Mathematical Interlude: Using Exponentials to Solve Wave Equations

Starting with a known Lagrangian, the Euler-Lagrange equations provide a template for writing the equations of motion. The equations of motion are themselves differential equations. For some purposes, knowing the form of these equations is good enough. However, sometimes we would like to solve them.

Finding solutions to differential equations is a huge topic. Nevertheless, stripped down to its bare bones, the basic approach is as follows:

1. Propose (okay, guess) a function that might satisfy the differential equation.

2. Plug the function into the differential equation. If it works, you’re done.

Otherwise, return to Step 1.

Instead of racking our brains over a solution, we’ll just provide one that happens to work. It turns out that exponential functions of the form

are the main building blocks for wave equations. You may find it puzzling that we choose a complex valued function as a solution, when our problems assume that ϕ is a real valued scalar field. To make sense of this, remember that

where kx − ωt is real. Eq. 5.25 highlights the fact that a complex function is the sum of a real function and i times another real function. Once we’ve worked out our solution , we regard these two real functions as two solutions and ignore the i. This is easy to see if the complex function is set to zero. In that case, its real and imaginary parts must separately equal zero, and both parts are solutions.¹¹ In Eq. 5.25,

cos(kx − ωt) is the real part, and

sin(kx − ωt) is the imaginary part.

If we ultimately extract real functions as our solutions, why bother with complex functions at all? The reason is that it’s easy to manipulate the derivatives of exponential functions.

5.9 Waves

Let’s look at the wave equation and solve it. We already have the Lagrangian for ϕ,

However, I want to extend it slightly by adding one more term. The additional term, , is also a scalar. It’s a simple function of ϕ and does not contain any derivatives. The parameter μ² is a constant. Our modified field Lagrangian is

This Lagrangian represents the field theory analog of the harmonic oscillator. If we were discussing a harmonic oscillator and we called the coordinate of the oscillator ϕ, the kinetic energy would be

The potential energy would be , where μ² represents a spring constant.

The Lagrangian would be

This Lagrangian would represent the good old harmonic oscillator. It’s similar to Eq. 5.26, our field Lagrangian. The only difference is that the field Lagrangian has some space derivatives. Let’s work out the equations of motion that correspond to Eq. 5.26 and then solve them. We’ll start with the time component. The Euler-Lagrange equations tell us to calculate

for Eq. 5.26. It should be easy to see that the result is

This is the analog of the acceleration term for the harmonic oscillator. We get additional terms by taking derivatives of the space components of Eq. 5.26.

With these additional terms, the left side of the equation of motion becomes

To find the right side, we calculate . The result of that calculation is

Gathering our results for the left and right sides of the Euler-Lagrange equations, the equation of motion for ϕ is

Now let’s put everything on the left side, giving us

This is a nice simple equation. Do you recognize it? It’s the Klein-Gordon equation. It preceded the Schrödinger equation and was an attempt to describe a quantum mechanical particle. The Schrödinger equation is similar.¹² Klein and Gordon made the mistake of trying to be relativistic. Had they not tried to be relativistic, they would have written the Schrödinger equation and become very famous. Instead, they wrote a relativistic equation and became much less famous. The Klein-Gordon equation’s connection to quantum mechanics is not important for now. What we want to do is solve it.

There are many solutions, all of them built up from plane waves. When working with oscillating systems, it’s useful to pretend that the coordinate is complex. Then, at the end of the calculation, we look at the real parts and ignore the i. We explained this idea in the preceding Mathematical Interlude.

The solutions that interest us are the ones that oscillate with time and have a component of the form

e^−iωt.

This function oscillates with frequency ω. But we’re interested in solutions that also oscillate in space, which have the form e^ikx. In three dimensions, we can write this as

where the three numbers k_x, k_y, and k_z are called the wave numbers.¹³ The product of these two functions,

is a function that oscillates in space and in time. We’ll look for solutions of this form.

Incidentally, there’s a slick way to express the right side of Eq. 5.28. We can write it as

Where does that expression come from? If you think of k as a 4-vector, with components (−ω, k_x, k_y, k_z), then the expression k_μX^μ on the right side is just

−ωt + k_xx + k_yy + k_zz.¹⁴ This notation is elegant, but for now we’ll stick to the original form.

Let’s see what happens if we try to plug our proposed solution (Eq. 5.28) into the equation of motion (Eq. 5.27). We’ll be taking various derivatives of ϕ. Eq. 5.27 tells us which derivatives to take. We start by taking the second derivative of ϕ with respect to time. Differentiating Eq. 5.28 twice with respect to time gives us

Differentiating twice with respect to x results in

We get similar results when we differentiate with respect to y and z. So far, the Klein-Gordon equation has generated the terms

based on our proposed solution. But we’re not finished. Eq. 5.27 also contains the term +μ²ϕ. This has to be added to the other terms. The result is

At this point, it’s easy to find a solution. We just set the factor inside the parentheses equal to zero and find that

This tells us the frequency in terms of the wave numbers. Either +ω or −ω will satisfy this equation. Also, notice that each term under the square root is itself a square. So if a particular value of (say) k_x is part of a solution, then its negative will also be a part of a solution.

Notice the parallel between these solutions and the energy equation, Eq.

3.43, from Lecture 3, repeated here:

Eq. 5.30 represents the classical field version of an equation that describes a quantum mechanical particle with mass μ, energy ω, and momentum k.¹⁵ We’ll come back to it again and again.

1 Of course we could have many more than two, and they don’t need to be orthogonal Cartesian coordinates.

2 With too few dimensions, of course.

3 In the video, Leonard shows the field by using a colored marker to fill the region with “red mush.” Our diagrams don’t use color, so we have to settle for “imaginary red mush.” Just think of your least favorite school cafeteria entrée.–AF

4 Or, equivalently, V(ϕ) = 0.

5 To avoid clutter, I’m using only one space coordinate.

6 F(x) is not strictly arbitrary. However, it represents a broad class of functions, and we can think of it as arbitrary for our purposes.

7 The symbol ∇²ϕ is shorthand for . See Appendix B for a quick summary of the meaning of and other vector notation. You could also refer to Lecture 11 in Volume I of the Theoretical Minimum series. Many other references are available as well.

8 Substitute your favorite expletive for the word single.

9 It might be better to express A_ν as a row matrix. However, the summation convention described in the next section minimizes the need to write out matrices in component form.

10 Eq. 4.7 also contains a potential energy term −V(ϕ), which we’re ignoring for now.

11 Confusingly, the imaginary part of a complex function is the real function that multiplies i.

12 The Schrödinger equation only has a first derivative with respect to time, and includes the value i.

13 You can think of them as three components of a wave vector, where 14 It turns out that (−ω, k_x, k_y, k_z) really is a 4-vector, but we haven’t proved it.

15 The equation should also include some Planck’s constants that I’ve ignored.