The scattering rate by emission of phonon is given by following formula;

$$\frac{1}{\tau_{emi}}\propto\int_0^{\omega_{max}}d\omega\ \omega^2[n_B(\omega)+1]\propto T \ for\ 1>>\beta\hbar\omega\ \ T^3 \ for\ 1<<\beta\hbar\omega$$

where $n_B$ is distribution function for bosons. I cant prove that the integral is proportional to the expressions given at left.


1 Answer 1


In the low temperature regime a Taylor expansion of the exponential Term in the Bose-Einstein distribution is justified and we obtain

$$\frac{1}{\tau_{emi}} \propto \int_{0}^{\omega_{max}}d\omega\ \omega^2[\frac{kT}{\hbar\omega} + 1] = const_1. + T *const_2.$$

For the high temperature limit we substitute $x = \beta\hbar\omega$ to get

$$\frac{1}{\tau_{emi}} \propto \frac{1}{(\beta\hbar)^3}(\int_{0}^{x_{max}}\ x^2 \ n_b(x) \ dx + const_3.)$$

since $x_{max} = \beta\hbar\omega_{max} >> 1$ and the bose distribution decays exponentially for large x we can safely extend the upper Integral limit to infinity. This immediately gives

$$\frac{1}{\tau_{emi}} \propto (\frac{kT}{\hbar})^3(\int_{0}^{\infty}\ x^2n_b(x) \ dx+ const_3.) \propto T^3$$

which is the desired result.


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