# Flying Relativistic Man

Several months ago, there was a week in which I spent all my time thinking about how to find the charge distribution of conducting disc(I end up with the distribution of cylindrical symmetric ellipsoid, that can’t be specialize into a disc, with sphere and needle as the two extreme case). Then I discuss with my friend hoping to find out another way to solve it, but we end up discussing about relativistic electrodynamics instead, which is triggered by the fact that if a sphere moving with relativistic velocity it will turns into an ellipsoid(without flight time effect) . During this discussion, somehow we found a  simple and very interesting paradoxical case of relativistic electrodynamics.

As shown below there are two point particles with charge q each, the distance between them for this instance is L . Then a relativistic human, for some reason, using his hands, starts moving particle b away from a with constant relativistic velocity $v$, while holding the particle a at rest .

At this instant, the force on b due to a is obviously

$F_{ba}=\frac{kq^2}{L^2}$

Now comes the problem, the force on a due to b is different

$F_{ab}=\frac{kq^2}{\gamma^2L^2}$

Which means that the relativisticman receives different force on his left and right hand. Which means…

The relativisticman can fly away like a superman just by moving his hands!!!

How can it be possible assuming that the relativisticman exists??

The explanation is as follows

The electric field due to a alone at a random point p is

$\vec{E_{a}}=\frac{kq\vec{r_{a}}}{|r_{a}|^3}$

The electric field due to b alone at a random point p is

$\vec{E_{b}}=\frac{kq\vec{r_{b}}}{\gamma^2|r_{b}|^3(1-(\frac{vsin\beta}{c})^2)^\frac{3}{2}}$

b will also produce magnetic field

$\vec{B_{b}}=\frac{\vec{v}\times\vec{E_{b}}}{c^2}$

$\vec{B_{b}}=\frac{kq(\vec{v}\times\vec{r_{b}})}{\gamma^2c^2|r_{b}|^3(1-(\frac{vsin\beta}{c})^2)^\frac{3}{2}}$

using geometry relation $r_{a}sin\alpha=r_{b}sin\beta$, $\vec{v}\times\vec{L}=0$ and $\vec{r_{a}}=\vec{r_{b}}+\vec{L}$, we can eliminate the dependence of $r_{a}$ in $B_{b}$

$\vec{B_{b}}=\frac{kq(\vec{v}\times\vec{r_{a}})}{\gamma^2c^2(r_{a}^2+L^2-2r_{a}Lcos\alpha-(\frac{r_{a}vsin\alpha}{c})^2)^\frac{3}{2}}$

The total poynting vector of 2 charges is

$\vec{S}=\frac{1}{\mu_{0}}[(\vec{E_{a}}+\vec{E_{b}})\times\vec{B_{b}}]$

the second term $\frac{\vec{E_{b}}\times\vec{B_{b}}}{\mu_{0}}$ is constant over time, so we are not interested in it. By neglecting the this term, the momentum hidden in the electromagnetic field becomes

$\vec{p}=\mu_{0}\epsilon_{0}\int \vec{S}dV$

$\vec{p}=\epsilon_{0}\int (\vec{E_{a}}\times\vec{B_{b}})dV$

substituting $\vec{E_{a}}$,$\vec{B_{b}}$, and $dV=r_{a}^2 sin\alpha d\phi dr_{a} d\alpha$ we end up with

$\vec{p}=\frac{\epsilon_{0}k^2q^2v}{\gamma^2c^2}\int \int \int \frac{r_{a}sin\alpha[\vec{r_{a}}\times(\vec{v}\times\vec{r_{a}})]}{(r_{a}^2+L^2-2r_{a}Lcos\alpha-(\frac{r_{a}vsin\alpha}{c})^2)^\frac{3}{2}}dr_{a} d\theta d\phi$

by symmetry we know that the momentum will be in the same direction with $\vec{L}$, so we can put $\vec{r_{a}}\times(\vec{v}\times\vec{r_{a}})=sin^2\alpha \hat{L}$ , substituting this and $\int d\phi=2\pi$ the integral becomes

$\vec{p}=\frac{2\pi\epsilon_{0}k^2q^2v}{\gamma^2c^2}\int_{0}^{\infty} \int_{0}^{\pi} \frac{r_{a}sin^3\alpha\hat{L}}{(r_{a}^2+L^2-2r_{a}Lcos\alpha-(\frac{r_{a}vsin\alpha}{c})^2)^\frac{3}{2}}dr_{a} d\alpha$

The integral of $r_{a}$ has the form of

$\int_{0}^{\infty} \frac{x}{(ax^2-bx+c)^\frac{3}{2}}dx$

Which can be solved without much effort by using wolfram alpha

http://www.wolframalpha.com/input/?i=%5Cint+x%2F%28ax^2-bx%2Bc%29^%283%2F2%29

After LOTS of mess, the integral becomes

$\vec{p}=\frac{2\pi\epsilon_{0}k^2q^2v}{Lc^2}(\int_{0}^{\pi} sin\alpha d\alpha+\int_{0}^{\pi}(\frac{sin\alpha cos\alpha}{\sqrt{1-(\frac{vsin\alpha}{c})^2}})d\alpha)\hat{L}$

We can see that the second integral is odd, then we get a surprisingly simple result

$\vec{p}=\frac{4\pi\epsilon_{0}k^2q^2v}{Lc^2}\hat{L}=\frac{kq^2v}{Lc^2}\hat{L}$

the rate of change of the momentum is

$\frac{d\vec{p}}{dt}=\frac{-kq^2v^2}{L^2c^2}\hat{L}$

which is equal to $F_{ab}-F_{ba}$. So it means that the force difference exerted by the Relativisticman is only used to change the momentum of electromagnetic field. Since the mechanical momentum is not changing at all, the Relativisticman must receives the same change in momentum but in the opposite direction. Thus the man can fly away like superman!

# Light in Matter

Recently I start to (re-)wondering about a question “What really cause light to appear slower in media?”

We know that if we solve the maxwell equation, we will end up with the phase velocity of light is related to the permeablity and the permitivity of the material. But this is not what I’m interested in, I want to go deeper than that. We know that the real speed of photons is actually not changing, the decrease in speed is just apparent. Material is mostly empty, the light will still travel with c in the spacing between atoms. The rare atoms will disturb the light in some way. So the problem is how exactly do the atoms affect the light.

Some textbooks that I read explain it in a way kind of like this:

In a material the photons is absorbed by an atom and then re-emitted a short time later, it then travels at $c$ a short distance to the next atom and get absorbed&emitted again and so on. How quickly the atoms in a material can absorb and re-emit the photon and how dense the atoms decides the speed of light in that material. So the light appears slower because it has a smaller “drift speed”.

It seems very convincing right? It makes the probability of the thought of questioning it comes to my mind pretty small. But an hour ago I luckily found out that this thingy is not as simple as that, it turns out to be pretty cool.

I realize an alternative “wave” explanation:

Atoms respond to the light by radiating electromagnetic wave. This “new light” interferes with the “old light” in some way that result in delayed light (advanced in phase),this can easily shown by using simple phasor diagram. Consequently effectively the light covers a smaller phase each second. Which gives the impression of a lower phase velocity $c/n$. However the group velocity is changing in a weird and difficult to imagine way.

I think that the first explanation does not explain the change in phase velocity of light. if we consider light travelling into a slab of non-dispersive negative refractive index material, let’s say the light is directed perpendicular to the slab. The phase velocity’s direction will be flipped, but group velocity’s direction in the material will not change. Only the second explanation can explain the flipped phase velocity direction. I guess that the velocity that we get in the first explanation is actually belongs to the group velocity. It makes sense to me that the front most of the photon stream determines the first information that the light delivers. Negative index of refraction means that the light get delayed in phase more that how much it gains when it travels between atoms.

And also, I still don’t really understand about the detail of the absorption-emission process for small light’s wavelength compare to the distance between neighboring atoms( for large lambda compare to the atoms spacing, the photons will be absorbed by atoms as a group( phonons)). The dispersion relation that we know is continous and also osme material is non-dispersive, therefore the absorption process must accurs in all frequency for a certain range. So definitely it doesn’t involve the atomic transition, otherwise it will be quantized. My guess is that or the relevant absorption process is the stimulated absorpsion/emission which makes a lot of sense to me, it agrees with my classical explanation.

But I’m not sure yet, I will correct everything as soon as I get some more convincing arguments.