# $2$'nd Law of thermodynamics in a relativistic gas?

## Question

I'm trying to understand in the context of a gas if a particular boundary condition can be used be to be the source of the $$2$$'nd law of thermodynamics: "Given an isolated system the entropy will increase or remain the same."

Is the attempt correct?

## Background

Entropy is a statement of the likelihood of a configuration. If that is reflected in the action as well I think I have a chance?

I try to make use of the fact it is only in thermal equilibrium (maximum entropy) that collisions can be neglected. Usually, the Hamiltonian of a gas at thermal equilibrium does not include a collision term which would imply at a collision:

$$\underbrace{\cdot}_{A} \rightarrow \leftarrow \underbrace{\cdot}_{B}$$

$$\underbrace{\cdot}_{B} \leftarrow \rightarrow \underbrace{\cdot}_{A}$$

They actually go through each other.

## My attempt

Consider a relativistic gas (point) particles with a $$2$$ particles $$A$$ and $$B$$ in a box and the only collide once.

The line element of the $$A$$and $$B$$ before a collision is given by $$ds_i^2$$ where $$i=A$$ or $$i=B$$. Similarly, the action is given by:

$$S_i = - m_ic \int_P d s_i$$

Where $$P$$ is the world line before the collision $$s_A^\mu = s_B^\mu$$ where $$s_i^\mu$$ is the four position vector. After the collision, we know the momentum $$p^\mu$$ is conserved:

$$p_A^\mu + p_B^\mu = {p'}_A^\mu + {p'}_B^\mu$$

where $${p'}_i^\mu$$ denotes the momentum after the collision. Differentiating with respect to $$\frac{d}{ds_i}$$ and using $$\frac{d p^\mu_i}{ds_i} =0$$ then:

$$\frac{d p_j^\mu}{ds_i} = \frac{d {p'}_A^\mu}{ds_i} + \frac{d {p'}_B^\mu}{ds_i}$$

with $$j \neq i$$

After the collision the action is given by:

$$S'_i = - m_i c \int_{P'} d s'_i$$

where $$P'$$ is the world line after the collision and $$d s'_i$$ is defined by $$\frac{dp'_i}{ds'_i} = 0$$.

Let us write $$ds_A$$ in terms of $$ds'_A$$. We proceed with:

$$\frac{d {p}_A^\mu}{ds_i} + \frac{d {p}_B^\mu}{ds_i} = \frac{d {p'}_j^\mu}{ds_i}$$

Using the chain rule:

$$\frac{d {p}_A^\mu}{ds_i} + \frac{d {p}_B^\mu}{ds_i} = \frac{d {p'}_j^\mu}{ds'_j} \frac{d s'_j}{ds_i} = \frac{\frac{d {p'}_j^\mu}{ds'_j}}{(\frac{d s'_j}{ds_i})^{-1}}$$

And as the L.H.S above is finite:

$$\frac{d {p'}_j^\mu}{ds'_j} \to 0 \implies (\frac{d s'_j}{ds_i})^{-1} \to 0$$

Using L' Hopital Rule:

$$\frac{d {p'}_j^\mu}{ds_i} =\frac{d {p}_A^\mu}{ds_i} + \frac{d {p}_B^\mu}{ds_i} = \frac{\frac{d^2 {p'}_j^\mu}{{ds'_j}^2}}{\frac{d}{ds'_j}(\frac{d s'_j}{ds_i})^{-1}}$$

Or:

$$\frac{d}{ds'_j}(\frac{d s'_j}{ds_i})^{-1} \frac{d {p'}_j^\mu}{ds_i} = \frac{d^2 {p'}_j^\mu}{{ds'_j}^2}$$

The above should be solvable as we have $$2$$ boundary conditions (conservation of momentum and $$s_i = s_j$$). Hence:

$${ds_i} = {d s'_j} \int \frac{d^2 {p'}_j^\mu}{{ds'_j}^2} (\frac{d {p'}_j^\mu}{ds_i})^{-1} {ds'_j}$$

Hence, the action of particle $$i$$ is:

$$S_i = - m_ic \int_P {d s'_j} \Big ( \int \frac{d^2 {p'}_j^\mu}{{ds'_j}^2} (\frac{d {p'}_j^\mu}{ds_i})^{-1} {ds'_j} \Big) - m_i c \int_{P'} d s'_i$$

One can take sum over $$i$$ to find the net action.

## Heuristically?

Since, the action depends on where the gas collides. It is possible to use this as a source for the second law in the limit of an infinite collisions?

• This question would fit better physics.stackexchange.com. But I don't see how the action could be used for the statistical quantity entropy. Jan 5, 2020 at 22:41