Interpretation of temperature in liquids/solids

[Fra]

Robert Alicki and Michal Horodecki in "Information-thermodynamics link revisited" (https://arxiv.org/abs/1905.11057v1):

"The discussion of above suggests the following picture. If by information we mean the stable information i.e. one that can be reliably stored and transmitted, rather than information carried by fast changing random congurations of physical systems, we expect that the following holds"

As I suspected that sounds like the picture of classical information, encoded in the commmon classical/macroscopic environment. Ie "objective information" that can be communicated between observers.

To me instrinsic information, is the what single physical agents/observers represent. And can't clone their states, without participating in an interaction that changes something. So alot of their premises put me off. Also what is "random" is in the eye of the beholder, random is just a generic word for that the agent fails to distinguish it from noise. Which may very well be due to limiting computational power.

/Fredrik

 

[Philip Koeck]

I gave you a recipe to compute the entropy increase (over time) in the general case. You can certainly apply it to the case of a bar placed between 2 reservoirs at different temperatures, where the thermal conductivity of the bar is anistropic. We can keep it simple by considering a 2D case and a crystal orientation such that kappa is a 2x2 matrix like so:
$\kappa = \begin{bmatrix} \kappa_{xx} & 0\\ 0 & \kappa_{yy} \end{bmatrix}$
You can quickly get an analytical formula for the general expression for $T(x,y)$ and you'll see that this temperature distribution depends on both $x$ and $y$ and that the thermal gradient doesn't align along $\hat x$.
Now onto the entropy question, the continuity eq. is $-\dot S =\nabla \cdot \vec J_S=-\nabla \cdot \left( \frac{\kappa}{T} \nabla T \right)$. That's the entropy production. You have everything in hands, and you can work out every details for the problem you set up. As you can see, there is an entropy production due to the thermal gradient, this is why I mentioned that there's an entropy production term due to Fourier's conduction term, earlier in this thread. The "external work being done on the solid" might be the entering heat flux, i.e. energy, into the system.
To answer Lord Jescott's question, yes, the driving force is 1/T (or T, depending on your preferred convention).

If I understand you correctly then the heat flow is actually from hot to cold, just not in the direction of the steepest temperature gradient.
If that's true then there's no problem with the second law and the temperature is the driving force.

 

[fluidistic]

If I understand you correctly then the heat flow is actually from hot to cold, just not in the direction of the steepest temperature gradient.
If that's true then there's no problem with the second law and the temperature is the driving force.

Hmm, I wouldn't say that's what I'm saying. I'm saying that if you connect 2 reservoirs at different temperatures via an anisotropic (in the sense that kappa is anisotropic) material, then the heat flux, at any point in the bar, will not point in the direction of "hot reservoir to cold reservoir". It will point in a direction such that there will be a component towards the cold reservoir, but also a component inward/outward the bar itself. This will disturb the temperature distribution in a way that T(x) if you will, will not be linear in x. In fact it would also depend on y, with parameters involving $\kappa_{xx}$, $\kappa_{yy}$, $L$ (length of the bar), $l$ (width of the bar), and even $\kappa_{xy}$ if the material (crystal) making up the bar crystal isn't oriented with a crystallographic axis along the bar itself.

I'm also saying that it's possible to have a bar in which heat flows from cold to hot. This can happen if you don't ignore thermoelectricity. In this case the "energy input" required to make this happens is also the heat flux entering the hot side of the bar.

In all cases, I am claiming that no thermodynamics law is broken, yes. You can convince yourself by computing $\dot S$ the way I suggested above. And saying that heat flows from hot to cold is not correct (unless you stick to simple cases). As a general statement, this is false.

 

[Philip Koeck]

I'm also saying that it's possible to have a bar in which heat flows from cold to hot. This can happen if you don't ignore thermoelectricity. In this case the "energy input" required to make this happens is also the heat flux entering the hot side of the bar.

Interesting!
Let's be very specific. If the bar is only in contact with the hot and cold reservoir at its ends and otherwise thermally insulated from the surroundings, is it possible for heat to flow from the cold to the hot reservoir without any sort of work form outside (such as a current from a battery)?

 

[fluidistic]

Interesting!
Let's be very specific. If the bar is only in contact with the hot and cold reservoir at its ends and otherwise thermally insulated from the surroundings, is it possible for heat to flow from the cold to the hot reservoir without any sort of work form outside (such as a current from a battery)?

Yes it is. I am in a hospital for a few days so no time to write a latex answer. The short answer is that the heat flux from the hot reservoir provides the energy for the thermoelectric material to operate, even though it is only a bar. If you do the math you get that if ZT, the figure of merit, is greater than 1, and if the TE material is short circuited with itself so as to allow an electric current through itself thanks to the voltage generated by the Seebeck effect, then heat can flow from cold to hot (depending on the sign of S).

 

[Philip Koeck]

... if the TE material is short circuited with itself so as to allow an electric current through itself thanks to the voltage generated by the Seebeck effect, then heat can flow from cold to hot (depending on the sign of S).

Could you explain what you mean by S? Is it entropy or change of entropy or something else?
If it's entropy, the entropy of what?

Without doing any calculation what you're describing in effect is that current is flowing due to the thermoelectric effect and at the same time heat is flowing from cold to hot. All this without an external driving force, for example a battery to produce the current.
Is that what you are saying?

 

[Hillbillychemist]

Please, lets get away from the "temperature is kinetic energy" high-school narrative.

The simple picture is

While this addresses two substances at different temperatures it does not explain the actual temperature of either which is their respective thermal kinetic energy. This is not a "high school narrative".

 

[fluidistic]

Could you explain what you mean by S? Is it entropy or change of entropy or something else?
If it's entropy, the entropy of what?

Without doing any calculation what you're describing in effect is that current is flowing due to the thermoelectric effect and at the same time heat is flowing from cold to hot. All this without an external driving force, for example a battery to produce the current.
Is that what you are saying?

Yes, except that the driving force exists, it is the heat flux from the hot reservoir into the bar, this inputs energy into the heat.engine, which is the bar itself.
The S I referred to in your quote is the Seebeck coefficient. (Which I considered as a scalar number for simplicity but which is a tensor in the most general case).

 

[DrClaude]

While this addresses two substances at different temperatures it does not explain the actual temperature of either which is their respective thermal kinetic energy. This is not a "high school narrative".

Where is the kinetic energy in spin temperature?

 

[Hillbillychemist]

Don't you think you are getting a little far afield from

"Interpretation of temperature in liquids/solids"?​

 

[DrClaude]

Don't you think you are getting a little far afield from

"Interpretation of temperature in liquids/solids"?​

Spin temperature is relevant in some solids.

In any case, to understand temperature in systems other than ideal gases, one must go beyond the high-schoolish "temperature is kinetic energy".

 

[Hillbillychemist]

Spin temperature is relevant in some solids.

In any case, to understand temperature in systems other than ideal gases, one must go beyond the high-schoolish "temperature is kinetic energy".

Only if the situation requires it.

 

[Philip Koeck]

Yes, except that the driving force exists, it is the heat flux from the hot reservoir into the bar, this inputs energy into the heat.engine, which is the bar itself.
The S I referred to in your quote is the Seebeck coefficient. (Which I considered as a scalar number for simplicity but which is a tensor in the most general case).

I'm still missing something here.
Assume again the thermoelectric bar with its ends touching a hot and a cold reservoir, but otherwise thermally insulated.
The heat current from the hot reservoir drives the TE-current (assuming we've short circuited).
The heat current has nowhere else to go other than to the cold reservoir since the bar is insulated everywhere else.
This is a kind of heat engine, just like you say.
Now the electric current produces a heat current due to the Peltier effect.
I guess this second heat current could partly cancel the initial heat current that's due to the temperature gradient, but never completely.
If it did you would get unlimited electricity for free.

So in total, I would say, heat flows only from hot to cold even in a TE-device.

 

[fluidistic]

I'm still missing something here.
Assume again the thermoelectric bar with its ends touching a hot and a cold reservoir, but otherwise thermally insulated.
The heat current from the hot reservoir drives the TE-current (assuming we've short circuited).
The heat current has nowhere else to go other than to the cold reservoir since the bar is insulated everywhere else.
This is a kind of heat engine, just like you say.
Now the electric current produces a heat current due to the Peltier effect.
I guess this second heat current could partly cancel the initial heat current that's due to the temperature gradient, but never completely.
If it did you would get unlimited electricity for free.

So in total, I would say, heat flows only from hot to cold even in a TE-device.

You are missing that the heat engine can do work, so not all the heat from the hot reservoir goes into the cold one.
It may be hard to visualize without a sketch, but here is a possible set up. Place a Bi2Te3 bar between 2 reservoirs kept at different fixed temperatures. The Seebeck effect is going to produce an emf (roughly) worth $S\Delta T$. This comes from the generalized Ohm's law $\vec J=-\sigma \nabla \overline{\mu}/e-\sigma S \nabla T$ (in open circuit condition, so J equals 0). So it's like a battery in some way. Now you attach a Cu wire (whose Seebeck coefficient S is near 0V/K compared to the Bi2Te3 one) near the top and the bottom (say near the 2 reservoirs, but you're not obliged too. At the junctions there will be heat released/adsorbed), to close the electric circuit. You can assume this Cu wire has a resistance Rload. The electric current going through the TE material and the Cu wire is $I=S\Delta T/(R_{TE}+Rload)=A|\vec J_e|$ where A is the cross section area.
Now if you look at the heat flux $\vec J_Q=-\kappa \nabla T +ST\vec J_e$ in the Bi2Te3 material, and look for the condition where it flows against the $-\nabla T$ direction, you'll get that (in the case where Rload is negligible compared to $R_{TE}$, which is feasible in practice) $ZT>1$ where Z is the usual thermoelectric figure of merit. Bi2Te3 satisfies this condition near room temperature. I have no time to show you the math, but it's extremely easy to do.

Note that there is no Peltier effect right at the reservoirs/Bi2Te3, since there is no electric current through the reservoirs. The Peltier effect manifested as heat released/absorbed is at the Cu/Bi2Te3 junctions.

Things are in reality severely more complicated though, the end result is that one can make almost anything one can imagine (to open a can of worms and wormholes, one should consider the Thomson effect whenever a current and a thermal gradient coexist, as in our case, and we can tune the temperature gradient to our likings, and hence the Thomson heat released/adsorbed locally to our liking. But that's only the visible scratch. Researchers have overlooked other thermoelectric effects that exist, but are not reported in the literature yet. I will not go there, this is not allowed in this forum).

 

[Philip Koeck]

Researchers have overlooked other thermoelectric effects that exist, but are not reported in the literature yet. I will not go there, this is not allowed in this forum).

Is that because those effects would lead to perpetual motion machines?

 

[fluidistic]

Is that because those effects would lead to perpetual motion machines?

I honestly do not know the reason(s). Of course not.
Here is what I can think on the top of my head:
I would guess that early scientists (Thomson, Bridgman, Altenkirch, Callen, Domenicali and Ioffe amongst others) considered cases that made sense with the kinds of materials available at their times. They did not consider more general cases, or "exotic" cases that we have nowadays. And scientists from nowadays are still missing, due to inertia (and possibly the lack of good up to date textbook on thermoelectricity), what they can achieve with thermoelectricity. Some of them (e.g. Uchida https://aip.scitation.org/doi/full/10.1063/5.0077497, and Snyder's https://journals.aps.org/prb/abstract/10.1103/PhysRevB.86.045202) are hinting at nice novelties, but they are still missing a lot (that I can think of).

Also, what you may be missing is that maybe you shouldn't focus on heat (or heat flux/transfer) only when you deal with the thermoelectric case, but maybe you should look at the energy flux $\vec J_U = \vec J_Q+\overline{\mu}/e\vec J_e$. Energy as in the "internal energy U" of thermodynamics. I'll repeat myself hopefully a last time, there's a very quick way to see that heat can flow from cold to hot just by looking at the generalized Fourier's law that appear in most thermoelectric books (and Wikipedia), $\vec J_Q =-\kappa \nabla T + ST \vec J_e$. Where $S$, the Seebeck coefficient can have either sign, $\vec J_e$ can have any direction since it's the current density and you can engineer the current to go in any direction you want and also any reasonable magnitude. It is clear that you can make a system where $ST \vec J_e$ outweights $-\kappa \nabla T$. When this happens, heat flows from cold to hot (or in another direction if you want it to be so and arrange it that way). If you are unable to convince yourself by looking at this simple relationship, there's nothing else I can do to convince you. This doesn't break thermodynamics laws. There's still an increase in entropy over time, and no energy is created out of thin air.

 

[Lord Jestocost]

The seebeck coefficient is a proportionality factor between a temperature gradient and a thereby caused gradient of the electrochemical potential of charge carriers when there is no electrical current flowing.

 

[fluidistic]

The seebeck coefficient is a proportionality factor between a temperature gradient and a thereby caused gradient of the electrochemical potential of charge carriers when there is no electrical current flowing.

Not sure to whom this is directed. What you state is just a special case of.the expression I have written in post #49 for.instance, as generalized Ohm's law, with Je set to 0.
Your statement is not universally true, as I,wrote earlier, this quantity is in fact a tensor in the most general case. But your statement is mostly correct.

 

[Philip Koeck]

... there's a very quick way to see that heat can flow from cold to hot just by looking at the generalized Fourier's law that appear in most thermoelectric books (and Wikipedia), $\vec J_Q =-\kappa \nabla T + ST \vec J_e$. Where $S$, the Seebeck coefficient can have either sign, $\vec J_e$ can have any direction since it's the current density and you can engineer the current to go in any direction you want and also any reasonable magnitude. It is clear that you can make a system where $ST \vec J_e$ outweights $-\kappa \nabla T$. When this happens, heat flows from cold to hot (or in another direction if you want it to be so and arrange it that way).

Could you send the Wikipedia link with the generalized Fourier law you mention above?

The electric current density J_e in the second term on the right of this equation, isn't that due to the Seebeck effect if we exclude external sources, such as a battery?
In that case it should be proportional to S, which means that the second term on the right doesn't change sign when you change the sign of S. (I'm thinking of a scalar S now, but a similar argument should hold for a tensor).
To me it looks like the second term cannot be varied as freely as you're saying.
Have you checked in detail that the second term can actually exceed the first and reverse the direction of J_Q?

 

[fluidistic]

Could you send the Wikipedia link with the generalized Fourier law you mention above?

The electric current density J_e in the second term on the right of this equation, isn't that due to the Seebeck effect if we exclude external sources, such as a battery?
In that case it should be proportional to S, which means that the second term on the right doesn't change sign when you change the sign of S. (I'm thinking of a scalar S now, but a similar argument should hold for a tensor).
To me it looks like the second term cannot be varied as freely as you're saying.
Have you checked in detail that the second term can actually exceed the first and reverse the direction of J_Q?

You're right, I have overlooked that the sign of S is directly linked to the direction of Je. I cannot.do the math now, no time, but you can,check it the details yourself of what happens to Jq.

 

[Lord Jestocost]

I meant the second example where you discuss anisotropic materials.
("... consider anisotropic materials, where kappa is a tensor. Write down Fourier's law under matricial form and you'll get that it's possible for a material to develop a transverse thermal gradient (and so a heat flux in that direction) even though it isn't the hot to cold direction.")

The example with an external current is understandable. It's just like a heat pump.
But in the above case I don't see any external work being done on the solid.
If heat flows spontanously in a direction that's not hot to cold I would expect an increase in entropy somewhere, for example a phase change.

@Philip Koeck

Your understanding is correct. Thermal energy is naturally tranfered in materials in one direction: from hot toward cold - unless people interfere.

 

[fluidistic]

@Philip Koeck

Your understanding is correct. Thermal energy is naturally tranfered in materials in one direction: from hot toward cold - unless people interfere.

I... interfere. I'm actually glad this discussion occurred, as I didn't think deeply enough of the case of no external energy source, so I will take the time to do it later.
However, what I am certain about, is that the direction of $\vec J_Q$ (the heat flux) needs not be the one of $-\nabla T$ (direction of hot to cold), especially in the case of anisotropic materials, which is ironically the text you quote. It's as if you discarded everything I wrote about it.

Similarly for thermoelectrics. What apparently cannot be done is to reverse the direction of the heat flux if one uses no external energy source. But you can certainly alter its direction (though not reverse it entirely). The claim that heat flows from hot to cold does not hold universally.

 

[Demystifier]

So in the first case an increase of average potential energy is related to an increase in temperature whereas in the second it's not, I would say.

I also get the impression that there's a qualitative difference between the potential energy in vibrations and that due to altitude, for example.

You are, of course, wrong. When the gas is cooled down, it falls down loosing its gravitational potential energy. That's, for example, how black holes are formed when massive stars become colder.

 

[Demystifier]

However I do see the difficulties with defining T as average kinetic energy.
What's not clear to me is how small the objects have to be that have this energy.
For example if you had a mechanical construction with billions of small steel balls on springs, how small would the steel balls have to be so that you would assign a temperature to the vibration of these balls?

They don't need to be small at all. For example, the balls in the lottery machine have a Maxwell distribution of velocities. However, the "temperature" of this ball velocities is different from the "temperature" of the atom velocities. This means that the system has two temperatures, one for microscopic degrees and another for macroscopic degrees, i.e. that the system is not in the full thermal equilibrium.

 

[Philip Koeck]

You are, of course, wrong. When the gas is cooled down, it falls down loosing its gravitational potential energy. That's, for example, how black holes are formed when massive stars become colder.

This idea is very strange for me.
So two volumes of helium with the same average kinetic energy at different altitudes in a helium atmosphere would have different temperatures?
How would a thermometer be able to measure the potential energy of the helium atoms in the gravitational field?
Also there's the problem that temperature has a definite zero point whereas potential energy in a gravitational field doesn't.
I just don't get it.

 

[Demystifier]

This idea is very strange for me.
So two volumes of helium with the same average kinetic energy at different altitudes in a helium atmosphere would have different temperatures?
How would a thermometer be able to measure the potential energy of the helium atoms in the gravitational field?
Also there's the problem that temperature has a definite zero point whereas potential energy in a gravitational field doesn't.
I just don't get it.

The thermometer measures the kinetic energy, but kinetic and potential energy are distributed according to the same temperature, see my post #21.

Note, however, that the formula in #21 is just a proportionality, one also needs to fix a constant (not depending on $v$ and $z$) in front of it, which is to be determined from the normalization of the probability distribution. So when you redefine the zero of the potential energy, it is just absorbed into this normalization constant.

Note also that in that formula the probability function is a product of the velocity-dependent and position-dependent function, which means that velocity is statistically independent from position. Hence, the average kinetic energy does not depend on position, so if two volumes have the same average kinetic energy at different altitudes, then they also have the same average kinetic energy at the same altitudes, which just means that they have the same temperature.

 

[dextercioby]

Thank you for this discussion. @fluidistic Can you name some books which helped you build knowledge on this topic of heat transfer? Thanks!

 

[Philip Koeck]

The thermometer measures the kinetic energy, but kinetic and potential energy are distributed according to the same temperature, see my post #21.

I can't make sense of what you write.

Let's assume a monoatomic ideal gas without gravity. The probability for any atom to have a kinetic energy between E and E+dE will be proportional to √E e^-βE.
In an isothermal atmosphere (with gravity) this should still be true.

However, the pressure in an isothermal atmosphere is proportional to e^-βE_pot.
And pressure is proportional to the number of atoms per volume in an ideal gas with constant T, which should be proportional to the probability of finding an atom with a potential energy between E and E+dE

 

[Demystifier]

I can't make sense of what you write.

Let's assume a monoatomic ideal gas without gravity. The probability for any atom to have a kinetic energy between E and E+dE will be proportional to √E e^-βE.
In an isothermal atmosphere (with gravity) this should still be true.

However, the pressure in an isothermal atmosphere is proportional to e^-βE_pot.
And pressure is proportional to the number of atoms per volume in an ideal gas with constant T, which should be proportional to the probability of finding an atom with a potential energy between E and E+dE

The temperature is related to the total average energy, which is the sum of average potential and average kinetic energy. Roughly,
$$T \propto \langle E_{\rm kin} \rangle + \langle E_{\rm pot} \rangle$$
The average is taken over all phase space, so $\langle E_{\rm kin} \rangle$ and $\langle E_{\rm pot} \rangle$ do not depend on velocity or position of the particle. Furthermore, if additionally $\langle E_{\rm kin} \rangle \propto \langle E_{\rm pot} \rangle$, then both
$$T \propto \langle E_{\rm kin} \rangle
\;\;\; {\rm and} \;\;\;
T \propto \langle E_{\rm pot} \rangle$$
are true, so there is no contradiction. Indeed, when the potential energy is quadratic in canonical positions, then $\langle E_{\rm kin} \rangle \propto \langle E_{\rm pot} \rangle$ is guaranteed by the equipartition theorem. The gravitational potential energy is not quadratic, but for general potential energies there is a generalized equipartition theorem with similar properties.

 

[fluidistic]

Thank you for this discussion. @fluidistic Can you name some books which helped you build knowledge on this topic of heat transfer? Thanks!

Not many books. There is "Physical Properties of Crystals: Their Representation by Tensors and Matrices" by Nye. I think it's a very nice book, it even has an accurate part on the Bridgman effect (a manifestation of thermoelectricity as a heat generation/adsorption in materials in which S is anisotropic). It is very rare, even in books which are solely about thermoelectricity.
There is Carlsaw and Jaeger's "Conduction of heat in solids", as a reference.

There is "Non equilibrium thermodynamics" by De Groot and Mazur. This one is hard to grasp, but it contains a lot of valuable information about fluxes and thermoelectricity.

Then there is a plethora of papers. There is a famous one by Callen (1948) https://journals.aps.org/pr/abstract/10.1103/PhysRev.73.1349.And what I consider a masterpiece (very informative and accurate): Domenicali's https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.26.237Then solving problems with finite elements. Compare the solutions with analytical results when possible before trusting the results. I use FEniCSx Python's "library" for this task. I find it extremely powerful and accurate.

 

[Philip Koeck]

I can't make sense of what you write.

Let's assume a monoatomic ideal gas without gravity. The probability for any atom to have a kinetic energy between E and E+dE will be proportional to √E e^-βE.
In an isothermal atmosphere (with gravity) this should still be true.

However, the pressure in an isothermal atmosphere is proportional to e^-βE_pot.
And pressure is proportional to the number of atoms per volume in an ideal gas with constant T, which should be proportional to the probability of finding an atom with a potential energy between E and E+dE

I noticed the last sentence above is ambiguous.

What I meant was:
Pressure is proportional to the number of atoms per volume in an ideal gas with constant T and the number of atoms per volume should be proportional to the probability of finding an atom with a potential energy between E and E+dE.

What I wanted to point out was that the distribution of atoms among kinetic energy levels doesn't follow the same mathematical function as the distribution among potential energy levels in a gravitational field.
So there's a qualitative difference somewhere (although both do contain a Boltzmann factor with the same T).
I'll continue in another reply.

 

[Demystifier]

the number of atoms per volume should be proportional to the probability of finding an atom with a potential energy between E and E+dE.

I have no idea how you arrived at that conclusion.

What I wanted to point out was that the distribution of atoms among kinetic energy levels doesn't follow the same mathematical function as the distribution among potential energy levels in a gravitational field.

Yes it does, according to Boltzmann and Gibbs.

 

[Philip Koeck]

The temperature is related to the total average energy, which is the sum of average potential and average kinetic energy. Roughly,
$$T \propto \langle E_{\rm kin} \rangle + \langle E_{\rm pot} \rangle$$
The average is taken over all phase space, so $\langle E_{\rm kin} \rangle$ and $\langle E_{\rm pot} \rangle$ do not depend on velocity or position of the particle. Furthermore, if additionally $\langle E_{\rm kin} \rangle \propto \langle E_{\rm pot} \rangle$, then both
$$T \propto \langle E_{\rm kin} \rangle
\;\;\; {\rm and} \;\;\;
T \propto \langle E_{\rm pot} \rangle$$
are true, so there is no contradiction. Indeed, when the potential energy is quadratic in canonical positions, then $\langle E_{\rm kin} \rangle \propto \langle E_{\rm pot} \rangle$ is guaranteed by the equipartition theorem. The gravitational potential energy is not quadratic, but for general potential energies there is a generalized equipartition theorem with similar properties.

That's what I would have said too. Degrees of freedom are related to quadratic canonical coordinates.

Potential energy in a gravitational field is something different.
Interesting information about the generalized equipartition theorem!

The way I've seen it so far:
If I don't allow any work to be done (keep the volume of a gas or solid constant, for example) then all heat added goes into the available degrees of freedom, such as translation, rotations and vibrational potential energy. This is directly related to a change in temperature.
I can't see the same direct connection to temperature if the gas changes it's position in an external potential such as gravity

 

[Philip Koeck]

I have no idea how you arrived at that conclusion.

I'm approximating the potential energy as mgh so that E_pot is proportional to h and therefore dE is proportional to dh.

That should mean that the number of atoms per volume in a slab of thickness dh at some height h must be proportional the probability of finding an atom with a potential energy between E and E + dE (if E = mgh).

Yes it does, according to Boltzmann and Gibbs.

The distribution among kinetic energies contains the factor √E, whereas the barometric formula doesn't.

 

[Demystifier]

Degrees of freedom are related to quadratic canonical coordinates.

No, they are related to all canonical coordinates, quadratic or not.

all heat added goes into the available degrees of freedom, such as translation, rotations and vibrational potential energy.

The heat also goes to the ability of particles to go up in the gravitational field.

I can't see the same direct connection to temperature if the gas changes it's position in an external potential such as gravity

What if the gravitational potential energy was proportional to $z^2$, would you see the connection in that case?