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Why is Entropy a concave function of internal energy?

by Wentu
Tags: entropy, microstates, temperature
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Jan14-13, 07:20 AM
P: 12

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:

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

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Jan14-13, 07:27 AM
Sci Advisor
P: 3,593
Quote Quote by Wentu View Post
If I add another dE, the possible microstates should be N + N(N-1) = N^2 .
Where do you get the first N on the left hand side from?
Jan14-13, 07:49 AM
P: 12
The first N is for a single particle having 2*dE energy and all other particles ground energy

Jan14-13, 08:08 AM
Sci Advisor
P: 3,593
Why is Entropy a concave function of internal energy?

Ah, ok. Shouldn't the second term should rather read N(N-1)/2?
Jan14-13, 08:14 AM
P: 12
You are right... I was considering distinguishble particles but this isn't enough, so yes, the term should be divided by 2. I wonder if this is enough to change the behaviour from linear to less-than-linear... I think the number of microstates still increases as a power with the increasing of E... but again, i could be all wrong


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