[SOUND] To illustrate the idea of what does it mean, transformation to an arbitrary coordinate system, let me make a transformation to a simplest non-inertial reference system. I mean I want to, in flat Minkowski spacetime, I want to describe a transformation to a accelerated reference frame. Simplest accelerated reference frame is the one which has a linear acceleration along the line. One has to bear in mind that three dimensional acceleration cannot be constant in a relativity system, because a squared a vector squared is not a Lawrence invariant. Hence, we have to consider four dimensional acceleration to be constan. We know that three velocity, what does it mean for acceleration? Four acceleration is just first derivative of the four velocity with the respect to the proper time. Not that in my lectures I do not distinguish interval from the proper time, because we have set the speed of light to be 1, and there is no distinction between interval and proper time. And for velocity, Is the differential of the world line of the particle with respect to proper time. I write the world line of the particle, I denote it by z nu to distinguish this z nu from the coordinate, to distinguish them from the coordinates in space-time. So this guy is just a function of s, and it maps, Range of validity of s, which is some interval, to the space-time. 3,1, it has three spatial dimensions and one time dimension. And it's image is a world line of the particle. So this is just to set up the notations. So one can see, one can see that this four acceleration is just the second derivative of that. So, let me stress one thing, that for velocity u to this, u to each definition Is just vector of a unit length, and because its square is positive, this vector is timeline. If we differentiate both sides of this equality, we obtain that W mu times U mu is equal to zero. It means that the four vector of four acceleration is octagonal to the four velocity. As a result, this vector is space like, because this one is time-like, this one should be space-like. It means that it's square should be negative. As a result, we assume that it is -a squared, where a squared is obviously some positive constant. So this is an equation specifying a motion with a constant for acceleration. Let us now plug this expression for the four acceleration into here. Namely let me specify reference system in space such that the particle accelerates along the first direction. As a result this equation acquires the following form that d2 zed 0 divided by ds squared square minus d2 of zed 1 of ds squared square equals 2 minus A squared. Obviously now, one can see from this equation that components of four acceleration combine a hyperbola. This is an equation of the hyperbola. Then, one immediately guesses the solution of this equation. That zed 0 is 1 over a hyperbolic sign of (a s), while zed 1 is 1 over a hyperbolic cosine of (a s). Indeed, if one will differentiate this, zed 0 with respect to s, two times, and zed 1 with respect to s two times. You will obtain this equation, due to the fact that hyperbolic cosine squared minus hyperbolic sine squared is equal to 1. Let me make a small change, for future convenience. A small change in zed one, as follows. -1. It will become apparent, why I did this change, in a moment. So now, one can see from this equation that zed 1 and zed 0 themselves obey the following equation that zed 1 plus 1 over a squared minus zed 0 squared is equal to 1 over a squared. It means that the particle which moves with constant for acceleration, it[s world line looks as a hyperbola, let me draw it. So it's t, x, they're asymptotes of this hyperbola as follows. It's a light like line, two light like lines, two light like lines composing, this angle is 45 degrees. And the hyperbola along which the particle is moving has a falling form, just looks like this. And I see equations specifying the asymptotes are as follows, that zed 1 is equal to plus, minus zed 0 minus 1 over 8. These are equations specifying the asymptotes of the hyperbola. So, let me describe in the standard terms what does it means that there is a motion with the constant four acceleration. Again, let me say that the world line of a particle which moves with constant full acceleration looks as follows. It is given by the following equation, that zed one plus one over a squared plus zed zero squared equals minus one over A squared. This is an equation of a hyperbola, so a let me just write a whats a clean surface of zed of 0 is just 1 over A hyperbola, a sorry plus sign here of course, 1 over a hyperbolic sine (as) is that 1 is 1 over a hyperbolic cosine (as)- 1. And so the picture, how the world line looks like is as follows, T, X1 and we have to life like asymptotes whose angle is 45 degree with respect to this axis and to the other axis of course. And the world line looks as follows. It's hyperbola, so one can check straightforwardly that three acceleration, three dimensional acceleration has the following form. First of all, the definition of the three acceleration is the second derivative of zed one with respect to zed 0, and from these formulas, one can find it to be equal as follows. A over hyperbolic cosine of (as) 1 minus hyperbolic tangents (as), And this is always greater than 0. What does it mean? It means that this part of the trajectory of the world line is a deceleration. And this part is actually acceleration. So the velocity of the particle here is just 0. Because the tangential line to the hyperbola is just vertical, so vertical has zero velocity at this point. And let us stress that at this point, the motion is just the standard three dimensional acceleration. Let us see that. When (as) is much less than 1, so when the particle's velocity is not big enough. So particle's just basically starts to accelerate around this point, we have the following situation. That zed 1 is just a zed 0 squared over 2, approximately. Which is a equation defining just the standard three dimensional acceleration. And when the particle becomes very, acquires a very high velocity due to this acceleration, it's where a line Is changed according to this equation. In fact this equation can be obtained from this equation by easy means. You see that zed 1 from here is just a square root of 1 over a squared plus zed 0 squared minus 1 over a. And if this guy is much smaller than this guy, we can taylor expand the square root, and that's how we get this expression. This is another way to obtain the same equation. So obviously we obtain accelerated motion. Now we want to transfer to the reference system of this kind of motion. To the reference system of a homogeneously constantly accelerating observers. How do we do that? Well that's what we're going to do right now, but let me just stressed that motion along this hyperbola from the point of view of the one who's moving along this hyperbola is absolutely homogeneous. The one who moves along this hyperbola cannot distinguish its point from any other point. It's indistinguishable. It's only from this picture this point is different from the other etcetera. But if you are sitting in the reference system, which moves with constant acceleration, this point is no different from this one or from this one. In fact, because like if you change your reference system, which is inertial and moving with instantly moving with this particle at this point. Well the hyperbola just turns around, it transforms and this point is transferred to this one and corresponds to the particle, which is having zero velocity. But again the motion along this hyperbola is homogeneous. And it's invariant on the proper time translations, and time reversal. And as a result, we expect that the metric in the co-moving non-inertial reference frame will be stationary. So let us see that explicitly. Inspired by these formulas, by these equations, now we will write our information. Let us write the full and current information. Namely, let us take x 0 to be low, hyperbolic sin of tau and x one to be rho hyperbolic cosine of tau. Rho and tau our new coordinates and x zero and x one our original coordinates, because of that, we can obtain the following that dx0 is just d rho hyperbolic sine of tau + rho d tau hyperbolical sign of tau. And also dx1 = d rho hyperbolical sign of tau + rho d tau hyperbolic sign of tau. Now, squaring this and subtracting from dx, the x zero is just t. To not to spoil the notations. So squaring this and minus square of this, we obtain the following, it's an easy exercise. It's an easy exercise following from here, that dx 0, or dt squared, sorry, dt squared minus dx1 squared dx1 squared is equal to, after opening these brackets and canceling equivalent terms, we obtain the following. D tau rho squared d tau squared minus d rho squared. [SOUND]