# Thermal wavelength

#### Quasi Particle

Hiya!

Somewhere in thermodynamics, the thermal wavelength
$$\lambda = \frac {h}{\sqrt{2 \pi m k T}}$$
appears, apparently out of nothing (well, out of the canonical state integral).

Does anyone know what the physical meaning of this wavelength is?

Also, what is the physical meaning of the fugacity $$z = e^{\beta \mu}$$ ?

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#### Claude Bile

For those of us who have never encountered this equation, could you specify exactly what the symbols in the equations indicate.

Claude.

#### Quasi Particle

Sorry!!
there is

h... Planck constant, k... Boltzmann constant, T... Temperature
m... mass of a particle (of the substance)

beta = 1/(kT)
mu ... chemical potential

M

#### MalleusScientiarum

I would guess that the thermal wavelength would be a length scale defining your thermodynamic system, as in you probably can't look at a system thermodynamically on a smaller length scale.

#### Quasi Particle

I wonder where the waves come in, in the first place. If it were a length scale defining the resulotion why would you talk about a wavelength?

1. it might be the wavelength of heat radiation of this gas at a given temperature - but:
The number of particles is related to the thermal wavelength by
N = zV/lambda^3
where z is the fugacity and V is the volume. I don't understand what fugacity is (see my first post), but it doesn't look like anything having to do with heat radiation.

2. it might be the wavelength of the matter waves of the particles - but the calculations are all classical. But this interpretation comes close to the length scale interpretation.

Does anyone have an idea about it?

#### Quasi Particle

great, that helps. cheers

#### Galileo

Homework Helper
My statistical mech. book's got some info about it. If you take the quantum mechanical ideal gas where the wavefunctions are subject to perdodic boundary conditions (period L) the single-particle eigenstates are plane-waves of momentum $(p_x,p_y,p_z)=\frac{h}{L}(n_x,n_y,n_z)$ where $n_x,n_y,n_z$ are integers the average occupation of one of the eigenstates is

$$\bar N_p = \frac{1}{\exp(\alpha+p^2/2mkT) \pm 1}$$
where the + is taken for fermions and the minus for bosons. The thermal wavelength comes into play in the paragraph of the classical limit.

If $e^\alpha gg 1$ then the $\pm 1$ in the above expression is neglible, giving:

$$\bar N_p \approx e^{-\alpha}e^{-p^2/2mkT}$$
which is the classical Maxwell distribution.

One way to interpret $e^\alpha \gg 1$ is to note that it implies that the average occupation of every momentum state is much less than one. In order to express $e^\alpha$ in terms of the macroscopic parameters of the system, let's assume that it is valid and evaluate the normalization sum for the qm-distribution.
$$\sum_{\vec p}\bar N_p = \sum_{\vec p} \approx e^{-\alpha}e^{-p^2/2mkT}=N$$
where N is the # of particles.The sum is taken over all momentum eigenstate values. The density of momentum e.v.'s in momentum space is $(L/h)^3=V/h^3$, where V is the volume of the system, the sum over the discrete value of p may be converted to an integral over a continuous vector variable.

$$\frac{Ve^{-\alpha}}{h^3}\int \exp(-p^2/2mkT) d^3\vec p = N$$
This is a standard Guassian integral (the product of three actually, one for each component):
$$\int \exp(-p^2/2mkT) d^3\vec p=(2\pi mkT)^{3/2}$$
Now the book defines a \emph{thermal deBroglie wavelength} $\lambda$ by:
$$\lambda \equiv \frac{h}{\sqrt{2\pi mkT}}$$
then the normalization gives:
$$e^{-\alpha}=(N/V)\lambda^3$$
Therefore, $\e^{-\alpha}$ is equal to the average number of particles in volume $\lambda^3$. If that number is much less than one, the classical approximation is justified. $\lambda$ is the deBroglie wavelength of a particle with energy $\pi kT$, which is about twice the average energy of the particles in a gas at temperature T.

Well, I hope that was at least a little bit enlightening. I guess the reason for the definition should be gotten from:

$$\lambda = \frac{h}{p}$$
where
$$p=\int \exp(-p^2/2mkT) d p$$

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#### Quasi Particle

I found an explanation for the fugacity $$z = e^{\beta \mu} = e^\alpha$$
It says that it replaces the pressure in the state equation of an ideal gas to make it the state equation of a real gas. It is the tendency of a fluid to expand or escape isothermally.

If lambda is the deBroglie wavelength of a particle, how can there be particles IN this volume? Or does it relate the "volume" of the particle to the free space, and if there are very few particles i.e. the spacings are big enough, the interactions are negligible and I can treat the gas classically?
Or am I thinking too classically?

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