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GR8677 #67
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Statistical Mechanics$\Rightarrow$}Partition Function

The problem gives three non-degenerate energies, so one can just directly plug this into the canonical(?) partition function to get,

$Z=\sum_i e^{-\frac{\epsilon_i}{kT}}=1+ e^{-\frac{\epsilon}{kT}}+ e^{-3\frac{\epsilon}{kT}},
$

where $k$ is the Boltzmann constant, $T$ is the (absolute) temperature.

Since,

$U = Nk T^2/Z \frac{\partial Z}{\partial T} = Nk^2 \left(\frac{\epsilon}{kT^2} e^{-\frac{\epsilon}{kT}}+ \frac{3\epsilon}{kT^2} e^{-3\frac{\epsilon}{kT}} \right).
$

For $\epsilon >> kT$, one can expand $e^x \approx 1+x$, and thus,

$\frac{U}{Nk^2} \approx \frac{\epsilon}{kT^2}(1-\epsilon/kT) + \frac{3\epsilon}{kT^2}(1-3\epsilon/kT) \approx 4\epsilon,
$

where one throws out the higher order $\epsilon$ terms.

($Z\approx 1$ in denominator) The average energy of each particle is $U/3=\frac{4}{3}\epsilon$, as in choice (C).

Alternate Solutions
 walczyk2011-03-04 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 "long-way" 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 $U=\frac{NkT^2}{Z}\frac{\partial Z}{\partial T}$. The nifty approximation I didn't remember is $e^x \approx 1 + x$ and it really makes this quick work. First $Z \approx 3 - \frac{4\epsilon}{kT}$, then churn out $U\approx N\frac{4\epsilon}{3\left( 1 - \frac{4\epsilon}{3kT} \right)$. The problem is asking for average energy per particle so that's $\frac{U}{N}$. I think you can figure out the rest,but I'll write out the proper way anyway. The other nifty approximation is $\frac{1}{1 -x} \approx 1-x$, and if we use it we get $\frac{U}{N}\approx \frac{4}{3}\epsilon - \frac{16\epsilon^2}{9kT}$. Cough, drop the second term, and you're done.Reply to this comment kroner2009-10-08 09:47:38 If $\epsilon$ is the difference in energy between two states, then as $\epsilon/kT \to 0$, the ratio of the probabilities of being in each state $e^{\epsilon/kT}$ goes to 1, so the states become equally likely (as jakevdp and others point out). Then the average energy of each particle is $(0 + \epsilon +3\epsilon)/3 = 4\epsilon/3$.Reply to this comment
Ronald
2017-10-25 06:21:51
Notice \"A large $\\bf{isolated system}$â€! This is microcanonical ensemble.\r\nSo the probability for each state is same! (equal a prior theory)\r\n\r\nThus $\\left(0 + \\epsilon +3\\epsilon \\right)/3 = (4/3)\\epsilon$\r\n\r\nChoose (C)
varsha
2017-01-17 11:10:51
How did epselon become >> than kT??....please explain,
mhendrix61
2014-08-22 13:33:23
Average is sum over N. 0+1+3/3=4/3
walczyk
2011-03-04 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 "long-way" 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 $U=\frac{NkT^2}{Z}\frac{\partial Z}{\partial T}$.
The nifty approximation I didn't remember is $e^x \approx 1 + x$ and it really makes this quick work.
First $Z \approx 3 - \frac{4\epsilon}{kT}$, then churn out $U\approx N\frac{4\epsilon}{3\left( 1 - \frac{4\epsilon}{3kT} \right)$.
The problem is asking for average energy per particle so that's $\frac{U}{N}$. I think you can figure out the rest,but I'll write out the proper way anyway.
The other nifty approximation is $\frac{1}{1 -x} \approx 1-x$, and if we use it we get $\frac{U}{N}\approx \frac{4}{3}\epsilon - \frac{16\epsilon^2}{9kT}$. Cough, drop the second term, and you're done.
 walczyk2011-03-04 13:09:22 a sign error messes things up right at the end: $\frac{1}{1-x} \approx 1 + x$ so for a final answer you get $\frac{U}{N} \approx \frac{4}{3}\epsilon + \frac{16\epsilon^2}{9kT}$
 jgramm2014-09-23 22:40:56 $\frac{1}{1-x} = \sum_{i=0}^\infty{x^i}$, not $1-x$. A counter-example to your approximation is $\frac{1}{1-.5} = 2 \neq 1-.5 = .5$
kroner
2009-10-08 09:47:38
If $\epsilon$ is the difference in energy between two states, then as $\epsilon/kT \to 0$, the ratio of the probabilities of being in each state $e^{\epsilon/kT}$ goes to 1, so the states become equally likely (as jakevdp and others point out).

Then the average energy of each particle is $(0 + \epsilon +3\epsilon)/3 = 4\epsilon/3$.
 chemicalsoul2009-11-03 21:23:51 This is it ! the average value of a three sided dice. No need for lengthy derivation.
Jeremy
2007-11-13 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 $U$, the $k^{2}$ should be replaced by $k T^{2}$. (3) $\lim_{t \to \infty} Z=3$. Also, I think the solution is cleaner using $\beta$ ($\beta \equiv \frac{1}{k T}$).

$Z=1+e^{-\beta \epsilon}+e^{-3\beta \epsilon}$

$E=-\frac{1}{Z} \frac{\partial Z}{\partial \beta}=\frac{\epsilon (e^{-\beta \epsilon}+3 e^{-3\beta \epsilon})}{Z}$

As $t \to \infty$, $\beta \to 0$. Therefore the limit causes all exponentials to become $1$.

$E=\epsilon \frac{1+3}{3}=\frac{4}{3} \epsilon$.
 FortranMan2008-10-28 11:10:05 So the absolute temperature is when T $\rightarrow \infty$? Is that what they mean by absolute temperature?
 eliasds2008-11-03 21:48:34 I think by absolute temperature, they are referring to temperature in degrees kelvin.
 physicsisgod2008-11-05 21:04:47 No, Jeremy is just taking kT >> 1, which makes $\beta \rightarrow$ 0
hefeweizen
2006-11-30 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
2006-03-18 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
 eliasds2008-11-03 21:47:43 I think by absolute temperature, they are referring to temperature in degrees kelvin.
jakevdp
2005-11-01 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
 nitin2006-11-16 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 $kT>>\epsilon$, 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 $\frac{0+\epsilon+3\epsilon}{3}=\frac{4}{3}\epsilon$.
 grae3132007-10-07 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.
 Richard2007-10-31 18:03:39 she...
 physicsisgod2008-11-05 21:07:32 My long, elaborate, innappropriate and silly calculation shows that nitin is a douche.
 Herminso2009-09-01 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.
 Setareh2011-10-26 07:55:39 Can anyone tell me why if kT>>e, entropy is near maximum?

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 "long-way" 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 $U=\frac{NkT^2}{Z}\frac{\partial Z}{\partial T}$. The nifty approximation I didn't remember is $e^x \approx 1 + x$ and it really makes this quick work. First $Z \approx 3 - \frac{4\epsilon}{kT}$, then churn out $U\approx N\frac{4\epsilon}{3\left( 1 - \frac{4\epsilon}{3kT} \right)$. The problem is asking for average energy per particle so that's $\frac{U}{N}$. I think you can figure out the rest,but I'll write out the proper way anyway. The other nifty approximation is $\frac{1}{1 -x} \approx 1-x$, and if we use it we get $\frac{U}{N}\approx \frac{4}{3}\epsilon - \frac{16\epsilon^2}{9kT}$. Cough, drop the second term, and you're done.
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