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Speed of sound is related to the derivative of pressure with respect to density:

$v_s=\sqrt{(\frac{\partial P}{\partial \rho})_S}$

where S tells us the derivative must be taken while keeping entropy constant.

For an ideal Bose gas, $P$ and $\rho$ can be expressed in terms of Bose-Einstein functions:

$g_v(z) = \frac{1}{\Gamma(v)}\int \frac{x^{v-1}}{z^{-1}e^x -1}$

where in the given context $z=e^{\frac{\mu}{T}}$ and $\lambda = \lambda (T)$ is the thermodynamic wavelength

Those expressions are:

$ P = \frac{T}{\lambda^3} g_{\frac{5}{2}}(z)$

$ \rho = \frac{m}{\lambda^3} g_{\frac{3}{2}}(z)$

Having all this in my hands, I still fail to perform the differentiation. Could you please hint what should my first step be?

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    $\begingroup$ You have $P=P(z,\lambda)$ and $\rho=\rho(z,\lambda)$, whence $dP=P_z dz + P_\lambda d\lambda$ and similarly for $\rho$. Now the condition $S=\text{constant}$ presumably relates $dz$ and $d\lambda$. You'll need to work out that constraint. Then just divide $dP/d\rho$. Your other option is to work out the change of variables $z,\lambda \to \rho, S$ to make $P=P(\rho,S)$. Either way it looks like it will get messy. $\endgroup$ – Michael Brown Oct 15 '13 at 1:02
  • $\begingroup$ Thanks for clarifying the situation! However, the simple form of the answer $v_s^2 = \frac{5T}{3m}\frac{g_{\frac{5}{2}}(z)}{g_{\frac{3}{2}}(z)}= $ possibly hints to an easy way around.. $\endgroup$ – user2077647 Oct 16 '13 at 20:57
  • $\begingroup$ $v_s^2 = \frac{5T}{3m}\frac{g_{\frac{5}{2}}(z)}{g_{\frac{3}{2}}(z)}= \frac{5}{3} \frac{P}{\rho}$ $\endgroup$ – user2077647 Oct 16 '13 at 21:03
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    $\begingroup$ So I tackled the problem from another angle, and discovered that for an adiabatic process $P*\rho^{\frac{-5}{3}}=Const$. This basically solves the problem. Pretty happy that I didnt try to brute-force that derivative! $\endgroup$ – user2077647 Oct 16 '13 at 21:48

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