I am trying to calculate the average thermal energy of a vibrating string that has mass M, under tension F, its boundaries are fixed a distance L apart $y(0) = y(L)$. The energy of the string for an instantaneous displacement y(x,t) is $$E[y(x,t)]= \frac 12\int_0^Ldx\left(\frac ML\left(\frac {\partial y}{\partial t}\right)^2 + F\left(\frac {\partial y}{\partial x}\right)^2\right)$$ By writing y(x,t) as Fourier series $$y(x,t) = \sum_{n=1}^N A_n(t)sin(\frac{n\pi x}{L})$$ I manage to reduce the energy equation (as asked by the question) using property of orthogonality to $$E[y(x,t)] = \sum_{n=1}^N\left(\frac M4 \dot A_n^2 + \frac{n^2\pi^2F}{4L}A_n^2 \right)$$ Then, to get the average thermal energy I attempted to evaluate the partition function as follows $$Z = \sum e^\frac{-(E[(y,t)]}{k_BT}= \sum_{\dot A_1=0}^\infty\sum_{\dot A_2=0}^\infty ... e^{-\frac M{4k_bT}(\dot A_1 +\dot A_2+...)}\sum_{A_1=0}^\infty \sum_{A_2=0}^\infty...e^{-\frac {\pi^2F}{4Lk_bT}(A_1 + 4A_2+...)}\\=\sum_{\dot A_1=0}^\infty e^{-\frac {M\dot A_1}{4k_b T}} \sum_{\dot A_2=0}^\infty e^{-\frac {M\dot A_2}{4k_b T}}...\sum_{A_1 = 0}^{\infty}e^{-{\frac{\pi^2FA_1}{4Lk_b T}}}\sum_{A_2 = 0}^{\infty}e^{-{\frac{4\pi^2FA_2}{4Lk_b T}}}...\\ =\prod_{l=0}^N {\frac{1}{1-e^{-\frac{M\dot A_l}{4k_bT}}}}\prod_{l=0}^N {\frac{1}{1-e^{-\frac{l^2A_l\pi ^2F}{4Lk_bT}}}}$$

Next, my thought is that I can write the Helmholtz free energy $F = -k_BT lnZ$ and assume ${N\to \infty}$ $$F=-k_BTln\prod_{l=0}^\infty {\frac{1}{1-e^{-\frac{M\dot A_l}{4k_bT}}}}\prod_{l=0}^\infty {\frac{1}{1-e^{-\frac{l^2A_l\pi ^2F}{4Lk_bT}}}}\\= k_bT\left(\sum_{l=1}^\infty ln(1-e^{-\frac{M\dot A_l}{4k_bT}})+ln({1-e^{-\frac{l^2A_l\pi ^2F}{4Lk_bT}}})\right)$$

Up to this point, I find myself somewhat stuck; in fact, all these steps are tailored for computing the average energy of a harmonic oscillator. After obtaining the free energy $\left(\frac{\pi^2}{6}(\frac{k_bT}{\hbar w})^2 \hbar w for\,quantum\,harmonic\, oscillator/string\right)$, we should be able to get the entropy and the average thermal energy.

This problem is from Introductory Statistical Mechanics by Bowley, Roger Chapter 8, problem 10.Full problem

  • $\begingroup$ And where is the question? It seems that in you screen there is typo because $\sum_{n>0}n^{-2}=\pi^2/6$, not $\pi^2/8$. You can also write down the expression for $\dot{A}_i$ from the equation of motion and see that it can be epxressed via $A_i$ (it is indeed like harmonic oscillator) $\endgroup$ Commented Nov 16, 2023 at 17:20
  • 1
    $\begingroup$ @ArtemAlexandrov I have attached the hyperlink to the full problem at the end, I think there is a typo in the hint given by the author. The question is asking for the average thermal energy and the mean square displacement of the string (ans for mean square displacement is $\frac{3k_B T}{\frac{16F}{L}}$ and that is the only answer provided. In the problem, the author suggests us to treat $\dot A_i $ and $A_i$ as 2N generalized coordinates, I am not sure whether expressing one coordinate in terms of the other helps or not. $\endgroup$
    – user281667
    Commented Nov 17, 2023 at 1:31

1 Answer 1


You can use the equipartition theorem since in fourier space the hamiltonian is a sum of uncoupled quadratic harmonic oscillators: $$E[y(x,t)] = \sum_{n=1}^N\left(\dfrac{M/2}{2}\dot A_n^2 + \frac{n^2\pi^2F/(2L)}{2}A_n^2 \right)$$ Thus $M/2\langle\dot A_n^2 \rangle= k_bT$, $n^2\pi^2F/(2L)\langle A_n^2\rangle = k_bT$ and cross terms are 0.

You also have: $$y^2 = \sum_{n=1, m = 1}^N A_mA_n\sin(\frac{n\pi x}{L})\sin(\frac{m\pi x}{L})$$ The average is easily done because $\langle A_n A_m\rangle\propto k_bT\delta_{n, m}$ : $$\langle y^2\rangle = \sum_{n=1}^N \langle A_n^2\rangle \sin(\frac{n\pi x}{L})^2 = \dfrac{Lk_bT}{2\pi^2F}\sum_{n=1}^N \dfrac{\sin(\frac{n\pi x}{L})^2}{n^2}$$

Specifically, in $x = L/2$: $$\langle y^2\rangle = \dfrac{Lk_bT}{2\pi^2F}\sum_{n=1}^N \dfrac{\sin(n\pi/2 )^2}{n^2}$$

The sin square takes the values 0, 1, 0, 1, ... with increasing n, thus the sum can be rewritten as: $$\langle y^2\rangle = \dfrac{Lk_bT}{\pi^2F}\sum_{n=0}^N \dfrac{1}{(2n + 1)^2} = \dfrac{Lk_bT}{2\pi^2F} \dfrac{\pi^2}{8} = \dfrac{Lk_bT}{16F}$$

This was in 1D, in 3D you'd have three times more (uncoupled) modes so: $$\langle x^2 + y^2 + z^2\rangle = \dfrac{3Lk_bT}{16F}$$


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