Express entropy density in terms of energy density (Ashcroft/Mermin)

[EE18]

Homework Statement

I am trying to solve Problem 2a in Chapter 2 of Ashcroft and Mermin. I'm not asking about the whole problem, but just the start. Essentially, one needs to expression the entropy density $s$ in terms of the energy density $u$ and, as far as I can tell from other methods to solving this problem, I should arrive at $s(T) = u(T)/T$.

Relevant Equations

$s(T) = u(T)/T$

The starting point is the identity
$$\left(\frac{\partial u}{\partial T}\right)_n = T\left(\frac{\partial s}{\partial T}\right)_n.$$
I then try to proceed as follows:
Integrating both with respect to $T$ after dividing through by $T$, we find
$$ \int_0^T \left(\frac{\partial s}{\partial T'}\right)_n = s(T) - s(0) = s(T) $$
$$ = \int_0^T \frac{1}{T'}\left(\frac{\partial u}{\partial T'}\right)_n = \frac{u}{T'}\biggr\rvert_0^T + \int_0^T \frac{1}{T'^2}u ...$$
where we have used an integration by parts, the fundamental theorem of calculus, and the third law of thermodynamics (in (1)), and where we recall that $u_0$ is the energy density in the ground state (see AM (2.79)).

But I can't seem to go any further. I found a solution which does something I totally can't understand in the last line, and I'm hoping someone can either clarify that solution and/or show me how to go further. The problem is obviously the last integral on the right-hand side.

 

[DrClaude]

I found a solution which does something I totally can't understand in the last line, and I'm hoping someone can either clarify that solution and/or show me how to go further. The problem is obviously the last integral on the right-hand side.

Compare to the line $s=\ldots$: you have the same integral on the RHS (with a factor 1/2), so you can combine the integrals together:
$$
\int \ldots = x + \frac{1}{2} \int \ldots
$$
so
$$
\frac{1}{2} \int \ldots = x
$$

 

[EE18]

Compare to the line $s=\ldots$: you have the same integral on the RHS (with a factor 1/2), so you can combine the integrals together:
$$
\int \ldots = x + \frac{1}{2} \int \ldots
$$
so
$$
\frac{1}{2} \int \ldots = x
$$

Thanks for helping out! Where is that factor of $1/2$ coming from though?

 

[DrClaude]

Thanks for helping out! Where is that factor of $1/2$ coming from though?

From the integration by parts (integral of $T^{-2} dT$).

Edit: To be clear, I moved the factor in front of the integral, but it appears in $1/2T$ in the image you posted.

 

[EE18]

From the integration by parts (integral of $T^{-2} dT$).

Edit: To be clear, I moved the factor in front of the integral, but it appears in $1/2T$ in the image you posted.

Maybe I'm going crazy, but isn't $-\int 1/x^2 dx = 1/x$? I don't see where the 1/2 comes from?

 

[vanhees71]

Of course!

 

[DrClaude]

Sorry, I was sloppy when I tried to reproduce what was is the image. I can't make sense of it now.

 

[strangerep]

Sorry, I was sloppy when I tried to reproduce what was is the image. I can't make sense of it now.

I begin to think that image is rubbish. It's essentially doing 2 successive integrations by parts (the 2nd one incorrectly, imho). When done correctly, it should just return (uselessly) to its starting point.

Btw, since at least one other variable is being held constant, maybe you just treat the first line as if it were $$ \frac{du}{dT} ~=~ T \; \frac{dS}{dT} ~,~~~~~~ \Rightarrow~~~
du ~=~ T dS ~.$$Of course, you've got to be careful about what depends on what, and what's being held constant.

 

[PhDeezNutz]

I think the authors were lazy in their notation of partial derivatives.

You were given the following.

$\left(\frac{\partial U}{\partial T} \right)_n = T \left( \frac{\partial S}{\partial T} \right)_n$

It should read (I think)

$$\left(\frac{\partial U}{\partial T} \right)_{n,total} = T \left( \frac{\partial S}{\partial T} \right)_{n,total}$$

$$U\left(S\left(U,T\right), T\right)$$

$$S\left(U\left(S,T\right), T\right)$$

There are more than one way that $U$ or $S$ depend on T

$$\left( \frac{\partial U}{\partial T} \right)_{n,total} = \frac{\partial U}{\partial S} \left(\frac{\partial S}{\partial T} \right)_{n,direct} + \left(\frac{\partial U}{\partial T} \right)_{n,direct}$$

$$T\left( \frac{\partial S}{\partial T} \right)_{n,total} = T \frac{\partial S}{\partial U} \left(\frac{\partial U}{\partial T} \right)_{n,direct} + T\left(\frac{\partial S}{\partial T} \right)_{n,direct}$$

Setting them equal

$$\frac{\partial U}{\partial S} \left(\frac{\partial S}{\partial T} \right)_{n,direct} + \left(\frac{\partial U}{\partial T} \right)_{n,direct} = T \frac{\partial S}{\partial U} \left(\frac{\partial U}{\partial T} \right)_{n,direct} + T\left(\frac{\partial S}{\partial T} \right)_{n,direct}$$

Comparing like terms

$$\frac{\partial U}{\partial S} = T \Rightarrow U = ST$$

$$T\frac{\partial S}{\partial U} = 1 \Rightarrow S = \frac{U}{T}$$

This is my take on it. I could be wrong of course. I know in principal there are more terms since $n$ might not be constant.