Thermal expansion (anharmonic effect)

[cavalier3024]

i have a question regarding calculating the thermal expansion parameter $$\alpha$$
it is given by:
$$\alpha = \frac{1}{L} \frac{\partial L}{\partial T} = \frac{1}{3B} \sum \frac{\partial(hw)}{\partial V} \frac{\partial n}{\partial T}$$

now, I am not quite sure how i calculate $$d(hw)/dV$$. how is $$w$$ related to $$V$$? the only thing i can think of is that $$w$$ is related to $$a$$ (the lattice constant) and $$a$$ is related to $$V$$ according to the lattice. is that right? and is that the only connection or there are other factors i should take into account?

 

[kanato]

Yes, that's correct. It looks like your equation has assumed a material with cubic symmetry, so that derivatives with respect to lattice constant are just 1/3 of a derivative wrt volume. Things are more complex if the material has lower symmetry.

 

[cavalier3024]

thanks a lot! i hope this will help me to pass my final exam two days from now

btw this formula is taken from ashcroft, they didnt say there that it applies only to cubic symmetry but I am not sure. do you have the general form of this formula, or can refer me to a site\book?

 

[kanato]

The first part $$\alpha = L^{-1} dL/dT$$ is the general formula. If you have a crystal with tetragonal symmetry (for example), then you will have two expansion coefficients, $$\alpha_a = a^{-1} da/dT$$ and $$\alpha_c = c^{-1} dc/dT$$. The rest of the derivation should be basically the same, but without the substitution that dV = 3a^2 da.