# Null Riemann tensor implies no gravity proof?

It's easy to show that no gravity implies a null Riemann tensor, but how could I prove that a null R. tensor implies no gravity?

• By the phrase “no gravity”, do you mean $\Gamma _{\alpha \beta }^{\sigma }=0$? – Bert Barrois Jun 29 '18 at 11:31
• @BertBarrois careful! The Christoffel symbols can be zero even in flat spacetime. For example in polar coordinates there are non-zero Christoffel symbols. It just means the coordinates are curved. – John Rennie Jun 29 '18 at 12:06
• I guess it depends on exactly how you want to define no gravity. Actually that's a surprisingly subtle question since even in flat spacetime you could be using curved coordinates. – John Rennie Jun 29 '18 at 12:13
• @JohnRennie: I think 'flat spacetime' is usefully equivalent to 'no gravity'. Choice of coordinate systems can't affect anything physical. (But now you're going to point out I'm wrong and I'll feel silly :-)) – tfb Jun 29 '18 at 12:53
• @tfb Isn't acceleration gravity according to the equivalence principle? – safesphere Jun 29 '18 at 13:02

A curved spacetime can be measured by the geodesic deviation. You set up a family of closely spaced geodesics indexed by a continuous variable $s$ and parameterized by the proper time $\tau$ (massive particle) or an affine parameter $\lambda$ (massless particle, e.g. photon).
You define a tangent vector to the geodesic $T^\mu = \partial x^\mu (s, \tau) / \partial \tau$ and a deviation vector $X^\mu = \partial x^\mu (s, \tau) / \partial s$, which is the displacement of two objects travelling along two infinitesimally separated geodesics.
The relative acceleration $A^\mu$ of the two objects is defined as the second derivative of the separation vector $X^\mu$ as the objects advance along their respective geodesics. Specifically you take the directional covariant derivative of $X$ along $T$ twice: $A^\mu = T^\alpha \nabla_\alpha (T^\beta \nabla_\beta X^\mu)$.
If you work out the formula you get the geodesic deviation equation: $A^\mu = R^\mu_{\nu \rho \sigma} T^\nu T^\rho X^\sigma$, where $R^\mu_{\nu \rho \sigma}$ is the Riemann tensor.