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Multiplicity of an einstein solid 
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#1
Mar3012, 02:41 PM

P: 123

Okay, I am not sure if this is the right subforum, but here goes:
An einstein solid is a solid composed of N quantum harmonic oscillators, which can store evenly spaced energy units q. Now suppose we have an einstein solid of N oscillators with q energy units, where q>>N. My book wants to derive an expression for the multiplicity of this solid, but I can't quite follow the approximations leading to the final expression: We have in general that that the multiplicity, Ω, of an einstein solid is given by the binomial: Ω = C(q+N1,q) , where C() is the binomial coefficient. I wont verify this expression, because it is the approximations of it that bother me: In general we have from the definition of the binomial coefficient: Ω = C(q+N1,q) = (q+N1)!/((N1)!*q!) My book approximates this to: Ω = (q+N)!/(q!*N!) with the argument that: "The ratio of N! to (N1)! is only a large factor (N). I dont see what this has to do with the approximation. The multiplicity above gets far bigger with the approximation because it all depends on q. For instance, say q=100 and N=2. Then you get: Ωreal = 101!/100! = 101 Ωapprox = 102!/2! ≈ 50*Ωreal or i general for N=2: Ωapprox = q/2 * Ωapprox So in general how is this approximation valid? Am I plugging the numbers in the wrong way? 


#2
Mar3012, 05:07 PM

PF Gold
P: 963

In the cases we're usually interested in, it's not just the number, q, of quanta that is huge, but the number, N, of oscillators as well. That's why it's permissible not to worry about the '1'.



#3
Mar3012, 08:08 PM

P: 1,005

hmm yes but the approximation assumes that q>>N. So even though N is large wont the factor still be quite large because q is very large?



#4
Mar3112, 03:13 AM

PF Gold
P: 963

Multiplicity of an einstein solid
(1) Where is the assumption made that q>>N?
(2) Omitting a factor of N doesn't matter, as long as the percentage difference made to ln Ω is negligible – which it will be, if N is huge. The use of Stirling's formula, for both N! and q! is also justified by this criterion, provided N and q are both huge. There is no need, as far as I know, for q>>N. 


#5
Mar3112, 06:57 AM

P: 1,005

This is just the first part of the approximation. The next part exactly uses that q>>N to simplify further. But indeed you are right that while q is a VERY LARGE NUMBER N is also just a large number. What difference does it make? Maybe your are actually dividing q/N! so is that why it becomes negligible?
I just don't understand the explanation: "The ratio of N! to (N1)! is only a large factor (N)." 


#6
Mar3112, 11:39 PM

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As Philip pointed out, what you're eventually interested in is ln(Ω), and even this is going to be a large number. Large, in fact, compared with ln(q) and ln(N).
A factor N in Ω turns into an addition of ln(N) on ln(Ω), and the effect of that is small. 


#7
Apr112, 08:02 AM

P: 1,005




#8
Apr112, 09:02 AM

PF Gold
P: 963

Can't quite see what you're driving at here.
(1) Still can't see why you think the ratio q/N is important. You say your book goes on to make this assumption, but it isn't required in order to neglect the 1, nor to apply Stirling's formula. (2) All approximations fail hopelessly if you take N as 2. The approximations start to become plausible (in my opinion) for N greater than 50 (and q greater than 50), though with N and q as large as they usually are in the cases we deal with, the approximations are hardly approximations at all. See attached sheet for more detail. 


#9
Apr112, 12:07 PM

P: 1,005

hmm... I definately must be making some kind of basic mistake, because I can't get the approximation to make sense. So let us try to do it for two large numbers for q and N and you can point out what I do wrong:
We have: Ωreal = (q+N1)!/((N1)!*q!) Ωapprox = (q+N)!/(N!*q!) Let N=10^{23}and q=10^{100}. These are both large numbers with q>>N. Now plug in to the two expressions: Ωreal = (10^{100}+10^{23}1)!/((10^{23}1)!*10^{100}!) Ωapprox = (10^{100}+10^{23})!/((10^{23})!*10^{100}!) = 10^{100}/10^{23} * (10^{100}+10^{23}1)!/((10^{23}1)!*10^{100}!) = 10^{77} * Ωreal How is this factor just a negligible term? And what has this term got to do with the reason ""The ratio of N! to (N1)! is only a large factor (N).", which is how the author of my book defends the approximation..? 


#10
Apr112, 01:29 PM

PF Gold
P: 963

Thank you for explaining the problem so clearly. The answer, I'm afraid is still more or less the same as before, but at least you've enabled me to explain it better...
You might think that [itex]10^{77}[/itex] is a large number. Compared to Ω it is tiny. The easiest way to see this is by taking logs. Taking logs is important anyway, because it turns out that ln Ω is what matters in statistical mechanics. Because your q is so vastly larger than N (I still don't know why you insist on this, but I'll just live with it!) we need to do some preliminary work on your top factorial. Can you see that since q>>N, then [itex](N+q)! = (q^N) q![/itex] to a very good approximation? This means that your approximate Ω becomes [itex]\frac{q^N}{N!}[/itex] Using stirling's formula, the log your approximate Ω is [tex]ln \Omega =N ln q – N ln N + N = 10^{23} (230 – 53 + 1) [/tex] This is very much larger than ln 10^{77}, which is a mere 177. [I almost put an exclamation mark at the end of the last sentence!] 


#11
Apr112, 03:47 PM

P: 1,005

Hmm yes, I see it now :) Thanks...
Let me ask you something else though: You say ln are what matters in statistical mechanics. And indeed entropy etc. are defined as lnΩ etc., which was kind of odd in the beginning and still is. Are these definitions merely made because ln is much easier to work with or is there more to it? 


#12
Apr112, 04:05 PM

PF Gold
P: 963

There's more to it; the ln function arises naturally in considering thermal equilibrium. [I do sometimes catch myself wondering how nature knows about the log function, but this is probably a pretty silly worry.]
Mandl ('Statistical Physics') sets out the framework of statistical mechanics cleanly and clearly, though there are plenty of other books to choose from! 


#13
Apr112, 04:19 PM

P: 123

It arises naturally? I can't see why at this point:
If we say that we define something call entropy to make it easier to work with the multiplicity then so far: S = klnΩ In thermal equilibrium of two systems we want ∂S/∂U to be the same for the highest multiplicity. This allows us to define the temperature: 1/T = ∂S/∂U But all these are definitions, so what comes naturally? I admit the k is kind of weird, I never understood that. Also I have heard people say S = klnΩ is a postulate but I don't see why it can't just be a definition. Either way I guess Ima go check if the library at uni has that book. 


#14
Apr112, 06:05 PM

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#16
Apr212, 01:58 AM

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p(A,B)=p(A)p(BA)
Independence p(A,B)=p(A)p(B) Logs convert multiplication to addition log(pA.pB)=log(pA)+log(pB) http://en.wikipedia.org/wiki/R%C3%A9nyi_entropy http://www.cnel.ufl.edu/courses/EEL6...is_entropy.pdf 


#17
Apr212, 08:06 AM

P: 1,005

But are all the following formulas definition because additivity and small numbers are practical:
S = klnΩ 1/T = ∂S/∂U P = T ∂S/∂V or do they arise naturally? I think your individual answers are kind of talking against each other. Also can someone tell me what the k in front of lnΩ is supposed to mean? 


#18
Apr212, 08:15 AM

Sci Advisor
P: 3,628

Also the second and third formula was known before. The factor k in the first equation is just a conversion factor for statistical entropy to have the same units as thermodynamical entropy. 


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