**Questions **

*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:

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* = 1cm^{2}

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?

**Answers**

In considering the function* f*(*x*), saying that the curvature at a certain point *x _{0}* is

*r*means that a circle of radius

*r*can be drawn tangent to

*x*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

_{0}*f.*

We can call n_{1}(*x*) the normal to the function at the point *x _{0}* and we know from basic analytical geometry that it has equation:

*n*_{1}(*x*) = f(*x _{0}*) – (

*x- x*) /

_{0}*f*’(

*x*).

_{0}(*f’* means derivative of *f *w.r.t. *x*)

A neighbouring normal, at the point *x _{0 }+ *d

*x,*will have the equation:

*n*_{2}(*x*) = *f*(*x*_{0 }+ d*x*) – (*x- x*_{0 }- d*x*) / *f*’(*x*_{0 }+ d*x*)

All neighbouring normals should cross at the centre of the circle tangent at *x _{0 }, *at least to first order in

*dx.*Taylor expanding

*n*

_{2}(

*x*):

*n*_{2}(*x*) ≅ *f*(*x _{0}*) +

*f’*(

*x*) d

_{0}*x*– (

*x- x*d

_{0 }-*x*) / [

*f'*(

*x*)

_{0}*+f’’*(

*x*) d

_{0}*x*].

and requiring that the lines cross at the centre of the circle, point *x _{c}*, gives:

*x _{c = }x_{0 }*-

*f*’

*/f*’’ (

*f*’

^{2}+1)

and the *y*-coordinate of the centre is

*n*_{1}(*x _{c}*) =

*f*(

*x*) + (

_{0}*f*’

^{2}+1)

*/f’’*.

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

*r*^{2} = (*x _{c} - x_{0}*)

^{ 2}+ (

*y*)

_{c}- y_{0}^{ 2}

= (*f’*^{2} +1)^{3}/*f’*’^{2 }

(Note :: *y _{0}* =

*f*(

*x*

_{0})). Curvature is the inverse of the radius, so:

*H* = 1/*r *= *f’’ */ (*f*’^{2} +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).

This gives:

From the data supplied we have

*q _{max}/q_{min}* = 1 cm/0.2 nm = 10

^{-2}/0.2 x 10

^{-9}= 5 x 10

^{7}

giving

<*h*^{2}>= 32 x 10^{-20 }m^{2}

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