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?  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?  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? 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.  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.  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.  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.  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.  You see this, there's an actual black hole somewhere there. And it's pulling a matter.  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?  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.  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.  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 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.  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.  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 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.  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.  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?  But r is r prime plus r g by 2.  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.  So plus 2.  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


Everybody force Jake and Jeff here got 23 24 Bowman Chrome. You basketball delights six box number four Look at that. Yeah, my 15. Good luck. I ran him seems good luck y'all magic one by 15 by five and a half 11 on our random Bet Kyle got free pizza. We pizza hack. Let's go. All right names one two three four five six seven eight nine ten and eleven Braden K down to Barry G in 11 times on the teams one two three four five six seven eight nine ten and eleven Oregon down to Iowa State All right, Braden K Oregon Andrew T Indiana Providence Weber State Anthony C Purdue UCLA Louisville Braden K Michigan State Chuck F Illinois Memphis K State Raymond D Houston Jai with Auburn, Florida State Baylor Braden K Yukon Barry G USC Scott V Arkansas Jonathan L BC Pitt Notre Dame Braden with LSU Iowa Sung Yun H with Maryland Braden has Tennessee Colorado, Alabama, Michigan Miami St. John's FAU Duke Barry with Kentucky Zeng S with Texas Norfolk State, Oklahoma State Zung Yun H UNC Benjamin T Marquette Braden K Kansas JD with Stanford Benjamin T Creighton Gonzaga, Florida Barry G Ohio State, Arizona Chiyuti Watt Villanova Virginia Virginia Tech Barry G Iowa States Results like it's in chat guys get one-on-one filled up Tonight one-on-one dude you said over five got a lot of work Go ahead and guess super guys 30 seconds five four three two and One The legal doesn't want to talk to us outside of just guessing the super he doesn't want to talk to us. He's avoiding it Refractors of Johnny broom Dingle royal and Caitlin Clark sharpshooters Like these the throwback auto L Marko Jackson to 99 for Kansas Kansas nice hit Michigan State first one auto Xavier Booker and Lava for Kansas El Marco Jackson auto to 199 to El Marcos Great boxer Kansas that we got blue to 199 and send me a wheelchair for St. Johns 399 hardaway Memphis 299 Angel Reese for LSU Let's go. I think I got my bed in time. Let's go Kyle That's Rikia Jackson 399 and at a 150 is Malik all Dose Marcos. That's right illegal. That's right. Stay thirsty my friends So if we're not worth it and get it probably should have kicked the field goal What a drop But probably should have kicked the field go up 14 right there Fractures of Stromer diamond Johnson Cody Williams and a big kahuna El Marco Jackson It's the same product There's just like it's like a different configuration So like if you hit a super in this say we hit like say this diamond Johnson was Superfractor auto That would be the only diamond Johnson first one with Superfractor auto. Yeah, there's no like the light supers and hobby supers There's no like We go diamond Johnson or folks eight nice hits Lay them blocker refractor auto to 250 for Arkansas. Yeah, it would be the only one in the entire product Yeah, it's the same product it's different configuration.  You're just guaranteed better things in the delights yellow to 75 Ashton hardaway for Memphis. Yeah, there's just three. There's an extra auto and there's no base There's probably a couple more numbered cards, too. It's hardaway to 75. All right green and 99 Tyler Burton Villanova Aqua wave Burton to 299 Nova Hoggard at a 399 Michigan State Hardaway to 199 Memphis and purple Jeremy fears out of 399 Michigan State What's okay saying Well, oh Correct. Yeah Garrison Dylan Jones Tyrese proctor and Cameron Brink first woman auto Kentucky DJ Wagner It's okay to say Larry Marquette first woman auto I Used to wonder that at first to the fracture auto 250 Omaha billy you for Iowa State I don't think so 250 billy you 399 is Burton for Nova to 150 Trey Norman Marquette Yeah, Cody Williams out of 199 for Colorado. You're good to go to love it, dude You can man someone else definitely if there's a question someone's asking someone I get I get your the no questions a bad question 399 table Williams UCLA and Colac out of 299 for Marquette Now this isn't the case of this product But with like I thought I was just gonna say with like prism H2 folder prism There's like prism choice. I will have like a nebula for a one-on-one rather than like a black fine. I Yes But you can't pull Like it's like totally different parallel sets, but the same like they said So if you guys a little differently, I think Where's the tops or some crossover with like bowman draft choice and bowman draft like jumbo But there's also different parallels that are exclusive to each one Sack leader Subjective opinions are Just like grating might get the 95 year old bifocals might get the kid that's got higher all better Chapman brink and love Wait, Chris, you don't love or hate Brock Hardy. I dare you do God. Oh Jared McCain You must choose He's very good. He has limitations, but You don't win this many games and Have those numbers not being good Brock. Pretty is just in there to try to execute the offense He's not gonna do the outlandish things athleticism like the big dogs like my home Middleton green auto to 99 Ohio states He's supposed to just keep it simple. I'll leave a auto over BC. Oh my god. He caught it. Oh He caught it He caught it off the ricochet and a 75 yellow auto Hannah Hidalgo for He caught it off the ricochet. I said are you govers? Should have known way Ray. We have culprit under 100 for Creighton and I coordinate this. Oh Wow, should have been a should have been a pick great past you That's a touchdown. Whoo.  Yeah, cuz he was off his feet Yeah, it's 199 Baycott. You guys coming down.  I'm trying to understand the rule It's a 299 Louisville. Yeah, exactly Where he's touched as long as you make the athletic play going down. It's about to be a CMC touchdown, but George Kittle They know what it is. Oh He got the short. Yeah, he was right. Shoulda went to Kittle so obvious short the Jared first round I Fracture Roberts Wagner hardaway and Cody Williams aqua to 150 Cal Caleradu The Mari Burnett refractor out of the 250 Michigan base auto Oregon Would be Kwame Evans. You got to go the cafe again right here It's too obvious.  Oh Oh Terrible and gold on Marco Jackson the 50 first one out of her Kansas. How many Marcos now? Chase Marcos watch. Oh, watch. Oh, Marcos Purple the 399 Van Lith LSU color match Aqua wave Hawker to 299 Michigan State the 399 Hawkins, Illinois To 199 mom's to love it's for Iowa State and the Mari Burnett to 99 Michigan Indeed There's a tiny a Uke a Uke baby Shoulda kicked that field goal Fractures of Venters Vescovy Co-leck and an orange big kahuna of Riley Cougal Florida color match That's out of 25 16 to 25 Looks sick with the orange the border of the card is fire 250 refractor auto Z Lowry Marquette Are you Are you very good blocker auto, Arkansas? Lava Chapman auto for Ohio State Over to 199 399 a Holloway for Auburn to 99 Filipowski Duke It's all go number to 299 Notre Dame Proctor of 399 Duke To 199 to Sean Harris Smith, Maryland's and the 75 Dylan DC for Texas That's a fix Do a recap here the radio international cards? That's not pretty sure. All right, so we had Chapman blocker Evans Burnett's a Boswell a lig bay Lowry McCain Wagner Lowry Billy you she'd blocker booker El Marco Jackson El Marco Jackson Middleton Hardaway Hidalgo and El Marco Jackson That'll do it everybody. Thanks again moving on now to some football


What's going on everyone? Sam here with late in sports cards working on 2022 Paninia maculate collegiate football hobby five box full case break number five random team style. Here is a look again at your results. So you can find these on the website under the results tab as well as the Lincoln chat currently. Let's get started here. See what we are going to find. Got a couple more spots coming in. Thank you guys so much. Appreciate it. Appreciate it. We're down to 15 on the next one. Here we go. Let's get into it though. There's only three boxes. Oh boy. Go get Lane. I would say probably one minute. Unbelievable. That's great. One second. We're going to get Lane in here. Yeah, this was in here. They showed us two boxes, Eric. Two boxes. There's supposed to be five. You could see. Five boxes per case on here. Five boxes per case. Sounds good. I don't know how that happened. This one feels like there's more in here. So hopefully maybe there's six boxes in this one. There's even a piece of... I know. It's like they meant to. That was weird. Yep. There we go. There's five in this one. All right. Here we go. All right. We've got Josh Allen going to the Bills. 92 out of 99. The case only had three boxes. And then had this in it too. Unreal. So that's from Harper. Oh yeah? That's nasty. Aiden Hutchinson, 18 out of 25. There you go. Lions. Grats on that one. Lions. That's cool. Who is that? Nice. Scridge going on here too. Heck yeah. We've got going to the Eagles. Nine out of 25. That's AJ Brown. Here we go, Philly. Let me go see where this guy is. More spots coming in. Thank you guys so much. I believe he's still in college. Yep. This is gonna be a random to all at the end. That is Phil Jerkovec for Boston College. He is still in college now, obviously. So that will be a random to all at the end. 14 to go on the next break. Breaking next. We've got Traylon Burks, 24 out of 25. Going to the Titan. Congrats, Titan. Nice one there, Tennessee. And to finish it off, going to the Giants. Three out of five. Kavon Tibido. Very nice one there. Congrats, Giants. That's sick. Congrats, Giants. Let's do the next one here. We've got Roger Svoback. That is at 52 out of 99. Going to the Cowboys. This one. Where's Charleston Rambo? This will be going to the Panthers. He was signed as an undrafted free agent to the Panthers. That is Charleston Rambo. Here you go, Carolina. The only team he has been on. Next up, nice one here. How about 15 out of 49? Drew Brees. Heiko New Orleans. Justin H. Congrats, buddy. We've got, let me go do another looksie on this guy. Another free agent. This will be going to the Steelers. This is Matteo Durant. Signed as an undrafted free agent on April was released in August. That will be going to the Steelers. There you go, Pittsburgh. We've got up next, going back to the Titans. Traylon Burks. Nice one there. Congrats, Titans. That's nasty. And see what this is gonna be. I'm assuming this is gonna be a random to all. We've got Joe Burrow, Jamar Chase, Justin Jefferson, Clyde Everett-Hallaire. So that's Bengals twice. And if the owner of the Bengals also owns the Vikings or the Chiefs, they will get it. Has the Bengals. This will be a random to these three teams at the end. All right, box number three. So with that one, theoretically, when I set the random up, will it be Bengals, Bengals, Vikings, Chiefs? Okay. Cool, cool, cool. Empty. One of the Vikings. That's Justin Jefferson. Nice one there. Three out of 99. This will be going to the Bills. 19 out of 99. That is Khalil Shakir. That's pretty sick, actually. This next one. We've got 63 out of 99. How about the Fridge? William the Refrigerator Perry. Nice one there, the Bears. On the Fridge. We've got, let me do a quick check on this one. Can you go on the next one, y'all? 14 to go. Only 90 bucks a spot. It's gonna be going to the Packers right here. Tyler Goodson. Monster. That is a monster. Holy cow. Who's got that team? That team is Amy B. Amy B. One of one. K-Von Tibodeau, boom! With the Pack 12 patch. Wow, what a card. Holy cow, man. That is wild. That is an absolutely wild card right there. We're going directly into a mag for you, Amy. Congrats again. Then we've got 52 out of 88. Sam Howell going to the Commanders. There you go, Washington. Such a safe patch on that. Any one second, guys? 14 to go on the next one. 14 to go. 14 to go. Let's do it. For the Patriots, that is Mac Jones. I think this guy's still in college. Let me take a look really quick. I'm pretty sure he still plays. Oh, no, no. He got drafted, number nine. He's Niners. Yes, going to the Niners. That is 41 out of 99, Brock Purdy. There you go, San Fran. Nice hit for you guys there on the Brock Purdy Iowa State Cyclone. We've got going to the Eagles. 32 out of 99. That is Nicoby Dean on the Rookie Auto. Nicoby Dean. We've got Hassan Haskins. That is 25 out of 99. There you go, Titans on the Hassan Haskins. That a Michigan, go blue. Got, wow, that's nasty. That is three out of 10. How about Garrett Wilson with the bull patch on it. Going to the Jets. There you go Jets. That's so sick. Congrats on that one. New York Jets, New York football Jets. And we've got 12 out of 25 going to the Titans again. That is Hassan Haskins. Congrats on that one. Again guys, 14 to go on the next one. You obviously see the stuff that comes out of here.  That breaks next. That breaks next. Last box of the case. Here we go guys. Bushbox came in for the, for a mixer. Thank you, Eric, Dan Marino. That is, for the dolphin. Let me check on this one. That's Coulter Bills, if we got it. That is Indianapolis Colts. That is 86 out of 99, 96 out of 99. Devante Price. There you go, Bill. Gotta check this one too. This'll be a random to all at the end. 45 out of 49. That is Will Shipley of the Clemson Tigers. He is still a Clemson Tiger. Good luck on that one everybody. We've got another RPA. That is Hassan Haskins. Five out of 49. Wow, man. Another nasty one. Got this team. We've got, where is Houston? Logan S. Two out of five, Derek Stingley Jr. With the laundry tag auto. Congrats on that one, Logan. That is absolutely disgusting.  Congrats on that one, Logan. So awesome. And then to finish it off, we've got Kona Seahawks. We've got 56 out of 99. Kenneth Walker, the third. There you go, Seattle. Another spot just came in. All right guys, got a couple multi-randoms to do. Let me set them up really quick and then we'll get going on it. So we've got, we'll do a recap after. That was a killer case. All breaks right after this. The sponsor on sale, 90 bucks for a full case break. Random team, I'm seeing to go. Drop in the link now here for you. Thank you, thank you. Let me copy these teams in. All right, guys, doing the multi-random. We've got this Will Shipley to 49. We've got Phil Jerkebeck to 49 and then we've got this Burrow Chase, Jefferson, Clyde Edward, Tulare Quad. So we're gonna go ahead and roll the dice. Gonna go the same amount of times on each. We've got four times, we're going four times on each. We'll start with the Shipley. One, two, three, four. That'll go to the 49ers. We'll be getting the Shipley. The 49ers going four times on the Jerkebeck. One, two, three, and four. That's going to the Cowboys. And we're gonna go four times on the Quad as well. Whatever team is on top, we'll receive it. One, two, three, four. Going to the Bengals. Going to the Bengals. Quick recap here for you guys. Got Kenneth Walker, Seahawks. Hassan Haskins, Titans, Sam Howe, Commanders. K Von Thibodeau, Giants. Hassan Haskins, Titans. Devonta Price, that is Colts. Hassan Haskins, Titans. Nicoby Dean for the Eagles. Sorry, Brock Purdy, 49ers. Tyler Goodson is going to the Packers. William Perry going to the Berks. Got Khalil Shakir, Bills. Traylon Berks, Titans. Mateo Durant, Steelers. Charleston Rambo, Panthers. Got Traylon Berks, Titans. AJ Brown for the Eagles. Ian Hutchinson going to the Lions. Nice one there. Got our Magged Cards. We had Drew Brees, 15 out of 49 for the Saints. We had Garrett Wilson, that is a three out of 10. Patch for the Jets. We had Derek Stingley Jr., two out of five going to Houston on the Laundry Tag Patch Auto. And how about the Cave-on-Tibidot, one of one Jumbo Patch Auto going to the Giants. Congrats on that one again, Emi with the Giants. That'll do it for the break. Thanks again, everyone. Next up, more iMac.


Hello and welcome to NewsClick, I am Samedha. On this episode of the Daily Roundup, we give you a glimpse of some of the most important stories from India and around the world. Take a look at our top stories today. Story of the day is straight out from Karnataka, where we are focusing on the Sante Benyur area, which falls in the Devnagari district. The state is all set to go to polls on the 23rd of April. However, a huge instance has come out to our attention today, where a 17-year-old Dalit boy was attacked by upper caste men in Karnataka. He was not even attacked once but thrice for swimming in a river which is supposedly to be used by upper caste men. Not just this, following this, a mob of about 200 men targeted multiple Dalit households and children and women were beaten up brutally and ruthlessly by the upper caste men. A child who was as young as 6 years old was also beaten up by a bicycle chain. The anti-caste activist and poet, Hustangi Prasad was also attacked when he tried to mediate with the upper caste upara men and he had to also be hospitalized with head and back injuries. He also suffered pain in his kidney area and his finger was also fractured. The instance actually happened on the 12th of April, which is just two days before the celebrations of the Ambedkar Jayanti and 10 days to go for the polls in the state. So this instance has come out and Prasad was also attacked in 2011 earlier by right-wing goons for daring to write about caste. He was then threatened that if he uses the anti-Hindu rhetoric, he will be attacked again and he was warned by upper caste men that his fingers would be broken. And in 2019, now his fingers have been fractured for daring to plan celebrations for the Ambedkar Jayanti and for taking on upper caste men. At Newsclick, we had previously had the fortune of speaking to Uchangi Prasad earlier on the issues of caste violence in Karnataka and a lot more. Have a look at this conversation. Moving on, we are going on to a second very important story which is straight from Kashmir. Kashmir is going to polls tomorrow and now we want to show you an exclusive ground report by Bhasah Singh called The Pulse of the Nation. Living true to its name, she talks to multiple people from different backgrounds to understand what are the Kashmiris wanting in this election, what are their issues, what are their aspirations. Take a look at our ground report. The matter is better, but what was going on was a bit more serious than what was going on. Kashmir's situation has become worse than that. The situation of Kashmir in the last 4-5 years has been very bad. Very bad. How much difference has been made to Kashmir since then? Many people have been behind the situation, not even in the future. Many people have died. This is the MP of Delhi. What is the difference?  You say, give us our vote. Vote for us. They don't come here. They don't come here. They come here, we take someone to the dark house, give us money for this work. We are very late, so we didn't give the IPPS for this. On the daily Nound Up, we've extensively covered issues of the election violence. We had previously reported how on the 11th of April in the first phase of the elections in Tripura, multiple instances of violence and boot capturing had come out. In connection with this, the CPIM had led multiple protests in the state as well as in Delhi and had also submitted a memorandum to the Election Commission to ensure that the second phase of the election is more safe and no violence is witnessed in the second phase of the election. Following the multiple pressures from the CPIM and the demands on the ground, the Election Commission has gone on to state that polls in Tripura for the second phase will be postponed to the 23rd of April instead of being conducted on the 18th of April. Similarly, the President Ramnath Kovind has also given a nod to cancel elections in Vellore following multiple cash caesars which had happened in the month of March. Cash as much as 200 crores was seized in the month of March and following this, elections in Vellore also stand cancelled. We will keep following updates on more stories relating to election violence and the change in the dates of the polls as well. Moving ahead, another big story that had gained attention yesterday was the story of the IT raids on the house of K. Kanemoi. Following the raids, nothing has currently been found but what is important in the story is to remember that this is not the first IT raid. In fact, it was the 12th IT raid that had happened in this election season. It is also important to remember that the Election Commission had warned the IT department to ensure that the raids were taking place in a neutral manner and also that the commission was informed of such raids. Moving ahead to another big story which is of Sadri Pragya Thakur, she has actually gone ahead to announce that she will be contesting elections from Madhya Pradesh from the Bhopal area against Congress's Digvijay Singh. It is important to remember that she is the same woman who was accused in the 2008 Maligaw case. Allegedly, her motorcycle was used in the blast and it has also been said by multiple activists and legal experts that allegedly members of the BJP tried to sabotage the case and members who are affiliated to the Sanatan Sunstar were also allegedly involved in the case. Now we are moving on to the international section where we are following up on the arrest of Ola Bini and speaking on this, we have Prashant with Kiran Chandra who is the General Secretary of the Free Software Movement of India. Take a look at that discussion. Kiran, to start with, could you talk a bit about why various tech organisations across the world have joined together in this campaign to free Ola Bini? See, the arrest of Ola Bini has been very sensitive to all the privacy lovers of the world and also the technologies. The way that he has been portrayed in certain sections of the European media is something that is really worrisome and it is unprecedented for him. He has been portrayed as someone with some criminal motives and it is very unfortunate if you look at who Ola Bini has been. It is an important question that the society beyond the tech community needs to really look at. Ola Bini has been a developer for more than one and a half decades and he has been contributing significantly to the society. In the last one year itself, we were to really quantify and look at the magnitude of the work that he has done for the global common good. In the last one year itself, he has committed close to 1200 times to the different free software repositories that are publicly available. It comes down if you assume that Ola was working for 10 hours a day and he was contributing in one year for every three hours he has been contributing code to the society. You have just blocked him, put him behind the bars and the way in which he has been portrayed is he really worrisome. What did Ola Bini do? It's a question that is pondering us. His contribution was very significant in the dimensions of privacy enhancement technologies in the area of free software development and also in the lines of multidisciplinary subjects that include drug discovery. And I think the way in which he's been criminalized, the way in which a techie programmer who is a Swedish national living in Ecuador with valid work permits and he has made Ecuador his home for the last few years. All the developers of the world were having a different perception about Ecuador. And now Ecuador illegally detains him in violation of all international laws, did not allow him to speak to his lawyers, detains him for 30 hours and then produces him in front of the judge that too very late in the night and even in the absence of any interpreter, he is forced to speak with the judge. And all these are serious issues of concern and he has been taken to preventive custody. He has been put into detain as a cause of prevention. Think with all valid permits who has been doing it and the way in which he has been criminalized, he has been picked up from the flight as if that he was fleeing the country. If you look at Ola Binis blogs or if you look at Ola Binis tweets, we all very well understand that he is a martial arts passionate person and he announced much in advance that he was leaving for Japan to pursue his martial arts events. And I think the way in which he has been pursued, he has been called as a Russian while he's a Swedish national. The Swedish embassy has not been communicated until his parents have been approaching this Swedish government. And the Swedish Ministry of Foreign Affairs needs to protect its state-owned citizens. The way in which they're attributing Ola Binis makes us feel that being a good programmer and working for the society is a crime. And I think this is precisely the reason why the whole global tech communities are organizing ourselves for the release of Ola Binis. He's one amongst us, we are one amongst him and this is going to be one for all, all for one among the developer factor. That's all that we have for you today on this episode of The Daily Roundup. To follow these stories and many more, log on to our website, www.newsclick.in, go to our YouTube channel and follow us on Facebook. Thank you for watching.