Why is Entropy a concave function of internal energy?by Wentu Tags: entropy, microstates, temperature 

#1
Jan1413, 07:20 AM

P: 10

Hello
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/ckfind...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(N1) = 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 



#2
Jan1413, 07:27 AM

Sci Advisor
P: 3,376





#3
Jan1413, 07:49 AM

P: 10

The first N is for a single particle having 2*dE energy and all other particles ground energy




#4
Jan1413, 08:08 AM

Sci Advisor
P: 3,376

Why is Entropy a concave function of internal energy?
Ah, ok. Shouldn't the second term should rather read N(N1)/2?




#5
Jan1413, 08:14 AM

P: 10

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 lessthanlinear... I think the number of microstates still increases as a power with the increasing of E... but again, i could be all wrong
W. 


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