PDA

View Full Version : Wave equation and collimated motion of light



ASEI
2007-Apr-28, 12:53 AM
I have a quick question (I'll probably be working it out myself as I go). I was looking at the basic scalar wave equation (laplacian(Field) - d^2Field/dt^2 = 0), and I was playing around with different scenarios on a computer sim. (I successfully set up situations for diffraction/reflection).

However, one thing has been bugging me. The acceleration of the field amplitude is proportional to the laplacian. And as long as there is some difference between the amplitude at two points, there should be divergence of the gradient.

So how do you get a collimated beam propogating, such as in light? It seems to me that points to either side of the beam path should themselves be experiencing divergence, oscillating, and spreading the beam.

Is this because I'm currently only looking at a simple scalar wave? (Maxwell's equations sim to come soon....) Could there be, for example, a collimated sound wave?

papageno
2007-Apr-28, 12:11 PM
So how do you get a collimated beam propogating, such as in light? It seems to me that points to either side of the beam path should themselves be experiencing divergence, oscillating, and spreading the beam.

My gut feeling is that you are right, but the point is on which scale this spreading occurs.

Laser beams, for example, are typically well collimated on laboratory scales (meters or tens of meters), but the spreading becomes observable over much larger length-scales (hundreds or thousands of meters).

Try to look up the Lunar Laser Ranging experiments.




Is this because I'm currently only looking at a simple scalar wave? (Maxwell's equations sim to come soon....) Could there be, for example, a collimated sound wave?
I don't see why not: a wave is a wave.
But again it is a matter of scale.

swansont
2007-Apr-28, 03:28 PM
Lasers are often Gaussian beams. You can't perfectly collimate them, and it's easier to get good collimation with bigger beams.

http://en.wikipedia.org/wiki/Gaussian_beam

Ken G
2007-Apr-28, 09:26 PM
Yes I think the only way to avoid spreading is to set up a "soliton" in a nonlinear medium-- then you can get the nonlinear behavior to cancel the spreading effect, maintaining the soliton without spreading. But I'm no expert on solitons or nonlinear waves.

publius
2007-Apr-29, 06:49 PM
I vaguely recall looking at this very question in an EM class years ago. Note that a wave guide, or even a dieletric "cable" like a fiber-optic cable can give you the same general behavior, keeping an EM wave confined to a region, essentially a planar area normal to the direction of propagation. There, it's the boundary conditions that strongly govern what happens.

In a vacuum, however, there are no hard boundaries.

Here's how I think you can approach this. Consider just a scalar wave (which could be say, E_x, an electric field that points in just one direction). Separate out the time-dependence by assuming it's a sinewave. Look at the resulting spatial equation. Now, drop one dimension for simplicity.

Specify you want it want confined to some region, say along the y-axis. IOW, E_x = 0 for y outside of some region. The would be roughyl a piece of a plane wave propagating in the z direction confined to a narrow region along the y-axis (it would just go to infinity in the x direction)

Look at what that says in the equations. IIRC, you can separate out the spatial part, then do some Fourier splitting, ie the function along the y-axis is some some of sine waves along the y-axis. You would get something like:

E_x(z, y) = E_0*f(z)*g(y), where f and g are expanded in Fourier series (or integrals in the full case). Now, f(z) has to "go along" with the time dependence for a wave propagating in the z direction, here. Via Fourier expansion, you can make g(y) be some narrow pulse thing in the y-direction initially.


Now, look at what the wave equation say about that, along with the full Maxwell. IIRC, you can show that something like that in vacuum must spread out. It's complex, and I'm not exactly sure how the time dependence really works out.

But I do remember looking at it. :)

-Richard

publius
2007-Apr-29, 08:37 PM
http://en.wikipedia.org/wiki/Gaussian_beam

That's what I was talking about. :) Note the above was expressed in cylindrical coordinates. The propagation direction is the z direction, and you're looking at how it spreads in the r direction -- it can be independent of O, the angle. This describes a beam of circular cross-section.

The intensity is modelled as a Gaussian function of 'r'. You'll note how that function has to spread out (and the amplitude decreases as it spreads, conserving energy). You can see why that has to be from the wave equations/Maxwell.

Anyway, you can show that with no hard boundary conditions on the edges, a "beam" solution has to spread. How fast depends, but it has to spread.

-Richard

ASEI
2007-Apr-30, 02:55 AM
Thank you everyone. I'll look into these equations later on this week.