# How to understand typo in MIT OCW chapter on Poynting vector?

• zenterix
zenterix
Homework Statement
While reading a chapter about Poynting vector I reached a passage with what seems like a typo.
Relevant Equations
If it is a typo, I am still not sure how to fix it.

If it is not, then I am not sure what exactly was done in the algebra.
Here is a chapter from MIT OCW's 8.02 Electromagnetism course.

At the end of page 14 is section 13.6 "Poynting Vector". The calculations I am interested in are on page 15.

There is a passage that seems to have a typo in it. Let me try to show why despite recognizing a typo I am unsure of what the correct version would be.

Consider a plane wave passing through the infinitesimal volume element below

The total energy in the EM fields in the volume element is

$$dU=UAdx=(U_E+U_B)Adx=\left (\frac{1}{2}\epsilon_0E^2+\frac{1}{2\mu_0}B^2\right )Adx\tag{1}$$

$$=\frac{1}{2}\left (\epsilon_0E^2+\frac{B^2}{\mu_0}\right )Adx\tag{2}$$

The rate of change of energy per unit area is

$$S=\frac{dU}{dt}\frac{1}{A}=\frac{c}{2}\left (\epsilon_0E^2+\frac{B^2}{\mu_0}\right )\tag{2a}$$

where I have used the fact that the EM wave is traveling with speed ##c## and so ##dx=cdt##.

It can be shown that ##\frac{E}{B}=c=\frac{1}{\sqrt{\epsilon_0mu_0}}##, the speed of light.

The chapter then rewrites (2) but the expression seems to contain a typo. Here is the exact passage as it appears in the chapter

$$S=\frac{1}{2}\left (\epsilon_0E^2+\frac{B^2}{\mu_0}\right )=\frac{cB^2}{\mu_0}=c\epsilon_0E^2=\frac{EB}{\mu_0}\tag{3}$$

When I rewrite (2) I get

$$S=\frac{1}{2}\left (c\epsilon_0E^2+c\frac{B^2}{\mu_0}\right )$$

$$=\frac{1}{2}\left (\epsilon_0\frac{E^3}{B}+\frac{EB}{\mu_0}\right )$$

What am I missing?

Last edited:
Where did the added factor of c come from in your rewrite of equation (2)?

zenterix said:
$$S=\frac{1}{2}\left (\epsilon_0E^2+\frac{B^2}{\mu_0}\right )Adx=\frac{cB^2}{\mu_0}=c\epsilon_0E^2=\frac{EB}{\mu_0}\tag{3}$$

First of all, you have brought along an ##A\, dx## here that should be divided by ##A\, dt## in the first expression. Try to keep away from such mistakes as they lead to confusion as displayed in post #2.

zenterix said:
When I rewrite (2) I get

$$S=\frac{1}{2}\left (c\epsilon_0E^2+c\frac{B^2}{\mu_0}\right )$$

$$=\frac{1}{2}\left (\epsilon_0\frac{E^3}{B}+\frac{EB}{\mu_0}\right )$$

What am I missing?
You are missing that ##E = cB## so take ##E^2 = c^2 B^2## in the first term and it also becomes the same as the second term - resulting in cancelling the 1/2 in front.

The reason to write it as a product of ##E## and ##B## and not any other powers is that the Poynting vector is proportional to ##\vec E \times \vec B##. The Poynting vector is more general than what they are showing in this particular passage and you cannot get it directly from this argument (although you can use the argument to verify that the Poynting vector indeed describes the expected energy current for the particular situation).

zenterix
@phyzguy I forgot one step. It is there now, equation (2a).

Orodruin said:
You are missing that E=cB so take E2=c2B2 in the first term and it also becomes the same as the second term - resulting in cancelling the 1/2 in front.
Ah yes. I need to get some sleep that is what I need to do. Then maybe I won't miss such things.

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