Re: [Phys-l] Temperture profile in a graviational field

["Folkerts, Timothy J" <FolkertsT@bartonccc.edu>]

John doesn't seem to agree with my conclusions, and I usually hold his opinion in high regard.  I had another thought experiment that I thought was strong, but now I can see a hole in it too.

"Put just a single molecule in the container (or such a sparse atmosphere that the molecules rarely collide).    Hold the top and bottom of the column at the same temperature -- say 300 K.  Put that molecule near the bottom of the column -- let's assume it is heading downward.  It will hit the bottom and (on average) end up with KE = 3/2 kT.  If it is going slowly, it might never reach the top before gravity stops it and pulls in back down. If it is going faster, it might reach the top, but at a significantly reduced speed and KE.  But when it hits the top, it can exchange energy.  Since KE < 3/2 kT for the molecule, it will on average gain energy from the collision with the top.  When it gets to the bottom, it will have lost PE and gained KE.  So now when it hits the bottom the gas molecule will have on average KE > 3/2 kT and it will lose energy in the transfer.   So every round trip carries (on average) energy from the top to the bottom.  "

It sounds logical, but now that I think about it, I can imagine that the distribution of speeds would affect this argument.  For instance, many slower molecules would never make it to the top, so the molecules actually hitting the top will have a very skewed distribution.   Perhaps their average is STILL 3/2 kT even though the molecules started with an average of 3/2 kT and every single molecule did indeed lose energy on the way up.  

This would imply that if you took the MB distribution, cut off the particles with energy less than some amount dE, and then shifted the distribution down by an amount dE, the average would still be the same as before (just with a considerably different shape).   

(The  trip down is harder to explain. It seems the particles would have a MB distribution at the top (after colliding with the top and starting their downward journey)with an average KE of 3/2 kT.  Every single molecule would gain KE on the way down, so every single molecule should have (on average) more than 3/2 kT when it hits the bottom, so it seems every single molecule would (on average) transfer energy to the bottom when it hit.) 


Hmmm ... more to think about.



-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of John Denker
Sent: Tuesday, January 17, 2012 8:50 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Temperture profile in a graviational field

On 01/17/2012 07:04 AM, Folkerts, Timothy J wrote:
> 2) If this is true, how can this best be squared with the 2nd law of
> thermodynamics?   If the top and bottom of the column were held at a the
> same temperature, there would be a continuous flow of energy from top 
> to bottom, even though they are the same temperature. Even if the top 
> were slightly cooler than the bottom, there would be a downward flow.  
> This would violate a standard statement of the 2nd law, since we have 
> spontaneous heat from cool to warm.

It cannot be squared with the laws of thermodynamics.

> I've been trying to think of a good way to explain that this is not 
> indeed a violation.

It is a violation.

> I suspect the best explanation will have to involve the more 
> fundamental statement of the 2nd law -- that entropy tends to a 
> maximum.  The adiabatic lapse rate leads to an isentropic gas and a 
> constant potential temperature.

In equilibrium, the gas will have constant temperature ... not constant "potential temperature" but rather constant plain old temperature.  This is quite a fundamental result, flowing directly from the definition of equilibrium, the definition of temperature, and the specification of the physical situation (i.e. the fact that various parcels of air can equilibrate by exchanging energy).
See below(*).

The _stratosphere_ is more-or-less in equilibrium and does indeed exhibit constant temperature.

Meanwhile, the troposphere is not in equilibrium.  It is more nearly adiabatic than isothermal, because it is being vigorously _stirred_.
Thunderstorms contribute a great deal to the stirring, and it is no coincidence that the height of the typical thunderstorm is comparable to the height of the tropopause.  This is discussed in Feynman volume II chapter 9 : "Electricity in the Atmosphere".


=================

(*) Here's the proof that equilibrium is isothermal:

Assume the system as a whole is isolated from the rest of the universe.

Divide the system in to two parcels, #1 and #2.  Assume the parcels can exchange energy, and have been exchanging energy long enough to reach thermal equilibrium.  Now consider a fluctuation that moves energy from one parcel to the other:

     ΔE1 = - ΔE2           by conservation of energy
     ΔS1 = - ΔS2           because (a) the total entropy
                           can never decrease, and (b) at
                           equilibrium it cannot increase,
                           because it is already maximal.

Therefore dS1/dE1 = dS2/dE2 in equilibrium.

This quantity dS/dE is obviously important.  It is so important that if it didn't already have a name, we would need to give it a name.  In fact it is conventionally called the inverse temperature β.  For nonzero β the temperature is defined via kT = 1/β.

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