When I learned anharmonic model of crystal, I read that considering anharmonic oscillations and Boltzmann distribution for the "atoms" of crystal we can get the dependence of distance between the "atoms" from a temperature as

$$ \langle r \rangle = r_{0} + \alpha T. $$

As I understood the words below, it's like

$$ \langle r \rangle = \frac{\int \limits_{0}^{\infty}re^{-\frac{U}{kT}}dr}{\int \limits_{0}^{\infty}e^{-\frac{U}{kT}}dr} \approx \frac{\int \limits_{0}^{\infty}re^{-\frac{U_{0} + a(r - r_{0})^{2} + b(r - r_{0})^{3} }{kT}}dr}{\int \limits_{0}^{\infty}e^{-\frac{U_{0} + a(r - r_{0})^{2} + b(r - r_{0})^{3} }{kT}}dr} = |x = r - r_{0}| = \frac{\int \limits_{0}^{\infty}(x + r_{0})e^{-\frac{ax^{2} + bx^{3}}{kT}}dr}{\int \limits_{0}^{\infty}e^{-\frac{ax^{2} + bx^{3}}{kT}}dr}, $$

and then - $$ \langle r \rangle \approx r_{0} + \frac{\int \limits_{0}^{\infty}xe^{-\frac{ax^{2} + bx^{3}}{kT}}dr}{\int \limits_{0}^{\infty}e^{-\frac{ax^{2} + bx^{3}}{kT}}dr}. $$

What can I do on the next step?

  • $\begingroup$ Ron's answer is fine, and knowing how to do that integral yourself is impressive and useful; but an easier way would be to look at a table of integrals. Even I'm the age of Mathematica, there is no substitute for Gradshteyn & Rhyzik. I'm certain G&R contains this integral. $\endgroup$ – Colin K May 18 '12 at 2:26
  • $\begingroup$ @ColinK: Did you check? I don't think so. Anything in G&R will be in Mathematica, and there are things not in either. But here, there is no analytic form here, it has an essential singularity at $b=0$, so no (non-asymptotic) power series. There is no chance that it is in a table of integrals. $\endgroup$ – Ron Maimon Jun 14 '12 at 0:06

You expand the top and bottom integral in a power series in b, and keep the lowest order term in $b$:

For the top integral:

$$ \int x e^{-ax^2} (1 + bx^3) = b\int x^4 e^{-ax^2} = b {d^2\over da^2} \int e^{-ax^2} = b{d^2\over da^2} \sqrt{\pi\over a} = -\sqrt{\pi\over a} {3b\over 4a^2} $$

For the bottom integral

$$ \int e^{-ax^2} (1 + b x^3) = \sqrt{\pi \over a} $$

So the quotient is

$$ {3b\over 4a^2}$$

The coefficients $a$ and $b$ absorb the $kT$ on the bottom in your expression, so replace $a$ by $a\over kT$ and b by $b\over kT$. The above becomes:

$$ kT {3b\over 4a^2} $$

and it is linearly proportional to T, as you expected.

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  • $\begingroup$ Thank you! But you considered that limits of integration are from -infinity to infinity. Is that correctly? Can r be negative? $\endgroup$ – user8817 May 18 '12 at 0:17
  • $\begingroup$ @Maxim_Ovchinnikov: Yeah, the limits are as you gave them, $-\infty$ to $\infty$, the sign of the correction depends on the sign of b, if the anisotropy goes the wrong way, the material will contract under heat (this will never happen for the potentials of pure atomic repulsion, because the anisotropy makes the force curvature weaker as you go out and stronger as you go in from the minimum, because the force blows up very strongly as the molecules start to overlap each other). $\endgroup$ – Ron Maimon May 18 '12 at 0:34

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