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Delta <=> dX

  1. Dec 6, 2012 #1

    Reading a book on thermodynamics and the guy often uses something like this :

    ∫1/T dQ = ΔS

    and then he says "this in differential form" :

    dQ/T = dS

    I kind of get the idea visually that one slice of "integral" will be dQ and you can think of it this way.
    But my question is how do you mathematically express this conversion between delta <==> dX (both ways).
    Some Examples would be nice.

    thank you
  2. jcsd
  3. Dec 7, 2012 #2


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    hi mraptor! :smile:
    integrate dQ/T = dS, and obviously you get …

    ∫1/T dQ = ∫ dS :wink:

    then write ∫ dS = ∆S :smile:
  4. Dec 7, 2012 #3


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    While I do not know much Physics

    The integral of dQ/t in a reversible system is independent of path. This means that the integral defines a differentiable function on the thermodynamic phase space. This function I think is called entropy and its derivative is dQ/T.

    The mathematical fact that you want is that when you have an integral that is path independent - AKA conservative - starting from any point one get a well defined function by integrating along any path from the starting point to another point. You should try to prove for yourself that this function can be differentiated.
  5. Dec 7, 2012 #4


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    In other words, dQ is not an exact 1-form, it needs an integrating factor (1/T) to make it exact. Caratheodory re-wrote known (as of 1910) thermodynamics from the point of view of differential geometry (as known to him).
  6. Dec 7, 2012 #5
    I thought :

    ∫ dS = S

    not ΔS ! that is what I can't get..
    and also what about the reverse conversion..
  7. Dec 7, 2012 #6


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    ΔS is the change in S between the endpoints of the integral.
  8. Dec 7, 2012 #7
    So you are saying that he is implying that he is doing definite-integral, rather than indefinite-integral..
    That is why he is getting ΔS, rather than S
  9. Dec 7, 2012 #8


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    yup! :biggrin:
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