Microstates and oscillators help

AI Thread Summary
The discussion focuses on calculating the increase in the number of microstates when an additional quantum is added to a system of N oscillators with Q quanta of energy. The formula for the number of microstates is given, and the goal is to show that this increase is approximately (1 + N/Q) when both N and Q are large. The solution involves simplifying the ratio of microstates W(N,Q+1) to W(N,Q) and applying the binomial expansion to approximate the result. Clarifications are provided regarding the cancellation of factorials and the reasoning behind the approximations used in the calculations. The participants express gratitude for the explanations that clarify their understanding of the problem.
karnten07
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Homework Statement


The number of microstates of a system of N oscillators containing Q quanta of energy homework is given by

W(N,Q) = (N+Q-1)!/[(N-1)!Q!]

Show that when one further quantum is added to the system the number of microstates increases by a factor of approximately (1+N/Q), provided that N,Q>>1.


Homework Equations





The Attempt at a Solution



So W(N,Q+1) = (N+Q)!/[(N-1)!(Q+1)!]

My problem class leader showed us how to do the question but I am unsure of how he did it, so this is what he did:

W(N,Q+1)/W(N,Q)= [(N+Q)!(N-1)!Q!]/[(N-1)!(Q+1)!(N+Q-1)!]

= [(N+Q)!Q!]/[(Q+1)!(N+Q-1)!] by cancelling

= N+Q/Q+1 = 1+N/Q

I think he got to the last step by approxiamtion since N,Q>>1 but i don't see how that works. Also i don't know why there aren't factorial signs there, whether he or i forgot to write them in or whether they disappear for some reason.

Any explanations would be so helpful, thanks
 
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karnten07 said:

Homework Statement


The number of microstates of a system of N oscillators containing Q quanta of energy homework is given by

W(N,Q) = (N+Q-1)!/[(N-1)!Q!]

Show that when one further quantum is added to the system the number of microstates increases by a factor of approximately (1+N/Q), provided that N,Q>>1.


Homework Equations





The Attempt at a Solution



So W(N,Q+1) = (N+Q)!/[(N-1)!(Q+1)!]

My problem class leader showed us how to do the question but I am unsure of how he did it, so this is what he did:

W(N,Q+1)/W(N,Q)= [(N+Q)!(N-1)!Q!]/[(N-1)!(Q+1)!(N+Q-1)!]

= [(N+Q)!Q!]/[(Q+1)!(N+Q-1)!] by cancelling

= N+Q/Q+1 = 1+N/Q

I think he got to the last step by approxiamtion since N,Q>>1 but i don't see how that works. Also i don't know why there aren't factorial signs there, whether he or i forgot to write them in or whether they disappear for some reason.

Any explanations would be so helpful, thanks


First, notice that (Q+1)!/Q! = Q+1 Do you see why?

In general, if you have (X+1)!/ X! , this is equal to X+1 (where X can be anything)

Now, note that \frac{N+Q}{Q+1} = \frac{1 + N/Q}{1+1/Q} \approx 1+N/Q - 1/Q - N/Q^2 \ldots where I have used the binomial expansion and have neglected the corrections of order 1/Q^2, N/Q^2 and higher .
 
Last edited:
kdv said:
First, notice that (Q+1)!/Q! = Q+! Do you see why?

In general, if you have (X+1)!/ X! , this is equal to X (where X can be anything)

Now, note that \frac{N+Q}{Q+1} = \frac{1 + N/Q}{1+1/Q} \approx 1+N/Q - 1/Q - N/Q^2 \ldots where I have used the binomial expansion and have neglected the corrections of order 1/Q^2, N/Q^2 and higher .

Yes i see it now, thankyou so much.
 
karnten07 said:
Yes i see it now, thankyou so much.

I made a few typos in my post (now corrected)

I meant to say (Q+1)!/Q! = Q+1

and (X+1)!/X! = X+1

Sorry for the typos.

And you are welcome.
 
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