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My course guide gives the following derivation for change in entropy w.r.t. energy, where I don't understand a step:

\begin{align} E & = E_1 + E_2 \\ S & = S_1 + S_2 \\ S(E,E_1 ) & = S_1 (E_1) + S_2(E-E_1) \end{align}

Then the question is asked, which $E_1$ the new S wouldn't change. I don't truly understand what the premise is of this question. But I understand that it is a maximum, thus you can find it with setting a derivative to zero as follows:

$$\frac{\partial S}{\partial E_1} = 0 =\frac{\partial S_1}{\partial E_1} + \frac{\partial S_2 }{\partial E_1} \tag{ chain rule}$$

Then the guide just says: "Use the following relation". I don't get how to find this relation?

$$ \frac{\partial S_2}{\partial E_1} = -\frac{\partial S_2}{\partial E_2} $$

Where then this follows:

$$ \frac{\partial S_1}{\partial E_1} = \frac{\partial S_2}{\partial E_2} $$

Central question: $$ \frac{\partial S_2}{\partial E_1} = -\frac{\partial S_2}{\partial E_2} $$ How to find this. When I differentiate $S$ to $E_2$ I just get:

$$\frac{\partial S}{\partial E_2} = 0 =\frac{\partial S_1}{\partial E_2} + \frac{\partial S_2 }{\partial E_2} \tag{ chain rule}$$

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  • $\begingroup$ Differentiate $S_2(E_2) = S_2(E-E_1)$ $\endgroup$ Apr 28, 2022 at 15:02

3 Answers 3

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It's not unreasonable to be confused about this. Let's say you have a function $f$ of one variable and $f'$ is its derivative. Then we have $$\frac{d}{dx} f(c-x) = \color{red}{-}f'(c-x)$$ via the chain rule. If we have a function $g$ of two variables, then we might similarly write $$\frac{\partial}{\partial x} g(c-x,y) = \color{red}{-} \big(\partial_1 g\big)(c-x,y)$$ where $\partial_1g$ is the function obtained by differentiating $g$ with respect to its first entry. It is extremely common to simply call $\partial_1 g$ the same thing as $\frac{\partial g}{\partial x}$, but that only works when we assume that $x$ is the thing which we plug into the first slot of $g$. When that isn't the case - e.g. here - that notation is bad in my opinion.


In our case, we have the following expression: $$S(E,E_1) := S_1(E_1) + S_2(E - E_1)$$ $S_1(\epsilon)$ is the entropy of the first system when it has energy $\epsilon$. $S_2(\epsilon)$ is the entropy of the second system when it has energy $\epsilon$. $S(E,E_1)$ is the entropy of both systems together when they have total energy $E$, and when the first system has energy $E_1$ (so the second system has energy $E_2= E-E_1$).

If we wish to maximize this with respect to $E_1$, we would differentiate and set the result to zero: $$\frac{d}{dE_1} \big(S_1(E_1) + S_2(E-E_1)\big) = S_1'(E_1) \color{red}{-} S_2'(E-E_1) = 0$$ $$\implies S_1'(E_1) = S_2'(\underbrace{E-E_1}_{=E_2})$$

This is what your course guide is trying to say.

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The point is that $E$ is fixed, so that if you raise $E_1$ by some amount $\delta x$ you have to lower $E_2$ by the exact same amount $\delta x$. This means that $E_1$ and $E_2$ can not be varied arbitrarily, but must always change like this $$ E_1 \to E_1 + \delta x $$ $$ E_2 \to E_2 - \delta x $$

This is reflected in the signs of the derivative because the derivative is: $$ \frac{df}{dx} = \lim_{\delta x \to 0}\frac{f(x+\delta x) - f(x)}{\delta x} =\lim_{\delta x \to 0}\frac{f(x) - f(x-\delta x)}{\delta x}\;. $$

We can also multiply the above equation by $-1$ to see that: $$ -\frac{df}{dx} = \lim_{\delta x \to 0}\frac{f(x-\delta x) - f(x)}{\delta x}\;. $$

Finally, we can drop the limits above and rearrange to see that: $$ f(x+\delta x) \approx f(x)+\frac{df}{dx}\delta x $$ $$ f(x-\delta x) \approx f(x)-\frac{df}{dx}\delta x \;, $$ where the approximation is exact as $\delta x$ goes to zero.

The total entropy $S_1(E_1) + S_2(E_2)$ is to be maximized subject to the above constraint (i.e., that the variation in $E_1$ is always the opposite of the variation in $E_2$).

At the maximum the first order variation in the total entropy $\delta S$ must be zero. As $E_1$ changes to $E_1+\delta x$, we have that the total entropy $S$ changes like: $$ S\to S+\delta S = S_1(E_1 + \delta x) + S_2(E_2 - \delta x) $$ $$ \approx S_1(E_1)+\frac{dS_1}{dE_1}\delta x + S_2(E_2) - \frac{dS_2}{dE_2}\delta x $$ $$ \approx S + \delta x (\frac{dS_1}{dE_1}- \frac{dS_2}{dE_2}) $$

Or, to first order in $\delta x$: $$ \delta S = \delta x(\frac{dS_1}{dE_1} - \frac{dS_2}{dE_2}) $$

The first order variation with $\delta x$ is zero at the maximum, so the maximum entropy condition is: $$ \frac{dS_1}{dE_1} - \frac{dS_2}{dE_2} = 0 $$ Or, rearranging: $$ \frac{dS_1}{dE_1} = \frac{dS_2}{dE_2} $$

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In my opinion the notation is very bad. $S_2$ is a function of a single variable (at least in this context here) and it simply does not make sense to compute partial derivatives with respect to two different energy variables. However, we can define a function:

$$\mathscr S := S_2 \circ Q_E \quad , $$ with $Q_E: E_1 \mapsto E-E_1 $ and hence $\mathscr S(E_1) = S_2(E-E_1) = S_2(E_2)\quad .$

By the chain rule, we thus obtain

$$\mathscr S^\prime (E_1) = - S_2^\prime(E_2) \quad , $$ where prime denotes the derivative.

So what this all means it that on the one hand, you have the entropy of system $2$ as a function of the energy of this system, namely $S_2$ and on the other hand you have the very same entropy as a function of the energy of system $1$, namely $\mathscr S$. Let me stress that while both yield the same physical quantity, i.e. the entropy of system $2$, these are different functions, explaining my comment at the beginning.

I guess in your case it is meant that \begin{align} \frac{\partial S_2}{\partial E_1} (E_1) &:= \mathscr S^\prime (E_1) \\ \frac{\partial S_2}{\partial E_2} (E_2) &:=S_2^\prime(E_2) \quad . \end{align}

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  • $\begingroup$ It's fairly standard Leibniz notation that is used all over physics. Yeah, it's "bad," but I'm pretty sure your notation is going to be even more confusing to OP. $\endgroup$
    – hft
    Apr 28, 2022 at 17:00
  • $\begingroup$ @hft Of course, this might be a matter of taste, and I did not claim that the notation in the question is useless per se, but to learn things, at least in my opinion, it is the best to be as precise as possible. $\endgroup$ Apr 28, 2022 at 17:07
  • $\begingroup$ Precision is not always a good thing. $\endgroup$
    – hft
    Apr 28, 2022 at 18:10
  • $\begingroup$ @hft As far as I can see, OP was concerned with the issue arising from the completely abstruse notation in the question...Be it as it may, you're of course free to disagree with my answer/perspective. $\endgroup$ Apr 28, 2022 at 18:18

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