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!