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SM:Parastatistics, grand canonical ensemble function

  • Thread starter binbagsss
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  • #1
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Homework Statement


Hi
I am looking at the question attached.

parastatistics.jpg


Parts c and d, see below

Homework Equations




The Attempt at a Solution




First of all showing that ##<N> ## and ##<n_r>## agree

I have ##Z=\Pi_r z_r ##, where ##Z## here denotes the grand canonical ensemble.
So therefore we have ## \log(Z) = \sum\limits_r log (z) ## and since ##<N>=\frac{1}{\beta}\frac{\partial}{\partial \mu} \log Z ## the calculation seems pretty trivial unless there are some subtleties I am missing. The only subtlety I can think of however is the requirement of convergence such that the derivative can be taken inside the summation and then the derivative of the summation= sum of the derivative of each term, please see below.

So I have:
##z_r= \sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } ##
so ## <n_r> =\frac{1}{\beta}\frac{\partial}{\partial \mu} \log ( \sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } ) = \frac{1}{\beta} \frac{\sum\limits_{n_r=0}^{n_r=p} n_r \beta x_r^{n_r }}{\sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } } ## (1)

And

##<N>= \frac{1}{\beta}\frac{\partial}{\partial \mu} \sum\limits_r \log ( \sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } ) = \frac{1}{\beta} \sum\limits_r \frac{\partial}{\partial \mu} \log ( \sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } ) = \frac{1}{\beta} \sum\limits_r \frac{\sum\limits_{n_r=0}^{n_r=p} n_r \beta x_r^{n_r }}{\sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } } ## (2)

Where the 2nd equality has assumed convergence over the summation ## \sum\limits_r ##.

QUESTION 1- 2ND EQUALITY USING CONVERGENCE
So the question asks to use (1) and ## <N>= \sum\limits_r <n_r> ## to show that (2) is consistent with the latter. It is clear that the two agree. HOWEVER, I am unsure how to justify convergence over the summation of ##r##. Here ##r## is a single particle state, as far as I'm aware, (although most likely physically unlikely) this can be infinite, an ##n_r ## is the number of particles occupying a single ## r ## state. And so the summation over ##r## is infinite. If I ##E_r > \mu ## then this is an infinite summation of the logarithm of a summation which I know converges...but I don't think this really helps me out? (since i have ## \sum\limits_r \log (\sum\limits_{n_r} x_r^{n_r }) ## where ##x_r = \exp^{-\beta(E_r-\mu)}##).

QUESTION 2- UNABLE TO USE GENERAL EXPRESSION TO RECOVER BOSON DISTRUBUTION
so for fermions ##p=1## for bosons ##p=\infty##.
Looking at (1), the numerator multiplies ##n_r ## (dropped down from the exponentiial derivative) and so clearly when ##n_r=\infty ## this expression is dodgy/invalid- I am unsure what is wrong, however, with my above working.
The best I can seem to do is to recover the boson distribution from the expression at an earlier step, more generalised, which was:

## \frac{1}{\beta} \frac{\partial}{\partial \mu}( \log (\frac{1}{1-x_r})) ## it its easy to show that I attain ## \frac{1}{x_r^{-1}-1} ## as expected. (where I have use infinite summation formula for this and the assumption ## E_r > \mu ##
However I imagine the question wants you to be verifying your final, simplified expression. I am unsure what is wrong with (1)...


the fermion distribution recovery is ok:
## <n_r> = \frac{1}{\beta} \frac{\sum\limits_{n_r=0}^{n_r=1} n_r \beta x_r^{n_r }}{\sum\limits_{n_r=0}^{n_r=1} x_r^{n_r } } ##
this gives:
## \frac{1}{\beta}( 0 + \frac{\beta x_r}{1+x_r} )= \frac{1}{x_r^{-1}+1}## as expected.

Many thanks in advance
 

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Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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I can't say for sure, but I suspect that for part (b) you are expected to evaluate the sum in ##z_r= \sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } ## and get a fairly simple result for ##z_r## in terms of ##x_r## and ##p##. Then you can use this in later parts of the problem.
 
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