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

[fluidistic]

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.

 

[vela]

I think the l's get sucked into what becomes dx since the derivative of l(l+1) is 2l+1.

 

[RUber]

$Z=\sum_{l=0}^\infty (2l+1) \exp \left [ -l(l+1) \frac{\omicron}{T} \right ]$ 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, $Z = \int_{l=0}^\infty \frac{T}{\omicron} \exp [-x] dx$.

 

[fluidistic]

Thank you guys. I missed the dx part indeed.

 

[paranormal]

I understand your confusion and I will try to explain it in a clear way. First, let's recall the definition of a partition function in statistical mechanics. The partition function Z is defined as the sum over all possible states of a system, each state multiplied by its Boltzmann factor (e^(-E/kT)). In this case, the states are represented by the quantum number l, and the energy is given by E = l(l+1)ω, where ω is a constant related to the energy levels of the system.

Now, let's focus on the change of variables that the person did in the exercise. By substituting x = l(l+1)ω/T, we are essentially changing the variable from l to x. This is a common technique in integrals, where we try to simplify the expression by changing the variable to something more manageable.

So, what happens to the sum when we make this change of variables? Well, let's write it out explicitly:
Z = ∑(2l+1)e^(-l(l+1)ω/T)
= ∑(2l+1)e^(-x)
= ∑(2l+1)dx/dl e^(-x)
= ∑(2l+1)dx/dl e^(-x)dx
= ∑(2l+1)e^(-x)dx
= ∫(2l+1)e^(-x)dx

In the last step, we have replaced the sum with an integral, since we are now integrating with respect to x instead of l. However, we still have the (2l+1) term outside the integral. But, if we look closely, we can see that (2l+1) is just a constant, and it can be taken out of the integral. Therefore, the integral becomes:
Z = (2l+1)∫e^(-x)dx

Now, we can evaluate this integral, since it is a simple exponential integral. The result is:
Z = (2l+1)(-e^(-x)) + C
= -(2l+1)e^(-x) + C

But, we are not interested in the constant C, since it will cancel out when we divide Z by Z. So, we can ignore it and just focus on the exponential term. If we substitute back x = l(l+