WannabeNewton said:
What gravity probe B measures is the geodetic precession, see here: http://physics.umd.edu/lecdem/services/refs_scanned_WIP/3%20-%20Vinit%27s%20LECDEM/D401/2/AJP001248.pdf
According to the Introduction of that paper, it measures both, which is consistent with what I've read in other sources about Gravity Probe B. Later on in the paper, when they derive the formula for geodetic precession, the derivation also includes the Thomas Precession term, so that is included as well.
yuiop said:
Not entirely sure what ##\Omega## is measured relative to. Here it seems to that ##\Omega## is the precession rate of a gyroscope axis measured relative to the instantaneous radial vector
Yes, that's correct; sorry for switching notation in mid-stream. There should be more forms of the Greek letter omega.
yuiop said:
The only issue is that some texts insist that the precession in a gravitational field cannot be broken down into a Thomas precession component because that only applies in flat space, but I think that is an interpretational issue.
As the paper WBN linked to makes clear (equations 46 and 47), both the geodetic and the Thomas precession have the same general form of some constant times ##\vec{v} \times \vec{a}##, where ##\vec{v}## is the orbital velocity and ##\vec{a}## is the "acceleration due to gravity" (the paper writes it as ##\nabla \phi_G##, the gradient of the Newtonian potential, which amounts to the same thing). So I think it is indeed an interpretational issue.
(Btw, when it comes to interpretational issues, normally I tend to come down on the side of *not* depending on analogies that go "here's what happens in curved spacetime, and here's what would have happened if spacetime were flat", since the "flat background" is unobservable. But in this case, conceptually, I think it helps--at any rate it helps me--to look first at the flat spacetime formula for Thomas precession, to get the ##\gamma^2## factor, and then look at how the formula changes in curved spacetime to see the geodetic effect due to the central mass. But ultimately that's more pedagogy than physics.)
yuiop said:
OK, so by this definition, the geodesically orbiting gyroscope axis (FWTB vector) is precessing relative to the LTSB vector at a rate of ##-\Omega## and in the geodesic orbit case is equal in magnitude to the orbital velocity ##\omega## in the Schwarzschild metric. Here ##\Omega## is relative to the instantaneous radial vector rather than the distant stars.
Yes.
yuiop said:
This is the problem I am having. The GBP experiment measuring how much the un-torqued gyroscopes in the perfectly geodesically orbiting spacecraft , precessed relative to a distant star and they expected (and measured) that the gyroscopes would not remain pointing at the distant guide star forever.
This suggests that the GPB experiment should only have detected L-T precession due to the rotation of the Earth and no geodetic effect due to orbiting (because of the self cancelling).
Well, the paper's result for the geodetic precession doesn't match the one we have been using, which came from the other paper WBN linked to, so something is going on. This paper's result is (equation 4 in the paper, and I'm switching units so that ##G = 1##)
$$
\Omega = \frac{3}{2r} \left( \frac{M}{r} \right)^{\frac{3}{2}}
$$
which can be rewritten, using ##\omega = \sqrt{M / r^3}##, as
$$
\Omega = \frac{3}{2} \frac{M}{r} \omega
$$
However, if we look at the details of the derivation (leading up to equation 48 in the paper), we see that the formula as I have just written it is actually generally applicable to any angular velocity, not just a geodesic orbit, at least in the weak field approximation (we substitute ##\omega r## for ##v## in the general formulas in the paper and use ##\nabla \phi = M / r^2## in the Newtonian limit). So this latter formula should be compared with the general formula from the other paper,
$$
\Omega = \omega \frac{1 - 3M / r}{1 - 2M / r - \omega^2 r^2}
$$
I'm not sure how to resolve this discrepancy.
yuiop said:
Nice blog entry. Very clear. It was nice to see that that with the apropriate transformation and change of sign convention, that your equation for the centrifugal force agrees with the one I gave in the old thread that was "inconclusive" due to excessive bickering. (
Last equation of this post).
Thanks! Feel free to link to the blog entry in future threads if this comes up.
