Fill This Form To Receive Instant Help
Fill This Form To Receive Instant Help
Homework answers / question archive / Consider a classical particle moving in a one-dimensional potential well u(x)
Consider a classical particle moving in a one-dimensional potential well u(x). The particle is in thermal equilibrium with a reservoir at temperature T, so the probabilities of its various states are determined by Boltzmann statistics.
(a) Show that the average position of the particle is given by
x =∫xe-βu(x) dx/∫e-βu(x) dx,
where each integral is over the entire x axis.
(b) If the temperature is reasonably low (but still high enough for classical mechanics to apply), the particle will spend most of its time near the bottom of the potential well. In that case we can expand u(x) in a Taylor series about the equilibrium point x0:
u(x) = u(x_0) + (x - x_0) du/dx |x_0 + ½ (x + x_0)2 d^2u/dx^2 |x_0 + 1/3! (x - x_0)^3 d^3u/dx^3 |x_0 + …
Show that the linear term must be zero, and that truncating the series after the quadratic term results in the trivial prediction x = x_0.
(c) If we keep the cubic term in the Taylor series as well, the integrals in the formula for x become difficult. To simplify them, assume that the cubic term is small, so its exponential can be expanded in a Taylor series (leaving the quadratic term in the exponent). Keeping only the smallest temperature-dependent term, show that in this limit x differs from x0 by a term proportional to kT. Express the coefficient of this term in terms of the coefficients of the Taylor series for u(x).
(d) The interaction of noble gas atoms can be modeled using the Lennard-Jones potential,
u(x) = u0[(x_0/x)^12 - 2(x_0/x)^6].
Sketch this function, and show that the minimum of the potential well is x = x0, with depth u0. For argon, x_0 = 3.9 Å and u_0 = 0.010 eV. Expand the Lennard-Jones potential in a Taylor series about the equilibrium point, and use the result of part (c) to predict the linear thermal expansion coefficient of a noble gas crystal in terms of u_0. Evaluate the result numerically for argon, and compare to the measured value α = 0.0007 K-1 (at 80 K).
The probability that the particle is in some state r is:
P_r = exp(-beta E_r)/Z
The probability that the particle is in some precise "classical state" specified by the position and momentum is obtained by putting:
E_r = u(x) + p^2/(2m)
So we have:
P[x,p] = exp[-beta u(x)]*exp[-p^2/(2m)]/Z (1)
The partition function Z makes this function properly normalized. The number of states between momenta p and p + dp and positions between x and x + dx is:
dx dp/h
The partition function is thus:
Z =integral over x and p of dxdp/h exp[-beta u(x)]*exp[-p^2/(2m)] (2)
To obtain the probability that the particle is between positions x and x + dx, we must sum P[x,p] over all states that are compatible with such a situation. The probability that the particle s in some precise state for which the position is x and momentum is p is P[x,p]. The number of states between x and x + dx and between p and p + dp is dx dp/h, so the probability that the particle is in this range of states is:
P[x,p]dx dp/h
If we integrate this over all momenta we get using (1) and (2):
exp[-beta u(x)]dx/Integral over x of exp[-beta u(x)]dx
This is the probability that the particle is at a position between x and x +dx. If we multiply this by x and integrate aver all x we obtain the average position:
<x> = Integral over x of xexp[-beta u(x)]dx/Integral over x of exp[-beta u(x)]dx (3)
The integrations here and in the following are from minus infinity to plus infinity.
At the minimum of the potential the derivative of u must be zero, so the linear term in the Taylor expansion must be zero at x_0. If we keep only the quadratic term then the integral in the numerator of (3) is of the form
Integral over x of A x exp[-beta (x-x_0)^2u''(x_0)/2]dx
where A = exp[-beta u(x_0)]
Define y = x - x_0. The integral then becomes:
Integral over y of A (y+x_0) exp[-beta y^2u''(x_0)/2]dy
The integral of yexp[-beta y^2u''(x_0)/2]dy = 0
because the integrand is an odd function. The other term is exactly x_0 times the integral in the numerator, so the average position is x_0.
To compute the average position in the presence of the cubic term, we evaluate the average of x - x_0. So instead of the factor x in the numerator we have a factor (x-x_0). We then change variables in all the integrations: y = x - x_0. The integrand in the denominator is (after Taylor expanding the cubic term in the exponential):
exp(-beta u(x_0))*exp(-beta u''(x_0)/2 y^2)*[1-beta u'''(0)/6 y^3]
And in the numerator this is multiplied by y:
y*exp(-beta u(x_0))*exp(-beta u''(x_0)/2 y^2)*[1-beta u'''(0)/6 y^3]
In the denominator the y^3 term vanishes after integration, because it is odd, and we only have to evaluate an integral of the form:
integral over y of exp(-ay^2)dy = sqrt(pi/a) (4)
In the numerator we only have to deal with:
integral over y of y^4 exp(-ay^2)dy (5)
this time the other integral vanishes.
To calculate (5) we can differentiate (4) twice w.r.t. a:
integral over y of y^4 exp(-ay^2)dy = 3/4 sqrt(pi) a^(-5/2)
If we divide the numerator and denominator we get the displacement:
3/4 a^(-2)*(-beta u'''(0)/6)
Inserting a = beta u''(x_0)/2 gives for the displacement:
-1/2 u'''(0) u''(x_0)^(-2) beta^(-1) = -1/2 u'''(0) u''(x_0)^(-2) kT
For the Lennard-Jones potential I find:
u''(x_0) = 72 u0 x_0^(-2)
u'''(x_0) = -1512 u0 x_0^(-3)
which implies a displacement of:
7/48 x_0/u0 kT
The average distance is thus:
<x> = x_0 + 7/48 x_0/u0 kT = x_0[1 + 7/48 kT/u0]
And we see that alpha is:
1/<x> * d <x>/dT = 7/48 k/u0
