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Question about total derivative/chain rule

  1. Mar 23, 2012 #1
    The following actually comes from Landau's 3rd Edit. Statistical Physics Part 1, Paragraph on Adiabatic Processes, Page 39.

    I have the following two equations, where [itex]\lambda=\lambda(t)[/itex]. I am not so sure about [itex]S[/itex] (which is somewhat my problem):
    [tex]\frac{\mathrm{d}S}{\mathrm{d}t} = \left( \frac{\mathrm{d}\lambda}{\mathrm{d}t} \right)^2[/tex]

    Which is supposed to mean:
    [tex]\frac{\mathrm{d}S}{\mathrm{d}\lambda} = \frac{\mathrm{d}\lambda}{\mathrm{d}t}[/tex]

    Now, I thought that what is essentially being done there is multiplying the first equation such that we get:
    [tex]\frac{\mathrm{d}S}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}\lambda} = \frac{\mathrm{d}\lambda}{\mathrm{d}t}[/tex]

    But if I now assume that:
    [tex]\frac{\mathrm{d}S}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}\lambda} = \frac{\mathrm{d}S}{\mathrm{d}\lambda}[/tex]
    doesn't that in turn mean that [itex]t[/itex] is a function of [itex]\lambda[/itex]? Mathematically this seams sounds (to me), but physically this does not make so much sense, if [itex]t[/itex] is the time.
  2. jcsd
  3. Mar 23, 2012 #2


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    welcome to pf!

    hi mSSM! welcome to pf! :smile:
    suppose λ is distance

    why shouldn't time be a function of distance? :wink:
  4. Mar 23, 2012 #3
    Re: welcome to pf!

    Thanks! :) Okay, I guess you could turn it that way. So essentially you say that we simply associate every instant of time with a certain "distance"... Since in my case the distance (=external condition) is a mechanical quantity I guess this would be sound.

    Would something like that still be acceptable for a macroscopic quantity?
  5. Mar 23, 2012 #4


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    i'm not sure i understand that sentence :confused:

    a quantity is a quantity

    why would length (or any other quantity) be unacceptable?
  6. Mar 23, 2012 #5
    Yeah, you are right. I can't think of a reason why that shouldn't work. :)
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