# Inverse square laws

1. Jul 29, 2005

### eep

do we know why gravitational and electromagnetic forces inversely proportional to the square of the distance? is there any sort of underlying principle as to why they must be this way?

2. Jul 29, 2005

### Danger

It's basically just due to spatial dispersion of the field. The farther it is from the source, the more area it has to be spread over. A crude example might be a light bulb. If you put a 40-watt bulb in your refrigerator, it will be too bright. The same bulb in a typical living room would be too dim, and if it were in the Astrodome, it would barely show up.

3. Jul 29, 2005

### arildno

Although the (brilliiant) idea of a fixed amount of "force-area" was the clue Newton found in order to formulate his gravitational law, whether this idea in its full generality (for a variety of phenomena) should be regarded as fundamentally "true" or how it is derivable from more fundamental ideas, is, I believe, still an open issue.

4. Jul 29, 2005

### Kazza_765

This is a question that has bugged me too. I always thought that if you imagine the force spreading out radially from a point it is like an enlarginng sphere. The surface area of the sphere is propotional to r squared. That's just the way I pictured it though, and like I said, it has always bugged me.

5. Jul 29, 2005

### Meir Achuz

If you start gravitation from Newton's law of gravity, and EM from Coulomb's law then the 1/r^2 seems ad hoc. But if gravity or EM are started with Gauss's law (which is an equivalent starting point) then the 1/r^2 comes out as a natural consequence of the 4pi r^2 area of a sphere.

6. Jul 29, 2005

### arildno

And why should Gauss' law be anything but an ad hoc fantasy?
Just because you start off with fancier maths doesn't resolve the issue.

7. Jul 29, 2005

### lurflurf

At some level the answer to these types of questions is "Because this equation matches experimental results well under certain conditions."

8. Jul 29, 2005

Gauss's law is true more generally than Coulomb's law. Coulomb's law would indicate that if a charge moves, then the $\vec{E}[/tex] field changes intantaneosly at every point in space. Gauss's law, in conjunction with the rest of Maxwell's equations, gives the correct result that the change propagates at the speed of light. This would seem to indicate that it is more fundamental. 9. Jul 29, 2005 ### DaTario I like to think like this, imagining the logic of the proportionality of what seems to be experimentally involved and considering what is being described. In this case, with two bodies situated in the points A and B, I would try the way of thinking the force as inversely proportional the the distance AB and inversely proportional to the distance BA. As an example let me introduce a game, paint ball. Let the player A have his gun as well as the player B. If both have fixed areas. If each one is aiming at the other, and if A only shot again after his last shot has touched B and reciprocally, it is natural to indentify the frequency of A hitting B as inversely proportional to the distance AB. Also the frequency of B hitting A is inversely proportional to the distance BA which is equal to AB (this, although aparently obvious, seems to me as fundamental to this issue). So, if we intend to compute the frequency of hearing an "ouch", no matter the author, A or B, the expression for this frequency would involve the inverse square law in distance, as it seems to me up to now. Last edited: Jul 29, 2005 10. Jul 29, 2005 ### LeonhardEuler Actually, that probabilty seems to be proportional to the inverse of the square of the distance. If A has a certain constant area, then the solid angle he substends from B's position is proportional to the inverse of the square of the distance. If we assume that B shoots in all directions with equal probability, then this angle is proportional to the probability of a hit. True that it would involve the inverse square, but over large distances the [itex]\frac{1} {r}[/tex] term would be more important. This can be seen because the probability of at least on hit is equal to 1 minus the probability on no hits. Assume the probability of A hiting B is [itex] \frac{k} {r}[/tex] which is the same as the probability of B hitting A. Then the probability of A not hitting B is [itex](1- \frac{k} {r})[/tex]. Therefore the probability of A not hitting B and B not hitting A is [itex](1-\frac{k} {r})^2=1-\frac{2k} {r} + \frac{k^2} {r^2}[/tex]. This means the probability of either A hitting B or B hitting A is [itex]1-[1-\frac{2k} {r} + \frac{k^2} {r^2}]=\frac{2k} {r} - \frac{k^2} {r^2}[/tex] 11. Jul 29, 2005 ### DaTario Sorry, I guess you have read the uneditted post. 12. Jul 29, 2005 ### LeonhardEuler Yeah, I did. The last part is right that way. 13. Jul 29, 2005 ### MalleusScientiarum I think it should be pointed out that the gravitational and electromagnetic forces are 1/r^2 forces in only certain special circumstances. Things get more complicated with GR and Electrodynamics. 14. Jul 29, 2005 ### DaTario What specific situations are you referring to ? 15. Jul 29, 2005 ### Igor_S One of the specific situations are, I guess, "extra dimensions" . If a world was D-dimensional (only space dimensions counted), the radial dependence of the force would be: $$F(r) \sim \frac{1}{r^{D-1}}$$, each new dimension introduces another factor 1/r. As far as current experimental results show, down to a milimeter (or 0.1 mm; something like that) scale, gravity follows $$1/r^2$$ law. Offcourse, this is only the beggining of the story with extra dimensions, but I wouldn't go any deeper :) . 16. Jul 29, 2005 ### Crosson Suppose you have a negative charge, and a small distance away you have a charge of equal but opposite magnitude. Then the electric field in the surrounding space will have a 1/r^3 dependence. I think that the situation of two charges near each other occurs more frequently then "extra dimensions" 17. Jul 29, 2005 ### LeonhardEuler In that case you still have a [itex]\frac{1} {r^2}$ force from each charge, but they tend to cancel to some extent over large distances and the total force is close to being proportional to $\frac{1} {r^3}$ at large distances. However Coulomb's law is not violated because the force from each charge is still proportional to $\frac{1} {r^2}$.

A simple real situation where Coulomb's law is violated is the following: a point is located 1 light year from a charge. The charge then moves at .5c for $\frac{1}{2}$ a year, so that at the end of this time it is .75 light years from the point in question. Coulomb's law would require that the $\vec{E}$ field at this point be stronger than it was $\frac{1}{2}$ a year ago, but we know that this is impossible because it would reqiure information (in the form of the $\vec{E}$ field) to travel faster than the speed of light. Gauss's law still holds in this case, though, because the $\vec{E}[/tex] field has not changed at any point on the sphere centered at the charges original position with radius 1 light year, so [itex]\oint\vec{E}\cdot d\vec{A}$ will still have the same value over this surface. Gauss's Law will actually hold over any surface, including those with part of the $\vec{E}$ field changed and the rest not.

edit:Something simmilar is true of gravity. Information in the form of an increased gravitational force can also not travel faster than c. Actually, it is my understanding that there is no gravitational "force" in general relativity, but certainly don't claim to know what I'm talking about when it comes to GR.

Last edited: Jul 29, 2005
18. Jul 29, 2005

### DaTario

Ok I agree, some other laws of nature may (and in fact do) disturb the Coulomb's law. But the ontology of the inverse square law is still strong. Note that if you calm down the situation (force the situation to be electrostatic of gravitostatic, the 1/(r*r) comes out again after a finite time. In this sense, it seems to be an interesting procedure to attribute to the inverse square law the status of existing thing inside the domains of Newton's Force Based Physical Theories.

19. Jul 30, 2005

### Igor_S

True , but I like more Euler's :) example.

Well, 1/r^2 dependence is not followed by either nuclear (or strong) force, or the weak force, but that's beyond Newton's simple theories.

20. Jul 30, 2005

### eep

thanks for the responses. it suddenly dawned on me one nite that the force laws must follow the inverse square law because of the fixed amount of "force-area" idea. now obviously the inverse square law doesn't make sense in terms of relativity, but the general idea just seemed to "click".