0
$\begingroup$

I was recently wondering why the field force between two objects is proportional to the square root of the distance (sometime you just take things for granted). This comes from the inverse-square law, in which fields are treated as field lines (started wondering why is that the case, but decided to take this for granted) and the "square" comes from the relation between the area of a sphere and its radius. Makes sense. This, however, made me ponder 1) if all fields must follow this law; 2) if in theory fields can be collimated in analogy to light beams (lasers)? After all doesn't QM allow for field particles or are these just useful but completely abstract mathematical creatures? I can't imagine how gravitons can be particles if gravity is a time-space distortion. What are your thoughts on the possibility of field beams and if there are fields that don't obey the $r^{-2}$ rule?

PS I'm not a physicist nor a fanatic fan of Star Wars (haha), just a fellow scientist from a different field.

$\endgroup$
8
  • 1
    $\begingroup$ A plane of charge produces a uniform field (on length scales smaller than the extent of the plane). Not sure if that would satisfy you. $\endgroup$
    – kaylimekay
    Jan 19 at 4:20
  • $\begingroup$ You are paying attention. Some very good observations, but a little coarse around the edges. Fields that support particles that decay into lower mass particles do not obey the inverse square law. The particles "don't make it" to the sphere, so the connection to the area of a sphere is not so simple. In what might be the simplest model, the particle (excitation of the field) decays exponentially. The interaction then goes as $e^{-\mu r}/r^2$, My memory is bad, but I think the "One Pion Exchange Potential" behaves this way, as does muon decay (but I'm reaching way back into my memory). $\endgroup$
    – garyp
    Jan 19 at 5:09
  • $\begingroup$ I was recently wondering why the field force between two objects is proportional to the square root of the distance. It isn’t. $\endgroup$
    – G. Smith
    Jan 19 at 5:16
  • $\begingroup$ What do you mean by "field force?" Do you mean the electric force, newtonian gravity, or something else? $\endgroup$
    – Sandejo
    Jan 19 at 6:01
  • 1
    $\begingroup$ What force is inversely proportional to the square root? $\endgroup$
    – G. Smith
    Jan 20 at 1:35
2
$\begingroup$

This is really just a pedantic detail, but only forces between monopoles fall off as $1/r^2$. For example the force between two electric dipoles falls as $1/r^3$, the force between two electric quadrupoles as $1/r^4$ and so on.

This detail aside, the obvious example of a force that does not obey the inverse square law is the strong force. The reason this happens is that the strong force is mediated by gluons, but gluons carry a colour charge so they interact with each other. Contrast this with the electromagnetic force, which is mediated by photons that are not charged and do not interact with each other. This means that with the strong force we do indeed get "force beams" of the sort you are thinking of. These "beams" are called flux tubes or QCD Strings, and as a result of the tube formation at long distances ("long" here means > $10^{-15}$ metres) the strong force between two particles bearing a colour charge tends to a constant value instead of an inverse square dependence.

While it isn't really in the spirit of your question, we can see flux tube formation and a non-inverse square behaviour even in electromagnetism when the field is propagating in a medium. For example if magnetic fields are propagating in a plasma then the fields interact with the charged particles in the plasma and again we can get flux tube formation and a non-inverse square behaviour.

$\endgroup$
1
  • $\begingroup$ Thanks. I guess I should have mentioned that I was thinking of a scenario in vacuum. What about the 1/r2 relation in gravity? If it is mediated by gravitons which have a mass wouldn't you expect a non 1/r2 relation of force to distance? $\endgroup$
    – MikeB
    Jan 19 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.