# existence of other forces obeying inv square law

Is there any restriction in what we know of physics to the existence of other type of forces that obey the inverse square law in 3 dimensions.

I mean other than electromagnetic and gravitational.

Also if there were some other kind of force ..would it follow that it would have a dual like electric and magnetic forces are duals of each other.

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The standard lore goes something like this: A $1/r^2$ force dependence is the signature of a massless force carrier, so we should have observed all such forces by now, at least their classical consequences. – Qmechanic Jun 20 '11 at 9:42
Ever heard ( :=) ) of sound? – Georg Jun 20 '11 at 10:10
Sound isn't a force. Sound waves exert forces, but those forces aren't inverse-square. The power density for a sound wave is what goes like $1/r^2$, not the force. – Ted Bunn Jun 20 '11 at 14:19

A $1/r^2$ force occurs because the particle mediating the force is massless. The only way there could be another such particle that had not yet been detected would be if the coupling constant for this force was small or zero for ordinary forms of matter. For example, suppose that we, our world, and our lab apparatus were all made out of neutrons. We might have a very hard time learning of the existence of the electromagnetic force. (Neutrons do have a magnetic moment, though.) I could imagine, for example, that dark matter possesses some new type of charge (not electric charge) that is zero for baryonic matter.

Re duals, these are general features of all fields due to the way special relativity works. For example, there is a standard, easy argument http://www.lightandmatter.com/html_books/0sn/ch11/ch11.html#Section11.1 that if we have electrical interactions plus SR, we must have magnetic interactions. However, the detailed nature of the dual depends on the nature of the field equations, or, in QFT terms, on the spin of the force-carrying particle. For example, the graviton has spin 2, not spin 1 like the photon, so we don't have a dual structure exactly like (E,B) in E&M. There are, however, some very nice analogies between the (E,B) structure and what you see in stationary fields such as that of the rotating earth. GR does have phenomena that people refer to as "gravitomagnetism," etc. But the analogy is not perfect, because otherwise GR would be the same theory as E&M. The fundamental reason for the difference is the spin of 2 instead of 1.

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I wouldn't say that fundamental reason for difference is spin 2. After all, linearized gravity is almost the same as EM while Yang-Mills theories (spin 1) are behaving very much like gravity in certain respects. Of course, spin 2 does make difference but it's just way more complicated than your simple statement. – Marek Aug 19 '11 at 15:52
Is there a simple derivation you can point me to online that shows that $1/r^2$ can only arise for massless arbitrator particles? – recipriversexclusion Aug 19 '11 at 19:01
@Marek: Gravity charge current is a rank-2 tensor because it's spin-2, and that even imposing symmetry you can't fit the 10 degrees of freedom into an $(E,B)$-like structure seems fundamental enough. Linearized gtr is equivalent to a rank-2 field on a flat background, GEM falls out only from a further degree-of-freedom reduction $\bar{h}_{ij} = 0$ in a certain gauge, and spin-2's ghost still haunts its Lorentz law. Moreover, effective charge (Coulomb part of large-scale interaction) scales with vel. $q\propto \gamma^{s-1}$, which is a more experimental reason spin-1 will inevitably fail. – Stan Liou Aug 19 '11 at 23:27
@Stan: why are you addressing me? I didn't argue with anything you have to say. My point was that huge difference is already caused by the non-abeliannes (or non-linearity if you wish) of the theory and this isn't unique to spin 2 interactions but present already in Y-M theories. Of course, gravity is different but by now I am merely repeating my first comment (which you haven't addressed at all...). – Marek Aug 20 '11 at 10:24
@Marek: Because I disagree with your evaluation of the relevance of spin-2. If you have an spin-2 field with infinite range coupling to some charge, then the charge-to-mass ratio of every object will be the same, for unpolarized case (ala Einstein-Cartan is another poss). Can be shown by a plane wave in the weak-field/slow lmt; by Lorentz invar. the source is just rescaled $T^{\mu\nu}$: the field is experimentally indistinguishable from gravity. Spin-2 is a feature fundamental to gravity, and is enough stall gravitomagnetism as defined by context of this post (and in the literature). Problem? – Stan Liou Aug 20 '11 at 14:44

The $1/r^2$ behavior is characteristic of a massless particle mediating the interaction. For electromagnetism, that particle is the photon, while for gravity it's the graviton. When an interaction is mediated by a massless particle, then it is of infinite-range. For example, two electrons will repel each other regardless of how far they are (as long as they are in causal contact of course). Another force that behaves like $1/r^2$ would have to also be of infinite-range, and so very easy to observe, unless it was extremely weak. So, most likely, there are no other $1/r^2$ forces.

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No, there's no reason that the electromagnetic and gravitational forces necessarily have to be the only ones with $\frac{1}{r^2}$ dependence. Or rather, there's no reason that we know of, and I don't think there are any theories being seriously considered that do predict those to be the only two inverse-square-law forces. They are simply the only ones that have been observed. (Note that gravity is actually not quite an inverse-square force once you take general relativity into account.)

If another inverse-square force did exist,it would not necessarily have a dual force, the way the electric and magnetic forces do.

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Thanks David..but let me rephrase..Suppose you use this hypothetical new kind of force(non dual and non gravitational) to signal...wont a non dual arrangement imply instantaneous communication across space?? – Ajay Jun 20 '11 at 10:54
For any force that obeys an inverse-square law in static situations, you have to work out what it does in situations where the sources are moving. The solutions for arbitrarily-moving sources are pretty much bound to involve waves of some sort, propagating at a speed of at most $c$, but they won't necessarily be the same as electromagnetic or gravitational waves. In particular, it's possible to imagine a scalar field obeying an inverse square force law. The solutions for this case would propagate as waves but would not have anything like an electric-magnetic split. – Ted Bunn Jun 20 '11 at 14:29
@Ajay: no it would not. As Ted explained, disturbances in the new force would propagate as waves, most likely at the speed of light, but definitely no faster. That has nothing to do with the force having a dual. – David Z Jun 20 '11 at 16:04
Lets assume that signal propagation in vacuum is exactly c(I dont think there is any other option but that maybe another topic). – Ajay Jun 21 '11 at 2:56
In unification of theories at very high energies,in gauge theories, when everything is massless, all the exchanged particles are 0 mass before the symmetry break. Thus I expect that QCD and the weak interaction should be going like 1/r^2, bar general relativity terms entering of course. That gives us another two forces :). – anna v Jul 20 '11 at 12:35

I would add that in Grand Unified Theories where the unification happens at very high energies, there exist a number of mediating particle X Y etc for forces consistent to the SU(2)XSU(3)xU(1) low energy structure which in the high energy limit before symmetry breaking are massless and similar to the photon, i.e. of 1/r^2 behavior. Even gluons themselves. More so if one includes supersymmetry in the brew.

The verification of the existence of these forces is an aim of the HEP experiments.

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