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 $$F=\frac{Qq\left(1+{\mu}r\right)}{4{\pi}{\varepsilon_0}r^2e^{{\mu}r}}$$ and $$\mu=\frac{m_{\gamma}c}{\hbar}$$ with $Q$, and $q$ being the two electric charges, $r$ being the distance between the two electric charges, $\varepsilon_0$ being the electric constant, $m_{\gamma}$ being the rest mass of the photon, $c$ being the speed of massless particles, $\hbar$ being the reduced planks constant, and $F$ being the electric force between the two electric charges.


This equation can also be expressed as $$F={D_r}\left(\frac{Qq\left(sinh\left({\mu}r\right)-cosh\left({\mu}r\right)\right)}{4{\pi}{\varepsilon_0}r}\right)$$ $$F=\frac{Qq\left(D_r\left(sinh\left({\mu}r\right)-cosh\left({\mu}r\right)\right)r-\left(sinh\left({\mu}r\right)-cosh\left({\mu}r\right)\right)D_r(r)\right)}{4{\pi}{\varepsilon_0}r^2}$$ $$F=\frac{Qq\left(\left({\mu}cosh\left({\mu}r\right)-{\mu}sinh\left({\mu}r\right)\right)r-\left(sinh\left({\mu}r\right)-cosh\left({\mu}r\right)\right)\right)}{4{\pi}{\varepsilon_0}r^2}$$ $$F=\frac{Qq\left({\mu}rcosh\left({\mu}r\right)-{\mu}rsinh\left({\mu}r\right)-sinh\left({\mu}r\right)+cosh\left({\mu}r\right)\right)}{4{\pi}{\varepsilon_0}r^2}$$ $$F=\frac{Qq\left({u}rcosh\left({\mu}r\right)-sinh\left({\mu}r\right)-{\mu}rsinh\left({\mu}r\right)+cosh\left({\mu}r\right)\right)}{4{\pi}{\varepsilon_0}r^2}$$



I'm not sure if the force law in higher dimensions would be $$F=D_r\left(\frac{Qq\left(sinh\left({\mu}r\right)-cosh\left({\mu}r\right)\right)}{S{\varepsilon_{0}}r^{n-3}}\right)$$ or if it would be $$F=\frac{Qq\left(1+{\mu}r\right)}{S{\varepsilon_{0}}r^{n-2}e^{{\mu}r}}$$ 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 $\varepsilon_0$ 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, $$f = \frac{1}{r^2}$$ 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 $$\nabla^2 \varphi(\mathbf{x}) = 0$$



Solving $\varphi$ in free space will produce the potential and hence the force. By doing a Fourier transform, $$\varphi(\mathbf{x}) = \int \frac{\mathrm{d}^d k}{(2\pi)^d} \frac{ e^{i \mathbf{k}\cdot \mathbf{x}}}{k^2}$$


Solving $\varphi(r)$ $$\varphi = \frac{1}{(2\pi)^d}\int \mathrm{d} k \; k^{d-3} \mathrm{d}^{d-1}\Omega {e^{i k r \cos \theta_1}}$$ where $\mathrm{d}^{d-1} \Omega$ is the $(d-1)-D$ angular element. Now this integral involves the integral of one azimuthal angle $\theta$. We can parametrize the coordinate in $d-D$ spherical coordinate as: $$x_1 = r \cos\theta_1$$ $$x_2 = r \sin\theta_1 \cos\theta_2\cdots$$ $$x_d = r \sin\theta_1 \sin\theta_2\cdots \sin\theta_{d-2}\cos\phi$$ Then the surface element becomes: $$\mathrm{d}^{d-1} \Omega = \sin^{d-2}\theta_1 \sin^{d-3}\theta_2 \cdots \sin \theta_{d-2} \mathrm{d}\theta_1 \mathrm{d}\theta_2 \cdots \mathrm{d}\theta_{d-2} \mathrm{d} \phi$$ The integral over angles except $\theta_1$ is just the surface area of a $(d-2)-D\space hypersphere$ (Which I will assume you know, and so I am just stating the formula) $$S_{d-2} = \frac{2 \pi^{\frac{d-1}{2}}}{\Gamma\left( \frac{d-1}{2} \right) }$$


So $$\varphi = \frac{ S_{d-2}}{ (2\pi)^d r^{d-2}} I_{d}$$ where $I_d = \int_0^\infty \mathrm{d}\xi \; \int_0 ^\pi \mathrm{d}\theta \; \xi^{d-3} \sin^{d-2}\theta \exp\left[ i \xi \cos\theta \right]$


We can do the $\theta$ integral first(the other way around results in a tan integral with an argument of $\pi /2$, and is very very big and bad), which yields (I am using mathematica, please forgive me...): $$I_d = \int_0^\infty \mathrm{d}\xi \; \sqrt{\pi} \Gamma\left( \frac{d-1}{2} \right) \frac{{ }_0F_1\left( \frac{d}{2}, -\frac{\xi^2}{4} \right)}{\Gamma\left( \frac{d}{2} \right)} \xi^{d-3} = 2^{d-3} \sqrt{\pi} \Gamma\left( \frac{d-2}{2} \right) \Gamma\left( \frac{d-1}{2} \right).$$


Therefore, $$\varphi = \frac{ S_{d-2}}{ (2\pi)^d r^{d-2}} 2^{d-3} \sqrt{\pi} \Gamma\left( \frac{d-2}{2} \right) \Gamma\left( \frac{d-1}{2} \right) = \frac{\Gamma\left( \frac{d-2}{2}\right)}{4 \pi^\frac{d}{2} }\frac{1}{r^{d-2}}$$


$-\ Massful\ calculations\ :\ $


Now that we have easily solved for an intermediary massless boson, time to give it some mass. The Poisson equation becomes $$(\nabla^2 - m^2) \varphi(\mathbf{x}) = 0$$


By doing Fourier transform, $$\varphi(\mathbf{x}) = \int \frac{\mathrm{d}^d k}{(2\pi)^d} \frac{ e^{i \mathbf{k}\cdot \mathbf{x}}}{k^2+m^2}$$ Apply the same technique as the massless counterpart
$$\varphi(r) = \frac{ (m r)^{\frac{d}{2}-1} K_{\frac{d}{2}-1} (mr)}{(2 \pi)^\frac{d}{2}}\frac{1}{ r^{d-2}}$$ where $K_n(x)$ 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 :


$$K'_n(x) = - \frac{1}{2} [ K_{n-1} (x) + K_{n+1} (x)].$$


Putting this in should give you the answer you seek.


Cheers!!


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...