Along the geodesic of a stationary observer in Minkowski spacetime we have the following tangent vector $$t^\mu = (1,0,0,0)$$ We have that hypersurfaces of constant time along this are just 3D Euclidean spaces.
I need to calculate the trace of the extrinsic curvature of these hypersurfaces and since the tangent is orthogonal to these the extrinsic curvature should be simply $\nabla_\mu t_\nu$ and of course I'm tempted to say it's zero since the covariant derivative in flat spacetime reduces to the usual derivative and we are basically deriving constants.
The reason why I'm not certain is that I'm currently working with bitensors and trough methods concerning Synge's world function and the Van Vleck determinant one obtains (cf. arXiv: 1102.0529 [gr-qc]) $$\nabla_t \ln{\Delta}=\frac{d-1}{\sigma}-K.$$ Where $K$ is the grace of the extrinsic curvature (the quantity I need), $d$ is the number of spacetime dimensions, $\Delta$ is the Van Vleck determinant and $\nabla_t=t^\alpha \nabla_\alpha$ is the covariant derivative in the direction of the geodesic and $\sigma$ is the geodesic distance. Since in flat spacetime $\Delta=1$ This gives $$\nabla_\mu t^\mu=K=\frac{d-1}{\sigma}$$ where in this case $\sigma=t$ where $t$ is time.
In this case this seems nonsense to me.
Could someone explain where I'm going wrong?
