# EM Wave amplitude

1. Jan 16, 2009

### Emreth

Hi
I hope this is the right place to post.
I have some questions regarding EM wave amplitudes. There are some other posts in the past regarding this and some people replied by saying that amplitude of a light wave is equal/close to wavelength of the wave. This is deduced from the fact that light wave with a certain wavelength does not pass through holes smaller than the wavelength, so the amplitude should be somewhat near. If this is even approximately true, then linear wave equation should not be valid for light waves since its derivation assumes small displacements/gradients in the transverse direction. For large amplitudes relative to wavelength, the displacement gradient can reach large values.
Of course i must be missing something here, E0 and B0 do not have length units, or do they?
Any ideas?

2. Jan 16, 2009

### Emreth

Ok I have to rephrase this because its wrong the way I put it. There's no transverse displacement, but rather extent of E and B fields.
Let me say, E field gradient gets large or not?Prob not i presume if E0 is really small. Is that the case?Although the spatial extent of the field is comparable to wavelength, the actual field gradient is very small.Is that it?

3. Jan 16, 2009

### Staff: Mentor

Wavelength is measured in units of length (meters, etc.). The amplitude of an electromagnetic wave is measured either in volts per meter if you're looking at the electric part of the wave, or in tesla if you're looking at the magnetic part. Either way, the amplitude is a completely different type of quantity from the wavelength, and it makes about as much sense to compare their sizes as it makes sense to compare the size of a color and an odor.

Bingo!

4. Jan 16, 2009

### Emreth

ehe i corrected that before you posted, beat you by 3 mins

5. Jan 16, 2009

### Staff: Mentor

Yeah, now I see what you posted. That sort of thing happens a lot around here. :yuck:

So now you're thinking of the "width" of the wave in terms of the amplitude (as measured in appropriate units) falling off as you go away from the centerline or axis of the wave? In that case, from a classical point of view, that's basically determined by the source of the wave (size of the aperture in the source). It isn't connected with the wavelength except if you make the aperture small enough that diffraction effects become important. When you get to that point, the "width" of the beam spreads out over an angular range corresponding to the first minimum in the diffraction pattern.

If you're thinking of single photons, I don't think it's meaningful to consider photons as having a particular size, either along the direction of propagation or transverse to it. Photons are quanta of energy of the electromagnetic radiation field, and position doesn't enter into their description in QED (or so I understand from postings by others who know more about QED than I do).

6. Jan 16, 2009

### Emreth

No i'm not interested in photons. What i'm wandering is if its really valid to use the linear wave equation?

7. Jan 17, 2009

### Emreth

Anybody?

8. Jan 17, 2009

### Staff: Mentor

What do you mean by "linear wave equation?" If you mean the differential wave equation

$$\nabla^2 \vec E = \frac{1}{c^2} \frac{\partial^2 \vec E}{\partial t^2}$$

and the identical equation for $\vec B$, they can be derived from Maxwell's Equations, as was first done by Maxwell himself.

These are linear differential equations whose solutions obey the principle of superposition.