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A Bending of light formula

  1. Jul 26, 2017 #1
    Good morning everybody. I have a problem with this wikipedia passage https://en.wikipedia.org/wiki/Schwarzschild_geodesics#Bending_of_light_by_gravity 'cause it says "Expanding in powers of rs/r, the leading order term in this formula gives the approximate angular deflection δφ for a massless particle coming in from infinity and going back out to infinity:". I tried to use Taylor formula and integrate but i can't reach that formula even considering orders. Can someone help me please? I'm just going crazy on a stupid expansion in powers xD Thank you in advance.
  2. jcsd
  3. Jul 26, 2017 #2


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    Post the details of how you tried to work it out, and someone may be able to see where you went wrong.
  4. Jul 26, 2017 #3


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    That's the calculation of the null geodesic in the Schwarzschild metric. Just use the square form of the Lagrangian
    $$L=\frac{1}{2} g_{\mu \nu} \dot{q}^{\mu} \dot{q}^{\nu}$$
    and use the conservation laws, i.e., it's a planar motion ##\vartheta=0## from isotropy (rotation invariance), the Lagrangian=Hamiltonian is conserved, and also the canonical momentum of ##q^0=c t## is conserved, because the Schwarzschild metric is static. Using this you can derive an equation for ##r(\varphi)## which can be solved up to an integral, leading to some elliptic function. You can solve it exactly for when expanding in powers of ##r_S/r \ll 1## (for, e.g., our Sun).
    The calculation is naturally very similar to what you are doing in Newtonian mechanics when solving the Kepler problem. You find the details in an exercise, we once gave at a cosmology lecture (however in German):

    http://th.physik.uni-frankfurt.de/~hees/cosmo-SS12/blatt01.pdf (problem set)
    http://th.physik.uni-frankfurt.de/~hees/cosmo-SS12/lsg01.pdf (solution)
  5. Jul 26, 2017 #4
    Perfect, that's exactly what i needed! Thank you very much!
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