[SOUND] So, so far we have described gravitational waves, small gravitational waves, which are solutions of the linearized approximation to the Einstein equations. And now in the remaining part of the lecture, we're going to discuss exact solutions of the Einstein equations that are describing some kind of gravitational shockwaves. But our discussion will be not very detailed, more concise in this respect. The lacunas in our arguments remain as exercises for the readers. So consider Schwarzschild metric. So the Schwarzschild metric, as you remember, is as follows, (1 minus rg over r) dt squared minus dr squared, 1 minus rg over r minus r squared d omega squared. And let us make the following coordinate transformation in this metric, r = 1 + rg over 4 rho squared times rho. If one makes this transformation, to do it remains as a home exercise for those who listen to these lectures, one obtains the following metric. The result of the coordinate transformation is just 1-rg over 4 rho divided 1- rg over 4 rho + squared dt squared minus 1 + rg over 4 rho to the 4th times d rho squared + rho squared d omega squared. Okay, so why did we make this transformation? Because the obtained metric up to this conformal factor is flat, this partial part of it is flat. So this metric is flat, and we can make the following coordinate transformation. If we represent rho squared as x squared + y squared + z squared. So we can make in this partial part of this metric transformation to the Cartesian coordinate. So we can make the following change such that this d rho squared + rho squared times d omega squared is equal to dx squared + dy squared + dz squared. Now, once we made this transformation, we have the following situation. We can make a Lorentzian boost along the z direction. So make the following transformation. At t bar is gamma t + gamma vz. And z bar is equal to gamma vt + gamma z, where gamma is just 1 square root of 1 minus v squared. Of course, the metric is not Minkowskian. Hence, this transformation is not an isometry of this metric. And, as a result, the metric will change somehow. How remains another exercise for those who listen to these lectures. Well, it's not hard to see how it transforms. But then, we want to make the following limit. So we want to take the following limit. So we want to take gamma to infinity. So v to 1, so take the velocity to the speed of light. Yeah, the result of this transformation is of course black hole which moves with velocity v. So this was a black hole at rest. After this transformation, we obtain black hole which moves along the direction with velocity v. Now, we want to take this black hole moving with the speed of light. So we take this limit such that the mass of the black hole goes to 0, but rg times gamma over 2 is fixed constant. So we take gamma to infinity, rg to 0, but we keep this quantity fixed. This quantity we call as rg times gamma over 2 is kappa times p, where p is the modulus of the momentum in this limit. So what do we obtain? We obtain a light-like wave as a result of taking of this limit should be light-like particle, like-like force, moving with the speed of light with momentum p, finite momentum p, this is kept fixed. So the result after this transformation and taking this limit, this metric is transformed in the layering limit, it's transformed to the following expression. It's dudv + 2H, H is a function of u and x perpendicular. What is x perpendicular I will explain right now. Minus dx perpendicular squared. So this is the resulting metric after taking this limit. So let me explain the notations. x perpendicular vector is just x and y. So this is just dx squared + dy squared. Then u is just t-z over square root of 2 and v is t+z over square root of 2. And H of u and x perpendicular is just the following function. It's kappa times p, this kappa and p, kappa is Newton's constant, times delta function of u, times log of modulus of x perpendicular. So that's what we obtained taking this limit. The other way to see that this metric is a solution of GR equations of motion is to use these answers for the metric and calculate that the only nonzero component of the Ricci tensor for this metric is just Laplacian along x perpendicular, acting on H as a function of (u, x perpendicular). Where the Laplacian perpendicular, x perpendicular, is just dx squared + dy squared. So the Laplacian in the perpendicular directions to the direction of motion of this body. So using the fact that the Laplacian of log x perpendicular is actually 2 pi times two-dimensional delta function of x perpendicular. So basically, a log is the two-dimensional Green function. Using this fact in here, we obtain that this metric solves the Einstein equations with the following source, Tuu, is just p times delta of u times delta two dimensional of x perpendicular. So this is the other way to see that the obtained metric solves Einstein equation. This solution is called Penrose parallel plane wave, or pp-wave. It describes so-called shockwave whose world line, according to this, is described by equation u0 and x perpendicular equals to 0. So the world line of this guy is u = 0, x = 0, where the source is, it travels with the speed of light, because this equation just tells us that the wave travels with the speed of light. So the momentum of this wave is given by p, according to this expression. And this metric is defined beyond this point, beyond the point where the source is. Because where the source is, we have a singularity. Beyond this point, this metric is defined. And it is not asymptotically flat, according to this expression, and it has innate singularity, as one can see. Now we're going to describe properties of this metric. [SOUND] [MUSIC]