Bloch Theory of electrical resistivity - Derivation

1. Mar 12, 2012

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?

2. Mar 13, 2012

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 writen 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 :).