# How two prove the second sound velocity is $1/\sqrt{3}$ times than the first sound [closed]

From Landau two hydrodynamics model in superfluid, we have the result $c_1^2=\frac{\partial P}{\partial \rho}|_T$ and $c_2^2=\frac{\rho_s s^2 T}{\rho_n c}$. In the zero temperature limit, how to relate those two quantities, to get the conclusion that $c_1^2=3c_2^2$?

Let's consider gas of particles which move with constant velocity $V$. In such case we will derive speed of sound. $c_s^2=\frac{\partial P}{\partial \rho}$ where $P$-is a pressure and $\rho$ is a density of gases. Set p is a momentum, then the momentum which gas give the wall from small time $\Delta t$ is $\Delta p=2 m V_x \frac{n}{2} V_x \Delta t S$ than $P=\frac{F}{s}=\frac{\Delta p}{\Delta t S}=m\rho V_x^2=\frac{\rho V^2}{3}$ and the speed of sound has following form $c_s^2=\frac{V^2}{3}$
The same picture is occur for second sound. There are two type of excitations in super fluid phonons and rotons. In case of small temperature the main contribution is given by phonon because rotons contribution suppressed via energy gap $n_{rot}\sim e^{-\frac{\Delta}{T}}$. The phonons spectrum has the following form $E(p)=u p$, where u is the speed of sound(constant).(It means that all particles moves with the same module of velocity $|u|$ ) In your notation $u=c_1$ The second sound is a temperature waves in gas of particles wich move with the same velocity.
Using argument in the first paragraph is is easy to obtain that $c_2^2=\frac{c_1^2}{3}$
• Yeah, it's a fantastic picture in this answer. But I wonder how can I relate the second sound with the parameter $V$ in this model. In other words, why I should use particles with velocity $c_2$ to calculate the pressure. And why the ratio seems $\frac{1}{3}$ in the superfluid but $3$ in this model. – Simon Nov 20 '14 at 14:40