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- Thread starter eep
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Danger

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arildno

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Meir Achuz

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arildno

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Just because you start off with fancier maths doesn't resolve the issue.

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lurflurf

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LeonhardEuler

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eep said:

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

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LeonhardEuler

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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).

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: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.

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]

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Sorry, I guess you have read the uneditted post.

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LeonhardEuler

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Yeah, I did. The last part is right that way.

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MalleusScientiarum

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MalleusScientiarum said:

What specific situations are you referring to ?

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DaTario said:What specific situations are you referring to ?

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:

[tex]F(r) \sim \frac{1}{r^{D-1}}[/tex],

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 [tex]1/r^2[/tex] law. Offcourse, this is only the beggining of the story with extra dimensions, but I wouldn't go any deeper :) .

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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.

I think that the situation of two charges near each other occurs more frequently then "extra dimensions"

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LeonhardEuler

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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.

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].

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.

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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.

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

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Crosson said:I think that the situation of two charges near each other occurs more frequently then "extra dimensions"

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

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

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.

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arildno said:

Just because you start off with fancier maths doesn't resolve the issue.

i dunno if you have gotten your question/challenge answered yet, but Gauss' Law is the only law that can work in a 3-dimensional (3 spatial dimensions) world and with conservation of energy or whatever "flux" is coming out of the point source. do you see why, given a 100 watt light bulb radiating omni-directionally, that the intensity (measured in watts/m^2) must have to be [tex] \frac{100 w}{4 \pi r^2} [/tex] at a distance of [tex] r [/tex] from the center of the light bulb (modeled as a point source)? is that ad hoc fantasy?

r b-j

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