# I have a question regarding the Painlevé-Gullstrand (PG) metric with factor 2

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?

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

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

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:

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

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

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

$$g_{11}=1$$

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