Taking classical limit question (statistical mechanics )

[binbagsss]

1. Homework Statement

Question attached.

I am looking at the second line limit $\beta (h/2\pi) \omega << 1$

2. Homework Equations

above

3. The Attempt at a Solution

Q1)In general in an expansion we neglect terms when we expand about some the variable taking small values of the function because the other terms are so small and so we get a good approximation from e.g first order terms. so in this case i see in the exponential expansion we have gone up to $O(x=\beta (h/2\pi) \omega )$ and the '1s' have cancelled. However, to me this makes sense if we were looking at $\exp^{x}$ but instead we are looking at $e^{-x}$ and when we raise to a negative exponent between $0 < x <1$ this is large, so why do we expand out?

for e.g i the log term in the 2nd term must be tending $\to -\infty$ and I thought this is the reason we don't expand it out? or why do we not expand it out, I know in general you can't expand out $\log (0)$ and it seems like we have a limiting case of this, but..thanks

 

[Dick]

1. Homework Statement

Question attached.

View attachment 222448

I am looking at the second line limit $\beta (h/2\pi) \omega << 1$

2. Homework Equations

above

3. The Attempt at a Solution

Q1)In general in an expansion we neglect terms when we expand about some the variable taking small values of the function because the other terms are so small and so we get a good approximation from e.g first order terms. so in this case i see in the exponential expansion we have gone up to $O(x=\beta (h/2\pi) \omega )$ and the '1s' have cancelled. However, to me this makes sense if we were looking at $\exp^{x}$ but instead we are looking at $e^{-x}$ and when we raise to a negative exponent between $0 < x <1$ this is large, so why do we expand out?

for e.g i the log term in the 2nd term must be tending $\to -\infty$ and I thought this is the reason we don't expand it out? or why do we not expand it out, I know in general you can't expand out $\log (0)$ and it seems like we have a limiting case of this, but..thanks

I don't see what you are worried about. If $|x| << 1$ then $e^x \approx 1+x$. It doesn't matter whether $x$ is positive or negative. And, of course, you can't Taylor expand $\log$ around $0$.

 

[binbagsss]

I don't see what you are worried about. If $|x| << 1$ then $e^x \approx 1+x$. It doesn't matter whether $x$ is positive or negative. And, of course, you can't Taylor expand $\log$ around $0$.

No I don't understand because $e^{-x}$ for $x<<1$ is large, and so by neglecting any term you are neglecting a significant contribution? (whereas for $e^x$ ofc you are not neglecting a significant contribution but smaller terms which makes complete sense)..

 

[mjc123]

No, for x << 1 , e^-x ≈ 1-x
Do you really think e^0.001 and e^-0.001 are very different?