How to obtain orthonormal tetrad basis for an infalling observer?

Question

An eternal Schwarzschild spacetime in Painlevé-Gullstrand coordinates reads as

$ds^2 = -\left(1-\dfrac{2m}{r}\right)c^2~dT^2 + 2\sqrt{\dfrac{2m}{r}}c~dTdr + dr^2 + r^2\left(d\theta^2+\sin{^2\theta}~d\phi^2\right)~$

for $~~m=\dfrac{GM_{Sch}}{c^2}$

In this paper, the authors have written the orthonormal tetrad basis for a static observer $(2m<r<\infty)$ as

$e_0 = \dfrac{1}{c\sqrt{1-\frac{2m}{r}}} \dfrac{\partial}{\partial T} ~,~~ e_1 = \sqrt{1-\dfrac{2m}{r}} \dfrac{\partial}{\partial r} + \dfrac{\frac{2m}{r}}{c\sqrt{1-\frac{2m}{r}}} \dfrac{\partial}{\partial T} ~,$

$e_2 = \dfrac{1}{r} \dfrac{\partial}{\partial \theta} ~,~~ e_3 = \dfrac{1}{r\sin{\theta}} \dfrac{\partial}{\partial \phi}$

which is very clear from the fact that it satisfies $g_{\mu\nu}~e^\mu_\alpha~e^\nu_\beta = \eta_{\alpha\beta}$.

Now if one wants to calculate the orthonormal tetrad basis for a radially infalling observer $(0<r<\infty)$, then how to do that?

In the same paper, those infalling tetrad basis have been written as (without much arguments or justification)

$\hat{e}_0 = \dfrac{\varepsilon-\sqrt{\frac{2m}{r}}\sqrt{\varepsilon^2-1+\frac{2m}{r}}}{\left(1-\frac{2m}{r}\right)c} \dfrac{\partial}{\partial T} - \sqrt{\varepsilon^2-1+\dfrac{2m}{r}} \dfrac{\partial}{\partial r} ~,$

$\hat{e}_1 = \varepsilon \dfrac{\partial}{\partial r} + \dfrac{\varepsilon\sqrt{\frac{2m}{r}}-\sqrt{\varepsilon^2-1+\frac{2m}{r}}}{\left(1-\frac{2m}{r}\right)c} \dfrac{\partial}{\partial T} ~,$

$\hat{e}_2 = \dfrac{1}{r} \dfrac{\partial}{\partial \theta} ~,~~ \hat{e}_3 = \dfrac{1}{r\sin{\theta}} \dfrac{\partial}{\partial \phi}$

where $\varepsilon = E/c^2$ for $~E~$ being the constant energy per mass of a test particle.

But how do these tetrads are coming?

Answer 1

Let me give you the general idea. When constructing an orthonormal frame of an observer, you want to start with the zeroth leg, which is the four-velocity of the observer.

In this case you are looking for a radially infalling observer. This means you are looking for an observer who has only the $T,r$ components of the zeroth leg. Furthermore, you want this to be the four-velocity of the same geodesic as it moves forward in proper time (this does not need to be the case, the integral curves of the zeroth leg do not need to be geodesics). How to deal with this? There is a couple of integrals of motion that characterize the motion of a geodesic in Schwarzschild. You have the angular momentum, which is zero for a radial infall, and then there is energy $\varepsilon \equiv - u_t$. Once you fix the value of $\varepsilon$ for the congruence and apply normalization of four-velocity, the expression for the zeroth leg is fixed uniquely.

Then you have to deal with the other legs. These only have to be orthogonal to the zeroth leg, and normalized to one. You have the full freedom of the $SO(3)$ group of rotations at every point to define them, so you have to make some choices. One of them is to align one leg along $\phi$ ($\partial/\partial \phi$), and the other one along $\theta$ ($\partial/\partial \theta$), because it turns out in that case they are automatically orthogonal to the zeroth leg, you just need to normalize them.

There is one leg left to find then, and it is natural for it to point in the $r$ direction somehow, since then it is orthogonal to the other two spatial legs. You can notice that it won't be that simple though. This last leg will need a $t$-component in order to be orhogonal to the zeroth leg. You solve for it, normalize, and then you are done.