A Does the EP imply that a free-falling observer follows a geodesic?

[Kostik]

TL;DR Summary

Does the (strong) Equivalence Principle imply that a free-falling observer follows a geodesic?

From https://www.astro.gla.ac.uk/users/norman/lectures/GR/part4-screen.pdf:

"It is possibly not obvious, but the Strong EP also tells us how matter is affected by spacetime. In SR, a particle at rest in an inertial frame moves along the time axis of the Minkowski diagram – that is, along the timelike coordinate direction of the local inertial frame, which is a geodesic. The Strong EP tells us that the same must be true in GR, so that this picks out the curves generated by the timelike coordinate of a local inertial frame, which is to say: Space tells matter how to move: Free-falling particles move on timelike geodesics of the local spacetime."

I am finding this explanation a little murky. Where does the geodesic equation appear? In the spirit of the "comma-goes-to-semicolon rule", the way to apply the EP should be:

1. Make a statement about a physical law in an inertial frame (flat spacetime);
2. Write it as a tensor equation, using the fact that the Christoffel symbols vanish in flat space with rectilinear coordinates;
3. Argue that a tensor equation is the same in all coordinate systems, so the covariant equation (this should be the geodesic equation!) is correct in any coordinate system.

Can anyone take a stab at a better explanation?

EDIT: I'm guessing the geodesic equation should be written as a tensor equation: $${v^\mu}_{;\sigma} \, v^\sigma = 0$$ which is the same as $$\frac{dv^\mu}{ds} +\Gamma^\mu_{\nu\sigma}v^\nu v^\sigma \, .$$

 

[Kostik]

Never mind, it's obvious. In an inertial reference frame, a "free-falling particle" is simply one not acted upon by any forces, hence its equation of motion is $$\frac{dv^\mu}{ds} = v^\mu_{,\sigma}\, v^\sigma=0 \, .$$ Just write this as a tensor equation $$ {v^\mu}_{;\sigma}\, v^\sigma=0 \, .$$ This is a tensor equation so it holds in all coordinate systems -- done.

 

[JimWhoKnew]

I'm guessing the geodesic equation should be written as a tensor equation: $${v^\mu}_{;\sigma} \, v^\sigma = 0$$ which is the same as $$\frac{dv^\mu}{ds} +\Gamma^\mu_{\nu\sigma}v^\nu v^\sigma\, . $$

Note that
$$\frac{dv^\mu}{ds} +\Gamma^\mu_{\nu\sigma}v^\nu v^\sigma =0$$
already appears in SR when you express the Newtonian $\vec{a}=0$ in arbitrary coordinates.

Also note that
$${v^\mu}_{;\sigma} \, v^\sigma = 0$$
has the pathology discussed in your previous thread.

Edit: corrected equation.

 

[PAllen]

Looking at the link in the OP, the cited reference is using a different definition of the strong equivalence principle than other sources I’ve seen (Clifford Will in his living review article on testing GR gives a more common set of definitions of different strengths of EP). Since there is no universally accepted formal definition of EP, I guess this is fine. The OP link is actually defining the strong EP as meaning that you can simply replace derivatives with covariant derivatives in any vector or tensor law as expressed in an inertial frame in SR. Thus the claim becomes trivial with that definition - inertial motion in SR has no proper acceleration. Changing this statement to covariant differentiation just gives the geodesic equation for inertial motion.