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.eep said: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?
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.DaTario said:It is natural to indentify the chances of A hitting B as inversely proportional to the distance AB (consider solid angle in the anallysis). Also the chances 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).
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]DaTario said:So, if we intend to compute the chances of hearing an "ouch", no matter the author, A ou B, the expression for tis chance would involve the inverse square law in distance, as it seems to me up to now.
What specific situations are you referring to ?MalleusScientiarum said: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.
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:DaTario said:What specific situations are you referring to ?
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.What specific situations are you referring to ?
In that case you still have a [itex]\frac{1} {r^2}[/itex] 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 [itex]\frac{1} {r^3}[/itex] at large distances. However Coulomb's law is not violated because the force from each charge is still proportional to [itex]\frac{1} {r^2}[/itex].Crosson said: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.
LeonhardEuler said: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 [itex]\frac{1}{2}[/itex] 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 [itex]\vec{E}[/itex] field at this point be stronger than it was [itex]\frac{1}{2}[/itex] a year ago, but we know that this is impossible because it would reqiure information (in the form of the [itex]\vec{E}[/itex] field) to travel faster than the speed of light. Gauss's law still holds in this case, though, because the [itex]\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}[/itex] will still have the same value over this surface. Gauss's Law will actually hold over any surface, including those with part of the [itex]\vec{E}[/itex] 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.
True , but I like more Euler's :) example.Crosson said:I think that the situation of two charges near each other occurs more frequently then "extra dimensions"
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.DaTario said: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