# Origin of spatial metric

1. Aug 21, 2008

### 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.

2. Aug 21, 2008

### 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.

3. Aug 21, 2008

### DrGreg

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 $d\ell^2 = dQ_{\alpha}dQ^{\alpha}$.

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

4. Aug 21, 2008

### 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

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5. Aug 21, 2008

### 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.