What is the Origin of the Spatial Metric in Non-Inertial Frames?

[jdstokes]

Can anyone explain to me the origin of the spatial metric for measuring distances in non-inertial frames?

$d\ell^2 = [(g_{0i}g_{0j})/g_{00} - g_{ij}]dx^i dx^j$.

I've heard it quoted but never seen it derived. I believe it works on the assumption that distance is half the proper time for return of an em signal (whatever that means).

Thanks.

 

[jdstokes]

My best guess:

$ds^2 = g_{00} c^2 dt^2 + g_{ij}dx^{i} dx^{j} + 2g_{0i} cdt dx^i$

$c^2 d\tau^2 = g_{00} dt^2$

On null geodesics

$0 = g_{00} c^2 dt^2 + g_{ij}dx^{i} dx^{j} + 2g_{0i} cdt dx^i \implies$
$c dt = -\frac{g_{0i} dx^i}{g_{00}} \pm \frac{\sqrt{g_{0i}g_{0j}dx^i dx^j - g_{00} g_{ij} dx^i dx^j }}{g_{00}}$ (quadratic equation)

Now $d\ell =c\frac{1}{2}d\tau = c\frac{1}{2}\sqrt{g_{00}}(dt_++dt_-) = \frac{g_{0i}dx^i}{g_{00}} \implies$

$d\ell^2 = \frac{g_{0i}g_{0j}dx^i dx^j}{g_{00}}$

which is wrong.

 

[DrGreg]

Can anyone explain to me the origin of the spatial metric for measuring distances in non-inertial frames?

$d\ell^2 = [(g_{0i}g_{0j})/g_{00} - g_{ij}]dx^i dx^j$.

I've heard it quoted but never seen it derived.

I've never seen this before, but a little thought suggests the following.

I think we have to assume that the curves $dx^1=dx^2=dx^3=0$ represent worldlines of observers and I think the metric you quoted measures distance within the surfaces that are orthogonal to those worldlines.

If $U^{\alpha}$ is parallel to the 4-velocity of such an observer, with components (1, 0, 0, 0), consider decomposing $dx^{\alpha}$ as

$$dx^{\alpha} = dP^{\alpha} + dQ^{\alpha}$$​

where $dP^{\alpha}$ is parallel to $U^{\alpha}$ and $dQ^{\alpha}$ is orthogonal to it. The orthogonality ensures that

$$dx_{\alpha}dx^{\alpha} = dP_{\alpha}dP^{\alpha} + dQ_{\alpha}dQ^{\alpha}$$​

The required projection is given by

$$dP^{\alpha} = \frac{U_{\beta}U^{\alpha}}{U_{\gamma}U^{\gamma}}dX^{\beta}$$​

and then I think it all follows from that, noting that

$$U_{\beta} = g_{\beta \alpha} U^{\alpha} = g_{\beta 0}$$​

and the answer you want is $d\ell^2 = dQ_{\alpha}dQ^{\alpha}$.

I believe it works on the assumption that distance is half the proper time for return of an em signal

That would be "radar distance" which amounts to the same thing infinitesimally, but not over larger distances (except in flat spacetime).

 

[Mentz114]

$$d\ell^2 = [(g_{0i}g_{0j})/g_{00} - g_{ij}]dx^i dx^j$$

It is the radar infinitismal spatial distance. There's a good derivation in appendix A of the attached.

DrGreg's derivation is along the right lines but I haven't checked it.

M

 

[jdstokes]

Hi DrGreg,

You're correct, the metric is actually the same thing as the projection operator $U^\alpha U^\beta - g^{\alpha\beta}$, written in the frame where the observer has vanishing 3-velocity.

The error in my derivation of radar distance was to add $dt_+,dt_-$ when I should have subtracted them. This is because one of them gives a time into the past which is negative.