Lecture 7: Symmetries and Conservation Laws

Lenny had trouble reading maps. It always seemed like whichever way he was facing must be north. He wondered why he had more trouble with NSEW than he did with up and down. He could almost always get up and down right.

Preliminaries

The relationship between symmetries and conservation laws is one of the big main themes of modern physics. We’re going to begin by giving some examples of conservation laws for some simple systems. At first, the fact that certain quantities are conserved will seem somewhat accidental—hardly things of deep principle. Our real goal, however, is not to identify accidental conserved quantities, but to identify a set of principles connecting them to something deeper.

We’ll begin with the system that we studied at the end of Lecture 6 in Eq. (16), but let’s free it from the interpretation of particles moving on a line. It could be any system with two coordinates: particles, fields, rotating rigid bodies, or whatever. To emphasize the broader context, let’s call the coordinates q instead of x and write a Lagrangian of similar—but not quite identical—form:

(1)

The potential is a function of one combination of variables, namely (q₁ − q₂). Let’s denote the derivative of the potential V by V'. Here are the equations of motion:

(2)

Exercise 1: Derive Equations (2) and explain the sign difference.

Now add the two equations together to see that the sum p₁ + p₂ is conserved.

Next, let’s do something slightly more complicated. Instead of the potential being a function of (q₁ − q₂), let’s have it be a function of a general linear combination of q₁ and q₂. Call the combination (a q − b q₂). The potential then has the form

(3)

For this case the equations of motion are

It seems that we’ve lost the conservation law; adding the two equations does not give the conservation of p₁ + p₂.

But the conservation law has not been lost; it just changed a little bit. By multiplying the first equation by b and the second by a and then adding them, we can see that b p₁ + a p₂ is conserved.

Exercise 2: Explain this conservation.

On the other hand, suppose the potential is a function of some other, more general combination of the q’s, such as q₂ + q₂². Then there is no conserved combination of the p’s. So, then, what is the principle? What determines whether there are conservation laws and what they are? The answer has been known for almost 100 years from the work of the German mathematician Emmy Noether.

Examples of Symmetries

Let’s consider a change of coordinates from q_i to a new set q_i'. Each q_i' is a function of all of the original q coordinates:

q_i' = q_i' (q_i').

There are two ways to think about a change of coordinates. The first way is called passive. You don’t do anything to the system— just relabel the points of the configuration space.

For example, suppose that the x axis is labeled with tick marks, x = . . ., −1, 0, 1, 2, . . . and there is a particle at x = 1. Now suppose you are told to perform the coordinate transformation

x' = x + 1.

(4)

According to the passive way of thinking, the transformation consists of erasing all the labels and replacing them with new ones. The point formerly known as x = 0 is now called x' = 1. The point formerly known as x = 1 is now called x' = 2, and so on. But the particle is left where it was (if it was at x = 1, then the new labeling puts it at x' = 2); only the label has changed.

In the second way of thinking about coordinate transformations, which is called active, you don’t relabel the points at all. The transformation x' = x + 1 is interpreted as an instruction: Wherever the particle is, move it one unit to the right. In other words, it is an instruction to actually move the system to a new point in the configuration space.

In what follows, we will adopt the active point of view. Whenever I make a change of coordinates, it means that the system is actually displaced to the new point in the configuration space. In general, when we make a transformation, the system actually changes. If, for example, we move an object, the potential energy—and therefore the Lagrangian—may change.

Now I can explain what a symmetry means. A symmetry is an active coordinate transformation that does not change the value of the Lagrangian. In fact, no matter where the system is located in the configuration space, such a transformation does not change the Lagrangian.

Let’s take the simplest example: a single degree of freedom with Lagrangian

Suppose we make a change in the coordinate q by shifting it an amount δ. In other words, any configuration is replaced by another in which q has been shifted (see Figure 1).

Figure 1: Shifting the coordinate of a point, q, by δ.

If the shift δ does not depend on time (as we will assume), then the velocity does not change, and—most important—neither does the Lagrangian. In other words, under the change

q → q + δ,

(5)

the change in the Lagrangian is δ L = 0.

In Eq. (5) the quantity δ can be any number. Later, when we consider transformations by infinitesimal steps, the symbol δ will be used to represent infinitesimal quantities, but for now it doesn’t matter.

We could consider a more complicated Lagrangian with a potential energy V(q). Unless the potential is a constant independent of q, then the Lagrangian will change as q is shifted. In that case there is no symmetry. The symmetry of moving a system in space by adding a constant to the coordinates is called translation symmetry, and we will spend a lot of time discussing it.

Now look at Equations (2). Suppose we shift q₁ but not q₂. In that case the Lagrangian will change because the potential energy changes. But if we shift both coordinates by the same amount so that q −₁ q₂ does not change, then the value of the Lagrangian is unchanged. We say that the Lagrangian is invariant under the change

(6)

We say that the Lagrangian is symmetric with respect to the transformation in Equations (6). Again this is a case of translation symmetry, but in this case, to have a symmetry we must translate both particles so that the distance between them is unchanged.

For the more complicated case of Eq. (3), where the potential depends on a q₁ − b q₂, the symmetry is less obvious. Here is the transformation:

(7)

Exercise 3: Show that the combination a q₁ + b q₂, along with the Lagrangian, is invariant under Equations (7).

If the potential is a function of a more complicated combination, it is not always clear that there will be a symmetry. To illustrate a more complex symmetry, let’s revert to Cartesian coordinates for a particle moving on the x, y plane. Let’s say this particle is under the influence of a potential energy that depends only on the distance from the origin:

(8)

It’s pretty obvious that Eq. (8) has a symmetry. Imagine rotating the configuration about the origin by an angle θ (see Figure 2).

Figure 2: Rotation by θ.

Since the potential is a function only of the distance from the origin, it doesn’t change if the system is rotated through an angle. Moreover, the kinetic energy is also unchanged by a rotation. The question is how we express such a change. The answer is obvious: Just rotate coordinates

(9)

where θ is any angle.

Now we come to an essential point about the transformations of translation and rotation. You can do them in small steps—infinitesimal steps. For example, instead of moving a particle from x to x + 1, you can move it from x to x + δ. Now I am using δ to denote an infinitesimal. In fact, you can build up the original displacement x → x + 1 by many tiny steps of size δ. The same is true for rotations: You can rotate through an infinitesimal angle δ and, by repeating the process, eventually build up a finite rotation. Transformations like this are called continuous: They depend on a continuous parameter (the angle of rotation), and, moreover, you can make the parameter infinitesimal. This will prove to be a good thing, because we can explore all the consequences of continuous symmetries by restricting our attention to the infinitesimal case.

Since finite transformations can be compounded out of infinitesimal ones, in studying symmetries it’s enough to consider transformations with very small changes in the coordinates, the so-called infinitesimal transformations. So let’s consider what happens to Equations (9) when the angle θ is replaced by an infinitesimal angle δ. To first order in δ,

(Recall that for small angles, sin δ = δ and , so the first-order shift in the cosine vanishes and the first-order shift in sine is δ.)

Then the rotation represented by Equations (9) simplifies to

(10)

You can also see that the velocity components change. Just differentiate Equations (10) with respect to time:

(11)

Another way to express the effect of the infinitesimal transformation is to concentrate on the changes in the coordinates and write

(12)

Now it’s a simple calculus exercise to show that the Lagrangian does not change to first order in δ. The variation is δ_v.

Exercise 4: Show this to be true.

One thing worth noting is that if the potential is not a function of distance from the origin, then the Lagrangian is not invariant with respect to the infinitesimal rotations. This very important point should be checked by examining some explicit examples. A simple example is a potential that depends only on x and not on y.

More General Symmetries

Before we get to the connection between symmetries and conservation laws, let’s generalize our notion of symmetry. Suppose the coordinates of an abstract dynamical system are q_i. The general idea of an infinitesimal transformation is that it is a small shift of the coordinates, that may itself depend on the value of the coordinates. The shift is parameterized by an infinitesimal parameter δ, and it has the form

δv qi = fi(q) δ.

(13)

In other words, each coordinate shifts by an amount proportional to δ, but the proportionality factor depends on where you are in configuration space. In the example of Equations (6) the value of f₁ and of f₂ are both 1. In the example of Equations (7) the f-functions are f₁ = a and f₂ = −b. But in the more complicated example of the rotations of Equations (12), the f’s are not constant:

If we want to know the change in velocities—in order, for example, to compute the change in the Lagrangian—we need only to differentiate Eq. (13). A little calculus exercise gives

(14)

For example, from Equations (12),

(15)

Now we can re-state the meaning of a symmetry for the infinitesimal case. A continuous symmetry is an infinitesimal transformation of the coordinates for which the change in the Lagrangian is zero. It is particularly easy to check whether the Lagrangian is invariant under a continuous symmetry: All you have to do is to check whether the first order variation of the Lagrangian is zero. If it is, then you have a symmetry.

Now let’s see what the consequences of a symmetry are.

The Consequences of Symmetry

Let’s calculate how much changes when we do a transformation that shifts q_i by the amount in Eq. (13) and, at the same time, shifts by the amount in Eq. (14). All we have to do is compute the change due to varying the and add it to the change due to varying the q’s:

(16)

Now we do a bit of magic. Watch it carefully. First, we remember that is the momentum conjugate to q_i, which we denote p_i. Thus the first term in Eq. (16) is . Hold on to that while we study the second term, . To evaluate terms of this type, we assume the system is evolving along a trajectory that satisfies the Euler-Lagrange equations

Combining the terms, here is what we get for the variation of the Lagrangian:

The final piece of magic is to use the product rule for derivatives:

Thus we get the result

What does all of this have to do with symmetry and conservation? First of all, by definition, symmetry means that the variation of the Lagrangian is zero. So if Eq. (13) is a symmetry, then δ L = 0 and

But now we plug in the form of the symmetry operation, Eq. (13), and get

(17)

That’s it: The conservation law is proved. What Eq. (17) states is that a certain quantity,

(18)

does not change with time. In other words, it is conserved. The argument is both abstract and powerful. It did not depend on the details of the system, but only on the general idea of a symmetry. Now let’s turn back to some particular examples in light of the general theory.

Back to Examples

Let’s apply Eq. (18) to the examples we studied earlier. In the first example, Eq. (1), the variation of the coordinates in Equations (6) defines both f₁ and f₂ to be exactly 1. Plugging f₁ = f₂ = 1 into Eq. (18) gives exactly what we found earlier: (p₁ + p₂) is conserved. But now we can say a far more general thing: For any system of particles, if the Lagrangian is invariant under simultaneous translation of the positions of all particles, then momentum is conserved. In fact, this can be applied separately to each spatial component of momentum. If L is invariant under translations along the x axis, then the total x component of momentum is conserved. Thus we see that Newton’s third law—action equals reaction—is the consequence of a deep fact about space: Nothing in the laws of physics changes if everything is simultaneously shifted in space.

Next let’s look at the second example, in which the variations of Equations (7) imply f₁ = b, f₂ = −a. Again, plugging this result into Eq. (18), we find that the conserved quantity is b p₁ + a p₂.

The last example—rotation—is more interesting. It involves a new conservation law that we haven’t met yet. From Equations (12) we obtain f_x = y, f_y = −x. This time the conserved quantity involves both coordinates and momenta. It is called l, or angular momentum. From Eq. (18) we get

l = y p_x − x p_y.

Again, as in the case of translations, there is a deeper thing involved than just the angular momentum of a single particle:

For any system of particles, if the Lagrangian is invariant under simultaneous rotation of the position of all particles, about the origin, then angular momentum is conserved.

Exercise 5: Determine the equation of motion for a simple pendulum of length l swinging through an arc in the x, y plane from an initial angle of θ.

So far, our examples have been very trivial. The Lagrangian formulation is beautiful, elegant, blah blah, but is it really good for solving hard problems? Couldn’t you just use F = m a?

Try it. Here is an example: the double pendulum. A pendulum swings in the x, y plane supported at the origin. The rod of the pendulum is massless, and the bob (weight at the end) is M. To make it simple, let the rod be 1 meter in length and let the bob be 1 kilogram in mass. Next, take another identical pendulum, but suspend it from the bob of the first pendulum, as shown in Figure 3. We can study two cases: with and without a gravitational field.

Figure 3: The double pendulum.

Our goal will not be to solve the equations of motion. That we can always do, even if we have to put them on a computer and do it numerically. The goal is to find those equations. It’s a tricky problem if you try to do it by F = M a. Among other things, you have to worry about the forces transmitted through the rod. The Lagrangian method is much easier. There is a more or less mechanical procedure for doing it. The steps are the following:

1.Choose some coordinates that uniquely specify the configuration of the components. You can choose them however you like—just make sure that you have just enough to determine the configuration—and keep them as simple as possible.

In the double pendulum example, you need two coordinates. I will choose the first one to be the angle of the first pendulum from the vertical. Call it θ. Next, I have a choice. Should I choose the second angle (the angle of the second rod) also to be measured from the vertical, or should I measure it relative to the angle of the first rod? The answer is that it does not matter. One choice may make the equations a little simpler, but either will get you to the answer. I will choose the angle α to be measured relative to the first rod rather than to the vertical.

2.Work out the total kinetic energy. In this case it is the kinetic energy of the two bobs.

The easiest way to do this is to refer temporarily to Cartesian coordinates x, y. Let x₁, y₁ refer to the first bob and x₂, y₂ to the second bob. Here are some relations among the angles θ, α and x, y: For bob 1,

and for bob 2,

Now, by differentiating with respect to time, you can compute the Cartesian velocity components in terms of the angles and their time derivatives.

Finally, work out the kinetic energy for each bob and add them. It should take a couple of minutes. Remember that we have chosen the masses and rod lengths to be 1.

Here is the result: The kinetic energy of the first bob is:

and the kinetic energy of the second bob is

If there is no gravitational field, then the kinetic energy is the Lagrangian:

If there is gravity, then we have to calculate the gravitational potential energy. That’s easy: For each bob we add its altitude times m g. This gives a potential energy

3.Work out the Euler-Lagrange equations for each degree of freedom.

4.For later purposes, work out the conjugate momenta for each coordinate,

Exercise 6: Work out the Euler-Lagrange equations for θ and α.

There is more that you may want to do. In particular, you may want to identify the conserved quantities. Energy is usually the first one. The total energy is just T + V. But there may be more. Finding symmetries is not always a mechanical procedure; you may have to do some pattern recognition. In the double pendulum case without any gravity, there is another conservation law. It follows from rotation symmetry. Without a gravitational field, if you rotate the whole system about the origin, nothing changes. This implies conservation of angular momentum, but to find the form of the angular momentum, you have to go through the procedure that we derived. That involves knowing the conjugate momenta.

Exercise 7: Work out the form of the angular momentum for the double pendulum, and prove that it is conserved when there is no gravitational field.