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Twice Newton

  1. Nov 23, 2008 #1

    I am reading a popular physics book. It discusses the test of Einsteins theory by Eddington at the eclipse. "The deviation of the light was double that predicted by Newton's physics"

    Why does classical physics predict any deviation of light by gravity, did light have a mass in classical physics and if so how was it estimated (to allow for the deviation to be predicted)???

    Many thanks in advance

  2. jcsd
  3. Nov 23, 2008 #2


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    He meant Newton's law of gravity combined with the idea that light consists of particles with energy equal to Plancks constant times the frequency, and that you can define a "mass" of such a particle by setting [itex]h\nu=mc^2[/itex]. If you use Newton's law of gravity with this mass, you get the wrong result by a factor of two.
  4. Nov 23, 2008 #3

    Jonathan Scott

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    You don't have to assign any mass to light to predict its Newtonian acceleration; you just assume that light in a beam accelerates in the same way as anything else.

    Einstein's theory introduces an extra factor of (1+v2/c2) into the coordinate acceleration in this case, which means that the deflection is doubled for light and similarly increased for anything else moving at relativistic speeds.
  5. Nov 23, 2008 #4
    Thank you, very clear (now!)
  6. Nov 23, 2008 #5
    Another way to think about curvature of light: half is classical (per Newtons laws) and the other half is relativistic due to the curvature of space and time itself.

    Had Eddington's experiment been conducted a few years earlier, Einstein's career might have suffered a major blow since he originally predicted the classical degree of curvature and only later when working general relativity discovered an additional amount. Dr Kaku just mentioned this on a 2 hour TV show now airing "EINSTEIN"....HISTORY CHANNEL I believe.
  7. Nov 23, 2008 #6


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    I actually forgot that that when we're dealing with gravity we can eliminate the mass simply by dividing both sides of F=ma with m. But we can't divide with m when m=0, so we have have to assume either that m>0 or that there's a law of gravity for massless particles that works in this particular way.
    Last edited: Nov 23, 2008
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