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Predicting angle of deflection of light

  1. May 2, 2015 #1
    Hello,
    I was recently reading relativity the general and special theory by Albert Einstein. When I came across the equation to calculate the angle of deflection for light as it approaches a massive body. I was just wondering if someone could explain why this works and also what "1.7 seconds of arc" means, as it is part of the equation.
     
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  3. May 2, 2015 #2

    mfb

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    The theory was made to match observations. What exactly do you mean with "this"?

    Small angles are described with minutes and seconds of arc. 1 degree = 60 minutes = 3600 seconds.
     
  4. May 2, 2015 #3
    Oh what I mean is what do the parts of the equation mean.
     
  5. May 2, 2015 #4
    Oh yeah thanks, so is this what used to predict the angle of deflection?
     
  6. May 2, 2015 #5

    mfb

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    That depends on the equations, there are different ways to derive that result and I don't know which set of equations you are looking at. I also don't know how much you know about general relativity.
    This is used to quote the result. It is just a unit, like meters for length, seconds for time or degrees for angles.
    Instead of 1.7 seconds of arc, you can also say "an angle of 1.7/3600 degrees".
     
  7. May 2, 2015 #6
    I'm talking about a=1.7 seconds of arc over delta
     
  8. May 2, 2015 #7

    Janus

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    The formula you are talking about comes from the formula

    [tex] \theta = \frac{4GM}{rc^2}[/tex]

    Where r is the distance from The center of M that the light passes. The answer will be in radians

    Basically if you use the mass of the Sun for M, set delta equal to 1 solar radius for r, and convert it to give an answer in second of arc rather than radians, you get

    [tex]\theta = \frac{1.7}{\Delta}[/tex] seconds of arc.

    For the angle of deflection of light passing at a distance of delta from the center of the Sun where delta is measured in sun radii.
     
  9. May 3, 2015 #8
    Ok yeah that makes sense, thanks for the answer.
     
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