 We have enslaved a hot chain rule to recover this formula here, so what we are trying to do is find the wave functions that represent the states of well defined orbital angular momentum. I explain what the strategy is for doing that and that strategy involves knowing what Ies â oes oedd yr waith cyfrindwyr sy'n golygu'r gwahanol a judacesio Cyfrindwyr yn y reliant am gyfrindwyr. Yn rhai fawr hwnneud ar gweithio ar y cyfrindwyr sunshine, yn ddyloch i fynd i ffordd gylir gael tua cyfrindwyr sy'n gylir gael tua cyfrindwyr sy'n gylir gael tua cyfrindwyr. Mae'n meddwl am y dyma'i ddau cyfrindwyr wedi gweld bethent i fynd y rai sy'n cerddoedd gael eu bobl mae'n gael môl yn gweld ar y peirwyr mewn gwaddodol. We are now in a position to find these wave functions. Let's call it a psi lm. This will be some function of r, theta, and phi, because we are looking at this in spherical polar coordinates. So this of course is r, theta, and phi lm. Now the radial dependence of this wave function is going to be completely unspecified because we are only going to require, all we're going to require is that lz on lm is equal to m lm. This thing has got to be an eigenfunction of this operator with the eigenvalue m, which we now know to be an integer. And we are going to require similarly that l squared on lm is equal to l l plus 1 of lm. We're not going to direct, and we will find, we haven't yet calculated this operator, what l squared looks like is a differential operator, we will get to that. But it will also turn out to involve only derivatives with respect to theta and phi. So these operators, none of them involve anything about radius, and so this function is an arbitrary function of radius and all we're going to be able to discover is what its angular dependence is by imposing these requirements here. OK, so what can we say? We can say, first of all, this equation is going to imply, put into the position representation, it says that minus i d by d phi of psi is equal to m of psi, which we, of course, immediately recognise is telling us that psi at r theta and phi is equal to e to the i m phi times of psi at r theta and nothing, if you see what I mean, and nothing, and nothing, and nothing. So there's some kind of a constant here, right? This thing doesn't depend on phi. If you differentiate this with respect to phi, you bring down an i m, the i's make another minus sign that cancels this, and you end up with m times whatever it is. So what we know is, this should have its subscripts, I suppose, so what we know is that psi l m is equal to some function of r in theta times e to the i m phi. I guess we kind of already knew that. Now we're going to, oops, and we're going to, well, yeah, we're now going to impose the condition that l plus on a psi l l is equal to naught, because this operator would create a state in which m, we've put here the value of m equal to l, which is its largest value. This would try and make a value make m even bigger than l. We know that's not possible, so we have this equal zero. So copying down what that is, this equation is the position representation e to the i phi, sorry, e to the, yes, e to the i phi, d by d theta plus i cotangent theta d by d phi, operating on e to the i l phi times this function k, which depends on r and theta, especially it depends on theta, the r dependence we don't care about, it depends on how much loperate is here with respect to r. So this term looks only at that and brings us down an i l, right? So what's that, sorry, this is going to be equal to zero. So we get two terms, we're getting at this d by d theta term looks only at that. So we discover that dk by d theta minus cot theta, sorry, l cot theta, we're bringing down an i l k equals naught. Then there's a factor of e to the something or other phi which we can cancel away, it's not interesting. So here we have, this is a linear first-order differential equation, the friendliest kind of differential equation, so we solve it with an integrating factor. The integrating factor is e to the integral of this here, e to the integral of minus l, oops, minus l cot theta d theta, but I think that the integral of cot theta d theta is log sin theta, so this becomes e to the minus l log sin theta or e to the log of sin to the minus l theta or simply sin to the minus l theta, that is the integrating factor of this equation. In other words the equation states that d by d theta of the integrating factor which is sin to the minus l theta times the function is equal to naught, in other words this thing is equal to a constant, in other words k is equal to a constant, she's obviously going to be some kind of normalising constant times sin to the l theta, and we have discovered this constant in principle depends on r. You can have any r dependence you like, so what we've discovered is that psi ll is any function of r you like times sin to the l theta e to the i l phi. This is an important kind of result. And now we're in a position to calculate anything else because if we want to find what psi l, l minus 1 is, then it's equal to l minus, divided by an appropriate normalisation factor which happens to be l l plus 1 minus l l minus 1. Remember these ladder operators come with square root e normalising factors, that was the case in harmonic oscillator, that's the case also with the angular momentum operators, operating on a psi ll, which we now know what it is, sin to the l theta e to the i l phi times the unspecified function of radius, and this l minus, that's roughly speaking put in what it is, no maybe we do it on the next board because we want to be able to see those magic formulae, right, there they are. This tells me that upside l minus 1 is equal to, I think this is just a square root of 2l, so it's function of radius over the square root of 2l, all being well, times e to the minus i phi times, there's probably an overall minus sign coming from that formulae at the top there, d by d theta minus i cot theta d by d phi working on the function we first thought of, which is sin to the l theta e to the i l phi and what are we going to get? This d by d phi will again bring down an l etc, and then this exponential will take one off that, so we'll end up with something that goes like e to the i l minus 1 phi, this will differentiate sin to the l and produce l, sin to the l minus 1 times a cosine, this cotangent multiplying that because this is cos over sin will again produce me a cos times sin to the l minus 1, so the whole thing is going to be minus, unspecified function of radius over the square root of 2l, times, everything is going to go like e to the minus i l minus 1 phi and then from here we're going to get an l, from here differentiating that we're going to get an l, well there's going to be a factor sorry, of sin to the l minus 1 theta cos theta and how many of them, from here we'll have an l and from here we will have, we're going to bring down an l, well a minus sorry an i l, they will cancel so I think it'll be plus another l of the same, same stuff, so you see that we have something like the square, minus the square root of l over 2, whatever your unspecified function of radius was, e to the minus i l minus 1 phi, times sin l minus 1 cosine theta and we could now apply l minus to this again and get the next in sequence, right? We're not going to do it because life gets very boring, l minus 2, but it's just a matter of differentiating, well the thing to pick up is that when we do this next, when we differentiate this, this thing is going to become more complicated because we're going to be doing a derivative of this with respect to theta, so we will get a term that goes like sin to the l minus 2 times cos squared and then differentiating this we'll get our sin to the l back, so we'll get two different terms and then when we differentiate again to get psi, so this is going to be an amount of, this will, this will, it's going to be amount of sin l minus 2 times cosine squared, differentiating this we'll get sin to the l minus 2 and then we'll get a cosine which goes on to that and we will also have, from differentiating this, plus an amount, call it b of sin to the l theta, so there'll be two terms and it'll all be times e to the minus l minus 2 phi and when we differentiate this again in order to get psi l minus 2 it'll get more Byzantine and it'll generate me an l sin to the l minus 3 times cos cubed, this will get back what we had here and so on and so forth, you get more terms, you get a longer thing coming in front of the exponential. So what are these things actually turning out to be? It turns out that what this is is a normalising constant times p l m of cos theta times e to the minus, sorry, no, no, no, make that l m, if we just keep going, this will turn out to be a normalising constant times the associated Legendre function p l m of cos theta times e to the minus, sorry, that's an i, not a minus, isn't it? Yep, this should have been a plus, e to the i m phi. So this thing, I think you may have met this in Professor Essler's lectures, this is an associated Legendre function and probably derived from Solution in Series by using Frobenius' method, I'm not sure. Is that right? But fundamentally, I don't think this is very helpful knowing this is an associated Legendre function, I think it's much more helpful knowing how to do it this way. The normalisation factors take care of themselves if we put in these square root animals and we start with this thing correctly normalised. How do we normalise this traditionally? What we do is we say, psi l m is proportional, is equal to some function of radius to be discussed, times y l m of theta, where this thing, the spherical harmonic, where this thing, the spherical harmonic, is a multiple of p l m times e to the i m phi normalised so that if you integrate d theta, sin theta d phi over the sphere of y l m mod squared, you get precisely one. So the y l ms are correctly normalised, so if you mod square them and integrate them over the sphere, they come to one. The p l ms have a daft normalisation and that's why I don't think you should bother with p l ms. They're just stupid functions. Historically they've been defined in a bad way. The y l ms are things to go on, but the y l m is actually one of these functions of cos theta times e to the i m phi. So it has a very simple phi dependence, this animal here. So now let's just summarise what we have. So these things, y l m, theta and phi are the wave functions, essentially they're the wave functions theta phi l m. They're the wave functions belonging to states of well-defined orbital angular momentum. That is to say, if in the position representation you apply l z to y l m, you get m times y l m, the trivial result because this thing goes like e to the i m phi. And if you apply l squared to y l m, you get l l plus 1 of y l m. So if you have an electron, here's the nucleus, if you have an electron in orbit around the nucleus, it seems reasonable to say that it's reasonable to ask what does the system look like, what does the wave function of the electron look like if the electron has well-defined orbital angular momentum? The answer is that its wave function is going to be a function of r, which we'll see, we'll tell you how much it's oscillating in radius as it goes round and round, times one of these y l m things. So these y l m things should give us, we should be able to understand them in terms of orbits at some level. So let's address ourselves to that. What can we understand about these mathematical functions y l m in terms of what we understand intuitively about how an electron should go in an orbit around its nucleus? So the place to start is not when l is large, because when l is large is when we're approaching the classical regime for which we have some grip, and the pictures at the top here, these are contour maps of the real part of y l m. So y l m is an inherently complex thing, right? Y l m consists essentially of p l m, some real function of cos theta times e to the i m phi. So by focusing on the real part of that complex function we've got that p l m times cos m phi. And these ones at the top are all for l equals 15, this is for m equals 15, this is for some intermediate one, m equals 7 and this is for m equals 2. So what's the physical interpretation of these? This thing, and y l m is a function on the sphere, right? It assigns a complex number to each point on the sphere. So the real part is a real number on the sphere and what's being plotted here are contours of constant value of this real number on the sphere. So you have to imagine that these are pictures of spheres. So what do we see here is that around the equator we have, so dotted contours mean negative values of the real part and full contours mean positive values of the real part. So the large values of the real part are around the equator here and that's apparent from this maths because we know that this is y l l for l equals 15 so in fact it's sine to the 15th theta e to the 15i5, right? That's what this thing here is and if sine theta is 1 on the equator and less than 1 everywhere else if you take a number that's less than 1 and raise it to the 15th power you have quite a small number. So you were expecting that the number gets small quickly as we go away from the equator. That makes exactly what we expect on physical grounds because the state l equals 15, m equals 15 means you've got 15 units of angular momentum broadly speaking and they're all of them parallel to the z axis. So this thing is an electron that's orbiting in a plane, classically, if you're orbiting in a plane the equatorial plane that was perpendicular to the z axis. So where do you expect to find the particle? Do you expect to find the particle in the equator where does the wave function peak in aptitude in the equator and nowhere else? Why is it segmented like this? Like an orange, right? We have sort of waves going around the equator here. It's big, small, big, small, big, small. That makes perfect sense because the change in the... because p, the momentum is minus i h bar d by d position, right? So if you have something with a large momentum it's to do with a large gradient or a large rate of change of the wave function. Now the amplitude of the wave function does not change one i ota as you go around the equator because this thing has amplitude which is sine to the 15th power of theta. So it's completely constant at one around the equator but the phase of this wave function is changing like crazy as you go around the equator because it's e to the 15 i phi. And that is expressing the fact, according to this, that the momentum of the particle is directed tangentially around the equator. It's rushing around the equator, it's in the equator and it's rushing around the equator. What else would you expect? That's exactly what should be the case. So let's go now to this case which is, oops, I lost it, the extreme right one, m equals 2. So we've still got sine, sorry, we've still got l equals 15 but we have m equals 2 so we've got a particle which has 15 units of angular momentum but only two of them are parallel to the z axis. So classically what this amounts to is that here's our sort of notional sphere and you think that the orbital plane classically would be tilted like this, well even more so, sort of like this-ish, very highly inclined so that the spin axis of the orbit was pointing almost in the xy plane. So what we're expecting is that the motion is mostly from the northern hemisphere down into the southern hemisphere and back up again. So we expect the contours on which the phase of the wave function changes rapidly. Oh, fiddle sticks this. So annoying. The direction in which the phase varies should be from north to south and lo and behold it is. So now instead of having an orange peel plan we have sort of rings going around almost on which the phase is. And if indeed we put m equal to zero we would have a wave function which had no variation as you went around the sphere. It would all be variation as you go from the northern hemisphere to the southern hemisphere which corresponds to the fact that the particle is moving this way. Now this particle has most of its angular momentum in the xy plane but the thing is because we know the angular momentum in the z direction and Lz does not commute with Lx we do not know how much angular momentum it has in the x direction. Most of its angular momentum is in the x and y directions but we don't know whether it's positive or negative. So that means that we cannot in this picture see an orbital plane. The probability of finding the particle is sort of large all the way down here and all the way down there and if m was zero and the angular momentum vector were exactly in the xy plane we would have absolutely no variation in probability to find the particle as we went around and around the sphere. When in fact even now we have no probability to go around and around the sphere. So what this thing is is a function of theta times e to the i to phi. So the phase is varying as we go around the sphere but in fact the amplitude is not varying as we go around the sphere. The amplitude to find the particle is constant as you go around the sphere on small circles. That is associated with the fact that we do not know we are not allowed to know it is forbidden to us to know which way this angular momentum vector is pointing. But where are we most likely to find the particle? Are we likely to find the particle most likely in a given patch on the equator or most likely to find it on the pole? Well this wave function is largest at the poles the north pole and the south pole because it is going around this particle is going around over the poles in a plane which is of unknown orientation. So there is great uncertainty there are many places where it could cross the equator but what we are sure of is it goes close to the pole. So that's why the probability there is a sort of crowding of the imagine a bunch of circles for a polar orbit going around the sphere at different orientations they all pass through the pole and there will be a great crowding of the circles near the pole and that generates the high amplitude to find the particle at the pole a relatively low amplitude to find it at the equator but not a vanishing amplitude to find the equator because it does cross the equator twice in each cycle. So this amazingly this is an intermediate case M equals 7L equals 15 this curious mess of squares in which you can see the real part is alternatively positive and negative the contours are dotted and full this represents the situation where the orbital plane in classical physics the orbital plane will be just moderately inclined at 45 degrees or 30 degrees or something to the equator and there is absolutely no orbital plane visible there and this is where we come to a key point that if you want an orbital plane to be visible and after all the orbital plane of the earth is entirely visible and the earth presumably moves according to these principles too we have to have how do we get an orbital plane to emerge the way we get an orbital plane to emerge is by quantum interference between many states that look rather like this and have a patchwork of pluses and minuses if you have several of those patchworks of pluses and minuses you can get the amplitude to cancel most places except in some inclined orbital plane so it's uncertainty in the angular momentum which will generate for you if you want it some degree of certainty in the location of the orbit going around the sphere it's the old uncertainty principle over again so those are the classical this is almost the classical regime up here and of course as the earth goes around the sun its angular momentum is 10 to the 50 hbar or something it's simply I haven't worked it out it's some staggering number so you would have to imagine 10 to the 50 patches here of pluses and minuses or maybe it's 10 to the 50 squared 10 to the 100 patches of pluses and minuses and then you can by taking a number of those maybe you take 10 to the 34 of those with 10 to the 50 patches you'll be able to arrange for exquisite the pixels to cancel everywhere except in some extremely narrow band which is the inclined orbital plane of the earth so atoms don't live in that regime up there of L equals 15 atoms live in this regime, this Tarsan regime down here this is where am I, this is L equals I've lost it, this is L equals 1 and these are the three things for L equals 2 so this is Y11 so that means you've got one unit of angular momentum and well it doesn't actually right because what does L equals 1 mean? L equals 1 means that L squared has answer 1 1 plus 1 equals 2 so the total angular momentum the square root of L squared has answer root 2 which is distinctly bigger than 1 so we've got as much angular momentum along the z axis in this 11 cases we can which means the particle definitely is going around the equator and you can see that it's going around the equator well I can't from this angle but I hope you can in the sense that the thing isn't constant the wave function has gradient as you go around the equator there's a gradient on the other hand there is not a very high probability of finding it in the this is only the real part of the wave function if we would look at the imaginary part of the wave function well how does this one, this one goes like sine theta not like sine to the 15th theta this function here is sine theta times e to the i phi so as you, in the equatorial as you go away from the equatorial plane the amplitude to find the particle falls but only falls like sine theta so it's really quite likely not to be the equatorial plane and that's associated with the fact that though we've done our best to get the angular momentum along the z axis it isn't along the z axis because it's total angular momentum is is 1.4 something times h bar and only one of those units is along the z axis so it's in some sense inevitably inclined and this is the case we have no angular momentum along the z axis so this is the case of polar orbits the amplitude to find the particle is greatest at the two poles smallest at the equator et cetera et cetera et cetera but the whole picture is less clear cut and I won't bore you by talking about these but it's worth thinking about the the n equals 2 case to see to what extent you can make sense of these physical sense of these pictures here okay so now we should address an important topic which is the parity this is practically an important topic the parity of ylm so remember the parity operator p working on a psi makes a state whose amplitude to be at x is minus is the amplitude to be at minus x if you were in the state of psi that's the definition of the parity operator and these states of well-defined angular momentum turns out have well-defined parity that's what we're about to show and what's more the parity is minus 1 to the l so states of different angular momentum have alternating parity some are even parity some are odd parity that's what we want to show okay so what we do now is so this is sort of imagined in Cartesian coordinates we need since our ys are all defined in terms of polar coordinates we need to translate the operation of going from x to minus x into spherical polar coordinates so as x, as we go from x to minus x it's easy to check that what happens is that theta goes to pi minus theta and phi goes to phi plus pi so this reflection action you need a picture really well we can just about show it I suppose I hate three dimensional pictures because it's a three dimensional picture here's theta the spherical polar coordinates theta and what you do what we have to do is take this point and move it down here and what we do is we move this point down to here that's the theta goes to so theta this theta goes to pi minus theta and then having got it down here we rotate it through out of the board and back into the board through pi and phi and that's how we get it down here so these are the changes in polar coordinates that are associated with that now ylll oh yeah well what else can we say when if so theta goes to pi minus theta what does that have to say about sin theta sin theta goes to sin pi minus theta and sin pi minus theta it's easy to check from a variety of arguments is actually equal to sin theta so sin theta doesn't change and e to the i l phi what happens if you add pi to e to the i l phi well you're adding e to the you're getting an extra factor e to the i l pi which is minus one to the l times e to the i l phi is that right ok now ylll is a constant rather a yucky constant so I won't bore you with it times sin we've proved this sin to the l theta e to the i l phi so this thing does not change sin or it doesn't change at all right so we can say now that ylll goes to this doesn't change sin and this one changes sin so it goes to minus one to the l of ylll that's under x goes to minus x so this means the parity of ylll is even if l is even an odd that's a very important result and moreover it generalizes because we have we have that ylll minus one is l minus over some square root that's really boring well it turned out to be 2l so we'll put it in times ylll and what about this what's l minus is lx minus i ly in the position representation what is this this is minus i h bar of y d by d z minus z d by d y plus minus who knows h bar it doesn't much matter the key thing is that we're going to have here a z d by d x minus x d by d z and when we change x to minus x y to minus y and z to minus z these things we get we get a change of sign here and a change of sign here a change of sign here a change of sign there so l minus and also as a matter of fact l plus is unchanged by p the strict mathematical statement is that the parity operator commutes with either of these animals indeed all the angular momentum operators commute with the parity operator basically because they contain products of position positions or if you like ratios of positions whatever they don't change so what that means is that what that means is that this is going to have the same parity as this because if you apply the parity operator to this you're pairing the parity operator to this those can swap in order this turns to minus itself the minus sign can be taken out and therefore we've shown that that leads to the conclusion that this thing has the same parity as this let me just write that argument down perhaps so we have that p l minus sorry p on y l l minus 1 is equal to p l minus upside y l l over some square root that's not interesting is equal to l minus p y l l over the square root which is equal to minus 1 to the l times p times sorry this thing produces y l l so we have l minus y l l over the square root but this is y l l minus 1 so it's equal to minus 1 to the l of y l l minus 1 so we conclude that y l l m has parity minus 1 to the l for all m this is a very important fact because it enables you to set to 0 all sorts of integrals which would otherwise be very tiresome to work out how are we doing there's one other thing so which we've unfortunately lost is it coming back? no what I wanted to do was show you the forms of the y l m the first few you need to have some sense of how they go so y nothing nothing is 1 over the square root of 4 pi y which order they put in so we want y 1 nothing is basically cos theta there happens to be some factor of root 3 over 4 pi y 1 1 is of course sin theta times some normalising factor e to the i phi y 1 minus 1 would be the same thing with a minus sign here so the point is that the y 1s go like cos theta and sin theta and the y 2s go like cos theta sin theta so y 2 nothing is equal to a normalising factor that happens to be 5 over 16 pi that's not so interesting times 3 cos theta minus 1 and y 2 1 is minus the square root of 15 over 32 pi which is not so interesting what's important is it goes like sin 2 theta which can also be written as sin theta cos theta and the other one goes like of course sin squared theta this one has e to the 2i sorry e to the i phi and this has e to the 2i phi so what do we need to remember is that obviously y 0 0 has no angular dependence so the machine has finally come back to life and the correct formula here that's what I wanted to show you the y 1s have a cos or a sin the y 2s have a width well that you can either think they have a strong width of cos 2 theta and sin 2 theta right because cos squared theta is something like a half of 1 plus cos 2 theta so there's a width of this could be a rearrange to involve cos 2 theta here we have a sin 2 theta and this sin squared has a width of cos 2 theta about it because we know that sin squared theta is a half of 1 minus cos 2 theta something like that so we have these double angle formulae and so it would go on if we were looking at y y 3 these things would have dependencies that look like cos 3 theta and sin 3 theta that's the pattern but you don't need to know about the pattern beyond here but these patterns you're expected to be able to sense so that when you're given a function of theta which is made up a linear combination of these things you need to be able to unscramble it and write it as the right linear combination of those ys right the next topic so in preparation for work on atoms we need to get an important formula for how kinetic energy can be expressed in terms of l squared and this finally obliges us to face up to the tedium of calculating what l squared is what differential operator represents l squared in the position representation so we start by observing that l squared can be written as l which way around do I want to write it plus minus I want to write it either way but I'm going to do it consistently like this well let's see what we're going to have to add to this to make l squared this is lx plus i ly lx minus i ly what's that going to come to that's going to come to lx squared plus ly squared plus well minus ilx comma ly commutator that's what this thing multiplies up to if we want to get l squared we'd better add here's a good start on l squared let's add lz squared but we need to get rid of this lx comma ly is i lz so we've got here what with this minus sign plus lz we'd better take an lz away in order to square the books so that's what this should be sorry this should be put equal to plus lz squared so what we do now is we write down l plus l minus which we have floating up there in the stratosphere so we have that l squared is equal to e to the i phi d by d theta plus i cotangent theta d by d phi that should operate on l minus which is minus I'll take the minus inside the bracket e to the minus i phi d by d phi d theta sorry this minus sign was up there outside the bracket I think plus because I've propagated the minus inside the bracket i cotangent theta d by d phi so this disgusting mess is that product and then we have to add lz squared this thing is minus lz is minus i d by d phi so with that minus sign we get a plus i d by d phi and this is going to be minus d by d phi squared so the name of the game is to differentiate out this piggy mess and find out what it simplifies to some parts of it are easy we're going to have for example the end of the day we will have terms where this is multiplying this and this is multiplying this and these two exponentials have killed each other off so we will have a term like d2 by d theta squared these i's will generate sorry there'll be a minus d2 by dc minus because one of these has got a minus sign these two will create me a minus cot squared d2 by d phi squared that's the easy part now the mess there's going to be some mess because this differential operator is going to bang into that and generate a minus i times what will kill this off so we'll have a minus i oops oh no but then it's times this so the minus i that we're getting from here will meet this and generate a plus 1 so we have cotangent theta that's this cotangent theta times this bracket so that's the result of this differential operator seeing this when this differential operator sees this bracket or we get actually sorry we get a mixed derivative term we get two terms we get one term that we've already written down and we get a term d2 by d theta d2 by d phi but that is going to be cancelled by a term that comes from here when this differential operator looks at that we'll deal with the differentiation of this in a moment so I'm not going to write down those mixed derivative terms otherwise we have dealt with the action of this on that now what about this one we've got the operation of this on this we've got the operation of this on this I've just said that that's cancelled away what we haven't got is this when that differential operator meets this we get the differential of cot is cusock squared I think so I think what we have is plus i cusock squared theta d by d phi now the sign should be checked at this point because because signs are a pain right I think it must be that the derivative is minus cusock squared both can be taken afterwards right so that's the derivative of this on this and then I claim that these brackets are dealt with and all we have to do is write down the training terms here which is a plus i d by d phi and a minus d by d phi squared now we need to consolidate our various terms we have three terms one two three which are just d by d phi terms and God be praised they all add up to nothing because we have a trig identity which is cot squared minus cusock squared is minus one so we have the cot squared theta minus cusock squared theta is minus one and here's our cot squared there's our cusock squared and I'm missing and this should have had an uh I I I I I we have an I problem right these have to be all no no no I'm not trying to mess with that one I'm not trying to mess with that right I'm going to have a cot squared here with an associated attendant I I've got a cusock squared with an I and I have here a one so let us buy that causes those all to add up to nothing then I also can use this identity to consolidate this double derivative and this double derivative so we have a cot squared and a one and I can trade it in for a cusock squared according to that formula there right so we end up with minus d2 by d theta squared um it's going to be cot squared minus cot squared so it's going to be um we've got a cot squared and a thing they both carry minus signs which means I have to have them on the other side so we get a minus cusock squared according to this I'm slightly worried by this uh so I'm going to end up with a cusock squared d2 by d5 squared and I strongly suspect that sign is wrong but that's what I've honestly got uh so that's this dealt with and the only this is so this has been dealt with this has been dealt with this has been dealt with I think we're all we're all tickity boo we're not are we uh what have I lost to believe this is the easiest way to do it it's hard to believe isn't it but it is it is um this one right so so that remains cot theta db theta thank you so we now consolidate this all being well into 1 over minus 1 over sin theta sin theta d by d theta and this should be I think a minus that sign is wrong a 1 over sin squared theta this has usually written d2 by d5 squared so I've screwed up on the sign there somehow um so when you when you differentiate this we get a cos which cos of a sin is cot so that's this term here we have the double derivative sin et cetera et cetera et cetera and what is this this is r squared times the angular part of del squared not on that note it's time to to leave we're not quite finished with the calculation but that's the important bottom line that l squared is is actually with a minus sign a minus r squared times the angular part of del squared and we'll push that forward into the kinetic energy tomorrow no on Wednesday