR is, for a dry ideal gas atmosphere, a steady state condition resulting from convection”

Is the case for air parcels moving up or down at constant speed?, changing temperature at 9K/km

“an equilibrium condition is an interesting question.”

Is the case for still air? => no temperature change

Both these cases satisfy the hydrostatic equilibrium condition and known as the neutral atmosphere.

AFAIK, the hydrostatic equilibrium condition is dp/dz = rho*g. It says nothing about temperature.

Perhaps it means something else. If so, please describe further or, much better yet, provide math.

Steady state and equilibrium are not the same thing. For simplicity, let’s take a dry ideal gas atmosphere and two cases

1) The planet is heated by radiation. At steady state, heat transfer in the atmosphere is dominated by convection and convection requires the vertical movement of air and this air must follow the DALR. Put another way, the planet surface is hot, the top of the atmosphere (OK, the troposphere, but let’s keep it simple) is cold, convection works as hard as it can to equalize those temperatures, but it can only drive the profile down to the DALR because it must also undergo isentropic compression and expansion as it convects.

2) At equilibrium, the planet is not heated by the sun. There is no weather. No air movement. There is, however, a gravity imposed pressure gradient. Over time, will the temperature gradient relax to what? The DALR. Or will it become isothermal. That, as the bard said, is the question.

I tend to think the DALR is a steady state condition, not equilibrium. But I can’t prove it and I think it is an interesting question. Granted, The first case is most interesting in the real world. But the question posed by case 2 is not irrelevant. The thought experiment I posed above address that question. the system should spontaneousy move to the highest entropy state.

Once we get that settled, then we can move on to how this might change onces understanding of the GHG effect. But that is for later.

kdk33 says:

“AFAIK, the hydrostatic equilibrium condition is dp/dz = rho*g. It says nothing about temperature.”

The hydrostatic equilibrium condition combined with an adiabatic atmosphere thermodynamics gives

dT/dZ = – g/Cp

You must have come across the derivation as Hans has posted it here.

Other than that there’s quite a nice derivation on the NASA site.

I’m particularly interested in the still air condition which, a Harvard publication tells me is quite stable at night.

The reason for my interest is that with little or no convection the only heat transfer methods left are radiation and conduction(diffusion)

This seems a good starting point to look for a GHE.

Brian,

Please look 2 posts up where I have derived the adiabatic lapse rate starting with an isentropic assumption on a dry ideal gas. This is actually the subject of my post.

I think you and I agree that hydrostatic equilibrium applies to pressure and not temperature.

Otherwise, I’m not sure what you are asking. Could you clarify?

kdk33 says:

“The dry adiabatic lapse rate is derived several ways, each of which assumes an ISENTROPIC NOT ADIABATIC atmosphere.”

I have looked at several derivations of the DALR all of them assume that the atmosphere is ADIABATIC.

I have never once come across a derivation that starts with an ISENTROPIC assumption explicitly stated.

An adiabatic atmosphere for sure assumes an adiabatic atmosphere.

Look at all the derivations of the DALR they stress the STATIC condition – no unbalanced force.

So even convection is not assumed as Hans has pointed out.

Bryan,

Yes, people are saying adiabatic. They are wrong, adiabatic is an incomplete description of the physics. The proper description is ISENTROPIC. And the assumption they are actually making isentropic, they just don’t seem to know it. You can google isentropic expansion; there is a lot of information out there.

You have now seen a derivation asuming an isentropic atmosphere. More than one. You have now seen mathematical proof that what they are actually assuming is constant entropy, not adiabaticity. I have provided these for you. Do you see any error in my math?

Hans says “The definition of an adiabatic energy situation is given by ∆E = 0”

I don’t believe this to be true. The following is from wiki (yea it’s not the greatest, but in this instance it is correct)

Adiabatic heating occurs when the pressure of a gas is increased from work done on it by its surroundings

(Please look at Wiki under adiabatic process, you will find it very interesting)

Work is energy. If work is done on or by the air parcel, it’s energy changes.

kdk33 says:

“You have now seen a derivation asuming an isentropic atmosphere.”

Yes I have no problem in agreeing that the DALR is also isentropic and that gravity is a conservative force.

I agree that the expansion/contraction work is stored/returned by the atmosphere.

The derivation makes the claim of a GHE of 33K seem a pure fantasy.

kdk33 says:

January 31, 2012 at 3:11 pm

Hans,

“It is not my intention to be disrespectful. I do mean to be clear. I am certain that the DALR satisfies the ISENTROPIC condition. I am also certain that adiabatic means no energy lost as HEAT.

The dry adiabatic lapse rate is derived several ways, each of which assumes an ISENTROPIC NOT ADIABATIC atmosphere.”

There is not need at all the use the concept of isentropy. I haven´t seen any derivation of DALR including the concept of isentropy. As you see I don´t nee it ín the head either. Furthermore i don´t care about the semantics about “adiabatic” as long as it means that it considers an air parcel inolated from its souroundings. This in turn means that ENERGY cannot enter or leave it which also makes it possible to apply the first law of thermodynamics to the situation.

The 2:nd law of thermodynamics alsoe relates to energy (all forms of energy) TB showed a version of that law which did not include heat at all. It only stated energy (in whatever form).

In my mind it seems that isentropy is a bad chose relating to my model atmosphere since friction and dissipation are processes that make sure that the equalising of energy per mass unit happens as time goes and before the STATIC DALR has formed. I only found isentropy in the context of processes. The static adiabatic situation in my model is an energy state, namely where all types of processes is available to maximize entropy to secure the evolution of DALR. Hope this has clarified your claims might not be correct.

If you do get familiar with the meteorological derivation of the DALR = g/Cp you cn imagin the vertical velocity getting closer and closer to zero and in the limit you will get an approximate Static DALR. This was once my way of thinking befor I created the versions in the head and in the E&E article.

Wikipedia says:

“An isentropic flow is a flow that is both adiabatic and reversible. That is, no heat is added to the flow, and no energy transformations occur due to friction or dissipative effects. For an isentropic flow of a perfect gas, several relations can be derived to define the pressure, density and temperature along a streamline.”

Hans

Bryan says:

January 31, 2012 at 8:39 pm

“The derivation makes the claim of a GHE of 33K seem a pure fantasy.”

Brian I really appreciate your contributions in this thread and also at WUWT. However you don´t seem to grasp why this is essential for the value of GE (33K).

First there are several definitions of the Greenhouse effect. I don´t favour the “circular” definition that claims GE is what is caused by greenhouse gases.

I consider GE to be confirmed by observations both on earth and on other planets. I accept the NASA definition as a fair approximation.

The static DALR show that there has to be a temperature increase in any atmosphere that can achieve or be near DALR in the troposphere.

Hence, the question arises why a static DALR should be of interest when discussing the energy flow in a real atmosphere.

My reasoning is simple:

A. Observational evidence shows that an approximate DALR can be found on Venus, Titan, earth, Jupiter, Saturn, Uranus and Neptune.

B. Can it be theoretically understood?

Consider my model atmosphere and take evenly spaced 1000 willis suns and let the energy enter with equal power (W/m^2). At the same time let the same IR power leave the system evenly spaced.

You have created a steady state system where the energy situation inside will be constant.

Any planetary atmosphere is in a long term steady state. It has a similarity to the adiabatic closed system in my model.

My theoretical reasoning plus observational evidence thus makes me refuse your statement that a “GHE of 33 k is a pure fantasy”. The correct statement should be. “The bulk of GHE is created by the tendency for all physical processes in the atmosphere to try to reach an energetic equilibrium. How well this is reached depends on a number of existing physical processes. This was actually the debate I wanted to see evolve when writing my article in 2003. Since then I have met much defaming remarks and inaccurate accusations. Just read the WUWT january 28, 2012 at 2:33 pm by Robert Brown. He is a professor and says that my E&E paper does not contain any mathematics making the point that it has to be wrong also for that reason.

He is wrong there is one equation. But the humor is that his comment covers 5 pages without a single equation and he hasn´t tried to falsify my conclusions in this paper. He is just stating his personal opinions. This time there are 5 equations, all very simple so professor Brown should be able to get the message, But despite 5 pages he is not coming close.

Hans

Hans says he cannot agree with my statement

“The derivation makes the claim of a GHE of 33K seem a pure fantasy.”

Hans says

“First there are several definitions of the Greenhouse effect. I don´t favour the “circular” definition that claims GE is what is caused by greenhouse gases.”

Yes and its for this reason that I find the term greenhouse effect not a particularly useful term.

The overwhelming majority understand by that term they mean IPCC science GHE.

This they say is entirely caused by the greenhouse gases.

The derivation in Unified Theory of Climate by Nikolov and Zeller of a properly implemented radiative temperature for the Earth confirms the G&T derivation of around 140K.

What IPCC science misses out is the energy storing capacity of the Earth surface and Sea as well as the adiabatic atmosphere.

The trouble with putting numbers to these storage systems is they generate differential equations as they are mutually linked.

I find I agree mostly with the analysis of Joseph Postma.

He is working on a simulation model of a rotating planet with storage and oscillating energy input from the Sun.

This he models with an RC circuit with a single diode to remove half of an AC signal.

It will be interesting to see how this matches up with Nikolov and Zeller

Hans,

I disgaree completely. You are wrong a the most fundamental level.

You claim that adiabatic means energy does not enter or leave the system. This is incorrect. Adiabatic means there is no heat transfer.

It isn’t just semantics, your phsics is wrong. When an air parcel expands, it does work on its surroundings. When an air parcel is compressed, work is done on it by its surroundings.

Work is a form of energy. Heat and work are (as limited by the second law) interchangeable – otherwise power plants would not work and I couldn’t have toasted my bread this morning. As an air parcel expands it loses or gains energy according to the work done on it, or by it. This is basic thermodynamics.

Let’s again refer to wiki:

**Adiabatic heating occurs when the pressure of a gas is increased from work done on it by its surroundings

**Adiabatic cooling occurs when the pressure of a substance is decreased as it does work on its surroundings

Furthermore, you are attributing changes in molecular energy to changes in elevation through the gravity field. This is incorrect. If I fill a jar with and ideal gas and take it into an airplane the temperature will not drop. Nor will temperature rise when I land.

You are conflacting the gravity imposed pressure gradient with an interchange of energy at the molecular level. The gravity imposed poressure gradient acts through the density.

I am very familiar with the meteorological derivation of the DALR. If you do not recognize my derivation as equal in every way to the meteoroological derivation, then I would encourage to review. The meteorological derivation introduces the pressure gradient dp/dz = rho*g, in exactly the same way I do.

Entropy, along with temperature, pressure, and volume, is one of the four fundamental thermodynamic variables. Thermodynamics cannot be defined seperate from entropy and atmospheric physics cannot be understood without a firm grasp of thermodynamics.

I’ve mentioned two excellent texts. I would again, highly reocmmen Smith and Van Ness.

kdk33: I wouldn’t be surprised if Hans doesn’t bother to reply to that. It is argument by repetition, assertion, and appeal to authority. Apart from the sciency bit about air packets doing work as they expand and having work done on them as they contract, which you have got wrong anyway.

Here’s why:

Any packet of air moving up has to be replaced by an equal sized packet moving down to maintain pressure in a neutral atmosphere. Thus any energy imbalance is canceled. You are considering single packets in isolation, and this is the wrong way to view the atmosphere.

Furthermore, providing Hans is correct about the lapse rate matching g/Cp in that neutral atmosphere in his gedanken experiment where there are no energy inputs or outputs from the system, the air packets won’t have any energy imbalance as they move up or down annd expand/cool, or contact/warm anyway.

So it comes back to the question of the total energy being composed of gravitational potential and kinetic energy. To understand why this 120 year old controversy is as yet unresolved, you need to read and spend some time thinking about the content of the Loschmidt thread.

Hans Jelbring and I, along with a lot of other people, are with Loschmidt, Lagrange and Laplace on this. On the other side you have Maxwell (who also argued by assertion) Boltzmann (who tried but failed to prove his assertion using probability theory) and Willis Eschenbach and Robert Brown plus acolytes.

My advice to you is to stop saying “you are wrong” and start saying “In my opinion you are mistaken because…”

In fact, I’m going to insist on it.

TB,

I take extreme exception to my your description of my argument. I have provided rigorous mathematical proofs and pointed you to top-notch texts – and other commenters have agreed that I am correct.

The atmosphere follows the same thermodynamics as any other ideal gas system – it is not special. There is no question, whatsoever, that the air packet gains and loses energy as it rises. There is no question that the proper assumption (and the one actually requried to derive the DALR) is constant entropy. There is no question that adiabatic does not mean equal energy.

I am interested in Hans’s work because I think the equilibrium versus steady state question is very interesting. I would like to know the answer. I have set up a thought experiment to do that. And I can’t get anything resembling a meaningful answer form Hans.

You can shut me down and run your site like Real Climate: No disagreement allowed. Fine.

I am on your side, and I am trying to help you get the science right. Please show me a single error in my equations . You can’t. Nor can you even engage me at a meaningful leave. But you can bluster like Gavin Schmidt because you have a PhD that says what you want to hear and no other opinions need apply.

Good luck and goodbye.

kdk33,

I think I covered most of what you are talking about in the Loschmidt thead here:

https://tallbloke.wordpress.com/2012/01/04/the-loschmidt-gravito-thermal-effect-old-controversy-new-relevance/#comment-12990

Let me know if that helps clear things up – or not.

Also, do you realize that the hydrostatic equation that you are using, dp/dz = rho*g, applies to a system composed of a non-compressible fluid and assumes a constant volume (V) – thus a constant density (rho) – and a constant temperature (T)? Be very careful when trying to apply this to a compressible gaseous system such as the atmosphere.

And what do you mean by your statement: “The gravity imposed poressure gradient acts through the density”? I’m confused. Density is just a result, not a cause.

Bill

kdk33 says:

February 1, 2012 at 10:04 pm

“There is no question that adiabatic does not mean equal (total/HJ) energy.”

Please show me why?/HJ

“I am interested in Hans’s work because I think the equilibrium versus steady state question is very interesting. I would like to know the answer. I have set up a thought experiment to do that.”

1. First, tell me what is wrong in my paper at the head.

2. Tell what is wrong with Williams derivation “William Gilbert says: February 2, 2012 at 2:01”

3. Write down your proof in a way similar to what I and William have done.

4. Stop repeating statements that you cannot prove or are challenged by me and William in our papers. Just point out exactly where we are wrong.

5. Stop accusing TB of keeping you out from here, It is your own choice. If you produce 3 I am quite sure TB will publish it on a thread if it keeps up to normal scientific standard or he might do so even if it doesn´t.

6. Before doing 3 study Williams paper very carefully since you seem to ignore mine. His paper might match your background knowledge better than mine even if mine should be simpler to understand in my opinion.

7. When you produce 3 I will be happy to directly comment on what you are writing without making new thought experiments for you to explain before I comment on your potential future paper.

Thanks for your cooperation and interest

Hans

Kdk33 is right on two points:

“You claim that adiabatic means energy does not enter or leave the system. This is incorrect. Adiabatic means there is no heat transfer.”Indeed so. The classic Carnot cycle heat engine is described as having alternating adiabatic and isothermal stages. At each stage work is either being done on the gas, or the gas is doing work. Clearly an energy exchange.

William Gilbert says:

“Also, do you realize that the hydrostatic equation that you are using, dp/dz = rho*g, applies to a system composed of a non-compressible fluid and assumes a constant volume (V) – thus a constant density (rho) – and a constant temperature (T)?”No, the equation stated is simply a local force balance. None of those restrictions apply.

William:

dp/dz = rho*g. this applies to any fluid, incompressible or not. for water (which is actually slightly compressible, but let’s set that aside) the math is simple. For a gas, the math is a bit complicated – you have to integrate and you also need to know the temperature profile (you need both p and T to know rho).

I am using this relationshipe correctly.

As for density, please see my comments to Hans. Upcoming, assuming my comments get past TB.

[Reply] I’ll take no more false accusations or innuendo before kicking you from the site. So keep it polite.William: you say this

If we assume that the atmosphere is at static or dynamic equilibrium

I am not sure what you think the differences are. In my terminology there is equilibrium (which may be your static equilibrium) and there is steady-state (which may be your dynamic equilibrium). I do not think the criteria for these two is the same.

This is, in fact, the question that I find interesting. The criteria for the two need not be the same – they might turn out to be the same, but physics does not require it. It seems to me that the DALR is the correct criteria for steady state. I do not think it is the correct criteria for equilibrium. But I cannot prove it to mysellf. Hence I find the question intersting.

Nick and kdk33,

As written, dp/dz = rho*g, assumes a constant “rho” which requires a constant “V” and “T” (and to be correct the RHS should have a negative sign). This works fine for calculating dp/dz, but if you want to derive conclusions about temperature or density, this equation is not very useful. This link talks a little about it:

http://mintaka.sdsu.edu/GF/explain/thermal/hydrostatic.html

You can do all kinds of things to modify the equation, but in the end, outside of the pressure gradient, you only get approximations for the real atmosphere. That is why I said to be careful how you use it.

Bill

William,

dp/dz = rho*g does not assume constant density. For an ideal as pv=rt and rho = 1/v so p/rt = rho. then dp/dz = p/rt * g. R is the gas constnat, g is the gravitational constant. p is pressure and there is no requirement that it be constant. t is temperature, and there is no requirement that it be constant; you do, however, need another equation that describes the temperature profile.

As further clarification, let us take the isothermal case, where t is constant. I can now separate variables

dp/p = g/rt dz, and then solve ln(p1/p2) = (g/rt)*(z2-z1). And this descrbes pressure as a function of position for a compressible, but isothermal gas. If the temperaure is not constant then I would need to solve the above equation simultaneously with the equation that describes temperature as a function of position.

If you review the standard meterological derivation of the DALR, you will find that they introduce dp/dz = rho*g exactly as I have. If dp/dz did not equal rho*g for a compressible fluid, the DALR would not be g/Cp. You may not like Wiki, but they do present the standard derivation here:

http://en.wikipedia.org/wiki/Lapse_rate

BTW, the negative sign depends on your convention for z, so it is arbitrary.

In summary dp/dz = rho*g for any fluid. Compresible or not.

Hans,

I am trying my honest, absolute best to be clear about where I think you are wrong and to provide rigorous proofs to back up my assertions. I believe I have done both and am very confused by your responses. Let me try agian.

You write: This derivation considers two air parcels of equal and suitable mass (a billion molecules) which have to carry an equal amount of total energy regardless of their altitude if an adiabatic condition is assumed.

I believe that this statement is wrong. Two air parcels do not carry an equal amount of energy if they are adiabatic. In fact, their energy changes because they either do work on the surroundings (expand) or have work done on them by their surroundings (compress). I believe the correct assumption is constant entropy.

This has to be the case. Let us consider a spherical air parcel. Inside the sphere there is a certain pressure. The force on the boundary of the sphere is the pressure divided by the surface area of the sphere F = P/Sa. When the sphere expands a force is acting through a distance and that is, by definition, work. Hence, the sphere loses energy in the form of work. And vice-versa if it compresses.

I believe that the proper assumption is isentropic. Not adiabatic.

Below I will provide 3 seperate proofs of my assertion. Two are mathematically rigorous.

Proof 1: A packet of dry air rises. Let’s first assert that it is adiabatic (hopefully not controversial), as it rises it expands and does work on it’s environemnt. But that work is reversible. Thus the process is adiabatic and reversible, hence isentropic. By definition.

Proof 2: We know from calculus that dx/dy(w)*dy/dw(x)*dw/dx(y) = -1. this can be found in any undergraduate calculus book. I learned it as the triplet rule.

Let us apply this identity to dT/dp(s), the temperature pressure relationship at constant entropy. We will find that dT/dp(s)*dp/ds(T)*ds/dT(p) = -1, and rearranging dT/dp=-dS/dP * dT/ds.

The second RHS term is T/Cp by the definition of entropy. The definition of entropy is ds = dQ/T, but at constant pressure for an ideal gas dQ = CpdT, therefore ds/dT = Cp/T and dT/ds = T/Cp.

The first RHS term is = -dV/dT (which for an ideal gas is R/p), by a Maxwell relationship, Maxwell relationship follow from the calculus identity that d2x/dxdy = d2x/dydx, the order of integration doesn’t matter. Applying this to differential form of the first law leads to the maxwell relationships. More detail is shown here: http://en.wikipedia.org/wiki/Maxwell_relations. You wiill find the same derivations in either of the thermodynamic texts I’ve recommended.

Making the substitutions: dT/dp = RT/pCp = V/Cp, becuase RT/p = V for an ideal gas.

Lastly we recognize that the gravity imposed pressure gradient is dP/dz = rho*g, but rho is 1/V, so when we multiply: dT/dp * dp/dz = V/Cp * g/V = g/Cp.

Proof 3. The state of an ideal gas is fully specified when two variables are specified. We shall choose T and p. The total entropy differential is: dS = dS/dT * dT + dS/dp * dp. at constant entropy, dS=0, then dT/dp = dS/dp * dT/dS and we proceed as above.

Now, Hans, once we get past this point, we can move to how I believe you are incorrectly applying potential energy and then we can move to my true interest: steady state versus equilibrium. But let us please, pretty please, settle this first.

Many people would like to ignore entropy; it’s a bit of a tricky concept. But the differential form of the first law for the four energies is:

dU = TdS – PdV

dH = TdS – VdP

dA = -PdV – SdT

dG = VdP – SdT

The four fundamental variables are S, P, T, and V. You simply cannot describe the energetics of the atmosphere without entropy.

So, Hans, my question is this: do you agree with me that constant entropy is actually the correct assumption for the DALR. If not, where am I wrong.

kdk33,

After pondering I disagree.

The expansion is doing no work. There is no energy change, if work was done there would be energy change.

“Adiabatic processes are those in which there is no net heat transfer between a system and its surrounding environment (e.g., the product of pressure and volume remains constant).”

“In adiabatic heating and cooling there is no net transfer of mass or thermal exchange between the system (e.g., volume of air) the external or surrounding environment. Accordingly, the change in temperature of the air mass is due to internal changes.”

This help anyone?

Entropy of an adiabatic process

http://www.physicsforums.com/showthread.php?t=259956

kdk33

“I believe that the proper assumption is isentropic. Not adiabatic.

Below I will provide 3 seperate proofs of my assertion. Two are mathematically rigorous.”

We must at all times remember the physical reality.

Just how rigorously adiabatic or isothermal is the real atmosphere?

K&T diagram 2009 tells us that the atmosphere absorbs one joule of solar radiation for every two that the Earth surface absorbs.

CO2 is radiatively active for the whole process.

Laboratory conditions are certainly not replicated.

Just how rigorously adiabatic or isothermal is the real atmosphere?

Shoud read

Just how rigorously adiabatic or isentropic is the real atmosphere?

Bryan,

Agreed that we are here dealing with a model system (a dry ideal gas atmosphere). But I think it is very useful for understanding what happens in the real world.

It has proven very useful in the past, and it is from this model that the DALR and MALR are derived. And they are useful.

TChannon,

There are two forms of energy: heat and work. they are, as limited by the second law, interchangable. A power plant takes heat energy generated by burning fuel and converts it into work – typically via steam generation and a turbine. The key is that energy comes in two forms, work and heat.

When the air parcel expands, it’s energy changes. It has to, please see my description of the work it performs.

Please carefully read the quotes you have provided. Adiabatic means no HEAT transer. It says nothing about work. Adiabatic cooling and heating occur becuase energy is lost or gained in the form of work. And it is the expansion compression that represents the work.

As an aside, I will point out that I am assuming, as does everyone else, that the expansion/compression work is reversible. This is a special kind of work that means the entropy doesn’t change. That’s why entropy is very important for understanding what is going on.

[Moderation note] kdk33 states: “There are two forms of energy: heat and work.” This is incorrect. There is also gravitational potential energy, and chemical energy.kdk33 says:

February 2, 2012 at 5:32 pm

“The four fundamental variables are S, P, T, and V. You simply cannot describe the energetics of the atmosphere without entropy.#

“So, Hans, my question is this: do you agree with me that constant entropy is actually the correct assumption for the DALR. If not, where am I wrong.”

I really hope you are serious which you for sure hasn´t showed yet.

No, I don´t agree. I will state in way similar to what you do, you simply cannot describe the energetics of the atmosphere without gravity involved. If you do you must be wrong.

The atmosphere is a system where potential energy is involved which seldom is the case in experiments where this variable is insignificant. If you move an air parcel aidabatically [adiabatically? –T] potential energy (gravity) has to be involved. Total energy has to be conserved. Hence, temperature has to change.

At the moment you inclose an insulated system its contained energy is CONSTANT. Still, the energy can be distributed “arbitrarily” at that moment. Its entropy can change but it can only change in one direction. It can only increase. This is the meaning of the second law of thermodynamics. Reactions can and will happen spontaneously without change of the energy content in the system. But it is an irreversible energy flow on its way to maximum entropy.

When the entropy is at maximum no further change in energy distribution can take place. When that has happened any EQUAL mass (a billion molecules) has to carry an EQUAL part of the constant energy inclosed in the system (regardless of its vertical position).

A look at the Wikipedia below tells about 2 systems where gravity can be dismissed. The third has an inadequate explanation. It does not mention that an “ideal” dynamic vertical adiabatic motion doesn´t require any (or very very little) energy to move.

Air expansion + mechanical potential energy + molecular kinetic energy stays constant. (approximately in a real atmosphere)

My way of interpreting an adiabatic atmosphere is that the inclosed energy has to reach a state of energetic equilibrium. That means delta E = 0 as in my proof. Just read it. There is no STATIC equilibrium before all winds and temperature inversions have disappeared in my model atmosphere.

Since my proof of a STATIC atmosphere gives a result equal to what in meteorology is proved for a “dynamic” dry adiabatic temperature lapse rate they both have to be at a (local in the latter case) maximum entropy state.

So in my vocabulary STATIC energetic equilibrium is when the entropy is at maximum. In that state DALR = -g/Cp. I don´t consider it necessary to use more concepts than what I have done. You can use your scientific language but now I have translated it to an easier one as in my proof which you refuse to discuss. Hope this helps.

Hans Jelbring

“Adiabatic heating and cooling (from Wikipedia)

Adiabatic changes in temperature occur due to changes in pressure of a gas while not adding or subtracting any heat. In contrast, free expansion is an isothermal process for an ideal gas.

Adiabatic heating occurs when the pressure of a gas is increased from work done on it by its surroundings, e.g. a piston. Diesel engines rely on adiabatic heating during their compression stroke to elevate the temperature sufficiently to ignite the fuel.

Adiabatic heating also occurs in the Earth’s atmosphere when an air mass descends, for example, in a katabatic wind or Foehn wind flowing downhill. When a parcel of air descends, the pressure on the parcel increases. Due to this increase in pressure, the parcel’s volume decreases and its temperature increases, thus increasing the internal energy.”

E2 = E1 + m0 g ∆z + m0 Cv* ∆T + (P1 V1 – P2 V2) ( 1)

I see a couple inconsistencies here right off the bat with your first equation.

1) you are using Cv (heat capacity at constant volume), yet the next term includes a change in volume. Is your m1 parcel of air at constant volume? or constant pressure? Neither?

2) more importantly, you leave out the buoyant force of the atmosphere. Your (m0 g ∆z) would be the work done on m1 (by my hand for example) when the gas is lifted from z1 to z2 — but only in a vacuum! The gas has more energy (when lifted in a vacuum) because my hand has done net work on the parcel of gas. (But note, this energy is entirely gravitational PE, so it does not warm the gas.)

In hydrostatic equilibrium, however, the upward buoyant force matches the downward gravitational force, and my hand doesn’t have to apply ANY net force to move m1 upward. No net force = no net work. The (m0 g ∆z) does not belong when the object is neutrally buoyant.

Or put another way, if I raise m1 up to where m2 was (increasing the PE of m1), then OTHER air has to get out of the way, with a net result of moving an equal amount of air down from z2 to z1, leaving the net energy constant. Your analysis leaves out this second half of the situation.

So, if the air column is in hydrostatic equilibrium, then I can raise or lower any bit of air I want with no net work done on the m1 and no net change in the potential energy of the air parcel. If the air is warmer at the bottom, I can raise that bit of air with no work done, then let it cool when it is higher up (warming the upper atmosphere a bit). I can take that cool air back down, and let it warm up, (cooling off the lower atmosphere a bit.) I can continue this as long as I like, doing NO WORK from the outside. Eventually, I could get the atmosphere to a uniform temperature profile.

At this point, what possible process would disturb this isothermal condition?

kdk33,

Your derivations in response to Hans are creative and entertaining. And it is much more elaborate than the standard derivation for the dry adiabatic lapse rate. (If you will notice, all these derivations, including yours, have a constant volume term that cancels out at the end). But I prefer my derivation from the first law that I described in the above link – it is much simpler and more straight forward with respect to the thermodynamics (as is Hans’ derivation above).

With respect to your emphasis on thermodynamic potentials and the corresponding natural variables, there is one problem. These relationships only hold when they are not under the influence of external fields. Unfortunately the gravitational field is one of those pesky fields and must be included in any thermodynamic equations that we work with. The classical thermodynamic potential relationships do not always hold in a gravitational field. See the first paragraph in this Wiki link:

http://en.wikipedia.org/wiki/Thermodynamic_potentials

Our atmospheric system is such that the working fluids encompass the vertical distance comparable to that between Mt. Everest and the Mariana trench. Gravitational potential energy must be included in all thermodynamic derivations. Just remember that, based on the 1976 Standard Atmosphere, approximately 30 % of the atmospheric internal energy is gravitational potential energy at 10 km. This cannot be ignored.

With respect to your question: Is the DALR indicative of an isentropic or adiabatic process? The answer is YES. It is both isentropic and adiabatic (it is reversible).

You also seem to be stuck with the concept of PV work energy and the first law. I thought I explained that in the above link which is here:

https://tallbloke.wordpress.com/2012/01/04/the-loschmidt-gravito-thermal-effect-old-controversy-new-relevance/#comment-12990

Have you had a chance to read it? Yes the parcel does work on the surroundings upon expansion (which is allowed in an adiabatic process). But the surroundings also does work to the parcel as it rises by replacing the lower density buoyant parcel with higher density parcels. PV work energy is thus transformed into gravitational potential energy and the total energy per mass unit is conserved. It is isentropic (and adiabatic).

You also seem to be interested in the difference between static equilibrium and steady state or dynamic equilibrium. I think you know, but let me try anyway. In both cases total internal energy (U) is conserved:

U = CpT + gz = Constant

And

dU = CpdT + gdz = 0

The difference is that no energy is entering or leaving the system in the static state but an equal amount of energy is both entering and leaving the system in the dynamic state (PV work energy in this case). But both systems will demonstrate the DALR when in equilibrium. The important point to remember is that molecular motion (e.g., convection) is not necessary to “

create” the DALR; it is necessary to “maintain” the DALR. For example, the formation of inversions (for many reasons) may disrupt a static (or dynamic) DALR, but daytime convection will restore it. The DALR is the isentropic equilibrium state in both a static and dynamic system.If I have missed anything, please let me know.

Bill

Hans,

I very much appreciate your thoughtful and kind reply.

—————————————————————————————————————

You say: you simply cannot describe the energetics of the atmosphere without gravity involved. If you do you must be wrong..

I say: I absolutely agree with you. You will notice that gravity enters my derivation through the pressure gradient. Specifically dp/dz = rho*g. g is the gravitational constat. I’m sorry that you missed it, perhaps I should have pointed it out more explicitly.

——————————————————————————————————————–

You say: Reactions can and will happen spontaneously without change of the energy content in the system. But it is an irreversible energy flow on its way to maximum entropy.

I say: I agree with you completely. If you wiill revisit the thought experiment I proposed above, you will see that I am applying the exact same criteria. I am asking which of a temperature profile equal to the DALR or an isothermal profile results in the most entropy and I am holding the mass and total energy of the system constant.

————————————————————————————————————

You say: If you move an air parcel aidabatically [adiabatically? –T] potential energy (gravity) has to be involved. Total energy has to be conserved.

I say: Yes gravity is involved. Through the pressure gradient. Total energy does not have to be conserved. This is not a requirement for adiabaticity, and is not consistent with adiabatic heating and cooling.

—————————————————————————-

You say: When the entropy is at maximum no further change in energy distribution can take place.

I say: Yes, I agree

——————————————————————————

You say: When that has happened any EQUAL mass (a billion molecules) has to carry an EQUAL part of the constant energy inclosed in the system

I say: No, this is not true and is not required by physics, it is not a general equilibrium criteria. I do not know which energy you are talking about (there are four, as I described above) and they cannot all be equal.

——————————————————————————

Hans, you dismiss the wiki section for reasons that I don’t understand. And adiabatic expansion is an adiabatic expansion under any circumstances. If you do not like that passage please read here where they are deriving the DALR. They derive it correctly and get the right answer, BTW. They say this:

Under these conditions when the air rises (for instance, by convection) it expands, because the pressure is lower at higher altitudes. As the air parcel expands, it pushes on the air around it, doing work (thermodynamics). Since the parcel does work but gains no heat, it loses internal energy so that its temperature decreases.

http://en.wikipedia.org/wiki/Lapse_rate

————————————————————————————–

You do not agree that the atmosphere is at constant entropy. I have provided multiple proofs. Please re-read them and point out my error. Show me a mistake in my math. Please take note of how gravity is introduced.

I am on your side Hans. I really am.

——————

I would much prefer that our discussion proceed in three parts

part 1: let us agree that the atmosphere is isentropic. This seems quite obvious to me, and I am befuddled that you find my proofs lacking. Never-the-less I shall persist. For a while.

part 2: let us discuss how potential energy is at work in the atmosphere and the mechanism by which it affects an air parcel.

part 3: let us discuss the difference between steady state (this may be your dynamic equilibrium) and equilibrium (which may be your static equilibrium).

With those issues clarified, we can design and experiment and perform the necessary calculations to determine which of these is consistent with the DALR and which is consistent with an isothermal atmosphere. (you can guess that I think my thought experiment does just that, but I think we need to go step by step).

William,

I very much appreciate your kind and thoughtful reply.

Stuck on PV work? I simply claim that such work exists and because it exists it changes the energy of the air parcel. I guess I don’t follow what you are saying. I have read your post.

Any system that is isentropic is, by defintion, adiabatic. I think on this point you and I agree.

I am pleased you find my derivation entertaining. Since you agree that the atmosphere is isentropic, then you must also agree that my derivations are correct – which is all that matters. Perhaps you could formulate your reasoning in a way that appeals to Hans.

If we can all three agree on that much, then we can move on to potential energy.

Yes, I like your definitions of steady state and equilibrium. We are saying the same thing; I would describe it differently. At steady state, the system is static in some coordinate systems, but not otheres. At equilibrium the system is static in all coordinate system (at a bulk not molecular level).

The important point to remember is that molecular motion (e.g., convection) is not necessary to “create” the DALR; it is necessary to “maintain” the DALR.

Ok, I was going to not take this bait, but I’ll do it anyway.

You’ve jump to the crux of the matter. I’m not sure I agree with you that convection isn’t necessary to “create” the DALR, but I don’t think it’s a very important point. Intuitively, I do agree with you that convection is necessary to maintian the DALR. This occurs at steady state, so I think the DALR is the correct steady state temperature profile.

OTOH, at equilbrium, I intuitively do not think convection will continue, and without it, I don’t think the DALR will be maintained, I think the system will become isothermal. Intuitively, I think the equilibrium condition is isothermal. But I can’t prove it to myself.

Hans starts his paper thusly: The ”static” dry adiabatic temperature lapse rate is derived for a hypothetical energetically isolated model atmosphere lacking advection and convection.

I believe he has actually derived the steady state temperature profile (dyanamic equilibrium) and not the equilibrium (static equilibriu) profile. And from your comments above, I think you agree.

But, since I cannot prove it to myself, I can’t be sure.

And that is why I find his work interesting.

But I want to get there step by step.

OK, I’m not necessarioly a patient guy. I’m going to jump ahead to the next step: potential energy. I’m quite worried that this will derail what I am trying to teach, but what the heck… One too many beers, maybe.

Here we go.

In Hans derivation of the DALR he converts potential energy into heat.

He says: E2 = E1 + m0 g ∆z + m0 Cv* ∆T + (P1 V1 – P2 V2)

And this is, in fact, a correct energy balance. Unfortunately, he is mixing two kinds of energy, and their interchange is a bit subtle. It requires one to wrestle with entropy.

Quick review. There are two kinds of energy: work and heat. Work is, more or less, bulk energy. Heat is, more or less, molecular energy. And it is an interesting interface. The interchange of work and heat is governed by the second law of thermodyanmics. This interchange is not, in general, 100% effecient. In fact, it is commonly much less; power plants, whose very business is the conversion of heat into work, struggle to exceed 60%.

Now, as an air parcel falls through a potential field, a force is exerted on that parcel that is proportional to itself and the proportionality constant is g, the gravitational constant. We all know this as F=mg. As the parcel falls, that force acts through a distance and does work – not heat, but work. dW = FdX, the general equation for the work. It does not make heat. And this is a critical point.

Now, if the parcel were falling through a vacuum, the parcel’s velocity would increase; potential energy would be converted into kinetic energy. It’s temperature and pressure woudl not change. We could accelerate that parcel to a tremendous velocity. It would carry tremendous energy. It’s temperature and pressure would not change. If we suddenly removed the gravity field, the parcel would continue to travel with tremendous velocity and carry tremendous energy and its temperature would and pressure would never change. Forever and ever.

The temperature and pressure of the parcel only change when that work is converted into heat, and that ihterchange is governed by the second law and that means… entropy.

A convecting parcel of air in a compressible fluid in a gravity field undergoes reversible expansion and compression. That expansion and contraction interchanges work and heat. This is unique. Reversibility is not common. Isentropic means adiabatic and reversible. Adiabatic is an incomplete description. The atmosphere is adiabatic and reversible. It is isentropic.

Why does this matter?

Because, as the air parcel falls through the gravity field the resulting force attempts to accelerate it; it is doing work. Simultaneously, the parcel is expanding against the other air particle that surround it. Reversibly, The reversible expansion allows the work done by the gravity field to be converted 100% to heat – the particle does not gain velocity (or not much, anyway). This feature, reversibility, is the essential property that allows Han’s derivation to work. The reversible conversion of work to heat via the expansion of the parcel against the parcels that surround it.

Earlier, I said that the gravity field acts throught the density. A better statement would have been the density of a compressible fluid. I hope my description above makes clear what I mean. The conversion of potential energy to heat requires that the parcel expand and do work on it’s surroundings. It only does work when it is surrounded by other parcels. It only does work because of the density and compressiblity of it’s surroudings.

Every derivation of the DALR requires an isentropic atmosphere, but very few people recognize it. In Han’s case, he converts potential energy to heat with 100% effeciency. That is only possible becuase of the reversible nature of thte contraction and expansion in the atmosphere. And that reversibility means constant entropy. if the atmospher were not isentropic, the interconversion would be less than 100% effecient and Hans derivation would not work. He would have to include an effeciency. That effeciency would be found via the second law. He would get a different value for the DALR.

In summary, Hans derivation works, but he is combining two steps. The step he is not explicitly stating is that potential energy can be converted 100% to heat and that requries a reversible expansison contraction, and that requires constant entropy. The atmosphere is isentropic. That is an essential, not trivial point.

Well, I’m not really sure, right now, why I care. But I certainly hope that somebody out there cares enough to get it. I would really like to move on the part 3: steady state versus equilibrium.

[Moderation note] Thank you for the improved tone of your foregoing replies. This is just a gentle note to point out that there is no Teacher-Pupil relationship here, so “what I am trying to teach” is misplaced. This is a scientific discourse between knowledgeable people who will treat each other with mutual respect under the assumption of equality on this site. It’s also very interesting, so keep up the good work.Hans… have you read this? Might be a good book to have… the authors agree with you it seems.

http://wattsupwiththat.com/2012/01/24/refutation-of-stable-thermal-equilibrium-lapse-rates/#comment-883144

[Reply] Thanks Wayne, I’m reproducing the relevant part the comment from WUWT contributor ‘Trick’ here for those whose browsers/computers struggle with loading that enormous thread:

“My copy of Bohren & Albrecht “Atmospheric Thermodynamics” arrived today. I have had a delightful time reading thru about half of it so far. Thanks to Rodrigo Caballero’s on-line text for pointing the way.The authors have many humorous anecdotes to relate & the chapter problems given are very thought provoking. I heartily recommend it to any poster interested in the topic of this thread.

On-topic for the top post, the authors develop the fundamental application of the thermo laws with the proper & correct detail algebra for Figure 1 in the top post – a GHG-free adiabatic ideal gas column in the presence of gravity. This gas column is isolated from its surroundings, neither heated or cooled by radiation nor by interaction with adjacent air or ground. There is no condensation or evaporation of water. (The same as that discussed in Verkley paper part b.)

The authors derive entropy maximization requires the equilibrium temperature profile to decrease as height increases. Thus they prove the hydrostatic temperature profile of top post Fig. 1 at equilibrium is non-isothermal which is the max. entropy point. ANY deviation from that thus requires a decrease in entropy which is not 2nd law physical.

Why isn’t the equilibrium temp. profile isothermal as Robert Brown writes (repeatedly)? The authors have a long explanation of why the temp. profile is non-isothermal. It needs to be read in context. They give a reason Robert Brown’s understanding here has been impeded – their reasoning is the conduction in solids impedes understanding of the ideal gas column w/gravity field non-isothermal temperature profile in equilibrium in top post Fig. 1.

I have edition 1 of that fine text book and, for interested posters, the non-isothermal stable equilibrium temperature profile derived from maximization of entropy of the GHG-free adiabatic ideal gas column top post Fig. 1 derivation is in Chapter 4.4 pp.164-168 for realistic pressures (holds for ~80% of pressure range of earth’s atmosphere).

Furthermore the authors point out only in the absence of gravity will the adiabatic ideal gas column equilibrium temperature be isothermal.

The authors elegantly show the top post statement is incorrect “..an ideal gas in a gravitational field in thermal equilibrium because, as is well known, thermal equilibrium is an isothermal state..” and thus refute the refutation of a stable equilibrium isothermal temperature profile for Fig. 1 in the top post. This non-isothermal profile being consistent with poster Rodrigo Caballero’s referenced 2004 Verkley paper part b which shows experimental evidence for the ideal non-isothermal temperature profile in their Fig. 2.”The battle of ‘authoritative textbooks’ continues.🙂I’ve uploaded a copy of the Verkley paper here.

Wayne,

Your comment is kind, thoughtful, and fascinating. thank yhou.

In your book: The authors derive entropy maximization,

You will note that this is exactly the thought experiment I have proposed. In my initial proposition I say: Which case has the most entropy? It seems to me that if you answer that question and you will know which is the thermodynamically stable condition.

and in my exchange with Hans, I agre with Han’s statement: When the entropy is at maximum no further change in energy distribution can take place.

So we all seem to agree that maximum entropy is the equilibrium condition.

Perhaps you would be so kind as to outline the authors method? Could you compare and contrast their approach to the thought experiment I proposed above? Perhaps you could show us their math?

Of course, this completelyh short circuits my 3-part master plan. But, oh well.

Well, since my 3-part plan is in the tank….

It is interesting to compare the various mathematical derivations of the DALR and then relate that to the underlying physics. I’d like to discuss three: mine, wikis’s, Han’s.

All three begin with a dry ideal gas in a gravity field, and we will make it so.

———————————————————————————————————-

In my derivation I begin with dT/dP(S), the temperature pressure relationship at constant entropy. I am explicitly requiring my atmosphere to be isentropic. After some math and a couple of identities I arrive at: dT/dp = RT/pCp = V/Cp, I then introduce gravity and potential energy through the pressure gradient, which is dp/dz = rho*g = g/V. multiplying these two I arrive at dT/dz = dT/dp*dp/dz = g/Cp.

—————————————————————————————————————–

In the wiki derivation (found here: http://en.wikipedia.org/wiki/Lapse_rate) they take the same approach, in a bit of a round about way. They start with PdV = − VdP / γ, where gamma is the ratio of heat capacities, Cp/Cv. Now, they say that this is for an adiabatic system. They are wrong. It is for an isentropic system. As proof, let us complete their math…

I separate variabales to find gamma*dV/V = dP/P. Upon integration I get: gamma*ln(V2/V1) = lnIP2/P1), then taking exponential we find: (V2/V1)^gamma = P2/P1. This is the isentropic pressure volume relationship, as most will recognize. Funnily, wiki has a section on isentropic expansion where they call this relationship out correctly (it can be found here: http://en.wikipedia.org/wiki/Isentropic_process). The inconsistency is striking, but who knows who writes this stuff…

As an aside, it is worth mentioning that the state of an ideal gas is completely specified once two variables are fixed. I choose T and P; Wiki chooses V and P; it makes no difference.

Anyway, wiki then introduces gravity and potential energy exactly as I do, through the pressure gradient.

I won’t repeat all of their math – the reader can easily follow the link – I’ll simply show that their second equation, CpdT = Vdp, can be rearranged to dT/dp = Cp/V. And then they proceed exactly as I do above.

The key point is this: they say adiabatic, but their math only holds for an isentropic system. If it were not isentropic, work and heat would not interchange at 100% efficiency and their pressure volume relationship would be different and they would find a different value for the DALR

————————————————————————————————————————-

Now to Hans derivation… He lumps the two step process into one by combining heat energy, PV work, and potential energy. He explicitly places potential energy into the initial energy balance, but does not explicitly require an isentropic atmosphere. However, the assumption is implicit.

Now I could simply assert that, since Hans find the lapse rate to be g/Cp, then this atmosphere must be isentropic, and refer the reader to the two derivations above. I would be correct, but that would be no fun.

Instead, lets do the following, Hans penultimate equation is g*dz = Cp*dt, which leads directly to the lapse rate. Let us show that this is equivalent to constant entropy under the requirement that dp/dz = rho*g (which is how I and wiki introduce gravity). We proceed as follows dT/dp = dT/dz * dz/dp = g/Cp * 1/(rho*g) = 1/(rho*Cp) = V/Cp. So dT/dp = V/Cp. And this is the isentropic temperature and pressure relationship derived above.

Now, I’m hoping that we can all be one big happy isentropic atmosphere family and we can move on to discussing what is really interesting: steady state versus equilibrium and how one might address those (But… I’ve already given away how I think that should be done, and a previous commenter may be in possession of the solution I seek. In a way that will be nice, but it kinda takes away a lot of the fun).

I can always hope, and hope springs eternal (insert smiley face).

Hey, let me out of jail. What did I do this time?

Kdk33: Sorry, out for a friday pint.

🙂

In my reading of Wayne’s comment (or is it ‘trick’). I did not notice the last line. I now have the Verkley paper. It seems that everyone agrees that maximum entropy is the equilibrium criteria. But there is another layer to the onion…

Folk apply different constraints. I will have to think on this. The obvious constraints seem to be constant mass and energy. But, apparently it is not so simple. Just as I thought I was closing in on the answer… The fun continues.

The Verkley paper in Table 2 contains an intermediate figure, presumably arbitrary which is close to a magic number, Golden ratio.

Tim, the intermediate figure is not arbitrary. It is derived using both the H and L constraints and leads to an alpha which matches the observed atmospheric profile. An interesting paper. On first reading it looks like a compromise of sorts may be possible between the isothermal and isentropic positions, which relates closely to the real atmosphere.

I’m not so sure the alpha value at 1.686 is all that close to the golden ratio of 1.618 but what is noticeable is that the observed value between the isothermal and isentropic limits as a fraction of H and L is close to half the alpha value.

kdk33: I’ve taken you off moderation; please don’t give me cause to regret it. Thanks.

kdk33 says:

February 3, 2012 at 2:33 am

“In summary, Hans derivation works, but he is combining two steps. The step he is not explicitly stating is that potential energy can be converted 100% to heat and that requires a reversible expansion contraction, and that requires constant entropy. The atmosphere is isentropic. That is an essential, not trivial point.

Well, I’m not really sure, right now, why I care. But I certainly hope that somebody out there cares enough to get it. I would really like to move on the part 3: steady state versus equilibrium.”

Thanks for your effort, curiosity and will to understand. It forces me to think and expand my concepts, too. I agree that “steady state versus energetic equilibrium” is highly interesting and should be discussed.

About your first paragraph above I have tried to recapitulate my minor knowledge in the chemical language. As far as I can understand “isentropic” means constant entropy. This is true about my model atmosphere but also a maximum entropy is needed which is a stronger claim. My claim to consider and compare ANY two air parcels with equal mass and claiming delta(E) = 0 requests that a maximum (energy) entropy situation has been established.

My equation (1) in the head means that potential energy can be 100% converted to heat AND PV work if delta (E) = 0 (Adiabatic system by definition that have reached a maximum entropy state according to the second law of thermodynamics operating spontaneously with energy redistribution in a gravity field. Hopefully I got it right).

tallbloke says:

February 4, 2012 at 9:58 am

“An interesting paper. On first reading it looks like a compromise of sorts may be possible between the isothermal and isentropic positions, which relates closely to the real atmosphere.

I’m not so sure the alpha value at 1.686 is all that close to the golden ratio of 1.618 but what is noticeable is that the observed value between the isothermal and isentropic limits as a fraction of H and L is close to half the alpha value.”

I am really surprised that Verkley (a meteorologist) doesn´t know that the total energy per kg in any real atmosphere column is lowest at the surface and then increases in a

~~monoton~~[monotonic? –T] way upwards.This is a consequence of hydrostatic balance.A real atmosphere that will be inclosed as my model atmosphere will always transport energy towards the surface after inclosure before reaching an energetic equilibrium.

I like that Verkley is using Holton as a reference. Holton said after deriving DALR “Hence, the dry adiabatic lapse rate is approximately constant throughout the lower atmosphere.” (dt/dz = -g/Cp page 49, second edition)

kdk33, the credit all goes to ‘Trick’. Don’t remember that coined name at WUWT before but whomever he or she may be, I’m sure at least half of the commenters really appriciate that information returned. 😉

Hans,

I very much appreciate your reply.

I think we agree on many, many things. I’m strapped for time at the moment (family weekend obligations), so have time for only a short note. (besides, I’m still wrestling with potential temperature).

I recognize maximum entropy as the proper equilibrium configuration [of] the atmosphere. That may, in fact, be equivalent to your equal energy concept. But I do not recognize equal energy as a general equibirum criteria.

For example, If I have a container with 1gm of water and 1gm of steam at 100C and 1 atms, these are in equilbirum, they have equal Gibbs eneriges, but they have different enthalpies, entropies, and internal energies. If I isolate the container from it’s surrounding, the water and steam will remain as such. Forever and ever.

It could be that once the proper atmospheric relevant constraints are applied the maximum entropy state is the equal energy state, but It isn’t clear to me at the moment. If you have some math that shows that I would very much like to see it.

I think that your equal energy is also equal entropy, but I don’t see how to show that equal entropy is the maximum entropy state.

_____________________________________________

I’m toying with another concept: Given that the atmosphere is completely reversible (isentropic), is it the case that any perturbation (for example a little heat transfer from hi to lo temperature) would cause a reversible response that would restore the system imediately to its constant entropy state. In this case, is the atmospheric specific entropy invarient to any change?

If so , then dS/dx = 0 where x is any perturbation we care to define (allowable under the appropriate atmospheric constraints, of course). This would be half of the mathematical criteria for maximum entropy. Is there a way to further show that d2S/dx2 < 0, which would be other half. But if S is invarient, then d2S/dx2 = 0, so is an inflection point, so the atmospher is in a "quasi-equilibirum" state. Maybe inflection point equilibrium is the correct equilibrium in a potential field… You see my ideas here are poorly formulated.

I think this more general demonstration of maximum entropy would be more satisfying than my original thought experiment, or the ones in the Verkely paper. I'm looking for something more general now that I know people get different answers applying different constraints (but I still need to sort through what these different constraints mean exactly).

If you have any thoughts on how to show that constant entropy is the maximum entropy state, I would love to hear them.

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More later. I very much appreciate your response.

Bryan;

“An adiabatic atmosphere for sure assumes an adiabatic atmosphere” You might like to rephrase that. So it doesn’t sound so, you know, inane.

[Reply] The correct word for describing a tautology is, you know, tautological.😉kdk33,

Before you go too far in approaching the system from the standpoint of entropy, it may be a good idea to define entropy in the presence of an external gravitational field. You can’t just use the classical derivations for entropy. Gravity and entropy is not that well understood, at least by me anyway. I have seen some analyses that treat gravity as negative entropy. That is why I have had trouble using free energy to determine the equilibrium state for a system influenced by a gravitational field.

I would welcome a discussion on this issue. I have played around with the concept but have not gotten serious about it.

Bill

Brian H says

“You might like to rephrase that.”

You have to read the tautology in the context of kdk33s posts that preceded it.

kdk33 wants to derive the adiabatic condition from the entropy no change condition.

This is as far as I know a unique kdk33 derivation of the DALR.

All other derivations directly apply adiabatic conditions hence the name adiabatic atmosphere.

Which came first, the chicken or the egg?

More partially formed thoughts….

I’m wondering if the following approach would get anywhere. Begin by noting that, absent a potential field, the state of an ideal gas is completely specified once two variables are fixed. If I explicitly account for the potential field in the energy balance there will be an mgz term, and it will take three variable to completely specify the gas – I will choose as the new variable g= gravitational proportionality (it need not, in general, be constant, eg. in a centrifuge).

The two situations that are of interest to me are the isentropic case and the isothermal case. The isentropic allows for a temperature gradient at equilbirum, the isothermal case does not.

Now, I define two new energy functions E* and E** such that E*(H,S,g), E**(H,T,g). So E* is a function of enthalpy, entropy and the potential field. E** is a function of enthalpy, temperature, and the potential field. I can then considre two equilbirums. For the first, I require E* to be equal everywhere under the condition that H,S, and g be equal everywhere. In the second case I require E** to be everywhere equal under the condition that H,T, and g be everywhere equal.

I’m thinking that, in this way, a very general equilibrium can be defined for the isentropc and isothermal cases and then one can compare to determine which gives the greatest total entropy…

OK, there’s probalby lots wrong with that. But I’m starting to like that approach.

Enthalpy is the wrong choice; it was a dumb idea. I think the better choice is P0=pressure at the planet surface. But I still like the approach above. So, I want to define 2 energies

E* = E*(P0,S,g) and E**=E**(P0,T,g), then I define a third, fictitous, energy E***=eps*E* + (1-eps)*E**, where 1>eps>0. *E**. Then write S as S(eps) and find what value of eps yield the highest entropy.

Procedure: define, mathematically, the first two energies above, then require that both dE=zero when their independant variables are constant, then join these two through eps, then find the value of eps that maximizes S, entropy.

BTW E* lends itself to describing an isentropic atmosphere, or one that follows the DALR. E** lends itself to describing an isothermal atmosphere. Eps, lets us move back and forth between the two cases. As an aside: setting P0 constant is setting the mass of the atmosphere constant, then P0 is found from dP=rho*g…

Anybody wanna help with the math?

My first post on this forum, and sorry for joining the discussion late. I have been trying to find a theoretical foundation for how a static DALR might form, and have so far failed to find one. Hence I was interested in the above paper. However, Eqn (1) is incorrect. The problem is that temperature is an ensemble property over all the molecules at a particular altitude. When considering a higher altitude, one has to account for the fact that not all the molecules at a lower level can reach the higher level, as many won’t have enough KE to climb that high. If one considers an ensemble with mass m0 that climbs to a higher altitude, one cannot just choose any molecules that make up the ensemble m0 – only the hotter ones can do it. It is then true that these hotter molecules slow down and cool down. BUT the temperature at each altitude is not determined by a special subset of molecules. Hence T1 and T2 used in Eqn(1) to make DeltaT do not refer to the temperature at each altitude, but only to the temperature of a special subset of molecules. The actual temperature at each altitude, taking an ensemble representing *all* molecules at each altitude, remains isothermal under the ideal gas laws under gravity (which are all that are assumed in the above paper). This has been shown rigorously many times. Comments?

br1 says: March 14, 2012 at 11:18 am

” The actual temperature at each altitude, taking an ensemble representing *all* molecules at each altitude, remains isothermal under the ideal gas laws under gravity (which are all that are assumed in the above paper). This has been shown rigorously many times. Comments?

What I have shown in my calculation is that you are wrong in your statment above. My equation (1) is correct. Most calculations don´t account for the impact of gravity which has to be included in the total energy of any atmospheric mass and also the delta (PV) term has to be understood, too. Just show me ONE of the derivations you consider correct. To refer to a multitude of works others have done is no valid argument.

In case my previous post got lost (I received no ‘awaiting moderation’ notice), please have a look at http://www.slideshare.net/brslides/maxwellboltzmann-particle-throw and http://www.slideshare.net/brslides/walton-applied-tovanderwaalsgas where I have written a simulation and derived some maths to find the temperature vs height of a gas under gravity, even including intermolecular attraction and the Joule-Thompson effect. How can you justify a static DALR from a microscopic point of view?

Ah, I see my other post did get lost. Anyway, I had written that the only way I can see of getting a DALR is by convection, which gives the particles a non-Maxwell-Boltzmann distribution as the particles receive an extra vertical velocity boost. This non-MB distribution can indeed cool down, but it acts as an energy transport mechanism and so must either bring separate system components to thermal equilibrium or else be available as a perpetual motion machine by extracting heat from a heat bath. In any case, you do not wish to deal with this as your condition 2 states that there are no winds. The problem (as I see it) with your Eqn(1) is that the mass term is taken over different ensembles – which must be the case seeing as the density drops with height. If I could get a microscopic justification for a static DALR, I would be very happy!

br1

Thanks for putting such effort in – very interesting.

Only point I would add is that hydrostatic equilibrium (the condition for the DALR) satisfies Newtons First Law.

This has a stationary and a constant speed solution.

Meteorologists call this the Neutral Atmosphere.

Its very similar but not the same as convection.

Parcel’s of air moving at a constant speed (no unbalanced force) may give a non isothermal distribution.

How easy would it be to test this condition?

br1,

In your Maxwell-Boltzmann slideshow, as I understood it, your initial boundary condition is that all particles are in thermal equilibrium and originating at the surface. Those particles with the higher velocity distribution will of course reach a higher altitude. But this initial boundary condition guarantees an isothermal distribution. Change your initial boundary condition to a state where all particles are in thermal equilibrium and originate at altitude. I believe your results will be quite different. The next step would be to set the initial boundary condition at thermal equilibrium where the particles are evenly dispersed between the surface and altitude.

I was unable to read the paper in your Walton slideshow so this may have been covered there. But let me know if this helps you get to a microscopic justification for the DALR.

Bill

Twiddle, how about this http://polymer.chph.ras.ru/asavin/teplopr/ttk98pre.pdf

Bryan says: March 22, 2012 at 6:57 pm

(My text in your comment for clarity in this way /HJ)

Thanks for putting such effort in – very interesting. (Thanks/HJ)

Only point I would add is that hydrostatic equilibrium (The condition for the DALR) satisfies Newtons First Law. (The hydrostatic equation is not enough for the conventional derivation of DALR. You also have to consider the adiabatic condition which is an air parcel which keeps its total energy constant/Hj)

This has a stationary and a constant speed solution. (Yes, and I consider one of my contribution to science is to prove that the static and dynamic solution is the same dt/dz = -g/Cp /HJ)

Meteorologists call this the Neutral Atmosphere. (It is very important since it occurs quite often in nature meaning any planetary atmosphere/HJ)

Its very similar but not the same as convection. (Unclear statement/HJ)

Parcel’s of air moving at a constant speed (no unbalanced force) may give a non isothermal distribution. (I assume you mean vertical speed where very little or no energy is added to the air parcel. It has to move “along” the DALR both up and down and it is -9.8 c/km in that case/HJ)

How easy would it be to test this condition?

(Very easy by observational evidence. Just one example: It is shown be any the temperature profiles at Koorin expedition data for 30 consecutive days between 1200 and 1600 fromn 0 – 1000 m above surface/HJ)

William Gilbert says: March 22, 2012 at 7:07 pm

Nice to hear you are still interested. I would like to get more of your balanced contributions.

Hans,

I am still interested but I may not be up to date with the various threads that Tallbloke has going on this subject. I just happened to receive an email notice of the br1 post at this site (because I had previously posted at this thread). If there are interesting discussions ongoing at other threads, please let me know.

Bill

Bryan:

“Only point I would add is that hydrostatic equilibrium (the condition for the DALR) satisfies Newtons First Law. This has a stationary and a constant speed solution. Meteorologists call this the Neutral Atmosphere.”

Very interesting observation – I will have to think about this some more! My present thoughts are that in a finite volume (such as an atmosphere or a box), a constant speed solution will look like a convection cell, as what leaves one place has to be filled in from the surroundings. I think this is indeed a long-term solution of Newton so long as you have perfectly reflecting walls, as then there is nothing to dissipate the curl of the flow. It will probably also give a long-term temperature gradient, and as you say, is not really the same thing as convection, because it doesn’t need the gradient to drive it. BUT the problem I see is that this is basically like a sustained electric current in a superconductor (which doesn’t need a constant voltage applied), but as soon as you have some ‘resistance’ (such as radiation loss), or scattering (such as thermalisation with the ground or the walls of the container), or mixing (where the different flows in the pseudo-convection mix), then the flow will eventually cease. At that stage you are back to a gas with no currents, and then as far as I can tell my slideshow analyses apply again.

But still a very interesting idea that I will definitely consider.

William:

“But this initial boundary condition guarantees an isothermal distribution”

Well, this was the question I wanted an answer to – will it or won’t it. And if it guarantees an isothermal distribution, where can a temperature gradient come from? And how does that square with Hans’ analysis?

“Change your initial boundary condition to a state where all particles are in thermal equilibrium and originate at altitude. I believe your results will be quite different.”

Afraid not – as the distribution is isothermal, then it doesn’t matter which height they start from.

“The next step would be to set the initial boundary condition at thermal equilibrium where the particles are evenly dispersed between the surface and altitude.”

I’m not sure I follow this – i take it that you suggest to start the distribution in a way that doesn’t satisfy the barometric formula, and see how the distribution progresses? In the above simulation, the particles always hit the ground and then thermalise – when they do this once, they completely lose memory of their initial conditions. So the initial distribution gets lost very quickly.

“I was unable to read the paper in your Walton slideshow so this may have been covered there. But let me know if this helps you get to a microscopic justification for the DALR.”

There is a ‘fullscreen’ button at the bottom left – I apologise that the text is not clearer, but I could read it on fullscreen. Thank you for your interesting suggestions, but I’m afraid I still don’t get it.

tchannon: “Twiddle, how about this http://polymer.chph.ras.ru/asavin/teplopr/ttk98pre.pdf”

yes, that is the paper I used.

Hans:

Did you read the presentations I posted at

http://www.slideshare.net/brslides/maxwellboltzmann-particle-throw

and

http://www.slideshare.net/brslides/walton-applied-tovanderwaalsgas ?

What am I missing?

I tried to convert your Eqn(1) into the microscopic realm (which I visualise more easily), but I keep getting that it is only valid for rising air parcels or convection, because terms such as PV depend on gas volumes, which implies that the mass refers to the mass of that air parcel, which must move as a whole from height 1 to height 2. But you claim that there is no convection or winds, so I think there is a contradiction.

br1 says: March 23, 2012 at 9:37 am

“Hans: Did you read the presentations I posted at

http://www.slideshare.net/brslides/maxwellboltzmann-particle-throw and

http://www.slideshare.net/brslides/walton-applied-tovanderwaalsgas ?”

“I tried to convert your Eqn(1) into the microscopic realm (which I visualise more easily), but I keep getting that it is only valid for rising air parcels or convection, because terms such as PV depend on gas volumes, which implies that the mass refers to the mass of that air parcel, which must move as a whole from height 1 to height 2. But you claim that there is no convection or winds, so I think there is a contradiction.”

“What am I missing?”

Probably that you stick to treating microscopic events when you should consider macro properties.

temperature is not defined for single atoms, at leas not in the ideal gas law.

DALR for an adiabatic air parcel moving up and down has been derived since long by meteorologists and is well known. You need the ideal gas law and the hydrostatic equation plus the first law of thermodynamics (applied to energy). See Holton, Introduction to Meterology, page 47-49.

The way which I persuaded myself that the “dynamic” DALR was equal to the static one was quite simple. Move the adiabatic parces very slowly upward and in the limit the dynamic ones gets “static”.

This thinking made me investigate the energetic situation of the atmosphere and I wrote my E&E paper 2003 and the alternative derivation at https://tallbloke.wordpress.com/2012/01/25/hans-jelbring-an-alternative-derivation-of-the-static-dry-adiabatic-temperature-lapse-rate/

Constant temperature in an atmosphere is energetically unstable and will turn into an equal total energy per mass unit if insolated according to the 2:nd law of thermodynamics applied to energy.

This statement is confirmed by massive observational evidence in physical situations where the atmosphere get well mixed for different reasons, a situation that applies approximatly well to the troposphere on any planetary atmosphere with a dense atmosphere but not above that level.

Hans: before I reply to your post, I’ve set myself to reading the comments more thoroughly. I got as far as Genghis’ comment and had to reply to that:

Genghis:

“Is there anything wrong with my thought bubble of single (perfectly elastic) atom bouncing up and down in a perfectly insulated, frictionless tube?

At the top of the tube the atom comes to rest (momentarily) and has zero kinetic energy, zero temperature and a lot of potential energy. Gravity accelerates the atom down to the bottom of the tube where it has maximum kinetic energy and heat and zero potential energy. Since heat and temperature are a measure of kinetic energy alone, that establishes the lapse rate, zero temperature at the top, 50% kinetic energy halfway down and 100% kinetic and temperature at the bottom.”

But that is just what I simulated in http://www.slideshare.net/brslides/maxwellboltzmann-particle-throw . The reason this doesn’t produce a temperature gradient is because the atoms are thrown up with a *distribution* of velocities. It is the velocity distribution at each height that determines the temperature at that height, and one finds, as if by magic, that the velocity distributions (and hence the temperatures) are always the same, no matter what the height! Provided you start off with a Maxwell-Boltzmann distribution, that is.

I didn’t really believe this until I did it for myself…

Hi br1 and thanks for bringing your well presented analysis here.

Just to pick up on a point Hans raised above about considering the macro situation.

Isn’t it the case that an air parcel will expand as it rises? It seems to me that this is connected to the reason why air is less dense at high altitude. So two parcels of air with identical mass as Hans specifies will occupy different volumes. Since there is only room for so many parcels at high altitude, the overall energy contained in a spherical ‘slice’ of atmosphere at the higher altitude is going to be less than a ‘slice’ of equivalent volume at low altitude.

Therefore the heat capacity of the upper atmosphere will be less than that of the near surface atmosphere, even if the temperature of individual molecules were the same.

Does that make any difference to the consideration of the system?

Hans:

“The way which I persuaded myself that the “dynamic” DALR was equal to the static one was quite simple. Move the adiabatic parces very slowly upward and in the limit the dynamic ones gets “static”.”

OK, that clears up where the equations come from – they *do* represent rising air parcels. But then we are left with the problem that in a static atmosphere, there aren’t really any such things as air parcels, and they don’t rise at all, never mind slowly.

I must confess I am not comfortable with the whole air parcel thing – and in my very brief (I’m new to this so far) reading of atmospheric discussions, a lot of people don’t seem comfortable with them either. They seem to be very ill-defined, hence people try to make them clearer by putting them in glass jars or latex balloons. The balloon concept I can sympathise with – the entire volume of gas inside the balloon rises as the balloon rises (no particles are dropped away with altitude or mixed in with the ‘surroundings’). This makes for a clear object of analysis. In a static atmosphere, balloons will not rise, and given infinitely thin and light balloon wall material which doesn’t apply pressure to the contained gas (oops, these idealisations do tend to creep in!) the balloon will just hang in mid air. Yet we know that molecules can zip around an atmosphere even if the atmosphere is static, so the air parcel concept seems quite inappropriate in that context. Trying to make air parcels smaller and relying on Brownian motion to flick them around doesn’t help as I don’t believe a 1mm cubed air parcel has any real meaning – it will simply dissociate before it does anything. So in a static atmosphere you are just left with the individual molecules, and as far as I can see, they will behave as I described and give an isothermal distribution. Do you have any suggestions as to how to convert your analysis into the microscopic domain?

I am familiar with the idea of equipartition of energy across the degrees of freedom of a system – unfortunately I’m not sure what the implications of this are for temperature vs altitude, I’ll have to think about this before making a comment.

Tallbloke:

Glad to contribute! And I’ve already learned a thing or two.

You said: “the overall energy contained in a spherical ‘slice’ of atmosphere at the higher altitude is going to be less than a ‘slice’ of equivalent volume at low altitude.”

I think this is something we can all agree on! The heat capacity in total does appear to be less, though I presume that the heat capacity per kg would remain the same. In any case this wouldn’t affect an isothermal atmosphere because there are no temperature differentials to transfer heat around. As for air parcels, I feel there are no such things in a static atmosphere, never mind rising ones. In a dynamic atmosphere, you can have a (very) large block of air that all seems to move in one direction (due to buoyancy, convection, etc) which can be modelled as air parcels. I’m sure you’re very familiar with how the story goes, my favourite link on this at the moment is http://farside.ph.utexas.edu/teaching/sm1/lectures/node56.html

The next complication is for an even more non-equilibrium state where the ‘surrounding’ air does not follow the DALR and one gets buoyancy effects, but apparently one can still treat things adiabatically, hence the heat capacity of the surroundings doesn’t seem to matter. I guess when the air parcel finds its ‘correct’ level it ‘loses its identity’ and simply mixes with the ‘surrounding’ air. So if the temperatures of the air parcel and ‘surroundings’ are different when the mixing happens, the heat capacity of the air makes a difference to the final temperature. I put ‘surrounding ‘ in quotes, because the distinction is far from sharp! But in dynamic steady state, the air parcel has the same temperature as it’s surroundings when it disperses, so again the heat capacity doesn’t seem to enter into it.

That said, I’m still getting to grips with what an air parcel actually is and what goes on at its boundaries – yeuch!

Hans:

OK, I think I can comment about equipartition now. All the references I read made it clear that equipartition applies to the *average* energy per DoF. The word average is important – how does this apply in practice?

Let’s take a box of gas, no gravitational field for simplicity, with the gas at a finite temperature. According to standard physics, the molecules will have velocities according to a Maxwell-Boltzmann distribution. The average of this will be a particular value, and you can relate that to ‘the’ temperature of the gas. But now look at any one molecule – what will its velocity be? Well, it can be almost anything – sometimes it will be fast, sometimes it will be slow. It is almost certain that its actual velocity will NOT be the average velocity! Nor will it likely be the same as any other molecule. But if you follow it over time, it will undergo collisions with other molecules and the walls, and will sometimes speed up and sometimes slow down due to these collisions. Over a long time, its average velocity will be the same average velocity as any other molecule in the gas. And this is what is implied by the equipartition theorem.

Now let’s go back to the presentation I made on throwing particles vertically upwards. After running the sim for a long time at a particular temperature, there will be an average height reached. Will all the molecules reach that average height? Certainly not. Some will overshoot, some won’t reach that high. The ones that overshoot clearly have a higher starting energy than the ones that can’t reach that high. So the molecules you find at a high altitude clearly have higher energy than a typical molecule you find at low altitude! If you take the total energy (including gravitational PE) of say a million molecules that reach a high altitude, this will certainly be higher than the total energy of a million molecules that are found at low altitude, even though they will have equal mass and the same temperature. It is like the gravitational field is filtering out high and low energy molecules (though I can’t see how to make this into a Maxwell Demon, as the temperatures are the same – the high energy molecules lose kinetic energy as they rise and cool down to the initial temperature!). My simulation agrees with the equipartition theorem, because it generates a MB distribution which is stable in time, hence any average over a large enough ensemble will converge to a stable value. So I’m afraid the statement that equal masses at different altitudes must have the same energy is false.

In an atmosphere, this implies that molecules high above average height won’t stay there very long – they must fall to below average height in order that their average height can be at the atmospheric average. Similarly with energy instead of height, molecules at high altitude (which have higher than average total energy) must lose that energy when they fall down (through collisions) so that they can get back to average energy.

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