Bloch Theory of electrical resistivity - Derivation

[kanonear]

In "Electrons and Phonons" by Ziman which you can find here, I have a problem with equation (9.5.14) on page 360 and (9.5.17) on page 361. I don't understand how from the first equation where you have double surface integral you obtain the second equation with only one integral over q.

I don't think it's a mathematical transformation, but something with physical assumptions.

Can anybody help me with that?

 

[kanonear]

Ok. I figured it out. I'll write my solution here so that it will be easy to find :).

ρ_{L}= \frac{9 π \hbar}{2e^{2}mNk_{b}T}\cdot \frac{1}{R^{2}σ^{2}}\int\int \frac{(K \cdot u)^{2}(K \cdot e)^{2}ζ(K)}{(1-e^{-h\nu /k_{b}T})(e^{h\nu /k_{b}T}-1)}\frac{dσ}{\upsilon}\frac{dσ\;'}{\upsilon\;'} =

As it is stated in the book:

(K \cdot u)^{2}=\frac{1}{3} K^{2}
(K \cdot e)^{2}= K^{2}

Of course K = k' - k , and for Normal Process we can write that K=q.

K^{2}= 2R^{2}(1-\cos ( \theta))
K\;dK= R^{2}\sin ( \theta) d\theta

In the spherical coordinates (in k space) the surface element can be written as:

d\sigma = R^{2} \sin (\theta)d\theta d\varphi

where R correspond to the Fermi Sphere radius.
Now we rewrite the equation for resistance using equations above:

ρ_{L}= \frac{9 π \hbar}{2e^{2}mNk_{b}T}\cdot \frac{1}{R^{2}σ^{2}}\int\int\int \frac{\frac{1}{3}K^{2}K^{2}ζ(K)}{(1-e^{-h\nu /k_{b}T})(e^{h\nu /k_{b}T}-1)}\frac{R^{2} \sin (\theta)d\theta d\varphi}{\upsilon}\frac{dσ\;'}{\upsilon\;'} = \frac{9 π \hbar}{2e^{2}mNk_{b}T}\cdot \frac{1}{R^{2}σ^{2}}\int \int \int \frac{\frac{1}{3}K^{4}ζ(K)}{(1-e^{-h\nu /k_{b}T})(e^{h\nu /k_{b}T}-1)}\frac{K\;dK d\varphi}{\upsilon}\frac{dσ\;'}{\upsilon\;'}

Now let us take \upsilon = \upsilon \; ' = \upsilon _{F} and K=q :

ρ_{L} = \frac{9 π \hbar}{2e^{2}mNk_{b}T}\cdot \frac{1}{R^{2}σ^{2}}\int \frac{\frac{1}{3}q^{5}ζ(K)\;dq }{(1-e^{-h\nu /k_{b}T})(e^{h\nu /k_{b}T}-1)}\frac{1}{\upsilon_{F}^{2}}\int d\varphi \int dσ\;'

Integral over d\varphi give us 2\pi, and the last integral is just \sigma.

ρ_{L} = \frac{9 π \hbar}{2e^{2}mNk_{b}T}\cdot \frac{1}{R^{2}σ^{2}}\cdot \frac{1}{3} \frac{1}{\upsilon_{F}^{2}}2\pi \sigma \int \frac{q^{5}ζ(q)\;dq }{(1-e^{-h\nu /k_{b}T})(e^{h\nu /k_{b}T}-1)}

Now we can substitute \sigma = 4\pi R^{2}:

ρ_{L} = \frac{3 π \hbar}{4e^{2}mNk_{b}T R^{4}\upsilon_{F}^{2}} \int \frac{q^{5}ζ(q)\;dq }{(1-e^{-h\nu /k_{b}T})(e^{h\nu /k_{b}T}-1)}

and that is what we were looking for :).