0:13

Now let us continue with a classical experimental approvals of general serial.

Â From the mutant seminal solution of Kepler's problem,

Â It is known that planets orbit around the sun along ellipsoidal trajectories.

Â And we want to see now how this behavior is changed if we take into

Â account corrections from general serial [INAUDIBLE].

Â To do that we want to find the r at the function of phi, so

Â the curves which covered by the particles during the planets or

Â particles point like objects which orbit around [INAUDIBLE] body.

Â So to do that, let us do the following.

Â So let us consider dr over ds and

Â write it as a dr over d phi times d phi over ds and

Â this guy can be expressed with the use of,

Â so this is dr over d phi.

Â And this can be expressed with the use of the conservation law,

Â as l divided by mr squared.

Â And then, this can be written as- l over m.

Â du over d phi where we have introduced

Â the notation that u is equal 1 over r.

Â 1 over r.

Â So then using this new definition,

Â the equation for the geodesic.

Â Equation for the geodesic, which was using dr over ds is, can be written as follows.

Â As du over d phi squared plus u squared,

Â times 1- r g times u = E

Â squared- m squared divided

Â by L squared plus m squared

Â divided by L squared rgu.

Â 3:05

Now, now let us differentiate.

Â Let us take differential of both sides of this equation over d over d phi.

Â So take differential of this equation over d over d phi.

Â And divide it after obtaining the expression divided du over d phi.

Â So we divide by this, both sides of this equation.

Â As a result we obtain the following equation,

Â du squared over d phi squared plus u

Â 4:02

In this equation, one has to bear in mind

Â that rg over r, which is nothing but

Â rg over u is much less than 1, because in fact,

Â we want to consider this equation

Â as a deviation from their non-relativistic limit.

Â 4:45

So this is a Newtonian limit

Â of this equation when this is much less than 1.

Â Let us see that explicitly.

Â The Newtonian limit of this equation is as follows.

Â Well, we're going to do the following,

Â we're going to consider U expanded as a perturbation.

Â u0 + u1 + etc.

Â So this is a lading term, which solves this equation with all this term.

Â And this is a perturbation over this, due to corrections coming from this term.

Â So let us consider lading contribution, Newtonian equation, Newtonian limit.

Â So in this limit, this equation reduces to the following one.

Â du 0 over d phi squared + u0 is equal to m

Â squared rg divided by 2L squared.

Â So this equation is oscillator type of equation,

Â because this is just in homogeneous part.

Â If it were 0 on the right-hand side, we would obtain just the oscillator equation.

Â With non zero this guy, we have just homogeneous perturbation.

Â 6:03

So then the solution of this equation is nothing but the following.

Â We can easily solve the oscillator equation.

Â u0 which is by definition Is 0 where

Â subscript 0 corresponds to the Newtonian limit, when we dropped off this term.

Â So, in this limit we have the following situation, that m square rg

Â divided by 2L squared, 1 + e cosine 5.

Â So where e is kind of an integration constant,

Â which follows from this equation.

Â 6:47

So this is amplitude.

Â Anyway if e, which is called eccentricity,

Â if e is less than 1, if e is less than 1,

Â this e is less than 1, then this thing is nothing but a ellipse.

Â If e = 1, we have a parabola.

Â If e is greater than 1 we have a hyperbola.

Â So these are three types of motion in a central radial,

Â central Newtonian force in classical mechanics.

Â 7:21

But the close trajectories which are of interest for

Â us correspond to e less than 1.

Â In this case, this is just ellipse.

Â Indeed then we see that when we drop off this term,

Â reproduce the Newtonian limit which corresponds to this situation.

Â 7:39

Now we want to consider the perturbation for this.

Â So for this reason we want to [INAUDIBLE] this thing, this expression,

Â approximate expression here and find an equation for u1.

Â So to find the equation for u1, in this term,

Â we have to take into account of only u0.

Â Then for u1 we obtain the following equation, d squared.

Â Well taking into account this equation obeyed by u0.

Â For u1 we obtain the following equation.

Â That d u1 / d phi squared + u1

Â = 3/2 rg times u0 squared.

Â 8:24

So this is easy to see.

Â We took into account u0 in this term only.

Â So this is the equation we obtained for u1.

Â Well this is a relativistic general theory

Â of relativity correction to Newtonian trajectory will follow from here.

Â We want to find the biggest contribution to u1 following from this equation.

Â This equation is very similar to the following one.

Â So we have oscillator plus external force.

Â So external force may cause a resonance on the left-hand side.

Â That will be the biggest resonance contribution to u1.

Â So our goal is to solve this equation and find biggest resonance contribution to u1.

Â So we are solving this equation,

Â d squared u / d phi squared +

Â u = m squared rg / 2 L squared

Â + 3 rg / 2 u squared.

Â We're solving this equation which is exact in the longer relativistic approximations.

Â So consider this term as a perturbation.

Â And so we are looking for the solution of this equation as u0 + u1 + dot, dot, dot.

Â And at this level we restrict ourselves to these first

Â terms where u0 solves this equation.

Â D squared u0 / d phi squared + u0

Â = m squared rg / 2 L squared.

Â And then for u1 we obtain the following equation.

Â It is just d squared u1 / d phi

Â squared + u1 = 3 rg / 2 u0 squared.

Â So we want to look for the solution of this equation when this is considered as

Â an external source, and we want to see for the biggest solution of this equation.

Â So let us see what is this.

Â And the solution of this equation

Â as we found is just u0 = m squared

Â rg / 2 L squared (1 + e cosine phi).

Â So having this, let us look at this term.

Â So this term, 3 rg / 2 u0 squared.

Â If we substitute u0 here, this guy, and reshuffle it a little bit.

Â This guy is equal to

Â 3 rg cube m to the 4th

Â / 8 L to the 4th,

Â (1 + e squared /

Â 2) + 3 rg cube m

Â to the 4th / 4 L to

Â the 4th, e cos phi +.

Â So let me, + 3 rg cube m to

Â the 4th / 16 L to the 4th

Â e squared cosine(2 phi).

Â 12:03

So what does this term do?

Â If we add this to u0,

Â this just leads to the shift of this term in this equation.

Â So this is just a correction to this term, or to this term, in this equation.

Â This correction, what does it do?

Â It changes just the biggest axis in the ellipsoidal motion of the planet.

Â This axis is not measured

Â with such a precision that we can observe the presence of this term.

Â So we just consider this term as irrelevant and neglect.

Â 12:42

Now, this term and this term are behaving as external force in this equation.

Â As external forces in this equation.

Â And we want to find the biggest solution of this equation.

Â The biggest is the resonant due to the external force.

Â But then this term due to the mismatch of the frequency between 2 and 1 here.

Â Due to the mismatch doesn't lead to the biggest resonant solution, while this

Â term does lead to the biggest resonant solution because this force is an exact.

Â The frequency of this force is an exact

Â agreement with the frequency of the oscillator.

Â So our goal then is to find the resonant solution of the following equation.

Â D squared u1 / d phi squared

Â + u1 approximately = 3

Â rg cube m to the 4th / 4

Â L to the 4th e cosine pi.

Â So we want to find the resonant solution of this equation.

Â 14:31

The equations for A and B are as follows.

Â So A double prime = B double prime = 0,

Â where double prime is just the second differential with respect to phi.

Â And we have the following equation,

Â that 2 A prime cosine phi-

Â 2 B prime sine phi = 3 rg cube

Â m to the 4th / four L 4th e cosine phi.

Â 15:44

And one can see from here that this is not a resonant solution.

Â If b is constant, this is just an oscillator solution.

Â From here, we find that A should be linear function of phi.

Â Then, while substituting here, we have a resonant, exactly resonant solution.

Â 16:06

Solution of the equation.

Â That's what we have been looking for.

Â This is the biggest contribution to u1.

Â So as a result, we obtain that u, this approximation,

Â which is u0 plus u1, so it's u0 plus u1,

Â which is just 1 over r in the 0 plus 1st

Â contribution is approximately equal to

Â the sum of this plus this contribution,

Â because I have phi that we have found.

Â So we obtain that this is m squared rg

Â divided by 2L squared 1 plus e cosine phi.

Â This is just u0 plus u1,

Â following from here, the resonant one,

Â 3rg cubed m to the 4th e divided

Â by 8L to the 4th phi times sine of phi.

Â But note, this term in our approximation is much smaller than this one.

Â So this is just a perturbation.

Â So then this expression, let me write it like this.

Â This expression can be approximately equated to,

Â so we, approximately, well,

Â let me write that from this follows

Â that r0 plus 1 is approximately

Â equal to m squared times rg divided

Â by 2L squared times 1 plus e cosine

Â phi minus 3rg squared times

Â m squared over 4 L squared phi.

Â 18:17

So this term is much smaller than this term on the cosine.

Â So if we Taylor expand the cosine,

Â from here we just obtain this.

Â So from the expansion of this term, we obtain this term.

Â So this expression is just approximately equal to this one.

Â But when we have written this correction in this form,

Â we can make the following observation.

Â 18:45

Before this correction,

Â the trajectory was closed like this.

Â It was r0 at phi was equal to = r0 at phi plus 2 pi.

Â Well, that's easy to see from this expression.

Â u0 at phi is equal to u0 plus 2 pi, phi plus 2 pi.

Â So r0 is closed, but now due to this extra correction,

Â the trajectory is closed under the following condition.

Â 19:20

Under the condition that r0

Â plus 1 of phi is approximately

Â equal to r0 plus 1 of phi plus 2

Â pi plus extra contribution due

Â to this term plus 3 pi rg squared

Â divided by 4 L squared m squared.

Â So there is a precession, so

Â the particle doesn't go back, but a little bit shifted.

Â 20:01

Shifted by a small angle following from here.

Â So the angle follows from here.

Â So let me remind you, we have arrived at the following statement,

Â that u0 is equal to 1 over r0,

Â 20:18

which is 1, which is m squared

Â rg 2 L squared 1 plus e cosine phi.

Â At the same time, u0 plus u1,

Â which is 1 over r0 plus 1, so

Â it's a 0th level plus first correction,

Â is approximately equal to m squared

Â rg divided by 2L squared 1 plus e

Â cosine of phi minus 3 rg squared m

Â squared 4 L squared times phi.

Â 21:19

So this is closed elliptic trajectory.

Â This guy corresponds to not closed trajectory.

Â So we have r0 plus 1 of phi is equal

Â approximately to r0 plus 1 of phi plus

Â 2 pi plus small correction due to this term,

Â which is 3 pi rg squared m squared 2 L squared.

Â So we have a precession, so the trajectory is not quite closed.

Â So there is a small rotation by an angle falling from here.

Â The rotation angle is just 3

Â pi rg squared m squared over 2 L squared.

Â 22:19

So in the case of circular orbit,

Â if we have circle instead of elliptic orbit,

Â L just in non-reduced limit, just mvR where m the mass

Â of the planet, v is its velocity, R its radius.

Â And we also have that v squared is equal to rg over 2 R.

Â 22:48

So then, for this case, from here we find that delta phi

Â is approximately 3 pi rg over R.

Â So this is for the circle orbit, while for the elliptic orbit,

Â 23:27

And then we expect, because for Mercury,

Â although it is much greater than rg, but

Â still has closest value to rg among all the planets.

Â So the Mercury is expected to have the biggest contributions of this sort.

Â Then, if we represent, substitute for the values of this expression for

Â the Mercury, we find that, for the Mercury, delta phi is

Â approximately tenth digit of the second, angle of second.

Â So it's 0.1 of angle a second.

Â But this is a circular effect, so it grows in time, grows in time.

Â And during a century, during 100 years, it goes up to 43 seconds.

Â And this is already a measurable effect.

Â And it is measured and is in perfect agreement with the discrepancy

Â that was observed for the Mercury more than 100 years ago.

Â [SOUND]

Â [MUSIC]

Â