Energy in EM Waves: Is E-field Approach Equally Valid?

[davidbenari]

I wanted to know if my reasoning is considered sound, and if not please tell me the loopholes you can observe.

The energy density in an EM field is (Its not necessarily a plane wave we're talking about here).

$u= \frac{\epsilon E^2}{2}+\frac{B^2}{2\mu}$

The relationship $E=cB$ is supposedly "general" so then the energy density is:

$u = \epsilon E^2$

the intensity (power per unit area) is then

$I = \epsilon c E^2$

and this is a general result as well.

Now suppose I have two plane waves coming in at different angles towards a point on which they intersect. I want to know the average intensity at that point.

I could proceed in two ways: one is to find the Poynting vector by adding the E and B fields and averaging out in time.

Or I could add the E fields and average out in time obtaining

$<I> = c \epsilon <E^2>$

and these should be equal to one another (of course I'm taking about the magnitude of the Poynting vector).

Is the finding-the-E-field approach equally as valid as finding the Poynting vector?

Thanks.

 

[Dale]

The relationship E=cB is supposedly "general"

No, it isn't. It is violated in many instances, like in a static field.

 

[davidbenari]

Is it valid for the case I mentioned?

edit: Also I was referring to its generality in an electrodynamic context. The sources I've read Griffiths and Fitzpatrick say the relationship is "general".

 

[Dale]

Suppose that you have two coherent plane waves, one in the x direction and one in the y direction. Assume further that they are linearly polarized with the E field in the z direction for both.

What is the total E field and the total B field? Does the relationship hold?

 

[davidbenari]

No it doesn't seem like it... So there's no way around having to calculate the Poynting vector then?

I believe I've seen experiments in interferometry where they only consider the amplitude of the electric field squared. But maybe that was assuming the same polarization for all incident beams (which in my case is not required).

 

[davidbenari]

Fowles optics, on his part about interference says the irradiance for two random plane waves (that coincide at P) is given by (aside from some factors) $|\mathbf{E}|^2$. This has me baffled. Why is it acceptable to not take the Poynting vector here? Just taking the modulus squared is going to give something different than the Poynting vector, I'm sure.

Any ideas?

 

[Dale]

Are they assuming the paraxial approximation?

 

[davidbenari]

No but the next section is about Young's interferometer so maybe they're assuming the same polarization even if they didn't explicitly mention this. :/

 

[davidbenari]

Sorry, I just read they're not assuming equal polarization.

For them

I=$|\mathbf{E}|^2=I_1+I_2 + 2 \mathbf{E_1}\cdot\mathbf{E_2} \cos\theta$ with $\theta$ being the phase difference.

If the polarizations are orthogonal you would have just I=I1+I2

 

[davidbenari]

https://books.google.com.mx/books?id=SL1n9TuJ5YMC&pg=PA57&source=gbs_toc_r&cad=4#v=onepage&q&f=false

pages 57 to about 59

 

[Dale]

For them

I=$|\mathbf{E}|^2=I_1+I_2 + 2 \mathbf{E_1}\cdot\mathbf{E_2} \cos\theta$ with $\theta$ being the phase difference.

I haven't worked it out, but that sounds plausible for plane waves.

 

[davidbenari]

http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide14.pdf

They do the same thing here.

I agree with you that it sounds plausible. Would the proportionality factors be $c\epsilon_0$ (and other factors having to do with taking an average)?

Any ideas on how I could prove this? Evidently as you said $E=cB$ isn't a general relation, so that couldn't be it.

I keep hearing one can derive energy relations from Maxwell's Equations. Where could I find a derivation of this sort? Advanced EM texts? Jackson?

Thanks.

 

[Dale]

Any ideas on how I could prove this? Evidently as you said E=cB isn't a general relation, so that couldn't be it.

E=cB isn't valid in general, but it is valid for an individual plane wave. So in this problem $E=E_1+E_2$ and similarly for B. Expand the field energy in those terms then make the substitutions $E_1=cB_1$ etc.