 Solution

ID: past-exam-of-the-soft-condensed-matter-course-of-the-university-of-cambridge/2022/section-b/5/d/solution

Put

k = βF

and

q = k δ

A = j_{0} e^{q} / (D k)

, hence

p (x) = \frac{j _{0}}{D k} [e^{k (δ - x)} - 1] .

This is nonnegative throughout the allowed interval. Normalizing the probability density gives

1 = \frac{j _{0}}{D k} [\frac{e ^{q} - 1}{k} - δ] = \frac{j _{0}}{D k ^{2}} (e^{q} - 1 - q) .

Therefore

j_{0} = \frac{D β ^{2} F ^{2}}{e ^{βF δ} - βF δ - 1}, A = \frac{βF e ^{βF δ}}{e ^{βF δ} - βF δ - 1} .

Equivalently the normalized density is

p (x) = \frac{k [ e ^{k (δ - x)} - 1 ]}{e ^{q} - 1 - q} .

The denominator is positive for

q > 0

, since

e^{q} > 1 + q

. The dimensions are consistent:

j_{0}

has units of inverse time,

A

and

p

of inverse length, and

q

is dimensionless.

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