 Hello. So as I said, we will begin a new chapter as an applications to celestial mechanics. The motion of the heavenly bodies has fascinated mankind since remote antiquity. There are references to this in Shakespeare's famous play, Merchant of Venice, when young Lorenzo has eloped with Jessica and they have come to the Porsche's garden and Lorenzo tells Jessica, look Jessica, how the floor of heaven is thick and laid with patterns of bright gold. There is not the smallest orb that thou beholdest, but in its motion sings to the young-eyed cherubins. I couldn't resist giving this quotation from Shakespeare. Here you see on the slide an epigram in Ptolemy's Almagust. I know that I am mortal by nature and ephemeral, but when I trace at my pleasure the windings to and fro of the heavenly bodies, I no longer touch the earth with my feet. I stand in the presence of Zeus himself and take my field of ambrosia. Almagust was a book that appeared in AD 145 and it has been one of the most influential books for 13 centuries in the western world. See the interesting article by Gingyarish was Ptolemy a Fraud, the quarterly journal of Royal Astronomical Society, volume 21, 1980, page 253 to 266. So let us begin with this chapter. In this chapter we shall discuss the original problem that led Bessel in 1824 to introduce the functions that bear his name. Wilhelm Bessel was an astronomer at Königsberg and his chief interest was a study of orbits of comets. We begin by recalling three basic laws of planetary motion enunciated by Johannes Kepler. The discovery of these laws forms an interesting culmination of classical astronomy. Equipped with calculus, Isaac Newton with his laws of dynamics was able to explain the motion of the planets, the precision of the equinoxes, the formation of tides. Astronomy which until now had been an empirical science transformed in the hands of Isaac Newton into a dynamical science. So, Kepler's laws of planetary motion. The first two laws were enunciated in 1609 in de Motibus Stele-Martis. Kepler's first law, the planets revolve around the sun in elliptical orbits and the sun lies at one of the two four sides. The second law says that the radius vector joining the sun and the planet sweeps out equal areas and equal intervals of time. This is just a restatement to the law of conservation of angular momentum. The third law, which is an approximate law neglecting the masses of the planet, appeared much later in 1619 in Harmonesis Mundi. Kepler's third law says that the square of the period t is proportional to the cube with a semi-major axis of the orbit. The constant of proportionality was determined by Gauss in his influential book on astronomy, Theoria Motors-Corporum Celestium and is today known as a gravitational constant. So, let's begin the problem. We first set the Kepler problem against the general backdrop in the theory of non-linear ODE's. We begin with the general ideas in the theory of ODE's. So, in equation 8.1, you see a system of non-linear autonomous system of ordinary differential equations, dyj by dt equal to fj, y1 by 2, yn, j running from 1 to n. Now, these functions fj's are assumed to be smooth functions and for simplicity, let us assume that they are independent of time or in other words, the system is an autonomous system. Now, what does it mean to say that we solve the equation 8.1? It means that for a given set of initial conditions that is yj of 0 equal to cj, j equal to 1, 2 up to n, we have to find a solution of 8.1 which means we have to find a trajectory, y1t, y2t, ynt and this can be thought of as a curve in Rn. So, idea is to find a specific curve in Rn. Now, what is the difficulty? To explain the difficulty, let us look at a very familiar system and even a linear system, the harmonic oscillator which we study from the time when we are in the 12th standard. So, what are the harmonic oscillator? x dot equal to y and y dot equal to minus x equation 8.2 and immediately we can see from 8.2, multiply the first equation by x, the second equation by y and add up and the right hand side becomes 0, xx dot plus yy dot equal to 0. So, x squared plus y squared must be a constant. What is the value of the constant? Remember, initial conditions are going to be given, x of 0 is given to you, y of 0 is given to you. The constant can be evaluated and you know that the function x squared plus y squared obtained above is constant and such a thing is called a first integral. So, let us define it in general terms. A non-constant smooth function phi y1, y2, yn is said to be a first integral for the system of OD is 8.1 if the function is constant along the trajectories. Namely, if I evaluate the function phi along the solution curve y1, t, y2, t, r, r, y, n, t, then the value remains constant. In other words, stated differently, you look at the level set phi equal to c. The level set phi equal to c is a surface in rn and the trajectory, the solution curve lies on the surface for all times. There are several different ways of looking at it. Another equivalent way of looking at it is that the gradient of phi dot with f equal to 0 where f is your f1, f2, fn. How do I know this? Remember that phi of y1, t, y2, t, r, r, yn, t is constant. Differentiate with respect to time, you are going to get 0 on the right hand side. How do you differentiate this composite function? Del phi by del xj dyj by dt, but dyj by dt is fj and so you get del phi by del xj fj summation j from 1 to n is 0. That is the last display in this slide. So, there are several different ways of understanding a first integral. So, what is the use of a first integral? The first integral tells you that the level set is a hypersurface. It is an n minus 1 dimensional hypersurface and our search for the solution is no longer the n dimensional space rn, but the n minus 1 dimensional entity, namely the locus 8.4 or we have reduced the problem to a n minus 1 dimensional problem. In physics books, you will say that the number of degrees of freedom has been reduced by 1. That is another way of saying the same thing. So, what happens for the harmonic oscillator? What was the first integral again? x squared plus y squared equal to constant and everybody knows that x squared plus y squared equal to constant is a circle in the plane and you know the circle. Maybe perhaps the initial conditions were such that this constant is 1, in which case you know that the circle is x squared plus y squared equal to 1 and the search for the solution has reduced from r2 to a circle, the unit circle. But remember that the unit circle can be parameterized in a variety of different ways. x squared plus y squared equal to 1 has several different parameterization, cos t comma sin t, sin t comma cos t minus sin t comma cos t, 1 minus t squared upon 1 plus t squared comma 2t by 1 plus t squared and so on and so forth. But if I give you x of 0 and if I give you y of 0, then the x t and y t are uniquely determined and so the solution of the original system of differential equation is 1 and exactly one of these infinitely many different parameterizations of the circle. So, the question is which one of these infinitely many different parameterization is the correct one. Now, remember that if I take a point x comma y on the circle, I could write x of t as cos of capital F t, y of t as sin of capital F t, where x t comma y t is the solution of our initial value problem. So, what we have to determine is the appropriate parameterization namely we have to find f of t. So, now let us differentiate this equation, what is x dot? x dot will be minus sin f of t times f dot t, but sin of f of t is y and but x dot equal to y because our first ODE is x dot equal to y. So, this cancels out and we are left with this condition f dot of t equal to minus 1. Moment you know that f dot of t equal to minus 1, then f of t must be alpha minus t for some constant alpha and we get the solution x equal to cos of t minus alpha y equal to minus sin t minus alpha as expected. We know that this is the solution of the harmonic oscillator not some other parameterization like 1 minus t squared upon 1 plus t squared 2 t by 1 plus t squared that is not the correct thing. This is the correct thing. So, to determine the correct parameterization we simply stipulate that since the point is lying on the circle x squared plus y squared equal to 1 the trajectory x t y t which is the solution of the differential equation I can write it as cos of f of t sin of f of t f has to be determined and we have determined f and we got the answer. Now, let us take a more complicated example the case of the harmonic oscillator was probably too simple and you are probably not convinced that the problem is non trivial. Let us consider Euler's equations for a spinning top. What are Euler's equations for a spinning top which ignores the effect of gravity a x dot equal to b minus c y z b y dot equal to c minus a x z c z dot equal to a minus b x y equation 8.5 that you see displayed in the slides. We immediately see that two first integrals say multiply the first equation by x second by y third by z and add up we get a x x dot plus b y y dot plus c z z dot equal to 0 or a x squared plus b y squared plus c z squared is constant and so that is a first integral multiply the first equation by a x second by b y third by c z and add and you will realize that a squared x squared plus b squared y squared plus c squared z squared is also a first integral and we know that the solution curve if I prescribe x of 0 y of 0 and z of 0 the solution to 8.5 is a certain curve in R 3 which curve in R 3 we do not know we have to search the whole of R 3, but no we can find out the constant values for these two first integrals and we know that the trajectory has to lie on both these ellipsoids and so the trajectory must lie on the intersection of these two ellipsoids. The intersection of these two ellipsoids is a curve. Now the gradients of these two functions a x squared plus b y squared plus c z squared a squared x squared plus b squared plus c squared z squared are linearly independent. So these two ellipsoids do cut along a curve and that curve is the solution curve that we are looking for, but not the actual parameterization. Again this intersection of these two is a curve which will admit infinitely many parameterizations which one of these infinitely many parameterizations qualifies to be called the solution of the ODE satisfying the given initial conditions. That is not obvious unlike the case of the harmonic oscillator where we could write our solution x y as x t equal to cos of f of t y t equal to sin of f of t. What is the corresponding procedure here? That is not so clear. Now let us do the following. The solution is x of t y of t z of t. Let us call x squared plus y squared plus z squared as u. So u is a function of t. So now we got a x squared plus b y squared plus c z squared which was alpha that was constant a squared x squared plus b squared y squared plus c squared z squared is beta that was also a constant. And now I am saying x squared plus y squared plus z squared is u that is not constant. So now I got three equations for x squared y squared and z squared. So solve them take square roots. I can write down x explicitly in terms of u y explicitly in terms of u z explicitly in terms of u where f, g and h are some known rational functions or with a square root thrown on it. And now let us substitute this into the system of differential equation the Euler's equation for the spinning top. Then we will get if you substitute these in the equation 8.5 then we will get f prime u du by dt equal to b minus c by a g u h u divide by f prime of u call g u h u upon f prime of u and this constant b minus c by this complicated thing you simply call psi u right. So now we got a first order variable separable ordinary differential equation for u and we can certainly solve this first order ordinary differential equation for u. What do we get t equal to integral from u 0 to u d s upon psi s equation 8.10. Now of course the question arises whether we can compute this integral. So there are two things left for us to do. We have to integrate this in classical literature we call it the problem of quadratures. So there is a problem of quadratures and the other problem which is equally difficult is we have got t as a function of u what we want is u as a function of t. If we can find u as a function of t then we can substitute it here and we would have obtained our solution the actual solution of the differential equation actual solution of the initial value problem x of t y of t z of t. So the problem involves two stages a quadrature and an inversion t is given here as a function of u you want to invert it you want u as a function of t. Already we see the innocent looking problem in three dimensions requires a substantial bit of work. So let us not underestimate the depth of the task the degree of difficulty. In the case of planetary motion it turns out that it is this last phase of inversion that calls for Fourier analysis. So it is here in the very last lap of the journey that Fourier analysis begins to enter and the equation that we need to invert the analog of 8.10 for the two body problem that is called the Kepler equation and the problem is of inverting the Kepler equation. Generally if we have a system of n differential equations and you have a system of n minus 1 first integrals with linearly independent gradients along their common locus then the intersection of these level sets is a curve and this is a solution curve except for parameterization. Finding the correct parameterization involves one integration and one inversion. So how do we get these first integral? Who is going to give us the first integrals? Physics will provide you a supply of first integrals, the law of conservation of energy, the law of conservation of momentum, linear momentum, angular momentum. Geometry may sometimes provide us with first integrals. When a system of ordinary differential equations exhibits symmetries then there are first integrals, there are conserved quantities. The harmonic oscillator x dot equal to y, y dot equal to minus x has a one parameter group of symmetries namely it is invariant under rotations. It is this one parameter group of symmetry that gave us the first integral x squared plus y squared. Now it is not surprising why x squared plus y squared was the first integral. For the case of the Euler spinning top we got two first integrals A x squared plus B y squared plus C z squared. What is this? It is twice the kinetic energy remember in chapter 6 we already encountered this rigid body problem. So A x squared plus B y squared plus C z squared is twice the rotational kinetic energy. ABC are the principle moments of inertia for the rigid body that is Euler's spinning top. What about A squared x squared plus B squared y squared plus C squared z? It is a square of the angular momentum. There is no external torque so that is going to be conserved. So we got two conserved quantities and in general if you can find n minus 1 conserved quantities we can expect to complete the problem of integration. Now let us come to the two body problem. The two body problem consists of two bodies a sun and a planet. In R3 the system is governed by Newton's laws of motion which is a system of six second-order ordinary differential equations for the instantaneous positions. Six second-order differential equations because Newton's law of motion gives you a second-order OD. We can convert it into a first-order system. When you take a second-order differential equation and convert it into first-order system we will get two first-order equations. So you see six second-order ODs three for the sun and three for the planet would mean 12 differential equations first-order ODs. So our system of ODs which we started out with 8.1 dyj by dt equal to fj j equal to 1 2 3 up to n. Here the n is 12. So we need a set of 11 first integrals in order to carry out the program. There are no external forces acting on it. So linear momentum is conserved. So v1 and v2 are the velocities of the sun and the planet. m1 and m2 are the masses of the sun and the planet. So what is m1 v1 plus m2 v2 upon m1 plus m2 that is going to be conserved. That is the conservation of linear momentum right. The left-hand side is linear momentum. But moment you have m1 v1 plus m2 v2 upon m1 plus m2 is conserved what are the left-hand side d dt of m1 x1 plus m2 x2 upon m1 plus m2 that is constant where x1 and x2 are the positions. So what happens is that m1 x1 plus m2 x2 upon m1 plus m2 is ct plus d where c and d are constant vectors. So it may say that the center of mass which is this left-hand side moves along a straight line in R3. So choose the origin to be the center of mass. So we got a moving coordinate system but it is an inertial coordinate system alright because the center of mass moves along a straight line with constant velocity. So if we choose the origin at the center of mass and if we choose this as a moving coordinate system the differential equations decouple into two systems of six equations and these two equations are identical except for a scaling factor. So if you solve one system then the other system you can simply read off the solutions. So now we have got a system of six differential equations. How many integrals do we need? We need five more integrals. Now there is no external torque. So angular momentum is going to be conserved and that will give you three more first integrals. Energy is going to be conserved that gives you one more first integrals. So four and the earlier six, three from the linear momentum and three by integrating the linear momentum equation c and d. So all in all I got ten first integrals. Where is the eleventh one? The divergence of the system is going to be zero. The f1, f2, fn that you see that has zero divergence and there is a classical result of Jacobi which says that we can then find one more first integral. So we are exactly in the situation where we can have the solution curve but not the parameterization. The parameterization is still going to evade us. Finding this parameterization leads us to the problem of inverting the Kepler equation and that is exactly what we are going to derive now. We are going to derive the Kepler's equation. Instead of carrying out these first integrals, these first ten first integrals that we got that assumes the following form that gets subsumed as the Kepler's laws of planetary motion. The Kepler's laws of planetary motion is a corollary of these first integrals and so what we will do is that we will derive the Kepler equation using the Kepler's laws of planetary motion because the Kepler's laws of planetary motion itself can be derived from these ten first integrals. So we short circuit this job and the second law basically is a law of conservation of angular momentum and the third law arises from the energy equation. So now our next job is to use the Kepler's laws of planetary motion and derive the Kepler equation and then turn to the problem of inverting the Kepler equation. Inverting the Kepler equation is exactly the problem that is going to lead us to Fourier analysis and we are going to use Bessel's functions and Fourier series both things that we studied in the first part of the course and so we see that as we go along we keep returning back to the earlier parts of the course time and again and I think this is a very good place to stop this lecture and we will continue this and finish this problem in the next capsule. Thank you very much.