Understand Poynting Density of Flow of Energy w/Weak Physics Background

[Mike Karr]

TL;DR

I have searched all over the web for Poynting density of flow of energy and density of energy . I can't find the formulas divided by $\pi$.

My physics background is weak. My search found lots of $E \times B$ and $E^2 + B^2$, often associated with $\mu_0$ and $\epsilon_0$, but never divided by $4 \pi$ and $8 \pi$, respectively. Could someone provide a reference? Or a derivation? Thanks.

 

[ergospherical]

welcome to the joys of electromagnetic units
(it's a bit of a cluster, sorry to say)

 

[hutchphd]

I suggest finding a tablelisting SI vs CGS and just getting confused for a while..

 

[Mike Karr]

Well, I have been inducted into the joys of electromagnetic units, and I have certainly been confused for a while. However, if I ignore all that and use only formulas in MTW, I can derive:

$\nabla \cdot \frac{E \times B}{4 \pi} = - \frac{\partial}{\partial t}(\frac{E^2 + B^2}{8 \pi}) - E \cdot J$

So I have my $\frac{E \times B}{4 \pi}$ and $\frac{E^2 + B^2}{8 \pi}$. What to do next? Integrating over a volume $V$ and using the divergence theorem leads to:

$\int_{\partial V} \frac{E \times B}{4 \pi} \cdot N dS = \int_V \nabla \cdot \frac{E \times B}{4 \pi} dV = \int_V (- \frac{\partial}{\partial t}(\frac{E^2 + B^2}{8 \pi}) - E \cdot J) dV$

That's as far as I have figured out. I still don't know how to "calculate the Poynting density of the flow of energy and the density of energy". I am not even sure that that means.

 

[Mike Karr]

Well, I have been inducted into the joys of electromagnetic units, and I have certainly been confused for a while. However, if I ignore all that and use only formulas in MTW, I can derive:

$\nabla \cdot \frac{E \times B}{4 \pi} = - \frac{\partial}{\partial t}(\frac{E^2 + B^2}{8 \pi}) - E \cdot J$

So I have my $\frac{E \times B}{4 \pi}$ and $\frac{E^2 + B^2}{8 \pi}$. What to do next? Integrating over a volume $V$ and using the divergence theorem leads to:

$\int_{\partial V} \frac{E \times B}{4 \pi} \cdot N dS = \int_V \nabla \cdot \frac{E \times B}{4 \pi} dV = \int_V (- \frac{\partial}{\partial t}(\frac{E^2 + B^2}{8 \pi}) - E \cdot J) dV$

That's as far as I have figured out. I still don't know how to "calculate the Poynting density of the flow of energy and the density of energy". I am not even sure that that means.

that that => what that

 

[ergospherical]

This is really a problem of classical electromagnetism. Try and absorb this derivation - apologies in advance for SI units.
https://en.wikipedia.org/wiki/Maxwell_stress_tensor#Motivation

 

[PeterDonis]

So I have my $\frac{E \times B}{4 \pi}$ and $\frac{E^2 + B^2}{8 \pi}$. What to do next?

Why do you need to do anything else? You've got the two quantities you wanted.

Integrating over a volume $V$ and using the divergence theorem

Is irrelevant since you're not looking for global quantities, you're looking for local densities. That is all the MTW exercise you refer to in the title of this thread is talking about.

 

[PeterDonis]

I still don't know how to "calculate the Poynting density of the flow of energy and the density of energy"

You already did. The wording of the MTW exercise is a little misleading: it's not actually asking you to calculate these quantities in terms of something else besides $E$ and $B$, it's just telling you that if you know $E$ and $B$, you can calculate these quantities, using the formulas given. (Weird quirks of wording like this are unfortunately fairly common in MTW; I suspect they are due to Wheeler since similar quirks appear in other writings of his.)

Then it's asking you to transform those quantities into a different frame using the Lorentz transformation; that's what you actually need to calculate.

 

[Mike Karr]

Why do you need to do anything else? You've got the two quantities you wanted.Is irrelevant since you're not looking for global quantities, you're looking for local densities. That is all the MTW exercise you refer to in the title of this thread is talking about.

 

[Mike Karr]

PeterDonis, thank you! It's nice when the problem is easier than it seems. And especially for the general tip about "weird quirks." I'll be on the lookout.