[MUSIC] Now let us continue with a classical experimental approvals of general serial. From the mutant seminal solution of Kepler's problem, It is known that planets orbit around the sun along ellipsoidal trajectories. And we want to see now how this behavior is changed if we take into account corrections from general serial [INAUDIBLE]. To do that we want to find the r at the function of phi, so the curves which covered by the particles during the planets or particles point like objects which orbit around [INAUDIBLE] body. So to do that, let us do the following. So let us consider dr over ds and write it as a dr over d phi times d phi over ds and this guy can be expressed with the use of, so this is dr over d phi. And this can be expressed with the use of the conservation law, as l divided by mr squared. And then, this can be written as- l over m. du over d phi where we have introduced the notation that u is equal 1 over r. 1 over r. So then using this new definition, the equation for the geodesic. Equation for the geodesic, which was using dr over ds is, can be written as follows. As du over d phi squared plus u squared, times 1- r g times u = E squared- m squared divided by L squared plus m squared divided by L squared rgu. So this equation is equivalent to the one that we had, dr over ds squared. Or let me write it explicitly. dr over ds = (E / m) squared- (1- rg / r) (1 + L squared / m squared r squared). So this equation is equivalent for this one, in the new notations. Now, now let us differentiate. Let us take differential of both sides of this equation over d over d phi. So take differential of this equation over d over d phi. And divide it after obtaining the expression divided du over d phi. So we divide by this, both sides of this equation. As a result we obtain the following equation, du squared over d phi squared plus u is equal to m squared rg, divided by 2 L squared, plus 3 of g over half times u squared. So this is the equation we're going to work with. In this equation, one has to bear in mind that rg over r, which is nothing but rg over u is much less than 1, because in fact, we want to consider this equation as a deviation from their non-relativistic limit. So this is non-relativistic limit when this is much less than 1. In this limit this term is small in comparison with the rest of the terms in the equation. So in this limit we can neglect this term. So this is a Newtonian limit of this equation when this is much less than 1. Let us see that explicitly. The Newtonian limit of this equation is as follows. Well, we're going to do the following, we're going to consider U expanded as a perturbation. u0 + u1 + etc. So this is a lading term, which solves this equation with all this term. And this is a perturbation over this, due to corrections coming from this term. So let us consider lading contribution, Newtonian equation, Newtonian limit. So in this limit, this equation reduces to the following one. du 0 over d phi squared + u0 is equal to m squared rg divided by 2L squared. So this equation is oscillator type of equation, because this is just in homogeneous part. If it were 0 on the right-hand side, we would obtain just the oscillator equation. With non zero this guy, we have just homogeneous perturbation. So then the solution of this equation is nothing but the following. We can easily solve the oscillator equation. u0 which is by definition Is 0 where subscript 0 corresponds to the Newtonian limit, when we dropped off this term. So, in this limit we have the following situation, that m square rg divided by 2L squared, 1 + e cosine 5. So where e is kind of an integration constant, which follows from this equation. So this is amplitude. Anyway if e, which is called eccentricity, if e is less than 1, if e is less than 1, this e is less than 1, then this thing is nothing but a ellipse. If e = 1, we have a parabola. If e is greater than 1 we have a hyperbola. So these are three types of motion in a central radial, central Newtonian force in classical mechanics. But the close trajectories which are of interest for us correspond to e less than 1. In this case, this is just ellipse. Indeed then we see that when we drop off this term, reproduce the Newtonian limit which corresponds to this situation. Now we want to consider the perturbation for this. So for this reason we want to [INAUDIBLE] this thing, this expression, approximate expression here and find an equation for u1. So to find the equation for u1, in this term, we have to take into account of only u0. Then for u1 we obtain the following equation, d squared. Well taking into account this equation obeyed by u0. For u1 we obtain the following equation. That d u1 / d phi squared + u1 = 3/2 rg times u0 squared. So this is easy to see. We took into account u0 in this term only. So this is the equation we obtained for u1. Well this is a relativistic general theory of relativity correction to Newtonian trajectory will follow from here. We want to find the biggest contribution to u1 following from this equation. This equation is very similar to the following one. So we have oscillator plus external force. So external force may cause a resonance on the left-hand side. That will be the biggest resonance contribution to u1. So our goal is to solve this equation and find biggest resonance contribution to u1. So we are solving this equation, d squared u / d phi squared + u = m squared rg / 2 L squared + 3 rg / 2 u squared. We're solving this equation which is exact in the longer relativistic approximations. So consider this term as a perturbation. And so we are looking for the solution of this equation as u0 + u1 + dot, dot, dot. And at this level we restrict ourselves to these first terms where u0 solves this equation. D squared u0 / d phi squared + u0 = m squared rg / 2 L squared. And then for u1 we obtain the following equation. It is just d squared u1 / d phi squared + u1 = 3 rg / 2 u0 squared. So we want to look for the solution of this equation when this is considered as an external source, and we want to see for the biggest solution of this equation. So let us see what is this. And the solution of this equation as we found is just u0 = m squared rg / 2 L squared (1 + e cosine phi). So having this, let us look at this term. So this term, 3 rg / 2 u0 squared. If we substitute u0 here, this guy, and reshuffle it a little bit. This guy is equal to 3 rg cube m to the 4th / 8 L to the 4th, (1 + e squared / 2) + 3 rg cube m to the 4th / 4 L to the 4th, e cos phi +. So let me, + 3 rg cube m to the 4th / 16 L to the 4th e squared cosine(2 phi). So what does this term do? If we add this to u0, this just leads to the shift of this term in this equation. So this is just a correction to this term, or to this term, in this equation. This correction, what does it do? It changes just the biggest axis in the ellipsoidal motion of the planet. This axis is not measured with such a precision that we can observe the presence of this term. So we just consider this term as irrelevant and neglect. Now, this term and this term are behaving as external force in this equation. As external forces in this equation. And we want to find the biggest solution of this equation. The biggest is the resonant due to the external force. But then this term due to the mismatch of the frequency between 2 and 1 here. Due to the mismatch doesn't lead to the biggest resonant solution, while this term does lead to the biggest resonant solution because this force is an exact. The frequency of this force is an exact agreement with the frequency of the oscillator. So our goal then is to find the resonant solution of the following equation. D squared u1 / d phi squared + u1 approximately = 3 rg cube m to the 4th / 4 L to the 4th e cosine pi. So we want to find the resonant solution of this equation. To solve it we're going to look for the solution in the following form. As A phi, A(phi) sine(phi) + B(phi) cosine(phi). After substitution where we can see that this is slow functions of phi. So this is the answers for the solution of this equation. We're going to substitute here and find the equation for A and B. That's not a very hard exercise. Not a very hard exercise. The equations for A and B are as follows. So A double prime = B double prime = 0, where double prime is just the second differential with respect to phi. And we have the following equation, that 2 A prime cosine phi- 2 B prime sine phi = 3 rg cube m to the 4th / four L 4th e cosine phi. So from this equation, from these equations, one can easily see that B prime should be 0 and A prime should be 3 rg cube m to the 4th / 8 L to the 4th e. So as a result, b is constant. B is constant. And one can see from here that this is not a resonant solution. If b is constant, this is just an oscillator solution. From here, we find that A should be linear function of phi. Then, while substituting here, we have a resonant, exactly resonant solution. Solution of the equation. That's what we have been looking for. This is the biggest contribution to u1. So as a result, we obtain that u, this approximation, which is u0 plus u1, so it's u0 plus u1, which is just 1 over r in the 0 plus 1st contribution is approximately equal to the sum of this plus this contribution, because I have phi that we have found. So we obtain that this is m squared rg divided by 2L squared 1 plus e cosine phi. This is just u0 plus u1, following from here, the resonant one, 3rg cubed m to the 4th e divided by 8L to the 4th phi times sine of phi. But note, this term in our approximation is much smaller than this one. So this is just a perturbation. So then this expression, let me write it like this. This expression can be approximately equated to, so we, approximately, well, let me write that from this follows that r0 plus 1 is approximately equal to m squared times rg divided by 2L squared times 1 plus e cosine phi minus 3rg squared times m squared over 4 L squared phi. So this term is much smaller than this term on the cosine. So if we Taylor expand the cosine, from here we just obtain this. So from the expansion of this term, we obtain this term. So this expression is just approximately equal to this one. But when we have written this correction in this form, we can make the following observation. Before this correction, the trajectory was closed like this. It was r0 at phi was equal to = r0 at phi plus 2 pi. Well, that's easy to see from this expression. u0 at phi is equal to u0 plus 2 pi, phi plus 2 pi. So r0 is closed, but now due to this extra correction, the trajectory is closed under the following condition. Under the condition that r0 plus 1 of phi is approximately equal to r0 plus 1 of phi plus 2 pi plus extra contribution due to this term plus 3 pi rg squared divided by 4 L squared m squared. So there is a precession, so the particle doesn't go back, but a little bit shifted. Shifted by a small angle following from here. So the angle follows from here. So let me remind you, we have arrived at the following statement, that u0 is equal to 1 over r0, which is 1, which is m squared rg 2 L squared 1 plus e cosine phi. At the same time, u0 plus u1, which is 1 over r0 plus 1, so it's a 0th level plus first correction, is approximately equal to m squared rg divided by 2L squared 1 plus e cosine of phi minus 3 rg squared m squared 4 L squared times phi. So while this guy corresponds to closed trajectories. So r0 phi is equal to r0 plus phi plus 2 pi. So this is closed elliptic trajectory. This guy corresponds to not closed trajectory. So we have r0 plus 1 of phi is equal approximately to r0 plus 1 of phi plus 2 pi plus small correction due to this term, which is 3 pi rg squared m squared 2 L squared. So we have a precession, so the trajectory is not quite closed. So there is a small rotation by an angle falling from here. The rotation angle is just 3 pi rg squared m squared over 2 L squared. So in the case of circular orbit, if we have circle instead of elliptic orbit, L just in non-reduced limit, just mvR where m the mass of the planet, v is its velocity, R its radius. And we also have that v squared is equal to rg over 2 R. So then, for this case, from here we find that delta phi is approximately 3 pi rg over R. So this is for the circle orbit, while for the elliptic orbit, in similar manner, one can find that delta phi is approximately equal to 3 pi rg divided by R(1 minus e squared). So let us, the closest planet to the sun is Mercury. And then we expect, because for Mercury, although it is much greater than rg, but still has closest value to rg among all the planets. So the Mercury is expected to have the biggest contributions of this sort. Then, if we represent, substitute for the values of this expression for the Mercury, we find that, for the Mercury, delta phi is approximately tenth digit of the second, angle of second. So it's 0.1 of angle a second. But this is a circular effect, so it grows in time, grows in time. And during a century, during 100 years, it goes up to 43 seconds. And this is already a measurable effect. And it is measured and is in perfect agreement with the discrepancy that was observed for the Mercury more than 100 years ago. [SOUND] [MUSIC]