▶️ Watch on 3Speak - Telegram - YouTube - Summary - Hive Notes - Sections playlist - MES Links --- In this video, I go over the actual spherical harmonics equations, which exclude the radial terms ...
Transcript
spherical harmonics, which is the solution to the angular Laplacian, but again, we pretty much have that already derived. I'll just discuss it, and so on. So the solid spherical harmonics includes the radial terms R. So you can see these R there, but there's also R everywhere here, so if we get rid of those when we do the angular or Laplacian. So while the spherical harmonics in general, typically are just in terms of the angular terms theta and phi. So there's the theta and there's the phi. and so on, so if we recall, so this is the Laplacian, and then here's the solution, F of R, of theta, and phi equals to E, phi, R of R, and then P theta, and this equals to this giant thing in here, there's the associated Lezondra functions. We have these two power ones, this is the E, this is the R, this is the P, and that is just this one here, I just got Grock A, I just do, write it in a formula form, so this one right here, and we got this exactly the same setup there. So we can combine the radial terms as a function Y. So that's typically what's written. So if we write F of R and Theta and Phi, this equals to, we'll call this R of R times it by Y. So we just combine those two together. Or we'll put it all together first, just to show you, just for completeness. So here's the R of R. p theta e of phi, so this equals to combine these into y of theta and phi. So this equals to R of R, and then y of, this is going to be y of theta and phi. So in other words, so this one is just going to be this one. Actually, no, not the R, it's going to be just this one. and this one. So that's going to be the solution. So in other words, we're going to have y of theta and phi. This will just equal two, and this will just be just this first part. So C1, C1, E to the I am phi, plus C2, E to the negative I am phi. and then times it by the associate leisandra function p n of m of cosine theta so that is just that so that is this is very much the spherical harmonics which excludes the rr function there so we're not counting for that is essentially we could graph these on any sphere just to assume the radius is r equals to one All right, yeah, so that's pretty much it, but just for more info and contact. So this Y, this function Y, is typically considered as the solution to the angular, to the angular Laplasian or Laplace Bell-Trami operator on the sphere, which can be denoted by this one, this wedge squared, or denoted as this one right here, Laplacian S squared. That's interesting, but this one is the typical one there that I like using. All right, yeah, so the next setup is, we'll go over right here, because again, this is Laplacian, angle Laplacian, but then there's an R squared, there we got to get rid of that. So I'm going to do is I'm going to copy this, and then paste it here, just to save some time. Yes, we have this, and now I'm going to go times it by R squared. At times the whole thing by R squared, so we basically cancel out this, cancel this out, and we're going to get, now we're going to get an R squared of this Laplacean equals to partial over partial r, and then r squared, partial F over partial R, and then plus 1 over sine Theta, then partial over partial theta, then sine, then partial F over partial theta, and then plus 1 over sine squared theta. partial, yeah, the second partial derivative of f in terms of phi like that, this equals zero. And again, just going over just typically how this angular, the spherical harmonics are written as there's solutions of it, because when you exclude the R term. So we have this, yes, we have this, and now note this whole side here, this one doesn't have any R terms, this one has R, this one doesn't have any R terms, and now this is the angular Laplocyan. This is going to be written with that wedge set up there, and this is of F. So that is their angular Laplacian. So we could just write it all like that, then, so we can go partial over partial R of R squared, partial F over partial R. Yeah, we have that then plus, this is the Laplacian, F equals to 0. And while we know that F, so F is going to be equal to our new setup now, F equals to R, is the number RPE, but then we're going to just change this into R. and y, and y. So that's what we're going to write it as, because again, we know that R is just, yeah, F is this function of, that's, yeah, F is a function of R, theta, and it equals to R, um, times it by, yeah, Y, we're just going to use this y equals to this, uh, p times theta, I mean, p times e. So that's what we have there, and also to get this, uh, partial derivative, well, we know that, um, this is going to be now the partial derivative of f. in terms of R, this equals to, well, these all gets treated as constants, this and, well, the Y. So in other words, that's just equals to as if it's nothing there, it's just y times R prime. Yeah, R prime, I got. Yeah, which is just, yeah, Y, D, P, D R over, over DR, I got. So what we get is, yeah, so note, I'll put this note in here. Thus, what we get is, now this is going to be derivative, D over DR, because we don't need to, because it's all just terms of R. And this one's like a concept, we just take it all out, just take the Y out. Just take the Y out and move it. And then we have the inside, it's going to be R squared, and then this is going to be, yeah, DR over R, like that. And then plus the Laplacian, well, there's no R terms in here, so we can just take that out. Yeah, because it could just be whenever they're inside the F, we just factor it out as if it's a constant, that's just going to be RY. So this is going to be R. Laplacian of y equals to 0. Next set up is, we'll divide both sides by RY, as we did before. Again, we already solved this in the other format and solved spherical marks. We divide it out, we're going to get a D over DR of R squared dR over R. and this is 1 over R, like that. Plus, then this R cancels, we're going to get a, yeah, we're going to get a Y, and the Laplacian over Y, over Y. Yes, we're going to get Laplacian of Y divided by Y, equals to zero. And now, typically, yeah, what you're due is, the exact same as thing as before, set this equal to a constant, because none of these are, these are only dealing with y terms, so the angular theta or phi. This one's only dealing with R. So then, but if it all is zero, then in other words, they have to both be constant. This one is chosen to be, this equals to n times n plus 1, as we did before in the Euler account G formula set up. This one's going to equal to negative n times n plus 1. All right, and again, if we scroll all the way back up, to the Euler-Kaushi formula setup, that is this part right here, it's exactly the same thing we did right here. So we had this equation, this is the R squared R double prime, it's the same thing we have, we just do apply chain rule, and we set it equals to n times n plus 1, and then we ended up getting this whole setup here, this right here, C3RN plus C4R, power of n minus 1, power of negative n plus 1, and that's going to be, yeah, this is set up here. So that's this, like that. That's what we do. And got that, and again, I remember that before, it was just a chain rule, you just apply here. So if you take the derivative of this, R squared, it's going to be R squared, and then R double prime, and then plus 2R, and then R prime. So again, that's the same thing. But we're only worried about this, we're only focused on this one, because typically how it's written. I'm only going over this, just because of the context of how it's written, like, why is the solution of this angular little Placian setup? So we get this part right here, the angular of y, divided by y, equals to negative n times n plus 1. Or you could write it as, multiply this over there, then move it over to this other side, so we get the typical equation involving equaling the 0. So we're going to get a laplasian of y plus n times n plus 1. times y equals 0. So that is typically how it's written. And then Y is a solution. Yeah, why is the solution to this, where the angular laplocene of y is equal to this setup. But then we just, that's angular plosin of F. So we copy this and change the Y, the F2 and a Y. And paste this. All right, so that's what we got, and then this becomes a Y, a Y, fascinating stuff. Yeah, and here I'm just going to circle this whole thing, so it's pretty, should I circle it or not, but anyways, put it like that. Yeah, actually here, I circled it, yeah, so that's that, so that's that, so because how it's written, and then where Y is it can be a solution to it, and I'll just circle this right here. So, and then equals to that. I'll say Y is a solution. Y is a solution. Alright, yeah, and here I just typed out and said, so Y is a solution to the above equation involving the angular or Laplacian, and maybe, this might be capital L, maybe. Anyways, and then we have y of theta n phi equals to this setup right here. There's the C1, a constant, e to the power of iM5 plus c2, e to power of I.m, a negative I.m. 5 plus, and then times it by the... associated lezondro function p n of m uh p n of m uh p n of cosine theta so yes a fascinating fascinating something this is this one actually i'm not going to circle this uh that's just the yeah that's the angular of laplacian there this is the solution we solve i mean this is the equation we have to solve and that's what we get