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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?

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    $\begingroup$ This question would fit better physics.stackexchange.com. But I don't see how the action could be used for the statistical quantity entropy. $\endgroup$
    – md2perpe
    Jan 5, 2020 at 22:41

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