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A Perihelion precession in GR using Robertson expansion

  1. May 25, 2017 #1
    In his book Gravitation and cosmology, Weinberg derives the perihelion precession of Mercury in the Robertson expansion. The final formula is
    [tex]\Delta\phi =\frac{6\pi M G}{L} \frac{2+2\gamma-\beta}{3}[/tex]
    The second term is one for GR (β=γ=1).
    I have two questions regarding this formula:
    1. The pre-factor for the second-order term of dt² in the Robertson expansion is (β-γ); the pre-factor for the dr²-term is γ. In GR, β-γ=0. So is it correct to say that the perihelion precession is due to the spatial curvature?

    2. In the Newton limit (β=γ=0), the second term is 2/3, whereas Newtons theory should not predict any precession at all. Why does setting β=γ=0 not recover the Newton result?
     
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  3. May 25, 2017 #2

    pervect

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    I assume you're talking about the PPN formalism when you're talking about ##\beta## and ##\gamma##. The PPN formalism can handle any metric theory of gravity, but Newtonian gravity is not a metric theory. Newton-Cartan theory is a geometric reformulation of Newtonian gravity, but it doesn't have a metric IIRC, though it does have a connection.

    My text taks about the slow-motoin weak field approximation as background, and it does claim that PPN formalism yields empty flat space-time to zero order, and Newtonian gravity to first order, and that second (and sometimes high order) terms will provide a framework that will apply to any metric theory of gravity.

    But it appears to me that ##\beta=\gamma=0## isn't really quite exactly equivalent to Newtonian gravity.
     
  4. May 26, 2017 #3
    @pervect
    Thanks for the insights - yes, Newton-Cartan does not have a metric according to MTW (IIUC, this is due to the fact that there is no way to compare lengths to times in Newton theory).
    However, the PPN formalism (called Robertson expansion in my books in the context of the Schwarzschild metric) is claimed to recover the newton theory for β=γ=0 (and low velocities). To be more precise, what this does is giving a metric with a coefficient of 1 for dr²:
    [tex]
    ds^2 = (1- 2GM/r) dt^2 - dr^2 - radial\ term [/tex]
    which yields the Newton limit.
    So I am confused why the perihelion shift would not come out to be zero.
    On searching further, I found the following remark in the german book by Fließbach:
    "In Newtons Theorie verschwindet die Periheldrehung. Dieses Resul-
    tat erhält man aber nicht durch Einsetzen von γ = β = 0 in (27.23). Dies liegt
    daran, dass (27.23) nicht nur die relativistischen Effekte des Gravitationsfelds ent-
    hält, sondern auch die der Bewegungsgleichung."
    Roughly translated:
    In Newton's theory, the perihelion shift vanishes. This result is not recovered by setting γ = β = 0 in (27.23). [My equation from the original post] This is due to the fact that (27.23) does not only contain relativistic effects of gravitation, but also of the equation of motion."

    However, I'm not sure what is meant by that. Relativistic efects of the equation of motion seems to imply effects due to special relativity; but if that were the case, wouldn't a perihelion shift already be found when combining special relativity with Newtons law (something I never heard of and cannot really believe)?
     
  5. May 26, 2017 #4

    vanhees71

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    If you solve the formal Kepler problem in Special Relativity you get also a "perihelion shift". It's the equation of motion of a charged particle in an electrostatic Coulomb field, neglecting radiation. Historically that was an important calculation in the Bohr-Sommerfeld theory of atoms. By a funny coincidence Sommerfeld got the correct fine structure from this totally wrong model without incorporating spin!

    Schrödinger, who investigated the relativistic wave equation (we now call Klein-Gordon equation) for this problem and got the wrong fine structure (as to be expected since it describes a charged scalar particle rather than a spin-1/2 particle, for which you need the Dirac equation). That's why he first investigated the non-relativistic case, and that's why his famous equation refers to the non-relativstic wave function rather than the relativistic equation for scalar particles.
     
  6. May 26, 2017 #5

    PeterDonis

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    That's not the same as "recovering Newton's theory". The "Newton limit" in this case just means the weak field, slow motion limit of the Schwarzschild metric. It doesn't mean the exact same theory as Newton's laws. Those laws are incompatible with a metric theory of gravity, as has already been said.

    You can't combine SR with Newton's laws as they stand, because Newton's laws are not Lorentz invariant. The simplest change to Newton's laws to make them Lorentz invariant is putting in the "retarded distance" (heuristically, the distance one light-travel time ago) into the gravitational equation ##G m M / r^2## instead of the instantaneous distance (since the latter is frame dependent, while the former is not). Einstein tried this and quickly found that this theory makes predictions grossly contradictory to observation--for example, it predicts that planetary orbits are unstable on fairly short time scales. I believe this theory also predicts a perihelion shift much larger than those that are observed, but I don't think much attention was paid to that because it was so obvious that this theory was unviable.
     
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