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I have a question regarding the Painlevé-Gullstrand (PG) metric.

If we have the line element in a radial fall we get:

$$d\theta = d\phi = 0$$

$$ds^2 = -dT^2 + \left(dr+\sqrt{\frac{r_s}{r}}dT\right)^2.$$

Writing out the binomial formula we obtain:

$$ds^2 = -dT^2 + dr^2 + 2 \sqrt{\frac{r_s}{r}} dr dT + \frac{r_s}{r} dT^2.$$

If we now want to write down the metric tensor, we should obtain:

$$g_{\mu\nu} = \begin{pmatrix} 1 & 2\sqrt{\frac{r_s}{r}} \\ 2\sqrt{\frac{r_s}{r}} & \frac{r_s}{r} -1\\ \end{pmatrix}.$$

So am I right, that the factor 2 also comes into the metric?

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No, in general the differentials are not really "multiplication" as such (I can go into it if you'd like). So when you expand, you need to write

$$ds^2 = -dT^2 + dr^2 + \underbrace{\sqrt{\frac{r_s}{r}} dr dT}_{g_{r T}} + \underbrace{\sqrt{\frac{r_s}{r}}}_{g_{Tr}} dT dr + \frac{r_s}{r} dT^2 .$$

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You can write the line element with this general ansatz:

$ds^2=\begin{bmatrix} dT & dr \\ \end{bmatrix} \begin{bmatrix} g_{00} & g_{01} \\ g_{01} & g_{11} \\ \end{bmatrix} \begin{bmatrix} dT \\ dr \\ \end{bmatrix} \qquad (1)$

The metric $g_{\mu\nu}$ must be symmetric!

from equation (1) we obtain :

$ds^2=g_{00} \,dT^2+2\,dT\,dr\,g_{01}+g_{11}\,dr^2$

Now compare the coefficients with your line element:

$g_{00}=\frac{r_s}{r}-1$

$2\,g_{01}=2\sqrt{\frac{r_s}{r}}$

$g_{01}=\sqrt{\frac{r_s}{r}}$

$g_{11}=1$

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No, the off-diagonal metric components are $g_{rT}=g_{Tr}=\sqrt{\frac{r_s}{r}}$ without a factor of 2.

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