▶️ Watch on 3Speak - Telegram - YouTube - Summary - Hive Notes - Sections playlist - MES Links --- In this video, I go over the derivation of the solid spherical harmonics, which are solutions to t...
Transcript
solid spherical harmonics, and again, there are solutions to the Laplace equation in spherical harmonics. So solid spherical harmonics are the solutions to the Laplace equation, spherical harmonics? Let's get that, I emphasize that. And I say solid because it includes the radial term, like these R's, there, so that even though solutions aren't a sphere, but the R changes. So the R, you can have it. So basically, it's a solid, it goes within and without, it goes, basically, the R is going to vary, so it's going to be, basically, if you have a sphere, it's the entire solid in and out, that's what solutions are for. And regular spherical harmonics are just on the sphere where you basically set R equals to 1, essentially. Now, but these are the general solutions. So, anyways, so Laplace equation is just the Laplace operator, or Laplacean set to equals zero. So all you do is take this and just set equals zero. So that's the Laplace equation. and you're dealing with harmonics because when it's equal to zero, it's in harmony, and nothing's in and out, it's always equal to zero, so that is that. So from the Laplocene and spherical coordinates, R theta-5, we can write Laplace equation as follows, and that's just basically this. I'm just going to copy our solution. So there's our solution. So we could write it, first I want to write this down using the Laplacian delta f sign, equal also equals using this inverted Nablo sign, or inverted triangle. It's the same thing as writing that, and this equals to zero. So in other words, we get, is this giant formula, right here, this right here equals to, or actually, I'll make this better, I'll move it down a bit. Yeah, this equals to zero. Yeah, so that's whole thing. I'll put it like that. I'll just put it like that, yeah. So that is what we get. So this whole thing equals to zero. And again, recall the spherical harmonics, just because I'll be putting this in section, so might as well just refresh it as well, because this video's pretty giant. So there's the x, y, axis, this is the z, and then we have an r like this, this is our distance, and this is the theta, and now if this goes down like that, that is going to be our horizontal distance, and this is the angle phi, and this is r theta. Phi, like that. And this is right angle, right. And also recall that, again, from above x is equal to r cosine, uh, cosine phi, sine theta. Y is equal to r sine phi, uh, Z is equal to, yeah, Z is just going to equal to, you look at this right here, R cosine, theta, and then R is equal to square root, just the distance formula, X squared, plus Y squared, plus Z squared. Yeah, so that's what we have there, and again, we have it all the way, or not, when I derived this right here, we already went over this, go all over back up. Yeah, so I already called, solved these ones for X. Y, Z, and R, so those are it. And there's the derivations, you could watch that earlier in the video, or in the separate video on Laplastian in spherical coordinates, which I'll upload as a separate video after I upload all these, after I finish this video, and so on. Yeah, so that's what we have. All right, and now that we have, basically, this Laplastian, the next step is we can apply the separation of variables to obtain. and basically we're going to basically try to find our function f, and we're going to assume that f could be written as a multiplication of three functions, that e, a function of, this is a phi, times it by p is a function of theta, and capital r is a function of r. So we're assuming that the function F could be written like this, and I'm using the terminology in the YouTube channel, the Mike guy, on YouTube, uses this terminology, I'll just follow that along. All right, so we have that. Now, next step is Geter Laplacian in spherical coordinates. I'm going to copy this. I'm going to go right here, paste. All right, so I'm going to move it like this. This is going to be paste here. All right, so we have this, and now the next step is, well, actually, before I even do that, I'm going to make it smaller. So make it smaller, and then multiply both sides by R. So, anyways, so now we're going to multiply this by R squared. Yes, so do that. All right, so if we multiply this out, we're going to get some cancellations here, the R squares cancel. That's why we're doing that. and this will cancel, or just have a left an r there, and this one's going to be an r square there. So in other words, we're going to get, so this becomes, we're going to get an r squared, is the first one, so r squared, and now, instead of writing partial derivative, a second partial derivative of f in terms of r, well, yeah, this right here, well, we'll just put the F inside. This just is the same thing, because these are functions, these are just functions of phi and theta, so that means that if we take the second derivative of F as a, in terms of R, so second partial derivative, these ones aren't included, so this just equals two, and I'll put our F now, this is going to be our F is going to be equal to E times p, times it by, R double prime. So this is just a, yeah, this is just a second derivative of R there, multiplied by these, these are treated as constants when we're taking partial derivative of F in terms of R. So we just set the other ones as constant. So put this R like that. So that is what that equals to. And now this part here, we're going to get a plus two, and this is the first partial derivative of F in terms of R, that's just R prime. So this 2R, R prime. and you actually just don't make a mistake so it's going to be e times p yes so e times p times r prime because those are going to have to include those constants for that partial derivative of yeah so they're not actually consulate but they only are when you take the partial derivative so now the next setup this is this one's gone so we're going to have a plus a one over sign Theta. The point of her sign, and also, before we go further, so the guy on YouTube, he does it this way. So first, so we would, instead of using this way, we'll take out, write it as the form, partial, part, yeah, partial of, divert over partial theta of, and this would just be, this just be sine, theta, partial F over partial theta. And you can see this is the case because this is just chain rule. So take the chain rule, so derivative of sine is going to be cosine theta, then times by partial F over partial theta, and then plus sine times derivative of this, right here. It's going to be a second partial derivative of F in terms of theta there. And again, that's just, again, if you go back to the Wikipedia set up here, so this giant thingy, the same thing as it's partial over partial theta of sine theta, partial f over partial theta like that. All right, and so the reason they do this, well, the YouTube channel, the mic guy, it's just a bit simpler, so you have less F over there. So now we're going to get a, and now it's going to be sine theta, and then times it by this one is a partial derivative of F over partial, yeah, it's basically a partial derivative of F over partial theta. That's just going to be the derivative here. We're going to write now E. and then E p prime, so we're going to get a times by E p prime times by r. So those are the constants now. The p is a derivative there, supposed to the R, like that. And, yeah, I've referred the partial derivative, so this will move it over. I'll move this over here, so this is going to be, and I'll put a square bracket, we need to include this part right here. So that's going to be partial over partial theta of this. Like that, yeah. Actually, we don't need to do a squared bracket, but we just keep it like that. Yeah, I thought we're going to use a second one anyway. So that's what that is. And then the last one is this part right here. That's just going to be plus. Put it all here, and then I'll shrink it. Yeah, this one's going to be, put a plus like that. This is 1 over sine squared theta. and now this is the second derivative, so everything else constant, this is of the phi, so that just equals to E, Phi, double prime, and then we have PR, and then the whole thing is equal to zero. Yeah, all right, so here's the strunk it, and we have it all over here, so now the next step is, then this is gonna be like this, what I'll do is, I'll put it like this, next step is gonna be, we're gonna divide every single thing out by, I'm just going to, instead of putting it there, I'm going to put it like the whole thing. Actually, I'll just do it over here. So now we multiply the whole thing by 1 over, 1 over E, yeah, EPR. So we get fractions on each thing, and we cancel all these extra terms. So that's why we do that. And here is quickly shrunk everything in here. So we have all that, so then this ends up equaling to, so we're going to get an R squared. and then EP cancel, so then these cancels with this, so again, that's the whole idea, is you just cancel everything out, so we're going to get an E-p, e-dotp, and then over R, yeah, this E-P, oh, no, EP is cancelled, yeah, I mean it's cancelled, so we're going to get an R-prime, and then over R, so R is not cancelled, so we have that over there, so R over R, and then likely, likewise, we're going to get an R2R squared, E-P, So basically all the ones that the R is not there, so EP is gone, so we get an R prime over R. Yeah, this is over R, so in other words, we can combine, these are EP, those cancel, put these all together, actually. So this will be a plus 2R, R prime, over R. And now the next setup is going to be 1 over sine theta. and then we have, there's one over sine theta, and then this one here, these are constants, so it doesn't matter, we could take those out, the E P is our constants, and then those cancel out, they're constants within this partial derivative, so those just cancel out, we're gonna get a partial, over partial theta of sine, and then E, E's gone, so now we just have a P prime, and then the only thing divided by is gonna be this P, like that. Yeah, just divide that whole thing there. We just put it over everything. Like this, over P. All right, here, just fix that up. And then lastly, we're going to get a plus, this one right here, 1 over sine squared theta. And now this one here, the E, the P's are gone. So we're just going to get an E double prime over E like that. Just put it like that. All right, so that's what you have. we could yeah divide it out or just leave it like that so anyways e all primary and this whole thing equals to zero yeah because then we multiply the zero one doesn't change anything so just zero all right and here is a strunk and move it over all into one line here so now the next step is well we can now set the equations equal to their corresponding uh separation constants and again these all equal to zero So in other words, these are the E functions are in here, there's the P1s in here, and this R1s aren't there. So, yeah, these are three separated functions, but they all equal 0 at all times. In other words, these are all equal constants. So this one's a constant, constant, etc. This one's a constant, this one's a constant, these are all C1, I'll say, C2, etc. So these are all constants. So in other words, we can just set these all equal to their own corresponding separation constants. and to ensure periodic solution with the middle function to be set as the difference between the constants, thus ensuring the overall equation sums zero. So basically, we just choose this one as a constant, this one as a constant, and this one as the difference, yeah, and then this one as the difference between the two. All right, and the constants specifically picked like this, so for the first one, we'll look at this part right here. So we'll just look at the, in this one here's, Yeah, yeah, this one, typically is written just like these two parts right here, so we'll ignore the sign squared for now. So basically, this one is written like this, E double prime of E, and this one is, these are both functions of phi, remember that, so E is a function of phi, going back here, so he's a function of phi, and then this is over a function of phi. Yeah, so now this one's equal to negative m square, and we'll go over that in a bit, and now the next one is this part right here. So this whole setup is, this is going to equal to R squared, R prime, and this is going to be, again, its function of R. Let's put them all just for a minus, and then plus 2R, and then this is going to be R, yeah, R prime. The other one's R double prime, and this is going to be a function of R. divided by R, and for completeness is this function of R, and now this equals to, the function that this one set up as is equal to n, then n plus 1, and the reason is so we get the associated legeron, legera function. Yeah, all right, we'll explain this in a bit, right? All right, here, it's really shrunk that in. So if we have these set to these constants, and then this one here is going to be the difference between the two, or just, yeah, this one equals to the negative of that. Yeah, so that equals to, let's put this whole thing there. Actually, I'll just copy and paste this whole thing. And just leave it like that. So I'm going to copy this paste there. Yeah, this one here. And move it here. Yeah, so this one here, this just equals to the negative of those. So it's going to be equal to, so it's a negative, well, we'll put this negative here there first. So we'll take the end. n plus 1, so put a negative n plus 1, and then the next one's going to be, well, minus, but it needs to all equals 0, so we have to include this 1 over sine squared, because this one is the M, is the negative m. So basically, it's going to be negative m, but then we're negative that, so it's going to be positive m. Positive m squared over sine squared, like that, sine squared theta. All right, so we have that, so thus, and again, the reason is so that thus, we'll have on this, this left side here is going to be n plus 1, so this is so they all equal to 0. N plus 1, now we're going to have the middle term, it's going to be negative n, n plus 1, then plus m squared over sine squared theta. So again, this is the middle term, like that, let's put it like this. I'll just leave it like that, this the middle one, and then the last one is just going to be this part right here, 1 over sine squared. and be minus, because it's one that's going to be negative m squared. So that's going to be negative m squared. So negative m squared over sine squared. Yeah, so it equals here. So again, these cancel, these cancels. Again, that's the whole idea. That's the whole idea of it. All right, so now we have this. So just go over some notes here, and also solving all of this equation. So the first constant, negative m squared, where m is an integer, is selected such that we have a periodic solution. So this one right here. So if we take this and write it down, so if we have E to the Phi over a double prime of five, the second derivative, over phi, I mean, E to the phi equals to negative m squared. What happens here now, we could write this as, we'll multiply this over to this side, and then drag it over, what we end up have, yes, move it to there, then drag it over, we get an equation like this. We have an E double prime, so we have, again, partial differential. I think it's just a differential equation or whatnot. It's going to be, this is going to be now m squared e phi, this equals to zero. All right, now the way to solve these, and I haven't done too much of this in my actual video series, but I'll probably do that for later in the future, but nonetheless to solve this, what we'll do is assume, yeah, so assume that E of phi, so you can find a solution that's an exponential equals E to the R-fi. So in other words, we're actually, instead of using the R, because we already have R for here, I'll just call it A then, E to the A, phi. So now, if we have the solution is like this, so assume that that is a solution, and we can actually prove that it is, then find A. So we just have to find this A variable or constant or whatnot. So anyways, all right, so we have this, well, if we take the derivative of this, derivative E to the E prime of phi, this equals to, well, again, just the derivative of exponential function is just itself. So it's going to be, well, and also, but then you put the A down. That's afterwards. So this is going to be, well, E to the A, phi, and then do the chain rule derivative of this A5. It's going to be A. Like that. Yeah, because derivative, yeah, the derivative of phi would just be one, and there's just the constant there. So that is that. Now, the second derivative is going to be a double prime of phi. We do the same exact thing. So this is a constant, doesn't change anything. Just leave it A, then E, a, uh, Phi chain rule, put that down, put the A down, so squared. So that's what we have there. So thus, so that's what we get now is, uh, now we'll take this entire thing. So E double prime, yeah, double prime of phi, that equals to A squared like that. So A squared, E, A, phi. And then this one here, M squared, E of phi, equals zero. Plus, m squared, and then we'll have a E to the, this is just going to be, yeah, this is E, A, phi, equals zero. Alright, now the next step is, well, I'm going to divide both sides, and the right side 0 doesn't change anything, E, a phi. So then, and what ends up happening is these just cancel. So that's all that happens. So we're going to get a a squared plus m squared equals 0. Yes, in other words, we get A, yeah, A is going to be, well, let's say, yeah, a equals to, This is going to be moving over to the other side, so we're going to get a negative m squared, and then yes, be squared, negative m squared, yeah. Actually, yeah, before we even write that, I'll just write this down. We're going to get an a squared equals to negative m squared, and the only way to do that is using complex numbers. So in other words, we're going to get an a equal, and then this equals to plus or minus squared, because again, you can have minus times minus, it's going to be positive, and plus positive times positive to be minus. So this is going to be negative m like this, yeah. This is going to be m squared. Yeah, so in other words, this equals to, well, keep going, plus or minus, negative 1, yeah, it's negative 1. M. Yeah, this is square root of that. This is just i. Yeah, this is i equals i. So, imaginary number i, which is defined as a square root of negative 1. So in other words, we get a is equal to plus or minus i. Yeah, so thus, so thus we have the solution E of phi, and this is a solution. So E of phi is equal to E, and then this is going to be, yeah, this is going to be E, A, which is plus or minus. So A is this right here, E, A, phi, it's going to be E, plus or minus, phi. So there it is. So we replaced, that's what A equals, and then we just replace this A with IM Phi. Yeah, and with IM, I mean, plus or minus I am. So that is the solution. That is actually just one solution, but then, or now a general solution. General solution is actually a linear combination of the two. Because this is plus or minus, so the general solution is, you can actually, this is a linear combination of this, we're going to get, E a phi is equal to, I'll put a C1, as a constant, E to the power of positive I m phi, plus C2, E to power of negative I am phi. So basically you take a linear combination of the two solutions, so that's plus or minus that, and then, yeah, and then I'll just box this in, so we box this in, so fascinating stuff. Alright, so now we have that going further. So again, yes, you want to see to our content. So note that the general solution allows for writing every possible solution as a linear combination of two basis functions. And this complex exponential function right here, again, that's complex exponential equivalent to trigonomic…