[MUSIC] Thus we have derived the following equation for the Geodesic in Schwartz's space time, dr over ds squared is E divided by m squared minus 1 minus r g divided by r 1 plus L squared over 2 m, sorry. No 2 m squares times r squared. No 2. Where e is energy and L is angular momentum. One should have in mind that this equation is applicable for r greater than rg because we're using Schwarzschild coordinates. And if one wants to derive geodesics below the horizon, under the horizon, one has to use a different coordinate system which is not singular at the horizon. That's one should bear in mind. Now, we want to use this equation. >> To start with, we want to show some of the statements that we have made during the previous lectures. For example, let us consider radial of a massive particle to the black hole. So mass is not zero, but because we have a radial motion L equals 0, angular momentum equals 0. So this equation simplifies to dr/ds squared equals E/m squared- (1- rg/r). So without loss of generality, we want to consider the following situation. That as r goes to infinity, dr over ds goes to zero. So, it means that when particle started its free forward to infinity with zero velocity. So from this equation, this goes to 0 as r goes to infinity. Which means that, from this fact, it means that e equals m. So the energy of the particle is just equal to its mass. As a result, this equation, this is one so this equation is simplified to the falling one. So, without loss of generality, we can consider this kind of motion. Of course, if we energies is not equal to one so, the partial studied this motion with known zero velocity and infinity, none of the conclusions that we're going to make is going to change. But in the circumstances that we consider, the equation is very simple, so, now we want to solve it to find how much proper time it takes for a particle. So our goal now is to find how much of the proper time it takes for a particle to free fall from the radius, R big, to the horizon, to r equals rg. So then, from this equation, we can find that proper time necessary to do this ref free-fall, is equal to minus R, rg, r over rg. One half, dr. So taking this integral, we find this time to be, this is a continuation of this formula, 2rg/3 [(r/rg)] three half minus one. So it takes finite proper time to make this free fall as we promise in the previous lecture. Yeah, here the minus sign is present because, for the free fall down the black hole, dr < 0, so the radius is decreasing, while ds > 0. So as time increases, radius decreases. That's the reason we have the minus sign here. So here we, taking square root, we can have two signs, so we do the minus sign. So, at the same time, from the fact that, remember that we have conservation law that (1- rg/r) dt over ds is equal to e divided by m. In our case, this is 1. So as a result, we have the following relation. We have a relation between coordinate time and proper time in the Schwarzschild spacetime. Using this, and this, we can find that dr over dt is equal to minus Rg over r to the one half, one minus rg over r. The minus sign, again, here for the same reason as the minus sign here. Now, for the same expression, we can find how much coordinate time it takes for a particle to make free fall. From my radius, r, to a radius very close to the horizon. So it's not rg, but some horizon. Epsilon is very small. So then, coordinate time necessary, to fall from the radius R to this rg plus epsilon). First of all, this is just by definition is minus R, rg plus epsilon (r divided by rg) one-half R dr, over r minus r g. So, one can see that as epsilon goes to zero, the leading contribution to this integral is proportional to Rg times log R divided by epsilon. So, the time goes to infinity as epsilon goes to zero. So it takes infinite time recording time for particle to fall from the radius r to the vicinity of the horizon, it also was observed during the previous lecture, but for more generic situation. Because it's not only the Schwarzschild observers which matter this time, do not see anybody crossing the horizon. Any observer which stays outside of the Schwarzschild horizon C is never seize anything crossing the horizon as we explained in the Pendrels diagrams during the previous lecture. [MUSIC]