# Time synchronization in arbitrary reference frames.

1. May 23, 2013

### center o bass

I'm having a bit of trouble with digesting an argument for which the conclusion is that it's not possible, in general, to Einstein synchronize clocks around closed curves. The argument goes as follows;

(Beginning)
Consider a four-velocity field $e_{\hat{0}}$ of the reference particles in a reference frame R.
In an arbitrary basis $\{e_\mu\}$ where we choose $e_0$ parallel to $e_{\hat{0}}$ the vectors $e_i$ need not be parallel to $e_0$. We thus introduce

$$e_{\perp i} = e_i - \frac{e_i \cdot e_0}{e_0 \cdot e_0} e_0$$

and define

$$\gamma_{ij} = e_{\perp i}\cdot e_{\perp j} = g_{ij} - \frac{g_{i0}g_{j0}}{g_00}$$

where $g_{\mu \nu}$ is the components of the metric tensor corresponding to the original general basis. Written in terms of the time orthogonal basis $\{e_{\hat 0}, e_{\perp,i}\}$ then we get

$$ds^2 = - d\hat t^2 + \gamma_{ij}dx^i dx^j.$$

We can thus write

$$d\hat t^2 = \gamma_{ij}dx^i dx^j - ds^2 = \gamma_{ij}dx^i dx^j - g_{\mu \nu} dx^\mu dx^\nu = [(-g_{00})^{1/2}(dx^0 + \frac{g_{i0}}{g_{00}} dx^i)]^2.$$

Therefore $d\hat t=0$ corresponds to

$$dx^0 = - \frac{g_{i0}}{g_{00}} dx^i$$

which is not a perfect differential. Thus the line integral of the coordinate time around a closed curve is in general not zero. The author then concludes that since $d\hat t = 0$ corresponds to simultaneity on Einstein synchronized clocks it is not in general ($g_{i0} \neq 0$) possible to Einstein synchronize clocks around closed curves.
(End)

Questions:
(1) Why does $d\hat t = 0$ correspond to Einstein synchronization of clocks?
(2) Is so bad that the line integral of the coordinate time is different from zero? Does this generally implies something other than a bad choice of time coordinate?