Monday, 25 November 2019

electromagnetism - What would be the force law for the electric force between two electric charges in higher dimensions in the case of photons that have rest mass?


I was reading about if photons have rest mass https://galileospendulum.org/2013/07/26/what-if-photons-actually-have-mass/, and found that if photons have mass the force law for the electric force between two electric charges, in 4+1 dimensions, would be F=Qq(1+μr)4πε0r2eμr and μ=mγc with Q, and q being the two electric charges, r being the distance between the two electric charges, ε0 being the electric constant, mγ being the rest mass of the photon, c being the speed of massless particles, being the reduced planks constant, and F being the electric force between the two electric charges.


This equation can also be expressed as F=Dr(Qq(sinh(μr)cosh(μr))4πε0r) F=Qq(Dr(sinh(μr)cosh(μr))r(sinh(μr)cosh(μr))Dr(r))4πε0r2 F=Qq((μcosh(μr)μsinh(μr))r(sinh(μr)cosh(μr)))4πε0r2 F=Qq(μrcosh(μr)μrsinh(μr)sinh(μr)+cosh(μr))4πε0r2 F=Qq(urcosh(μr)sinh(μr)μrsinh(μr)+cosh(μr))4πε0r2



I'm not sure if the force law in higher dimensions would be F=Dr(Qq(sinh(μr)cosh(μr))Sε0rn3) or if it would be F=Qq(1+μr)Sε0rn2eμr or if it would be something else with, n being the total number of dimensions and S, being what the radius of a sphere is multiplied to get the hyper-surface area in the number of dimensions minus one. Also the units of ε0 would depend on the number of dimensions.


What would be the force law for the electric force between two electric charges if photons are massive?



Answer



[ WARNING:This answer uses natural units   Use dimensional analysis to bring back any constants you need]


Let's jump right into the workings - which is basically an assumption and some integrations.


In 3+1 dimension (without that time component, we wouldn't have a magnetic force, which will be bad and so we will talk about d+1 dimensions - although magnetism becomes complicated for higher dimensions, but that's a story for another day), Coulomb's law takes the form of an inverse-square law, f=1r2 with proper definition of the source and distance.


Jumping dimensions :


Time for our assumptions! We assume that the Lagrangian stays the same form in d+1 dimension.


 Massless calculations : 


It means, the Maxwell equations hold; or equivalently, the Poisson equation holds 2φ(x)=0



Solving φ in free space will produce the potential and hence the force. By doing a Fourier transform, φ(x)=ddk(2π)deikxk2


Solving φ(r) φ=1(2π)ddkkd3dd1Ωeikrcosθ1 where dd1Ω is the (d1)D angular element. Now this integral involves the integral of one azimuthal angle θ. We can parametrize the coordinate in dD spherical coordinate as: x1=rcosθ1 x2=rsinθ1cosθ2 xd=rsinθ1sinθ2sinθd2cosϕ Then the surface element becomes: dd1Ω=sind2θ1sind3θ2sinθd2dθ1dθ2dθd2dϕ The integral over angles except θ1 is just the surface area of a (d2)D hypersphere (Which I will assume you know, and so I am just stating the formula) Sd2=2πd12Γ(d12)


So φ=Sd2(2π)drd2Id where Id=0dξπ0dθξd3sind2θexp[iξcosθ]


We can do the θ integral first(the other way around results in a tan integral with an argument of π/2, and is very very big and bad), which yields (I am using mathematica, please forgive me...): Id=0dξπΓ(d12)0F1(d2,ξ24)Γ(d2)ξd3=2d3πΓ(d22)Γ(d12).


Therefore, φ=Sd2(2π)drd22d3πΓ(d22)Γ(d12)=Γ(d22)4πd21rd2


 Massful calculations : 


Now that we have easily solved for an intermediary massless boson, time to give it some mass. The Poisson equation becomes (2m2)φ(x)=0


By doing Fourier transform, φ(x)=ddk(2π)deikxk2+m2 Apply the same technique as the massless counterpart
φ(r)=(mr)d21Kd21(mr)(2π)d21rd2 where Kn(x) is the Bessel function of the second kind.


To go from the potential to the force, you will need to differentiate this equation. This shouldn't be much of a problem, except the Bessel function.



The differentiation of the Bessel function yields :


Kn(x)=12[Kn1(x)+Kn+1(x)].


Putting this in should give you the answer you seek.


Cheers!!


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