GR8677 #67



Alternate Solutions 
walczyk 20110304 04:08:25  Other people have remarked about the errors in the posted solution, which there are a couple big ones, and while there are posts about solving it by approximating by max entropy which is the fastest method, I still thought an instructive "longway" solution was needed.
I didn't remember the internal energy in terms of partition function at all so I had to look it up. Its .
The nifty approximation I didn't remember is and it really makes this quick work.
First , then churn out .
The problem is asking for average energy per particle so that's . I think you can figure out the rest,but I'll write out the proper way anyway.
The other nifty approximation is , and if we use it we get . Cough, drop the second term, and you're done.   kroner 20091008 09:47:38  If is the difference in energy between two states, then as , the ratio of the probabilities of being in each state goes to 1, so the states become equally likely (as jakevdp and others point out).
Then the average energy of each particle is .  

Comments 
Ronald 20171025 06:21:51  Notice \"A large ”! This is microcanonical ensemble.\r\nSo the probability for each state is same! (equal a prior theory)\r\n\r\nThus \r\n\r\nChoose (C)   varsha 20170117 11:10:51  How did epselon become >> than kT??....please explain,   mhendrix61 20140822 13:33:23  Average is sum over N. 0+1+3/3=4/3   walczyk 20110304 04:08:25  Other people have remarked about the errors in the posted solution, which there are a couple big ones, and while there are posts about solving it by approximating by max entropy which is the fastest method, I still thought an instructive "longway" solution was needed.
I didn't remember the internal energy in terms of partition function at all so I had to look it up. Its .
The nifty approximation I didn't remember is and it really makes this quick work.
First , then churn out .
The problem is asking for average energy per particle so that's . I think you can figure out the rest,but I'll write out the proper way anyway.
The other nifty approximation is , and if we use it we get . Cough, drop the second term, and you're done.
walczyk 20110304 13:09:22 
a sign error messes things up right at the end: so for a final answer you get

jgramm 20140923 22:40:56 
, not . A counterexample to your approximation is

  kroner 20091008 09:47:38  If is the difference in energy between two states, then as , the ratio of the probabilities of being in each state goes to 1, so the states become equally likely (as jakevdp and others point out).
Then the average energy of each particle is .
chemicalsoul 20091103 21:23:51 
This is it ! the average value of a three sided dice. No need for lengthy derivation.

Almno10 20101112 00:09:12 
thats bad ass

  Jeremy 20071113 12:21:20  A few things about the long way (official solution)... (1) We don't want the average energy of the system, we want the average energy of each particle. (2) In the first equation for energy , the should be replaced by . (3) . Also, I think the solution is cleaner using ().
As , . Therefore the limit causes all exponentials to become .
.
FortranMan 20081028 11:10:05 
So the absolute temperature is when T ? Is that what they mean by absolute temperature?

eliasds 20081103 21:48:34 
I think by absolute temperature, they are referring to temperature in degrees kelvin.

physicsisgod 20081105 21:04:47 
No, Jeremy is just taking kT >> 1, which makes 0

  hefeweizen 20061130 12:42:40  i think there is a typo 
epsilon << kT
beta*epsilon << 1
and you can just use:
sum(E__i*exp(beta*epsilon))/Z
  Andresito 20060318 20:31:05  jakevdp, thank you for posting your alternate solution. It seems that that is the shortest and the one ETLS wants you to think of.
Yosun, when you have zeta*(Dzeta/Dtemperature) how come that in the remaining exponential terms (second equation) do not have a factor of 2 in their powers?
I think there should be exp(2 epsilon/kT) and exp(6 epsilon/kY
eliasds 20081103 21:47:43 
I think by absolute temperature, they are referring to temperature in degrees kelvin.

  jakevdp 20051101 11:19:24  Alternately, you can realize that at kT>>e, Entropy is near maximum, thus each particle has roughly equal probablility of being in any of the three states. Thus, average energy is simply (0+e+3e)/3 = (4/3)e
nitin 20061116 11:17:24 
Yosun
This long and elaborate calculation is truly inappropriate and silly when it comes to answering a GRE MCQ. As jakevdp pointed out, since , all 3 possible nondegenerate energy states are equally likely to be occupied by any of the N particles. Therefore, the average energy of each particle would be .

grae313 20071007 18:09:55 
nitin, if you look at all the answers Yosun gives, one could only conclude that he(?) is presenting the rigorous derivation of each solution where applicable, for those who want to see and study the physics behind the solution. I doubt he thinks it is the best solution or the way to approach the ETS exam. This is for studying only.

Richard 20071031 18:03:39 
she...

madfish 20071102 16:24:24 
it. j/p

physicsisgod 20081105 21:07:32 
My long, elaborate, innappropriate and silly calculation shows that nitin is a douche.

Herminso 20090901 12:35:18 
kT>>e means that Entropy is near maximum, and remember that the maximum for the entropy is achieved when the system reach the equilibrium thermodynamic. The Postulate of Equal Priori Probability: When a macroscopic system is in thermodynamic equilibrium, its state is equally likely to be any state satisfying the macroscopic conditions of the system (k. Huang 2ed pag 129).
That justify why each particle has roughly equal probability of being in any of the three states, just as jakevdp did.

Setareh 20111026 07:55:39 
Can anyone tell me why if kT>>e, entropy is near maximum?

 

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