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Entropy change with angular momentum

  1. Nov 12, 2007 #1
    Hello,

    I have a artist friend who designs spin tops. As we were discussing the physics (and the magic) of angular momentum and gyroscopic effects, the conversation drifted on entropy.

    We came up with this question: is the entropy of a spin top different when the top is at rest and when it is spinning. In other word, is entropy dependant on angular momentum?

    Can somebody shed some light on this?

    Thanks!
     
  2. jcsd
  3. Nov 12, 2007 #2
    Not really. It might have more rotational energy, but since it's just one classical rigid body you can completely convert that energy back into work.

    On the other hand, if you have a collection of many spinning tops, then you would say there is less entropy if none are spinning. You might want to research the heat capacity of gases with different numbers of degrees of freedom.
     
  4. Nov 12, 2007 #3
    Thanks CesiumFrog,

    From your answer:
    I gather that energy that is 100% convertible to work without loss (through friction I guess) can contribute to the entropy of the system. This is from a thermodynamic perspective. Am I correct?

    Now, when you say
    are you refering to Entropy related to the information theory, i.e. that it is easier to describe the system if all the rotation speed (RPM) are the same? i.e. the entropy would be the same, providing RPM(i) = RPM (j) =...= RPM(n) (with the possibility of RPM=0, but not necessarily)?

    And finaly, from a thermodynamics stand point, in a system of 2 spins (spin(1) and spin(2)), where both spin are initially at rest, if spin(2) is given rotational energy, its entropy S2 does not change (S2(rest) = S2(spinning)), from what you previously said. Now, lets suppose spin(1) is still at rest. Why would the entropy of the system (spin(1) , spin(2)) increases? I would have thought S(system) = S1 + S2(rest) = S1 + S2(spinning). Isn't the entropy of a system the sum of the entropy of each of its elements? And if so, then, that would differentiate the 2 kind of entropy (information theory, and thermodynamics)?

    My souvenir of heat capacity of gases is that this is related to statistical thermodynamic. Can (or should) a classical solid object like a spin be described by statistical thermodynamics?

    Or am I totally confused....?
     
  5. Nov 13, 2007 #4
    How would you define entropy, or any other statistical/thermodynamic quantity for the simple system you're referring to? And why would it be meaningful in any way?

    Assaf
    Physically Incorrect
     
  6. Nov 13, 2007 #5
    ozymandias :

    I guess you could define a top entropy as the energy that could not be used to produce work. And that would conform to cesiumfrog's answer.

    I did not want to consider statistical thermodynamic. Therefore, the top would be a 0K with only one set of microstates. W=1, i.e. S=LogW=0

    My initial question was purely theretical i.e., is entropy dependant on rotational energy for a macrospopic body (my physics classes happened 25 years ago). I think it is answered. For a pure iron top spinning at 0K in a perfect vaccuum, assuming there is no other movement than the overall rotation (i.e. no vibration, no rotation, no translation of Fe atoms within the latice), then its total entropy remains null if we assume that all this rotational energy can be converted into work. Correct?

    And to answer the second part of your question, I guess you are right, entropy is not meaningful if one does not allow any degree of freedom at the microscopic level, even though there is one degree of freedom at the macroscopic level.

    Thanks for stiring up.
     
    Last edited: Nov 13, 2007
  7. Nov 13, 2007 #6
    Hi Lou,

    Hey, how can I be right if I'm not even sure what I'm trying to say? :)
    I think what I meant by asking about the meaning of entropy in your system is the following. The fact that you have a particular system which is in a definite state is not enough to calculate the entropy of your system. In fact, entropy seems to be a property defining not a particular system, but a realization of an ensemble of systems (I believe the word "Ergodic" should be in that sentence somewhere :) ).
    Here is an example: suppose you have a two-level system, with levels A and B. Now suppose you have an atom at level A. What is the entropy of the atom? I don't know. I can only talk about the entropy of the system (and even then, only when probabilities - associated with temperature, usually - are given).
    This seems to me to be true whether I use the thermodynamic, statistical-mechanic or information-theoretic definition of entropy.

    Hope I've made some sense.

    Assaf
    Physically Incorrect
     
  8. Nov 14, 2007 #7
    I am not sure about that, ozymandias. Absolute entropy of a system at equilibrium at temperature T(K) is said to be the integral of Q/T for T varying between 0K and T(K). So, you could, in theory, work it out once your system (e.g. a top, is at equilibrium at T(K)) is in equilibrium. But you don't even need to warm up your top from 0 to T(K) : the entropy of a top of mass M, at 298K, made of pure Iron, can be easily calculated from a table giving the standard molar entropy of Fe.

    You could similarly calculate the entropy of a similar top spinning without friction inside a vacuum using the same standard molar entropy. The result would be exactly the same number.

    Therefore I would be tempted to conclude that the rotational energy of a non elastic solid (e.g. a top) has no effect on its entropy if this energy does not affect the vibrational, rotational or translational motion of its atomic or molecular constituents (by friction or internal deformation). In other words, would the entropy of a Fe atom inside a top increase just because the top is spinning? Probably not. On the other hand, one could argue that on the macroscopic scale, centrifugal forces inside the solid create enough pressure on the metal lattice to induce a slight rise in temperature, therefore changing the entropy of the system.

    What about if the top is traveling linearly with constant velocity relative to an other identical top at rest. Would the entropy of the two tops be equal? I would guess : yes.

    Does entropy depend on the referential chosen? And would entropy of a closed system be affected by it accelerating? I would guess : no, if its constituents do not "feel" the effect of that acceleration. And what about relativistic thermodynamics?... :eek:)
     
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