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Let $\epsilon_i$ be the internal energies of a microstate. We are supposed to show that the equation $$\mathscr E = \frac{\sum_{i=1}^{N}\epsilon_i\mathrm e^{-\beta \epsilon_i}}{\sum_{i = 1}^{N}\mathrm e^{-\beta \epsilon_i}}$$ has a unique real solution for $\beta$. We can assume the following restriction: $$\text{min}(\epsilon_i) \leq \mathscr E\leq \frac{\sum_{i = 1}^{N}\epsilon_i}{N}.$$

Alright, I am not quite sure yet how to finally solve this task, but here are two ideas of mine:

  1. I first multiplied by the denominator and applied the logarithm on both sides, but since there is no rule for $\ln(a+b)$, this isn't fruitful, I guess.

  2. Maybe it suffices to show that $\mathscr E(\beta)$ is a strictly monotonic function? But then the problem is that the first derivative looks 'horrible'.

Here is the first partial derivative: $$\frac{\partial\mathscr E}{\partial \beta} = \frac{ \sum_i \mathrm e^{-\beta \epsilon_i} \cdot \left(-\sum_i \epsilon_i^2 \mathrm e^{-\beta \epsilon_i} \right) - \sum_i\epsilon_i \mathrm e^{-\beta\epsilon_i} \cdot \left(-\sum_i\epsilon_i \mathrm e^{-\beta\epsilon_i}\right) }{\left(\sum_i \mathrm e^{-\beta\epsilon_i}\right)^2} = \frac{ \sum_i \mathrm e^{-\beta \epsilon_i} \cdot \left(-\sum_i \epsilon_i^2 \mathrm e^{-\beta \epsilon_i} \right) + \left(\sum_i\epsilon_i \mathrm e^{-\beta\epsilon_i}\right)^2 }{\left(\sum_i \mathrm e^{-\beta\epsilon_i}\right)^2}$$

I am sure that ansatz (2) must work, but how I don't know yet. I guess that the trick is now to show that the first derivative is strictly positive or negative.

EDIT: I found the following solution online (under a password-protected site): enter image description here

I don't even understand the second line, where does $U_i^2$ suddenly come from? ("Betrachte Zähler" is German for "Consider the numerator".)

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  • $\begingroup$ You computed the first derivative and it's manifestly $\leq 0$, so the function is monotonic. Why are you re-expanding it? $\endgroup$ Commented Dec 12, 2019 at 10:12
  • $\begingroup$ @Hans: I computed the first derivative wrongly, sorry for that. See my edit. $\endgroup$
    – user248824
    Commented Dec 12, 2019 at 10:15

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I have a little bit of a glib argument. You want to prove that for $\mathcal{E}$ in a certain range, there exists a $\beta$ such that $$ \mathcal{E} = \langle E \rangle $$ where the expectation value of course depends on $\beta$. You have computed that

$$- \frac{d \langle E \rangle}{d \beta} = \langle E^2 \rangle - \langle E \rangle^2.$$

But in any statistical average, such a difference is positive, since $$ \langle E^2 \rangle - \langle E \rangle^2 = \langle (E - \langle E \rangle)^2 \rangle \geq 0$$ so it follows that $\langle E \rangle$ is monotonically decreasing with $\beta$. (It can only be constant if there is no variance in $E$ i.e. all $\epsilon_i$ are identical.)

Finally you have to check the limits of $\langle E \rangle$ as $\beta \to 0,\infty$ and you'll see that they coincide with the question.

A mathematician could have used a different argument: starting from the fact that $x \mapsto x^2$ is a convex function, she would appeal to Jensen's inequality. This is more general and applies to any convex or concave function, not just the function $x \mapsto x^2$.

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  • $\begingroup$ Thanks for the time you took. However, I cannot really follow your arguments. What is $\mathcal E$, and why is $\mathcal E =\langle E\rangle$? Further clarification would really be appreciated! $\endgroup$
    – user248824
    Commented Dec 12, 2019 at 21:01
  • $\begingroup$ With $\mathcal{E}$ I just meant the curly $E$ from your question. With $\langle X \rangle$ I mean the expectation value of the observable $X$ in the ensemble, i.e. $\langle E^n \rangle = Z^{-1} \sum_i \epsilon_i^n \exp(- \beta \epsilon_i)$, and of course $Z = \sum_i \exp(- \beta \epsilon_i)$. $\endgroup$ Commented Dec 12, 2019 at 23:23
  • $\begingroup$ Okay, thanks for that as well. But why do I want to prove that there exists a $\beta$ such that $\mathcal E = \langle E\rangle$? I mean, I don't even know what $\langle E\rangle$ is, do I? $\endgroup$
    – user248824
    Commented Dec 13, 2019 at 13:29
  • $\begingroup$ We know what $\langle E \rangle$ is: it's the RHS of your first equation. Maybe you prefer to call it $\mathscr{E}$ or whatever, fine. To not overload the notation, we would say that we want to prove that for any $y \in [\ldots,\ldots]$ there exists a $\beta$ such that $y = \langle E \rangle$. $\endgroup$ Commented Dec 13, 2019 at 18:59
  • $\begingroup$ Oh, you had written that $\langle E\rangle=\mathcal E$ ... Sorry for that. But anyway, I don't think I have proven that $-\frac{\partial \mathcal E}{\partial\beta} =\mathcal E^2-\mathcal E^2$ (here I am not sure how to write my expression), just look at my partial derivative ... $\endgroup$
    – user248824
    Commented Dec 13, 2019 at 22:04

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