Newtonian Mechanics, Degrees of Freedom and Phase Space

Counting degrees of freedom

So let us start now the formal study of the first topic we would like to look at in this course, which is classical dynamics — by which I mean Newtonian mechanics. And then, after we finish the study of classical dynamics in the conventional way, we will go on to see how it merges with other areas of physics.

The first topic we would like to take up is the idea that the mechanics of a particle, which you study in physics one, can be generalised to systems of particles — many, many particles taken together. And the first concept we would like to get clear is that of degrees of freedom. So let me illustrate what I mean by the degrees of freedom of a physical system, such as a collection of particles, or an object, or rigid bodies obeying Newtonian mechanics.

I start with a single particle which can move in three-dimensional space. I need three coordinates — three independent coordinates — to specify its position at any given instant of time, and so it has three degrees of freedom to start with. What happens if I have two particles moving in space? What is the number of degrees of freedom? I need six. And if I have N particles, I need 3N degrees of freedom — provided these particles move in space and there are no constraints among them at all. They are, of course, subject to interaction; they interact with each other, they move under each other’s influence and so on. But they move in three-dimensional space, and each of them requires three coordinates.

Background — what “degrees of freedom” means

The number of independent numbers you must give to say exactly where everything is. A bead threaded on a wire needs one; a coin sliding on a tabletop needs two; a fly in a room needs three. That is all the phrase means.

The subtlety he is building towards is that tying parts together can remove some of these numbers — but not always, and not one for one.

Fuller explanation: degrees of freedom

It does not matter whether these are Cartesian coordinates or spherical polar coordinates or cylindrical polar coordinates. It does not matter. What matters is the number of coordinates. An N-particle system has 3N coordinates, and I am going to call these generalized coordinates.

The reason is that they do not have to be Cartesian coordinates, and they do not necessarily have to be angular coordinates. Some of them could be Cartesian, some angular, some in other coordinate systems altogether; it does not matter. What matters is the number. And we are going to use a certain symbol for these coordinates — I am going to use little q.

Constraints, and what they take away

But before I do that, I would like to introduce the idea of constraints. If I tell you that a particle is constrained to move on a plane such as this, then of course it has only two degrees of freedom; and if you confine this particle to a line, then it has a single degree of freedom.

What is important to recognise is that you must count the number of degrees of freedom before you start solving the equations of motion. So first you have to specify what the independent degrees of freedom are.

Now, we have seen that for N particles you have 3N coordinates, 3N degrees of freedom. But suppose I start putting in constraints. Suppose I tell you that particle one and particle two are connected up in such a way that the distance between them is fixed — it cannot be changed. Then how many independent degrees of freedom are there? There are five, because you have six coordinates between these two particles, but you also have a constraint that r₁₂ equals a constant. The distance between 1 and 2 is fixed; it is given to you as part of the problem. In which case you subtract one independent coordinate out, and you have five left.

Therefore the lesson is this: to find the number of independent degrees of freedom, you must first take the total number of degrees of freedom and subtract out all those you can eliminate on account of the constraints.

Suppose I have three particles, the third one here, and I tell you that the distance between 2 and 3 is also constant. How many independent degrees of freedom do we have now? We have seven. And if I join all three and tell you that all these three distances are fixed, then these three particles could be imagined to be at the vertices of a rigid triangle, and you have six independent degrees of freedom left.

Green board 1 · reconstructed by Claude

The paradox of redundant constraints

So now what is the general story? I have N particles, 3N coordinates, and now I tell you that all the distances between particles are constant: r_ij = constant for 1 ≤ i, j ≤ N. How many independent degrees of freedom do we have now — and how many constraints are there? I tell you that all the pairwise distances are constant. The number of constraints is the number of pairs, N(N−1)/2. So how many independent degrees of freedom are there?

Well, let us try that out. Number of independent degrees of freedom equals — and I put a question mark here, I am not altogether convinced — 3N − N(N−1)/2. What happens if we put N equal to 4? What happens if we put N equal to 5? What happens if we put N equal to 8? It becomes negative — but that is not physical. It is obvious that you cannot have a negative number of degrees of freedom. This cannot go on increasing; it is very clear from here, because this increases linearly in N and that increases like the square of N, so pretty soon that is going to overtake it. So what would you say is wrong? What should we do?

StudentIf we have a structure like a triangle, some constraints become redundant.

Some of the constraints become redundant. It is quite clear that you do not need all these constraints; just some of them are enough to hold the object rigidly. How many would those be? Well, in the case of a triangle, you have…

StudentTwo.

No, it is not two — because if you have these two, you can still move this around, you can still do this, and the distances are still fixed. But if there is a third one, then it is rigid.

So how many independent degrees of freedom do we really have? Now this is mimicking a rigid body. A rigid body in Newtonian mechanics is defined as one where the distance between any pair of points is fixed once and for all. So it is quite clear that this cannot be a formula — this cannot be the number of independent degrees of freedom, simply because it increases like a quadratic function of N whereas this is linear. Where does this stop?

StudentSix.

Six degrees of freedom. A rigid body has six independent degrees of freedom. No matter how large N is, you have just six; the remaining constraints are completely redundant.

Six degrees of freedom, counted two ways

And what are the six degrees of freedom? Now we can start counting them — and when you compute degrees of freedom, the way to check whether something is right or wrong is to do it in two different ways. Compute the number of degrees of freedom in two different ways, and if the answers match, then you know you are on the right track.

We have said that a rigid body has six degrees of freedom. I should like to count these and tell you what they correspond to. You can choose coordinates in different ways, but I would physically like to understand what these coordinates correspond to.

So here is a rigid body — let us take something like a cube — and this has six independent degrees of freedom. Here is one way of counting them. I need three degrees of freedom to tell you where the centre of mass of this object is in space. With respect to some fixed coordinate system, the centre of mass is at some point with position vector R, which of course has three coordinates associated with it, and that takes care of three degrees of freedom.

There are three more, which tell you what the orientation of the object is in space. How do you tell what that is? There are several ways of doing this, but a simple one is the following. Here is my fixed coordinate axis in space; now associate with the body a coordinate system of its own — a body-fixed coordinate system. Then the orientation of the body in space, relative to the fixed system, depends on how this coordinate system is twisted or turned with respect to that one.

Now, what this means is that you start with a space-fixed coordinate system and you want to go to a body-fixed coordinate system; and to go from this set of coordinates to that, you have to do three Euler rotations. You rotate about three different axes, and you need three angles for this purpose. Therefore you have three coordinates for the centre of mass, and three angles which specify the manner in which one coordinate system is turned to reach the other. That is three more degrees of freedom, and therefore you have six.

Green board 2 · reconstructed by Claude

Another way of looking at it is this. Here is a rigid body, and here is its centre of mass — so three coordinates are gone there. Then, to reach the orientation of this coordinate system from that one, I take the coordinate system and turn it about some axis through a certain angle. To specify the axis of rotation I need two angles, because the axis of rotation in three-dimensional space is a unit vector, and a unit vector requires two numbers to specify — the sum of the squares of the three numbers is unity. So I need a latitude and a longitude, if you like, on the surface of a unit sphere to specify an axis; and about that axis I can rotate through any angle from 0 to 2π. That is one more degree of freedom.

Background — unit vectors, and why an axis costs two numbers

A unit vector is an arrow of length exactly 1, used when only the direction matters. It has three components, but they are not free: the length condition forces x² + y² + z² = 1. One equation among three numbers removes one, leaving two.

The globe is the picture. Every direction from the centre of the earth meets the surface at one point, and one point is fixed by latitude and longitude — two numbers. A third number then says how far you have twisted about that axis, which is where his 2 + 1 comes from.

Euler angles, in his other counting, are three angles that between them can bring an object to any orientation at all — pitch, roll and yaw are the everyday version.

Fuller explanation: unit vectors · Euler angles

So we have computed this in two different ways, and each time we discover that for a rigid body you have six degrees of freedom. That is a general rule: the number of independent degrees of freedom equals six, for any N.

Molecules as a check

And of course you are used to this from spectroscopy, when you study the spectra of diatomic molecules. You are told there are three translational degrees of freedom and two rotational ones. The reason is that in the diatomic molecule you have two atoms connected by an invisible bar, and there is no moment of inertia for rotations along that axis; therefore you have two rotational degrees of freedom and three translational.

The moment you have a triatomic molecule you have six, and so on — provided the molecule is not deformable. If it is deformable, and the distances between the atoms change, then you have nine degrees of freedom, and depending on the energy you can excite any number of these modes. But for a rigid body, as far as we are concerned, we have six degrees of freedom: three which I will call translational, because they tell you where you should move to take its centre of mass from a fixed set of coordinates, and three orientational, which tell you how the system is oriented.

So henceforth what we are going to do is to pretend that we have always eliminated the constraints and computed the number of independent degrees of freedom.

When a constraint removes nothing

Now, you have to be a little careful here, because a constraint does not always eliminate a degree of freedom. For instance, if I told you that you have a particle moving along on this table, and the table has boundaries, and you cannot get out of those boundaries — then what you are doing is confining the particle to a certain region. It has two degrees of freedom because it is on a plane, but it has inequalities: its x coordinate cannot exceed this value and its y coordinate cannot exceed that value. Those are inequalities, but they do not reduce the number of degrees of freedom. If I told you that a particle moves on the xy plane in the first quadrant only, it still has two degrees of freedom; nothing is eliminated.

So, whenever you can eliminate degrees of freedom, these are called holonomic constraints, or integrable — never mind why I call them integrable for the moment. And then you have non-holonomic constraints. A non-holonomic constraint cannot in general be used to eliminate a degree of freedom. In general, non-holonomic constraints would be inequalities of the kind I have just illustrated, or they could involve velocities as opposed to positions alone, in which case it is not at all clear that you can integrate matters out to eliminate a degree of freedom. We will come across some examples of this as we go on, but by and large, in this course, we will restrict ourselves to holonomic constraints.

Dropping the chalk: position and force are not enough

Now, I made the point that you must first identify the number of independent degrees of freedom, and then start writing down the equations of motion and solving them — not the other way about. A very simple example is the following. I imagine this piece of chalk to be a point particle moving in space under the action of the earth’s gravity. I hold this particle here, at this point. So I know its initial position, and I know the force on it due to gravity. Can you predict its future motion from that?

Well, I take this chalk and I drop it from rest, and of course it drops straight down. On the other hand, I give it a little horizontal velocity, and then it moves in a parabolic path — strictly speaking part of an elliptic path. I give it a slightly higher velocity and it goes into orbit round the earth. I give it an even higher horizontal velocity and it escapes in a hyperbolic orbit.

So how could you say that if I tell you the initial position and the initial force, the future is predictable? It is not; it looks like it is not. Is there sufficient data? Does it suffice to tell you what the initial positions and the initial forces are?

StudentYou need the initial velocities also.

Newton’s equations are second order in time. You have two constants of the motion in the one-dimensional case, and you need to know both the initial position and the initial velocity to predict what is going to happen next.

Background — why knowing the position and the force is not enough

Newton’s law fixes the acceleration, so the equation you must solve has two derivatives in it — it is second order. Undoing each derivative costs one piece of information, so two facts must be supplied before the answer is definite.

That is exactly what the chalk shows. Same starting point, same gravity: dropped it falls straight, flicked it arcs, flicked hard enough it orbits. The missing fact was the initial velocity.

Fuller explanation: second-order equations · why two initial conditions

So it is immediately clear that dynamics occurs in a space which is not just the coordinates, but also the velocities. This space is called phase space. We are going to do a lot about phase space, but it is good to get this idea right at the beginning — that the equations of motion require you to specify both the initial coordinates and the initial velocities, the slopes of the trajectories, not just the point on the trajectory, for you to be able to solve these equations uniquely. And this was a simple example of it. That immediately tells you that dynamics is not happening in real space; it is happening in something called phase space, which we will study in greater detail. But it is a lesson.

Count first, solve later

And secondly, let me ask you — how many degrees of freedom does this particle have? It has three. You must not make the mistake of saying that if I drop it from rest it moves in a straight line and therefore has one degree of freedom. You should not say that; it is not true, because you cannot count degrees of freedom after you solve the equations of motion. You have to count the independent degrees of freedom before you solve the equations of motion. You must then identify the corresponding velocities as well, and then the equations of motion are written down in terms of these coordinates and velocities, and solved with some specific initial conditions. The idea is that once you specify the initial conditions, the motion is in principle solved.

So this is the idea behind dynamics. We are going to do this in great generality — in much greater generality than even mechanics itself, as we go along. But you have to get this idea right at the beginning: that the number of independent degrees of freedom has to be determined first, and the corresponding generalized velocities have also to be added on to the set of dynamical variables.

So let me write that down here: generalized coordinates q₁ … q_n, and generalized velocities. I am going to use an overhead dot for the time derivative of any dynamical variable, and these are the so-called generalized velocities. This is the notation I am going to use for a system — and I do not care what kind of system it is — which has n independent degrees of freedom, which I label q₁ to q_n. Some of them could be Cartesian coordinates, some could be angular coordinates, they could be mixtures of the two; we do not care. But this is the general framework.

And then, to solve the equations of motion, you need to know a certain amount of initial data. You need to know the values of these generalized coordinates and the generalized velocities at some initial instant of time. And then the task is to write the equations of motion down and to solve them. And you are guaranteed, under suitable conditions, that the solution is unique. It is in this sense that Newtonian dynamics is deterministic: you give me the initial data and the equations of motion, and the future is uniquely determined.

Newton’s equation, and what the force may depend on

Let us do this for the very simple case of a single particle moving along the x-axis, just in one dimension; there is just one degree of freedom. A particle of mass m moving along the x-axis has a single degree of freedom. So — one-dimensional motion of a particle, and let me call the x-coordinate of this particle just q, to get used to the notation. What is the equation of motion of this particle, if it is subject to some force F(q)? It is mq̈ = F, the force on this particle.

Background — the dots, and F = ma

A dot means “rate of change per second”. So q̇ is the velocity and q̈ the acceleration. A prime is different: V′(q) means the rate of change of V as you move along q — the slope of the potential, nothing to do with time.

So mq̈ = F reads: mass times acceleration equals force. The part worth pausing on is that force sets how the velocity changes, not the velocity itself. Everyday life suggests otherwise only because we are surrounded by friction — which is his point about bacteria in a viscous fluid.

Fuller explanation: dot notation · Newton’s second law

And this force could in fact be quite general. It could depend on where the particle is; certainly it could do that. So in general it is a function of q. Could it depend on q̇ as well? Can you give me an example of a force which actually depends on the velocity of the particle?

StudentViscous force.

Anything else?

StudentMagnetic force.

Force due to a magnetic field — it is velocity dependent. So certainly, in general, this would also have a q̇ in it. Could it depend on time explicitly? Yes or no? Yes. What is to stop me from pushing this particle up and down by some prescribed external force which changes with time? So in general it could depend on time as well. Certainly it could.

StudentIf the force depends on q and t, does it not account for q̇?

He says that if the force depends on q and t it accounts for q̇. Is that true? It is a good question. You have to appreciate the fact that q and q̇ are independent dynamical variables, because I could specify them independently at the start. After you have solved an equation of motion under certain conditions, then of course it turns out that q is a function of t, and q̇ is found just by differentiating it. But that is after you solve the equation of motion, for a specific motion. It is not true in general.

So again you have to get used to this idea, that coordinates and velocities are actually independent dynamical variables — because the equation of motion depends on the acceleration. That is Newton’s equation of motion; that is the way it is. We will see why as we go along. But accepting Newton’s equations for the moment, this is an equation for the rate of change of the velocity, and it is a second-order equation in the coordinates. So positions and velocities are independent dynamical variables, because initially you could specify them independently of each other. You must not make the mistake of saying that q̇ is determined by giving q; it is determined only if you give me q already as a function of t, which happens only after you solve the equation of motion.

That is a very good question — why can F not depend on higher derivatives? On q̈, for example, or on the third derivative, and so on. Well, our experience has been — and first of all, Newton’s equations do not say anything about what kind of force you have here at all. You could in principle have very complicated forces; they could depend on the history of the particle as well. We are making an assumption that it does not do so, that it depends only on q, q̇ and t, and that is an assumption based on experience to start with.

There are situations where the force depends on q̈. I give you a simple example, which we will not consider much at the moment. If you have a charged particle subject to external electric and magnetic fields and it moves in space, then if it accelerates it radiates; and once it radiates, that electromagnetic radiation could act back on the particle and produce what is called radiation reaction, leading to radiation damping. That force indeed turns out in classical physics to depend on the acceleration of the particle. So you could have situations with q̈ here as well. This is possible — but these are the simplest instances most of the time.

Now, already this is not an easy problem as it stands, because the force in general is coordinate dependent, velocity dependent and time dependent. And when it is time dependent, this implies that you are changing the rules of the game as time goes on. I am applying a force which is explicitly time dependent, and such a dynamical system is said to be non-autonomous. If it did not have this time dependence, I would say it is autonomous. Of course, non-autonomous systems are all round us. If, for instance, I took a single charged particle and applied an electric field, and changed that field as a function of time, then it would be a non-autonomous system. We will come to non-autonomous systems; I just tell you the possibility exists.

The oscillator as two first-order equations

But in the simplest instances, when you do not have explicit time dependence, and you do not have magnetic fields or viscous forces and so on, you have just F(q); nothing more than that. Now let us look at a simple harmonic oscillator and ask what it does. So you have the x-axis, this is 0, and you have a particle oscillating back and forth. What is the equation of motion, if the natural frequency with which it oscillates is ω? It is mq̈ equal to… q̈ = −ω²q. So moving the m across, this is −kq, where k was the spring constant, and you could write this as −ω²; just to recall.

What is special about this force? It is directed towards the centre of attraction — but what else is special about it? It is a conservative force, in the sense that it is derivable from a potential.

Background — conservative forces and the minus sign

A force is conservative if moving out and back costs no net energy — springs and gravity qualify, friction does not. For such a force you can define a potential energy V, and then F = −dV/dq: the force is minus the slope of the potential.

The minus sign is the physics. It says the force points downhill, towards lower potential energy — a ball on a slope rolls down. Where the potential is flat the slope is zero, so the force is zero, and that is an equilibrium point.

Rewriting the one second-order equation as the pair q̇ = v, v̇ = −V′(q)/m is pure bookkeeping: give the velocity its own name and the same physics splits into two equations, each needing one starting value.

Fuller explanation: conservative forces · first-order form

So as soon as you have a force that is conservative, we know that mq̈ is F(q), and if this is a conservative force, this implies the force is −dV/dq — the derivative of a potential. Now what is the potential here? V(q) is ½mω²q². In a general sense, if I took an arbitrary potential V(q) and said I have this particle moving on the x-axis under the influence of a conservative force derived from that potential, then the equation of motion is mq̈ = −dV/dq.

But now you would say, look, this is a second-order equation, and to solve this single equation you need two initial conditions. What should those be?

StudentPosition and velocity.

For instance, I could choose q(0) and q̇(0); I could specify them independently and solve this in principle. Therefore it suggests that what we should really do is the following. We should write q̇ = v, and v̇ = −V′(q)/m — I use the prime for a derivative with respect to the argument. This is the way I should write these two equations. That is a set of two first-order equations, and it explicitly tells you that the independent dynamical variables are q and v, or q and q̇. And then of course this, plus the initial conditions q(0) and v(0), together implies a unique solution.

It turns out this is the most fruitful approach in mechanics: we write the whole thing down in terms of first-order differential equations, always. And what is the advantage of writing this set of first-order equations? They are coupled to each other, because the q equation involves v and the v equation involves q; but they are first-order differential equations, and therefore specifying the complete set of initial conditions leads in principle to a unique solution.

The phase plane

Now let us pursue this in the case of our harmonic oscillator. We know the solutions; we can write the solution down completely. It is quite trivial to solve — you get cosines and sines as functions of t, and then you fit in the initial conditions and you have the solution uniquely. But this is not as interesting as finding out what the general kind of solution looks like. In this problem it is periodic, you know that; but what is the most general kind of solution one can write down?

You do not even need to solve this problem. The reason is that I could take the attitude that I plot q versus v — that is q̇ on the vertical axis. Since the initial conditions imply specifying a point on this plane, some initial q(0) and v(0), as time goes along q changes to q(t) and v changes to v(t); this point moves on the plane, and this plane is called the phase plane. A point on this plane specifies the system completely, and this point changes as a function of time, and it traces a trajectory called the phase trajectory. So what we do is start at this point — the initial point — and move as a function of time in this way.

Why trajectories cannot intersect

Now tell me, can this trajectory cut itself? This is the phase trajectory. This point here is q(0), v(0), and at any arbitrary instant of time this is q(t). I am interested in looking into the future. So I start with an initial condition and I ask what happens as a function of time, as things go along. But surely the initial condition itself — had the motion been going along forever — would have been reached from some earlier time, and therefore it is really part of a half trajectory. If I had started the whole thing at t equal to minus infinity, there would have been a trajectory which comes along, reaches this point, and then keeps going further down.

Now, can this trajectory do the following? Can it do that? Why not?

StudentIf the point of intersection is an initial condition, the future is supposed to be determined uniquely.

Very good. If the point of intersection is an initial condition, the future is supposed to be determined uniquely from that initial condition and the equation of motion. So you start with this point, and you have two outward arrows, and therefore the future is not unique — because you could have started with that point of intersection as the initial state. And then you are told the future is uniquely determined; it cannot branch off into two different trajectories. Therefore the phase trajectories of such a system cannot intersect themselves.

Green board 3 · reconstructed by Claude

And this is true in any number of dimensions; it is true in general for dynamical systems, provided you do not change the rules in between — in other words, provided you do not have a non-autonomous system. If you have a non-autonomous system then this is possible, because then it distinguishes in time, it is not time-translation invariant; the equations of motion are changing as a function of time. The rules are changing, and therefore there is nothing to stop the phase trajectories from intersecting themselves.

For autonomous systems, phase trajectories cannot intersect themselves, and they cannot intersect each other either. Two different phase trajectories corresponding to two different initial conditions cannot intersect. It is so important that we should write this down.

Is this possible? Is it possible that the system starts at some point, goes along, and comes back to its initial point? This is eminently possible — because once it reaches this point it has no choice but to retrace what it did earlier and come right back. And what do you call such motion, where the system comes back to itself after some time? Periodic motion. That is the only exception. So we say: a simple closed curve implies, and is implied by, periodic motion.

The ellipses of the harmonic oscillator

Let us look at the case of the harmonic oscillator once again and ask what happens. We know that in this case every initial condition corresponds to periodic motion of some kind. So I specialise to the harmonic oscillator: q̇ = v and v̇ = −ω²q, so that q̈ = −ω²q. What will the phase trajectories look like? You are going to integrate this. So let us do that.

We have to take these two equations, find out what the slope of the phase trajectory is like, and then integrate it. So I divide one equation by the other and I get dv/dq = −ω²q/v. That is the slope of the phase trajectory at any point, and I need to integrate it. What should I do? Well, the variables are separable, as you can see — it is very simple. So it says v dv + ω²q dq = 0. And if I integrate this, what happens? It says ½v² + ½ω²q² equals a constant of integration.

Background — separating the variables, and why the answer is an ellipse

Dividing one equation by the other and cross-multiplying gathers every v with the dv and every q with the dq. Each side can then be integrated on its own. That is all “separation of variables” means.

Notice what disappears: time. What survives is a fixed relationship between q and v — a curve the system can never leave — rather than a timetable of where it is when. This is why he can draw the picture without ever solving for q(t).

And ½v² + ½ω²q² = E is an ellipse because both squared terms are added: neither can grow large, so the curve closes on itself. Change that plus to a minus and you get a hyperbola, which runs off to infinity. Bounded or unbounded, repeating or escaping — one sign decides it.

Fuller explanation: separation of variables · ellipses and hyperbolas

What is the significance of this constant of integration?

StudentEnergy.

Almost. If I multiply this by m, then of course you immediately see that this is nothing but writing down the fact that the total energy of the system — and this, remember, is V(q), the potential energy — is equal to E. I choose the symbol E for this constant of the motion, which is very evocative, because I know it is the energy in the system.

Now what is this equation? What does it look like? In general it is an ellipse — unless, of course, the numbers are such that the coefficients of these two are equal. And what kind of ellipse? Where are the principal axes of the ellipse? The coordinate axes themselves. It is of the form x²/a² + y²/b² = 1, so it is an ellipse of this kind.

Which way round the ellipse

Wherever you start on this ellipse, whatever the initial conditions, you are going to remain on this ellipse and come back forever. But you should never draw a phase trajectory without drawing an arrow on it to show the orientation — namely, how the system moves on the phase trajectory as time increases. So what should I do? There are just two possibilities: either it goes clockwise or counter-clockwise. What does it do here? This physical problem corresponds to a harmonic oscillator oscillating with its centre of oscillation at the origin of coordinates, so you can determine what the direction of motion is. In which direction does it move? How do you say that?

Well, if this is the centre of oscillation, I am going back and forth on this axis, and this is 0. If I stretch this oscillator to the end of its amplitude and let go from rest, this corresponds to being here at this point: q is at its maximum, the velocity is 0. And then, as you can see, when I let go from rest the velocity directs it back towards the centre of motion, and therefore it is directed to the left — in other words the velocity goes negative. So it does this: goes negative, passes through the origin again going rapidly leftwards, goes to the leftmost extreme — the amplitude here — and starts moving to the right again, so the velocity becomes positive until it comes back here. Therefore this is the phase trajectory for a simple harmonic oscillator.

Green board 4 · reconstructed by Claude

We know, of course, that the amplitude of the oscillator determines the energy and vice versa. So it is quite clear that this point, the maximum value of q, is given in terms of E by the square root of 2E/mω². That is another way of remembering the fact that the energy of an oscillator is ½ka², where k is the spring constant and a is the amplitude; it is exactly the same statement. And what does this point correspond to — what is the semi-minor axis? The square root of 2E/m.

What happens if I start with an oscillator which has a slightly higher energy? What would the ellipse look like? It would be a concentric ellipse, exactly the same way — so that would be another ellipse, which will go like this. And therefore, as you can see, every time you specify a positive number E, the entire phase trajectory is determined. The system is confined to this ellipse; depending on the value of E, the plane is therefore laminated by these ellipses.

There is just one exceptional point — just one initial condition which does not fall into this picture. And what is that? If it starts at the origin. In other words, you do not stretch it; it is at the origin and it has zero velocity, and then it remains there forever. The origin in the phase plane is a trajectory all by itself. It does not move, it just stays; it is the equilibrium point. We will come back to the significance of this equilibrium point, because it is a special trajectory all by itself. It corresponds to putting E equal to 0, and then of course both v and q are compelled to be 0.

What is special about the harmonic oscillator

What is really special about harmonic oscillations, which are there all around us? The time period is constant — in what sense? Independent of the amplitude. Harmonic motion is the only motion where the time period is independent of the amplitude. There are other potentials for which you also have a time period independent of the amplitude, and we will come to that; they are related to the harmonic oscillator. But there is a simple way of showing that harmonic motion is the only one that is independent of the amplitude, or equivalently of the energy.

For the moment, I want to point out that our statements here are borne out: different phase trajectories do not intersect themselves, and each phase trajectory that is a closed orbit corresponds to periodic motion. That is it.

We would like to show that the time period is independent of the amplitude, and we would like to see if we can do that without doing any hard work. Is there a simple way? Of course you can solve the harmonic oscillator problem — it is very trivial the moment you solve it — but is there a simpler way? There is a simple way of understanding that there is no amplitude dependence in the time period. By the way, could you write down a formula for the time period from this phase trajectory?

Notice that I have not solved the equation of motion. I have not started writing cosines and sines or anything like that; I have just looked at the phase trajectory, and that is sufficient, actually. Because what it is telling me is that an oscillator which starts here, or here, or here, anywhere on this trajectory — they are just minor changes in initial conditions — the motion is exactly the same. So one of the primary advantages of looking at the phase trajectory is that you do not have to worry about specific initial conditions; we are not so worried about that.

More degrees of freedom, more constants of the motion

And in the case of one degree of freedom we were able to write down the phase trajectory immediately. We did not solve the equations of motion; we were able to integrate this directly. If you have a conservative mechanical system we know the total energy is constant. If, therefore, you write the total energy equal to a constant — ½mv² + V(q) = E — this specifies the phase trajectory already. Because to specify a curve on a plane I need one equation between two variables, and this provides it. There is nothing more to be done.

What happens if I have two degrees of freedom? I have a problem with q₁ and q₂. Then, as you can see, the independent degrees of freedom are q₁ and q₂, and associated with them there will be a q̇₁ and a q̇₂. What is the dimensionality of phase space in this problem? It is four. So I cannot draw a picture of this kind. But the phase trajectory, after I solve the equations of motion, would still be a one-dimensional object in a four-dimensional space; and the same thing goes through — the phase trajectories cannot intersect themselves, any closed phase trajectory is periodic motion, and so on.

But is it enough to find one constant of the motion in that problem? Would that suffice to tell you what the trajectory is? No. Between four variables, if you have one equation you do not specify a curve; you specify a three-dimensional surface. How many independent equations do you need before you can specify a curve? Suppose I say φ₁ = c₁ — I have discovered a constant of the motion, some function of q₁, q̇₁, q₂, q̇₂ which is constant, maybe the total energy. If I get an equation of this kind, it is going to specify a three-dimensional hypersurface in that four-dimensional space.

Background — counting dimensions in a space you cannot picture

The rule: each independent equation you impose removes one dimension. In ordinary space one equation leaves a surface (two-dimensional); a second leaves a curve (one-dimensional).

With two degrees of freedom, phase space has four axes. One constant of the motion gives one equation, removing one dimension and leaving a three-dimensional object — “hypersurface” is just the word for a surface in more than three dimensions. But a trajectory is a curve, so you must get down to one dimension, and that takes two more equations. Hence his φ₁, φ₂, φ₃.

You cannot visualise four dimensions and you do not need to. The counting is the argument.

Fuller explanation: counting dimensions · constants of the motion

But I need a phase trajectory. I therefore need more constants of the motion. I need a φ₂ equal to c₂, and I need a φ₃; I need all three, and the mutual intersection of these constant surfaces could give me the line, could give me the phase trajectory. So you begin to see already that it is not a trivial matter to solve problems with more degrees of freedom than one or two — already with two it becomes complicated. If it is of the order of Avogadro’s number, as in the gas in this room, it is a hopeless task. So our trick would not be to attempt to solve the equations of motion in general, except for simple systems.

But we need to go a little further with this, to see how we can find the constants of the motion and what we need to do to do so. But this should be clear already: that you need more constants of the motion, and the more you find, the closer you are to solving the problem. With one degree of freedom, if it is a conservative system, the fact that the total energy is a constant of the motion is enough — if you write this energy as a function of the coordinate and the velocity, the job is in principle complete.

The inverted parabola

We will write down phase trajectories and see what they look like. What if I change the sign of this potential? What if, for sheer perversity, I wrote this as minus one half? What would this motion look like? We have to be a little cautious.

In the case of the harmonic oscillator, if I plot q versus V(q) = ½mω²q², this was parabolic, and then the motion took place as if you had a particle moving in this potential well, back and forth. So for any given total energy, this was the amplitude of the particle and the particle moves back and forth here. When it is here it is all potential energy; when it is here it is all kinetic energy, because the potential is 0; and when it is here it is again all potential energy. It moves back and forth, and this was simple harmonic motion. And then we found the corresponding phase trajectories — so here is q and here is q̇ — and we found these were simply concentric ellipses.

What happens if I took this potential and inverted the sign? So this is an inverted parabola. What would this motion look like? Let us try to guess. Here is q̇, here is q. If the particle starts with q equal to 0 and q̇ equal to 0, it stays there, because there is no change any more. Remember the equations of motion, always. The equations of motion are q̇ = v and v̇ = −V′(q)/m. So if it starts at the origin with zero velocity — this is an extremum of the potential, the maximum of the potential at the origin — therefore this quantity is 0 and that is 0 to start with. Therefore neither q nor v can change with time, because both the derivatives are 0 to start with and remain 0. So the particle would remain here at this point.

On the other hand, if you start it with a slight positive velocity at this point, what would happen? It would just roll down this hill and escape to infinity. If you start it here with a slight negative velocity, it will roll down this hill and escape to minus infinity. Does this correspond to periodic motion? There is no periodic motion here at all. Therefore the trajectories cannot be closed phase trajectories.

So what would they look like? They would look like hyperbolas, and you can see that very easily, because in this case the equation of motion says that ½q̇² − ½ω²q² equals a constant — the total energy. That suggests immediately that this is a hyperbola. You know x² − y² = constant gives a family of hyperbolas, and they are not closed curves; they are open curves.

What would they look like here? Think physically. I start at this point at q equal to 0, but I give it a slight initial positive velocity. So I am really here, at this point, and then q increases as a function of time. What happens to q̇? It also increases in the forward direction. So where does this go? It goes off in this fashion. Now what happens if I start here and push it to the left? q decreases. Then what happens? So what should I draw? We are here, and then where does it go? It goes from here in this fashion.

What happens if I start at infinity and give it just enough energy to crawl up this well and reach this point? What would then happen? Well, it would crawl up here and end here — if it stops exactly at this point. If I gave it a little more energy, it would go over the barrier to the other side.

Green board 5 · reconstructed by Claude

What to work out before next time

So what does the complete set of phase trajectories look like? They are only half trajectories. You must look at all possible initial conditions. By the way, the motion on this goes in this direction. These would be getting completed, so you would have this as part of a half trajectory, and it radiates like this, in this fashion; it is a family of trajectories.

I would like you to tell me if this picture is correct, and to complete this phase portrait. The full set of phase trajectories, all the different kinds, is called the phase portrait. The phase portrait for the simple harmonic oscillator was very simple: it was just a family of concentric ellipses. The phase portrait for this inverted parabolic potential does not have any closed trajectories, no periodic motion; it consists of hyperbolas. But you have to tell me what these hyperbolas look like.

The other thing you have to tell me is whether you could have negative energies in this problem. In the oscillator there is no motion corresponding to a negative energy. Remember, if the total energy were negative, and this is the potential energy, this would imply the kinetic energy is negative, which is not possible, because that is the square of the velocity. So no physical motion happens for E less than 0; at E equal to 0 you are at the equilibrium point; and for E greater than 0 you have physical periodic motion.

In this problem, in contrast, you could have E less than 0 — and this implies that the particle could be anywhere in coordinate space except between these two points; because between these two points, if that is the total energy and that is the potential energy, the kinetic energy is negative, which is not allowed. So this would be the allowed region for you, either on this side or on that side. But for E greater than 0 you can be anywhere.

So this is what I would like you to complete. You need to complete, in this problem, a typical phase trajectory for E less than 0, the trajectories for E equal to 0, and the phase trajectory for E greater than 0. And this is not altogether trivial. There are three possible kinds — E negative, positive and zero — and when E is 0, remember, you can have a trajectory which corresponds to just this point. But you could also have other trajectories, because if you put E equal to 0 on the right-hand side, it says that q̇², apart from constants, is equal to q². Therefore q̇ is plus or minus q with a certain slope; these are lines which go through the origin asymptotically, and we will see what happens.

So already you begin to see that the phase plane analysis is much more powerful than trying to solve the equations of motion; and at the same time it tells you the difference between qualitatively different kinds of dynamical behaviour, some of which would be stable, some unstable, and so on.

Now, we take it from this point next time, where we will complete this phase portrait, and then see what happens with higher numbers of degrees of freedom. And the other thing I would like to do is to show you that in this particular problem, in the case of the single simple harmonic oscillator, there is a very simple dimensional argument which will tell you that the time period is independent of the amplitude. Of course, as a caution, there are other potentials which would do this, but they are much more complicated, and I will introduce a few of them as we go along.