# Why are so many forces explainable using inverse squares when space is three dimensional?

It seems paradoxical that the strength of so many phenomena (Newtonian gravity, Coulomb force) are calculable by the inverse square of distance.

However, since volume is determined by three dimensions and presumably these phenomena have to travel through all three, how is it possible that their strengths are governed by the inverse of the distance squared?

The gravitational force and intensity of light is merely 4 times weaker at 2 times the distance, but the volume of a sphere between the two is 8 times larger.

Since presumably these phenomena would affect all objects in a spherical shell surrounding the source with equal intensity, they travel in all three dimensions. How come these laws do not obey an inverse-cube relationship while traveling through space?

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–  Qmechanic Dec 17 '12 at 18:06
Er, because space is three dimensional. –  dmckee Dec 17 '12 at 21:31
Thanks @dmckee. –  Cayetano Gonçalves Dec 17 '12 at 21:37
Feynman had an interesting comment: nature tends to be described by laws that can be understood or mathematically expressed in qualitatively different ways. Take a look at Ch. 8 of lecture 2 of the Messenger Series (research.microsoft.com/apps/tools/tuva/#data=3|||). –  emarti Jan 12 '13 at 5:45
Possible Duplicates: physics.stackexchange.com/q/32779, physics.stackexchange.com/q/41109 –  Dimensio1n0 Jul 15 '13 at 16:31

This is not paradoxical and it is not necessary for any physical phenomenon to a priori have to obey any particular law. Some phenomena do have to obey inverse-square laws (such as, particularly, the light intensity from a point source) but they are relatively limited (more on them below).

Even worse, gravity and electricity don't even follow this in general! For the latter, it is only point charges in the electrostatic regime that obey an inverse-square law. For more complicated systems you will have magnetic interactions as well as corrections that depend on the shape of the charge distributions. If the systems are (globally) neutral, there will still be electrostatic interactions which will fall off as the inverse cube or faster! The van der Waals forces between molecules, for instance, are electrostatic in origin but go down as $1/r^6$.

It is for systems with a conserved flux that the inverse-square law must hold, at least at large distances. If a point light source emits a fixed amount of energy per unit time, then this energy must go through every imaginary spherical surface we think up. Since their area goes up as $r^2$, the power per unit area (a.k.a. the irradiance) must go down as $1/r^2$. In a simplified picture, this is also true for the electrostatic force, where it is the flow of virtual photons that must be conserved.

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For all systems, however, the monopole contribution dominates when you are far from the charge distribution, so long as it is not zero. –  Jerry Schirmer Dec 17 '12 at 18:08
Yes, but I'd say about half of all interesting systems have zero total charge and thus nil monopole contributions. –  Emilio Pisanty Dec 17 '12 at 18:41
(which, of course, simply defines "interesting".) –  Emilio Pisanty Dec 17 '12 at 18:41
You neglect to mention any of the benfits or ways of how you would use three dimensional information like he thinks you should use. On top of you say a simplified picture... do you mean using a plane to repersent you magnitudes? You should probably adress the point more clearly instead of the "your wrong I'm right" approach. –  Bored915 Dec 17 '12 at 19:08
Does this hint to the holographic principle at all? –  Cayetano Gonçalves Dec 18 '12 at 15:35

The field line picture known from school might be helpful with that:

The surface area of the surrounding sphere (and not it's volume) determines the density of the lines sourced by a point charge, corresponding to the field strength.

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These physical phenomena (gravity, Coulomb force) are forces caused by an object you can consider pointlike. That is, for the inverse square law to hold, the object emits the force uniformly in all directions from one point. That means that at any distance (call it R) from the object, you'll feel the same force as you would anywhere over the surface of a sphere whose radius is that distance. The surface of a sphere is 2 dimensional, not 3 dimensional, and its area goes like R^2. The larger the radius, the larger the surface of the sphere, and the further away you are from the source. So the strength of the source is inversely proportional to the surface area of the sphere.

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These kind of forces are coming from a system which is invariant under rotations, so under the SO(3) group (dim space : 3). Therefore, it should exists 3 generators of these rotations, thus 3 gauge transformations. Moreover, if your system is conserved in time, the energy is conserved and these generators are constant of motion.

When we are interested of interactions, we observe interactions which become really small at large distances, and in the case of gravity their is an attractive force.

Then, if you look at a force F = f(r), if I well remember, only in the case f(r) = 1/r^2 you can obtain such gauge generators which are known as 1 component of the angular momentum (imposing a plenary motion, so invariant under a rotation around the angular momentum) and two components of the Laplace-Runge-Lenz vector (imposing the axes of the ellipse to be constant, generating the 2 others rotations).

If you change the geometry of your system, you will study some other symmetries and thus obtain some other group leading to other generators. Then the allowed forces which will conserve the geometry of your problem will be different.

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## protected by Qmechanic♦Jan 12 '13 at 8:17

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