Statistical Mechanics-Limit in canonical ensemble

Click For Summary

Homework Help Overview

The discussion revolves around the application of approximations in statistical mechanics, specifically within the context of the canonical ensemble. Participants are examining the conditions under which the exponential function can be expanded in terms of a Taylor series, particularly in relation to low temperature scenarios.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the criteria for expanding the exponential function, questioning the conditions that define when the argument is considered "small." There is a focus on the implications of low temperature on the terms of the partition function, particularly regarding the significance of the ground state and first excited state contributions.

Discussion Status

Some participants have provided insights into the behavior of the exponential terms as temperature approaches zero, noting that for higher energy states, the exponential factors diminish rapidly. This has led to a discussion on which terms can be retained in the partition function for accurate approximations. There is an ongoing exploration of the physical constants involved and their relevance to the problem.

Contextual Notes

Participants are considering the implications of physical constants such as the moment of inertia, Planck's constant, and Boltzmann's constant in determining the smallness of the argument in the exponential function. There is an acknowledgment of the need for further information regarding these constants to fully assess the validity of the approximations being discussed.

binbagsss
Messages
1,291
Reaction score
12

Homework Statement



question attached.

rotsm.jpg
My question is just about the size of the limit, how do you know whether to expand out the exponential or not (parts 2) and 4))

Homework Equations



for small ##x## we can expand out ##e^{x} ## via taylor series.

The Attempt at a Solution



Solutions given here for parts 2) and 4) respectively. (not full sol but only as relevant as needed for the question):
rot2sol.jpg
rot4sol.jpg


For parts 2) and 4) below I would have used the same approximation ##1+3e^{x}##, but would have expanded out the exponential as ##1+x##.

I'm just wondering how you know whether or not to expand this out or not, going by 'low ##T## many thanks.
 

Attachments

  • rotsm.jpg
    rotsm.jpg
    44.2 KB · Views: 947
  • rot2sol.jpg
    rot2sol.jpg
    17 KB · Views: 805
  • rot4sol.jpg
    rot4sol.jpg
    26.8 KB · Views: 878
Physics news on Phys.org
binbagsss said:
For parts 2) and 4) below I would have used the same approximation ##1+3e^{x}##, but would have expanded out the exponential as ##1+x##.

I'm just wondering how you know whether or not to expand this out or not, going by 'low ##T## many thanks.
The approximation ##e^x \approx 1 + x## is only valid if ##x## is "small" (i.e., ##|x| \ll 1##).

For low ##T##, is the argument of the exponential small in any of the terms of ##Z## (other than the ##j = 0## term)?
 
TSny said:
The approximation ##e^x \approx 1 + x## is only valid if ##x## is "small" (i.e., ##|x| \ll 1##).

For low ##T##, is the argument of the exponential small in any of the terms of ##Z## (other than the ##j = 0## term)?

would you not have to know an approx magnitude of the moment of inertia to know this? or should we be looking at planks constant, and Boltzmann constant, and physically feasible magnitudes of the moment of inertia, i hae no idea what these would be , and know they arent comparable so x is not small?
 
The exponential factors in ##Z## have the form ##e^{-\beta E_j}##, where ##\beta E_j = \frac{\hbar^2 j(j+1)}{2 I k T}##.
As ##T## →0, ##\beta E_j## gets large (for any ##j > 0##). So, ##e^{-\beta E_j}## → 0 for any for any ##j > 0##. The approximation ##e^x \approx 1+x## doesn't apply here. The larger the value of ##j##, the faster ##e^{-\beta E_j}## → 0. So, as an approximation, you can keep just the ##j = 0## and ##j = 1## terms in ##Z##.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 309 ·
11
Replies
309
Views
17K
Replies
2
Views
4K