## Relativistic aberration and locally measured electric flux density

Does relativistic aberration apply to electric fields and magnetic fields? If a spaceship accelerated to a relativistic speed with respect to an initial inertial frame, in a constant direction with respect to a uniform electric field, do the effects of aberration affect the magnitude of electric flux density perceived by that spaceship?

If I have a point source of light of uniform, constant brightness, will the total amount of power received by a spherical shell surrounding that point source of light vary depending on the direction and relativistic speed of that trajectory, in a manner itself varying with the shell position with respect to that light source?
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 Blog Entries: 1 Recognitions: Science Advisor Answer to part 1: the transformation of the E and B fields is E∥' = E∥ B'∥ = B∥ E⊥' = γ(E⊥ + v/c x B) B⊥' = γ(B⊥ - v/c x E) In words, for a pure E field the component in the direction of relative motion remains unchanged while the transverse component increases by a factor of γ, so the magnitude increases while the direction of the field tends to line up perpendicular to the motion. You can call this 'aberration' if you like, but it's not related to the aberration of a light ray.
 Blog Entries: 3 Recognitions: Gold Member Part 2. I did the calculation in (1+2)D and found the sqare of the energy reaching the circle is invariant. If a point source of light is at the centre of a circle, and moving at velocity β wrt to the circle. The square of the energy falling on a unit element of the circle in unit time is proportional to $$h^2\nu^2\frac{1+\beta\cos(\theta)}{1-\beta\cos(\theta)}$$ where θ is the angle between the light ray and the direction of travel. To get the total we integrate for -∏ < θ < ∏ and multiply by 2. I get 4∏h2v2 which is independent of β. Maybe someone can extend this to 4D ? Spherical symmetry means we can just integrate the great circles of the sphere to get the same result in 4D. viz, energy flux through the surface of the sphere is independent of β.

## Relativistic aberration and locally measured electric flux density

 Quote by Bill_K Answer to part 1: the transformation of the E and B fields is E∥' = E∥ B'∥ = B∥ E⊥' = γ(E⊥ + v/c x B) B⊥' = γ(B⊥ - v/c x E) In words, for a pure E field the component in the direction of relative motion remains unchanged while the transverse component increases by a factor of γ, so the magnitude increases while the direction of the field tends to line up perpendicular to the motion. You can call this 'aberration' if you like, but it's not related to the aberration of a light ray.
Why is there this strange break in symmetry? What so different about the electromagnetic wave that makes it have a different kind of "aberration" than electric and magnetic fields?

Has any particle accelerator experiment been run to see whether or not the E-field really aberrates differently than light from the frame of a relativistic charged particle? I do not think that would be an easy experiment. It is unlikely to me that the idea has even been tested because particle accelerators are really designed to focus beams towards a target as opposed to letting them scatter before reaching their targets, and efforts would be engineered to ensure that anomalous deflections (such as that resulting from an aberrating E-field) would not materialize. Of course, I do not work at a particle accelerator facility, so these perceptions may or may not be valid.

 Quote by Mentz114 Part 2. I did the calculation in (1+2)D and found the sqare of the energy reaching the circle is invariant. If a point source of light is at the centre of a circle, and moving at velocity β wrt to the circle. The square of the energy falling on a unit element of the circle in unit time is proportional to $$h^2\nu^2\frac{1+\beta\cos(\theta)}{1-\beta\cos(\theta)}$$ where θ is the angle between the light ray and the direction of travel. To get the total we integrate for -∏ < θ < ∏ and multiply by 2. I get 4∏h2v2 which is independent of β. Maybe someone can extend this to 4D ? Spherical symmetry means we can just integrate the great circles of the sphere to get the same result in 4D. viz, energy flux through the surface of the sphere is independent of β.
Your integral does indeed work for (1+2)D http://www.quickmath.com/webMathemat...t&v3=-pi&v4=pi

For (1+3)D, I only have to integrate from 0 to pi, but I will have to multiply it by a weight of 2*pi*sin(θ) (the circumference at θ) before integrating.

$$h^2\nu^2\frac{1+\beta\cos(\theta)}{1-\beta\cos(\theta)}2\pi sin(\theta)$$

http://www.quickmath.com/webMathemat...2=t&v3=0&v4=pi

$$h^2\nu^2\left(2 ln\left(|\beta+1|\right)/\beta-2 ln\left(|\beta-1|\right)/\beta-2\right)2\pi$$

http://www.quickmath.com/webMathemat...3=1&v4=0&v5=40

The chart has a floating point error near the y-axis of the chart. It also has the sign wrong.

Microsoft Excel has floating point errors as well. For E-16 and up, I get positive numbers. An anomaly at 1.0E-16 returns an answer of 1.385103378, For E-17 and below, I get -4*pi, which is the correct answer. Microsoft Excel also has another problem when there is "0." followed by sixteen 9's:

Code:
0.0000000000000000000100000000000000	-12.56637061
0.0000000000000000001000000000000000	-12.56637061
0.0000000000000000010000000000000000	-12.56637061
0.0000000000000000100000000000000000	-12.56637061
0.0000000000000001000000000000000000	1.385103378
0.0000000000000010000000000000000000	13.94142997
0.0000000000000100000000000000000000	12.54628257
0.0000000000001000000000000000000000	12.56023405
0.0000000000010000000000000000000000	12.56720978
0.0000000000100000000000000000000000	12.56637269
0.0000000001000000000000000000000000	12.56637269
0.0000000010000000000000000000000000	12.5663713
0.0000000100000000000000000000000000	12.5663706
0.0000001000000000000000000000000000	12.56637062
0.0000010000000000000000000000000000	12.56637061
0.0000100000000000000000000000000000	12.56637062
0.0001000000000000000000000000000000	12.5663707
0.0010000000000000000000000000000000	12.56637899
0.0100000000000000000000000000000000	12.56720842

0.1000000000000000000000000000000000	12.65065269
0.2000000000000000000000000000000000	12.90975348
0.3000000000000000000000000000000000	13.36388311
0.4000000000000000000000000000000000	14.05227672
0.5000000000000000000000000000000000	15.04476775
0.6000000000000000000000000000000000	16.46811059
0.7000000000000000000000000000000000	18.57311471
0.8000000000000000000000000000000000	21.94755234

0.9000000000000000000000000000000000	28.54575323	1
0.9900000000000000000000000000000000	54.62315484	2
0.9990000000000000000000000000000000	83.03870702	3
0.9999000000000000000000000000000000	111.8963423	4
0.9999900000000000000000000000000000	140.821133	5
0.9999990000000000000000000000000000	169.7549757	6
0.9999999000000000000000000000000000	198.6899578	7
0.9999999900000000000000000000000000	227.6250772	8
0.9999999990000000000000000000000000	256.5602132	9
0.9999999999000000000000000000000000	285.4953492	10
0.9999999999900000000000000000000000	314.4304869	11
0.9999999999990000000000000000000000	343.3659035	12
0.9999999999999000000000000000000000	372.2968563	13
0.9999999999999900000000000000000000	401.2459489	14
0.9999999999999990000000000000000000	430.1810865	15
1.0000000000000000000000000000000000	457.7922249	16
1.0000000000000000000000000000000000	#NUM!	17
1.0000000000000000000000000000000000	#NUM!	18
1.0000000000000000000000000000000000	#NUM!	19
1.0000000000000000000000000000000000	#NUM!	20
1.0000000000000000000000000000000000	#NUM!	21
Google is no more precise at these small numbers:

At E-16

Positive numbers still emerge in the same range as that in Excel.

For $\beta=10^{-25}$:
-12.5663706 or about -4 pi (just like the (1+2)D case)

0.01, 0.02...0.08, 0.09

0.1, 0.2...0.8, 0.9
Regardless of the floating point errors where higher precision is demanded, it is clear that the integral of $$h^2\nu^2\frac{1+\beta\cos(\theta)}{1-\beta\cos(\theta)}2\pi sin(\theta)$$ is by no means a constant.