Specific heats

  • Thread starter Silviu
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Homework Statement


Show that ##\frac{\partial U}{\partial T}|_{N} = \frac{\partial U}{\partial T}|_{\mu} + \frac{\partial U}{\partial \mu}|_{T} \frac{\partial \mu}{\partial T}|_{N} ## (Pathria, 3rd Edition, pg. 197)

Homework Equations


##U=TS + \mu N - pV##

The Attempt at a Solution


I tried to take the derivative of U with respect with T at constant N and then at constant ##\mu## but I can't get the term on the right. I am not sure how can I get a product of 2 derivatives, because in taking derivatives in ##U=TS + \mu N - pV## I always have a derivative times a number (for example ##T\frac{\partial S}{\partial T}|_N##). What should I do? Is there another way other than using this formula?
 

Answers and Replies

  • #2
Charles Link
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Just a suggestion=perhaps the book is correct, but they appear to be using ## U=U(N,\mu,T) ##, so that in taking ## \frac{\partial{}}{\partial{T}} ##, both ## N ## and ## \mu ## should be held constant.
 
  • #3
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Just a suggestion=perhaps the book is correct, but they appear to be using ## U=U(N,\mu,T) ##, so that in taking ## \frac{\partial{}}{\partial{T}} ##, both ## N ## and ## \mu ## should be held constant. Likewise for the other partial derivatives.
OK, But does this leads to having a term with 2 derivatives. I might have stuff like ##\frac{\partial U}{\partial T}|_{N,\mu}## or something similar, but I still don't see how do I get a product of 2 derivatives. Thank you!
 
  • #4
Charles Link
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OK, But does this leads to having a term with 2 derivatives. I might have stuff like ##\frac{\partial U}{\partial T}|_{N,\mu}## or something similar, but I still don't see how do I get a product of 2 derivatives. Thank you!
I'm going to need to study it further. It looks like it might take a little work if it is correct.
 
  • #5
Charles Link
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I'm going to need to study it further. It looks like it might take a little work if it is correct.
With a little further thought, it seems taking a partial with only holding one other variable constant is ambiguous, because on the multi-dimensional surface, you can not precisely define how much the quantity ## U ## will change. It allows multiple paths depending on what you do with the 3rd variable. @Chestermiller Might you offer an input here?
 
  • #6
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With a little further thought, it seems taking a partial with only holding one other variable constant is ambiguous, because on the multi-dimensional surface, you can not precisely define how much the quantity ## U ## will change. It allows multiple paths depending on what you do with the 3rd variable. @Chestermiller Might you offer an input here?
But even if you keep 2 constant at each derivation, I am still not sure how to prove it.
 
  • #7
Charles Link
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But even if you keep 2 constant at each derivation, I am still not sure how to prove it.
If you look at the term on the left and the first term on the right, if you include the third variable in both of the partial derivatives to be held constant, these two are then the same. IMO the equation is mathematically unsound, but perhaps @Chestermiller can add to that and/or possibly concur.
 
  • #8
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U is typically a function of 3 parameters, usually S, V, and N. So two parameters need to be held constant in those partial derivatives. What is the other parameter?
 
  • #9
Charles Link
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U is typically a function of 3 parameters, usually S, V, and N. So two parameters need to be held constant in those partial derivatives. What is the other parameter?
The author is using ## N ## , ## \mu ##, and ## T ## as the 3 parameters, but he's only holding one constant at a time. To me this appears to be mathematically unsound.
 
  • #10
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The author is using ## N ## , ## \mu ##, and ## T ## as the 3 parameters, but he's only holding one constant at a time. To me this appears to be mathematically unsound.
I asked my professor and he told me that for ##C_n## N and V must be held constant and for ##C_\mu##, ##\mu## and V must be held constant in order to solve the problem. Does this help?
 
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  • #11
Charles Link
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I asked my professor and he told me that for ##C_n## N and V must be held constant and for ##C_\mu##, ##\mu## and V must be held constant in order to solve the problem. Does this help?
That is very helpful=I solved it now=I will post it momentarily. ## \\ ##
## U=U(N,T,V) ## and ## N=N(\mu,T,V) ## so that ## U=U(\mu,T,V) ## and ## \mu=\mu(N,T,V) ##. ## \\ ##
## (\frac{\partial{U(N,T,V)}}{\partial{T}})_{N,V}=(\frac{\partial{U(\mu,T,V)}}{\partial{T}})_{N,V}=(\frac{\partial{U(\mu,T,V)}}{\partial{\mu}})_{T,V}(\frac{\partial{\mu(N,T,V)}}{\partial{T}})_{N,V}+(\frac{\partial{U(\mu,T,V)}}{\partial{T}})_{\mu,V} ##. ## \\ ##
Yes, your professor's input was very helpful. (Note: Ten minutes after posting= I just edited the T, V subscript in the first term after the second "=" sign from an incorrect N,V subscript)
 
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  • #13
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@Silviu Please see the latest editing on the previous post.
Thank you so much for help. But I am a bit confused about the math behind it. Is this like a chain rule, when you keep some variable constants?
 
  • #14
Charles Link
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Thank you so much for help. But I am a bit confused about the math behind it. Is this like a chain rule, when you keep some variable constants?
Very much so=the chain rule. You go through the list of variables and take the appropriate derivatives, and skip over any variables that are held constant.
 
  • #15
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Very much so=the chain rule. You go through the list of variables and take the appropriate derivatives, and skip over any variables that are held constant.
OK, makes sense. Thank you so so much!
 
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