speed of sound and the potential energy of an ideal gas; Goldstein derivation

Question

I am looking the derivation of the speed of sound in Goldstein's Classical Mechanics (sec. 11-3, pp. 356-358, 1st ed). In order to write down the Lagrangian, he needs the kinetic and potential energies.

He gets the kinetic energy very easily as the sum of the kinetic energies of the individual particles (the sum going over to an integral in the limit). Let $\eta_i, i=1,2,3$ be the components of the displacement vector (each $\eta_i = \eta_i(x,y,z)$ being a function of position). So the kinetic energy density is $${\cal T}=(\mu_0/2) (\dot{\eta}_1^2+\dot{\eta}_2^2+\dot{\eta}_3^2),$$ where $\mu_0$ is the equilibrium mass density.

For the potential energy, he uses a thermodynamic argument, relying on the work done in a PV diagram, and using the equation $PV^\gamma = C$. His ultimate result, after several steps, is

$${\cal V} = -P_0 \nabla\cdot\vec{\eta}+\frac{\gamma P_0}{2}(\nabla\cdot\vec{\eta})^2$$

Here, $P_0$ is the equilibrium pressure, and $\gamma$ is the ratio of specific heats.

He later shows that the term $P_0 \nabla\cdot\vec{\eta}$ has no effect on the equations of motion, and so he drops it. So his final formula for the Lagrangian density is:

$${\cal L} = (1/2)\left(\mu_0\dot{\vec{\eta}}^2 - \gamma P_0(\nabla\cdot\vec{\eta})^2\right)$$

and the Lagrangian of course is the integral of this over all space.

Now in the case of an ideal gas (or better yet, a perfect gas), my understanding is that the internal energy is entirely kinetic. Naively, the statistical model is a bunch of non-interacting point particles racing around, bouncing off the walls of the container. (For simplicity, ignore gravity.)

This seems contradictory. Shouldn't we get the same results from a microscopic and a macroscopic viewpoint?

To put it another way, this suggests that in a gas made up of non-interacting point particles, with no external forces except for the hard-wall forces, sound waves could not propagate (since the Lagrangian density would reduce to the kinetic part). That doesn't seem right.

Answer 1

He gets the kinetic energy very easily as the sum of the kinetic energies of the individual particles (the sum going over to an integral in the limit). Let $\eta_i, i=1,2,3$ be the components of the displacement vector (each \eta_i = $\eta_i(x,y,z)$ being a function of position). So the kinetic energy density is ${\cal T}=(\mu_0/2) (\dot{\eta}_1^2+\dot{\eta}_2^2+\dot{\eta}_3^2)$, where is the equilibrium mass density.

Naively, the statistical model is a bunch of non-interacting point particles racing around, bouncing off the walls of the container. (For simplicity, ignore gravity.)

Just to clarify, it seems you understand it, the "particles" in these two cases are different. In the second they are just molecules, in the first they are macroscopic pieces of gas, small volumes.

As for your question getting from microscopic energy to the macroscopic. I'll show how do they related.

$E_\text{mic} = \sum_{i=1}^N \frac{1}{2} m \boldsymbol c_i^2$

In that approach we'll have to deal with $N$ molecules, that's like finding individual trajectories of all the molecules, which we don't want. There should be also the term to reflect either the interaction of molecules (collisions) either with each other or with the walls, but it doesn't matter right now. To reduce the number of degrees of freedom we introduce one-particle distribution function:

$$f(\boldsymbol r, \boldsymbol c, t)$$

The quantaty $f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol r d\boldsymbol c$ tells us how many molecules are there in the phase space element $d\boldsymbol r d\boldsymbol c$

$$E_\text{mac}(\boldsymbol r, t) = \int \frac{1}{2} m \boldsymbol c^2 f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol c$$

As you remember we want to express macroscopic energy in terms of macroscopic kinetic energy, i.e. we need to introduce macroscopic speed $\boldsymbol v$:

$$\rho(\boldsymbol r, t) = \int m f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol c$$

$$\rho \boldsymbol v(\boldsymbol r, t) = \int m \boldsymbol c f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol c$$

Then

$$ \begin{multline} E_\text{mac}(\boldsymbol r, t) = \underbrace{\int \frac{1}{2} m (\boldsymbol c - \boldsymbol v + \boldsymbol v)^2 f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol c}_{\text{Averaging microscopic kinetic energy}} = \\ \underbrace{\frac{1}{2} \rho \boldsymbol v^2}_{\text{Macroscopic kinetic energy}} + \underbrace{\int \frac{1}{2} m (\boldsymbol c - \boldsymbol v)^2 f(\boldsymbol r, \boldsymbol c, t) \; d\boldsymbol c}_{\text{Macroscopic potential energy}} \end{multline} $$

The part with $\boldsymbol v - \boldsymbol c$ was omitted for good reasons, but going into details would be too much, I've already essentially telling the kinetic theory. For the details you can consult appropriate books.

So, we've extracted the macroscopic kinetic energy, what is left is macroscopic potential energy, pressure if you want.

To put it another way, this suggests that in a gas made up of non-interacting point particles, with no external forces except for the hard-wall forces, sound waves could not propagate (since the Lagrangian density would reduce to the kinetic part). That doesn't seem right.

Ideal gas model does not assume particles to be non-colliding, it says that the potential energy of the molecular interaction can be neglected. Abrupt hard wall potential with delta functions have zero overall energy. You can view gas made of non-interacting particles only for the equilibrium case. And even then you have to consider collisions with the walls. For the non-equilibrium case you have to take into account collisions. If they were not interacting, indeed, there would be no sound.