Is Clock Synchronization Throughout a Time-Like Congruence Transitive?

[WannabeNewton]

"Global simultaneity surfaces"

For a long time I had basically taken for granted the usual interpretation of one-parameter families of orthogonal space-like hypersurfaces relativized to a time-like congruence as "global simultaneity surfaces" (c.f. MTW section 27.3), and left it at that. See also p.24 of this paper: http://arxiv.org/pdf/gr-qc/0311058v4.pdf and p.5 of this paper: http://arxiv.org/pdf/gr-qc/0506127.pdf But a past discussion with PeterDonis about operational and mathematical definitions of "synchronizable reference frames" given in the text "General Relativity for Mathematicians"-Sachs and Wu has me confused about some aspects of the surface-level identification of hypersurface orthogonality with "global simultaneity", all relative to a time-like congruence.

More specifically, consider a time-like congruence with 4-velocity field $\xi^{\mu}$. If the congruence is irrotational, that is, $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$, then there exists a one-parameter family of space-like hypersurfaces $\Sigma_t$ everywhere orthogonal to $\xi^{\mu}$ in which case we can write $\xi^{\mu} = h \nabla^{\mu} t$ for some positive scalar field $h$ (technically $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$ only guarantees such a foliation locally in general, but we can consider cases wherein it holds globally for simplicity).

Sachs and Wu define a "synchronizable reference frame" to be a time-like congruence for which $\xi^{\mu} = h \nabla^{\mu} t$ (section 2.3). If $h = 1$ identically then they call this a "proper time synchronizable reference frame"; obviously this implies that $\nabla_{[\mu}\xi_{\nu]} = 0$ which itself holds if and only if the congruence is irrotational and geodesic. Physically, clocks at rest in a "proper time synchronizable reference frame" are freely falling and so undergo no gravitational time dilation whereas clocks at rest in a "synchronizable reference frame" will, in general, undergo gravitational time dilation of different factors due to different proper accelerations.

In section 5.3 they then give these definitions operational meaning (see attachments below). However I have a couple of problems with their discussion of clock synchronization throughout $\xi^{\mu}$, assuming that $\nabla_{[\mu}\xi_{\nu]} = 0$.

(1) Doesn't the congruence need to be rigid, that is, $\nabla_{(\mu}\xi_{\nu)} = 0$? Take, for example, the Painleve congruence in Schwarzschild space-time; this congruence is both geodesic and irrotational but it is not rigid. Because of gravitational tidal forces exerted by the self-gravitating source, two Painleve observers that are infinitesimally separated radially will have a non-vanishing relative radial velocity between them. Therefore they can't synchronize their clocks because their clocks tick at different rates due to their relative Lorentz factor.

(2) Say that $\nabla_{(\mu}\xi_{\nu)} = 0$ does hold. Sachs and Wu's discussion of clock synchronization throughout $\xi^{\mu}$ makes no explicit use of $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$. Where then in their operational method of synchronizing all the clocks following orbits of $\xi^{\mu}$ do they implicitly make use of the irrotationality of the congruence? More precisely, they take two clocks A and B infinitesimally separated and Einstein synchronize them (call this "local Einstein synchronization"). Then B is locally Einstein synchronized with a third clock C and so on. By doing this we get a space-like hypersurface $t = \text{const}$ which is everywhere orthogonal to $\xi^{\mu}$ and represents the set of events that are simultaneous with respect to all the clocks in this congruence. But clearly this only works if all the clocks in the congruence are synchronized with one another. In other words if clock A is locally Einstein synchronized with clock B and clock B is locally Einstein synchronized with clock C then we need clock A to be synchronized with clock C i.e. we require the synchronization to be transitive, which Sachs and Wu don't seem to mention at all.

If we consider the family of clocks laid out around a rotating ring in flat space-time, then two infinitesimally separated clocks on the ring can be locally Einstein synchronized but as we know all the clocks on the ring cannot be synchronized with one another i.e. the synchronization fails to be transitive. All the clocks on the ring experience the same time dilation factor and $\nabla_{(\mu}\xi_{\nu)} = 0$ but $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} \neq 0$.

So how does one prove that $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0 \Rightarrow$ local Einstein synchronization is transitive? Otherwise I don't even see how this is a clock synchronization procedure throughout $\xi^{\mu}$ and hence I don't see how the identification of hypersurface orthogonality with "global simultaneity" can be interpreted as the set of events that are simultaneous with respect to all the clocks in this congruence.

Thanks in advance!

 

[WannabeNewton]

In case anyone is interested, this paper explains everything beautifully using gauge theory: http://arxiv.org/pdf/gr-qc/0204063v2.pdf

Section 5.6 of this paper has a more concise explanation, also using gauge theory: http://arxiv.org/pdf/math-ph/0602018v2.pdf

 

[WannabeNewton]

Actually I'm still confused about something regarding the above.

Say we have a family of ideal clocks at rest with respect to one another laid out throughout space and as above let $\xi^{\mu}$ be their tangent field in space-time. Pick out a closed one-parameter subfamily of clocks $A_{\lambda}$ from this family and a clock $A_{\lambda_0}$ that we can call our "initial" clock. Now we can always Einstein synchronize two clocks at rest with respect to one another, regardless of their kinematical properties, if they are "neighboring" meaning they are infinitesimally separated from one another. So take a clock $A_{\lambda_1 = \lambda_0 + d\lambda}$ neighboring $A_{\lambda_0}$ and Einstein synchronize them. Now take a clock $A_{\lambda_2 = \lambda_1 + d\lambda}$ neighboring $A_{\lambda_1}$ and Einstein synchronize these two. Repeat this process until we come back to $A_{\lambda_0}$ after going full circle around the closed curve of clocks.

Now let $p_{\lambda_1}$ be an event in the vicinity of clock $A_{\lambda_1}$ simultaneous with an event $p_{\lambda_0}$ in the vicinity of $A_{\lambda_0}$. Next let $p_{\lambda_2}$ be an event in the vicinity of $A_{\lambda_2}$ simultaneous with $p_{\lambda_1}$ and so on until we go full circle and come back to $A_{\lambda_0}$ . Doing this we get a closed curve $\gamma(\lambda)$ of events in space-time bounding $\xi^{\mu}$ such that $\gamma(\lambda)$ is simultaneous with $\gamma(\lambda + d\lambda)$ according to the clocks $A_{\lambda}$ and $A_{\lambda + d\lambda}$.

Finally, go to a coordinate system $(t,x^i)$ in which the clocks are all at rest; let $dx^{\mu}$ be the tangent vector to $\gamma$. Then $\gamma(\lambda)$ is simultaneous with $\gamma(\lambda + d\lambda)$ according to the Einstein synchronized clocks $A_{\lambda}$ and $A_{\lambda + d\lambda}$ if $dt = -\frac{g_{0i}dx^i}{g_{00}}$ for these two events. Hence the clock synchronization is transitive, thus giving rise to global simultaneity surfaces, if and only if $\oint_{\gamma} dt = -\oint _{\gamma}\frac{g_{0i}dx^i}{g_{00}} = 0$.

How then does one show that $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0 \Rightarrow \oint _{\gamma}\frac{g_{0i}dx^i}{g_{00}} = 0$? Note that $\xi^{\mu} = \delta^{\mu}_0$ in these coordinates so $\xi_{\mu} =g_{0\mu}$.

I can't find anything in either of the two papers above that proves this implication in full generality.

 

[PeterDonis]

(1) Doesn't the congruence need to be rigid, that is, $\nabla_{(\mu}\xi_{\nu)} = 0$? Take, for example, the Painleve congruence in Schwarzschild space-time; this congruence is both geodesic and irrotational but it is not rigid. Because of gravitational tidal forces exerted by the self-gravitating source, two Painleve observers that are infinitesimally separated radially will have a non-vanishing relative radial velocity between them. Therefore they can't synchronize their clocks because their clocks tick at different rates due to their relative Lorentz factor.

They can't synchronize their clocks using Einstein clock synchronization, no. But the definition of "synchronizable" doesn't require Einstein clock synchronization, does it? It only requires $\xi^{\mu} = h \nabla^{\mu} t$, which the Painleve congruence satisfies ($t$ is Painleve coordinate time). In other words, "synchronizable" only requires that there is a family of spacelike hypersurfaces which are everywhere orthogonal to the timelike worldlines in the congruence.

Or look at it this way: Painleve coordinate time is equivalent to proper time along any worldline in the Painleve congruence; therefore worldlines in the Painleve congruence can obviously use Painleve coordinate time to synchronize their clocks. This synchronization will *not* be equivalent to Einstein clock synchronization in any local inertial frame along any of the worldlines, but it still works fine as a synchronization.

 

[WannabeNewton]

Thanks! That I certainly agree with.

And you can ignore post #3. I made a really silly mistake and just now realized how obvious the implication is!

$\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0 \Rightarrow \xi^{\mu} = h\nabla^{\mu}t$ so $\gamma(\lambda)$ simultaneous with $\gamma(\lambda + d\lambda)$ according to the locally Einstein synchronized clocks $A_{\lambda}$ and $A_{\lambda + d\lambda} \Rightarrow \xi_{\mu}dx^{\mu} = 0 \Rightarrow g_{\mu\nu}\nabla^{\mu}t dx^{\nu} = g_{\mu\nu}g^{\mu 0}dx^{\nu} = dt = 0$ where $dx^{\mu}$ is the tangent vector to $\gamma$. Hence $\oint_{\gamma} dt = 0$ is trivially satisfied and transitivity of local Einstein synchronization is trivially met, giving rise to global simultaneity surfaces for the entire family of comoving clocks.