Locally non-rotating observers

[George Jones]

Why is it that we must contract $\hat{\eta^m}\nabla_m(h^{b}{}{}_{n}\phi^n) = 0$ with the spatial metric $h_{ab} = g_{ab} - \hat{\eta_a}\hat{\eta_b}$ (using the author's metric signature convention) in order to have that constancy?

It is not that $\hat{\eta^m}\nabla_m(h^{b}{}{}_{n}\phi^n) = 0$ is contracted with the spatial metric, it is that $\hat{\eta^m}\nabla_m(h^{b}{}{}_{n}\phi^n)$ can be non-zero even though the spatial projection of $\hat{\eta^m}\nabla_m(h^{b}{}{}_{n}\phi^n)$ is zero.

This morning I had a look at the simpler scenario in which $\gamma$ is the worldline of an observer in Minkowski spacetime (one dimension for time and one dimension for space is sufficient) who has constant proper acceleration. Consider an orthonormal frame $\left\{e_{0'} , e_{1'} \right\}$ with $e_{0'}$ tangent to $\gamma$ and with the frame Fermi-transported.

Should $e_{1'}$ be considered spatially constant with respect to $\gamma$?

Does $\nabla_{e_{0'}} e_{1'} = 0$?

What about the spatial projection of $\nabla_{e_{0'}} e_{1'}$?

 

[WannabeNewton]

In this case, $e_{1'}$ is already purely space-like and of unit norm so it can truly be considered a spatial direction and it should be considered spatially constant w.r.t. $\gamma$ since Rindler observers are non-rotating with respect to whatever global inertial frame (which are non-rotating by definition) we boosted from to get to the instantaneous Lorentz frame $\{e_{0'},e_{1'}\}$.

With $e_{0'} = \cosh g\tau \partial_t + \sinh g\tau \partial_x = gx \partial_t + gt \partial_x$ and $e_{1'} = \sinh g\tau \partial_t + \cosh g\tau \partial_x = gt \partial_t + gx \partial_x$, where $\{\partial_t,\partial_x \}$ is the original global inertial frame, I get $\nabla_{e_{0'}}e_{1'} = ge_0$. From here, using that $h = \eta + e_{0'}\otimes e_{0'}$ (where $\eta$ is the Minkowski metric), I do indeed get that $\langle h,\nabla_{e_{0'}}e_{1'} \rangle = 0$ after a bit of computation.

By the way, I phrased my question incorrectly before. What I meant to ask is, physically why is $h^{a}{}{}_{b}\hat{\eta^m}\nabla_m (h^{b}{}{}_{n}\phi^n) = 0$ the necessary condition for the spatial direction of $\phi^a$ to remain constant along $\gamma$, as opposed to $\hat{\eta^m}\nabla_m (h^{b}{}{}_{n}\phi^n) = 0$ being the necessary condition? $\hat{\eta^m}\nabla_m (h^{b}{}{}_{n}\phi^n) = 0$ is the rate of change of $h^{b}{}{}_{n}\phi^n$ along $\gamma$, and the author states that $h^{b}{}{}_{n}\phi^n$ is the spatial direction of $\phi^a$, so it would seem that requiring this quantity to vanish would ensure that the spatial direction remain constant along $\gamma$. On the other hand, $h^{a}{}{}_{b}\hat{\eta^m}\nabla_m (h^{b}{}{}_{n}\phi^n) = 0$ seems to be saying that the spatial component of the rate of change of the spatial direction of $\phi^a$ vanishes along $\gamma$ which doesn't exactly seem like what we need...? To relate this back to the example above: why is $\nabla_{e_{0'}}e_{1'} \neq 0$ even though $\nabla_{e_{0'}}e_{1'} = 0$ seems like the requirement we want in order to keep the spatial direction (which is exactly $e_{1'}$ in this case) constant?

Is the answer related to what MTW explains in the attachment below?

Thanks George!

 

[George Jones]

Is the answer related to what MTW explains in the attachment below?

Yes.

I don't know if the following helps at all. Taking a $\nabla_{e_{0'}}$ derivative of the orthogonality condition $0 = g \left(e_{0'}, e_{1'} \right)$ gives

$$ 0 = \nabla_{e_{0'}} g \left(e_{0'}, e_{1'} \right) = g \left( \nabla_{e_{0'}} e_{0'}, e_{1'} \right) + g \left(e_{0'}, \nabla_{e_{0'}} e_{1'} \right).$$

Since $g \left( \nabla_{e_{0'}} e_{0'}, e_{1'} \right)$ is non-zero, $g \left(e_{0'}, \nabla_{e_{0'}} e_{1'} \right)$ also is non-zero, and consequently $\nabla_{e_{0'}} e_{1'}$ has to have a component along $e_{0'}$. This part must be projected away if we want to say that say that $e_{1'}$ is spatially constant with respect $\gamma$, which is an integral curve for $e_{0'}$ (appropriately extended).

 

[WannabeNewton]

Gotcha, thanks again George!