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I may well be all wrong about this but I am trying to understand from a microscopic point of view why Entropy is a concave function of internal energy. I found this in the following .pdf:

http://physics.technion.ac.il/ckfinder/userfiles/files/avron/thermodynamics_potentials.pdf

I started from this wikipedia article and i understand why, if the particles composing the system have a limited number of available energy levels, then S(E) first increases and then decreases.

But saying that S(E) is concave should mean:

- when the temperature is T1, if i give a dE to the system its entropy increases of dS1

- when the tempereture is T2>T1, if I give the same dE to the system, its Entropy increases only of dS2 < dS1

I cannot see this with single particles.

If I have N particles in their lowest energy state there is only one microstate: all the particles are still.

If I give to this system the tiniest possible amount of energy, it will be taken by just one of the particle, so the possible microstates are N.

If I add another dE, the possible microstates should be N + N(N-1) = N^2 ... that is or one particle gets both dE or two different particles get it. Every time I add a dE I should increase the power of N.

Now, if the entropy is somehow proportional to the logarithm of the number of microstates, I should get S proportional to K ln(N^E), that is, something that is proportianl to E... taht is, no concavity

I am sure I am getting all this wrong... could you please help me understand this?

Thank You

Wentu

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# Why is Entropy a concave function of internal energy?

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