Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermal expansion (anharmonic effect)

  1. Dec 22, 2009 #1
    i have a question regarding calculating the thermal expansion parameter [tex]\alpha[/tex]
    it is given by:
    [tex]\alpha = \frac{1}{L} \frac{\partial L}{\partial T} = \frac{1}{3B} \sum \frac{\partial(hw)}{\partial V} \frac{\partial n}{\partial T}[/tex]

    now, im not quite sure how i calculate [tex]d(hw)/dV[/tex]. how is [tex]w[/tex] related to [tex]V[/tex]? the only thing i can think of is that [tex]w[/tex] is related to [tex]a[/tex] (the lattice constant) and [tex]a[/tex] is related to [tex]V[/tex] according to the lattice. is that right? and is that the only connection or there are other factors i should take into account?
     
  2. jcsd
  3. Dec 22, 2009 #2
    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.
     
  4. Dec 22, 2009 #3
    thanks alot! 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 im not sure. do you have the general form of this formula, or can refer me to a site\book?
     
  5. Dec 22, 2009 #4
    The first part [tex]\alpha = L^{-1} dL/dT[/tex] is the general formula. If you have a crystal with tetragonal symmetry (for example), then you will have two expansion coefficients, [tex]\alpha_a = a^{-1} da/dT[/tex] and [tex]\alpha_c = c^{-1} dc/dT[/tex]. The rest of the derivation should be basically the same, but without the substitution that dV = 3a^2 da.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook