Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Consider the following Hamiltonian:

$$ H = \sum_n \left[\dfrac{p_n^2}{2m_n} + U(x_l) + V(x_{l+1} - x_l) \right], $$

that corresponds to a 1-D system of particles with nearest-neighbor interactions (contained in the expression for $V$). $U(x_l)$ corresponds to an external potential.

Suppose that the particles in the opposite extremes are connected to heat reservoirs at unequal temperatures and that a steady state is reached.

Let's denote the steady state average of any physical quantity $A$ by $ <A> $.

Can you figure out why

$$ \left< \dfrac{dV(x_{l+1} - x_{l})}{dt} \right> = 0 \; ? $$

In words, it means that the average of the rate of change of the potential energy for all pair of particles is equal to zero. How can it be deduced? I supposed that due to the time invariance of the steady state this is an obvious consequence but not sure about the correct reasoning.

For reference, you can check this (page 6, before equation $12$) or that (page 12, before equation $26$).

share|improve this question
    
Reading assignment before the next meeting of class: "Virial Theorem". You'll also need to examine the modes by which the system does or does not lose energy to its surroundings. –  dmckee May 21 at 0:43
    
what do you mean with "reading assignment"? @dmckee –  dapias May 21 at 0:49
1  
It's a playfully sarcastic way of introducing a hint. I've been teaching in recent years and sometimes feel that asker would be better off hunting something up for themselves. Probably if you wait someone will be along to give you some more complete help. –  dmckee May 21 at 1:13
    
oh I see, I was thinking about virial theorem since you mentioned it but not sure that it leads to the answer. Thanks for answering @dmckee –  dapias May 21 at 1:30
    
By the way, you may not be applying the Virial theorem directly but making use of the same set of arguments that led to it in the first place. I haven't worked the exercises so I'm not sure exactly what will be necessary to romp home. –  dmckee May 21 at 3:08

1 Answer 1

Usually averages of any total time derivatives vanish. Easiest way to see this is to use the fact that steady state time averages and ensemble averages should be the same if the system is ergodic....therefore

$$ <dA/dt>= \lim_{T \to \infty} (1/T) \int_0^T dt dA/dt = \lim_{T \to \infty} (1/T) [A(T)-A(0)] =0 $$ unless $A(t)$ is unbounded.

share|improve this answer
    
oh, thanks for replying! It makes sense under the ergodic hypothesis. I was wondering about what the steady state average exactly means, not sure about the ensemble average in the non-equilibrium state (is it locally microcanonical due to the local equilibrium assumption?, how can we calculate that?). –  dapias May 22 at 14:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.