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Do rotational degrees of freedom contribute to temperature?

  1. May 31, 2015 #1
    I cannot find a simple answer to this question anywhere.

    What degrees of freedom contribute to the temperature of a gas?

    Let's say we have a box of ideal gas. The temperature is the average kinetic energy of the particles and only includes translational degrees of freedom: velocity.

    Now let's say we have a box of polyatomic gas instead, in which there are rotational degrees of freedom but no vibrational degrees of freedom. Is the temperature of gas still just the average translational kinetic energy, or do we have to add its average rotational energy?
  2. jcsd
  3. May 31, 2015 #2


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    If there are rotational degrees of freedom, then the energy will be the sum of rotational and kinetic energies. So yes, I agree with what you are saying. We probably have to also assume that the polyatomic atoms only interact in brief collisions, and don't have extended interactions or correlations.
  4. May 31, 2015 #3
    OK, if that's true, then why if we heat a polyatomic gas at constant volume would the temperature be less for the same amount of heat input compared with a monatomic gas? There is a huge inconsistency here.
  5. May 31, 2015 #4
    The rotational degrees of freedom contribute to the internal energy and not to the temperature.
    I am not even sure it make sense to say that "it contributes to the temperature".
  6. Jun 1, 2015 #5


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    If we heat a polyatomic gas at constant volume, the heat capacity is greater than for monatomic gas. Therefore, for the polyatomic gas, it takes more energy to increase the same amount of temperature. This makes sense because the polyatomic gas has more degrees of freedom, i.e. more places to put the energy.

    So yes, if we add the same amount of heat to a polyatomic gas and a monoatomic gas, the temperature increase of the polyatomic gas would be less than for monoatomic. This makes sense compared to your original statement because our polyatomic gas now must increase its translational and rotational energies, and conversely the monoatomic gas only had to increase its translational energy. So therefore the monoatomic gas will be able to have the greater temperature increase.
  7. Jun 1, 2015 #6


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    I think the OP'er is saying that the energy stored in the rotational degrees of freedom contribute to the temperature. Which I guess is sort of true, since the temperature and average rotational energy are related in a simple way in this case. But I agree, generally it is not a good way to talk about temperature.
  8. Jun 1, 2015 #7
    But you just said that rotational degrees of freedom contribute to the temperature? Therefore, if you give a polyatomic and monoatomic gas with the same amount of heat energy then the temperature should rise the same amount. So the only conclusion if what you say is true is that rotational degrees of freedom do not contribute to the temperature of a gas.

    I guess what I'm really asking is the following: if we put a thermometer against our box of gas, what does the thermometer measure? Only the translational kinetic energy, or all of them?

    If you say only translational - then how can we measure the temperature of a solid? A solid has little translational kinetic energy but mostly vibrational kinetic energy. If we put a cooler thermometer in contact with a hotter solid then the thermometers temperature will increase, therefore suggesting that vibrational kinetic energy DOES contribute to the measurement of temperature of the system.

    Can you help clear this up?
  9. Jun 1, 2015 #8
    One additional point:

    How does one measure the heat capacity (defined as the change in temperature with respect to the change in heat energy) of e.g. hydrogen and find the contribution of the rotational and vibrational degrees of freedom if they do not in fact increase the temperature?
  10. Jun 1, 2015 #9


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    uh... I think I see where we are getting mixed up. Each degree of freedom is independent of the others. And at a given temperature, each degree of freedom will contain the same energy as the others (due to equipartition of energy). So in this sense, temperature is proportional to the average kinetic energy. But temperature is also proportional to the average rotational energy. And temperature is proportional to the average total energy. The important point here is that the constant of proportionality for the average total energy will be double that of the constant of proportionality for just the average kinetic energy.
  11. Jun 1, 2015 #10
    I don't think we're getting mixed up. This is the heart of the problem.

    If you have a diatomic gas and you add some heat energy and a monatomic gas and you add the same amount of heat energy, and you measure the temperature of both after you've added the heat energy, you find that the temperature of the monatomic gas has increased more than the polyatomic gas.

    The only conclusion is that rotational degrees of freedom do not contribute to the measurement of the systems temperature.

    The equipartition theorem just tells you how the internal energy is distributed amongst the particles in the gas. So if you add heat energy, some of it goes into translational kinetic energy, some into rotational energy. That part is clear. But when you measure the temperature of the system, are we measuring the internal energy of the system or the translational kinetic energy of the system?

    If we are only measuring the translational kinetic energy, then there is no way to measure the temperature of a solid, since almost all of its energy resides in vibrational energy states, and not in translational kinetic energy states.

    So what is the truth?
  12. Jun 1, 2015 #11
    When you measure temperature you measure temperature and not any energy, translational or other kind.
    The fact that the temperature is related to the motion of the molecules does not mean that temperature is the energy of this motion.

    You are starting with some unsound assumptions in the OP.

    One is:
    "The temperature is the average kinetic energy of the particles"
    which is obviously wrong. Temperature and energy are different quantities, with different units.

    And then you say that "only includes translational degrees of freedom: velocity" which I suppose you mean to refer to the energy and not the temperature. (the sentence as written is confusing). Definitely the energy includes all degrees of freedom that are thermally excited, so it is not true if you mean it to refer to energy.
    And to say that "the temperature includes" does not make sense. So what do you actually mean?
    Is there some reference or definition that you are confused about?

    Sure, the average energy per degree of freedom is proportional with the temperature but this does not mean that each degree of freedom (translational or not) contributes to temperature. The temperature is not a sum of terms from various degrees of freedom, as the word contribution would (I think) imply.
  13. Jun 1, 2015 #12
    You've done a great job of telling me what temperature isn't, now tell me what it is.
  14. Jun 1, 2015 #13
    Temperature is defined in thermodynamics as the partial derivative of the internal energy U with respect to entropy S, at constant volume V:
    T = (∂U/∂S)_V

    The internal energy is the sum of the total kinetic energy of the atoms in the gas, as well as the potential energy due to the interaction between the atoms.
  15. Jun 1, 2015 #14
    Yes I understand that.

    But let's think of it like this:

    Let's say we want to measure the temperature of a solid, with a thermometer. We place the thermometer against the solid. What happens to the reading on the thermometer?

    Intuition tells us it goes up. But everyone is claiming that temperature only measures the translational kinetic energy of a system. What gives?

    Using the entropic definition, we say that two things have the same temperature if adding a quantum of energy increases their number of thermally accessible microstates by the same amount.

    So saying that temperature is a measure of the average translational kinetic energy is just completely false.
    Last edited: Jun 1, 2015
  16. Jun 1, 2015 #15
    Who's claiming that? In what context? What does even means "temperature measures" here?
    If I know the temperature I can tell the average energy of all the thermally excited modes, can't I? Is this what you mean that "temperature measures"?

    But this is true only if I have a good model of the system and I know that equipartition applies.
    What if I measure the temperature of a sample of copper at 50K? Is this a "measure" of the kinetic energy of the copper atoms?
    Can you tell the average kinetic energy of a copper atom just from the temperature measurement?
  17. Jun 1, 2015 #16
    OK, here is the crux of my problem:

    Several books and websites I have read have said something like the following:

    "Simply increasing the rotational and vibrational energy does not increase the temperature of the system"


    "If you provide the same amount of energy to a monatomic and otherwise identical diatomic gas, the temperature of the diatomic gas will increase less because some of the energy is spread as rotational energy"

    How can these statements possibly be true?
  18. Jun 1, 2015 #17


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    Please have a look at http://en.wikipedia.org/wiki/Equipartition_theorem
    The energy in each degree of freedom is equal to kT/2 in thermal equilibrium. So to answer the original question, the kinetic energy (in three dimensions) is equal to 3kT/2. This is true even if there are rotations are possible. Rotations also add another 2 degrees of freedom, so the total energy (translation + rotation) is equal to 5kT/2.

    So, a diatomic gas has higher heat capacity than a monotomic gas of the same molecular mass, since the thermal energy is spread out over more modes.
  19. Jun 2, 2015 #18


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    yes, I think Khashishi is right to put some actual numbers down. Maybe this is the best way to explain. So, for a monoatomic gas, the average kinetic energy of the gas is 3kT/2 and the average rotational energy of the gas is zero, therefore the average total energy is 3kT/2. And for a diatomic gas, the average kinetic energy is 3kT/2 and the average rotational energy is kT, therefore the average total energy is 5kT/2.

    So, if you add an amount of energy ##\Delta E## to the monoatomic gas, we have ##\Delta E = 3k \Delta T /2## gives the change in temperature. And for the diatomic gas, if we add the same amount of energy ##\Delta E##, we have ##\Delta E = 5k \Delta T /2##. You can see that the temperature will increase more for the monoatomic gas. In an intuitive sense, this is because the diatomic gas has more degrees of freedom to put the energy into.

    edit: this is ignoring the vibrational modes of the diatomic molecule, for simplicity.
    Last edited: Jun 2, 2015
  20. Jun 2, 2015 #19
    That has helped clarify it somewhat, but my feeling is that this confirms the original point: that rotational degrees of freedom do not contribute to temperature measurements!

    The only conclusion from that statement is that when you put a thermometer against a system, the only thing it measures is the translational kinetic energy. So I go back to my original point, if that is true, how can we measure the temperature of a solid, in which nearly all the energy resides in vibrational degrees of freedom?

    Let's put it another way. using your example. Let's say I have three different kinds of systems, such that each possess 3 different degrees of freedom:

    1) Ideal Gas: E_trans = 3/2kT
    2) A strange polyatomic gas with no translational: E_rot = 3/2kT
    3) An ideal solid: E_vib = 3/2kT

    In all cases, if you add an amount of energy delta E, then temperature increases by the same amount, even though the energy type in each case is different.

    So your statement "In an intuitive sense, this is because the diatomic gas has more degrees of freedom to put the energy into." only makes sense if you assume that these other degrees of freedom do not get counted when you measure temperature, which I think we established earlier is incorrect.

    So I'm saying that this intuitive "reason" that people and books always give us actually incredible incorrect.
  21. Jun 2, 2015 #20
    I have finally figured out a way to explain it, using the proper definition of temperature:

    [tex] \frac{1}{T} = \frac{\,dS}{\,dE} [/tex]

    Let's assume that because the diatomic molecule has more degrees of freedom, it has more energy levels (rotational energy levels) than a simple monatomic gas. This means that, for a small increase in energy, the change in multiplicity i.e. the number of accessible energy levels is greater in a diatomic molecule than a monatomic molecule, which means that [tex] \frac{1}{T_d} = \frac{\,dS}{\,dE_d} > \frac{\,dS}{\,dE_m} = \frac{1}{T_m} [/tex] and hence [tex]T_m > T_d [/tex]

    Of course, in the real world, this doesn't work, because of the large difference between translational and rotational energy levels: you need to get to a certain temperature before you can thermally access those rotational energy levels, hence the observed heat capacity vs T for hydrogen.

    Thanks all for helping me figure this out!
    Last edited: Jun 2, 2015
  22. Jun 2, 2015 #21
    I am not sure what you figured out. I still think that the original question relies on some premises that are either wrong or poorly defined.
    This is why I asked to give a reference to that statement that only translation levels "contribute to temperature". Whatever that means. It may be just a matter of interpretation of the context.

    It is true that in a gas only the translational KE modes are excited by thermal motion. Maybe this is what it was intended?

    However, when you say that a thermometer only measures translational energy is at least misleading. The thermometer does not measure energy anyway. But maybe the idea is how does a cold thermometer (or the wall of a container) put in contact with a gas gets hotter from the gas.
    Now if you think microscopically, the gas molecules are faster on the average and collisions with molecules of the wall will result in a net transfer of energy from the gas to the wall. But now, which component of the velocity of the gas molecules contribute to this? If the wall is in the plane xy, the z component will be perpendicular to the wall. What if a molecule has velocity in the x direction? Will this transfer energy to the wall? What if the wall is the thermometer? Will it measure only the energy of the z degree of freedom?
    Of course not. So the question as it is formulated does not make sense.
    It does not matter with what degree of freedom the thermometer interacts. As long as equipartition "functions" they all have the same average energy. You can say they all have the same temperature.

    Actually the key for equipartition is that the degrees of freedom interact so even though out of equilibrium they may have different average energies, they will transfer energy from between them until they reach equilibrium. I think there are cases when this does not happen, but this is not relevant here.
    So it does not matter which DOF is "heated" first or for each one you measure temperature, if the equilibrium between them is reached.
    Think about microwave heating. Do the microwaves increase (directly) the translational KE of the molecules in the sample? Is the sample's temperature increased?
  23. Jun 2, 2015 #22
    I figured out the answer to the original question, which is why does the temperature of a polyatomic gas increase less upon an addition of the same energy quantum than a monatomic gas.
  24. Jun 2, 2015 #23

    Mark Harder

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    Very good. The precise definition of temperature in terms of partial derivatives predicts behavior that the quasi-intuitive definition - temperature is a measure of microscopic kinetic energy - does not. I thought I understood temperature when I learned (sort of) it is the change that all those system molecules made when they banged into thermometer molecules. But when I was told that temperature can be negative, my mind reeled - does not compute, does not... But given the ∂U/∂S definition, we can see that temperature can indeed be negative whenever internal energy increases while entropy decreases, or vice versa. Placing systems in external fields, like paramagnetic salts in strong magnetic fields, can result in negative temperatures. There was a Scientific American article about 50 years ago that explained this experiment. I'm sure there are more recent, up-to-date ones.
  25. Jun 3, 2015 #24


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    No, the thermometer is not only affected by translational kinetic energy because of the reason nasu gave. The molecules in the gas will collide with each other, and will transfer their energy to and from kinetic and rotational. Therefore, the rotational energy can also be passed to the thermometer, if only indirectly. And nasu's other example, that if our thermometer only lies on a plane of constant z coordinate (for example), then it will be affected by all the translational kinetic energy, not just the z component.

    In our example, we had a diatomic gas, which had E_rot=kT. So the 3 systems do not all have the same number of degrees of freedom. The number of degrees of freedom is not always exactly the same. It depends on the system, and in this particular case, we can easily calculate the heat capacity of the system. (it is just a constant).

    edit: and in these simple cases, the heat capacity depends in a very simple way on the number of degrees of freedom of the system.
  26. Jun 3, 2015 #25
    I am not convinced at all that "a thermometer measures only the translational degrees of freedom".
    If one goes back to the basics, I think the matter is very clear.

    I have a box containing an ideal gas, at a volume, pressure, temperature.
    Now I contact it with a thermometer. They will reach thermal equilibrium (i.e. the entropy of the composite system, gas + thermometer will reach a maximum).
    At equilibrium, for such an ideal system, the Equipartion Theorem will be valid: hence all degrees of freedom will be equally excited, and each of them will "contribute" equally to the definition of temperature, allowing some loose language.
    PS: It was said temperature and energy have different units, this is not true at all (entropy is dimensionless).
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