Subsections


A.42 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.42.1 Basic definitions

The drop is assumed to be a sphere of radius $R_0$ when it is not vibrating. For a nucleus, $R_0$ can be identified as the nuclear radius,

\begin{displaymath}
R_0 = R_A A^{1/3}
\end{displaymath}

When vibrating, the radial position of the surface will be indicated by $R$. This radial position will depend on the spherical angular coordinates $\theta$ and $\phi$, figure N.3, and time.

The mass density, (mass per unit volume), of the liquid will be indicated by $\rho_m$. Ignoring the difference between proton and neutron mass, for a nucleus the mass density can be identified as

\begin{displaymath}
\rho_m = \frac{A m_{\rm p}}{\frac43\pi R_0^3}
\end{displaymath}

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:

\begin{displaymath}
\rho_c = \frac{Ze}{\frac43\pi R_0^3}
\end{displaymath}

It too is assumed constant.

The surface tension $\sigma$ can be identified as

\begin{displaymath}
\sigma = \frac{C_sA^{2/3}}{4 \pi R_0^2} = \frac{C_s}{4\pi R_A^2}
\end{displaymath}


A.42.2 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 ${\rm d}{S}$ of that chunk gobbles up an amount of volume while it moves given by $\vec{v}\cdot{\vec n}{\,\rm d}{S}$, where $v$ is the liquid velocity and ${\vec n}$ is a unit vector normal to the surface. But for a given chunk of an incompressible liquid, the total volume cannot change. Therefore:

\begin{displaymath}
\int_S \vec{v}\cdot{\vec n}{\,\rm d}{S} = 0
\end{displaymath}

Using the Gauss-Ostrogradsky, or divergence theorem, this means that $\nabla\cdot\vec{v}$ must integrate to zero over the interior of the chunk of liquid. And if it must integrate to zero for whatever you take the chunk to be, it must be zero uniformly:

\begin{displaymath}
\nabla\cdot\vec v = 0
\end{displaymath}

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 ${\skew0\vec r}_0$ of the infinitesimal volume elements ${\rm d}^3{\skew0\vec r}$ of liquid. The velocity field inside the drop is governed by Newton’s second law. On a unit volume basis, this law takes the form

\begin{displaymath}
\rho_m \frac{\partial \vec v}{\partial t} = - \rho_c \nabla\varphi - \nabla p
\end{displaymath}

where $\varphi$ is the electrostatic potential and $p$ is the pressure. It will be assumed that the motion is slow enough that electrostatics may be used for the electromagnetic force. As far as the pressure force is concerned, it is one of the insights obtained in classical fluid mechanics that a constant pressure acting equally from all directions on a volume element of liquid does not produce a net force. To get a net force on an element of liquid, the pressure force on the front of the element pushing it back must be different from the one on the rear pushing it forward. So there must be variations in pressure to get a net force on elements of liquid. Using that idea, it can be shown that for an infinitesimal element of liquid, the net force per unit volume is minus the gradient of pressure. For a real classical liquid, there may also be viscous internal forces in addition to pressure forces. However, viscosity is a macroscopic effect that is not relevant to the nuclear quantum system of interest here. (That changes in a two-liquid description, [39, p. 187].)

Note that the gradients of the potential and pressure should normally be evaluated with respect to the perturbed position coordinates ${\skew0\vec r}$. But if the amplitude of vibrations is infinitesimally small, it is justified to evaluate $\nabla$ using the unperturbed position coordinates ${\skew0\vec r}_0$ instead. Similarly, $\nabla$ in the continuity equation can be taken to be with respect to the unperturbed coordinates.

If you take the divergence of Newton’s equation. i.e. multiply with $\nabla\cdot$, the left hand side vanishes because of continuity, and so the sum of potential and pressure satisfies the so-called “Laplace equation:”

\begin{displaymath}
\nabla^2 (\rho_c \varphi + p) = 0
\end{displaymath}

The solution can be derived in spherical coordinates $r_0$, $\theta_0$, and $\phi_0$ using similar, but simpler, techniques as used to solve the hydrogen atom. The solution takes the form

\begin{displaymath}
\rho_c \varphi + p =
\sum_{l,m} c_{lm}(t) \frac{r_0^l}{R_0^l} \bar Y_l^m(\theta_0,\phi_0)
\end{displaymath}

where the $c_{lm}$ are small unknown coefficients and the $\bar{Y}_l^m$ are real spherical harmonics. The precise form of the $\bar{Y}_l^m$ is not of importance in the analysis.

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

\begin{displaymath}
\vec v = \sum_{l,m} v_{lm}(t) R_0
\nabla \left(\frac{r_0^l}{R_0^l} \bar Y_l^m(\theta_0,\phi_0)\right)
\end{displaymath}

where the coefficients $v_{lm}$ are multiples of time integrals of the $c_{lm}$. What multiples is irrelevant as the potential and pressure will no longer be used.

(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 $\nabla$ $\times$ $\vec{v}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.)

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

\begin{displaymath}
{\skew0\vec r}= {\skew0\vec r}_0 + \sum_{l,m} r_{lm}(t) R_0
\nabla \frac{r_0^l}{R_0^l} \bar Y_l^m(\theta_0,\phi_0)
\end{displaymath}

where ${\skew0\vec r}_0$ is the unperturbed position of the fluid element and the coefficients of velocity are related to those of position by

\begin{displaymath}
v_{lm} = \dot r_{lm}
\end{displaymath}

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 $r_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R_0$. That produces
\begin{displaymath}
R(\theta_0,\phi_0) = R_0 + \delta(\theta_0,\phi_0)
\qqua...
...elta = \sum_{l,m} r_{lm}(t) l
\bar Y_l^m(\theta_0,\phi_0) %
\end{displaymath} (A.266)

To be sure, in the analysis $\delta$ will be defined to be the radial surface displacement as a function of the physical angles $\theta$ and $\phi$. However, the difference between physical and unperturbed angles can be ignored because the perturbations are assumed to be infinitesimal.

The kinetic energy is defined by

\begin{displaymath}
T = \int {\textstyle\frac{1}{2}} \rho_m \vec v\cdot\vec v {\,\rm d}^3 {\skew0\vec r}_0
\end{displaymath}

Putting in the expression for the velocity field in terms of the $r_{lm}$ position coefficients gives

\begin{displaymath}
T = \int {\textstyle\frac{1}{2}} \rho_m
\sum_{l,m} \dot ...
...ne l}}^{{\underline m}}\right)
{\,\rm d}^3 {\skew0\vec r}_0
\end{displaymath}

To simplify this, a theorem is useful. If any two functions $F$ and $G$ are solutions of the Laplace equation, then the integral of their gradients over the volume of a sphere can be simplified to an integral over the surface of that sphere:

\begin{displaymath}
\int_V \left(\nabla F\right) \cdot \left(\nabla G\right) {...
...0 =
\int_S F \frac{\partial G}{\partial r_0} {\,\rm d}S_0 %
\end{displaymath} (A.267)

The first equality is true because the first term obtained in differentiating out the product $F\cdot\nabla{G}$ is the left hand side, while the second term is zero because $G$ satisfies the Laplace equation. The second equality is the divergence theorem applied to the sphere. Further, the surface element of a sphere is in spherical coordinates:

\begin{displaymath}
{\rm d}S_0 = R_0^2 \sin\theta_0 {\,\rm d}\theta_0{\rm d}\phi_0
\end{displaymath}

Applying these results to the integral for the kinetic energy, noting that $r_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R_0$ on the surface of the droplet, gives

\begin{displaymath}
T = {\textstyle\frac{1}{2}}\rho_m \sum_{l,m}\sum_{{\underl...
...}^{{\underline m}} \sin\theta_0{\,\rm d}\theta_0{\rm d}\phi_0
\end{displaymath}

Now the spherical harmonics are orthonormal on the unit sphere; that means that the final integral is zero unless $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline l}$ and $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline m}$, and in that case the integral is one. Therefore, the final expression for the kinetic energy becomes
\begin{displaymath}
T = {\textstyle\frac{1}{2}} \rho_m R_0^3 \sum_{l,m} l \dot r_{lm}^2 %
\end{displaymath} (A.268)


A.42.3 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 $\theta$ and $\phi$ as $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R(\theta,\phi)$. An infinitesimal coordinate element ${\rm d}\theta{\rm d}\phi$ corresponds to a physical surface element that is approximately a parallelogram. Specifically, the sides of that parallelogram are

\begin{displaymath}
{\rm d}{\skew0\vec r}_1 = \frac{\partial{\skew0\vec r}_{\r...
...partial{\skew0\vec r}_{\rm surface}}{\partial\phi}{\rm d}\phi
\end{displaymath}

To get the surface area ${\rm d}{S}$, take a vectorial product of these two vectors and then the length of that. To work it out, note that in terms of the orthogonal unit vectors of a spherical coordinate system,

\begin{displaymath}
{\skew0\vec r}_{\rm surface} = {\hat\imath}_r R
\qquad
...
...\hat\imath}_r}{\partial\phi} = \sin\theta {\hat\imath}_{\phi}
\end{displaymath}

That way, the surface area works out to be

\begin{displaymath}
S = \int\int
\sqrt{1
+ \left(\frac{1}{R}\frac{\partial...
...ial\phi}\right)^2}
R^2 \sin\theta{\,\rm d}\theta{\rm d}\phi
\end{displaymath}

Multiply by the surface tension $\sigma$ and you have the potential energy due to 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

\begin{displaymath}
R=R_0+\delta
\end{displaymath}

where $\delta$ is the small deviation away from the radius $R_0$ of a perfect spherical drop. This can be substituted into the integral, and the integrand can then be expanded into a Taylor series in terms of $\delta$. That gives the potential energy $V_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sigma{S}$ as

\begin{eqnarray*}
V_s &=& \sigma \int\int R_0^2 \sin\theta{\,\rm d}\theta{\rm ...
...right)^2
\right]
R_0^2 \sin\theta{\,\rm d}\theta{\rm d}\phi
\end{eqnarray*}

where the final integral comes from expanding the square root and where terms of order of magnitude $\delta^3$ or less have been ignored. The first integral in the result can be ignored; it is the potential energy of the undeformed droplet, and only differences in potential energy are important. However, the second integral is one problem, and the final one another.

The second integral is first. Its problem is that if you plug in a valid approximate expression for $\delta$, you are still not going to get a valid approximate result for the integral. The radial deformation $\delta$ is both negative and positive over the surface of the cylinder, and if you integrate, the positive parts integrate away against the negative parts, and what you have left is mainly the errors.

Why is $\delta$ both positive and negative? Because the volume of the liquid must stay the same, and if $\delta$ was all positive, the volume would increase. The condition that the volume must remain the same means that

\begin{displaymath}
\frac{4\pi}{3}R_0^2 = \int\int\int r^2 \sin\theta{\,\rm d}...
...textstyle\frac{1}{3}} R^3 \sin\theta {\rm d}\theta{\rm d}\phi
\end{displaymath}

the first because of the expression for volume in spherical coordinates and the second from integrating out $r$. Writing again $R$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R_0+\delta$ and expanding in a Taylor series gives after rearranging

\begin{displaymath}
- \int\int R_0^2 \delta \sin\theta {\rm d}\theta{\rm d}\phi
= \int\int R_0 \delta^2 \sin\theta {\rm d}\theta{\rm d}\phi
\end{displaymath}

where the integral of $\delta^3$ has been ignored. Now the integral in the left hand side is essentially the one needed in the potential energy. According to this equation, it can be replaced by the integral in the right hand side. And that one can be accurately evaluated using an approximate expression for $\delta$: since the integrand is all positive, there is no cancellation that leaves only the errors. Put more precisely, if the used expression for $\delta$ has an error of order $\delta^2$, direct evaluation of the integral in the left hand side gives an unacceptable error of order $\delta^2$, but evaluation of the integral in the right hand side gives an acceptable error of order $\delta^3$.

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

\begin{eqnarray*}
V_s &=& V_{s,0} - \sigma \int\int \delta^2 \sin\theta{\,\rm ...
...right)^2
\right]
R_0^2 \sin\theta{\,\rm d}\theta{\rm d}\phi
\end{eqnarray*}

Now $\delta$ can be written in terms of the spherical harmonics defined in the previous subsection as

\begin{displaymath}
\delta = \sum_{l,m} \delta_{lm} \bar Y_l^m
\end{displaymath}

where the $\delta_{lm}$ are time dependent coefficients still to be found. If this is substituted into the expression for the potential, the first integral is similar to the one encountered in the previous subsection; it is given by the orthonormality of the spherical harmonics. However, the final term involves an integral of the form

\begin{displaymath}
I = \int\int
\left[
\frac{\partial \bar Y_l^m}{\partia...
...{\partial\phi}
\right] \sin\theta{\,\rm d}\theta{\rm d}\phi
\end{displaymath}

This integral can be simplified by using the same theorem (A.267) used earlier for the kinetic energy. Just take $F$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r^l\bar{Y}_l^m$ and $G$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r^{\underline l}\bar{Y}_{{\underline l}}^{{\underline m}}$ and integrate over a sphere of unit radius. The theorem then produces an equality between a volume integral and a surface one. The surface integral can be evaluated using the orthonormality of the spherical harmonics. The volume integral can be integrated explicitly in the radial direction to produce a multiple of $I$ above and a second term that can once more be evaluated using the orthonormality of the spherical harmonics. It is then seen that $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 unless $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline l}$ and $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline m}$ and then $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l(l+1)$.

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

\begin{displaymath}
V_s = V_{s,0} +
\sum_{l,m} {\textstyle\frac{1}{2}} (l-1)(l+2) \sigma \delta_{lm}^2 %
\end{displaymath} (A.269)


A.42.4 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 $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R_0$ to the slightly perturbed value $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R_0+\delta$. The change in potential energy must be accurate to the order of magnitude of $\delta^2$.

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 $\delta'$ that is gradually increased from zero to $\delta$. Imagine that you start with the perfect sphere and cumulatively add thin layers of charged liquid ${\rm d}\delta'$ until the full surface perturbation $\delta$ is achieved. (With adding negative amounts ${\rm d}\delta'$ understood as removing charged liquid. At each stage, just as much liquid is removed as added, so that the total volume of liquid stays the same.) The change in potential energy due to addition of an infinitesimal amount of charge equals the amount of charge times the surface potential at the point where it is added.

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

\begin{displaymath}
\delta' = \sum_{l,m} \delta'_{lm} \bar Y_l^m
\qquad
{\rm d}\delta' = \sum_{l,m} {\rm d}\delta'_{lm} \bar Y_l^m
\end{displaymath}

The amount of charge added per unit solid angle ${\rm d}\Omega$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sin\theta{\rm d}\theta{\rm d}\phi$ will be called $\gamma'$. It is given in terms of the charge density $\rho_c$ and $\delta'$ as

\begin{displaymath}
\gamma' = \rho_c \left({\textstyle\frac{1}{3}} (R_0+\delta')^3 -{\textstyle\frac{1}{3}} R_0^3\right)
\end{displaymath}

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

\begin{displaymath}
{\rm d}\gamma' \sim \rho_c R_0^2 {\rm d}\delta'
\end{displaymath}

However, if ${\rm d}\gamma$ is written in terms of spherical harmonics,

\begin{displaymath}
{\rm d}\gamma' = \sum_{l,m} {\rm d}\gamma'_{lm} \bar Y_l^m...
...uad {\rm d}\gamma'_{lm} \sim \rho_c R_0^2 {\rm d}\delta'_{lm}
\end{displaymath}

then the coefficient ${\rm d}\gamma'_{00}$ is zero exactly, because the net volume, and hence the net charge, remains unchanged during the build-up process. (The spherical harmonic $Y_0^0$ is independent of angular position and gives the average; the average charge added must be zero if the net charge does not change.) The coefficient ${\rm d}\delta'_{00}$ is zero to good approximation, but not exactly.

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 $\delta'$ of liquid that has been laid down, (the removed liquid here counting as negative charge having been laid down.) For the spherical drop, to the needed accuracy

\begin{displaymath}
V_{0,\rm surface} \sim \frac{Ze}{4\pi\epsilon_0(R_0+\delta...
...e}{4\pi\epsilon_0R_0} - \frac{Ze}{4\pi\epsilon_0R_0^2}\delta'
\end{displaymath}

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 $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R_0$. The potential inside the shell can always be written in the form

\begin{displaymath}
V_{1,\rm inside} = \sum_{l,m} V_{lm} \frac{r^l}{R_0^l} \bar Y_l^m
\end{displaymath}

though the coefficients $V_{lm}$ are still unknown. The potential outside the shell takes the form

\begin{displaymath}
V_{1,\rm outside} = \sum_{l,m} V_{lm} \frac{R_0^{l+1}}{r^{l+1}} \bar Y_l^m
\end{displaymath}

where the coefficients $V_{lm}$ are approximately the same as those inside the shell because the shell is too thin for the potential to vary significantly across it.

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 $V_{lm}$. First, integrate Maxwell’s first equation over a small surface element of the shell. Since the shell has thickness $\delta'$, you get

\begin{displaymath}
\rho_c \delta' {\,\rm d}S =
({\cal E}_{r,\rm immediately\ outside} - {\cal E}_{r,\rm immediately\ inside}) {\,\rm d}S
\end{displaymath}

where ${\rm d}{S}$ is the area of the shell element. Note that ${\cal E}_r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\partial{V_1}$$\raisebox{.5pt}{$/$}$$\partial{r}$, and substitute in the inside and outside expressions for $V_1$ above, differentiated with respect to $r$ and evaluated at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R_0$. That gives the $V_{lm}$ and then

\begin{displaymath}
V_{1,\rm surface} = \sum_{l,m}
\frac{\rho_cR_0}{(2l+1)\e...
..._{lm}\bar Y_l^m
\qquad \rho_c = \frac{Ze}{\frac43\pi R_0^3}
\end{displaymath}

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

\begin{displaymath}
{\rm d}V_c =
\left[
\frac{Ze}{4\pi\epsilon_0R_0}
- \...
...\right]
{\rm d}\gamma' \sin\theta{\,\rm d}\theta{\rm d}\phi
\end{displaymath}

Substituting in $\delta'$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sum_{l,m}\delta'_{lm}\bar{Y}_l^m$ and ${\rm d}\gamma'$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sum_{l,m}{\rm d}\gamma'_{{\underline l}{\underline m}}\bar{Y}_{\underline l}^{\underline m}$, the integrals can be evaluated using the orthonormality of the spherical harmonics. In particular, the first term of the surface potential integrates away since it is independent of angular position, therefore proportional to $\bar{Y}_0^0$, and ${\rm d}\gamma'_{00}$ is zero. For the other terms, it is accurate enough to set ${\rm d}\gamma'_{lm}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\rho_cR_0^2{\,\rm d}\delta'_{lm}$ and then $\delta'$ can be integrated from zero to $\delta$ to give the Coulomb potential of the fully deformed sphere:
\begin{displaymath}
V_c = V_{c,0} - \sum_{l,m}
\frac{l-1}{2l+1}
\frac{Ze}{4\pi\epsilon_0}\rho_c\delta_{lm}^2 %
\end{displaymath} (A.270)


A.42.5 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 $r_{lm}$ as generalized coordinates, getting rid of the $\delta_{lm}$ in the potential energy terms using (A.266). But this is a bit of an overkill, since the only thing that the Lagrangian analysis really does is show that each coefficient $r_{lm}$ evolves completely independent of the rest.

If you are willing to take that for granted, just assume $r_{lm}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varepsilon\sin({\omega}t-\varphi)$ with $\varepsilon$ and $\varphi$ unimportant constants, and then equate the maximum kinetic energy to the maximum potential energy to get $\omega$. The result is

\begin{displaymath}
\omega^2 =
\frac{(l-1)l(l+2)}{3} \frac{C_s}{R_A^2m_{\rm ...
...2l+1}\frac{e^2}{4\pi\epsilon_0R_A^3m_{\rm p}} \frac{Z^2}{A^2}
\end{displaymath}

Note that this is zero if $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. There cannot be any vibration of a type $\delta$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta_{00}Y_0^0$ because that would be a uniform radial expansion or compression of the drop, and its volume must remain constant. The frequency is also zero for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. In that case, the potential energy does not change according to the derived expressions. If kinetic energy cannot be converted into potential energy, the droplet must keep moving. Indeed, solutions for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 describe that the droplet is translating at a constant speed without deformation. Vibrations occur for $l$ $\raisebox{-.5pt}{$\geqslant$}$ 2, and the most important ones are the ones with the lowest frequency, which means $l$ = 2.