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Relativistic aberration and locally measured electric flux density

  1. Jan 24, 2012 #1
    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?
     
  2. jcsd
  3. Jan 25, 2012 #2

    Bill_K

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    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.
     
  4. Jan 25, 2012 #3

    Mentz114

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    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
    [tex]
    h^2\nu^2\frac{1+\beta\cos(\theta)}{1-\beta\cos(\theta)}
    [/tex]
    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 ?

    [edit]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.
     
    Last edited: Jan 25, 2012
  5. Jan 25, 2012 #4
    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.
     
  6. Jan 26, 2012 #5
    Your integral does indeed work for (1+2)D http://www.quickmath.com/webMathema...(1+b*cos(t))/(1-b*cos(t))*2&v2=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.

    [tex]
    h^2\nu^2\frac{1+\beta\cos(\theta)}{1-\beta\cos(\theta)}2\pi sin(\theta)
    [/tex]

    http://www.quickmath.com/webMathema...(t))/(1-b*cos(t))*2*pi*sin(t)&v2=t&v3=0&v4=pi

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

    http://www.quickmath.com/webMathema...x-2*log(abs(x-1))/x-2)&v2=0.0&v3=1&v4=0&v5=40

    attachment.php?attachmentid=43145&stc=1&d=1327557425.jpg

    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 (Text):

    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
    https://www.google.com/search?q=2*p...(0.0000000000000001-1))/0.0000000000000001-2)

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

    For [itex]\beta=10^{-25}[/itex]:
    https://www.google.com/search?q=2*p...0000000001-1))/0.0000000000000000000000001-2)
    -12.5663706 or about -4 pi (just like the (1+2)D case)

    0.01, 0.02...0.08, 0.09
    12.567 https://www.google.com/search?q=2*pi*(2*ln(abs(0.01+1))/0.01-2*ln(abs(0.01-1))/0.01-2)
    12.570 https://www.google.com/search?q=2*pi*(2*ln(abs(0.02+1))/0.02-2*ln(abs(0.02-1))/0.02-2)
    12.574 https://www.google.com/search?q=2*pi*(2*ln(abs(0.03+1))/0.03-2*ln(abs(0.03-1))/0.03-2)
    12.580 https://www.google.com/search?q=2*pi*(2*ln(abs(0.04+1))/0.04-2*ln(abs(0.04-1))/0.04-2)
    12.587 https://www.google.com/search?q=2*pi*(2*ln(abs(0.05+1))/0.05-2*ln(abs(0.05-1))/0.05-2)
    12.597 https://www.google.com/search?q=2*pi*(2*ln(abs(0.06+1))/0.06-2*ln(abs(0.06-1))/0.06-2)
    12.608 https://www.google.com/search?q=2*pi*(2*ln(abs(0.07+1))/0.07-2*ln(abs(0.07-1))/0.07-2)
    12.620 https://www.google.com/search?q=2*pi*(2*ln(abs(0.08+1))/0.08-2*ln(abs(0.08-1))/0.08-2)
    12.635 https://www.google.com/search?q=2*pi*(2*ln(abs(0.09+1))/0.09-2*ln(abs(0.09-1))/0.09-2)

    0.1, 0.2...0.8, 0.9
    12.651 https://www.google.com/search?q=2*pi*(2*ln(abs(0.1+1))/0.1-2*ln(abs(0.1-1))/0.1-2)
    12.910 https://www.google.com/search?q=2*pi*(2*ln(abs(0.2+1))/0.2-2*ln(abs(0.2-1))/0.2-2)
    13.364 https://www.google.com/search?q=2*pi*(2*ln(abs(0.3+1))/0.3-2*ln(abs(0.3-1))/0.3-2)
    14.052 https://www.google.com/search?q=2*pi*(2*ln(abs(0.4+1))/0.4-2*ln(abs(0.4-1))/0.4-2)
    15.045 https://www.google.com/search?q=2*pi*(2*ln(abs(0.5+1))/0.5-2*ln(abs(0.5-1))/0.5-2)
    16.468 https://www.google.com/search?q=2*pi*(2*ln(abs(0.6+1))/0.6-2*ln(abs(0.6-1))/0.6-2)
    18.573 https://www.google.com/search?q=2*pi*(2*ln(abs(0.7+1))/0.7-2*ln(abs(0.7-1))/0.7-2)
    21.948 https://www.google.com/search?q=2*pi*(2*ln(abs(0.8+1))/0.8-2*ln(abs(0.8-1))/0.8-2)
    28.546 https://www.google.com/search?q=2*pi*(2*ln(abs(0.9+1))/0.9-2*ln(abs(0.9-1))/0.9-2)

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

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    Last edited: Jan 26, 2012
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