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
5,054 questions , 2,207 unanswered
5,347 answers , 22,720 comments
1,470 users with positive rep
818 active unimported users
More ...

  Violation of Virial theorem as indication to ergodicity breaking

+ 2 like - 0 dislike
804 views

Under which conditions the break of virial theorem implies break of ergodicity?

I've seen [this](https://physics.stackexchange.com/questions/253986/why-is-the-virial-theorem-not-a-special-case-of-the-ergodic-theorem-what-is-the) question, but it is very limited and not sufficient. To constrain the discussion I'm interested in 1D hamiltonians of the form (2 degrees of freedom)-

$$H=a p^2+V(x)\\
V_1(x)=bx^2+cx^4 \\
V_2(x)=-A\exp\left(-\frac{x^2}{2\sigma^2}\right)+A$$

That is, confining potentials with and without steady state for infinite time. Notice that partition function diverges for Gaussian trap, thus it has no steady state distribution for position.
The virial predicts essentially (for confined case)-
$$E_k=\frac{1}{2}\langle x\frac{\partial H}{\partial x}\rangle_t = b \langle x^2\rangle_t+2c\langle x^4\rangle_t$$
Here $\langle \cdots \rangle_t$ is time average. Assuming ergodicity, that is (for my purposes) $\langle \cdots\rangle_t = \langle \cdots\rangle$  with $\langle \cdots\rangle$ being ensemble average, we get the prediction that -
$$a\langle p^2\rangle=b\langle x^2\rangle+2c \langle x^4\rangle$$
under which conditions the break of this relation will imply break of ergodicity? Will any violation suffice? Does the fact that Gaussian potential has no steady state means we cannot apply virial to it, or the ergodic prediction is not valid in some sense?

Answers to part of the questions are welcomed too.

asked Jan 22, 2019 in Open problems by anonymous [ no revision ]

1 Answer

+ 2 like - 0 dislike

The virial theorem is based on the fact that if we pick any function $f(x,p)$ on phase-space such that it grows slower than linearly during the evolution, then the time-average of its time-derivative must go to zero

$$ \langle \frac{d f}{d t} \rangle_t = \lim_{T\to\infty} \frac{f(x(T),p(T)) - f(x(0),p(0))}{T} = 0$$

With the use of equations of motion we then generally have virial identities of the type

$$\langle \frac{\partial f}{\partial p} \frac{\partial H}{\partial x} \rangle_t = \langle \frac{\partial f}{\partial x} \frac{\partial H}{\partial p} \rangle_t$$

Your example can be generated by $f = xp/2$. The ergodic theorem tells us that if the system is ergodic, we can always (up to cases of measure zero) switch temporal and phase-space averaging.

Now I assume that you are in fact talking about an ensemble of particles placed in the potential $V_2(x)$, some of which will attain energies above $A$. For such particles the virial theorem is generated by the function $f = \sum_i x^i p_i/2$, where $i$ labels the particles. It is now easy to see that every $i$-contribution to $f$ for which the repective particle has energy above $A$ grows asymptotically linearly with time and the virial theorem breaks.

In other words, we can say that the breaking of the virial theorem implies a breaking of ergodicity only when we can guarantee that

$$ \lim_{T\to\infty} \frac{f(x^i(T),p_i(T))}{T} = 0$$

answered Jan 23, 2019 by Void (1,645 points) [ no revision ]

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:
p$\hbar$ysi$\varnothing$sOverflow
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
...