Question

Help requested to solve a question in electrostatic potential and spherical harmonics.

Answer 1

The question is shown enclosed ...
To a mathematician, I suspect this is a snip ... and it's the C4,4 and C4,-4 bits I want

Answer 2

The file hasn't come up for me ;-;

Answer 3

are you sure that a browser didn't block it as "potentially dangerous" download or something. It still downloads for me.
@osprey I've been trying to get that answer for a while now and I can't even get to the coefficients of 4, 4 because I get a solution claiming that only odd integers m exist and the others are 0. I'm not saying the solution is wrong, I'm probably messing up, it's been a while. Just saying you're not the only one having issues solving the task.

Answer 4

It just came up as blank.

Answer 5

try this one.

Answer 6

None work ;-;
Sorry

Answer 7

Hi ... there should be two pages to the presentation with the usual pile of algebra. the "show" may open in design mode, and usually F5 puts it into show mode. With luck, though, here, it doesn't matter.
@ljetibo To be a moron, someone has to be at the very left hand side of the so called IQ normal distribution. Basically, it's either the moron or the idiot that is dam' nearly totally dysfunctional. I think I saw some film footage of some morons in spain, europe some years ago. A very, very sad sight. So, you probably aren't a moron.
@Unavailabilityy Do I have to upload the thing again, or has it come through ?
Bon voyage to both of you ...

Answer 8

Before I start I've got to say, I don't know what that eq in your answers is supposed to represent, frankly it looks to me as a potential of a single charge (because of the first member that has just the one q) but is missing the 1/R in it. The radial part of the second part of the equation resembles the equation for the interior spherical multipole moment but then the spherical harmonics only depend on the angle theta but both of them have different indices A and B (marking different theta's in question?). Anyhow, I couldn't make heads or tails out of that eq. so I decided to start from the top.

Considering I have not done EM in a while I look up the formulas for multipole expansion in spherical coordinates for a general case and I found the following equations (below). As I said, considering I've not done EM or integrals like this I've not managed to solve it quite right, I'm missing a factor or two here and there because I'm out of practice and prone to stupid mistakes. The procedure seems to be correct though, because it produces the result (minus some of the constants as I said).

$$\Phi = \frac{1}{4\pi\epsilon_0}\sum_{l=0}^\infty\sum_{m=-l}^l\frac{4\pi}{2l+1}q_{lm}r^{-l-1}Y_{lm}(\theta,\phi) \\
q_{lm} = \int Y_{lm}^*(\theta ', \phi ')r^{l\prime}\rho(x')dx'$$
the prime variables (as in x', r'...) are not derivations, I just used the prime symbol to mark the variables that are being integrated over to distinguish them from i.e. a constant I might use that has the same symbol later on. The x' is implied to be the infinitesimal "chunk" of volume we're integrating over.

Since this is a formula for the most general case for a continuous distribution of charge we need to write our distribution of charges such that it mimics the point charge configuration given to us by the task. I just naively constructed a discrete point source distribution out of a continuous one by using dirac delta function as:

$$\rho(r, \theta, \phi) = \frac{q}{R^2}\delta(r-R)\delta(\cos\theta) \left[ \delta(\phi)+\delta(\phi-\frac{3\pi}{2})+ \delta(\phi-\pi)+\delta(\phi-\frac{\pi}{2})\right] \\
+\frac{q}{2\pi R^2}\delta(r-R)\left[\delta(\cos(\theta)-1)+\delta(\cos(\theta)+1)\right]$$
Because multipole expansion can depend on the choice of origin and coordinate system I have to note that I went with the coordinate system the task proposed. Origin is in the point of inversion symmetry with 4 charges lying in the xy plane each one lined up with an axis and two remaining charges are located on the z axis just as the task suggested. The only difference I made is that I'm using the azimuth angle theta from xy plane upwards and not z downwards as in your sketch (so that xy plane is in 0 theta to ease the calculations)

We can begin the integration now, but before going further notice that we can also separate the spherical harmonic $Y_{lm}$
$$Y_{lm}=\Phi_{m_l}(\phi)\Theta_{m_l}(\theta)\\
\Phi_{m_l}(\phi)=\frac{1}{\sqrt{2\pi}}e^{-im_l\phi}\\
\Theta_{m_l}(\theta)=\left[ \frac{2l+1}{2} \frac{(l-m_l)!}{(l+m_l)!} \right]^\frac{1}{2}P_l^m(\theta)$$
where $P_m^l$ is the Legendre polynomial. Back to integration:
$$q_{lm} = \int Y_{lm}^*(\theta ', \phi ')r'^l\rho(x')dx' = \int_0^R\int_0^{2\pi}\int_0^\pi Y_{lm}^*(\theta', \phi') r'^l \rho(r', \theta', \phi') r'^2sin\theta'dr'd\theta'd\phi'$$
We can see that once we input our charge density function in (rho) we can split the integral into two: one for the xy plane and one for the two z charges. For the xy plane:

$$I_1=\frac{qR^l}{\sqrt{2\pi}}\left[ \frac{2l+1}{2} \frac{(l-m_l)!}{(l+m_l)!} \right]^\frac{1}{2}P_l^m(0)\int_0^{2\pi}e^{-im\phi'}\left[ \delta(\phi')+\delta(\phi'-\frac{3\pi}{2})+ \\
+\delta(\phi'-\pi)+\delta(\phi'-\frac{\pi}{2})\right]d\phi'$$
since the integral is in the $xy$ plane which sets our $\theta$ and gets rid of the $\delta(\cos\theta)$ and only leave 1 solution for theta: that when theta=0 (xy plane). The $\Theta_{m_l}$ can get thrown in front of the integral as constant, as well as the $R$'s. The dirac delta (r-R) allows only for r=R solutions to survive so all r's become R's and go to front. Skipping a bit of calcs. I end up with:

$$I_1=\frac{qR^l}{\sqrt{2\pi}}\left[ \frac{2l+1}{2} \frac{(l-m_l)!}{(l+m_l)!} \right]^\frac{1}{2}P_l^m(0)\big( 1+i^m+(-1)^m+(-i)^m\big)$$
Noticing that $$(-i)^m = (-1)^mi^m$$ we can simplify:
$$I_1=\frac{qR^l}{\sqrt{2\pi}}\left[ \frac{2l+1}{2} \frac{(l-m_l)!}{(l+m_l)!} \right]^\frac{1}{2}P_l^m(0)\big(1+(-1)^m\big)\big(1+i^m\big)$$
From this point on we can see the solution to this part. The only allowed m's are even ones otherwise $$1+(-1)^m$$ will be 0, such that $$i^m = 1$$ because otherwise the $$1+i^m$$ would be zero. That rules out 2, 6, 10 and leaves us with 0, 4, 8, 12 etc... For the case of m=4, -4:
$$I_1=\frac{qR^l}{\sqrt{2\pi}}\left[ \frac{2l+1}{2} \frac{(l-m_l)!}{(l+m_l)!} \right]^\frac{1}{2}P_l^m(0)*4$$
We still have the second integral to solve.
The z plane (this is where I think I missed something stupid but can't exactly locate):
$$I_2 = R^{l+2}\int_0^{2\pi}\int_0^\pi Y_{lm}^*\sin\theta' \frac{q}{2\pi R^2}\left[\delta(\cos(\theta'-1))+\delta(\cos(\theta'+1))\right]d\theta'd\phi'=\\
= \frac{qR^l}{2\pi}\int_0^{2\pi}\int_0^\pi Y_{lm}^*\left[\delta(\cos(\theta')-1)\sin\theta'+\delta(\cos(\theta')+1)\sin\theta'\right]d\theta'd\phi'$$
Without even going as far as writing the full sph. harm. and doing the integral, I can see that because
$$\int_{-b}^b f(x)\delta(x-a) = f(a) $$
as long as -b < a < b (which is the case now) the integrals evaluate to zero. Both dirac deltas will only be 1 when sine is 0:
$$\cos\theta - 1 = 0 \\
\cos\theta = 1\\
\theta = \cos^{-1}1 = 0$$
and sin(0)=0. The case is the same with the other one except it's for pi. This is beyond weird for me. I had to make a mistake somewhere here already, am I miss-integrating with the dirac delta? Is it the source distribution formula? How can the two charges on z axis contribute nothing?

In any case I leave the error finding to you (with a hint in PS). We can continue on with solving the task. Now that we have the required multipole moments we need to send them back to the expansion formula.
$$\Phi = \frac{1}{4\pi\epsilon_0}\sum_{l=0}^\infty\sum_{m=-l}^l\frac{4\pi}{2l+1}q_{lm}r^{-l-1}Y_{lm}(\theta,\phi) \\
\Phi = \frac{1}{4\pi\epsilon_0}\sum_{l=0}^\infty\sum_{m=-l}^l\frac{qR^l}{\sqrt{2\pi}}\left[ \frac{2l+1}{2} \frac{(l-m_l)!}{(l+m_l)!} \right]^\frac{1}{2}P_l^m(0)*4*\frac{4\pi}{2l+1}q_{lm}r^{-l-1}Y_{lm}(\theta,\phi) $$
Task required us to show the expansion only for the lowest degrees (i.e. first two, the monopole and the next one (hexapole in this case?)). For l=0, m=0 we have:

$$\Phi = \frac{qr^0}{4\pi\epsilon_0R}\frac{1}{\sqrt{2\pi}}\left[ \frac{1}{2} \frac{1)!}{(1)!} \right]^\frac{1}{2}P_0^0(0)*4*4\pi Y_{00}(\theta,\phi) \\
\Phi = \frac{q}{4\pi\epsilon_0R}\frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt 2}*1*4*4\pi \frac{1}{2} \frac{1}{\sqrt\pi} \\
\Phi = \frac{4q}{4\pi\epsilon_0R}$$

Which is obviously wrong, since I'm missing two charges. But I think that if you're more careful with the integration for xy plane you might find the missing two charges. The next things to check are lm= 4, 4 and 4,-4 but you won't get the same constants as he does (i.e. just check out the list on wiki). It's easy to see, from my solution, that potential dies off as r^4/R^5 which is good, but you won't get the same constants (btw, honestly see the spher. harm on wiki, I can't imagine how he would get something like that from them without using some other convention on writing them [read the book prologue it usually says there]).

P.S.
From the point when I setup the coordinate system. It's easy to see that in spherical coordinates the distance of the charges from the origin is R - therefore dirac delta(r-R). Four charges in xy plane are found by setting the azimuthal angle to 0 in the first integral. I did this without thinking through it a bunch. This sets the theta angle up as if its measured from xy up and not z down. By that logic the z axis can't be found when cos is 1, -1 (angles 0 and pi) but instead when cos is 0 (angles pi/2 and 3pi/2 (or pi/2)). In this case the second integral (for the z axis) doesn't disappear as it did for me, but stays around. With a bit of luck and ingenuity you'll see how you'd solve it. If my hunch is right, the two extra members should not affect anything except the case when lm=0, 0 (they are the missing two charges for that case).

In any case not a particularly well suited task to get to know multipole expansion. It's more of a mathematical practice than anything. I planned on finishing the task today correctly but I have'nt got the time.

Answer 9

@ljetibo Very many thanks for your post. I'd be interested to know how long it took you to do that post. The problem comes from a UK text book on electricity and magnetism published by the "The Clarendon Press" at Oxford University. When I've recovered from the pleasant shock/surprise of seeing it, I'll post again.
Many thanks again.

Answer 10

@osprey I have quite a bit of practice with writing in latex so it wasn't that bad. Still took a while, a lot of indices and little things to get right with sup/superscripts. Not to proud about the clarity of some sentences either. You'll see that the integrals are setup so that they cancel out except for just couple of points which was a blessing.

Answer 11

ok, I guess the next question, aside from the actual question that is, is what is "latex" ? I use basic powerpoint and it is sort of ok, but I guess that squiggly symbols carry a penalty if you want to key them into a computer program. The facilities on this site though, I think are abysmal, or at least they are with the browser I'm using. I say abysmal relative to the details we're trying to get up on the screen.

Answer 12

latex is a typesetting system. People use it to write articles, books, etc...

It's got a lot of operating modes, one of which is called math mode. It's very useful for typesetting equations. MathJax is that system implemented in JavaScript so it works on webpages. You enter math mode by writing two dollar signs. (without spaces between them.). Every math environment has to be closed by two dollar signs too.
$ $ write math $ $
You can write math in it using commands (lot of them, read a bit online). Just some for example:
exponent: a^{exp}
index: a_{index}
vector: \vec{a}
fractions: \frac{top}{bot}
integrals: \int or with limits: \int_{bot}^{top}
sums are same as integrals....

There's a whole selection of letters and symbols too. Basically all the math you see in books was probably done like this.

Answer 13

I've tacked another question on to the electrostatics question. Anyone with any thoughts/expertise please pitch in ...
Grazie.

Answer 14

I have a bramble bush or two outside. Some of the Bs are loooonnnnnnggggggg. NOW, which is easier, to tangle with maths/algebra/science/technology OR to jump into a B bush and then TRY to get out. Or should I ask a "mouse in a maze" the same question ?
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