[MUSIC] >> Let us continue with white light particle motion and Schwarzschild background. So we have obtained the following geodesic equation for the geodesic in the Schwarzschild space-time, which is as follows. m^2 * rg / 2L^2 + (3rg / 2)u^2. So this is for the massive particle motion, but for the massless, this is 0. So we obtain the following equation for the massless particle. (d^2u / dφ^2) + U = (3rg / 2)u^2. Again, we consider this term as a perturbation. So we, in the Newtonian limit, the equation that we have, so in the Newtonian limit, the equation that we have is as follows. (d^2Uo / dφ^2) + Uo = 0. And the solution of this equation obviously is Uo, which is equal to 1 / ro. Again zero subscript corresponds to that, this is Newtonian limit, letting approximation. The solution of this equation is just (1 / a) cos (φ + φo), where a and φo are some constants. This expression defines as a line in the plane. So if we have a plane with a coordinatization which is ro and φ. So this defines a plane. Indeed let us consider it. So this is just a, this is just φo, and this is a situation that we have ro and φ. So this was just to picture it. In reality we have the following situation. So 1/ well, then one can see from this picture, from this rectangle, one can see this relation. And indeed one can see that as φ + φo goes to π, ro goes to ∞. So as this argument goes to π, ro goes to ∞. Now we want to find acting as in the case of the massive particle. We want to find U1 and the equation, so we have to substitute Uo to here and then we obtain the equation for U1. So d^2 U1 / d φ^2. So this is the first relativity correction. U1 = 3rg / 2a^2 cos^2φ. The solution, so we for simplicity put it φo to 0. So their original phase, initial phase. Then the solution of this equation, the solution of this equation is 1 / r1 which is U1. So this is U1 = rg / 2a^2 (1 + sin^2φ). So this is a solution for this equation. So for the light like particle, we have obtained that 1 / ro = 1 / a cosφ. But remember that we put φo to 0. Then this corresponds, I can draw the picture, this corresponds to the straight line. Straight vertical line. Where this distance is a straight vertical one. Now if we take into account relativistic correction from general serial, 1 / ro + 1 is approximately 1/a cos φ + r g / 2a^2(1 + sin^2φ). And now, we have the following thing. Well, when φ = π / 2 ±. We have our zero going to ± ∞. So that corresponds to going there and there. Now here we have a little bit different situation, ro + 1 goes to ∞ at φ = 2 ± π + some small angle alpha. So, this guy, well, sorry, 2 + ∞ of course here. So, ro becomes infinitely large. This goes to + ∞ where this small angle alpha solves equation falling from here. So -1 / ia sine of alpha. So I just flag this here. Sin alpha + rg / 2a^2 (1 + cos^2 of alpha) = 0. So alpha solves this equation. When this becomes 0, our 0 goes to infinity. Approximate solution of this equation is alpha ≈ rg / a. So we have the full length picture, light ray is deviated by the following situation. So, we have alpha deviation here, and alpha deviation here, alpha deviation here. So, as a result, like for example, if I have a star here and looking here we see the star as if it is present here. So the deviation is by the angle two alpha as a result, the total deviation is two alpha which is 2rg / a. Now, for the case of the sun, for the sun, we have that rg for the sun ≈ 3 kilometers, 3 to the 10^3 meters. At the same time, its actual radius is approximately 700,000 kilometers, which is 7 to the 10^8 meters. Fro here, we obtain that this is, for the case of the sun, this is approximately 1.75 angle of seconds. And this effect was observed in 1919. And that was the beginning of the era of the general theory of relativity when it was considered as being proved by this experiment. [MUSIC]