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Special relativity - transformation of angle

  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    Capture.jpg

    2. Relevant equations
    Gamma factor:
    $$\gamma = \frac{1}{\sqrt{1-\beta^2}} $$
    Lorentz contraction
    $$l'=\frac{l}{\gamma}$$
    Trig:
    $$ cos\theta = \frac{adjacent}{hypotenuse}$$

    3. The attempt at a solution
    20170321_151624.jpg
    20170321_151630.jpg

    I have all the quantities but the algebra doesn't seem to work out.

    Thank you in advance for any help
     
  2. jcsd
  3. Mar 21, 2017 #2

    PeroK

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    You've probably approached it the wrong way. Think about energy-momentum transformations for light instead.
     
  4. Mar 21, 2017 #3
    wow, that was soo easy. Thank you.

    However what was wrong with my method? I understand why your way works, but why didn't mine?
     
  5. Mar 21, 2017 #4

    PeroK

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    There's a relativity of simultaneity issue that you missed. You need to use a Lorentz Transformation on the x-coordiate, not simply a length contraction.
     
  6. Mar 21, 2017 #5
    Oh ok, thank you
     
  7. Mar 21, 2017 #6

    PeroK

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    I found my notes on this. If we do things in the Earth's frame first, with light emitted at ##t=0## with the star at ##(0,0)## and the Earth at ##(x,y)##, then ##\tan \theta = \frac{y}{x}##

    Note that the light reaches Earth at ##t = \frac{r}{c}## where ##r^2 = x^2 + y^2##

    In the star's frame, the light is emitted from ##(0,0)## at ##t'=0## and reaches the Earth at ##(x', y') = (\gamma(x-vt), y)## at some time ##t'## that isn't important.

    So, ##\tan \theta' = \frac{y'}{x'} = \frac{y}{\gamma(x-vt)} = \frac{y}{\gamma(x-vr/c)}##

    If you work through that and do a bit of trig manipulation, you should get the same answer.
     
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