One can find the sound speed for the acoustic waves as follows:

$\frac{\partial p}{\partial t} = -\nabla \cdot \rho \vec v$ (1)

$\rho (\frac{\partial \vec v}{\partial t} + (v \cdot \nabla)\vec v)= - \frac{\gamma p}{\rho} \nabla \rho$ (2)

eq. 1 is the continuity equation and eq. 2 is the the Navier-Stokes equation. Linearizing we can derive the dispersion relation for ion acoustic waves:

$\frac{\omega}{k} = \sqrt{\gamma kT/M}$

Then we can find $v_g = \sqrt{\gamma kT/M}$

Q: In the model we didn't use any information about collisions. In the book I am reading it says that "the waves propagate from one layer to another by collisions between molecules". But in the model we are using to derive there is no collisions. So how can this model give the correct $v_g$?


1 Answer 1


There are two issues here:

1) In using $c_s^2\sim T/m$ you assume an ideal gas equation of state. There are kinetic models based on the Boltzmann equation in which we have a finite elastic collision rate and an ideal equation of state, but these models are somewhat artificial. In real theories based on some underlying many body theory a finite collision rate will require corrections to the equation of state.

2) In using ideal fluid dynamics you assume perfect local equilibrium, which corresponds to zero mean free path and an infinite collision rate. Corrections to perfect equilibrium are encoded in dissipative transport coefficients like shear viscosity. The shear viscosity is roughly $\eta\sim \tau P$, where $\tau$ is the mean collision time. Viscosity has two effects: Damping and dispersion, that is a dependence of the speed of sound on the wave number.

You may think that simultaneously using zero (ideal equation of state) and infinite (ideal fluid dynamics) collision rates is a terrible idea. This is not the case. For typical gases and audible sound frequencies these approximations are quite reasonable.

Postscript (About hydrodynamics and the collision term) The equations of fluid dynamics can be derived by taking moments of the Boltzmann equation. For example, the Euler (Navier-Stokes) equation is obtained by taking moments with $p_i$. We get $$ \partial_0 (\rho v_i) + \nabla_j \Pi_{ij}=0 $$ with $$ \Pi_{ij} = \frac{1}{m}\int d^3p\; p_ip_j \,f_p(x,t) $$ where $f_p(x,t)$ is the distribution function. This result is independent of the collision term.

However, in order to turn this into a fluid dynamic equation I need to write $\Pi_{ij}$ in terms of thermodynamic variables like $P$, $\rho$ and the fluid velocity. For that purpose I need to assume that $f_p\simeq f_p^0 + \delta f_p$, where $f_p^0$ is an equilibrium distribution (paramterized by thermodynamic variables like $T,\mu$ and $v$) and $\delta f_p\ll f_p^0$. Why would this be true? Because in the limit of very short collision time the distribution is driven to a stationary point which makes the collision term vanish: $C[f_p^0]=0$. This equation is indeed solved by the Maxwell-Boltzmann distribution. Using $f_p^0$ I get the equations of ideal fluid dynamics. If $\tau$ is finite, $\delta f_p$ is not zero and I get non-ideal, dissipative, corrections. At leading order, this is the Navier-Stokes equation.

  • $\begingroup$ It is quite confusing. I can derive fluid dynamics equations taking moments of the collisionless Boltzman equation. $\endgroup$
    – Kirill
    Commented Feb 25, 2016 at 23:44
  • $\begingroup$ 1) The hydrodynamic limit is indeed subtle. 2) The hydro equations involve the conserved charges, so the corresponding moments of the collision term vanish. 3) To derive the hydro equations from the moment equations, we also need constituitive equations (we need to express the currents in terms of thermodynamic variables). The constituitive equations do depend on the collision term. $\endgroup$
    – Thomas
    Commented Feb 26, 2016 at 4:19
  • $\begingroup$ So your point is that even if in the hydro-equations there are no collisions we used them to derive the equations? I think to derive hydro-equations we just have to assume the Maxwellian distribution function. $\endgroup$
    – Kirill
    Commented Feb 26, 2016 at 11:37
  • $\begingroup$ @Kirill I added a postscript. $\endgroup$
    – Thomas
    Commented Feb 26, 2016 at 20:52

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