# Freely falling observer in Schwarzschild metric

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.

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
$e_2 = e'_2 = 1 / r \partial_\theta$
$e_3 = e'_3 = 1 / (r \sin \theta) \partial_\phi$

• I think there's something missing here. How do you get the "remember that gamma = (..)"? Shouldn't we specifically figure out what the magnitude of the boost is for us to go from the stationary observer frame to the moving observer frame? Furthermore, this basis does not seem to satisfy $u^{\prime\, m}=e^{\prime\,m}_{\,\,\mu}u^\mu=(1,0,0,0)$ when the $u^\mu$ is the 4-velocity of a freely falling observer, i.e., $u^\mu=(g^{-1},\pm (1-r_s/r),0,0)$. What gives? It does not seem like it's the correct solution.
– OTH
Oct 27, 2022 at 17:14
• AHH! FINALLY! I thought I was going crazy. I was trying to compute $e^\mu_{\,\,\,\,\nu}$ inverse and then multiplying it with the 4-vector, and it wasn't working. Turns out I was missing the transpose from my matrix calculation. Scratch the last part. However, I think it's still missing a part of the answer, in particular how the lorenz boost is related to the freely falling observer.
– OTH
Oct 27, 2022 at 17:24