# Energy in EM waves

1. Feb 18, 2016

### 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.

2. Feb 18, 2016

### Staff: Mentor

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

3. Feb 18, 2016

### 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".

4. Feb 18, 2016

### Staff: Mentor

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?

5. Feb 18, 2016

### 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).

6. Feb 21, 2016

### 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?

7. Feb 21, 2016

### Staff: Mentor

Are they assuming the paraxial approximation?

8. Feb 21, 2016

### 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. :/

9. Feb 21, 2016

### 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

10. Feb 21, 2016

### davidbenari

11. Feb 21, 2016

### Staff: Mentor

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

12. Feb 21, 2016

### 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.

13. Feb 21, 2016

### Staff: Mentor

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.