Geodesic equation without metricity

Question

If we relax the metricity condition, what change will be seen in the geodesic equation ? For instance, when the torsionless connection is relaxed while keeping the metricity intact, we see that the original geodesic equation is augmented with the contorsion tensor. Similarly what other changes can be expected if the metricity is also relaxed. I have heard that geometry without non-metricity is non-Riemannian. Is it true ?

Answer 1

I have heard that geometry without non-metricity is non-Riemannian. Is it true ?

Yes, this is true. The covariant derivative of the metric tensor $g_{\alpha \beta}$, $\nabla g_{\alpha \beta}$ vanished in the case of Riemannian Geometry whereas if it doesn't, and denoting it by $$\Xi_{\mu \alpha \beta} = \nabla_{\mu} g_{\alpha \beta}$$

is clearly a tensor field of order $3$ and measures the rate of change of the components of the metric along the flow of the vector field.

As an example, consider a spacetime including torsion that will generate a different Elecromagentic field, and test particles send in such conditions will deviate from the trajectory we would expect without torsion. It will affect the conservation of scalar quantities along geodesics, for instance, in the case of the mass

\begin{eqnarray} \frac{dp^2}{d\lambda}&=& \frac{d(g_{\mu\nu} p^\mu p^\nu)}{d\lambda} \\ &=& m^2 u^\alpha \nabla_\alpha (g_{\mu\nu} u^\mu u^\nu)\\ &=& m^2 u^\alpha (\Xi_{\alpha\mu\nu} u^\mu u^\nu + g_{\mu\nu} u^\nu \nabla_\alpha u^\mu + g_{\mu\nu} u^\mu \nabla_\alpha u^\nu) \end{eqnarray}

for geodesics, $u^\alpha \nabla_\alpha u^\mu = 0$, leaving

\begin{eqnarray} \frac{dp^2}{d\lambda}&=& m^2 \Xi_{\alpha\mu\nu} u^\alpha u^\mu u^\nu \end{eqnarray}

Meaning a free particle would change mass along its trajectory.