Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Precession of the LRL Vector in SR

  1. Sep 6, 2011 #1
    This is a continuation of https://www.physicsforums.com/showthread.php?t=526249", now that my confusion on Thomas precession has been clarified and my root question has been shown to not stem from this effect.

    In classical mechanics for any inverse square force, [itex]\vec{F}=-\frac{C}{r^2}\hat{r}[/itex], one has the Laplace-Runge-Lenz eccentricity vector. This is a constant of motion and has the magnitude of the eccentricity and points from the center of attraction towards the periapsis.

    [tex]\vec{e}=\frac{\vec{L}\times \vec{p}}{mC} + \vec{r}[/tex]

    For the argument's sake, let's consider a charged mass in orbit about a much more massive oppositely charged mass under Coulomb's law ([itex]C=kQq[/itex]), with a speed high enough that the SR effects are significantly larger than the effects of the curvature of space-time from the mass-energies involved.

    Q1) How much precession would such an orbit have in SR? Is this twice the excess in the angle [itex]\theta[/itex] given on http://en.wikipedia.org/wiki/Laplac...lizations_to_other_potentials_and_relativity" evaluated between two consecutive apsis?

    Q2) Is this the "Newtonian Model" that I have heard predicts half of the observed precession of Mercury (without clarification and attributed to Einstein), or does that model include a change to the inverse square law force due to the relativistic energy of the orbiting body? Half of the first order estimate of the DeSitter precession in GR would be [itex]\delta=\frac{3\pi C}{c^2rm_0}[/itex], [itex]C=GMm_0[/itex] and r being the semi-latus rectum of the precessing ellipse.

    Thoughts) My guess is that the first order SR precession defined above is [itex]\delta_{SR}=\frac{\pi C}{c^2rm_0}[/itex], with an additional [itex]\delta_{E=mc^2}=\frac{2\pi C}{c^2rm_0}[/itex] when the classical effects of Newton's theorem of rotating orbits is considered with the inverse square law adjusted by replacing the rest mass with the relativistic energy. I believe I have worked out the second half of this previously (http://farside.ph.utexas.edu/teaching/336k/Newton/node45.html" [Broken]), but need to relocate those notes. I may have the factor of two in the wrong component, or my guess might be completely off base.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 7, 2011 #2
    Is it safe to assume that with no comment on this post in over a month means that no obvious error exists in my assumptions?
  4. Nov 15, 2011 #3
    Speaking of that integral in Wiki, how do you solve an integral of the form:

    [tex]\theta = L \int^{u} \frac{du}{\sqrt{m^{2} c^{2} \left(\gamma^{2} - 1 \right) - L^{2} u^{2}}}[/tex]

    I would be content with the limit as [itex]\gamma[/itex] and L are nearly constant. Setting [itex]X=\frac{mc\sqrt{\gamma^{2} - 1}}{L}=\frac{mv\gamma}{L}[/itex] gives:

    [tex]\theta = \int^{u} \frac{du}{\sqrt{X^{2} - u^{2}}} = \sin^{-1}\left(\frac{u}{X}\right)\approx\sin^{-1}(1)=\pi[/tex]

    Now, if we have a perturbation from this, how does one determine whether to use the value from quadrant I or II?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook