How to get rid of units in Partition Function

In summary, the conversation discusses the partition function for a classical ideal gas trapped in a one-dimensional harmonic potential. The speaker is trying to make the partition function unitless but is struggling to find a characteristic length to cancel out the units. The other participant suggests using a normalization constant, but it is not necessary in classical statistical physics.
  • #1
"pi"mp
129
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 :)
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
"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.
 

1. How do I convert units in a partition function to get rid of them?

The most common way to convert units in a partition function is to use the Boltzmann constant, which has the units of energy divided by temperature. Multiplying the partition function by the Boltzmann constant will convert the units to energy, which can then be easily converted to the desired units.

2. Can I just ignore the units in a partition function?

No, the units in a partition function are important and cannot be ignored. They represent the physical properties of the system being studied and are necessary for accurate calculations and interpretations.

3. Are there any shortcuts for getting rid of units in a partition function?

There are no shortcuts for getting rid of units in a partition function. It is important to carefully consider the units and use the appropriate conversion factors to ensure accurate results.

4. How do I know which units to use in a partition function?

The choice of units in a partition function depends on the physical properties of the system being studied. It is important to use consistent units throughout all calculations and to ensure that the units are appropriate for the specific system.

5. What are some common mistakes when dealing with units in a partition function?

One common mistake is using inconsistent units throughout calculations, which can lead to inaccurate results. Another mistake is using incorrect conversion factors or forgetting to convert units altogether. It is important to carefully check all units and conversions to avoid these errors.

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