How to get rid of units in Partition Function

Click For Summary

Discussion Overview

The discussion centers around the formulation of the partition function for a classical ideal gas in a one-dimensional harmonic potential, specifically addressing the issue of dimensionality and the desire for a unitless expression. Participants explore the implications of normalization constants and the role of units in statistical mechanics.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the partition function as requiring two Gaussian integrations and notes that the resulting expression has units of (length)2 divided by (time).
  • Another participant suggests that the normalization constant can be chosen arbitrarily, specifically mentioning the use of Planck's unit of action in quantum systems.
  • A different participant questions whether they can simply posit a normalization constant with the correct units for the classical case and carry it through the computation.
  • It is noted that in certain quantities like energy and enthalpy, the normalization constant does not affect the results, while in others like free energy or entropy, it only contributes as an additive constant.
  • A later reply argues that the partition function does not necessarily need to be unitless, stating that its natural unit is a power of the unit of action, and that making it unitless is possible but not essential in classical statistical physics.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of a unitless partition function, with some suggesting it should be unitless while others argue that it does not need to be. The discussion remains unresolved regarding the best approach to handling units in this context.

Contextual Notes

There are unresolved assumptions regarding the choice of normalization constants and their implications for the dimensionality of the partition function. The discussion reflects varying interpretations of the role of units in classical statistical mechanics.

"pi"mp
Messages
129
Reaction score
1
Hi guys,
I'm studying a classical ideal gas trapped in a one-dimensional harmonic potential and I first want to write out the partition function for a single particle. This, I believe, requires two Gaussian integrations, like so:

[tex]Z=\int_{-\infty}^{\infty} d\dot{x} \int_{-\infty}^{\infty}dx \,\, e^{-\beta E(\dot{x},x)}[/tex]

However, we should like the partition function to be unitless. The above expression has units of (length)^2 divided by (time), as best as I can tell. Now, I know we want to divide by constant parameters of the problem to make it dimensionless. However, there is no characteristic length in this problem! The only constants we have are:

[tex]\omega = \sqrt{\frac{k}{m}}[/tex]

and I can't figure out how these can be combined to cancel out the (length)^2 units in Z. How does one figure this out. Thanks :)
 
Science news on Phys.org
The length scale is in deed quite arbitrary. That's why we chose the normalisation constant to be m/h with h being Plancks unit of action. In classical mechanics, this is nothing but an arbitrary constant.
 
Right, I knew I could use \hbar for quantum systems. So you're saying for classical ideal gas, I just posit some normalization constant A, claim it has the correct units, and carry it through the entire computation?
 
In quantities like energy, enthalpy, etc. the constant doesn't enter at all. In others like free energy or entropy, it enters only as an additive constant and has little relevance as we mostly measure entropy differences.
 
"pi"mp said:
However, we should like the partition function to be unitless...and I can't figure out how these can be combined to cancel out the (length)^2 units in Z. How does one figure this out.

The partition function does not have to be unitless. Its natural unit is a power of unit of action, since the integrations are over conjugated pairs ##q,p##. If you like, you can make it unitless by using arbitrary constant with appropriate dimensions, but there is little reason to do that in classical statistical physics.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K