Oh, I know nothing about the general relativity theory and what is gauge-invariant. Do you mean to say that my question about the momentary self-force does not make sense even in the classical limit? I find it hard to imagine how a momentary force, which is in principle a measurable quantity (at least in the classical limit), can be not defined. Or do you mean to say that this force will depend on the entire history of the previous motion?

Comparing with the lowest order $l=2$ term in Thorne‘s (4.20’), it appears that the fourth $I$ should be a $J$. I get $32/45$ for that coefficient. What happened to his $JJ$ term?

Is the $JJ$ term smaller by a factor of $1/c^2$?

@G.Smith Yes the last $I$ should have been a $J$. I'' check on the coefficient tomorrow.

Great. I would also really appreciate an explanation of why the $JJ$ is not relevant. Thanks.

There is a similar expression in electromagnetism, for the radiation reaction force, with a strange mix of multiplepole moments. It would be great to show it for comparison.

In electromagnetism, you get the following for the radiation reaction force acting on a non-relativistic distribution:\begin{equation}F_i = -\, \frac{2k}{3 \, c^5} \: (\, \ddot{\mathbf{p}} \times \ddot{\boldsymbol{\mu}} \,)_i - \frac{k}{15 \, c^5} \: \dddot{Q}_{ij} \, \ddot{p}_j,\end{equation} where $\mathbf{p}$ is the electric dipole moment, $\boldsymbol{\mu}$ is the magnetic moment and $Q_{ij}$ is the quadrupole moment. This formula comes from Landau-Lifchitz.

I rechecked and I do get 16/45 for the $IJ$ coefficient. Sorry, I previously just misevaluated a single factorial as a double factorial.

@Cham After working through the dimensions, I think the prefactor is $G/c^7$ if $I$ is a mass multipole, not an energy multipole.

@G.Smith, I agree. It must be $G/c^7$.

When I apply the leading terms in Thorne's equations to a circular orbit, I get agreement with the $dE/dt$ and $dL/dt$ in Wikipedia. But when I evaluate the reaction force, I don't find that $F\cdot v=-dE/dt$. (They differ not just by a numerical factor but by a factor of $(a\omega/c)^2$ .) This equality should hold, shouldn't it? Maybe I don't properly understand $J_{ij}$ in the case of a nonrel point mass. I think it is $J_{ij}=m[x_i(x\times v)_j+x_j(x\times v)_i]$.

@G.Smith Regarding the $JJ$ term. Each current multipole enters the expression for the radiation in the far field at a factor $1/c$ lower than the corresponding mass multipole. So as you suggested the first $JJ$ term comes in $1/c^2$ down from the leading order term. The leading $IJ$ term however comes in at the same order as the leading $II$ term because it starts at a lower multipole order.

@Mitsuko The problem comes about due to the fact that we cannot distinguish gravitational forces from "fictious" forces due to the choice of coordinates. As a consequence you can always apply local coordinate transformations that change the local gravitational force due to the emission of GWs. When the accelerating force itself is gravitational this becomes even worse.

@mmeent : Oh interesting. But what if we consider our planet orbiting around the sun? We can calculate the power of emission of gravitational waves, and if we divide that power over the speed of the Earth, we get some force. This is, obviously, the stopping force, the gravitational self-force, and the Earth must experience it, right? And now you claim that the gravitational self-force is not really defined. Then what is actually the ratio between the power of emission of gravitational waves by the Earth and the speed of the Earth? Is it something that makes no sense?!

@mmeent : What I suspect is that the "fictional forces" you are talking about vanish in the classical limit. So I suspect that if you properly take the classical limit in the lengthy expressions of the general relativity theory, your "fictional forces" will disappear in that limit (or, to be exact, will be higher-order corrections), and you will get some well-defined gravitational self-force in the classical limit.

@mmeent : After all, I've heard that deriving the expression for the Abraham-Lorentz force from the relativity theory by taking the classical limit requires quite a few mathematical tricks, so taking the classical limit is not an easy task at all and might be even more difficult in the gravitational case, but I suspect it is still doable.

@Mitsuko The average part of the GSF is well-defined, it is the oscillatory pieces that are extremely subtle. And no, fictious forces do not disappear in the classical limit. Think about the Coriolis force on Earth for example, or the fact that the gravitational acceleration on the equator is smaller due to the Earth's rotation.

@Mitsuko An interesting observation about the GSF (due to your countryman Mino) is that one can always use the gauge freedom to choose coordinates such that the GSF on an orbiting body is constant. No matter how eccentric the orbit.

@mmeent: Could you please specialize the formula for the case of near-Newtonian two body problem with one of the bodies much smaller than the other (Sun–Earth)? That way we (hopefully) also remove the ambiguities in definition of the force, since the heavier body would provide a natural reference frame.

@A.V.S. I can see if I can find some time to do that. It would do very little to reduce the ambiguity on the force due to its gauge freedom. That freedom remains in the small mass-ratio limit.

@A.V.S. I found the force in that case to be tangential, opposite the direction of motion, and of magnitude (1168/315)Gm^2a^5\omega^7/c^7.

@G.Smith I am assuming that is for circular orbits?

Yes

Can you look at my earlier comment about what I understand J_{ij} to be for a nobrel point mass? Is this correct?

nonrel

Dotting the force with the velocity doesn’t give the radiated power. So I don’t know whether I miscalculated or should not expect this relationship.

I didn’t want to investigate elliptical orbits until I can get a sensible result for a circuiar one.

Also, the results for radiated power and ang mom are typically averaged over an orbit. But averaging the radiated mom over an entire orbit will produce zero for a circular orbit or something of lower order for an elliptical one.

So the exp

@A.V.S. Thanks. I’ll take a look at them. Cliff was my first Ph.D. advisor at Stanford.