Temperature and degrees of freedom (split from Hypothesis about temperature)

[KJW]

One thing should be mentioned: Only the translational modes of molecular motion contribute to the temperature. Different substances have different heat capacities because they absorb energy in all their modes, but only the translational modes increase the temperature, thus the more modes that are available to the molecule, the more energy that is absorbed for a given increase in temperature.

 

 

[exchemist]

1 hour ago, KJW said:

One thing should be mentioned: Only the translational modes of molecular motion contribute to the temperature. Different substances have different heat capacities because they absorb energy in all their modes, but only the translational modes increase the temperature, thus the more modes that are available to the molecule, the more energy that is absorbed for a given increase in temperature.

 

 

How then is it possible for solids to have a temperature?

[swansont]

9 minutes ago, exchemist said:

How then is it possible for solids to have a temperature?

The translational modes, which are the vibrations mentioned elsewhere

[KJW]

22 minutes ago, exchemist said:

How then is it possible for solids to have a temperature?

I was actually thinking the same question. The answer I came up with was that vibrational motion within crystals is translational motion of individual molecules. The modes that do not correspond to temperature are rotations of individual molecules, vibrations within individual molecules, which are distinct from the vibrations between molecules within a crystal, and electronic states.

The way I see it is if one were to place a quantity of helium, which has only translational modes, within a sample of a given substance that has an arbitrary number of modes, then how much energy is transferred from the sample to the helium or visa versa? At thermal equilibrium, the two temperatures must be equal, but what is the distribution of energy?

 

9 minutes ago, KJW said:

The answer I came up with was that vibrational motion within crystals is translational motion of individual molecules.

I think the virial theorem is involved. I seem to recall that vibrational motion within crystals is not just translational motion but also potential energy. That is, two modes but only one corresponding to temperature.

 

Edited March 15 by KJW

[sethoflagos]

49 minutes ago, KJW said:

I was actually thinking the same question. The answer I came up with was that vibrational motion within crystals is translational motion of individual molecules. The modes that do not correspond to temperature are rotations of individual molecules, vibrations...

T = dq_rev/dS   (constant V, N)

What's going to your calculation of T if you decide to ignore some of the degrees of freedom that contribute to system entropy?

[KJW]

15 minutes ago, sethoflagos said:

T = dq_rev/dS   (constant V, N)

What's going to your calculation of T if you decide to ignore some of the degrees of freedom that contribute to system entropy?

All degrees of freedom do contribute to the entropy. But not all degrees of freedom contribute to the temperature. If it did, then given the equipartition theorem, all substances would have the same heat capacity, and the above equation you posted would not hold.

 

[sethoflagos]

34 minutes ago, KJW said:

All degrees of freedom do contribute to the entropy. 

Therefore all degrees of freedom contribute to equilibrium system temperature. This is the traditional thermodynamic definition of temperature.

[exchemist]

2 hours ago, KJW said:

I was actually thinking the same question. The answer I came up with was that vibrational motion within crystals is translational motion of individual molecules. The modes that do not correspond to temperature are rotations of individual molecules, vibrations within individual molecules, which are distinct from the vibrations between molecules within a crystal, and electronic states.

The way I see it is if one were to place a quantity of helium, which has only translational modes, within a sample of a given substance that has an arbitrary number of modes, then how much energy is transferred from the sample to the helium or visa versa? At thermal equilibrium, the two temperatures must be equal, but what is the distribution of energy?

 

I think the virial theorem is involved. I seem to recall that vibrational motion within crystals is not just translational motion but also potential energy. That is, two modes but only one corresponding to temperature.

 

I'm struggling with this. What does it means to say that rotational degrees of freedom don't contribute to temperature? Surely they do, since the heat capacity (Cᵥ) of a diatomic gas is 5/2R rather than 3/2R, once temperatures have been reached at which rotational levels are populated. And then it goes up to 7/2R once vibrations are excited. So these degrees of freedom affect the temperature of the substance for a given energy input. Is that not contributing to temperature?   

[KJW]

1 hour ago, sethoflagos said:

Therefore all degrees of freedom contribute to equilibrium system temperature. This is the traditional thermodynamic definition of temperature.

But this conflicts with the equipartition theorem. Each degree of freedom contributes an energy of ½kT. The more degrees of freedom, the more energy. But this requires that only the ever-present three translational degrees of freedom contribute to the temperature. Otherwise, the energy would not be proportional to number of degrees of freedom.

 

1 hour ago, exchemist said:

I'm struggling with this. What does it means to say that rotational degrees of freedom don't contribute to temperature? Surely they do, since the heat capacity (Cᵥ) of a diatomic gas is 5/2R rather than 3/2R, once temperatures have been reached at which rotational levels are populated. And then it goes up to 7/2R once vibrations are excited. So these degrees of freedom affect the temperature of the substance for a given energy input. Is that not contributing to temperature?   

The increase in heat capacity as the available degrees of freedom increases indicates that the energy is increasing faster than the temperature. Thus, the 5/2 R for the inclusion of rotations indicates that the increase of R from 3/2 R to 5/2 R is not causing an increase in temperature beyond the 3/2 R arising from translations. If all degrees of freedom contributed to the temperature equally, the heat capacity of all substances at all temperatures would be the same regardless of the number of available degrees of freedom.

 

 

Edited March 15 by KJW

[swansont]

1 hour ago, KJW said:

The increase in heat capacity as the available degrees of freedom increases indicates that the energy is increasing faster than the temperature. Thus, the 5/2 R for the inclusion of rotations indicates that the increase of R from 3/2 R to 5/2 R is not causing an increase in temperature beyond the 3/2 R arising from translations. If all degrees of freedom contributed to the temperature equally, the heat capacity of all substances at all temperatures would be the same regardless of the number of available degrees of freedom.

I think I understand what you mean, but IMO this isn’t the way to express it. Of course the other degrees of freedom contribute to the temperature; it’s right there in the equation. As you said, each degree of freedom has an energy of ½kT. If the other degrees don’t contribute, why does their energy depend on T? If you place an object with 3 degrees of freedom in contact with one with 5, they will still equilibrate at the same temperature. It’s just there are more modes in which you store the thermal energy in the second object.

If it’s not temperature, how is the energy accounted for in any of the thermodynamic potentials?

[sethoflagos]

1 hour ago, KJW said:

But this conflicts with the equipartition theorem.

No it doesn't. It just conflicts with your rather unusual interpretation.

 

1 hour ago, KJW said:

The more degrees of freedom, the more energy. But this requires that only the ever-present three translational degrees of freedom contribute to the temperature.

As @exchemist, points out, your unusual interpretation collapses for cases where there are no translational degrees of freedom. 

 

[KJW]

1 hour ago, swansont said:

If the other degrees don’t contribute, why does their energy depend on T?

Suppose you add thermal energy to a substance with 5 degrees of freedom. The equipartition theorem says that equal energy will be added to each degree of freedom. This means that 3/5 of the total added energy will be added to the 3 degrees of freedom corresponding to translation, leading to an increase in the temperature. The remaining 2/5 of the total added energy do not contribute to the increase in temperature, but because the increase in temperature due to the energy added to the 3 translational degrees of freedom is concomitant with the energy added to the 2 non-translational degrees of freedom, the energy of the 2 non-translational degrees of freedom will still depend on the temperature.

 

 

1 hour ago, swansont said:

If you place an object with 3 degrees of freedom in contact with one with 5, they will still equilibrate at the same temperature. It’s just there are more modes in which you store the thermal energy in the second object.

There will be 3 units of energy in the object with 3 degrees of freedom and 5 units of energy in the object with 5 degrees of freedom. But the two objects are at the same temperature. Therefore, only 3 units of energy in the object with 5 units of energy are producing temperature, the same temperature as produced by the 3 units of energy in the object with 3 units of energy.

 

 

Edited March 15 by KJW

[exchemist]

2 hours ago, KJW said:

But this conflicts with the equipartition theorem. Each degree of freedom contributes an energy of ½kT. The more degrees of freedom, the more energy. But this requires that only the ever-present three translational degrees of freedom contribute to the temperature. Otherwise, the energy would not be proportional to number of degrees of freedom.

 

The increase in heat capacity as the available degrees of freedom increases indicates that the energy is increasing faster than the temperature. Thus, the 5/2 R for the inclusion of rotations indicates that the increase of R from 3/2 R to 5/2 R is not causing an increase in temperature beyond the 3/2 R arising from translations. If all degrees of freedom contributed to the temperature equally, the heat capacity of all substances at all temperatures would be the same regardless of the number of available degrees of freedom.

 

 

That seems to me a rather perverse way of seeing it. Surely all that is happening is that there are now 5 pots to fill instead of 3, when energy is added, so the rate of rise of the level is slower, for a given rate of addition? 

What experiment could be imagined to show that rotations do not contribute to temperature? I'd have thought there would be no such experiment, since energy is rapidly equilibrated among all degrees of freedom. 

[KJW]

36 minutes ago, exchemist said:

Surely all that is happening is that there are now 5 pots to fill instead of 3

Why does the 5 pots have the same temperature as the 3 pots? Why does the 7 pots have the same temperature as the 3 pots and not the 5 pots?

 

One other thing: In the ideal gas law, pressure, which is proportional to temperature, depends only on the translational motion of the gas molecules. Neither rotational motion, nor vibrational motion affects the pressure and therefore the temperature of an ideal gas.

 

Edited March 15 by KJW

[swansont]

We don't break out the energy stored in rotational modes as a separate term when doing an energy balance

Q = mc∆T

If temperature is only the translational motion, you'd need an additional term to account for the energy in the other degrees of freedom, but you don't do this, or need to, because you already accounted for it in the heat capacity. 

Equipartition of energy means that you can't separate the translational from the rotational modes in terms of energy. If the temperature changes, all of the modes gain or lose energy, and the temperature is proportional to this.

The distinction shows up in calculations of the internal energy U = aNkT = aPV

Seems to me there's no unambiguous argument in either direction, because the equipartition force the energy to be shared between the modes. I see nothing in literature making the distinction that it's only translational KE in systems with additional degrees of freedom. Nothing shows up in the equations, and IMO it's confusing to make that distinction when it doesn't show up in or matter to the calculation. 

 

[sethoflagos]

21 minutes ago, KJW said:

One other thing: In the ideal gas law, pressure, which is proportional to temperature

Only in one very special case: that of a theoretical constant volume thermodynamic process.

More general cases show P, T dependence involving the ratio of specific heats, (~7/5 for diatomic gases) which per force requires consideration of all available thermal degrees of freedom, not just the three translational ones. 

 

3 minutes ago, swansont said:

Seems to me there's no unambiguous argument in either direction, because the equipartition force the energy to be shared between the modes. I see nothing in literature making the distinction that it's only translational KE in systems with additional degrees of freedom. Nothing shows up in the equations, and IMO it's confusing to make that distinction when it doesn't show up in or matter to the calculation.

Exactly!

[exchemist]

8 hours ago, KJW said:

Why does the 5 pots have the same temperature as the 3 pots? Why does the 7 pots have the same temperature as the 3 pots and not the 5 pots?

 

One other thing: In the ideal gas law, pressure, which is proportional to temperature, depends only on the translational motion of the gas molecules. Neither rotational motion, nor vibrational motion affects the pressure and therefore the temperature of an ideal gas.

 

Equilibration between degrees of freedom takes care of that. The pots are all connected together, in other words. Pressure, I grant you, is due to translational motion, (which is why a solid exerts no pressure), but I don’t follow your “and therefore the temperature”. Especially since solids do have a temperature.

[KJW]

On 3/16/2024 at 5:15 PM, exchemist said:

Pressure, I grant you, is due to translational motion, (which is why a solid exerts no pressure), but I don’t follow your “and therefore the temperature”.

If pressure is independent of non-translational motion, and temperature is directly proportional to pressure, then temperature is independent of non-translational motion.

One thing that was never made clear throughout this discussion is what temperature is. I felt that temperature was being conflated with energy. But I will agree that my view was somewhat perverse, and I'm not even sure how it arose. It is correct for gases, but questionable for solids and perhaps liquids as well. Each degree of freedom has an energy of ½kT. Therefore, temperature can be regarded in terms of the energy of a single degree of freedom. And the equipartition theorem ensures that the single degree of freedom can be any of the degrees of freedom, not just translation. This seems to be the argument against me. My argument was basically that it isn't all the degrees of freedom that contribute to temperature because it is only one degree of freedom whose energy specifies the temperature. But translations are quite ubiquitous, and the three translation modes always occur together. So instead of temperature being specified in terms of one degree of freedom with energy ½kT, temperature could be specified in terms of the three translational degrees of freedom with energy ³/₂kT. And given that only translations contribute to pressure, it would seem that translations are quite special compared to the other degrees of freedom.

However, I am also aware that temperature emerges as a Lagrange multiplier in statistical thermodynamics, so I'm not entirely wedded to the ideas I expressed in this thread.

 

[swansont]

21 minutes ago, KJW said:

If pressure is independent of non-translational motion, and temperature is directly proportional to pressure, then temperature is independent of non-translational motion.

In thermal equilibrium you can’t decouple these effects. You can’t create a situation where you have the translational motion without the non-translational motion. 

[KJW]

2 minutes ago, swansont said:

In thermal equilibrium you can’t decouple these effects. You can’t create a situation where you have the translational motion without the non-translational motion. 

No, you can't decouple translational from non-translational motion, but the point I'm making is that the energy that is in the non-translational motion doesn't contribute to either the pressure or the temperature of the (ideal) gas, except if you try to remove it.

[swansont]

1 hour ago, KJW said:

But translations are quite ubiquitous, and the three translation modes always occur together

There are situations where this isn’t true, though it’s not truly thermal equilibrium - you can cool a sample of a dilute gas with lasers, with the axes being independent. One- and two-dimensional cooling can take place. The atoms equilibrate with the laser light. The photons don’t interact with each other, so the velocity profile in each dimension can be different. The atoms are dilute and move very slowly so they scatter much less often with each other than with the photons. 

[sethoflagos]

2 hours ago, KJW said:

If pressure is independent of non-translational motion...

But it isn't. The 'boink' of each collision that creates the emergent property of pressure is in itself the vector sum of translational, rotational and vibrational momentum changes along the axis of point contact. The 'independence' you claim relies on the fact that statistically the rotational and vibrational transfers average out to zero over a sufficiently large number of collisions. It's a mathematical artefact.

Imagine being hit by a spinning dumbell. It makes a difference which end hits you.

Edited March 18 by sethoflagos
typo

[exchemist]

18 hours ago, KJW said:

If pressure is independent of non-translational motion, and temperature is directly proportional to pressure, then temperature is independent of non-translational motion.

One thing that was never made clear throughout this discussion is what temperature is. I felt that temperature was being conflated with energy. But I will agree that my view was somewhat perverse, and I'm not even sure how it arose. It is correct for gases, but questionable for solids and perhaps liquids as well. Each degree of freedom has an energy of ½kT. Therefore, temperature can be regarded in terms of the energy of a single degree of freedom. And the equipartition theorem ensures that the single degree of freedom can be any of the degrees of freedom, not just translation. This seems to be the argument against me. My argument was basically that it isn't all the degrees of freedom that contribute to temperature because it is only one degree of freedom whose energy specifies the temperature. But translations are quite ubiquitous, and the three translation modes always occur together. So instead of temperature being specified in terms of one degree of freedom with energy ½kT, temperature could be specified in terms of the three translational degrees of freedom with energy ³/₂kT. And given that only translations contribute to pressure, it would seem that translations are quite special compared to the other degrees of freedom.

However, I am also aware that temperature emerges as a Lagrange multiplier in statistical thermodynamics, so I'm not entirely wedded to the ideas I expressed in this thread.

 

Well temperature is proportional to energy so in a way it is just a matter of choice of units whether one talks about temperature or energy in this context. To my way of thinking the distinction between the role of modes is not real, since all degrees of freedom that are excited (at NTP in gases vibrational modes generally aren't) contribute 1/2kT each to the overall energy - which means temperature, in effect. Yes, pressure is proportional to the temperature (or energy)  in the translational modes, but it is also proportional to that in the non-translational modes too, as they are all equal.

One test of the idea that the translational modes are special might be if one could make a case that the flow of heat is transmitted only through translational motion. I am sceptical, since the modes all exchange energy. 

  

[KJW]

4 hours ago, exchemist said:

One test of the idea that the translational modes are special might be if one could make a case that the flow of heat is transmitted only through translational motion. I am sceptical, since the modes all exchange energy.  

An alternative is to consider a gas thermometer. The temperature defined by such a device is based on the kinetic energy of the translational motion of the gas and does not include non-translational motion that may occur in the gas. Thus, one could replace the gas with a different gas that has a different heat capacity, based on the universality of the ideal gas law.

 

[sethoflagos]

@KJW. I understand that you regard my last few posts as too much of a waste your precious time to be worth responding to. But I trust that in turn, you will not object to me regarding your silence, particularly wrt the non-conservation of linear momentum in general gas collisions, as concessions to the arguments presented. Live long and prosper.