# Maxwell's equations and plane waves

## Homework Statement

Show that the general relationship from Maxwell's equations for the conservation of energy

$\nabla \cdot \textbf{S} + \frac{\partial u}{\partial t} = 0,$

where

$u = \frac{1}{2} \epsilon _{0} \left| \textbf{E} \right| ^{2} + \frac{1}{2 \mu _{0}} \left| \textbf{B} \right| ^{2},$

holds for plane wave solutions to Maxwell's equations.

## Homework Equations

Plane wave solutions:

$\textbf{E} = E_{0} e^{i(\textbf{k} \cdot \textbf{r} - \omega t)}$

$\textbf{B} = B_{0} e^{i(\textbf{k} \cdot \textbf{r} - \omega t)}$

## The Attempt at a Solution

I need a starting point. I can use vector identities to try and derive the answer but I need to know what to start from. I've tried starting from various equations but I can't seem to end up with the conservation law. Any help is appreciated.

TSny
Homework Helper
Gold Member
Can you show a little of what you tried? Note that your equations below are inconsistent:

$\textbf{E} = E_{0} e^{i(\textbf{k} \cdot \textbf{r} - \omega t)}$

$\textbf{B} = B_{0} e^{i(\textbf{k} \cdot \textbf{r} - \omega t)}$

On the left side of the equations you have vector quantities, but on the right side you have scalar quantities.

Yes those should be vectors on the right hand sides of those expressions.

Here is what I'm doing:

$\textbf{S} = \frac{1}{\mu _{0}}( \textbf{E} \times \textbf{B})$

$\nabla \cdot \textbf{S} = \frac{1}{\mu _{0}} \nabla \cdot ( \textbf{E} \times \textbf{B})$

$=\frac{1}{\mu _{0}}( \textbf{B} \cdot (\nabla \times \textbf{E}) - \textbf{E} \cdot (\nabla \times \textbf{B})$

$= \frac{1}{\mu _{0}} (\frac{- \partial}{\partial t}(\textbf{B} \cdot \textbf{B}) - \frac{\partial}{\partial t}(\frac{1}{v^{2}} ( \textbf{E} \cdot \textbf{E})$

$= \frac{- \partial}{\partial t} \frac{1}{\mu _{0}} (B + \frac{1}{v^{2}} E)$

But I don't know how to get the vector lengths squared...nor do I know how to implement epsilon and mu...and how do I show this specifically for plane waves??

marcusl
Gold Member
You have left out the complex conjugation. Note that $$\textbf{S} = \frac{1}{\mu _{0}}( \textbf{E} \times \textbf{B}^*)$$ and so on. Then when you form B dot B you get the amplitude squared.
In fact, it's usually defined in terms of H instead of B. Many authors also include a factor of 1/2 to give power units with the peak amplitudes E0 and B0. Then it's $$\textbf{S} = \frac{1}{2}( \textbf{E} \times \textbf{H}^*)$$

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TSny
Homework Helper
Gold Member
$\nabla \cdot \textbf{S} = \frac{1}{\mu _{0}} \nabla \cdot ( \textbf{E} \times \textbf{B})$

$=\frac{1}{\mu _{0}}( \textbf{B} \cdot (\nabla \times \textbf{E}) - \textbf{E} \cdot (\nabla \times \textbf{B})$

$= \frac{1}{\mu _{0}} (\frac{- \partial}{\partial t}(\textbf{B} \cdot \textbf{B}) - \frac{\partial}{\partial t}(\frac{1}{v^{2}} ( \textbf{E} \cdot \textbf{E})$

$= \frac{- \partial}{\partial t} \frac{1}{\mu _{0}} (B + \frac{1}{v^{2}} E)$

But I don't know how to get the vector lengths squared...nor do I know how to implement epsilon and mu...and how do I show this specifically for plane waves??

##\textbf{B} \cdot (\nabla \times \textbf{E}) = \textbf{B} \cdot( \frac{- \partial}{\partial t}\textbf{B})=-\frac{1}{2}\frac{ \partial}{\partial t}(\textbf{B} \cdot \textbf{B}) = -\frac{1}{2}\frac{ \partial}{\partial t}B^2##, etc.

Your approach here is going to get the result in a general way rather than specifically for plane waves. If you want to show the result for plane waves, then use the plane wave expressions for ##\textbf{E}## and ##\textbf{B}## at the beginning. You can avoid issues with complex numbers by using just the real parts of your expressions for ##\textbf{E}## and ##\textbf{B}##.

You will need to keep in mind the relative orientations of ##\textbf{E}##, ##\textbf{B}##, and ##\textbf{k}## for plane waves. You can simplify things by orienting your coordinate axes in an appropriate way. You will also need to know the relationship between the magnitudes of ##\textbf{E}## and ##\textbf{B}## for plane waves.

Well as much as I'd love to use the real parts, I don't think I'd get full marks. I think we're expected to show this for all complex numbers. But it'd be so much easier using the real parts.

$\nabla \cdot \textbf{S} = \frac{1}{\mu _{0}} \nabla \cdot ( \textbf{E} \times \textbf{B})$

$= \frac{1}{\mu _{0}}\nabla \cdot((\textbf{ E }_{0} e^{i(\textbf{k} \cdot \textbf{r} - \omega t)}) \times(\textbf{ B }_{0} e^{i(\textbf{k} \cdot \textbf{r} - \omega t)}))$

$= \frac{1}{\mu _{0}}( \textbf{ B }_{0} e^{i(\textbf{k} \cdot \textbf{r} - \omega t)} \cdot (\nabla \times \textbf{ E }_{0} e^{i(\textbf{k} \cdot \textbf{r} - \omega t)}) - \textbf{ E }_{0} e^{i(\textbf{k} \cdot \textbf{r} - \omega t)} \cdot (\nabla \times \textbf{ B }_{0} e^{i(\textbf{k} \cdot \textbf{r} - \omega t)})$

Now can I substitute in some $\frac{\partial}{\partial t}$'s as I did before? I just don't know how I'm gonna get rid of those exponentials...

vela
Staff Emeritus
Homework Helper
Why are you ignoring what marcusl said above?

I didn't. He marcusl suggested I use the real parts. I addressed that. Then marcusl said to use the plane wave solutions from the beginning, which I am.

Obviously this stuff is very confusing. I think what I'm trying to say is that I really don't get this stuff...

vela
Staff Emeritus
Homework Helper
TSny said that. Marcusl pointed out you left out the complex conjugation, which would take care of the complex exponentials.

##\textbf{B} \cdot (\nabla \times \textbf{E}) = \textbf{B} \cdot( \frac{- \partial}{\partial t}\textbf{B})=-\frac{1}{2}\frac{ \partial}{\partial t}(\textbf{B} \cdot \textbf{B}) = -\frac{1}{2}\frac{ \partial}{\partial t}B^2##, etc.

Before I carry on, where does the factor of 1/2 come from when factor the partial differential operator?

ehild
Homework Helper
Before I carry on, where does the factor of 1/2 come from when factor the partial differential operator?

derivative of a square...

ehild

Oh right. Okay I am plugging away at things.

TSny
Homework Helper
Gold Member
I’m not seeing the relevance of the expression ##\frac{1}{\mu _{0}}( \textbf{E} \times \textbf{B}^*)## for this problem.

If the fields vary with time sinusoidally, then this expression is related to the time average of the Poynting vector through the relation

##\langle\textbf{S}\rangle## = Real part of ##\frac{1}{2\mu _{0}}( \textbf{E} \times \textbf{B}^*)##

For example, see here .

But I don’t think you’re working with time averages in this question.

marcusl
Gold Member
Using the time average gives the correct answer for harmonically varying fields. The expression u is the instantaneous field energy and the average field energy--they are the same for harmonic traveling waves. Similarly, the time-averaged power flux density equals the instantaneous power flux density in this case.

TSny
Homework Helper
Gold Member
The expression u is the instantaneous field energy and the average field energy--they are the same for harmonic traveling waves.

The contribution to the instantaneous u from the plane wave's electric field E(r,t) = E0cos(k$\cdot$r-ωt) is

uE =## \frac{\epsilon_0}{2}##E02cos2(k$\cdot$r-ωt)

This fluctuates in time. Likewise, for the magnetic field contribution. (The time averages of course are independent of time.)

For the instantaneous u, ##\frac{\partial{u}}{\partial{t}} \not= 0## at general times t.

My interpretation of the problem is that you want to show ##\nabla\cdot\textbf{S}+\frac{\partial{u}}{\partial{t}} = 0## for the instantaneous quantities.

marcusl
Gold Member
The energy conservation law $$\nabla\cdot\textbf{S}+\frac{\partial{u}}{\partial{t}}=-\textbf{E}\cdot\textbf{J}$$ is actually derived for a volume; the change in energy within the volume plus the energy removed by a sink must equal the integrated power entering (exiting) the closed surface. As Reitz and Milford point out,

"It is tempting to interpret E×H itself as the energy flow per unit time per unit area. The latter interpretation leads to certain inconsistencies; the only interpretation which survives scrutiny is that the integral of E×H over a closed surface represents the rate at which EM energy cross the closed surface."

Since we are talking about a volume, then we can take the spatially integrated instantaneous energy equal to the time-averaged energy. The time averaged quantity is therefore valid for this problem. In fact, Jackson uses time averaged quantities when writing Poynting's Theorem for harmonic fields (section 6.10, 2nd ed.).

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BruceW
Homework Helper
I'm not sure why you guys are talking about time-averaging and spatial integration. The equation:
$$\nabla\cdot\textbf{S}+\frac{\partial{u}}{\partial{t}}=-\textbf{E}\cdot\textbf{J}$$
Comes directly from maxwell's 4 laws (in differential form). The only possible tricky part might be using complex numbers instead of the reals.

Edit: now I read through the posts on the previous page more carefully, I don't understand the use of complex conjugates... The way I would do this question is to use the E and B given, plug into the above equation (with J=0) and show it is consistent.

Edit again: You can also show that (assuming J=0), the above equation (with J=0) is true. (even better than just 'consistent'). But you would have to use more than just the given form of the E and B fields. (i.e. you need to use Maxwell's equations, or the relation between the E and B fields for plane waves)

final edit: for clarity, when I say plug in E and B, I mean using the usual definition of the Poynting vector
$$\textbf{S} = \frac{1}{\mu_0} \textbf{E} \times \textbf{B}$$

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vela
Staff Emeritus