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As said by Peter Atkins,

Though two bodies are in thermal equilibrium,energy always flows in both direction, but the net energy is zero.

Now, how does it take place? I know the macrostate remains the same. But how does the transfer effect the microstates?? Please help.

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  • $\begingroup$ same argument as previous questions: stuff at equilibrium fluctuates at the microscopical level, so some components might tend to drift in some direction, causing a local rise of temperature which is then washed away by thermal gradients. on average though this fluctuation cannot be seen, so the process of averaging smears them out. this then defines equilibrium $\endgroup$
    – Phoenix87
    Commented Dec 30, 2014 at 18:02
  • $\begingroup$ Peter Atkins needs to get himself a basics physics textbook and start reading. Energy is the ability of a system to perform work. A system in thermodynamic equilibrium can not perform work. The reason why he thinks that the ideal gas he is thinking of can perform work microscopically is because he has inserted a classical microscopic observer with a temperature of 0K into the setup. That, of course, is not the original system but a greatly distorted one. $\endgroup$
    – CuriousOne
    Commented Dec 30, 2014 at 19:14

1 Answer 1

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Suppose your gas is made of only two "billiard ball" (no internal structure) molecules, each molecule will have an average energy of $3/2kT$. The total energy, the addition of the energy of the two molecules, will be constant $E_{total}=2*3/2kT=3kT$.

But, if the particles collide all the time, the velocity of each particle will change with time. The total energy will be the same, but the velocities before and after the collision will not be the same. The final speeds can be calculated from the initial speeds assuming conservation of momentum and energy. The result is that the system is at different microstates (a microstate is a specific configuration $(v_1, v_2)$ where $v_i$ is the velocity of each particle), but the same macrostate (same total energy, and same average individual energy) across time.

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  • $\begingroup$ Both balls will have the same average energy,right? $\endgroup$
    – user36790
    Commented Dec 31, 2014 at 8:33
  • $\begingroup$ @user36790 yes. $\endgroup$
    – user65081
    Commented Dec 31, 2014 at 15:57

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