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.

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 \(\beta\) tools

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

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,786 comments
1,470 users with positive rep
820 active unimported users
More ...

  Why is it impossible to have a constant in the action via classical mechanics?

+ 1 like - 0 dislike
596 views

Question
---
So I don't  understand why there seems to be a general consensus that it is impossible to have a constant in the action via classical mechanics? Is there some no-go theorem I'm missing?

My Attempt
---
Consider the Lagrangian $\mathcal{L_M}$ for a gas. Generally in the gas ideal model only the kinetic energies are considered but let us think of the potential energy of a collision and not assume the collision is an event in spacetime but has finite duration. The turning point can be thought as a consequence of regularisation. 

The potential experienced by $2$ objects when they collide is given by:
$$ V_{exp} = \frac{1}{2} \mu v_{rel}^2 $$

where $V_{exp}$ is the potential experienced, $\mu$ is the reduced mass and $v_{rel}$ is the relative velocity. The new action density when collisions are included is given by:

$$S(p) \to S(p) + S_c$$

where $p$ is the momentum, $S(p)$ is the action when only kinetic energies are considered and $S_c$ is the action contributed by the potential energy. Now, if I assume a short ranged interaction:

$$ S_c = \int L_c dt \approx  V_{exp} \tau $$
where $\tau$ is the collision duration. 
Now for a gas, the number of collisions per $4$ volume is given by:

$$ d N_c = \frac{1}{2}\rho^2 A |\langle v_{rel} \rangle | dt dx dy dz$$
where  the $\rho$ is the density, $A$ is the area of the molecule and $dt$, $dx$, $dy$, $dz$ are infinitesimals. 
Hence, the action density $\tilde S_c$ for the entire gas is given by:

$$ \tilde S_c  \approx \frac{1}{4} \rho^2 A |\langle v_{rel} \rangle |   \mu\langle  v_{rel}^2  \rangle \langle \tau \rangle  $$

Now while $\langle \tau \rangle$ should depend on the short range potential I see no reason it couldn't possibly have the term (after using the equation of state):

$$ \langle \tau \rangle = \frac{C}{\rho^2 |\langle v_{rel} \rangle | \langle v^{2}_{rel} \rangle  } + \dots $$

where $C$ is a constant?

asked Jul 16, 2022 in Theoretical Physics by Asaint (90 points) [ revision history ]
edited Jul 16, 2022 by Asaint

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):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
$\varnothing\hbar$ysicsOverflow
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
...