# How to get rid of units in Partition Function

1. Jul 29, 2014

### "pi"mp

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:

$$Z=\int_{-\infty}^{\infty} d\dot{x} \int_{-\infty}^{\infty}dx \,\, e^{-\beta E(\dot{x},x)}$$

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:

$$\omega = \sqrt{\frac{k}{m}}$$

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 :)

2. Jul 29, 2014

### DrDu

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. Jul 29, 2014

### "pi"mp

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. Jul 29, 2014

### DrDu

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. Jul 30, 2014

### Jano L.

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.