 Okay, we continue with Hamiltonian dynamics from where we left off. Recall the conditions we had for the integrability of a Hamiltonian system. Once again, given a Hamiltonian q comma p with f degrees of freedom, I think I used n, n degrees of freedom. We have Hamilton's equations q i dot is delta h over delta p i and p i dot is minus delta h over delta q i for i running from 1 to n. I pointed out that this system is integrable in the sense that you can explicitly write down time dependent solutions for all the q's and all the p's given any set of initial conditions provided there exist n constants of the motion f1 to fn in involution with each other such that the Poisson bracket of any fi with any other fj vanished identically. If there exist these constants of the motion then the system is integrable and the problem is completely solved and it was done by going to what are called action angle variables where the Hamiltonian h of q comma p was transformed by a canonical transformation a transformation with Jacobian equal to plus 1 and the Poisson bracket structure preserved to a new Hamiltonian in variables called action angle variables and this became a function of just the action variables and then one discovered that the problem was completely integrable completely solved. Now what is this imply in geometrical terms this is what we were trying to understand and I pointed out that the Poisson bracket condition this quantity vanishing here can be rewritten in terms of this matrix J we introduced which was a 2n by 2n matrix with zeros here zeros here the unit n by n matrix here minus the unit n by n matrix here and this condition simply became this quantity equal to the gradient of fi transpose J the gradient of fj in other words these vector fields formed by the gradients of these constants of the motion of these functions of q's and p's they are pseudo orthogonal to each other in the sense that the dot product of this row vector with that column vector is 0 this means that if the system is integrable in the Louisville Arnold sense there exist n vector fields formed by the gradients grad f1 grad f2 through grad fn which are independent of each other on the space of these i's and theta's completely that has a profound implication and it implies that you have in this 2n dimensional phase space a 2n-1 dimensional energy hyper surface the motion is restricted to that for any set of initial conditions over and above that the motion is actually restricted to a subspace of this 2n-1 dimensional energy space such that these n quantities are constants completely more over on this n dimensional subspace the n vector fields formed by these gradients are independent of each other now it turns out that there is a deep theorem mathematics which says that if you have a space which is compact and everywhere in this space you have n independent vector fields by compact I mean it does not go off to infinity it is unbounded there is a technical definition which will write down a little later then the only such space possible is isomorphic to something called an n dimensional torus and let me explain that in slow terms a one dimensional space which is compact and which has at every point a unique tangent vector would be something like a circle in the mathematicians who denote this by s1 this space is one dimensional and at every point on this space there is a unique tangent vector a straight line which runs from minus infinity to infinity tangent to this point and as you can see if you took this tangent and moved it along this entire circle when you come back here you come right back to the same point to the same tangent as before such a space is said to be parallelizable or developable in the sense that you can unroll it you can roll it on a sheet of paper and make a one-to-one mapping between a straight line segment and this circle without any kinks without any difficulty you can do exactly the same thing with a two dimensional donut a torus in which you can take two vector fields one of which is for example directed along these lines and the other one is directed along these lines this direction form a basis from these two vector fields and move it all over the space and come right back to the starting point to the original configuration itself you can't do that on a sphere this is by the way called the direct product of two circles it's just the Cartesian product of two circles and it's called the two torus the moment thought shows you that you can't do this with another two dimensional object the surface of a sphere embedded in three dimensions denoted by s2 it's not possible to find a unique tangent map which is not singular at any point on this sphere because if I start with some point here and draw the tangent plane to that point as I move it around it's clear that there is going to be at least one point where this direction of this tangent plane is undetermined there's a singularity what does it mean you could define this tangent plane by saying imagine like a tennis ball that there are fibers sticking out of this ball and you're trying to comb it and when you comb it it's clear that somewhere maybe at the North Pole there's a little cowlick there's a little point that sticks out a singularity of the vector field so the technical way of saying it is that there is no non-singular global tangent map to s2 like there is to s1 or t2 and the statement being made here is that the most general space which is compact and which is parallelizable in the sense that you can form a basis set of n vector fields pardon me could I explain why this imagine combing a ball imagine combing a tennis ball what happens you comb it down everywhere flat tangent what happens can you do this without a cowlick without a parting there's at least one point where there's going to be a singularity and the hair sticks out the direction which it's placed is indeterminate there's a singularity of this field there'll be a ball spot invariably that's not true for a torus you can comb it down completely so this is a basic difference in a property of a torus as opposed to a sphere and the statement being made here is that if you can find n independent vector fields on an n dimensional manifold which is globally applicable smooth everywhere then that space has to be an n dimensional torus it's a generalization of the two dimensional torus I can't draw an n dimensional torus here because I can't draw anything more than three dimensional where it applies here is the fact that integrable Hamiltonian systems integrable in the sense of Louisville Arnold the phase space on which the action takes place is eventually just n dimensional not to n dimensional it's reduced from to n to to n minus 1 by the constancy of the Hamiltonian itself now it's further reduced from to n minus 1 to just n by the fact that it's integrable now this is an abstract statement we'll look at specific examples and see how this works out so we take simple examples and I'll take the simplest of them all namely the harmonic oscillator and we see how this thing comes out how the two torus structure comes out for a couple of uncoupled for a pair of uncoupled harmonic oscillators so we'll do that step by step and before I do that let's give a few examples of what Hamiltonian systems look like so this was a bit of a digression but we come back so first let's look at the linear harmonic oscillator this is of course our simplest problem of all it's got one degree of freedom the Hamiltonian as a function of a single q and a p is one half it's p squared over 2m the kinetic energy plus the potential energy v of q which in this case is p squared over 2m plus one half m omega squared q squared where omega is the natural frequency of the oscillator and m is its mass of course we're going to get equations of motion which are just the simple harmonic oscillator equations of motion but let's go through the steps simply to see how this works out and we know that q dot is delta h over delta p which turns out to be just p over m this problem and p dot is minus delta h over delta q that's equal to minus m omega squared q we've just written Newton's equations now because the conventional Newton's equation would say q double dot the acceleration is equal to one over m times the force which would be the rate of change of the momentum so together it's clear that this implies the usual q double dot plus omega squared q equal to 0 which is the oscillator equation of motion but like I said we prefer to write everything down in phase space because that's where the dynamics is taking place and we have the set of couple the equations now let's go through the formal analysis of this this is a linear set of equations on the right hand side so there's no need to linearize the problem it's already linear where's the critical point of the system at the origin the right hand sides must vanish so the only critical point is at 0, 0 in the qp plane what about the matrix L which acts on the right hand side it's 0 1 over m minus m omega squared and 0 and what are the eigenvalues of this matrix plus or minus i omega so it immediately says lambda 1, 2 is plus or minus i omega this implies 0, 0 is a center is that stable unstable or asymptotically stable it's stable it's not asymptotically stable it's just a stable center and what do the phase trajectories look like in general their ellipses depending on the units you choose because there's just a single constant of the motion in this problem since n is 1 the set f1 through fn becomes just f1 and you need to find just one constant of the motion to integrate the system that constant of the motion is already given to you it's the Hamiltonian remember that for any Hamiltonian system which is autonomous the Hamiltonian is always a constant of the motion so we have a phase portrait in this case which is just a set of ellipses h of qp equal to constant and in this case the constant is simply the total energy of the system in which direction is the phase trajectory traversed would this be in the counter clockwise or clockwise direction why do you say that exactly if you pull this oscillator and let go from the right most point it moves back towards the left so p becomes negative at that point therefore if you start here the next instant it's here and that fixes the direction in which this thing is traversed therefore clockwise the critical point at the center at the origin is a stable center all motion is periodic no matter what the initial conditions are and every point in this plane lies on one and only one ellipse phase trajectories don't intersect themselves for autonomous systems and the entire plane is laminated by these concentric ellipses what's the time period of motion it's 2 pi over omega and it happens to be independent of the energy in this problem because it's a linear harmonic oscillator it turns out that this is one of the unique properties of the linear of the harmonic oscillator that the time period is independent of the amplitude of motion or of the energy of the motion there are other oscillators which are not linear the equations of motion of which are not linear for which this phenomenon occurs and we'll come up with an example very shortly a little later but this is a distinguishing feature of harmonic oscillators unless of course you look at a very special class of oscillators which are nonlinear but also isochronous anything where the time period is independent of the energy is called isochronous so in this problem the motion does occur as you can see on one dimensional tori on a torus which is essentially one dimensional namely this curve itself or on this curve and this magic happened simply because this problem had a potential which was quadratic and therefore it led to an equation of motion which was linear on the right hand side so the problem is exceedingly simple as you can see now let's take this a little more general again with one degree of freedom and see what we can say before we go on to two degrees of freedom suppose I have a general potential of this kind what would this equation of motion become on this side yes just the derivative – dv q over dq and of course that's the force – the gradient of the potential with respect to the coordinate so we recover Newton's equations of motion except that now the critical points of the system would be given by the vanishing of P and the vanishing of B' of q in other words the extrema of the potential these could be maxima these could be minima they could be inflection points at which the slope is 0 and then of course you would have to further examine the stability or otherwise of these critical points and you could in principle write the entire phase trajectory down the phase portrayed down simply because there is just one constant of the motion but notice one interesting fact right away so let's say where the CP CP is located at P equal to 0 and the roots of B' of q equal to 0 which correspond as we have said to just the extrema of the potential but notice an interesting fact right away that the phase trajectories are actually already known to you they simply given by P squared over 2 m plus v of q equal to constant since one equation between two variables P and q on a plane specifies a curve the phase trajectories are completely specified even without solving the equations of motion solving the equations of motion for a specific set of initial conditions will of course tell you how P and q change as a function of time explicitly but to write the phase trajectory is down you don't need that notice something else notice also that the phase curves are given by dividing this by this equation and you get dp over dq is equal to minus v' of q divided by P m times that m times that therefore whenever the phase trajectory intersects the horizontal axis the q axis it will generally do so at right angles because this quantity vanishes on the x axis on the q axis but this may not if it does then you have to examine the problem further and take limits but otherwise phase trajectories would intersect the q axis at right angles and that's indeed what happened in the harmonic oscillator example where you had this kind of behavior and these were at right angles that happened because the restoring force is not zero at those turning points at the end points of the motion but the momentum vanished at those points you could integrate this equation you get P dp plus m v' of q dq equal to zero and if you integrated it what would you get you would simply get P square over 2 m plus v of q equal to constant which we already know so in principle a one degree of freedom Hamiltonian system is always integrable you don't need any further conditions let's look at a potential which is a little more complicated than a linear one so let's suppose v of q and let's choose units conveniently so that I don't have to run into problems with writing these constants down so let's simply write p squared over 2 plus perhaps q squared over 2 so it's a harmonic oscillator without any extra terms but then I include a nonlinearity and make it q cube over 3 insuitable units what happens to the right hand side here this becomes p what happens here minus q because you need a minus sign there minus q squared that becomes equal to minus q times q plus 1 where are the critical points of the system well 0 0 is still a critical point but minus 1 0 is also a critical point and we can easily write down what the solutions are we should draw the phase trajectory or phase portrait but before we do that let's draw the potential so we get some physical idea of what it looks like so here's the q axis here's v of q so what does this potential look like just this alone sufficiently close to the origin the q squared dominates over the q cubed therefore it looks like a parabola so we certainly guaranteed that the potential looks like this here and then for large positive q it shoots off like q cubed goes off to infinity but then for large negative q this term dominates over this no matter how large you get once q becomes sufficiently large this becomes much bigger than this and then this curve has to come down in this fashion and not surprisingly this extremum is at minus 1 this is at 0 which is a minimum there and a maximum at minus 1 what do the phase trajectories then look like so if you permit me to draw it on the same curve on the same vertical axis but now I draw p here versus q this would be a critical point and this point here would be a critical point and it's quite apparent that this is a center about which you have stable oscillations small oscillations and what kind of critical point would this be it would certainly be unstable you'd have to linearize the equations of motion about the point q equal to minus 1 so you might want to set say u equal to q plus 1 and then shift the origin to q to q equal to minus 1 and see what happens in the vicinity of this point but we've already seen that for Hamiltonian systems there's no dissipation in the only critical points possible are centers and saddle points and this is a saddle point it's unstable and that's a center and that's stable what do the phase trajectories look like what would the phase portrait look like in this problem you'd have to specify now the initial conditions in other words you have to tell me the initial q and p or better still tell me the initial value of the energy and that remains constant because you're on curves in which this quantity is constant so what we're really doing is plotting the curve p squared over 2 plus q squared over 2 plus q cubed over 3 equal to a constant that constant could be positive or negative in this problem because this term could take on large negative values as well so what would these phase trajectories look like in general suppose I started with a value of the total energy that corresponded to some level like this at this level on this figure this is the zero of the energy of v of q suppose I had a total energy equal to this much where would the motion be this is my total available energy it's clear I can't go into this region because if I did so then this quantity v of q plus p squared should be equal to this number but v of q is already larger than this number which implies p squared should be negative that's not possible with real p so if this is the total energy the system doesn't have enough energy to get into this region it's restricted to this region and therefore it can never move to the right of this point now imagine you start with a little ball bearing here in this potential hill and let go from rest what would it do it would move away to q equal to minus infinity with increasing acceleration in which direction would this acceleration be to the left or to the right in which direction would the momentum be to be to the left to move further and further to the left so p would get more and more negative and q would get more and more negative and therefore this is what the trajectory would look like on the other hand imagine starting at minus infinity in q and shooting a ball up this potential hill with a fixed amount of energy equal to this much it's clear it can crawl up this hill this barrier up to this point where it's energy where it's kinetic energy goes to zero and then it rolls back what would that trajectory look like that half trajectory but it starts there and moves to the right but with smaller and smaller values of p of momentum till it reaches this point with zero momentum it would therefore be the other half of this curve and in principle if you shot something here you started off with something here with this much total energy at minus infinity it would crawl up this hill and fall down corresponding to this phase trajectory and as we know already since the restoring force at this point is not zero the slope is not zero there v prime of q is non zero at that point therefore it must intersect this line at right angles what happens if I have a little higher energy nothing much happens it follows another trajectory which does this what happens if I have a total energy equal to this much it's clear that the particle could move up to this point and this would be a trajectory but it's also clear that if the initial conditions permitted it to be inside this region to start with it would simply oscillate about that origin therefore for the same value of the total energy there exists another trajectory which would correspond to oscillations as I said a closed phase trajectory implies periodic motion and vice versa so for the same total energy there are two regions in configuration space where the particle could find itself one would be to the left of this point and the other would be in this well motion here would correspond to periodic motion motion here would correspond to open or unbounded motion but both these phase trajectories correspond to the same total energy same value of h equal to constant same constant these oscillations here would for sufficiently small energies above zero be essentially ellipses because you could neglect the effect of the q q term and then you have a harmonic oscillator but it's quite evident that as the amplitude increases this is no longer a parabola but it flattens out on this side and becomes cubic on that side and therefore it's non harmonic it's some kind of oval but it's still periodic motion the time period in general would depend on the energy except for very very small amplitude oscillations when the system looks like a simple harmonic oscillator what happens if you have an energy which is higher than the height of this barrier it's clear that the barrier no longer can trap this particle into oscillatory motion therefore this would be open trajectory of some kind an open trajectory imagine shooting the particle up it comes up here it certainly slows down because you have very high potential energy here but then it crosses this barrier falls into this well climbs up to that point and then goes right back and falls off in this fashion therefore I would expect this thing to come down go around and go off escape to infinity again crossing this at right angles that's what would correspond to an energy which is higher than the height of the barrier. So now you begin to see that there is one very special value of the energy where these two possibilities namely periodic open motion versus periodic motion they merge the boundary between the two which would correspond to a total value of the energy a value of the total energy which is exactly equal let me call that E sub s which would correspond to two different kinds of motion one of which would be remember this point by itself this point by itself is an unstable critical point it's a phase trajectory by itself if therefore you shoot a particle from here up this hill with just this critical value of energy so that it can barely reach it out there it's going to take an infinite amount of time to do so it would eventually as t tends to plus infinity go and stop there in this fashion that would correspond to trajectory which comes along like this and tends to this point as t tends to plus infinity had we started with a particle there and displaced it infinitesimally to the left it would fall off and go off to minus infinity here which would correspond to this had we started on this side up here and pushed it slightly to the right it would go up the barrier go down this well go up to this point turn back and come back and crawl back to this point the reason it would crawl back is because the slope is getting flatter and flatter the restoring force is getting smaller and smaller and therefore it's barely able to reach this top it would therefore do the following go here go around and come back and this point of course would correspond to that so it's quite clear that a lot of interesting things happen in this region and let's magnify that region and see what it looks like that region near the separate ricks is a saddle point here there's an unstable orbit coming out of it which eventually falls back tends back towards it and then there is a separate ricks which is flowing in and something which is flowing out I should really let these things tend to that point asymptotically but instead of that let me just draw it in this fashion so you can see that this is a limiting point this saddle point if you linearized about the saddle point you discover that the system has two Eigen values one of which is positive and the other is negative and the two Eigen directions or Eigen vectors of the linearized matrix L would correspond to these directions this is called the stable manifold of this critical point and that's called the unstable manifold of this critical point and as is typical of a saddle point two lines come in and two lines go out near this saddle point the system is hyperbolic and whole thing looks like hyperbolas the phase trajectories look like hyperbolas so you have behavior of this kind of course these trajectories would eventually flow off and this would go around and join up there and similarly inside here these would be parts of periodic orbits and these would be parts of open orbits but locally it looks like a saddle point should notice also that this tangency here is not at right angles this is the one case where this intersection at right angles doesn't happen and the reason is v prime of q also vanishes at this point and therefore you have to take the limit v prime of q over p as you approach the critical point both numerator and denominator vanish at that point and you have this typical saddle structure I leave it to you as an exercise to find out in this problem what this angle is what the angle subtended by the two separatrices are this trajectory which separates open motion of this kind from open motion of this kind is called a separatrix corresponding to the energy E sub s and that's the reason I used a subscript s there to show that its power energy corresponding to a separatrix this trajectory is a separatrix and this trajectory which separates open motion from periodic motion inside this closed loop is also part of the same separatrix this particular trajectory has an even greater significance there's a special name given to it because it's starting off from a saddle point moves off in the unstable part of the unstable manifold and it loops back and comes back to the same saddle point as part of the stable manifold such an orbit is called a homo clinic homo clinic orbits play a crucial role in the behavior of non-linear dynamical systems as you can see small changes in initial conditions around this separatrix is around this point can cause very different futures all together this is a lesson of some generality if I started here slightly above the separatrix and move down this way and move off there but if I started here I move off somewhere else similarly if I am here I move off altogether to infinity but if I am here just inside this loop I keep going around so it's clear that separatrix is play a very crucial role in the behavior of non-linear systems this is a non-linear system it's very clear because of this the equation of motion has become non-linear that's responsible for many of the things that we see here having seen what a typical phase portrait would look like for such a one dimensional problem a simple problem let's look at the very model of one dimensional problems of this kind the simple pendulum in the absence of dissipation once we do that we are set to look at higher degrees of freedom what I mean by a simple pendulum is a mathematical pendulum it corresponds to a bob of some mass m suspended without friction from by a light rod of some length L so this is the point of suspension which I take to be the origin and from that you have a light mass less rod of length L and a heavy bob of mass m and the motion of this pendulum is in a specified plane say the plane of the blackboard and the angular displacement about the vertical I call theta and the pendulum moves back and forth in this now the question is what sort of Hamiltonian does it have once again because we know there is no friction in this problem the degree of freedom that we have is one the dynamical variable which specifies the position of the pendulum at any point at any time is in fact the angular coordinate theta about the vertical so it's a function of theta and a conjugate momentum P theta which is nothing but the angular momentum of this pendulum the orbital angular momentum of this Bob about the origin and this is a theta squared twice ML squared since ML squared is the moment of inertia of this Bob about the origin so it's the square of the angular momentum divided by twice so it's the square of the angular momentum divided by twice the moment of inertia plus the potential energy which is the function of the angular displacement theta alone. Now let us assume that the potential energy is 0 when the bob is at its lowest position. Then when it is at an angle theta about the vertical the potential energy corresponds to raising the bob by this height here and therefore it is nothing but twice mL squared plus mgl times 1 minus cos theta. So you have to subtract this distance from that distance multiplied by mg and that gives you the potential energy. Now remember this is a light massless rod a rigid rod and therefore two kinds of motion are possible either the pendulum oscillates about its lowest point or else it rotates completely and both possibilities are included in this expression for the potential energy. So all we have to do is to plot this potential energy find out where the maxima and minima of the potential are and we have our face portrait. So let us do that let us write down what v of theta looks like we have to plot mgl times 1 minus cos theta that is simple it has a bunch of maxima and minima this is at 0 this is at minus 2 pi this is at plus 2 pi and so on this is at pi this is at minus pi this is at minus 3 pi and so on. Where are the critical points of the system let us write the equations of motion down theta dot is delta h over delta p theta which is equal to p theta over mL squared just corroborates the fact that the angular momentum is the moment of inertia multiplied by the angular velocity the dynamics is buried here p theta dot equal to minus delta h over delta theta what is that equal to it is equal to minus mgl sin theta I differentiate this I get a minus sin theta this minus goes against this minus cancels is this a linear system or a non-linear system highly non-linear highly non-linear because of this sin theta it is called all powers of theta in it all odd powers of course you can eliminate p theta completely by differentiating this a second time and substituting for p theta dot here what would you get get theta double dot plus g over L sin theta equal to 0 that is the famous pendulum equation and if I call g over L this quantity the square of the natural frequency for small oscillations then this simply says or theta double dot plus omega naught squared sin theta equal to 0 where I have set omega naught squared equal to g over L this equation is very famous there is a long history it is a non-linear second order ordinary differential equation it is called the sin Gordon for reasons we will not go into right here this started off this name was a joke to start with but then it stuck completely it is similar and formed to an equation which is known in other context for instance in relativistic quantum mechanics called the Klein Gordon equation and this non-linear equation has is related to the Klein Gordon equation and because it has a sign here it was as a joke initially called the sign Gordon equation and that name is stuck completely it is called a long and distinguished history very interesting properties it is a very non-linear equation but it has some very special solutions as we will see non-linear because of all powers of theta sitting here I might mention here that you can actually solve this equation in general the solution is in terms of elliptic functions and elliptic integrals which are not elementary functions they are not ordinary trigonometric functions they are a little more complicated than that but we are not going to do that we are not going to write the solution down we are going to look at the phase trajectories and see what the phase portrait looks like I remind you that in the small amplitude approximation where you can replace sin theta by theta this becomes the harmonic oscillator equation and then of course the time period of oscillation is just 2 pi over omega 0 that is only true for small oscillations the moment the oscillations become reasonably large in amplitude then the time period depends on the amplitude and in fact increases with the amplitude in a fairly complicated fashion now what do the phase trajectories look like where are the critical points so Cp is at P theta equal to 0 and theta equal to the zeros of sin theta which happens at all integer multiples now of course to cut a long story short we pretty much know what these critical points are going to be like so if I plot P theta here versus theta I know that this is a minimum of the potential and therefore there is a center here so is this there is a center here and so is this center here these points so centers occur at all at 0 and all even multiples of pi what sort of critical points do you have at odd multiples the maximum of the potential they are unstable and in this Hamiltonian system the only possibility is saddle points once again so you have a saddle point here a saddle point at this point a saddle point here a saddle point here and so on what would the phase trajectories look like well it is quite clear that in this problem the only allowed values of the energy are non negative of the total energy the moment you have a small positive energy the system could find itself trapped in either this well or this well or this well or this well and in each of those it would execute small oscillations looking like that a little higher energy and these oscillations would slightly bigger ovals these are not ellipses except for extremely small amplitude oscillations because this theta is not approximated the sin theta is not approximated by theta except for theta sufficiently small sufficiently close to a even multiple of 2 pi and then of course these are ellipses as you come closer and closer but after that they are ovals given by this quantity equal to a constant the entire Hamiltonian equal to some constant what would happen if the energy were larger than the maximum value here and the maximum of the potential this thing here corresponds to all these maxima are at exactly the same point same value this maximum corresponds to the separatrix energy which is twice mgl because that corresponds to theta equal to pi in which case the potential energy becomes twice mgl so if e is greater than twice mgl I would expect the motion becomes unbounded because instead of oscillating this way the amplitude keeps increasing and finally it is got enough energy to overcome the barrier to go all the way around and then it would be open motion to go this way or the other way and this would correspond to open trajectories right here or here but the interesting thing happens when you have an energy equal to this separatrix energy then of course you could for example start here at minus pi crawl down extremely slowly accelerate as you come down and go up and crawl up all the way to plus pi that would correspond to a trajectory it starts here and ends there and vice versa which would correspond to something doing this asymptotically similarly you could start here and go there which would correspond to a loop like that so now you have saddle points in which this is the unstable manifold and this is the stable manifold the tangents there if you linearize about these points but now you have a situation where these loops go from one saddle point with the other and back from the next back to this such a loop is called a heteroclinic orbit and of course they correspond to on this trajectory the energies is a mess and what about the open trajectories it is clear that you would have an infinite family of open trajectories which would look like this and on the other side and as the energy increases these things would get flatter and flatter so such a trajectory would correspond to counter clockwise rotation in this theta is going on increasing monotonically and the other one corresponds to clockwise rotation where theta becomes more and more negative monotonically the separatrix as before separates rotational motion from oscillatory motion the interesting point and that's it this is the phase diagram of the undamped simple pendulum the moment you put in damping the moment you have a first order term here a theta dot term which would correspond to a system which is not Hamiltonian then this entire picture changes and it's clear no matter where you start maybe it rotates a few times but eventually it comes to a halt it would oscillate and then damp out so the trajectories would look very different all together and the fate of any point on the phase plane wherever you start would depend on where you started which of these it goes gets attracted to because all these points would become stable spiral points asymptotically stable spiral points and which one it goes to depends on where you start what's interesting is that for very small amplitude oscillations the solutions are trigonometric functions the general solution of an equation with the theta here is simply cos or sin omega not t the solutions for larger amplitude oscillations are elliptic integrals as I mentioned but the solution for this critical value of the energy on the separate ricks is again expressible in terms of elementary functions once again it turns out that you don't need any elliptic integrals or anything like that if you set the total energy to be equal to twice mgl then those trajectories are actually simple to write down and the reason is on those trajectories and have h of p theta and theta equal to p theta squared over 2 ml squared plus mgl times 1 minus cos theta equal to twice mgl and of course we know what this term is it's nothing but one half ml squared theta dot squared and if I bring that down to this side or take this over to the other side what happens is 1 plus cos theta on the other side the m gets removed we took take the 2 there and the right hand side gets simplified what's this equal to this is twice cos squared so this becomes 4g over l cos square theta over 2 if you took this trajectory for example in which theta dot is positive then corresponding to that you have theta dot equal to twice root g over l cos theta over 2 so on that trajectory theta dot equal to twice square root of g over l but that's omega naught and this can be integrated because all you have to do is to rewrite this as d theta times sec theta over 2 and integrate it and it can be done in terms of elementary functions so I leave it to you to write down the explicit solution for an initial condition where at t equal to 0 you are at theta equal to 0 and as t tends to plus infinity you are going to approach theta equal to pi and at t equal to minus infinity you start off from here so I leave you to write the solution down and then we look at its special features and once we are done with this we can move on to understanding how two dimensional and higher degrees of freedom integrable systems would lead to the torus structure I mentioned earlier so we will do that next time any questions it's not with respect to time time has been eliminated completely so the point the reason you draw an arrow on face trajectories is to tell you in which direction the phase space point the representative point representing the system moves as time increases but time itself doesn't appear here it's clear that time has gone has been eliminated and what you have done is simply to say where does the point which is represented in the system which is represented by point in phase space in the space of all its coordinates and its moment where is it located and how does it move as a function of time yes one of these lobes okay that's a phase trajectory so actually what's happening here yes absolutely this is a different phase trajectory from this this is a different phase trajectory from this this point is a phase trajectory by itself so the statement is if you are here then as t tends to plus infinity you are going to flow towards this point if you started here and let t go backwards you would flow towards this point so that's the implication of what is meant by an unstable manifold and a stable manifold to a saddle point because that point alone is a solution to the systems equations of motion in which all the left hand sides vanish and since these are first hand first order differential equations if all the initial conditions are zero if all the derivatives are zero to start with then the system never takes off and it remains there so that corresponds to taking this Bob and balancing it at pi vertically up of course an infinitesimal displacement would cause it to move so it's an unstable equilibrium but it's an equilibrium point nevertheless the crucial thing to note is that these separatrices they actually qualitatively different kinds of motion are separated infinitesimally to the inside of it the motion is periodic infinitesimally to the outside of it the motion is completely open it's rotational motion as opposed to oscillatory motion so they play clearly a very very special role what really happens is that in a system which is perturbed and non integrable unlike this system the separatrices would really determine the fate of the system dynamical system in some sense a homoclinic orbit was one where you started at a saddle point made a loop and came back to the same saddle point a heteroclinic orbit consists of more than one separatrix where you started one saddle point flow into another you started that saddle point and flow back to the original one could be more than two saddle points involved in this loop but it's still a loop and these loops get perturbed very easily and that's how chaos appears in Hamiltonian systems.