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Some Background:

I've been trying to understand electromagnetic waves, how they travel, and how they're produced. After some Googling and Wikipedia(ing?) I've learned that we use the EM Wave Equations to model how they propagate. However, every single derivation I've seen online does something like this:

$$\nabla\times{E}=-\frac{\partial_t\mathbf{\vec B}}{\partial t}$$ $$\nabla\times{B}=\frac{1}{c^2}\frac{\partial_t\mathbf{\vec E}}{\partial t}$$

Take the curl of both sides

$$\nabla\times\nabla\times{B}=\nabla\times\frac{1}{c^2}\frac{\partial_t\mathbf{\vec E}}{\partial t}$$

Substitute for $\nabla\times E$

$$-\nabla^2B=\frac{1}{c^2}\frac{\partial_t}{\partial t}(\nabla\times E)=-\frac{1}{c^2}\frac{\partial_t^2\mathbf{\vec B}}{\partial t^2}$$

After a little rearranging, we now have the wave equation describing the magnetic component of the electromagnetic wave...

$$\frac{\partial_t^2\mathbf{\vec B}}{\partial t^2}=c^2\nabla^2B$$

While this derivation is super simple and elegant, because I'm new to this area of math and physics, I don't really understand the physical thought process around each action we take. For example, I can understand curl physically, but the curl of curl is a complete mystery to me.

For this reason, I've been trying to find a different derivation, one that makes it easy to follow the physical thought process behind each step. After a bit of messing around, I think I have something:

The Derivation:

$$\nabla\times{E}=-\frac{\partial_t\mathbf{\vec B}}{\partial t}$$ $$\nabla\times{B}=\frac{1}{c^2}\frac{\partial_t\mathbf{\vec E}}{\partial t}$$

These two of Maxwell's Equations describe the behavior of the electric and magnetic field in 3-D space (with no charges or currents). In general, they say that a changing magnetic field will have an electric field "rotating" or curling around it, and a changing electric field will have a magnetic field curling around it.

Now let's draw a diagram of a situation where we have a changing magnetic field. For simplicity, we'll assume that the magnetic field will only point up (y-direction), the electric field will point out of the screen (z-direction) and the wave will propagate in one dimension only (x-axis):

Diagram 1, with no fields

As a result of these simplifications, we can rewrite the two equations so that movement on the y & z-axis is removed:

Notation of curl ($\nabla\times F)$ according to Wikipedia:

Definition of Curl according to Wikpedia

Electric Component:

$$\nabla\times{E}=-\frac{\partial_t\mathbf{\vec B}}{\partial t}$$ Because the magnetic field is increasing only in the y direction, the x & z components of $\nabla\times E$ will be zero:

$$0i + \left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)j + 0k=-\frac{\partial_t\mathbf{\vec B}}{\partial t}$$

Because we are only considering movement on the x-axis, the term $\frac{\partial E_x}{\partial z}$ is removed and we are left with:

$$0i + \left(-\frac{\partial E_z}{\partial x}\right)j + 0k=-\frac{\partial_t\mathbf{\vec B}}{\partial t}$$

Looking at magnitude only: $$\frac{\partial E_z}{\partial x}=\frac{\partial_t B}{\partial t}$$

We can do the same for $\nabla\times B$ and then we have the two equations in 1-D form: $$\partial E=\frac{\partial_t B}{\partial t}\partial x$$ $$\partial B=\frac{1}{c^2}\frac{\partial_t E}{\partial t}\partial x$$

Let's say that there is a magnetic field at the origin, $B_0$, whose y-component is increasing at a rate of $\frac{\partial_tB_0}{\partial t}$.

Diagram 2, increasing B field at origin

The equation: $\partial E=\frac{\partial_t B_0}{\partial t}\partial x$, tells us that as we move an infinitesimal distance on the x-axis ($\partial x$) away from $B_0$, the electric field will increase by $\frac{\partial_t B}{\partial t}\partial x$. Meaning that this increasing magnetic field will induce a perpendicular, increasing, electric field $E_1$ which is equal to $\int{\partial E}$:

Diagram 3, induced electric field

The second equation, $\partial B=\frac{1}{c^2}\frac{\partial_t E_1}{\partial t}\partial x$, tells us that this increasing electric field will also induce a magnetic field:

Diagram 4, induced magnetic field

Now we have a nice picture of how the increasing magnetic field induced an increasing electric field and vice-versa. We also have these two equations describing the interaction between the two:

$$\partial E=\frac{\partial_t B_0}{\partial t}\partial x$$ $$\partial B=\frac{1}{c^2}\frac{\partial_t E_1}{\partial t}\partial x$$

Substituting $\int{\partial E}=\int{\frac{\partial_t B_0}{\partial t}\partial x}$ for $E_1$ we get how the magnetic field changes with time:

$$\partial B=\frac{1}{c^2}\frac{\partial_t}{\partial t}\left(\int{\frac{\partial_t B_0}{\partial t}\partial x}\right)\partial x$$

To get the wave equation, we simply take the derivative of both sides, eliminating the integral:

$$\partial^2 B=\frac{1}{c^2}\frac{\partial_t^2 B_0}{\partial t^2}\partial x^2$$ At very small distances, $B = B_0=B_2$, and after a little rearranging, we get the one-dimensional equation describing the magnetic component of the electromagnetic wave.

$$\frac{\partial^2B}{\partial t^2}=c^2 \frac{\partial^2B}{\partial x^2}$$

We can do the same for the electric component: $$\frac{\partial^2E}{\partial t^2}=c^2 \frac{\partial^2E}{\partial x^2}$$

My Question:

Is this derivation/explanation correct? Does it make sense and is the math correct in its steps? If so, is it useful in its explanation or are there other derivations that do a better job at giving an intuitive/conceptual idea of what's happening?

I'm hoping that it is, since although it's long, I feel it gives a nice picture of what is physically happening rather than just performing vector calculus operations on the Maxwell equations. For me, the hardest part was trying to visualize the propogation of EM waves, and every single derivation I saw would just skip the physical explanation and go to the math, which didn't give an intuative explanation. Any input would be greatly appreciated :)

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    $\begingroup$ Hi there! Welcome! It's sure you try to understand the physics and not only math. Very good. One thing though. You state to substitute $\int{\frac{1}{c^2}\frac{\partial_t E_1}{\partial t}\partial x}$ for $E_1$. But compare the units of these two...Somehow there is circularity in play. And I don't think you can say that at very small distances $B_0=B_2$ (+1, by the way). $\endgroup$ Aug 10, 2020 at 4:18
  • $\begingroup$ @descheleschilder Oof that was my bad, I made a typo. It should be: $E_1=\int{\partial E}=\int{\frac{\partial_t B_0}{\partial t}\partial x}$ I'm fixing that now. Thank you for your comment! $\endgroup$
    – nreh
    Aug 10, 2020 at 4:37
  • $\begingroup$ @descheleschilder That is one of the things about this derivation that seems "hand-wavy" and makes me question it. I can't think of anyway however to get from our diagram into a more general wave equation form. $\endgroup$
    – nreh
    Aug 10, 2020 at 4:49
  • $\begingroup$ I see. I deleted my comment because I was re-reading to make sure I didn’t miss you explaining your notation. Just be aware that this is non-standard notation that will likely raise questions. $\endgroup$
    – J. Murray
    Aug 10, 2020 at 4:51
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    $\begingroup$ A PDF deriving the speed of light in a similar way. $\endgroup$
    – user258881
    Aug 10, 2020 at 10:28

1 Answer 1

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Now let's draw a diagram of a situation where we have a changing magnetic field. For simplicity, we'll assume that the magnetic field will only point up (y-direction), the electric field will point out of the screen (z-direction) and the wave will propagate in one dimension only (x-axis):

You are assuming that $\mathbf E$ and $\mathbf B$ are perpendicular to each other and to the direction of propagation of the wave. At this point in your derivation, you have no justification for this.

Looking at magnitude only [...]

This is generally not correct, in the sense that $$\left|\frac{\partial}{\partial x}f\right| \neq \frac{\partial}{\partial x} |f|$$

Meaning that this increasing magnetic field will induce a perpendicular, increasing, electric field $E_1$ which is equal to $\int \partial E$

I don't know what the symbol $\int \partial E$ means. Also, the reason that the electric field is perpendicular is because you already demanded that be the case in the very beginning.

To get the wave equation, we simply take the derivative of both sides, eliminating the integral

You can't just eliminate the integral by taking a derivative. In particular, whatever the symbol $\partial x$ means, should it not also depart when you take the derivative?

$$\frac{d}{dx} \int_0^x f(u) du = f(x) \neq f(x) du$$


I think the spirit of your derivation is reasonable. You are essentially taking differential equations and turning them into finite difference equations. That's how a computer solves them (roughly).

However, I don't take this as an attack, but the mathematical reasoning is all over the map. Beyond using a fair bit of non-standard notation, you assumed quite a substantial amount of what you were trying to show in the very first step. Integral signs cannot simply be erased by saying that some derivative or another was taken.

From your comment,

I don't think so since $B_0$ is increasing only at the origin, and it is inducing/traveling a second magnetic field an infinitesimal distance away. If you think about it, if $B_2=B_0+\partial B$ then a third induced magnetic field, $B$ would imply $B_4=B_2+\partial B$ resulting in an infinitely increasing magnetic field. In short, the induced magnetic fields cannot be greater than the original.

You can't have a continuous magnetic field which is increasing only at a point. The magnetic field of an electromagnetic wave is constantly changing at every point, so this falls apart.

I won't continue to pick at things. You may find Feynman's physical reasoning about electromagnetic waves to be instructive - he's basically doing what you are doing, but with a bit firmer mathematical and logical footing. You can see his work here. In particular, you should start at the passage which begins

All of our electromagnetic fields satisfy the same wave equation, Eq. (20.8). We might well ask: What is the most general solution to this equation? However, rather than tackling that difficult question right away, we will look first at what can be said in general about those solutions in which nothing varies in y and z.

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  • $\begingroup$ First of all, thanks for reading through the whole thing, and writing a detailed answer. "You are assuming that E and B are perpendicular to each other and to the direction of propagation of the wave. At this point in your derivation, you have no justification for this." I assumed this because of the expanded form of $\nabla\times E$ which, if the magnetic field is increasing only on the y-axis, tells us that as we move an infinitesimal amount $\partial x$ on the x-axis, the electric field on the z-axis will increase.The y and z axis are perpendicular. $\endgroup$
    – nreh
    Aug 10, 2020 at 5:41
  • $\begingroup$ @Xertz What about the term $\frac{\partial E_x}{\partial z}$? $\endgroup$
    – J. Murray
    Aug 10, 2020 at 5:44
  • $\begingroup$ That is what i was refering to when i said as we move an infinitesimal amount ∂x on the x-axis, the electric field on the z-axis will increase. $\endgroup$
    – nreh
    Aug 10, 2020 at 5:45
  • $\begingroup$ @Xertz No, you said Because we are only considering movement on the x-axis, the term ∂𝐸𝑥∂𝑧 is removed[...] There is no justification for this. Why should $\frac{\partial E_x}{\partial z}$ be zero? $\endgroup$
    – J. Murray
    Aug 10, 2020 at 5:47
  • $\begingroup$ Oh my bad, I misread it. My "justification" was that we are just looking one dimension so this term can be removed. Though I admit this is one of the biggest concerns/problems I have with the explanation I wrote. $\endgroup$
    – nreh
    Aug 10, 2020 at 5:50

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