 As you can see more detail in this class, there is some effect of potential motion happening. So the planet in the other direction goes from maximum to maximum and comes back. We'll see this in more detail in this class. So we wanted to know how much does fine change as it comes. And it was this shh-shh-shh motion. And you don't forget that what we did was take this equation. And since a delta phi is equal to this object, there is r, phi, there is n, integrated over r, over a whole oscillation. Right? Then we massaged, we did a change of variables in this equation. I remember I used the order at which we were working. We said r prime is equal to r minus r g by 2. The point of the change of variables was that to the order at which we were working, this changed. At first thing, there was no change in the integration pressure. d r plus d r. Second e to the order at which we were working, this main, this object, let's see, there was the m to the n square is r into r minus r g. Right? And so if we write r as r minus r prime minus r g, if r minus r g by 2 is r prime. Because this is m square divided by r prime plus r g by 2 into r prime minus r g by 2. Okay? And the linear terminology goes away. So up to linear order, this is m squared over r prime squared. So the change of variables made the m dependence in the problem n square divided by r prime squared. The m dependence is crucial because it's a different stage with respect to n that gives you the order. Okay? So to the order at which we were working, this change of variables made the m dependence inside the action m squared by r prime squared. And then we kept track of what it did to the rest of the action, to the rest of the radial part of the action. And part of it was just giving you a constant which was a renormalised energy and the renormalised one by r, the renormalised new reinforcement. But there was an important other term which was a renormalised one by r square piece. I mean, a renormalised one by r square piece. We're new one by r square piece. A one by r square piece where there would have been none in Utah. Okay? So ignore the, you see, the renormalisation of the constant of the one by r square piece gives you a small renormalisation of the parameters of the ellipse. But retain the ellipse. You know, this equation here, phi is equal to del sr by del n would be the equation of an ellipse for any integral of the sort constant plus one by r plus n squared by 2. The parameters of the ellipse depend on what the constant is and what the one by r piece is. It gives you an ellipse. But it's a one over r square piece in the action that is not associated with n squared. Then you see the renormal, the precision. Changes this orbit from being a closed orbit to one that persists. Okay? And then you use some clearer techniques to compute how much the angle changed in a single orbit. Okay? The idea was that the change in an angle in an orbit was this much. Okay? But we know that if we don't take it on the general characteristic correction, that change in angle is too high. This is something which we studied in our last course, in the last course. Okay? So what we're interested in is the change in this change of angle. Okay? Because of the new term. And then we use the fact that the new term appears like a renormalization of the angular moment. Okay? To very cleverly compute this change in the change in angle, we're not ever doing it again. Because we get what was the change of angle. You know, we got what was a derivative of the effective angle in the upward motion. But that derivative we knew gave us a 2 pi. And then all that was left over was a change in what was the effective shift. What was the effective shift in angle, which we computed from this procedure. Okay? Just trying to remind you of what we did. Just to chock your memory enough to ask if you have any questions or questions. Okay? And when we went through this procedure in the end, some shift in angle, like put your result if you want. And this is the famous shift in angle that agreed with the position of the variable in a mercury. We got this delta phi is equal to 6 pi a mass of the sun over 6 pi k squared m squared times squared times c squared times q squared. So there are two masses. One is the mass of the sun and the other is the mass of the body. And there was the angular momentum. What is this? That's only 50 factors. Okay? Because this is a big problem. Let's start with the next one. Okay? Is there any questions or comments about this figure? Do you want to study it for motion? Motion in the center, what's the problem? Unless you have a question. That depends. That depends. This is because phi, as a function of r, is given by the same question. So now, suppose we've got some radial motions. As we will see, it's radial motions. Some sort of effect with potential. That's very important. But we want to know how much has phi changed? As we have undergone this radial. Since phi is given as a function of r by the same way, the change in phi is integrating. It's given by integrating over 1 hour. Is that clear? So what is the change in phi? It means the value of the function, the first value of r. It means the value of the function is the same. So now, of course, there's a multi-value here. So you do this. See how much phi changes. And you do that. Okay. So now what we're going to do is two or three more things about this central motion. The first thing I want to do is, so far what we've done is to study motion is the most neutral. We've looked at small corrections about neutral motion. It's very important to experiment with this. Okay. But, you know, when you've got a completely new theory, it's important to make contact with the old theory. But it's most interesting to us. We're completely new phenomena. Now we're going to ask, suppose we look at motion that is very far from neutral. What phenomena do we see in the center? Once again, use this effect direction. And of course, for any particular motion, we can integrate these equations as numerically or using elliptic functions. But what we're more interested in than any particular motion is the net. It is the qualitative feature. Okay. The first thing we're going to do is to try to write, you know, try to use, try to massage this formula into a form so that we can apply some Newtonian intuition to it. Okay. Find, from this formula, what are the function of times? W. What do we do? Together with this action, with respect to what? Good. Good. What is the basic equation of Hamilton and Jacobi? Once you've got the action, what do you do to find? Hobbits. Find the trajectories. The action is written as a function of new momentum. Hamiltonian and the new coordinates vanishes. Therefore, new coordinates are also constants. They have a time derivative with respect to the new momentum. Okay. So the basic equation of how should the Hamilton and Jacobi write what is generated by taking this whole action and differentiating it with respect to E, which is one of the two ways. These are the integration constants, right? Those are the two integration constants. The rule is take the action differentiated with respect to the integration constants and set what you get in the zero. So that's what we're going to do. What we're going to do is to write down, take this object and differentiate it with respect to E. Now, here, when we differentiate it with respect to E, we get minus E. And here, when we differentiate it with respect to E, we get what? We get E divided by 1 minus r of g by r in squared divided by square root E squared divided by 1 minus r of g by r in squared. You see, something I should say is the bottom line. We're using this Hamilton and Jacobi method because it's the most elegant thing we've done. Okay? It's not necessary. In fact, I'll give you a problem in the new problem section in which I'll ask you to derive many of these results. You didn't know anything to do with this. You know, if you find Hamilton and Jacobi confusing, it's more or less the bottom line. You should let that confusion make you think general relativity is complicated. Okay? These are two separate things. Um... Okay, minus m squared by r squared plus m squared by 1 minus r of g by r. Okay? So, I'm trying to follow this method because, you know, it's sort of nice to give you practice in Hamilton. And it's very fast to swing to the answer. But, you know, once again, what we have to do in principle is quite clear. You find the equations of motion for this particle. Do you know what I'm trying to write? And solve that, if you look like after each other. It'll always give you the same answer. Okay? So, anyway, so this is the equation of the end. If your r is a function of t. Now, if you have any particular initial conditions you're planning here, that's what you need in the order. But, we're not interested in a particular order. We want general features. So, what I'm going to do is unintegrate this particle. You see, what this equation has really seen very comfortably. Well, what this equation is representing is minus dT by dr. What does that mean, minus sign? So, I say goodbye. Thank you. Thank you. Cheers. So, this is what this equation is writing down here is that dT by dr is equal to integral e to r. Well, no integral. So, we differentiate both sides. Okay? So, e divided by 1 minus r of g by r square divided by square root. Okay? So, what? R here. No, there's no r. Now, we can take this and take square root square is equal to e square by 1 minus r of g by r for those people. Fourth, dr by dT over d square root. So, we get, okay. So, square root has these the square root has these denominations of 1 minus r of g by r over d square. So, it matters if we're taking two of these values. So, let's let's write. So, what we're getting is square d r by dT over d square is equal to what? The answer is equal to this. This is equal to 1 minus r of g by r over d square into this stuff with the denominators into d square minus m square by r square plus m square into 1 minus r. This equation is very much like an equation in, well, let me just say it a little bit. So, just follow me. Okay. Okay, good. Now, you see, this side here is like the kinetic energy of a bottle. So, e squared by, but e squared has twice the mass. If we think e squared has equal to twice the effective mass, that's what m vector. And this equation of the form kinetic energy is equal to some function of r. So, this function of r could be thought of as minus the potential for zero energy motion. So, suppose you had motion in a potential. This is equal to minus v of r. And e squared was twice m effective. This would be the motion of a particle moving with e effective. So, energy e effective equals zero. Moving in this potential. It's just a very basic equation. So, what I'm saying is that look in tutorial mechanics, we're trying to make a connection with our Newtonian equation. Suppose we have the one-dimensional motion of a Newtonian particle within a potential v of r, v of x. So, by conservation of energy, we would get m x dot squared by 2 minus v of, plus v of x is equal to the conserved energy. So, suppose we were looking at a fictitious model. Just a fictitious mathematical model where we've got a particle of mass e squared by 2. It's not clearly on the possibility. And it was moving with energy zero in the potential. This potential such that this was minus v of r. Then we would get exactly the same motion. General Newtonian intuition for zero energy motion for this effective potential. So, we can think of turning points and oscillations and so on, all of that. Just understand the qualitative nature of the motion while easily in pictures once we understand how this effective potential is. How the most important part. So, this effective potential has two pieces. There's this 1 times r of g by r squared into something. Now, the most important things about this potential are going to be it's zeros. You see, because we're looking at zero energy motion. So, that's it. So, what we're doing, we've got some energy and we call it potential. And if this potential behaves something like this, then we get bound state motion. So clearly the zeros of this potential are really distinguished objects. Now, as far as zeros go, as long as the motion is not going to go towards what she brings, r equals r g. The zeros of this object are the same as the zeros without this multi-planet. So, the first thing that we're going to do is to understand just diagrammatically how this object is really simple. Because it's e squared minus something else squared. So, the first term is just a constant piece. Just a constant e squared. So, in order to understand how the object that exactly behaves, where do I understand first how this object? This object is sometimes called u, the effective squared. And the circle green guy goes to zero. When u effective squared becomes e squared, or where u effective is equal to e. So, the points where u effective is equal to e are going to be the distribution order analysis. So, now let's just so, for the next that's what we're going to do. Just look at this u effective squared as a function of r and try to understand what the other function is. Just as a mathematical problem. Now, I can use that understanding to understand what central force is working at this point. Okay? So, let's write that down. It's not u effective squared. We have that is equal to m squared by r squared plus m squared into 1 minus rg. Potential. So, let's say rg. rg is as far as this problem is going, setting the state. We can even think of it as r. r equals rg is the distinction. But other than rg is 1, well, then there's some over, you know, it's this ratio of m's little m squared by capital M squared. That's where we set one other boundary. Okay? So, it's the ratio of m squared or g squared by e squared, little m squared. That sets, that's an effective parameter. By scaling our way, I can remove rg from the problem. But that would put an rg squared here. And then there's an overall constant outside. But there's this effective ratio of capital M squared by rg squared little m squared. Effectively, a 1-parameter set of effective potentials here. Parameter is where this one. 1-parameter capital M squared by rg squared little m squared. Okay? This effective potential could be a point different depending on what this 1-parameter is doing. Okay? So, let's first, before starting doing quantitative analysis, let's first just look at this effective potential. So first, let's look at two limits. Let's first look at this parameter 0. Well, this parameter 0 because capital M equals 0. So, very simple. It's constant x1 minus rg divided by r. Okay? So, so, yeah, so, so, let's see. Okay? So the scaling that I talked about, I'm going to do it as much as possible. So, let's look at u effective squared divided by capital M squared. Okay? And let me remind the variable r divided by rg into x. Okay? Then, let us see. So, what we get is m squared by little m squared rg squared 1 by x squared plus 1 in 10. This makes it clear that this is the only parameter. So, let me block this u effective squared by z squared by axis and x which is rg divided by rg. rg divided by x. Okay? So, let's first look at this guy being 0. Then, this is just 1 minus 1 by x. Okay? So, if we stop plotting at x equals 1, we don't want to block for x less than 1 because the funny things are associated with this parameter. Okay? We trust, physically we trust these parts. Okay? So, we stop plotting. We stop plotting at x equals 1. It's 0. And then, it asymptotes to 1 from middle. Let's look at an argument. x times u is the limit in which this parameter is very, very large. So, now what happens when this parameter is very, very large? When this parameter is very large, we have to go to very large values of x before this becomes relevant. Right? So, if this parameter I call alpha for x squared much, much less than alpha and alpha is very large, we can just forget about the 1. Okay? So, in that region, this is just alpha divided by x squared multiplied by something like that. If x is reasonably large, okay? If x is also reasonably large, we can forget about this term. This is just like a 1 by x squared. So, the potential comes now, please. However, when x becomes much larger than alpha, then, you can forget about this term, your best. And then, we have the same approach to 1 from below that we saw before. Okay? So, if we draw 1 here, it's clear this curve has to come somewhere. Okay? And then, approach this. Okay? Also, at x equals 1, the whole thing is 0. So, it's clear qualitatively because this parameter here is, we get two rather different kinds of behaviors. The first behavior is just monotonic increase load. The second behavior is bomb pop, come down and then approach 1 from below. This is more quantitative, right? But before that, we do that with physical implications of these two different kinds of behaviors. Um, you see, firstly, firstly, suppose we plotted you see, what we were interested in was that whole circle of n be equal to 0. That is the same as this potential be equal to e squared, remember? Okay? So, what we would like to do is a line of this potential is equal to e squared or this potential divided by m squared is equal to e squared by m squared. Somewhere in this graph. So, suppose we did that on this graph here and it went like this you know, a nice table of it. Okay, that's the situation, hasn't it? You could also have circular if e exactly touched the line. It would be a stable circle. On the other hand here, there's also an unstable circle out. That's another value. It would be there nice and circular but any little perturbation would disrupt. This kind of potential this kind of potential here has both stable circular or stable like circular. And also, it's unstable circular. Right? This kind of potential here has no extremum at all. So, just as no circular happens, if we do this kind of potential we just fall or maybe escape but if you can't stand, stay standing. It's not some steady orbit. Whereas, this kind of potential here we have the ability to orbit in a steady way. Make all this more more quantitative. We're going to try to understand this. Smaller than the larger ones. Smaller than the larger ones. Why is it a little better? Well, because angular momentum is better. So, if you express it in terms of just velocities velocity data angular momentum is very intense. Yes. So, in this second case I don't understand the motion of three points intersecting the curve. That's not part of it. Well, with three points intersecting the curve? Yeah. Basically, in this case there's a value. So, there are two possible kinds of motion. Yeah. You could have had a motion in which something came from. In this case, suppose it's like this. At this edge there are two kinds of motion. One kind of motion is just this. Another kind of motion is that the particle comes out of the black hole. Reaches a finite maximum distance and goes back in. At the same energy both things are opposite. So, what more those two intersects? The first two intersect. This and this? Yes. Now, this one doesn't intersect. Where was one? Let me draw this curve. Let's say one was this line. Suppose that one is this line. Let me draw it. This is one. So, then this thing does this. Okay. Now, depending on whether your energy is greater than the green line or less than the green line. Okay. Suppose because energy is less than the green line. Okay. Let me draw it. So, what about the third point? This point? Yes, this side. So, let's take two points over this. There's this and there's this. So, in this case, where energy is smaller than the particle has at infinity with that. Okay. So, you can have a bound state and you're also going to have the particle emerging from the black hole going to a maximum distance and going back here. Okay. At this point, where the energy is larger than the particle would have at infinity. This is going to be like positive energy for the new particle. Okay. So, you just go zooming off. Or again, you could have it at infinity. Now, there's a third point. Let's see how. If a particle was shot out of the black hole, however something is shot out of the black hole. Okay. If the particle was shot out of the black hole with that energy, it'll go all the way between. Is that possible? We'll be talking later. We'll be talking later. Okay. But let's get to the point so, it is to understand these curves better. In particular, I want to understand what value of this variable is there. This potential makes a transition between this kind of behavior and this kind of behavior. Okay. So, it looks like a good idea. It looks like it would be a good idea to study the derivatives of this. Because the distinguishing picture of this potential is that it's derivative of nowhere conditions. Whereas here, it manages two things. So, when you first start having solutions to managing a derivative of this potential, that's going to be the the two solutions, of course, will merge. This minimum and maximum will merge in the first one. Okay. And that's going to give you the distinguishing the dividing line between these two. Okay. So, let's compute the derivative of this potential. So, the derivative of this potential is 1. So, let's call the solving alpha here. So, that's alpha by x. I'm just going to write down the potential first. Minus alpha by x. Q. Plus 1 minus alpha by x. That's the potential. Okay. Let's differentiate it. The differentiating it gives minus 2 alpha by x. Q minus plus 3 alpha by x to the fourth. 1 is the derivative and plus 1 by x. Okay. I want to solve the equation that this is equal to 0. This is the, let your common power of x. So, this is equal to x squared. x squared. So, I've got a quadratic equation. So, let me write it like this. x squared here. So, solve with this. So, solve with the derivative equation. Okay. And now you will let me get this right. That's the x is equal to minus 2 minus 6 plus minus alpha by 6 alpha. That's nothing. Plus 4 alpha plus minus b squared. 2 alpha minus 4 is minus 12 over 2 a. minus 12 alpha. minus 12 alpha. minus 12 alpha. So, let's get to the 2 in the denominator. That's 2 alpha. That's minus 3 squared. Okay. So, we get alpha plus minus alpha squared minus 3 alpha as our solutions. Excellent. So, first question. Where do we first, when do we, when do we have solutions to this equation? Clearly, we should remember what alpha was. Alpha was even in the squared by RG squared. Okay. So, clearly we only have solutions to the equations. Have solutions. So, we've got this. When we take down this border line between these two variables, when alpha is less than 3, the graph looks like the first one. When alpha is greater than 3, the graph looks like the second one. Woop, woop. Not really a better sense of the physics. Let's study the circular orbits that exist when alpha is greater than 3. So, when alpha is greater than 3, we just determine the value of RG at the points where the derivatives match. So, you see, okay. So, there are two possible values. There's a smaller value, which will be young stable, and the larger value between the stable is the smallest possibility for the larger value. Okay. So, for stables, for x stables to alpha plus, the larger value, alpha squared minus 3 alpha. This is this function, minimum. What value of alpha is it? Okay. So, here it doesn't even exist for alpha star less than 3. And then when alpha increases to be greater than 3, this thing just increases. So, alpha equals 3 is where it's minimum. Okay. So, x stable is minimum and when that's the case, this square root is 0. So, at that point, x stable, we have already concluded something that's very important. We've concluded that stable circular orbits in the short sheet background do not exist for R less than 3 times the short sheet. The minimum value of x on a stable circular orbit R is equal to 3 times the short sheet. Now, this fact is dramatically seen in astrophysical examples in observations of black hole. One of the ways you see a black hole is this. You see this, there's an actual black hole somewhere there. And it's pulling a matter. So, this matter is swirling into the black hole and there's been an increase in disc. And there's an increase in disc because of course solving the equation of fluid dynamics. But it's not too bad an approximation to think of all the fluid forces being changed. Roughly speaking, each of the particles are just moving along a radius. Okay? So, you've got this equation this and then when you look at this black hole you suddenly see a big hole in that equation. It's because these particles are moving in roughly stable circular orbits and slowly losing energy due to this dissipation going as small as small as a circular orbit. Suddenly, they've got the 3 times Rg. And then there's not a stable circular orbit in the 3 times Rg. And then just whoosh! So, you see the secretion disc with the hole. This is a phenomenon that is just not there in the internal area. And every value of the distance from the Sun to the Earth is stable circular orbits. In general relativity we don't have stable circular orbits for a Schwarzschild black hole at less than 3 times the Schwarzschild radius. And this is a fact that there is laws of the observation of this. Okay. Now, please. Is there anything for muscles? For muscles molecules to be able to misstudy this? Okay. Now, the next question that we're going to ask the next question that we're going to ask is where were the unstable circular orbits? Do we have unstable circular orbits all the way? You see, it's clear from this graph that the unstable circular orbits have smaller values than your circular orbits. Okay. So, let's study if x is unstable. So, that's unstable. Okay. So, that is the alpha minus square root of... So, let's take the alpha root of 1 minus 3 by alpha. Now, where is this function? 0. That's nonsense. Because the square root doesn't exist as a real number. Remember, it starts at alpha equals 3. Okay. And alpha equals 3. x unstable and x stable are the same values. Okay. But you see from this graph that this guy moves to the left and that guy moves to the right. As you move away from... Okay. So, it's actually the... Actually, the other end maybe alpha equals infinity, but this guy exists. What is the value of this x unstable? Where an alpha is equal to x stable. Okay. You might think, where is it equal to? That's not right. It's alpha minus alpha because this is small, we'll expand our theories. 1 minus 3 by 2 alpha. The length of gas distance of gas is equal to 3 by 2. So, this is x unstable. It starts off at 3 when alpha is equal to 3 and then decreases monotonic to 3 by 2 when alpha is infinity. Okay. So, next thing to learn even stable circuit of this learning exists for x. Even unstable circuit of this learning exists for x. For x greater than, for r greater than 3 by 2 times the Schwarzschild radius there are no circuit orbits in the problem whether stable or unstable. There's a lot of factors unimportant observations in it because nothing innocence is an unstable circuit of any theory, but it's sort of an interesting point. And the final thing that we... The final thing that we can do is a final thing that's sort of interesting to do in this problem is to compute the energy of energies of the circuit. This is a very simple task because you just have to take this value here and plug that into effective potential square and that gives you the energy square. Okay. Let me just tell you what the answer is. Okay. So, there are no circuit... So, remember the potential. So... Yeah. So, for r less than 3 by 2 r g So, suppose you start... That means you're always below this point. So, you're always on this side of the line. Okay. So, there are two possibilities. You could be a value such that this is the case or you could be at the value such that you're alphas less than 3. In either case, it's quantity to be 7 because you're always on this side of the curve. So, you're always such as falling in. Falling in between 3 and 3 by 2 r g. Yes. Unstable, sir. Yes. Yes. No, there are no... There are no stationary points. So, the effective potential for... The effective potential could be of this form or it could be of this form. In either case, the lowest minimum here never goes below 3 by 2 r g. Never goes below 3 by 2 r g. Never goes below 3 by 2 r g. No. No. No. No. No. No. Never goes below 3 by 2 r g. Never goes below... Actually, even this guy. This point here never goes below 3 by 2 r g. So, here, this point is never, if we call this beta, beta is never smaller than 3 r g. So, if you start the 3 by 2 r g, you can't be stationary. You have to fall. Or you have to go up. But you can't sit stationary below 3 r g. There's no minimum below 3. Minimum. Is this clear? So, just to complete this, let's go to the values of the energies. So, the energy here is equal to the square root of 2 by r r g into 1 minus r g by r. And then you have to put it out by r g. This is just the base. What I've done is equate v squared over lambda v. V squared v squared. And just plug in the solution of alpha by r g. Plug that in. So, you could, if you wanted to find what the energies of the I want energy to have this minimum circular, minimum stable circular orbit and so on. It turns out for the for the minimum stable circular orbit e is equal to square root 8 by 9. We'll get that out of here. r by r g is 3. So, this should be equal to m into 2 divided by the polar r g square out here square root of 2 divided by 3 into 1 minus 1 by 3. We have alpha is equal to r by r g is equal to, no, we have x. x is equal to 3 mostly. x is equal to 3 which implies r by r g. We also have alpha is equal to 3 which implies m e squared by r g squared square. The energy was given by m 2 times r g in 1 minus r g by r. Let's write this in terms of first thing I want to do is write this as I want to put out an r g square. Good. So, that's equal to m divided by r g into 2 r g by r. r g by r is 1 by x. So, that's 2 by 3 into 1 minus r g by r that's 1 by 3 and we also have this term here. So, this is square root of 3 times capital. Let me let me clear this up I'll send you an email. More like this. The sum energy I was going to make so the sum energy comes over which which one you're going to make. Okay, so this ends our discussion of massive article motion in the short any questions or comments about it before we proceed. You know, you could ask what your orbits look like when we're at r g by r g. They look very weird because they come far from closing up, for instance. It's quite a miracle that for the 1 by r potential orbits are closed. You go up and down in this radial effect of potential and by the time you've done that you've gone through exactly 2 by 3. That was a miraculous property of the 1 by r potential. Shared by almost no r potential in the world. It's not shared by this potential. So, if you are orbiting in a in a caught up, caught up elliptic manner in this potential, it looks very complicated. You know, your orbit will look like this. You just keep performing that sort of orbit. Computing. You know, it's not. There's some other shape, other massacal shape that replaces the ellipse. It's just something else. Yes, the elliptic function is any rational multiple of 2 by r. It will eventually close. That is the parameters that we know. Do you remember the last part of classical mechanics course? However, in this extra conserved charge you see if it happens that delta phi is one of this rational multiple, then there's some extra conserved charges. It distinguishes different orbits you could be on in facelifts. It's having to do with an extra conserved charge in the problem. This Laplace-Rouge vector that was there in the 1 by r potential. But it generally is not there. So this will just be some s. Okay. So all this calculate r equals 0 is completely fixed. Right. Now, good. So, this is something that we will explore in some detail in some part of the course. You see, this metric looks like it's nonsensical r equals rg. Actually, that's not the case. Okay. As we will see in some detail what breaks down is the use of this particular coordinate system. Okay. Below r equals rg. Physics is completely well defined. And actually many of the conclusions that we will reach just using these formulas will also just go through what are less than rg. As we will see, but it requires some analysis to see. So I'm not going to try to explain that. The one thing I want to point out is that everything we said about circular orbits was self-consistent. The outside life was actually so all the conclusions we reached for those conclusions, we don't need to worry about that. But we come back to that. We plan this then. Okay. So we're almost there. But you know, in today's lecture I want to talk about the land-out, if you should stress energy and stuff. I want to talk about gravitational waves. But other than that the remaining part, of course, for our basic general relativity we're almost there. So let's go on. What we have not studied so far is the motion of light. That's the question. So let's now move to the study of the motion of light. The second classic test of general relativity which may or may not have been a fraud test was about the deviation of light as it skinned the surface of the sun. So let's first start trying to compute that deviation. Okay. Now, we're going to use the same formula as before. Because we saw that as from the point of view of the Hamilton-Jibbubi equation. The only difference between the motion of a master and a master particle is that we use the same Hamilton-Jibbubi action. It's an n equals 0. Okay. So that is one of the beautiful things about this. This master and master particle is in a demographic fashion. So that's once again right now in the action but this time studying energy. You remember we also spent a whole lecture on this that the action was actually the role of the action was actually played by the phase of the fine-gotten equation, the Maxwell equation to which your master's particle is a solution. Okay. There's a right action here that's on the right side and then we've got minus something times T and I'll call that omega because that's what it is when it's a phase. The regular change of phase with respect to time is an angular frequency. It's called the energy. It's very closely related to energy but it begins each month after omega. This is called the energy I call it. Okay. Now there's an m times 5 and at last it's called the psi r. Psi r was just the same formula but it was integral omega squared by the 1 minus r regime of omega squared minus m squared by r squared with no plus x squared. That's our point of 1 minus r. So this is the action, the radius of the action that describes the motion of a master's particle in the function. Okay. Let's try to what we're interested in. Okay. Let's first look at a sense of what's going on. Let's first look at our g. Suppose our g was equal to 0 but this is the motion of a master's particle in flat space. Okay. So we've got psi r is equal to integral omega squared minus m squared by r squared. Okay. And we're going to ask if we solve this problem. Of course we should be able to solve the function in flat space. It would be a disaster if we could. So what do I mean by solve the problem? I need to solve for the object. Not for the motion of a particle. So the motion is given by phi is equal to 0 to m. So it isn't equal to integral m divided by r squared square root of omega squared minus m squared by r squared. So I'm going to take this r squared out of here by omega times cos again. So then what's d r is m by omega cos square sine sine sine 2 minus sine sine sine minus sine sine. This is the problem. Let's call it nothing. You see, the point of taking the substitution is that the other way around should have a function. We should have omega. Then what do we get? So we get integral m divided by omega by m and so m squared cos theta m by m. So this is okay. And this is m divided by m by omega. Because it's cos theta. Then this was m by omega sine theta by cos theta. And then this bracket here becomes m sine theta. So omega is cancelled. m is cancelled. Cos theta is cancelled. Sine theta is cancelled. Oh, then that's because I look at this one. Sine theta is cancelled. This thing here was 1 by cos square theta. So that gives you a cos theta in the numerator and a sine theta. I'll do that again. The square root was into cos theta by m. Sine theta is cancelled. Cos theta is cancelled. So phi is equal to d theta after constant. Phi is equal to theta. Very good. So this what is the motion of this line look like? So now we have r of m is equal to m by omega cos phi. Make a particular choice for this constant to 0. So that implies that r cos phi is equal to m by omega. Now r cos phi in polar coordinates is just x. So whereas this orbit was just just a particular point of exit. Through your y's the next step of the text. Which of course is a straight line. So you'd expect that there's no limitation of theta but it goes to the straight line. So again that's our sum because 0 is 1. We have the photon coming. What happens to phi in this process? Phi starts on a minus phi by 2. So it changes by part. So the change in phi is fine. So that's some trivial exercise. Now let's put in the let's put in the first let's try to put in the first depends on the gravitational field of the sum. Yes. This is an artifact of the fact that we're in this particular straight line. We're working in a coordinate system in which that space is really just the expert. So can we say that when we write our modified metric that somehow that is the most natural? Inertial sort of thing. Well you see that because the test that you apply is that when you take M0 you get back flat space. But any coordinate redefinition of what we have that depends on the mass of the sum such that the coordinate redefinition goes to identity of M0 we also have that problem. So this is not a sufficient test. So now we're going to try to work intermingling the first of the including the first effects of the first non-trivial effects of the gravitation. Okay. So for that, as I thought of these formulas so for that as we did before when studying master's portion we're going to try to make the string of coordinates that sets this to horizon. So this M squared multiplies r to r minus r of g as per form. Once again it's a good idea to make the r minus r of g by 2 with the input of r prime. Well once again assuming that r g is smaller than r prime 2. You're going to impact that. Okay so I'm working first on that. Releasing all that. That's marks. So once again this will become M squared by r prime squared. And what will this become? We've done it before but let's do it again. This becomes the square root divided by that's why this is r squared into r minus r of g times r of g squared minus M squared by r prime squared. Now what's this? You see this guy r minus r of g by 2 is r prime minus r g by 2. But r is r prime plus r g by 2. So this is equal to integral omega squared into r prime plus r g by 2 squared over r prime minus r g by 2 squared minus M squared by r prime squared. Now what we'll do is to expand each of these strictly. sr is equal to a group omega squared. So the leading term has this one. The second term is what? clearly the factor of r g by r you get a plus 1 from here because there's a by 2 squared and we get now plus 1 from here minus by 2 squared. So plus 2. So plus omega squared 2 omega squared r g by r the same squared. The leading order the regular part of the action is given by this. Is this okay? Now what do we want to do? What we want to do is the same as before what we want to do is to the important point here is that we don't even need to keep the 1 by r squared pieces. If you remember in our analysis of orbital motion we kept the addition of this the precision of the parallel in the Mercury was proportional to r g squared because this trivial piece there was a trivial renormalization of the Newtonian kind of attraction. There was the Newtonian attraction but for the motion of light Newtonian theory is silent about our problems. The most straightforward prediction would be that light is just unmediated. So any deviation at all first prediction is the leading order effect in general. So we're just going to work on our experiment. Well, if you want to see how how much length baked what we want to do is to find the correction to the free prediction of change in the angle phi is pi the straight line in the x axis and phi changing by pi. What's the correction to that pi? Okay, so what we want is to conclude so let's first approximate this that s of r is equal to omega squared minus u squared by r squared plus now the derivative of this 1 by omega squared minus m squared by r squared into half and two back answers so omega squared r g by r. So to leading order this is what this is really about the actions. Okay, and here it's so simple that we don't need to resort to there's even less motivation to resort to trick, so let's just do the integral. So this is whatever it is we just compute it we just compute it this is whatever it is this is the free straight line this is the interesting piece so let's just compute that integral so that integral so delta s of r is equal to omega squared r g by r so let's simplify omega squared r squared minus okay, now this integral you all know the way to get this integral is to make the substitution r is equal to let's say omega squared r g divided by r squared minus m squared by now we make the substitution r is equal to m by omega posh whatever one omega remains so we make the substitution that r is equal to n by omega posh of an angle and then when we make the substitution we get a factor of m by omega out from here but a factor of m by omega is the measure as well and then the two sin h just gets okay so the solution is that delta s r is equal to r g omega that I use posh h inverse r divided by r omega to make some choice with it so now that we've got this change in the action in order to get the change in the angle all we need to do is to differentiate this change in the action with respect to m and compare, you know, look at what you get when you subtract what it is at the beginning and what it is at the end the change in the angle will be twice the change in the angle to come to the minimum position like this okay, so we could so the maximum position is let's say infinity the minimum position is the impact okay so what we have to do is take this differentiate so let's first differentiate so delta by delta m or d by dm of delta s r is equal to what okay, is equal to what so let's differentiate the cross-agent let's remember how it works h theta so that alpha is equal to theta then sine h theta d theta by dm well no, no, we just we have to differentiate the respect to m we have almost this expression we have almost this expression we have to get the 1 over x squared okay, yeah, okay, fine we have to use what we know the derivative of a cos h of a cos h inverse function so cos h inverse is what is argument squared minus 1 1 over squared okay, good so r of g omega then we have argument squared minus 1 over argument squared minus 1 r squared omega squared by m squared is 1 into the derivative of this so that's r omega by m squared okay, I hope you've done this right so that's r of g omega into r omega r squared omega squared minus 1 squared and then the whole thing right here so if I've done this right, that's r g omega squared by m r divided by r squared no, 2 because you do twice that's like r g times r we have r g times r as well divided by r squared, okay, we must write it as we take an extra omega r from here x of this omega divided by r squared minus r squared I think r squared is n squared r is n squared r is n squared good, good, good, good and it has 2 r to the power divided by r which is n by r perfect, okay, good excellent excellent, excellent, excellent okay, so the thing about this formula is that if you just set what we want to do is to compute the value the value of the deviation between r equals infinity and r equals infinity okay but if you just set of course r equals infinity minus r equals infinity you can see but this is one of these the problems with this thing are multi-valued functions you know, you have phi not on one stretch of the orbit from the other stretch of the orbit you know what I mean so the right answer is actually to effectively think of this as plus infinity minus minus infinity okay so it's just twice the value that you would have got by just taking the r equals infinity you understand what I mean it's an effective flip of sign right so that's not very easy to do let's just answer this rg omega so that's the change in delta phi this this to rg omega by m it sounds like a strange answer because what we are interested in is what is the change in angle as a function of the impact value you said the photon is it okay what is the change in angle as a function of the impact value but we can give, sorry impact value is there again impact value is there again so what we want to do is to find what omega is in terms of impact value but we have already done that by computing what the free motion was in terms of omega can somebody read out what we have this is the answer r times r times cos phi m this is the constant value so this is the impact value so impact value tell us what it will be okay so this that tells us that this is to rg divided by m that's the answer that makes sense impact value goes to infinity we get 0 when impact value becomes like the Schwarzschild radius it's very large but if the impact value is smaller compared to the Schwarzschild radius smaller compared to the Schwarzschild radius it's not a deviation in the case of the sun the Schwarzschild radius is something like 3 kilometers the radius of the sun is huge does anyone know that the radius of the sun is 6000 miles that's something I'm sorry why 10 to the 8 actually I don't know the earth sun distance is 1.5 million so earth sun distance is 10 to the 8 kilometers it's 1.5 to 10 to the 8 so I think it's less than 10 to the 8 kilometers it's something that's much bigger than 3 kilometers if you put it in you get a very small number so arc seconds according to the definition of arc seconds you get some arc seconds and you compare that excellent so we got this the grazing motion of light almost non-deviated motion of light in this background I'll give you a little problem asking you to find where the light can go the non-grazing motion of light in the short chain ok unfortunately oh there's one more thing I wanted to do I really made this a problem I wanted to include the absorption ok let me just say this ok let me just say this you know in Newtonian motion so how does it ask the question where is the absorption of a point like mass you got a point like mass you throw something at it what is the absorption cross section the cross section within which if you throw it it will be absorbed by the point you remember we dealt with absorption cross sections a class T mechanics course probability normalized by class ok and so what's the answer so what is the cross section that it built you send something out from infinity then it will be somehow captured this is the kind of question we ask the TI about interviews generally relativity has been used what are the three kinds of motions what for Newtonian motion the three kinds of geometric shapes reversed parabolic and elliptic if something starts at infinity what kind of motion can it be can it be elliptic no parabolic and elliptic are not captured like this so the absorption cross section sees you and you throw something from infinity it always comes like this is a white part of course it will hit the earth in that state it will be absorbed into the fire of the sun then I mean just the gravitational motion there's no absorption cross section however in in general relativistic motion that's no longer true in general relativistic motion if you throw something at the sun from infinity it could get captured by the black hole this is scary like the black hole is this region of space from which nothing emerges if you throw it near enough the black hole stays ok so there's a big difference between general relativity and Newtonian physics and one of the problems I wanted to work out here was to compute try to distinguish which trajectories is thrown from infinity would go and would be absorbed by the black hole which ones would come back to infinity ok but I'll just make this one let's say that we've completed our our discussion of planetary motion in the in the background there are a couple of things we have in the next problem one of them is this motion of light which is a strong motion of light the other one is the absorption cross section and another thing we're going to ask you to do is the volume I'm going to ask you to write down just the Lagrangian for a particle moving in this motion the Lagrangian and then check that we get the same the same orbiting ok this is actually easier to do than to compute the duly's equation because that requires computing the crystal this is just a simple variation but actually I can say that it's very square root ok just to give you a sense for these problems ok and any other questions in the next class what I'm planning to do is to discuss the Lagrangian's solubility stress energy solubility and then perhaps we can talk about the collapse the collapse process after which we talk about gravitational waves and then black holes and cosmology not quite far apart of course you know you should be please ask me questions about things about the general structure of the Lagrangian