# Free-falling stationary observer in curved spacetime?

Let us consider the pseudo-riemannian manifold $$(\mathcal{M},g)$$ with $$\mathcal{M}=\mathbb{R}\times\mathcal{N}$$ with $$\mathcal{N}$$ being a maximally symmetric, 3-dimensional riemannian manifold and

$$g = - d t ^ { 2 } + g _ { i j } d x ^ { i } d x ^ { j } = - d t ^ { 2 } + R ^ { 2 } ( t ) \underbrace { \left( \frac { 1 } { 1 - k r ^ { 2 } } d r ^ { 2 } + r ^ { 2 } d \Omega ^ { 2 } \right) } _ { = : h _ { k } }$$

where $$k \in \{\pm1,0\}$$.

Since $$\partial_t$$ is a unit vector, $$t$$ coincides with the proper time for observers with constant space coordinates. Moreover, such observers are freely falling: Indeed, if $$d x ^ { \mu } / d t \partial _ { \mu } = \partial _ { t }$$ then

$$\frac { d ^ { 2 } x ^ { \mu } } { d t ^ { 2 } } + \Gamma _ { \alpha \beta } ^ { \mu } \frac { d x ^ { \alpha } } { d t } \frac { d x ^ { \beta } } { d t } = \Gamma _ { 00 } ^ { \mu } = \frac { 1 } { 2 } g ^ { \mu \sigma } \left( 2 \partial _ { 0 } g _ { \sigma 0 } - \partial _ { \sigma } g _ { 00 } \right) = 0.$$

While I understand the mathematical justification for ths statement, I am uncertain about the physical interpretation. How can a stationary observer with fixed spatial coordinates be free falling in a curved geometry? My intuition would have said that a spacetime with non-vanishing curvature would imply that there are no freefalling stationary observers. As the sectional curvature for this metric is non-zero for $$k\neq0$$, this seems to be incorrect.

I would be very happy if you could offer some insight into this problem! Thanks

• The coordinates used in the metric are comoving coordinates i.e. they are chosen so comoving (freely falling) observers remain at fixed positions. – John Rennie Nov 16 '18 at 11:06