Sub­sec­tions


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 $R_0$ 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.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 ${\rm d}{S}$ 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)
\qquad
\delta = \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 r_...
...line 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_{{\underli...
...}}^{{\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.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 $\theta$ 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}_{\rm...
...\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
\f...
...{\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 R}...
...rtial\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 d...
...i}\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}r...
...\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 d...
...i}\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}{\partial\t...
...}}{\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.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 $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R_0$ 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
\qquad {\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'...
...Ze}{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)\eps...
...a'_{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}
- \frac...
...
\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.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 $r_{lm}$ 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 p}...
...{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.