# How to obtain orthonormal tetrad basis for an infalling observer?

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

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

In this paper, the authors have written the orthonormal tetrad basis for a static observer $$(2m 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, 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?

• Perhaps look for the so-called PPON frame, the frame parallel-propagated along geodesics. See arxiv 1011.3502 for the derivation of this frame in the Lifshitz case. Jun 1 at 16:58

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.

• Could you please write a step-by-step calculation, with generic metric potentials? Like, $gtt$, $grr$ and so on... Jun 28 at 14:37
• @M.N.Raia The point of questions with the homework and exercise tag is not to give full detailed answers but rather to give "hints" so that you can learn from working out the solution on your own.
– Void
Jun 28 at 15:22
• I'm struggling with tetrads for at least 3 years up to now. I simply understand the concept, but I not be able to do the explicit calculations besides for the static observer, or ZAMOs. I would like to see something like: For radially comoving observer the natural tetrad is $e_{0}=(f(g_{\mu\nu}),g(g_{\mu\nu}),0,0)$; $e_{1} = (h(g_{\mu\nu}),j(g_{\mu\nu}),0,0)$; $e_{\theta} = (0,0,r^{-1},0)$, $e_{\phi} = (0,0,0,r^{-1}sin^{-1}\theta)$ Jun 28 at 15:36
• I mean, I know and I can prove and perform all the calculations when the metric is diagonal and when we are seeking the frame of a static observer: is just a matter of divide every tangent vetor by the square root of the metric potentials. But, this generic view seems to fail in every case that I read. I understand that we have a infinite set of tetrads to choose. But, how can't exist a "most common" choice that works fine in most cases? Jun 28 at 15:40
• Can you appoint me a reference that says: "for the metric with $ds^{2}=-f(r)dt^2+-g(t,r)\big(dr^2 + r^{2}d\theta^{2} +r^{2}sin^{2}\theta d\phi^{2})$ the tetrad vector are: ...." ? Jun 28 at 15:43