Freely falling observer in Schwarzschild metric

Question

I was looking at this question about non coordinate basis: https://www.physicsforums.com/threads/noncoordinate-basis.102902/

In answer number 4 the orthonormal basis for a free-falling observer in the Schwarzschild metric is given, I'm trying to derive this basis but I haven't been able to do this. The basis for an observer with constant $r$, $\theta$, $\phi$ is also given by I already understand how to obtain this.

For the free-falling observer we have coordinates $t'$, $r'$, $\theta'$, $\phi'$ Since it's radially falling then we have $$\frac{\partial}{\partial t'} = \frac{\partial t}{\partial t'} \frac{\partial}{\partial t} + \frac{\partial r}{\partial t'}\frac{\partial}{\partial r}$$

We can evaluate $\frac{\partial}{\partial t'} \otimes \frac{\partial}{\partial t'}$ into the metric given in the link

$$\mathbf{g}=\left( 1-\frac{2M}{r}\right) \mathbf{dt}\otimes\mathbf{dt}-\left( 1-\frac{2M}{r}\right) ^{-1}\mathbf{dr}\otimes\mathbf{dr}-r^{2}\left( \sin^{2}\theta\mathbf{d\theta}\otimes\mathbf{d \theta}+\mathbf{d\phi}\otimes\mathbf{d\phi}\right).$$

We obtain then the equation: $$1 = \left( 1-\frac{2M}{r} \right) \left( \frac{\partial t}{\partial t'} \right)^2 - \left( 1-\frac{2M}{r} \right)^{-1} \left( \frac{\partial r}{\partial t'} \right)^2$$

And from the geodesic equation we can deduce

$$\frac{\partial t}{\partial t'} = -\frac{C}{\left(1-\frac{2M}{r} \right)} $$

From here I see that if I take $C = 1$ and then replace in the equation from before I obtain

$$\mathbf{e}_{0}^{\prime} =\left( 1-\frac{2M}{r} \right)^{-1} \frac{\partial}{\partial t}-\left( \frac{2M}{r} \right)^{\frac{1}{2}}\frac{\partial}{\partial r}$$

Then i'm stuck for the $r'$ coordinate since I don't have a geodesic equation for it. Also I'm not sure I can assume $C=1$. Is there an standard way to solve this? any help with this is appreciated.

Answer 1

Let us consider the Schwarzschild metric, assuming as signature convention the negative sign on the time-time component of the metric.
$ds^2 = -g dt^2 + g^{-1} dr^2 + r^2 (d\theta^2 + \sin^2 \theta d\phi^2)$
where:
$G = 1$
$c = 1$
$g = (1 - 2M/r)$

We have three reference frames:
Schwarzschild with coordinates $(t, r, \theta, \phi)$
Stationary observer with coordinates $(\tau, r_{stat}, \theta_{stat}, \phi_{stat})$
Free falling observer with coordinates $(t', r', \theta', \phi')$

Stationary observer at constant $r$, $\theta$ and $\phi$
The time basis is built on the four-velocity $U_{stat} = \partial_\tau = dt / d\tau \partial_t$, with $dt / d\tau = g^{-1/2}$, giving $e_0 = \partial_\tau = g^{-1/2} \partial_t$. It is normalized with squared norm = $-1$. The radial basis is built as $\partial_r$ then normalized with squared norm = $+1$, giving $e_1 = \partial_{r_{stat}} = g^{1/2} \partial_r$. That is
$e_0 = (1 - 2M/r)^{-1/2} \partial_t$
$e_1 = (1 - 2M/r)^{1/2} \partial_r$
The $e_0$ and $e_1$ are orthonormal.

Free falling observer from rest at infinity
It is a radial path, that is at constant $\theta$ and $\phi$. You can relate the free falling frame to the stationary frame with the Lorentz transformation
$\tau = \gamma t' + \gamma v r'$
$r_{stat} = \gamma v t' + \gamma r' $
$v = -(2M/r)^{1/2}$
The velocity $v$ is got by comparing the energy of the free falling as measured by the stationary observer calculated both as $E = -p_\mu U_{stat}^\mu$ and as $E = \gamma m$ with $\gamma = (1 - v^2)^{-1/2}$.
The relation between the partial derivatives is
$\partial_{t'} = \partial \tau / \partial t' \partial_\tau + \partial r_{stat} / \partial t' \partial_{r_{stat}}$
$\partial_{r'} = \partial \tau / \partial r' \partial_\tau + \partial r_{stat} / \partial r' \partial_{r_{stat}}$
where:
$\partial \tau / \partial t' = \gamma$
$\partial r_{stat} / \partial t' = \gamma v$
$\partial \tau / \partial r' = \gamma v$
$\partial r_{stat} / \partial r' = \gamma$
As you have
$e'_0 = \partial_{t'}$
$e'_1 = \partial_{r'}$
you can write
$e'_0 = \gamma e_0 + \gamma v e_1$
$e'_1 = \gamma v e_0 + \gamma e_1$
expressing against $\partial_t$ and $\partial_r$
$e'_0 = \gamma g^{-1/2} \partial_t + \gamma v g^{1/2} \partial_r$
$e'_1 = \gamma v g^{-1/2} \partial_t + \gamma g^{1/2} \partial_r$
and remembering that
$\gamma = g^{-1/2}$
$g = (1 - 2M/r)$
$v = -(2M/r)^{1/2}$
we have
$e'_0 = g^{-1} \partial_t + v \partial_r = (1 - 2M/r)^{-1} \partial_t - (2M/r)^{1/2} \partial_r$
$e'_1 = v g^{-1} \partial_t + \partial_r = -(2M/r)^{1/2} (1 - 2M/r)^{-1} \partial_t + \partial_r$
The $e'_0$ and $e'_1$ are orthonormal as well.
Note:
To complete the orthonormal basis, we have also
$e_2 = e'_2 = 1 / r \partial_\theta$
$e_3 = e'_3 = 1 / (r \sin \theta) \partial_\phi$