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Yes, that's it. How to find a particular metric of an observer in general relativity?

Let's say we have a static metric:

$$ds^2=-g_{00}(\vec{r})dt^2+d\vec{r}^2=-g_{00}(\vec{r})dt^2+g_{ij}(\vec{r})dx^idx^j$$

and an observer following a worldline $x^i=f^i(t)$ (or $\vec{r}=\vec{r}(t)$). How can we find a coordinate and the corresponding metric of this observer, such that it has a form:

$$ds^2=-g'_{00}(r')dt'^2+d\vec{r}'^2$$ It is ideal if we can find $t'=t'(t,\vec{r})$ and $\vec{r}'=\vec{r}'(t,\vec{r})$ (and when $d\vec{r}'=0$ I think we should restore $x^i=f^i(t)$)

anonymous67
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2 Answers2

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simply $$ g_{ij} dx^idx^j=g_{11}dx^2+g_{22}dy^2+g_{33}dz^2=dr^2 $$ so metric should be $$ g_{ij}=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} $$

aQuestion
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I'm not clear on the meaning of variables in the question, but in general here's a nice trick to define a new coordinate which is the proper time of an observer field. Label the 4-velocity / observer field $\mathbf u$. the observers. Take the dual vector $\mathbf u^\flat$, and write it in terms of the dual basis. e.g. Suppose in Schwarzschild-Droste coordinates we have 4-velocity: $$u^\mu = \bigg(\Big(1-\frac{2M}{r}\Big)^{-1},-\sqrt\frac{2M}{r},0,0\bigg)$$ Or writing in terms of the coordinate basis vectors: $$\mathbf u = \Big(1-\frac{2M}{r}\Big)^{-1}\partial_t -\sqrt\frac{2M}{r}\partial_r$$ The (negative of the) dual vector is: $$-\mathbf u^\flat = dt +\frac{\sqrt{2M/r}}{1-2M/r}dr$$ If we can write this as the differential $dT$ of a scalar $T$, then take this as a new coordinate in place of $t$. Check things mesh together well, by Frobenius' theorem. In our example, we now have Gullstrand-Painleve or "raindrop" coordinates! Martel & Poisson (2000) use this elegant approach.

Colin MacLaurin
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