# Steady state average of physical quantities

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 $26$).

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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 '14 at 0:43
what do you mean with "reading assignment"? @dmckee – dapias May 21 '14 at 0:49
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 '14 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 '14 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 '14 at 3:08

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