Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.
W3Counter Web Stats

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public β tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

208 submissions , 166 unreviewed
5,138 questions , 2,258 unanswered
5,414 answers , 23,090 comments
1,470 users with positive rep
823 active unimported users
More ...

  Deriving the thermodynamics of a gas from an action perspective?

+ 1 like - 0 dislike
679 views

Question


I was developing the following ideas for a gas and was wondering if these ideas were already in the literature?

The main ideas are:

  1. Using the action in thermodynamics

  2. Including the collision
  3. Using the mean free path to define temperature

I'm not sure how one would derive the intended formula at the end.

Background

Usually, the Hamiltonian of a gas at thermal equilibrium does not include a collision term which would imply at a collision:

A→←B


B←→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 Aand B before a collision is given by ds2i
where i=A or i=B. Similarly, the action is given by:

Si=micPdsi

Where P is the world line before the collision sμA=sμB where sμi is the four position vector. After the collision, we know the momentum pμ is conserved:

pμA+pμB=pμA+pμB

where pμi denotes the momentum after the collision. Differentiating with respect to ddsi and using
dpμidsi=0 then:

dpμjdsi=dpμAdsi+dpμBdsi

with ji

After the collision the action is given by:

Si=micPdsi

where P is the world line after the collision and dsi is defined by dpidsi=0.

Let us write dsA in terms of dsA. We proceed with:

dpμAdsi+dpμBdsi=dpμjdsi

Using the chain rule:

dpμAdsi+dpμBdsi=dpμjdsjdsjdsi=dpμjdsj(dsjdsi)1

And as the L.H.S above is finite: 

dpμjdsj0(dsjdsi)10

Using L' Hopital Rule:

dpμjdsi=dpμAdsi+dpμBdsi=d2pμjdsj2ddsj(dsjdsi)1

Or:

ddsj(dsjdsi)1dpμjdsi=d2pμjdsj2

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

dsi=dsjd2pμjdsj2(dpμjdsi)1dsj

Hence, the action of particle i is:

Si=micPdsj(d2pμjdsj2(dpμjdsi)1dsj)micPdsi

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

General case and more work

Now, there are an infinite number of collisions for N particles and the is temperature T1. We know the mean free path is a function of temperature. Hence,

Si=micPdsimicPdsi

T=f(Most Probable(Length(Pk))

Where Length(P) is the length of an arbitrary worldline Pk, Most Probable tells one the most probable function in the series of the action Si and f is an unknown function.

asked Mar 18, 2020 in Theoretical Physics by Asaint (90 points) [ revision history ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol in the following word:
pysicsOerflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...