- A.43.1 Basic definitions
- A.43.2 Kinetic energy
- A.43.3 Energy due to surface tension
- A.43.4 Energy due to Coulomb repulsion
- A.43.5 Frequency of vibration

A.43 Classical vibrating drop

The simplest collective description for a nucleus models it as a vibrating drop of a macroscopic liquid. To represent the nuclear Coulomb repulsions the liquid can be assumed to be positively charged. This section gives a condensed derivation of small vibrations of such a liquid drop according to classical mechanics. It will be a pretty elaborate affair, condensed or not.

A.43.1 Basic definitions

The drop is assumed to be a sphere of radius

When vibrating, the radial position of the surface will be indicated by

The mass density, (mass per unit volume), of the liquid will be
indicated by

The mass density is assumed constant throughout the drop. This implies that the liquid is assumed to be incompressible, meaning that the volume of any chunk of liquid is unchangeable.

The charge density is defined analogously to the mass density:

It too is assumed constant.

The surface tension

A.43.2 Kinetic energy

The possible frequencies of vibration can be figured out from the kinetic and potential energy of the droplet. The kinetic energy is easiest to find and will be done first.

As noted, the liquid will be assumed to be incompressible. To see
what that means for the motion, consider an arbitrary chunk of liquid.
An elementary element of surface area

Using the Gauss-Ostrogradsky, or divergence theorem, this means that

This is the famous “continuity equation” for incompressible flow. But it is really no different from Maxwell’s continuity equation for the flow of charges if the charge density remains constant, chapter 13.2.

To describe the dynamics of the drop, the independent variables will
be taken to be time and the unperturbed positions

where

Note that the gradients of the potential and pressure should normally
be evaluated with respect to the perturbed position coordinates

If you take the divergence of Newton’s equation. i.e. multiply
with

The solution can be derived in spherical coordinates

where the

Plugging the solution for the pressure into Newton’s second law
shows that the velocity can be written as

where the coefficients

(You might wonder about the integration constant in the time
integration. It is assumed that the droplet was initially spherical
and at rest before some surface perturbation put it into motion. If
the drop was initially rotating, the analysis here would need to be
modified. More generally, if the droplet was not at rest initially,
it must be assumed that the initial velocity is “irrotational,” meaning that

Since the velocity is the time-derivative of position, the positions
of the fluid elements are

where

What will be important in the coming derivations is the radial displacement of the liquid surface away from the spherical shape. It follows from taking the radial component of the displacement evaluated at the surface

To be sure, in the analysis

The kinetic energy is defined by

Putting in the expression for the velocity field in terms of the

To simplify this, a theorem is useful. If any two functions

Applying these results to the integral for the kinetic energy, noting
that

Now the spherical harmonics are orthonormal on the unit sphere; that means that the final integral is zero unless

A.43.3 Energy due to surface tension

From here on, the analysis will return to physical coordinates rather than unperturbed ones.

The potential energy due to surface tension is simply the surface tension times the surface of the deformed droplet. To evaluate that, first an expression for the surface area of the droplet is needed.

The surface can be described using the spherical angular coordinates

To get the surface area

That way, the surface area works out to be

Multiply by the surface tension

Of course, this expression is too complicated to work with. What
needs to be done, first of all, is to write the surface in the form

where

where the final integral comes from expanding the square root and where terms of order of magnitude

The second integral is first. Its problem is that if you plug in a
valid approximate expression for

Why is

the first because of the expression for volume in spherical coordinates and the second from integrating out

where the integral of

If this is used in the expression for the potential energy, it produces

Now

where the

This integral can be simplified by using the same theorem (A.267) used earlier for the kinetic energy. Just take

Putting it all together, the potential energy due to surface tension
becomes

A.43.4 Energy due to Coulomb repulsion

The potential energy due to the Coulomb forces is tricky. You need to
make sure that the derivation is accurate enough. What is needed is
the change in potential energy when the radial position of the surface
of the droplet changes from the spherical value

Trying to write a six-dimensional integral for the Coulomb energy would be
a mess. Instead, assume that the surface perturbation is applied in
small increments, as a perturbation

The surface radius perturbation and its differential change can be
written in terms of spherical harmonics:

The amount of charge added per unit solid angle

To first approximation, the incremental amount of charge laid down per unit solid angle is

However, if

then the coefficient

Now the surface potential is needed for the deformed drop. There are
two contributions to this potential: the potential of the original
spherical drop and the potential of the layer

For the surface potential of the laid-down layer, fortunately only a
leading order approximation is needed. That means that the thickness
of the layer can be ignored. That turns it into a spherical shell of
negligible thickness at

though the coefficients

where the coefficients

However, the electric field strength does vary significantly from one
side of the shell to the other, and it is that variation that
determines the coefficients

where

Multiplying the two surface potentials by the amount of charge laid
down gives the incremental change in Coulomb potential of the drop
as

Substituting in

A.43.5 Frequency of vibration

Having found the kinetic energy, (A.268), and the potential energy, (A.269) plus (A.270), the motion of the drop can be determined.

A rigorous analysis would so using a Lagrangian analysis,
{A.1}. It would use the coefficients

If you are willing to take that for granted, just assume

Note that this is zero if