Particles and Fields
5.2 Particle Affects Field
How does the particle affect the field? The important thing to understand is that there’s only one action, the “total” action. The total action includes action for the field and action for the particle. I can’t emphasize this enough: What we’re studying is a combined system that consists of a) a field and b) a particle moving through the field.
Fig. 5.1 illustrates the physics problem we’re trying to solve. It shows a region of spacetime, represented as a cube.2 Time points upward and the x axis points to the right. Inside this region, there’s a particle that travels from one spacetime point to another. The two dots are the end points of its trajectory. We also have a field ϕ(t, x) inside the region. I wish I could think of a clever way to draw this field without cluttering the diagram, because it’s every bit as physical as the particle; it’s part of the system.3
Figure 5.1: Particle moving through a region
of spacetime filled with
“red mush”—the scalar field ϕ(t, x).
To find out how this field-particle system behaves, we need to know the Lagrangian and minimize the action. In principle, this is simple; we just vary the parameters of the problem until we find a set of parameters that results in the smallest possible action. We wiggle the field around in different ways, and we wiggle the particle trajectory between its two end points until the action integral is as small as we can make it. That gives us the trajectory and the field that satisfy the principle of least action.
Let’s write down the whole action, the action that includes both the field and the particle. First, we need an action for the field,
where the symbol means “Lagrangian for the field.” This action integral is taken over the entire spacetime region, t, x, y, and z. The symbol d4x is shorthand for dtdxdydz. For this example, we’ll base our field Lagrangian on Eq. 4.7, a Lagrangian that we used in Lecture 4. We’ll use a simplified version that only references the x direction in space and has no potential function V(ϕ).4 The Lagrangian
leads to the action integral
This is the action for the field.5 It doesn’t involve the particle at all. Now let’s incorporate the action for the particle, Eq. 5.2 with the speeds of light removed. This is just
or
Although Actionparticle is the particle action, it also depends on the field. This is important; when we wiggle the field, this action varies. In fact, it is wrong to think of Actionparticle purely as a particle action: The term
Figure 5.2: Particle at Rest in Imaginary Red
Mush (black ink on paper).
is an interaction term that tells the particle how to move in the field, but it also
tells the field how to vary in the presence of the particle. From now on I will call this term and cease thinking of it as having to do only with the particle.
We’ll consider the simple special case where the particle is at rest at x = 0.
It’s reasonable to assume that there’s some solution where the particle is at rest
—classical particles do rest sometimes. Fig. 5.2 shows the spacetime trajectory of a resting particle. It’s just a vertical line.
How do we modify Eq. 5.5 to show that the particle is at rest? We just set , the velocity, equal to zero. The simplified action integral is
Because the particle sits at the fixed position x = 0, we can replace ϕ(t, x) with ϕ(t, 0) and write
Let’s take a closer look at . Notice that it only depends on the value of the field at the origin. More generally it depends on the value of the field at the location of the particle. Nevertheless when the field is wiggled, ϕ(t, 0) will wiggle, affecting the action. As we will see, this affects the equation of motion for the field.
The action for a field is normally written as an integral over space and time, but Eq. 5.6 is an integral that only runs over time. There is nothing wrong with it, but it’s convenient to rewrite it as an integral over space and time.
We’ll use a trick that involves the idea of a source function. Let ρ(x) be a fixed definite function of space, but for the moment not time. (I’ll give you a hint:
ρ(x) is something like a charge density.) Let’s forget the particle but replace it with a term in the Lagrangian that I’ll continue to call ,
The corresponding term in the action for the field is
This looks quite different from the action in Eq. 5.2. To make them the same, we use the trick that Dirac invented—the Dirac delta function δ(x). The delta function is a function of the space coordinates that has a peculiar property. It is zero everywhere but at x = 0. Nevertheless it has a nonzero integral,
Let’s imagine graphing the delta function. It is zero everywhere except in the immediate vicinity of x = 0. But it is so large in that vicinity that it has a total area equal to 1. It is a very high and narrow function, so narrow that we may think of it as concentrated at the origin, but so high that it has a finite area.
No real function behaves that way, but the Dirac function is not an ordinary function. It is really a mathematical rule. Whenever it appears in an integral multiplying another function F(x), it picks out the value of F at the origin. Here is its mathematical definition:
where F(x) is an “arbitrary” function.6 If you have some function F(x) and integrate it as shown, the delta function picks out the value of F(x) at x = 0. It filters out all other values. It’s the analog of the Kronecker delta, but operates on continuous functions. You can visualize δ(x) as a function whose value is zero everywhere, except when x gets very close to zero. When x is close to zero, δ(x) has a very tall spike. For our current problem we need a three-dimensional delta function that we call δ3(x, y, z). We define δ3(x, y, z) as the product of three one-dimensional delta functions,
δ3(x, y, z) = δ(x)δ(y)δ(z).
As we do elsewhere, we often use the shorthand δ3(x), where x represents all three directions of space. Where is δ3(x, y, z) nonzero? It’s nonzero where all three factors on the right side are nonzero, and that only happens in one place:
the point x = 0, y = 0, z = 0. At that point of space it’s enormous.
The trick in writing the interaction term as an integral should now be fairly obvious. We simply replace the particle with a source function ρ that we choose to be a delta function,
The action in Eq. 5.8 then takes the form
The point is that if we integrate this over the space coordinates, the delta function rule tells us to simply evaluate gϕ(x) at the origin, and we get something we’ve seen before:
Now let’s combine the field action with the interaction term. We do that in the simplest possible way, by just adding the two of them together. Combining the action terms of Eqs. 5.4 and 5.12 results in
or replacing the source function with the delta function,
Here it is, the total action as an integral over space and time. The expression inside the big square brackets is the Lagrangian. Like any good field Lagrangian, it has the field action (represented here by partial derivatives) along with a delta function term that represents the effect of the particle on the field. This particular delta function represents the special situation where the particle is at rest. But we could also jazz it up so that the particle moves. We could do that by making the delta function move around with time. But that’s not important right now. Instead, let’s work out the equations of motion, based on the Lagrangian
5.2.1 Equations of Motion
For convenience, I’ll rewrite Eq. 4.5, the Euler-Lagrange equation, right here.
Eq. 5.15 tells us what to do with in order to find the equations of motion.
The index μ runs through the values 0, 1, 2, and 3. The first value it takes is 0, which is the time component. Therefore our first step is to find the derivative
Let’s unwind this calculation step by step. First, what is the partial of with respect to ? There’s only one term in that involves . Straightforward differentiation shows that the result is
Applying to each side results in
That’s the first term in the equation of motion. It looks a lot like an acceleration. We started with a kinetic energy term that amounts to . When we worked out the derivative, we got something that looks like an acceleration of ϕ.
Next, we let μ take the value 1 and calculate the first space component. The form of these terms is exactly the same as the form of the time component
except for the minus sign. So the first two terms of the equation of motion are
Because the y and z components have the same form as the x component, we can add them in as well,
If those were the only terms, I would set this expression equal to zero, and we would have a good old-fashioned wave equation for ϕ. However, the Lagrangian depends on ϕ in another way because of the interaction term. The complete equation of motion is
and we can see that the last term is
Adding this final piece to the equation of motion gives us
Thus we see that the source function appears in the field equation for ϕ as an addition to the wave equation. Without the source term, ϕ = 0 is a perfectly good solution, but that’s not true when the source term is present. The source is literally that: a source of field that prevents the trivial ϕ = 0 from being a solution.
In the actual case where the source is a particle at rest, ρ(x) can be replaced by the delta function,
Let’s suppose for a moment that we’re looking for static solutions to Eq. 5.16.
After all, the particle is standing still, and there may be a solution where the field itself doesn’t change with time. It seems plausible that a standing-still particle can create a field that also stands still—a field that does not vary with time. We might then look for a solution in which ϕ is time-independent. That means the term in Eq. 5.16 would become zero. We can then change the signs of the remaining terms to plus and write
Perhaps you recognize this as Poisson’s equation. It describes, among other things, the electrostatic potential of a point particle. It’s often written by setting
∇2ϕ equal to a source (a charge density) on the right side.7 In our example, the charge density is just a delta function. In other words, it’s a high, sharp spike,
Of course, our current example is not an electric or magnetic field.
Electrodynamics involves vector fields, and we’re looking at a scalar field.
But the similarities are striking.
The third term of the Lagrangian (Eq. 5.14) ties everything together. It tells the particle to move as if there were a potential energy −gϕ(t, x). In other words, it exerts a force on the particle. This same term, when used for the equation of motion of the ϕ field, tells us that the ϕ field has a source.
These are not independent things. The fact that the field affects the particle tells us that the particle affects the field. For a particle at rest in a static field, Eq. 5.18 tells us exactly how. The parameter g determines how strongly the particle affects the field. The same parameter also tells us how strongly the field affects the particle. It makes a nice little story: Fields and particles affect each other through a common term in the Lagrangian.
5.2.2 Time Dependence
What happens if we allow the particle to move and the field to change with time? We’ll confine ourselves to a single
dimension of space. Eq. 5.18 becomes
∇2ϕ = gδ(x).
In one dimension, the left side of this equation is just the second derivative of ϕ. Suppose I wanted to put the particle someplace else. Instead of putting it at the origin, suppose I want to put it at x = a? All I need to do is change x in the preceding equation to x − a, and the equation would become
∇2ϕ = gδ(x − a).
The delta function δ(x − a) has its spike where x is equal to a. Suppose further that the particle is moving, and its position is a function of time, a(t). We can write this as
∇2ϕ = gδ(x − a(t)).
This would tell you that the field has a source, and that the source is moving.
At any given time the source is at position a(t). In this way we can accommodate a moving particle. But we still have one little wrinkle to deal with. If the particle is moving, we would not expect the field to be time independent. If the particle moves around, then the field must also depend on time. Remember the term that we zeroed out in the equation of motion (Eq.
5.16)? For a time-dependent field, we have to restore that term, resulting in
If the right side depends on time, there’s no way to find a solution where ϕ itself is time-independent. The only way to make it consistent is to restore the term that involves time.
A moving particle, for example a particle that accelerates or vibrates, will give the field a time dependence. You probably know what it will do: It will radiate waves. But at the moment, we’re not going to solve wave equations.
Instead, we’ll spend a little time talking about the notation of relativity.