Lecture on GR

50:00 But of course from their point of view it can be, it could be a very double-edged well compositionality and therefore immediately questions of, that lend themselves, so it seems to me very naturally to, expression in terms of properties of pullback in the appropriate category arise, that could in fact be just the right instrument to turn them aside from their objective idealist solutions and towards a correct scientific... Thank you very much for your work, for thinking about these problems. That's why I think you can compute in certain areas. Oh, exactly, exactly. That was my point. When they speak about, when they try and analyze this using these... In this frame of the Mariology, there are many various questions, such as questions of compositionality. And he replied that this was not the servant. Well, I reply that I'm not searching for something that explains facts about reality. There's a lot of stuff written on this blackboard, but I think it's still there since last September, and at certain times it's bound to be a meal or something like that.

52:30 And it has precisely that flavor of trying to decide by who it's going to be. Yes, yes, yes. Oh, I'm looking down? The idea of the category of objects is given to us in pure thought. But in advance of any physical structure or this kind of thing, there is the use of category theory by people who, you understand what I'm saying? Yeah, yeah, that's right. But that's a thing, isn't it? Well, I hope so. The way we derive this is by minimizing the action function. The action function is the scalar r times the square root of minus g, which is the metric density. We have the requirement now that if this represents the energy momentum density, then the left-hand side has to be symmetric. The answer is that this does, and the covariate divergence must vanish, which is true for this tensor.

55:00 It was just summarized. Along the way, we found that when we take this variation, there was an extra term that was the variation of r. I mean, this had to show that the theorem could get this equation up by virtue of the vanishing of this. I showed you three different ways that this response was here. One was in Einstein's computational calculation, and the other was Schrodinger's variation. Now, this variation means, with respect to the ten components of the gene we knew separately, are the variational parameters that exist. So that's what the variation refers to. Where he said, for more general expression, let's take for the variational parameters g, nu, n, all the coefficients of the anti-connection, a lot of extra variational parameters, in this way, in more general expression, Schrodinger was trying to use this method in order to generalize Einstein's equations. You need to find the outside variation first to the delta of the arm union. Yeah, when you say this is the delta of the arm union, it refers to the variation with respect to the consortium. So the arm union is the delta. Yeah, this is the delta of the arm union, but when I say this is the variation of the delta of the arm union, it refers with respect to the components of the convention. With respect to that. Yeah, right. So it's like a part of the theory. Yeah, but this is, remember in the picture we had, we had all the different pads between two four-dimensional surfaces.

57:30 We had all the different pads. And the variation takes you from one pad to the next. Only one of the pads corresponds to this equation. And the way you go from one pad to the next is by varying g mu. That's what it means by g mu. Now, if Calatini said, let's make a more general expression. By saying that not only is G blue, but also these two quotes are filled in to the argument that G is a more general form, but I'm just going to show you the same result that we find given and fixed during the experience of complex relationships. Yeah, this, well, and also we have to define, this, this, this, originally you say that you have an action issue. The S of G that is right on the left side plus the matter part of the action. And this is actually the thing that's here. So this implies that delta Sg is equal to minus delta Sm, and this is the equation on top of it. Minus delta Sm is equal to the variation of matter with respect to g. So by definition, this is just focusing on the left side of the equation.

1:00:00 And that's equal to the total delta for the square root of minus g, r, and u. In fact, this gives you the left side of the equation. Plus the square root of minus g, delta r, and u. If you take this variation, and if you get the term on the left side of the grade 14, you can say the variation of R we know is by virtue of the change in the ion connection. It's not by virtue of the metric tensor. So if you say that the change in the part of the action referred to as S of M is only by virtue of the change in the metric tensor, Then you have this equation that this actually relates to this part, not that... Under those conditions, you have the integral of delta sine of delta m, and you have the integral of delta r of u is zero, which means zero if there's no dependence in the variation on the line above on the ethnic connection, wherever. Delta, R, mu, is equal to... This just comes from the expression we had for the switching tensor. So this is what we derived last time, not simply... But last time what we did was to express the gammas in terms of the metric tensor. So this variable gamma was by virtue of the variation in G. But now this is a separate variation.

1:02:30 And if you have, for example, what is called derivative probability, then it would have terms, this would have terms, which go on this line. So the simple equations I'll write down in a few minutes don't have that in effect. So then you have two equations. So this variation is the respective, and this equation involves the variation respective of gamma. At this stage, you don't know, as we knew last time, how gamma depends on the metric tensor, because it's a separate variational parameter, it's just separate. So I'm just looking at this equation now, and now, this actually is the derivation of the Schrodinger's method, which is to integrate by a partial derivative with respect to the semicolons. The same rules as, these are rules to integrate by a partial derivative. The ordinary derivative is going to apply to anything. Covariant equations. So if you integrate by parts, with respect to the center components, then this is equal to the integral of g, u, v, r, that's the first term, just integrate by parts, then you have minus the integral of g, of course it's d4x.

1:05:00 And then, plus, E and E and U are delta A and 1. A and B are just two arbitrary spacetime equations. So this, of course, is zero, and this is zero because of the definition of the variational problem, that the variations vanish at the end points. They just have these two here. So we convert this equation into this. We have, then, an odd. I just took the second term and re-labeled it, so I put it together with the first term. If you look at this equation, see if mu is not equal to nu, and where this is zero, this is not arbitrarily zero, this is variation, so that implies that g is zero. Is that a game or? Where do you put the right arrow?

1:07:30 Okay. So if mu, if you look at this expression how mu is not equal to nu, this term is zero, Then we have a g, bar, nu, minus, plus and minus, we have both terms, and, uh, okay, so now, uh, What's the difference between those two? I'll tell you. Alpha's not mu. Alpha's not mu. This is zero. So you have the integral of this thing times this variation equals zero. So, because this is, uh, multiplied, integrated, this is not arbitrarily zero, If alpha is not new. If alpha is new, then I have both terms. So g-bar-new-new-alpha minus g-bar-new-new-alpha. This is good. This is good. I mean, it goes to terms the same. It's a different shape of the letters. They're multiplying this, though. Well, the operators? Well, they're multiplying this. They know they have a sum. Well, I'll write it out. If alpha is equal to new, then I'm going to just take... For example, new and beta equals 1 and 2. So I'll write this out.

1:10:00 On some scaling terms, there's weight, energy, momentum. Remember, this is geometrically learned, and this has to do with the number of equations, because the number of equations is one, and in other ways, if I multiply it by the sum of the views, then I have that new thing, one and a half, twelve, thirteen. There is a contraction over view of the aesthetic by definition of the physical, and that is if you take language as a new sound, that's the solution, but that's the vacuum system, but if you have a vacuum, you have a real vacuum, that means no matter what, that means it's zero, so the only solution to that equation would be flat space.

1:12:30 The solution is equal to 1-1-1-1. That's the recommendation. We call it this. So the only, even though there are critical, if you have a fraction of 2,000 vectors, you're having to do a new graduation. No, you're having to represent the coupling between gravitational and magnetic. The magnetic coupling. So this is, I mean, for an actual legal vacuum, where this is exactly true, for that situation, you think there are other solutions besides that, basically. Right. Besides, I mean, this is a trivial solution. Because this is the threshold that's being used, though. It's used as a constant automatic material. But R means 0 because that implies that theory. Right. Now, plasticity actually exists. R means 0. That's 0 everywhere. But that's a plasticity thing. So if you have this 0, it's automatically 0. But if this is not 0, we could have this 0. That would lead to a lot of confusion, I think. Because a lot of people are going to tell me that there are two equations. The next topic we're going to do is look at this equation in the context of Einstein's theory of gravitation, not by space, but based on curvature by virtue of gravitation and pieces of matter.

1:15:00 The fact that it works, and we'll find that some of the most effective, the success of Einstein's theory came from the conclusion that that equation applies to situations where you don't have a faculty, It didn't rise to the current state of the universe, but it did rise to the current state of the universe, but it didn't rise to the current state of the universe, but it did rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise to the current state of the universe, but it didn't rise Some particular solutions can't be solutions of the homogeneity theory. You can't just employ them in an homogeneity theory. It's not generally a solution, but that's what's important to find out. But I want to do some examples now. I think what mathematicians call Einstein manifolds is the case where t would be equal to zero, so that means that it is less than or equal to seven. If you look at the top line, it would mean that the action metric could be solved by the action metric expressed in terms of the curve. Thank you for your time, and I hope to see you again soon.

1:17:30 The fact that they are so, if you verify that they have the accuracy, that implies that the university of matter in the universe is very, very small. Let's say the universe is much more dense in terms of matter distribution. There's a small, little, infinite, little piece of the universe called the solar system, where it's been applied, and it's found that the solution of this equation is the observations, but the observations correspond to everything. So to me, it means it doesn't cut off.

1:20:00 What I'm saying is because the gravitational field is so strong that when we use the Schwarzschild solutions, or let's say, Bini-Krebs solutions, or Kerr or whatever, these vacuum solutions, how can we apply it to the physics and still get accurate results? So the thing is, the matter is then...

1:22:30 Subtracts of G, G is a part of the Euler equation, so this is the Euler equation, so this is the matter requirement for X, and that's equal by definition to the integral of lambda, Euler, delta, and G, and that defines, and so, this minus lambda, T, L, M... This is the variational derivative, so this is g rho. That's what g nu is. This term, this should be very mixed up. So that defines the energy momentum. First we look at this term, g plus lm.

1:25:00 This term before, well this, with respect to the covariant, because of the way we define minus g, remember, this minus g stands for the determinant of the covariant metric. So if you take this derivative, you just get one over two minus one over two. This derivative is simply the cofactor, and the cofactor is just the inverse of the covariance, so that's equal to minus whatever minus g over 2, so that's the thing that goes in here, so I just write this up. So you have the derivative of l m squared over minus g over something else. I have actually, what I worked out was, with respect to the code very well, right, so let me just write that again, that last line that I have there, right, d m squared over minus g a, right, equals g over 2 times, you have an inverse of it.

1:27:30 Covariant tensor, you have something that behaves like a contravariant tensor, which is this. So I can just invert this by multiplying by the contravariant tensor twice, so I can, this implies that e squared times g squared times g over 2, which is the ratio of those two units, I mean, lowering those two units is to covariant tensor. So therefore... You have that derivative of lm very much gtlm minus one half. So that's this part here. And the Lagrangians I'm going to talk about, there's no dependence on the derivative of the metric tensor, so that won't come in. But if there are general Lagrangians later that involve something you already know about, if you want to insert it just for generalization, then that would be an extra part in here. And, Brian Jane tells them that even if your energy momentum actually goes on the right side of the equation, provided that there's no dependence on the derivative of the equation. So, now let me just give some examples that you know of. We kind of come into this room in the sharpest period of time.

1:30:00 So, the Maxwell period, you know, is the functional, is the variational derivative. Due to the equations, like the Maxwell equations, when you take them with respect to the field solutions, the Maxwell Lagrangian is one over two times two in terms of E and H, that's just E squared minus H, and that's equal to F, right, and all of this. So, from this Lagrangian... In this form above, we find that T is 1 over 4 pi, FU minus 1 quarter, and the goal is 3. So you can show this yourself, it's easy to show about this Lagrangian. Now this is kind of very kind of from the last dimension when we talked about pressure relativity in the same form, except they didn't do all the down indices. Now this was the conker delta, this is delta. That's the Maxwell. And then, of course, the Grande-Euler equation, in terms of the F, gives you the equation of the F times the rho of rho, which is 4 pi j times the F. The Maxwell equation is concerned with Fourier, which I saw earlier, and this comes from the Grande-Euler equation of the variation with respect to the field F. And here we buckle the bridge with respect to that. So if you take this term, you see, and you put it on the right-hand side of the Einstein equation here,

1:32:30 then you have the effect of a electromagnetic field giving rise to a certain metrical field, a certain metric tensor. In fact, Wheeler, in his exuberance, called this the already unified field theory. There's nothing unifying. There's just nothing unifying. You can call it because it's an electromagnetic source, you can call it because it's an electrical field, you know, you solve that equation, and that's why we need mathematicians, like, to help you, because then you find out what is the precise, explicit form that you need to do that's due to the electromagnetic field. So what you do in principle, then, is you solve this equation, and put in the F here. That's now your source, here, when you solve this horrendous non-indirect differential equation. The trouble, of course, is that this equation involves G-mean-new-truth. It involves the covariance divergence, which, remember, we discussed in terms of kappa-connection, which is in terms of G-mean-new. But to solve this equation, you have to know the solution of that equation. To solve that equation, you have to know the solution of this equation. So you really have a pair of self-consistent equations in G-mean-new-truth and evidence. But you have to solve them all together. Why do you put them all together? Because the solution of each one depends on the other one. So, you know, one way to do it is to guess the solution of one equation, and then put it into the other equation, and then get a solution of the other equation, put that into the first equation, see how close you were. And if you were close, you couldn't find out. If you weren't, you couldn't find out. If you were close, just keep iterating them back.