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What is Feynman saying here? (entropy)

  1. May 22, 2010 #1
    So I bought myself the Feynman lectures and was reading up on his discussion on Statistical Mechanics and Thermodynamics (in volume I). The following is regarding Thermodynamics.

    So he defined entropy (at least the change in it from a to b) as delta S = int(Q/T) over a reversible path from a to b. He then proved that all reversible paths from a to b give the same result (at least in the framework of gases). However:

    "First, suppose that we do irreversible work on an object by friction, generating a heat Q on some object at temperature T. The entropy is increased by Q/T. The heat Q is equal to the work, and thus when we do a certain amount of work by friction against an object whose temperature is T, the entropy of the whole world increases by W/T."

    This makes me wonder about two things:
    1) I thought that for calculating delta S, you had to find a reversible path from start to end. Is it perhaps because of the following: to calculate the change in entropy for the floor alone, we can say heating it is in itself reversible (floors can cool down), so delta S_floor = Q_floor/T; identically for the object delta S_object = Q_object/T. This would give S = S_floor + S_object = Q/T. That does seem like a big jump to ignore in your reasoning (especially since it was his first example). But even if this is how it works, then that leads me to question 2
    2) How do you know that it is well-defined then? For example, imagine a model for friction force where if you were to pull an object faster or slower, it would result in a different Q (for example because if you were to pull really slowly, you'd continously have to fight the larger static friction, then if you were to pull it reasonably fast, just giving the normal (and lower) kinetic friction). Following the reasoning from 1, we get different entropies for the same change from a to b.

    All replies welcome,
    mr. vodka
  2. jcsd
  3. May 22, 2010 #2


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    A few thoughts that I hope will be helpful:

    1. "Reversible" has a very specific definition in thermodynamics. It doesn't mean merely that a system can be brought to its original state (as in your floor example). It means that no gradients exist and therefore that no entropy is produced. It means that any heat transfer is accomplished by an infinitesimal temperature difference. Reversibility is an idealization that isn't possible in real life, though it can be approached.

    2. Entropy is a state variable, so the entropy change from state A to state B is independent of reversibility. Once you've calculated if from a well-defined, reversible process, you can apply the same value to any irreversible process like friction.

    3. It makes a difference how fast one pulls against a frictional force. The faster one pulls, the more work one does. Thus, we should not be surprised to calculate different entropy increases.

    Does this help answer your questions?
  4. May 23, 2010 #3
    Thank you Mapes.

    a) Do you mean the way I calculated it in the first post is wrong? If it is, then how can you calculate the entropy change after sliding an object over a floor with heat Q in a temperature T? How did Feynman get Q/T? What reversible path did he choose?

    b) And I understand entropy should be a state variable, but is it obvious from the definition? Applied to this case: well first I don't understand why pulling an object faster will ask for more work, I would think less work. (Take in account that the distance from a to b is fixed). But anyway, the main thing is: going from a to b, you say, we can get different entropie increases, yet on the other hand, you say S is a state variable. Does this sentence not contradict itself?

    EDIT: as you say, if there is no temperature gradient, it is reversible. Is this why Feynman said delta S = Q/T, because there was no temperature gradient? (as he said we held T constant) But... obviously friction is irreversible, what do we think of this?
    Last edited: May 23, 2010
  5. May 23, 2010 #4


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    Feynman's reversible path was heat transfer into an object at constant temperature T. It is assumed that the object is large enough that the temperature change from the transfer of energy Q is minimal. It is also assumed that the friction does negligible mechanical work to the object, so that any energy transfer is predominantly in the form of heat. (But note that mechanical work was applied externally to cause movement in the first place.)

    No; if the object travels the same distance but you pull with more or less force, you're doing a different amount of work to the system, and thus the final entropy of the system is different. The physical endpoints a and b are the same, but the thermodynamic state B is different.

    The irreversibility is captured in the conversion from work to heat. Once you start dealing with the heat Q, you can consider the entropy increase to the rest of the object to be reversible. In other words, all of the entropy increase occurs in a localized (and poorly defined) region at the surface of the object through friction. We avoid detailed calculations of what's going on at this region (e.g., material deforming, bonds breaking) by assuming that the energy originally took the form of reversible work (which doesn't carry entropy), was entirely converted to heat (which does carry entropy), and is now propagating through an object at constant temperature (which is reversible and so transfers entropy but does not produce it).
  6. May 23, 2010 #5
    I see. Everything is clear now :) That was quick. (although I still think Feynman could've elaborated on his example)

    Thank you very much!
  7. May 23, 2010 #6


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    Feynman definitely moves very fast and loose in those lectures.

    Glad to help!
  8. May 23, 2010 #7

    Andy Resnick

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    There was a little misdirection here. He considers the transfer of energy from one object to another, but then talks about the entropy change "of the whole world".

    First- friction is modeled as a 100% dissipative process. That means all work is converted into use*less* heat, regardless of how much work is done, or how long or how fast the process occurs.

    So, since we have (deliberately) obscured the details of the transfer of energy, we can't calculate the change of entropy involved in that process for the two objects participating in the process. However, we invoke conservation of energy *for the whole world*, and can then talk about the change of entropy for the whole world- that work energy was completely and irretrievably lost to heat.
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