# Proving that test particles in GR, follow spacetime geodesics

My question is pretty much in the title. According to this paper, this is not exactly proven rigorously yet. What I dont understand is what exactly is not proven. If I'm not too wrong, a test particle is just a particle, which does not affect the ambient gravitational field.

According to this Paper, it can be easily shown that the eqn. Of motion for a point particle in a curved space (having metric $$g_{\mu \nu}$$), can be found by considering the action to be just, $$S = -mc \int ds \, ,$$ where $$ds^2 = -g_{\mu \nu}dx^{\mu}dx^{\nu}$$ This would give the geodesic eq. As the eq. Of motion.

Now, if we assume that the particle is somewhat massive, couldn't we just modify the metric linearly, by superposing a mass coupled metric as, $$g'_{\mu \nu} = g_{\mu \nu} + m h_{\mu \nu}$$

Considering the same form of the action above, we should get the same geodesic eq. Along with a subsidiary eq.(which has a mass coupling). In the limit $$m \rightarrow 0$$, the subsidiary term should vanish, leaving the original geodesic eq.

So what I want to know, is what exactly is the unsolved part of this problem ? I failed to understand it from the text itself. It would be great if someone can point that out.

• The google link is to a pdf file on someone's web site, with no information in the pdf on the author or title, and it doesn't really treat the full problem. The arxiv paper's references 52 and 53 are to papers from 1988 and 1986. There is more recent work on this by Ehlers and Geroch, arxiv.org/abs/gr-qc/0309074v1 . – user4552 May 15 '19 at 15:12
• Ok. Sorry for not researching properly before posting the question. But can you briefly summarise what exactly is the unsolved part of this problem, given that we can sort of extend the action as I mentioned above ? Or is finding such an action rigorously the difficult part ? – Lelouch May 15 '19 at 15:14
• Possible duplicate of Why do objects follow geodesics in spacetime? – user4552 May 15 '19 at 15:17
• I'd always been happy with combining the formula for the energy momentum tensor of a "dust cloud" $T^{\mu\nu}= \rho_{00} U^\mu U^\nu$, where $\rho_{00}$ is the proper density of proper mass and $U^\mu$ the four-velocity with $\nabla_\mu T^{\mu\nu}=0$ to deduce that $U^\mu \nabla_\mu U^\nu=0$, which is the equation for $U^\mu$ being the tangent to a geodesic. – mike stone May 15 '19 at 16:03
• @BenCrowell I don't think this is a duplicate. The issue here is how to make the concept of a test mass mathematically rigorous i.e. how to show the trajectory is a geodesic in the limit of no backreaction. – John Rennie May 15 '19 at 17:09