Quantum gases. The ideal Fermi gas

[Petar Mali]

Relations for an ideal Fermi gas:

$$\frac{P}{k_BT}=\frac{1}{\lambda_D^3}f_{5/2}(\lambda)$$

$$\frac{1}{\upsilon}=\frac{1}{\lambda_D^3}f_{3/2}(\lambda)$$

But in some book books I find


$$\frac{P}{k_BT}=\frac{g}{\lambda_D^3}f_{5/2}(\lambda)$$

$$\frac{1}{\upsilon}=\frac{g}{\lambda_D^3}f_{3/2}(\lambda)$$

where $$g$$ is degeneration of spin I
guess.
$$g=2s+1$$

Can you tell me something about this

$$\lambda_D=\sqrt{\frac{2 \pi\hbar^2}{mk_BT}}$$

$$f_k(\lambda)=\sum^{\infty}_{n=1}(-1)^{n-1}\frac{\lambda^n}{n^k}$$

$$\lambda=e^{\frac{\mu}{\theta}}$$ - fugacity

 

[azntoon]

The relations given in your book are for a Fermi gas with spin degeneracy g, where g is the number of possible spin states. In this case, the pressure and volume equations can be written as:\frac{P}{k_BT}=\frac{g}{\lambda_D^3}f_{5/2}(\lambda)\frac{1}{\upsilon}=\frac{g}{\lambda_D^3}f_{3/2}(\lambda)where \lambda_D=\sqrt{\frac{2 \pi\hbar^2}{mk_BT}} is the de Broglie wavelength and f_k(\lambda)=\sum^{\infty}_{n=1}(-1)^{n-1}\frac{\lambda^n}{n^k} is the Fermi integral. The fugacity \lambda=e^{\frac{\mu}{\theta}} is the ratio of the chemical potential \mu to the thermal energy \theta.