# Conductivity and Integration over Fermi-Sphere?

• A
Hi,
I am reading "An Introduction of Solid State Physics" from Ibach Lüth and don't understand the integration process.
They write $$\sigma=\frac{e^2}{8\pi^3 \hbar} \int df_{E}dE \frac{v^2_x(\bf{k})}{v(\bf{k})} \tau(\bf{k}) \delta(E-E_F)$$

$$= \int_{E=E_F}^{}df_{E} \frac{v^2_x(\bf{k})}{v(\bf{k})} \tau(\bf{k})$$

But why does ## k ## NOT become ##k_F## after the Integration of the Delta-function?

I would think that the integral becomes immedately after the dE-integration:
$$= \int_{E=E_F}^{}df_{E} \frac{v^2_x(\bf{k_F})}{v(\bf{k_F})} \tau(\bf{k_F})$$

THX
Abbi

#### Attachments

• Versuch.png
125.8 KB · Views: 301
Last edited:

DrDu
Yes, you are right, but this is only a matter of notation. k is still a vector and will have different length for different directions as long as the Fermi surface is not spherical.

I don't understand the notation ##dS_E##.
$$d\bf{k} = k^2 sin\theta d\theta d\phi dk$$
$$= dS_k dk_\bot$$ This means a surface-element times a radial distance-element perpendicular to the surface-element. Both in k-space.

In the book instead of ##dS_k## the expression ##dS_E## is used. They say ##dS_k= dS_E dk_\bot ##

What means exactly ##dS_E##?
If I think at the notation $$dS_k= k^2 sin\theta d\theta d\phi$$,
it should be $$dS_E=E({\bf k})^2 sin\theta d\theta d\phi$$

But then the units don't correspond anymore to k-space:
$$d{\bf k} =dS_E dk_\bot$$ which corresponds to $$\frac{1}{m^3}=\frac{eV}{1}\frac{1}{m}$$

Last edited: