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  Violation of Virial theorem as indication to ergodicity breaking

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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

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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 ]

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