Einstein Field Equations - for beginners!

Hello, today I am going to try to derive the Einstein

field

equations

. This will be a basic introduction, it won't be rigorous, it will cover the gist, but it will still be quite complex. If you want more details, I refer you to Professor Leonard. Susskind's excellent lectures that he gave from October to December 2008 at Stanford. I put a link to the YouTube version of these lectures in an annotation on this video. I'm going to follow his approach so it's familiar to you, but I'll do it. do it in a much more simplified version and let's always remember that it took Einstein ten years to develop these

equations

and he had access to many math teachers along the way, as usual.

I assume that you understand calculus and basic geometry and that you have also seen or know the information in my general relativity videos that are already on YouTube and, in particular, you know two key points: the first is what is called the equivalence principle and what What it says is that if you are in a box without windows, then you cannot distinguish between being in outer space and accelerating with an acceleration G or being in the same box on the surface of the earth stationary but subject to the gravitational force that comes with a acceleration. G, there's no experiment, let's put you in the box, there's no experiment you can do that distinguishes between those two situations, that's actually not entirely true, the difference of course is that if you're in space outside traveling with an acceleration G. then all parts of this box, including you, are accelerating with acceleration G if you are on Earth, then because the value of G varies depending on the height above the Earth, the value of G at this point will be slightly larger than the G value at this point, so there is a slight difference in the G value in this situation and if you can measure that, then you can tell the difference.

That difference is called tidal force, but other than that, Einstein said that was an equivalence principle. You can't tell the difference if you are accelerating in GE or subject to a gravitational force which is represented by the acceleration G. The second principle is that light bends in a gravitational

field

this was Einstein's reasoning if you take the person in the box that is accelerating through space with an acceleration G and here is the box and this time we are going to shine a light from this side of the box to the other side of the box, but of course that will take a finite, albeit small, time and if this box is accelerating with an acceleration G when the light reaches the halfway point, the box will have accelerated a a little further forward, so now it will be here, which means that the light will have reached that position there when the light reaches the right side, the box of course will have accelerated even more and now the light, which of course is going in a straight line, it will arrive here, so if you look at it, inside the box it looks like the light has gone from here. to a point lower to a point even lower in other words the general impression is that the light has gone from here to here to here and that is a curve, at least it should be if I have drawn it correctly, so Einstein says if that's what happens when you're accelerating through space with an acceleration G, since the equivalent principle says that you can't distinguish between that and being stationary on Earth subject to an acceleration G, then the same thing should happen in this scenario here in which light Now appears to bend in a gravitational field, of course, I could have been wrong, but this was in fact proven to be true during a solar eclipse.

What happens during a solar eclipse is that the Moon essentially blocks the Sun. He said that from Earth the Sun was blurred out and what was noticed from the position on Earth was that they could see a star slightly to the right of the Sun, but they knew that astronomers knew that that star was actually located behind the Sun and it should not have been possible to see it and the explanation was that what was actually happening was that the light from that star was being deflected in the Sun's gravitational field. and then it came towards us on Earth giving the impression that the star was here when in fact the star was. really here, but the light was bending in the Sun's gravitational field, now here was the problem with Einstein Newton's law of gravitation, which everyone had thought had served well for so long, says that gravitational force is equal to the gravitational constant times the mass of one body times the mass of another body divided by the distance between them R squared or divided by the distance squared and that's fine except in this situation here the light is bent in a field gravitational but photons of light have no mass, so these two masses In terms of at least the Sun has an S but the photon does not, so this form of Newton's law does not work if it is to be applied to the light.

Five photons have no math, so Einstein came up with a completely different approach to gravity. that gravity exists, he said that actually what is happening is that all forms of motion are represented by motion in a curved space-time and I can illustrate that this way suppose we take a trampoline, this is the surface of the trampoline and that represents space-time and we put a small walnut marble on that trampoline, the marble will have practically no effect, it might make a very small indentation on the trampoline, but now suppose I get on the trampoline if I get on the trampoline with my weight .

It's going to bend quite a bit and that's the equivalent of bent space-time or curved space-time. What is going to happen to this marble? Well, the marble is now going to fall towards me. The analogy is not brilliant, but it will serve our purposes now Newton said that this movement towards me was caused by gravitational attraction Einstein on the other hand says no what is happening is that the marble is moving along the shortest path in the curved space-time so that when the Earth revolves around the Sun there is the Sun and here is the Earth in a reasonable circular orbit around the Sun Newton says that there is a gravitational force between the Sun and the Earth that provides the centripetal force that keeps the Earth in orbit Einstein says no, that's not all that happens is the Sun being massive curves of space-time and the Earth just twists that curve into a straight line in curved space-time.

Now you might be wondering two things, firstly, what is spacetime, and secondly, why did Einstein think it would be a curved spacetime? In reality, time is just space and time, all thought of as a set of coordinates. We generally think that space has three dimensions that we represent by three so-called orthogonal axes. They are at right angles to each other, so we normally have the x axis. The y axis and the z axis or if you want we have three dimensions in space up and down, side to side and side to side, if you add a time dimension to it you get a four dimensional space time problem with a four-dimensional space-time. very difficult to draw because we can only conceive of three dimensions so I'm going to cheat and I'm going to draw a time dimension and a space dimension which is the x axis which I could have caused or another dimension this way but I'll leave it as one dimension of time and a dimension of space.

Now, if you draw a constant velocity in spacetime, you get a straight line because a straight line is distance divided by time, it's a constant distance divided by a constant time. and that is a constant speed, but if you accelerate in spacetime you get a curve, the reason of course is that acceleration means that the more time passes, the further you go, so there is a curve in the acceleration and because gravity causes acceleration, it causes a graph that is curved at these space-time coordinates and Einstein's view was that therefore somehow space-time must be curved if it is creating an acceleration which has the effect to create a curve in those space-time coordinates, well, that's the end. from the introduction, so now it's time to introduce you to the Einstein field equations that we're killing to try to derive and there they are, you may notice that there's only one equation, but actually that's not true because you'll notice that there are some indices arriving mu and nu arise in various places in the equations now mu and nu represent the dimensions of space-time no one two and three nothing is time one is the x axis two is the y axis and z is the sorry three is the z axis , so mu and nu can each take four values ​​between zero and 3 and the combination of mu and nu means that there are 16 variations of this equation with mu and nu having numbers between zero and 3, so you might think that There are 16 equations in four-dimensional spacetime, but in fact 6 of them are duplicated, so it actually reduces to 10, so you're looking at 10 Einstein field equations.

What do all these hieroglyphs mean? You'll be pleased to know that some of them you already know, you'll know what 1/2 is, you know what 8 is, you know what PI is, G is the traditional gravitational constant and C is the speed of light, so there's just a few things that we must solve first. we have this peculiar term R mu nu which is called the Ricci curvature tensor we will derive that they need two places we have this term G mu nu G mu nu is the metric tensor we will derive that then we have R but without indices which is called a scalar of curvature and we will show what es, then we have capital lambda which turns out to be the cosmological constant and we will explain the reason for that and finally we have T mu nu, which is the tension energy momentum tensor and we will try to explain what the central point of this equation is that it balances two things, everything on the left side is about the curvature of spacetime, everything on the right side is about mass and energy and, as some people have helpfully said, what basically what Einstein's field equations say is that mass tells spacetime how to curve and curved spacetime tells it more how to move, so here's our agenda for the rest of this video, we've done the introduction, next we'll do metric tensors, then we'll move on to Christoffel symbols, which are uppercase gamma terms.

You might be wondering why I need to do that because they don't seem to appear in the Einstein field equation, but they are heavily involved in the Ricci curvature tensor, so we'll need the Christoffel symbol, then we'll look at curvature itself, which will give us will allow us to derive the Ricci curvature tensor and the curvature scalar, then we will look at the strain energy momentum tensor. and then the cosmological constant and finally we'll put it all together, you might want to take a break in this video just to collect your thoughts before we move on to the next section, you might want to take a break now, okay, part two, the metric tensor . we are going to consider a field we could consider any field a magnetic field an electric field a temperature field I am going to consider one of the most basic fields of all a field with grass where cows graze here is my field and it is It is going to be a very field hilly, there are hills all over my field and here I am standing on one of the hills and the value of the field will be for the height above sea level, so we usually give the Phi value and everywhere.

We will agree in my field there will be a height above sea level that I can measure and that height will constantly change because my field is a very bumpy field now, how do I know where I am? I can impose a set of coordinates here is my x coordinate here is my y coordinate and then I'm at that position, that position on the x coordinate and that position on the y coordinate is a bit like a map reference, so I define my position like unique and for each position I take I can find a Phi value which is essentially the height above sea level of my position in the field the question I now want to ask is how does that height change if I move a little and you should answer immediately which depends on which direction I'm moving.

The reason for this is that let's assume that part of the field consists of a now ridged stretch where the slope goes down one side and down the other side, but at the top it's flat, so if I'm standing on top of the ridge and I move, let's say this is the X direction, if I move in the I'm simply walking along the top of the ridge and that's flat, so the direction of movement is important in the the height as I go? Well, let's consider a gradient here. a slope that I'm going to call a slope of one in ten, what that means is that for every 10 meters that I walk, I will go down 1 meter, so if I start here and walk 10 meters by the time I get to At the end I have not only walked 10 meters , I have gone down 1 meter and the way you would describe it mathematically is that you would say that the change in height as I walk is equal to the gradient that we express as challenging with the X the rate of change in height with distance that is what 1 in 10 is the rate of change in five thedistance multiplied by the distance you actually walk DX let's try for a moment that the change in height is equal to the gradient well, I told you that the gradient is 1 in 10 and let's say I walk 5 meters so that the distance DX is 5 meters, that means that D Phi will be equal to 5 over 10, which of course is half a meter, so what we have just calculated is that if you walk 5 meters along a slope of one in ten your height changes by half a meter but this is the general way you change height or you change height the value of the field is equal to the gradient multiplied by the distance you travel now clearly in my field I am not going to have clear gradients like this because it is a very bumpy and irregular field, but in a distance very, very small, that's why I put DX and D well, that will be true, but as I said, direction matters, therefore one would have to say that the change in height if traveling in the X direction is equal to the gradient in the X direction multiplied by the distance multiplied. by the distance you walk in the Of course it won't be the same if we were on the ridge, this would be a very steep slope in the . direction suppose instead of walking in the vectors and in my video on manipulating vectors I showed that when you want to add vectors, you do it through an end-to-tail process, so you actually take these two vectors dy and DX and put them. them from tip to tail it doesn't matter which way you do it so let's take DX is like this so dy goes the tip of the vector of the first and the tip of the last, so it becomes the resulting vector that we call d s and we can use pythagoras to calculate what d s is, of course, s squared is equal to DX squared plus dy squared, that of course is very simple if you want to add the vectors what it gives you is the magnitude but for the purposes of adding the vectors we simply say that D s and we put a line above it to indicate that it is a vector it is equal to DX + D Y or if we want to know what it is the change in height we can say that the change in height as we move along path s is equal to the change in height as we move along path X plus the change in height as we move along path why, but these two terms here D I'm going to replace here is that I'm going to give these curly DS gradient terms and the reason I'm going to do that is because that's what's called a partial derivative, it simply indicates that even though this is the derivative with respect to In this case a derivative with respect to Y is not the only derivative so it is called a partial derivative and is indicated as a partial derivative by virtue of having a curly D now I want to pause for a moment while I make a change in the thumbnail and tell you I promise it won't hurt until now I've been using the x and y coordinates now I'm going to change them wherever I used space?

We normally use XY and Z as our coordinates. I guess I need another coordinate. What letter am I going to use? After Zed, of course, I couldn't find another spare letter, but the letters of the alphabet tend to be used for different purposes in physics and it gets very confusing, whereas if I pass on thugs x1, x2 and x3, if I need another coordinate, it's a no-brainer, it's x4, in fact, I can even have a null X that I'll actually use for a time coordinate, but I find these extra dimensions of space. I can have X 4 which I'll use for a different purpose a little later, so now I'm going to rewrite this equation here in my new coordinates by simply replacing X with X 1 and Y with X 2 are exactly the same.

I just called them different names so if I do that I'll get that D Phi which is the overall change in The height as far as my field is concerned will be equal to the gradient defined by the X and now that's x1 multiplied by the distance I move DX 1 plus the gradient in the Y direction, which is now the DX 2 direction multiplied by distance I move in the X 2 direction and of course I could add more terms X 3 X 4 X 5 as many as I need. I'm just going to work in two dimensions to keep it simple and I think you'll see that this equation here can be simplified by saying that D Phi is equal to the sum over N of the gradient divided by n, sorry, divided by X n multiplied by X n , when n is 1, you get that term there, when n is 2, you get this term here and the addition term means that you just add them together, so this is a very useful way to summarize the change in the value of the field of okay with all these values ​​and you'll notice that n is the sum term you have to do this for all values ​​of n here I only have two dimensions, but no matter how many dimensions you have, you need that many terms and I'm going to call that equation number.

I hope you can see, let's write that again, equation number one. and I'll surround it with a box so that when we refer to it later you can see it. Now here comes the general problem that relativity is special relativity or general relativity. Any rule you come up with should be independent of the frame of reference you are using to make the observation. In my videos on special relativity which I will refer to during the course of this presentation, we show that different observers measure different distances and different times depending on of its relative speed which was called longitude. contraction and time dilation, so if you want to have something that is universally true, you have to make sure it is true in all frames of reference, so when I took my field where I was standing in the hills and I got into position P I established a coordinate system I call the XY coordinate system, but I later renamed it x1 x2 and said that was my map reference. x1 and x2 define my position, but someone else could come in and create another set of coordinates that we'll call y1 and y2. and now P, of course, I haven't changed my position.

I am still standing on the hill at position P, but as far as the reference frame y is concerned, the coordinates are quite different and y1 and y2 are quite different from x1 and x2 even though I am in the same place and you will notice that if I want to express y1 in terms of x coordinates, I have to express it both in terms of an So the question we have to ask is do we know if I know all the gradients. Remember that we describe this V Phi by DX in Reference Frame X, how can we find all the gradients in Reference Frame y?

And more specifically, how can we relate the two? And for this we use what is called the chain rule. We say /dy and I'm going to choose. the coordinate y1 you could choose y2, but I'm just going to choose Y 1 which is now equal by the chain rule which is equal to challenge times X 1 times DX 1 times dy 1 plus D Phi times DX 2 times X 2 times dy 1 , so you'll notice that we only stick to the Y coordinate 1, but we add the X coordinates 2 because remember I said that to identify the Y coordinate 1 you need to know all the other x coordinates. coordinates, this can be summarized as D Phi times D Y and now I'm going to call it D Y n because it could be either, it could be 1 or 2 in my two-dimensional system and that will be the sum over m of D Phi times DX.

M times DX M times d and n now let's look at this and compare it to this. What I did was put the general value of n for Y here I only had y1 but it could have been Y for you. to choose which one you want, it's one or it's two, but M is a sum term, you add n when N is 1 you get that term when M is 2 you get that term this will be equation number 2 and I put a box around that too , but let me repeat and emphasize that n is the value you choose, it is y1 or y2, you choose, but M is a sum term, you must add all the values ​​of X just to get one. of the values ​​with respect to And now I want to introduce you to the concept of tensors.

We mentioned that when we looked at Einstein's field equations there were several tensors. What is a tensioner? Well, let me start by talking about a scalar. A scalar is something that has magnitude but no direction, temperature would be a good example of a scalar, it only has a value of 32 degrees Celsius, a scalar is what is called a rank 0 tensor, now consider a vector, a vector, of course it has magnitude and direction, so it has a length and it has a direction which is often described as the angle of one of the axes, in this case, the x axis and a vector is described as a rank 1 tensor and How do you transform a vector well?

We can use equation 1 that we derived earlier and that is shown on the screen for you to recognize and we just make a slight adjustment so that instead of D Phi and D X we replace it with the vectors themselves, so we would simply say that the vector in the Y frame of reference y with the Coordinate n that we choose is equal to the sum over m where m is now the summation term of the gradient, so to speak, which is d and n over X M where m is the summation term multiplied by the vector in the x frame of reference M, so that's the way we and I will call that equation number 3, that's the way the vectors are transformed.

We are using equation 1 that we derived and modifying it slightly so that the vector on the wall is the length coordinate of the vector. in the reference frame Y is equal to the sum of this gradient term multiplied by M where m is added, of course, to the coordinate in the reference frame frame of reference, the tensors that we will be interested in are the rank two tensors and what a rank two time is is a combination of vectors where there is a fixed relationship between the two, so let's think about some system that has a combination. of vectors let's take a very simple one, place a block on the floor and we will apply a force to it to move the block a certain distance force, we have to take the component of the force that moves in the direction we want, so if that angle is alpha, then the component of the force that acts in the direction of movement is F. cosine alpha then the work done will be the component of the force that is in the direction we want to move multiplied by the distance we travel and you can think of F and a magnitude and a direction, the component at least of the force in the X direction has magnitude and direction, so we essentially have something that combines two vectors.

Now the question is suppose the force is in that direction, then. the angle alpha will be 90 degrees and the cosine of alpha is zero and therefore the work done will be zero because of course the block will not move if you push it down you will not get it to move in the X direction and here it is the question, but no matter what frame of reference you apply to this system, whatever frame of reference you apply, that block will not move, the value of the work done will be zero in all frames of reference and that is the value of a tensor a tensor is the relationship between two vectors and if the relationship if the tensor has a value of zero in one reference frame it must have a value of zero in all reference frames if that block does not move it it does not move, it does not move No matter what reference frame you apply to observe it, it does not move the tensors, therefore, our relations between fixed relations between two vectors such that if that relation is equal to zero in one reference frame, it is equal to zero in all frames reference and that is why tensors are so important in Einstein's field equations, so let's consider a combination of vectors, it doesn't matter what it is, for these purposes we are going to take the vector a and we are going to take the length coordinate of that vector remember that there are dimensions in space.

I'm actually using only two dimensions for this analysis.simplistic, so M is one or two, you choose which one you want it to be and we take another vector B and we will take the nth coordinate which can be one or two you choose it and we say that the combination of the two becomes a Rank tensor 2 which we indicate by saying that it has those two coordinates M or two indices m and N is equal to the vector a with the coordinate or dimension M and the chord and the vector B with the nth dimension, then obviously if there is, if I am talking about a two-dimensional space where M can be 1 or 2 and n can be 1 or 2, so T MN can There will be 4 versions of that tensor where m and M each have values ​​1 and 2 if you have three dimensions of space. m and n can each have values ​​1 2 3, so t MN has nine different versions if there are four dimensions of space. time once three dimensions t MN will have 16 different values ​​okay, now I'm going to use equation three that we derived earlier and that should be displayed on the screen to take the vectors a and B, so we're going to take a.m. in the y frame of reference and B M in the y frame. of reference and we're going to try to see what they are in the x reference frame using equation three.

Well, you can see that we simply have to replace V in equation three with a and then B, so let's do a first of all we need is some term. I can't use them anymore because it's already been used here, so I'm just going to plug in an R value. Remember this is a sum term, it's called a dummy variable, it doesn't matter what you put in and that's it. to be dy m times DX ah multiplied by a in reference frame X ah that makes vector a and now we need to make vector B in reference frame X which again comes from equation three.

Now I need another addition term called is and that's going to be dy n times X s multiplied by B in the reference frame the values ​​that you get choose for this particular equation, well let's see what this boils down to, first am BN, well that's just the tensor t MN in the reference frame and because that's what we select here and it's going to be equal to the sum of R and s. and that's not the right way to do it, but essentially you have some terms here that I'm just going to simplify here x dy M times XR and then you have a dy in DX s times a X R B X s that's that term and that term but what it is this?

Well, if you look at this term up here you'll see that it's just another tensor, in this case it's the tensor in the reference frame that we need it in the future and that is what it is called, it receives a technical term, it is called contravariant transformation, it is the transformation from the reference frame y to x. framework for tensors there is an equivalent form of tensor transformation called covariant transformation where the indices that instead of being up go down but then the dy times the X terms are changed to DX times the dy terms so let me write that you get the T and now it's in the Y reference frame, but the M&N goes down, so I'm going to put the M&N down here and, to be clear, I'm going to put Y in parentheses so that the MMTs in the y reference frame are the same.

Again you have this sum term over R and s and now these two terms just go backwards DX R times dy M DX s times dy n just reversing these terms here and then the tensor terms again go down but they are in X frame of reference and that is called covariant transformation and call that equation number five so this is called contravariant this is called covariant contravariant had the indices up covariant had the indices down contravariant had dy for the X terms covariant had DX for dy terms and it's actually the covariant form of transformation that will interest us now.

I want to manipulate something as basic as Pythagoras a little. Pythagoras applies to a right triangle and if this is dx1 using our framework. of reference in the reference frame take it plus DX 3 squared and so on and disagree with you that that is simply the sum of m of DX m times DX m in other words when m is 1 you get DX 1 squared that is the term when M is 2 you get DX M you get DX M is 2 DX 2 all squared, that's for that term, if M were 3 you would get the additional term DX 3 squared.

Now I'm going to make it a little more complicated, but you'll see why, of course. in a minute that means D s squared is equal to the sum of m and n what am I doing with DX m multiplied by DX n? Now you will say no, that cannot be correct because there is no combination of DX 1 DX 2 in Pythagoras. it's just DX 1 squared and DX 2 squared. If you do this you will get a term DX 1 DX 2 and we don't need that to be perfectly true so we add an additional term which is called Delta M N and that term is called the Kronecker Delta and what happens with the Kronecker Delta is which has the value of 1 if M is equal to in and has the value of 0 if M is not equal to n, then when M n you simply get the squared terms you want when M is not equal to M, then the entire term is returns 0 because Delta MN is 0, so if M is not equal to n, all the cross terms come out and we will need that approach, it's a very complicated way to describe Pythagoras Delta MN remember the Delta clinic. is equal to 1 if M is equal to n is equal to 0 if M is not equal to N and therefore avoids any of the dx1 dx2 terms that you don't need in Pythagoras, so I rewrote that formula that we just derived.

I simply put the chronicle. Delta at the beginning and not at the end, no matter where I go and now I want to remind you of equation one that we derived and that should be on the screen and I'm going to rewrite it in a slightly different way. I'm going to say that DX m is equal to DX m times dy R multiplied by dy bar and you'll notice the difference is that instead of Phi in equation one I've used X M instead of using M as a dummy addition term. I've used ah and changed the x and y coordinates, there's no reason why I shouldn't, so now I'm going to take the value of DX m, which is this according to equation 1, and substitute it in. in this equation here and then I'll do the same thing for DX n, so what I'll get is that D s squared is equal to Delta MN, which is this term here plus the addition term, which we don't need to worry about. too much about Einstein actually dropped it, he said it was implied, he didn't even bother to put it in, but I'll leave it in there to show that there is a sum and then instead of DX M, I'm going to Put this term here which is d X M for dy R multiplied by dy R and then we are going to multiply by dxn, which I am going to put here the equivalent version which will be DX and for dy now I need another addition term call it multiplied by dy s so this term is simply this term is equal to this this term is this modified term is equal to this and now I can rearrange it so I can say that D s squared is equal to Delta M n we need a sum term. just put the two gradients together DX M times dy r DX n times dy s and that leaves us with these two terms here dy r dy s and that's like it's a Pythagorean form, but this term here is called the metric tensor is G M N and if If you look , you will see that there is an equivalence with what we derive here as Pythagoras.

Pythagoras was simply the sum term with a chronic delta that made sure you didn't get any cross terms. Here we have exactly the same thing as us. We have dr squared equals the combination of these two elements here, but instead of just having a simple chronic delta we now have a more complicated metric tensor, so we can now write that d s squared is equal to the metric tensor g MN multiplied by dy r dy s and of course there is a sum a sh gM contains a sum term for flat space, it is quite obvious that g MN will simply reduce to the Kronecker Delta term which should be d m n Delta MN, so will be 1 when R is equal to s and will be 0 when R is not equal, so for flat space the metric tensor will be 1 or 0 depending on where R is equal to s or not, but here is the problem, suppose you are not in a flat space, it is perfectly true. that if that is DX 1 and DX 2 and that is d s, it is true that in flat space D s squared is equal to DX 1 squared plus DX 2 squared, that is Pythagoras in flat space, but that is not true if you take a sphere and you try to draw a triangle on the surface of a sphere on the surface then this formula that we derived which is simply that of Pythagoras, that is not true on the surface of a sphere and how do you make the correction for that surface the correction is made by introducing this metric tensor.

So if you want, you can think of the metric tensor as the device that makes corrections to Pythagoras when you have the right triangle in curved space instead of flat space, so that you can start to see the value of the metric tensor si Let's talk about curved spacetime, so here are the Einstein field equations or here are the Einstein field equations and the terms g mu nu are the metric tensors that we just derived , why do they have mu nu instead of em n? that you use mu nu when you also include spacetime, what I just derived was only associated with space m and n if you want to introduce time as well, then you say mu nu where mu can now be 0, meaning that is the component of time or 1 2 3, which are the three components of space, so that's the metric tensor and now you might want to take another break.

Well, now we proceed to the third part, the Christoffel symbols. Now you will remember that I said that the important thing about turnbuckles was. which represent a fixed relationship between two vectors and are independent of the coordinate system, in particular if a tensor is zero in one coordinate system, it will be zero in all coordinate systems, so suppose we have a tensor called W and We will take the coordinates N and M in the reference frame X and say that is equal to another tensor also with coordinates N and M in the reference frame the reference frames to take my field example right at the beginning if the height of the field is 2 meters, no matter what reference frame you are talking about, it will always be 2 meters, but here is the problem: The problem is that the Derivatives of tensors don't necessarily transform between reference frames and we're going to need derivatives of tensors, so we need to know what goes wrong, so let's take a tensor in reference frame let's say that is equal to the derivative of a vector V with respect to N in the reference frame X now the question is, is that same tensor in the reference frame y equal to the derivative of the vector M? with respect to Y in the Y reference frame, that's true, so the answer is no, unfortunately, and we'll find out why I go back to equation five, which should appear on the screen, but I'm going to rewrite it exactly. as it was that the tensor T m and a TM n in the y reference frame is equal to the XR times dy m times DX s times dy n times t RS in the addition term just like Einstein did and the s and our terms are dummy variables that you need to add over them, but let's leave out the addition term, well, let's rewrite what it will be.

I'm just going to rewrite this. terms here DX r times dy m multiplied by DX s multiplied by dy n but instead of TR s in the reference frame and s so this now becomes the above equation D V R X divided by or works with respect to DX s. I just wrote t RS in this form here a term DV for DX now if you look at that tone combination there We have what is essentially the inverse of the chain rule. This is a DX term. V. This is a DX term. In a sense, you can collapse this to just become DV R times D Y N and so you get T M and Y, which are. what we're trying to calculate so that T MN in the y reference frame is equal to this term here DX R times dy M multiplied by this contracted, which is just DB R times D Y, so these DV are in the frame reference x / devo dy n and the question we want to ask is is it equal to DVM times dy n in the reference frame y?

So is it equal to equal? question mark, is it the same? because we don't know if it is equal to DVM in the reference frame y times dy. n well, what is this term DVM times D Y n, let's write that V M times dy n? I'm going to use equation 5, which is actually a tensor term, but remember that with tensors you can reduce two vectors by simply removing one of the indices and if you only do it for a single index and you can see that dy DV M times dy n is actually equal to the derivative of with respect to Y n of DX r times dy m multiplied by VR in the reference frame with an index but this is a differentiation of a product and you should know that if we differentiate a product we differentiate a by B then the differential of a product will be a multiplied by D B plus B multiplied by D a, so if this is a and this is B, so hedifferential of that differential from to B will be equal, so now we have that DVM and times Y n, which is this term here, is equal to a DB which is DX R times D Y M multiplied by DB which is D V R in the box dy n which is d B plus B which is V R here is equal to this term here TM n and is equal to that term there and we were asking if that term there is equal to this term and this term is equal to yes we' We've got exactly the same answer that we wanted.

Those two are exactly the same, except there's an extra term here, so they're not actually the same. We have this extra thing here and that combination of differentials. That little combination is called uppercase gamma gamma. In reality the title has been given to the Christoffel symbol and we represent the indices that appear here by putting as there is an R above we will put an R above and as there is an N and an end below we will put an N and the name below so that symbol that symbol Christoffel's uppercase esegamma is simply a shorthand way of writing this double differential pair.

It is the additional term that says that those two things are not the same. In other words, let's be clear about what we're asking: Does TM n in the frame of reference y equal DVM times D Y N and the answer was no, it doesn't because there is that extra term and the way we avoid that extra term is that we say that t MN y is equal to a new concept that we will call the covariant derivative, that is what that symbol for M means so it is not the ordinary derivative that we show, we will not call it the covariant derivative and what is the covariant derivative?

Well, the covariant derivative will be the ordinary derivative, the VM times d and n, which is what we thought it was. but it wasn't more the Christoffel symbol plus the correction term multiplied by the arc ex and now we have a transformation where you have derivatives that have to include this Christoffel symbol, it's the extra term and I'm going to call that equation 7, so let's put a box around and now we can ask the question: what is this Christoffel symbol and is it the compensation for the fact that the ordinary derivative of a tensor in a reference frame does not transform, the covariant is needed? derivative to be able to obtain a transformation of a derivative of a vector that is part of a tensor and that is important because many of Einstein's field equations required derivatives of tensors, well, we did it using the vectors that we are going to need to expand that For the purposes of tensors, then let's take the covariant derivative.

I'm going to have to use P because I'm going to use M and in a moment because I'm going to do the covariant derivative of the tensor T M. and that will be if you look at the equation that we just derived, which is equation number 7, which essentially shows if I double that for two indices, I'm going to get the ordinary derivative t MN times T and PE, you'll see how that follows from equation seven, but now I'm going to need Christoffel symbols, so I'm going to need one for the element M and one for the element N , so this will be more another Christoffel symbol for the element M and that is the transformation of a tensor from one reference frame to another and we will call that equation number eight now what about the covariant derivative of the metric tensor that we derived before?

Let's remember what the metric tensor was that we started with. Pythagoras where this was you like, you could write that as the sum of DX m DX n multiplied by the Kronecker Delta M n where remember that the Kronecker Delta was 1 when M is equal to n and was 0 when M is not equal to n and we said that in flat space is essentially G MN for flat space, not true for curved space, in fact that is the reason you have the metric tensor G MN, correct Pythagoras for curved space. Now, what is the derivative? We will call it covariant derivative of G MN, let's say with respect to R in the reference frame the covariant derivative of a constant in a flat space is 0, but what did I tell you? about tensor derivatives, if you have a value of 0 in one reference frame, then you have a value of 0 in all reference frames, so the derivative of the metric tensor and the covariant derivative of the metric tensor will be zero in all frames of reference, which means that if we take this formula here, equation number eight and replace it, just write it again but put T, sorry, put G instead of all the t's, we will have a transformation or a value of the covariant derivative. of G in terms of all these terms, but the covariant of G, of course, is always equal to zero, so if we replace T with G, then everything becomes zero, so let's do that.

I'm going to take equation number eight and I'm just going to write it back exactly as it was, but this time I'm going to substitute G for all the t's, so I'm going to get DG m n times d YP plus the Christoffel symbol p.m. and R of G are n plus the Christoffel symbol P n R of G M R and well, now I can tell you that the covariant derivative is equal to zero, so I can set that as zero and now I have an equation that has the tensor metric in it and a derivative of the metric tensor and the Christoffel symbol, so you should be able to rearrange that equation so that you can express the Christoffel symbol in terms of the metric tensor.

Now for that you need a mathematician so move on to Your friendly math teacher and I dare say Einstein did the same thing and he will tell you that if you have a Christoffel symbol with just a B and a C then it will be equal to breathing deeply 1/2 G a from where D is an additional term, it becomes an additional sum term to the derivative of G DC / DX B or by DX B + remember that mathematicians came up with this, so, Naturally, we take their word for it. I hope this fits into d G BC times DX D and I'll call that equation 9 and what you have, courtesy of the mathematicians, is that you took this term that we managed to differentiate ourselves and you introduced it to have it because it is equal to zero, you have the gamma terms, the Christoffel terms expressed in terms of the metric tensor and the first derivatives of the metric tensor, so what can you say about this Christoffel symbol?

First of all, it is not itself a tensor, it is in the sense that it is a Christoffel correction term. The symbol will be equal to zero in flat space but will not be equal to zero in curvilinear spacetime or spacetime and you will also notice that it can be expressed entirely in terms of the metric tensor and the first derivatives of the large metric. tensor and you might wonder why we've spent all this time trying to find something called the Christoffel symbol when it doesn't even appear in Einstein's field equations. You won't see a Christoffel symbol in that equation.

In those equations, the answer is no. you won't, but that's because the Christoffel symbol is buried in the Ricci curvature tensor and that's what we're going to do next, but first I think it's possibly time to take another break, so now we move on to the curvature if we go. to talk about curved space-time have we ever understood what curvature means let's take a path in flat space flat space fell flat space-time my role is flat what you do is take a vector a vector of course will have magnitude and direction and what you do is what's called transporting it parallel around the path, that means you keep the length of the vector the same and you keep it oriented the same way so that it travels parallel to itself, so all those lines are intended to be my drawing I'm not making it too obvious, but they are meant to be the same length and all the vectors are parallel, so they stay parallel until you return to the starting point and of course when you return to the starting point of the vector. it will be the same length and will be in the same direction because it always moves parallel to itself, so it will return to the point where it started.

Okay, all well and good, but now let's take some code. Here's a cone I made. For illustrative purposes, as you can see, it's a curved cone, but what I can do is cut the side of the cone and open it up, so if I make a cut along a line, I can open up the cone and if it's flat and it has that kind of shape, so let me recreate the cone. Just put those two lines together when it's a cone, that edge is joined together, so in a sense this edge here and this edge here are the same line, the same point on the cone, so let's draw what the code looks like when you open it , you cut this point here, you open it up and you get a shape that looks like a circle, but with a portion removed, that's the cone, just a quick reminder. there's the cone and remember that the line o ay and the line OB when they're in a cone are the same line, it's that dotted line now I'm going to do a little parallel transport and to remember where the vector was, let me just remove that V of the path and put it there at B.

I'm going to have a vector that continues the line OB, so it's a straight continuation and now I'm going to parallel transport that vector around the cone. Remember that it is still the same length although mine may not always stay parallel and we continue until we reach a and we stop because a is point B remember that they are the same points on the cone but what has happened well the length of the vector is the same but the direction of the vector has changed, you can see that it has changed direction because I started it as an extension of the line OB and OB is the same as the line or remember it is this dotted line here, so if the vector was in the same direction it should end up like that, but it didn't, it ended up like that, it went through an angle, so even though the length stayed the same, the direction of the vector changed and that's a measure of curvature if you carry a vector in parallel. around a surface and you don't go back and you go back to the same position remember that position A is position B you go back to the same point but the vector has changed direction so you've turned around it happens that's how we are I'm going to define curvature, a angle change in the vector when you transport it parallel you transport it around a surface if it's a flat surface as we don't show any change in direction if there is a curved surface it changes direction now I'm just going to take a slight deviation. and a pause to address two definitions, the first is what is called a commutator tsar commutator written like this between brackets a and B and all it means is a B minus B a a multiplied by B minus B multiplied by a now, of course, If we are simply talking about pure numbers then that will be equal to 0 3 times 2 minus 2 times 3 is 0 the point about the commutator is that they are not always equal to 0 and particularly I am going to look at one which is the commutator of D times DX and some function of Remember that the commutator is B minus B a, so it's going to be those times. that minus that multiplied by that, so let's write that which will be d times DX of F as we did before we can say that D a B is equal to BD a plus a DB no matter which way you do it differential of a product BD a plus a DB where this is a and this is B, therefore this term here using this formula here will be equal to B, which is V times da, which is D FX times you'll see that that term and that term are the same, so one minus the other is zero and you're left with V times DF x times DX, well, V is something that we just plugged in, so now we can take that and now you have to the commutator is equal to DF I'm going to have a distance that I'm going to call DX mu and I'm going to have a distance like this that is new DX, so a distance DX mu in the view X coordinate system and a new DX distance in the new X Coordinate what I'm going to do is build a parallelogram from that, so essentially it's going to be a parallelogram where the length of these two is going to be DX mu and that's the same thing there and the length of these two sides D X again so far so good and now I'm going to parallel transport around this path we're going to go from a to B to C to D and back to again but I'm going to call the return point prime because I'm not sure we're going to get it back, the victor will be exactly the same when it comes to a prime, so the vector in a prime may not be the same as the vector in a, if we have traversed a curved surface, now let's take two paths, consider two parts, the first is VC minus VD, of that is essentially what we're talking about, the difference between the vector at point C and the vector at point D and compare thatwith the vector at point B and the vector at point a now, if we go around a flat surface, then C minus PD will be equal to VB minus VA and that whole term will be equal to zero, but what this really represents is the differences in the D mu direction because it's the or the DX mu direction because those are the directions that we're moving here similarly, if you take C minus V B, which is this path here minus VD minus VA prime, you get the difference in the new direction DX because that is that path there and that path there are parallel in flat space there would be no difference but in curved space there could be and now you are looking at the difference in the new direction D now let's look at the difference of the differences if we take this term here and subtract this term here, what do we get?

Well, let's take a look at the term VB, we have a VB minus and the VB minus minus which becomes a plus, so we get a minus. plus zero we don't get a resulting VB term, then we take a computer, we have a VC plus minus VC, the VC term becomes BD, we have a minus VD, minus VD, so it's plus minus plus the VD term as well. What's happening? VA we have a minus minus, so that becomes a plus, so we have VA and then we have for VA Prime, we have a minus minus minus, so threeminus makes a minor, so minus V becomes a prime and that is the difference of differences that will be the change in the vector as it is transported in parallel around that parallelogram, as I say, in flat space there is no problem, DV will be zero because there will be no change, it will just be transported in parallel and it will return to the same magnitude and the same direction, but it is in a curved space, there will be a difference not in the length but in the angle of the vector.

Now let's take this first term here VC. less VD, what is it really? Well, that's the change as the vector goes from C to D over a distance DX mu, so VC minus VD is simply the change in the vector over the distance X mu multiplied by the distance that you go through, which of course is DX mu multiplied, of course, by the vector itself, so what we're saying is that the change in the vector between C and D is the gradient. Remember we did this from the beginning with our field, gradient times distance times of course. the value of the vector, but we have just learned that we must be very careful with ordinary derivatives, when we talk about curves it is necessary to use the covariant derivative, so, to be more precise, we better call it the covariant derivative.

To use this gradient we are going to use the covariant derivative multiplied by DX mu, that is this term multiplied by the value of the vector. It's a much healthier way to do it because we just learned that you have to be very careful with ordinary derivatives. When you are transforming between different reference systems, what is this term BC - VD minus VB minus VA? That's the difference in differences on a line that's separated by the distance DX nu, so essentially what you're doing is you. We're looking at the difference of vectors with a separation of DX nu, which means you have to apply a new gradient to it, so VC minus VD minus VB minus VA, which is this term here, which is, like I said, the difference in the vectors in those two. lines separated by different distances DX, well we said that VC minus VD is the covariant derivative multiplied by the distance multiplied by all that, sorry I didn't know that V multiplied by the vector value, but the difference is essentially another differential and it is the differential in the new direction. so now we have to add the covariant derivative in the new direction multiplied by the distance distance, so that's effectively what that difference equals.

What's up with this one here? Well, this will just be the opposite, in other words, VC minus VB minus V. VA prime is going to be good, VC minus VB will be the covariant derivative, but this time in the new direction, so now we need the covariant direction in the new address multiplied by DX nu times the vector value, but this subtraction means you're looking. in the difference between this line and this line separated by DX mu, then you're essentially applying another differential, but this time in that direction mu, so now we have that term is essentially that term and this term is essentially this term and what did we say? that if you subtract that or if you subtract that term from that term you get VA minus V prime a prime that is the difference in the vector the difference between the vector at the initial point and the vector at the final point, so if if you subtract this term from this term, you will get DV, the difference in the vectors, so let's write it down.

DV will be equal to that term minus that term, which is gamma nu, sorry, a highly derived new covariant derivative, mu no. It doesn't matter what order we put them in DX nu DX mu V minus this term covariant derivative mu covariant derivative nu DX mu DX nu V and we can take out the DX nu DX mu and V because the author again doesn't matter here, so that becomes DX mu DX nu V in what's left, which is covariant new covariant mu minus covariant mu covariant nu and what you will recognize, of course, is simply the definition of commutator which, of course, is simply Delta Nu comma Delta mu not the derivatives Delta covariants off, but you will remember that from Equation seven the covariant derivative is equal to the ordinary derivative, by the way, when I write D nu like this, what I really mean is D times X nu, it's just an abbreviation plus a term gamma.

I won't write the full one. Details of this, but there was a gamma term. Remember that that was the corrective term, the Christoffel symbol term, so wherever we have this covariant form, we will remember from Equation 7 that the covariant form is the ordinary derivative plus this corrective term involving the Christoffel symbol. So on that basis, the commutator that we identified earlier is going to be equal to that times that minus that times that, that's what a commutator is, but remember that the covariant derivative is the ordinary derivative times the assortment plus the Christoffel term, so now we get that. that is equivalent to this term is d nu plus a Christoffel symbol associated with nu multiplied by this term which is d mu plus the Christoffel symbol associated with mu minus this term multiplied by this term I hope you can see that and let's multiply all of that then we have a term D nu multiplied by D mu, then we have a gamma nu multiplied by D mu, then we have a Dinu multiplied by gamma mu and finally we have a gamma nu gamma mu and that's all multiplied. from that we have to subtract this term, well let's multiply that, we have a d mu by d nu, those should be DS curly, by the way, all of these should be DS curly, plus we have a term D mu gamma nu, so we have we have a term gamma mu D nu and finally we have a term gamma mu gamma nu, well, let's do the subtractions D nu D mu minus D mu D nu is equal to 0 because for ordinary derivatives the order does not matter gamma nu D mu minus D mu gamma nu will allow a commutator to actually be equal to minus the commutator of D mu gamma nu gamma nu D nu is equal to minus the commutator of D nu gamma nu we have another commutator here D nu gamma mu minus gamma nu D nu well that's it just the commutator of D nu gamma mu so we have gamma nu gamma mu minus gamma mu gamma nu and that again will be the commutator of whoops the commutator of gamma nu gamma mu so just remember that was the commutator of the two covariant vectors here or two covariant terms here becomes this term here, but you may remember that when I took that detour to talk about the z' commutator I said that if you have a function like that, it is equal to the derivative of the term gamma with respect to X mu and this is becomes the derivative of the gamma term with respect to I can tell you that for our purposes it is also and can be considered as the Ricci tensor that we need and that is why we needed to derive the Christoffel symbols because as you can see the Ricci tensor contains Christoffel symbols. and derivatives of the Christoffel symbols and the Ricci tensor remember that was the first term in the Einstein field equations that I'm showing you now just to remind you, so what we just showed is that the change in the vector remember the angle of the vector instead of the length is equal to DX mu DX nu multiplied by V multiplied by the commutator covariant nu covariant Moo and that term we said was the Riemann tensor, but for our purposes it is also the Ricci tensor, So let's take stock of where we got to.

I just showed that the Ricci tensor is formed by Christoffel symbols and the derivatives of the Christoffel symbols, but our Christoffel symbols will go back to equation 9 and you will see that the Christoffel symbols are formed by metric tensors and the derivatives of tensors metric and What are metric tensors? They can be thought of as the device you need to correct for Pythagoras on a curved surface. Now what I can also tell you is that from the Ricci tensor arm you knew that you can derive what is called a scalar of curvature. simply a scalar, it is not a tensor, it is not a vector, it is just a scalar, like for example, if you take the scalar product of two vectors, you get the scalar, you can take the Ricci tensor and from it you can create a scalar of curvature about the point that is, if the scalar curvature is not zero, then the surface is not flat, well, let's review where we have we have covered the metric tensor G mu nu, we have covered the Ricci tensor R mu nu, which includes many Christoffel Symbols, we have covered the curvature scalar R and therefore we have covered most of the Stein field equations.

Only two more terms remain: the tension energy momentum tensor T mu nu and the cosmological constant uppercase lambda, but the first time for another pause. Okay, now we are ready to make the tension energy momentum tensor T mu nu What is the shortest distance between two? Answer a straight line. What is the shortest distance between two points? Let's say that on the surface of a sphere there are not two points on a circle but two. points on the surface of a sphere the answer is the shortest distance is what is called geodesic and in a sphere on earth for example the equator is a geodesic any line of longitude is a geodesic the shortest distance between two points where there are that If you go around the curve, you can't go in a straight line, it's called a geodesic and we can define what's called a tangent vector DX mu times D tau as a rate of change of the distance we travel with respect to time. tau because tau is what is called proper time, you can find out what proper time is in my videos on special relativity.

The important point about proper time is that it is the time at which all observers, regardless of frame of reference, can agree, so if we use tau we know that there can be no arguments about what we mean, so We are analyzing the rate of change of distance with time and that is called a tangent vector and what we mean is that the derivative of that tangent vector must be equal to zero. To find the minimum distance, generally speaking, if you want to find the minimum on a curve, you take the derivative, you take and set that derivative equal to zero and that will give you the minimum, it also gives you, incidentally, the maximum, but in In this case we are talking about the minimum, so if you take the derivative of the tangent vector and set it to zero you will obtain, so to speak, the path for the geodesic, the shortest distance.

Now when I say take the derivative, remember that we should always be careful with derivatives in general relativity, we have already learned that it can turn out bad if we take ordinary derivatives, we should go for the covariant derivative, so let's take the covariant derivative of this vector tangent DX mu times D tau and from Equation 7 that should be on the screen, you can see that it will essentially be the ordinary derivative plus a gamma term which I won't fill in, but there is a gamma term, remember that is the corrective term and what We can say that this term is for geodesics. must be equal to zero so you have this term Also this term is equal to zero let's see this term here this is essentially a double differential with respect to time is if you like D a X mu D tau squared what it looks like that is t that acceleration the rate of change of the rate of change of distance with time D to see that?

Let's remember that general relativity and Newton's law of gravitation should be the same when you're just talking about ordinary masses, ordinary velocities, low velocities, we're not talking about black holes because Newton's law of gravitation describes very well what appears to be gravity, etc. the simple case of general relativity and Newton's law must conform to the same thing. Newton's law says that force is mass multiplied by acceleration or if you prefer, acceleration is a force term divided by mass and here we have aacceleration equal to minus one gamma. term, then what we can say is that this gamma term is the equivalent of the force in Newton's laws and there will be a mass term as well, but the mass is constant, so let's forget that this gamma term, this Christoffel symbol, it has a kind of broad equivalence to the force now to save time I just rewritten the equation nine that we derived for the Christoffel symbol remember that we said that it consists of metric tensor plus derivatives of the metric tensor now let's consider a situation where we are talking about low gravity and low speed we are not in black holes, we are nothing like that, it is just ordinary space, slow motion, general relativity must reduce to not to Newtonian gravity, when that happens G becomes 1 and the derived terms are very, very small, with one exception and that is the The time term and time remember that mu and nu are equal to 0, so the time component DG 0 0 0 times DX is the only term in the derivatives that has some meaning, everything else when talking about small, low-speed masses, without black holes.

These other terms are very, very small, the only one that has any meaning is the time term and this G term becomes 1 and that gamma term reduces to 1/2 and G becomes 1, so we can ignore that multiplied by this whole lot which is simply DG 0 0 0 times DX and which we said was equivalent to a force term, so the force is roughly equivalent to this term here, but in Newtonian mechanics the force is minus the derivative of the potential, for example, if the potential energy is near the Earth. is equal to mg in one direction that X is measured in the other direction, so all I'm showing here is that in Newtonian mechanics the force is minus the derivative of the potential.

Now we just show that the uppercase gamma of the Christoffel symbol is the equivalent of a force term. and in ordinary space where we are not even remotely black holes that equals the derivative of the time component of the metric tensor with respect to see that if gamma is a force term in which these two terms are broadly equivalent and therefore you get that G zero zero is equal to 2 Phi plus a constant from the integration you need to do, but you can completely forget about the constant because as soon as we go back to differentiating that constant will go, there's probably a minus sign somewhere, but we don't need to worry too much about that.

We're looking at the principle here, what we're saying is that because the Christoffel symbol is equal to that and the Christoffel symbol has a kind of force equivalence and in Newtonian mechanics the force is equal to less than if Rachele of the potential energy, therefore, you can equate these two terms and if you equate those two terms essentially the metric tensor or the time value of the metric tensor is equal to two times the potential energy plus a constant now we said that the force is minus D Phi sorry, it's minus D Phi times DX but of course we really need to do that in three dimensions and that usually comes down to what's called the Phi divergence. where the divergence term is simply a D term times DX plus a D times dy plus a D term times DZ D and that of course should be minus there, but we also know from Newton's law of gravitation that the force is equal to minus G, the gravitational constant. multiplied by the mass of one object multiplied by the mass of the other object divided by the square of the distance between them, so let's consider an object of mass M and consider a distance R from mass M and ask what the force is. acting on a unit mass, put a distance R from M.

Well, if the unit mass, that means the unit mass is 1, the force will be minus G M over R squared because M is 1, which is, Therefore, the force capacity on that entire surface? sphere, well, that will be the integral of F da, where da is the surface area of ​​the sphere and that will be the integral of the force, which is minus GM over R squared multiplied by da, well, if you integrate that, of course, it's just going to be equal to minus GM over R squared times the full surface area of ​​that sphere of radius R which is 4 PI R squared the cancellation of R squared and then you'll find that that reduces to minus GM times 4 pi now there is a theorem there called the divergence theorem which I won't derive but you can look it up.

I'll write it first. It says that the integral of F Da over an area is equal to the integral over the volume of Dell F DV and What that means in simple terms is that the outflow through the area of ​​a sphere is equal to the integral of volume of the divergence of force. That's just a divergence theorem that can be derived and you can look it up. I also want to remember you. you know that density is mass divided by volume and therefore you can say that the mass of an object will be the integral of density with respect to volume so now I'm going to take F given that we calculate as this term and set equals this term, so now you have that F dot Da is four pi G times M or more exactly minus four PI G times M, but for M I'm going to use the density integral Rho DV times volume, that's F da or more exactly this is the integral of F dot Da and that will be the integral of the volume of DF DV or of F DV the divergence of F well, although it is not really appropriate to do so, you can cancel the D V essentially what you are left with is that of F the divergence of F is equal to four PI G Rev actually minus four PI G unfortunately I had a camera failure for the next section so I'm just repeating the flow of the argument that we've shown that the force is minus the derivative of the potential at Newtonian mechanics we have also just shown that the derivative of the force is equal to negative four PI G goes down consequently if the force is minus the derivative of Phi the derivative of the force is the derivative of minus the derivative of Phi and that is equal a minus four PI G Rho in other words minus the squared derivative of Phi B potential is equal to four PI G Rho the minuses on both sides cancel out before we show that the time component of the metric tensor G zero zero is equal to twice the potential plus a constant, which means the potential is half the metric tensor.

I have removed the constant and you will see that there is no problem without doing that because as soon as you do a derivative of Phi, squared derivative of Phi, which is this term here replacing Phi with 1/2 G zero zero, all the constants will disappear, of course , so now you have the squared derivative of Phi is four PI G Rho, so the squared derivative of half the time component is the metric. tensor is equal to four PI G Rho and if you multiply both sides of the equation by two you get that the squared derivative of the metric tensor is equal to eight pi gee ro the problem is that it is not a tensor equation and for general relativity we need equations tensors so what we're looking for is something that looks like this, but we're on the left side of the equation, you have a tensor and on the right side of the equation you have a tensor, so we want something that has the shape of G mu nu, which is actually called the Einstein tensor, is equal to 8 pi G, but instead of having the mass density term Rho, we want an energy tensor, something that contains all the energy stress pressure terms of mass that can be had. so essentially we have to find something for this and something for that.

Let's look at the T mu nu side first if you go back to my video on special relativity e equals mc-squared part 5 and you go to the point in the video that is 2 minutes and 10 seconds long, you will find that we derive what is called an impulse of four vectors, the four vectors, of course, one dimension of time, three dimensions of space and you'll find that we define that four vector momentum. as mass multiplied by X zero over tau X 1 over tau X 2 over tau and X 3 over tau and this is just, obviously, that's the time component, the of spacetime divided with tau, the right moment, the moment that everyone agrees and what we showed was that if you reduce this term, you essentially get a MC squared which is energy and is equal to MC squared and all these terms effectively become the momentum in the different coordinates, so MV in the x direction is MV in the y direction and MV in the same direction, so this for momentum reduces to the basic energy of the mass in rest plus the moments in the three coordinates of space, well that's fine, but of course it's essentially a vector, we need a tensor we don't need P that T mu nu and that means we need a four by four matrix and for save time I have drawn one and there it is, you will see that T has all the values ​​of mu and nu from zero to 3 and what do all these values ​​mean?

The top T nothing nothing that will be the time component of the energy tension energy moment tensor and that has this kind of equivalence is the energy part of the tensor of the three components along up here I want to call the parts of the flow of tensor energy the three components here I want to call the tensor moment density and the nine components that remain are essentially the tension pressure parts of the tensor moment flow now those are incorporating every part of the energy that we can imagine, since be it pure energy, rest, mass, energy, momentum, stress, everything somehow finds a place in this momentum tensor, stress energy, it's if you want a measure of energy per unit volume, now energy has the same units. as work, so we can think of this as work per unit volume work is force times distance distance is a length volume is a length cubed, so it's the equivalent force divided by a length squared, which is the force over area, which of course are the units of pressure.

This is how energy per unit volume can also encompass pressure or stress so we can't derive it but I'm trying to illustrate how we got to this point which has given us a stress energy momentum tensor with 16 terms different, some of which It turns out are duplicates, comes in matrix form, incorporates all forms of energy and is the strain energy moment tensor that we use on the right side of the Einstein field equation, so now we have on the right side. On the side we have the 8 pi G that we derived, but we're going to replace the density term that we came up with with this new stress energy moment tensor and we're going to have something on the left side, now what?

Could it be that this is like the mass term? On this side we need something that is the curvature term of spacetime, so Einstein thought the obvious candidate for it would be the Ricci curvature tensor, ah mutant, ooh, but there's a problem, the problem is. that we have something called conservation of energy energy cannot be created or destroyed so if you take the derivative of the energy tensor it should be zero there is no change in energy the problem is that if you take the derivative of the Ricci tensor , not equal to zero, so the equation cannot be true because the derivative of one side would be zero, the derivative of the other side would not take into account that we should always, when we talk about general relativity and field equations, never make derivatives ordinary. will come off, we should always use the covariant derivative, so if we have the covariant derivative on the energy side is zero, we need something on the left side whose derivative is equal to zero.

What Einstein actually found was that the covariant derivative of the Ricci tensor was not zero, but was actually half the covariant derivative of the metric tensor multiplied by the curvature scalar we talked about earlier. Do you have an equation like that? You can take everything to one side and essentially you get that the covariant derivative of the Ricci tensor minus half the metric tensor multiplied by the scalar of curvature is equal to zero and now we have a covariant derivative of the term that is equal to zero, we have the covariant derivative of t MV is equal to zero and the covariant derivative of this term is equal to zero, therefore that term and that term must be equivalent, so now we have the left side of the equation R mu nu minus 1 /2 G mu nu multiplied by R is equal to the right side of the equation when It's not just the equipment, you knew that, of course, it's 8 pi G T mu nu and we're almost there.

It turns out that for four-dimensional purposes you need to have a C in the fourth term here, but that's essentially just a constant for dimensional purposes, but Einstein realized that you had forgotten something, remember that earlier we showed that the covariant derivative of the metric tensor is equal to zero because you did it in one frame and therefore you must do it in all frames and therefore we could have included the metric tensor. in this equation here not only as a multiplier of R but byright and of course we could have included it with any constant in front of it, so Einstein decided that it was necessary to include this term in this equation again, so now you need to put together, you knew minus 1/2 G mu nu R, which is this system right here and now you need to include the fact that you could have had a metric tensor term here, so we're going to add that and we're going to have a constant that we're going to call capital lambda multiplied by the metric tensor and that's equal to 8 pi G T mu nu divided by C to the fourth this part, you will remember, was the Einstein tensor G mu nu what is this capital lambda term where it is called the cosmological constant and Interestingly, it is the constant that Einstein first thought of when he was trying to identify how space could be described in mathematical terms.

Now he was told at that time that space was fixed and unchanging, but that nothing moved in space if he looked at it. In a huge cosmological framework of time and space, obviously the Earth revolves around the Sun, but in general everything remains as it is, that's what people thought. Einstein said well, that can't be right because gravity would tend to make all the galaxies move closer together and if they tell me they're not coming together, there must be something that balances the gravity and pulls the galaxies apart so that the gravitational attraction is balance with what separates them and discovered that he could reintroduce that into his Einstein field. equations like the constant you multiplied by the metric tensor this term is a very small term and you will often find that it will fall outside the field of Einstein patience, it only becomes relevant when you talk about important cosmological scales and that's it.

We've sort of derived Einstein's field equations just to remind you that mu and nu are, of course, the time and space dimensions zero one, two, and three, and therefore there are sixteen equations there, but six of them They are identical to six others, so it is reduced to only ten. If you want more information, I remind you that Professor Leonard Susskind Leonard Susskind of Stanford University gave an excellent series in 2008 and I gave you the link at the beginning of this video, that That's it, we finished the Einstein field equations for

beginners

.