# Light Wave Amplitude

1. Feb 7, 2005

### DB

Maybe this should be under QM but it's a pretty simple question. How does the amplitude of a light wave varie (if it varies at all), if it doesn't what is the lenght?

2. Feb 7, 2005

### dextercioby

QM doesn't account for amplitudes of light waves.Classical electrodynamics does.

Depends on the situation.In the most common & simple cases of propagation through vacuum,it does't vary at all.But,as soon as the wave interacts,sure,the amplitude and the wavelength can vary.

Daniel.

3. Feb 8, 2005

### Gonzolo

The amplitude^2 of lightwaves is equivalent to the number of photons. It doesn't change spontaneously, it changes when it interacts with matter, where each photon either transmits, reflects, scatters, or gets absorbed.

4. Feb 8, 2005

### DB

Thanks.
So,
$$amplitude=\sqrt{number \\\ of \\\\ \gamma}$$

But isn't the # of photons a probability? How do you go about getting the # of photons?

5. Feb 8, 2005

### Gonzolo

No not equal, but "proportionnal to". By equivalent, I mean it's two ways of talking about the same thing. In the case of a monochromatic plane wave, if you know how many $$Js^-^1m^-^2$$ (beam intensity or power), well this number should be the same as $$Nhfs^-^1m^-^2$$, N being the number of photons.

Any good EM text should show how to relate beam amplitude to power (Poynting vector etc.).

6. Feb 8, 2005

### DB

So the strong the beam intensity the higher the amplitude? Why is this?

7. Feb 8, 2005

### Gonzolo

Because beam intensity is power (J/s/area), and beam amplitude is an electric field (or the often neglected magnetic field).

Your question now comes to : "the higher the electric field, the more energy there is?". Well, it certainly makes sense. I'll come back later...

8. Feb 8, 2005

### DB

this confuses me alot too...

9. Feb 8, 2005

### Gonzolo

When we say light is a wave, it's not some undefined wave of something we don't know about. Light is specifically an oscillating electric field, paired with a (perpendicular) magnetic field, and nothing else (according to classical theory). In a plane monochromatic wave, both are perpendicular to the direction of propagation. The optimal value the electric field attains in a half cycle is the wave's amplitude.

This is the classical wave theory that results from Maxwell's equations (which themselves follow from the electrostatics etc. you may be familiar with, Gauss's Law, Faraday induction, Coulomb's Force etc.)

In this theory, the energy of light can be calculated from its amplitude. With quantum theory, the same energy can be counted as Nhv. So that's why the number of photons is practically equivalent to the wave's electric field amplitude. The two different theories are compatible.

10. Feb 8, 2005

### Gonzolo

11. Feb 9, 2005

### DB

Thanks
Yes, I'm starting to get it now and I've seen the perpindicular electric and magnetic wave propagation diagram before, but the one thing I don understand is:
I dont understand how light is the occilation of an electromagnetic Field...
I've always understood it as electromagnetic radiation.

12. Feb 10, 2005

### Staff: Mentor

Why do you think the two concepts are incompatible? ("oscillation of electromagnetic field" versus "electromagnetic radiation")

The reason I'm asking is that I suspect that you don't completely understand one concept or the other. If you can state as specifically as possible, why you think there's a problem, it will be easier for someone to clear things up for you.

13. Feb 10, 2005

### DB

I didn't say they weren't compatible, I just don't understand how they are. But here's what I think.

2 charged particles at distance r create a force and electric/magnetic potential between them, through space. So this it the electromagnetic field. Now as soon as these particles begin to interact (exert Coulomb's force) then they occilate the electromagnetic field creating electromagnetic radiation?
Is that right?

Also, I'm guessing that the frequency of the wave is dependent on the strenght of the charge, but physically, why is it that as charged particles repell/attract, electromagnetic radiation is emmited?

14. Feb 10, 2005

### clive

Hi DB,

The amplitude of a wave can varie even in vacuum. Firstly, it depends on the source shape.

If you have a point source, then the primary energy transfered through any closed surface that contains the source inside must be the same. So,
$$A^2*4\pi R^2=const$$
and you obtain in this case
$$A prop \frac{1}{R}$$
(please read prop as "proportional" - I don't find its LaTex symbol now)
In the case of a plane wave, yes, the amplitude is constant in vacuum. The amplitude of a wave may change because of dispersion too. But there are 2 types of dispersions and I don't think your question is related with.

Last edited: Feb 10, 2005
15. Feb 10, 2005

### DB

thanks clive, though wat is R?

16. Feb 10, 2005

### clive

You're welcome DB!

$$R$$ is the distance from the source to the point where $$A$$ is evaluated. So,

$$A(R) \varpropto \frac{1}{R}$$

(I found $$\varpropto$$ )