# Inverse square law ?

1. Feb 20, 2004

### jlenard

Inverse square law ?

Does anyone have an absolute basic understanding of "first order principles of energy dispersion"

(A) Energy from a point source disperses spherically at the inverse square law.
(B) Energy from an infinite line source disperses cylindrically at ½ inverse square law.

Question?
What is the energy dispersion from a finite line source?
Can the question be explained with elementry geometry?

Measured 90deg from center of line source, at distance x, where transition from ½ inverse square law, back to inverse square law, takes place.

Can anyone throw 'light' on this question without diversions?
1 First order principles only.
2 In space, zero boundaries.
3 Line length magnitudes greater than any wavelength of energy dispersed.

Thank you.

2. Feb 21, 2004

### HallsofIvy

Staff Emeritus
It's not at all clear what you are asking. I don't know what you mean by "first order principles" or "zero boundaries".

Also "Measured 90deg from center of line source, at distance x, where transition from ½ inverse square law, back to inverse square law, takes place. " makes no sense. There is no such "transition".

It is possible to find the energy dispersion from a finite line source but very difficult (and might require a numerical integration). Working with either a point source or an infinite line source, we can use symmetry to reduce the problem to a single variable: the distance from the point and the distance from the line, respectively. With a finite line source, that is not possible: you have to keep all three coordinates as variables.

If your question, that I quoted above, applies to the finite line source, then, while there still is not definable distance where a "transition" occurs, it is true that very close to the line (close enough that the particle can't "see" the endpoints and so the line "looks" infinite) the field will be approximately that of an infinite line source. Also very, very far from the line (so that the finite line "looks" like a single point) the field will be approximately that of a point source.

There is still no "transition" distance because:
1) Even those extremes are only approximately like an infinite line or a point source.
2) For "almost all" of space neither of those are even approximately true. The "transition" occurs just about everywhere!

3. Feb 23, 2004

### HallsofIvy

Staff Emeritus
Actually, an "inverse square law" is just a "conservation" law. If you have something (gravity, light, electrons, space aliens) moving in 3 dimensions from a central point as long as the total measure remains the same ("conservation") the density must be "inverse square" of the distance. That's simply because the area they must cover increases as the square.

One of my teachers had a "paint-gun" that sprayed paint on surfaces at different distances. It had 4 rods running out from the barrel, holding targets at two different distances. The first target was 1 decimeter by 1 decimeter. The second target, at twice the distance was 2 decimeters by 2 decimeters and so was 4 times the area. In fact, it was divided into 4 sections of exactly the same size as the first target. It was easy to show, by "similar triangles", that the second target must be four times the size of the first target and that, in general, multiplying the distance by "x" multiplies the area by "x2". Of course, since the same amount of paint was spread over an area x2 times as great, its thickness was 1/x2 as great.

If, on the other hand, you have an infinite line source, you take it that your paint (or light, or energy, or whatever) is spreading out in a cylinder. The area of the surface of any "sub"-cylinder is 2&pi;rh (h is the height of the cylinder which does not change. 2&pir is, of course, the circumference of the expanding circle.) Since the area increases as r, the density of the paint (intensity of light, energy, etc.) decreases as 1/r so that the same amount is spread uniformly over the larger area.

"(B) Energy from an infinite line source disperses cylindrically at ½ inverse square law. "

No, that's wrong. It disperses as a 1/r law.