These questions relate to lectures 5 and 6.

1 Show that, for a 1D curve described using the Monge representation, the curvature is given by Equation 12.

2 Ignoring any effect from gravity, the amplitude spectrum of thermal fluctuations of the interface between two fluids, as a function of the wavevector q, is approximately:

Problem Equation 1

where γ is the interface tension and A is the area of the interface.

Given that:

the interface is in a square container with A = 1cm2

there is a small wavelength cutoff at the molecular scale

the molecular size is 0.2nm

the interface tension is γ = 38 x 10-3 N m-1

what is the average undulation amplitude at room temperature due to thermal fluctuations for the interface between water and hexane?


In considering the function f(x), saying that the curvature at a certain point x0 is r means that a circle of radius r can be drawn tangent to x0 and its immediate neighbouring points. Equation 12 can be proved if we construct such a circle, showing how its radius is related to the first and second derivatives of the function f.

We can call n1(x) the normal to the function at the point x0 and we know from basic analytical geometry that it has equation:

n1(x) = f(x0) – (x- x0) / f’(x0).

(f’ means derivative of f w.r.t. x)

A neighbouring normal, at the point x0 + dx, will have the equation:

n2(x) = f(x0 + dx) – (x- x0 - dx) / f’(x0 + dx)

All neighbouring normals should cross at the centre of the circle tangent at x0 , at least to first order in dx. Taylor expanding n2(x):

n2(x) ≅ f(x0) + f’(x0) dx – (x- x0 - dx) / [f'(x0) +f’’(x0) dx].

and requiring that the lines cross at the centre of the circle, point xc, gives:

xc = x0 - f/f’’ (f2 +1)

and the y-coordinate of the centre is

n1(xc) = f(x0) + (f2 +1)/f’’.

The radius of the circle must satisfy the equation of the circle:

r2 = (xc - x0) 2 + (yc - y0) 2

= (f’2 +1)3/f’2

(Note :: y0 = f(x0)). Curvature is the inverse of the radius, so:

H = 1/r = f’’ / (f2 +1)3/2

as in Equation 12.

2 The average amplitude is given by the integral over all wavevectors of the surface roughness. We proceed similarly to Equation 26, but for the case of a tension dominated interface (with no bending modulus).

Answer Equation 1

This gives:

Answer Equation 2

From the data supplied we have

qmax/qmin = 1 cm/0.2 nm = 10-2/0.2 x 10-9 = 5 x 107


<h2>= 32 x 10-20 m2

i.e. a root mean square roughness of 5.6 x 10-10 m (= 0.56 nm).