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Homework Statement
Please see attachment. The correct answer is supposed to be ##\alpha = .15 k_B/U_0##.
The Attempt at a Solution
The minimum of the potential is at ##x = a##. Expanded about this minimum we get ##U = -U_0 + 36\frac{U_0}{a^2}(x - a)^2 - 252\frac{U_0}{a^3}(x - a)^3 + ...##. The mean distance is ##\langle x \rangle = \frac{\int_{0}^{\infty} x e^{-\beta U} dx}{\int_{0}^{\infty}e^{-\beta U}dx}##. Here is a plot of the potential: http://en.wikipedia.org/wiki/Lenard_Jones_potential and as you can see it falls of steeply from infinity at ##x =0## to the minimum at ##x = a## and then rises slowly to zero as ##x\rightarrow \infty##. In order to insert the above series expansion of ##U## about ##x = a## into the expression for ##\langle x \rangle##, we would need to be able to show that the integrals in it are overwhelmingly dominated by the values of the integrand near ##x = a##; let's say that there is a characteristic width ##\delta a## that codifies being near ##x = a##.
As we move away from ##x = a## and towards ##x = 0##, we see that ##U/U_0## very, very rapidly becomes larger than unity so we can say to good approximation that ##U/U_0 \gg 1## when ##x \leq a - \delta a##. Furthermore ##\beta U_0 \gg 1## by hypothesis so ##\beta U \gg 1## and note ##U > 0## when ##x \leq a - \delta a##. Hence ##\int_{0}^{a - \delta a} x e^{-\beta U} dx \approx 0, \int_{0}^{a - \delta a} e^{-\beta U} dx \approx 0##.
Now we have to show that ##\int_{a + \delta a}^{\infty} x e^{-\beta U}dx \approx 0, \int_{a+ \delta a}^{\infty}e^{-\beta U}dx \approx 0##. This is where I'm not sure. We see that for ##x \geq a + \delta a##, ##U/U_0 \ll 1##, ##U < 0##, and ##U \rightarrow 0##. So we see that even if ##\beta U_0 \gg 1##, eventually ##\beta U \ll 1## i.e. for ##x \gg a + \delta a##, which from the diagram happens as soon as ##x \approx 2##, but ##U < 0## so ##e^{-\beta U} = e^{|\beta U|} \approx 0## for ##x \geq a + \delta a## and what's more ##e^{|\beta U|} \rightarrow 0## much more rapidly than ##x \rightarrow \infty## so we get the desired results.
Thus ##\langle x \rangle = \frac{\int_{0}^{\infty} x e^{-36\beta U_0 \frac{(x - a)^2}{a^2}} e^{252\beta U_0 \frac{(x - a)^3}{a^3}}...dx}{\int_{0}^{\infty} e^{-36\beta U_0 \frac{(x - a)^2}{a^2}} e^{252\beta U_0 \frac{(x - a)^3}{a^3}}...dx}##.
We know that ##\delta a/a \ll 1## so it would be natural to try working only to second order in ##\frac{(x - a)}{a}## since ##x - a \sim \delta a## in the integrals. To second order the integrals are trivial to evaluate but give the wrong result of ##\alpha = .5/T##. I don't know why. The next step then, albeit a very ad-hoc step, would be to try working to third order but beyond second order, the order at which we work for this particular problem seems quite arbitrary rather than natural. At this point I'm just trying the third order in order to get what the book says is the right answer. I don't have any physical justification for why the third order would be the right order to work in, as opposed to second order (which for some reason gives the wrong answer) or any higher orders.
Put in other words, for this problem how would I go about figuring out to what order in ##\frac{(x - a)}{a}## I should work in? I don't want to just keep trying higher orders until I get the right answer, there has to be a more physically intuitive way to figure it out. Could anyone shed some light on this issue? Does it have something to do with the size of ##\pm \beta U_0 \frac{(x - a)}{a} \sim \pm \beta U_0 \frac{\delta a}{a}## as opposed to just the size of ##\frac{(x - a)}{a}## itself? Thanks in advance.
P.S. I avoided posting the third order calculation in the OP itself because then the OP would just get way too long to read.
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