Consider a gravitational field caused by a central source object and a gyroscope in orbit around that source. It is well-known that the gyroscope experiences rotational frame-dragging at a rate 3/2 times the cross product of the velocity and the field, which is known as the(adsbygoogle = window.adsbygoogle || []).push({}); de Sitter precessionorgeodetic precession. It is also well-known that when a moving source passes a static gyroscope the gyroscope experiences rotational frame-dragging, but in this case the factor is 2 times rather than 3/2 times. Some time long ago I asked Clifford Will why the two effects were not equal, as it seemed that one could look at the orbit situation from the gyroscope point of view with the source moving past it. He very kindly pointed out that if the test object is in free fall and accelerating in a changing direction, there is an additional effect with a factor of -1/2 (forThomas precession).

This effect is shown even more clearly in Ciufolini and WheelerGravitation and Inertiaequation 3.4.38', giving an extended version of the de Sitter precession of which this is the relevant subset (assuming the appropriate factors specifically for GR):

$$ \dot\Omega = - \frac{1}{2} \mathbf{v} \times \frac{d \mathbf{v}}{dt} + 2 \mathbf{v} \times \nabla U $$

Here ##\mathbf{v}## is the velocity of the "gyroscope test particle" (which may be subject to non-gravitational forces as well as gravitational), the source is assumed to be at rest, and ##U## is the potential. If the particle is free to be accelerated by the field, then the acceleration is equal to ##\nabla U## and the factor of 3/2 is clear.

If we switch to a frame in which the test particle is initially at rest, I would expect the same predicted rate of frame-dragging. In that case the source is moving with velocity ##\mathbf{v}## in the opposite direction, and gives the usual factor of 2. However, in that frame the test particle is initially at rest and is accelerating towards the source, so the Thomas precession in the original form does not seem to apply.

What I'd like to know is the corresponding form for this equation for this second case, explaining where the 1/2 correction comes from to make it consistent. My guess is that for purposes of the rotational effect, the velocity here is effectively the relative velocity of the source and test particle, so the equation remains unchanged (assuming non-relativistic speeds), but clearly the explanation of the first term as Thomas precession no longer applies in that case.

I admit that in the past this should have been well within my own capabilities to calculate, assuming a weak field, the Schwarzschild solution and an isotropic coordinate system, but I never seem to have the time or focus now, so I'd appreciate it if anyone can point me to the answer, or can show me how to derive it. I've Googled a few references to gravitomagnetism without finding an answer in a form I can easily understand, and I know for certain that some of the Wikipedia stuff on the subject contains errors and inconsistencies (mostly related to factors of 2) because I spotted them and called attention to them myself some years ago.

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# Gravitomagnetic effect of moving source on test object

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