I don't understand an approximation in an expression in stat. mech.

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  • #1
fluidistic
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Homework Statement


Hello guys, I fail to understand a mathematical approximation I see in a solved exercise.
The guy reached a partition function of ##Z=\sum_{l=0}^\infty (2l+1) \exp \left [ -l(l+1) \frac{\omicron}{T} \right ]## and he wants to analyze the case ##T>> \omicron##.
He states that with the change of variables ##x=l(l+1)\frac{\omicron}{T}##, ##Z\approx \frac{T}{\omicron} \int _0^\infty e^{-x}dx##.
I really don't understand this last step.

Homework Equations


Sum becomes integral. The change of variables.

The Attempt at a Solution


Making the change of variables, I understand that the sum transforms into an integral and I also understand why the limits of the integral are 0 and infinity (because l goes from 0 to infinity and thus x too).
I am unable to perform the change of variables and get rid of l's outside the exponential.
I'd appreciate if someone shed some light. Thanks.
 

Answers and Replies

  • #2
vela
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I think the l's get sucked into what becomes dx since the derivative of l(l+1) is 2l+1.
 
  • #3
RUber
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##Z=\sum_{l=0}^\infty (2l+1) \exp \left [ -l(l+1) \frac{\omicron}{T} \right ]## and he wants to analyze the case ##T>> \omicron##.
He states that with the change of variables ##x=l(l+1)\frac{\omicron}{T}##, ##Z\approx \frac{T}{\omicron} \int _0^\infty e^{-x}dx##.
Notice that once you insert the substitution, ## Z=\sum_{l=0}^\infty (2l+1) \exp \left [ -l(l+1) \frac{\omicron}{T} \right ]##, ## =\sum_{l=0}^\infty (2l+1) \exp [-x] ##.
Then see that ## dx=(2l+1)\frac{\omicron}{T} ##.
So, ## 2l+1=dx\frac{T}{\omicron} ##.
Since ##T>>o ## the integral is a good approximation, since it is the limit as ## l(l+1)o/T \to 0 ##.
Putting this all together, ## Z = \int_{l=0}^\infty \frac{T}{\omicron} \exp [-x] dx ##.
 
  • #4
fluidistic
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Thank you guys. I missed the dx part indeed.
 

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