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Thinking about the equivalence principle, is there a nice, simple way to show that a local, freely falling frame in Schwarzschild spacetime is described by the Minkowski metric


I thought if I could describe a test body (moving radially inwards with an acceleration due to gravity) using the Schwarzschild metric

$$ds^{2}=\left(1-\frac{2GM}{c^{2}r}\right)c^{2}dt^{2}-\frac{dr^{2}}{1-\frac{2GM}{c^{2}r}}-r^{2}d\theta^{2}-r^{2}\sin^{2}\theta d\phi^{2},$$

the Minkowski metric would somehow pop out (with $d\theta = d\phi = 0$). I substituted $ar=-GM/r$ into the metric ($a$ is the acceleration due to gravity) to get


which, on Earth for example with $2ar\ll c^{2}$, is pretty close to the Minkowski metric. Is this valid/correct?

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I dont know if this Link is helpful to answer your question, but I will give it a try: – Hansenet Jul 12 '13 at 22:22
up vote 1 down vote accepted

Let's start with the Schwarzschild metric $$ \text{d}s^2 = \left(1 - \frac{r_\text{s}}{r}\right)c^2\text{d}t^2- \left(1 - \frac{r_\text{s}}{r}\right)^{-1}\text{d}r^2 - r^2\text{d}\Omega^2, $$ with $$ \begin{align} r_\text{s} &= \frac{2GM}{c^2},\\ \text{d}\Omega^2 &= \text{d}\theta^2 +\sin^2\theta\,\text{d}\varphi^2. \end{align} $$ Now let's introduce a new radial coordinate $\bar{r}$, defined as $$ r = \bar{r}\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^2. $$ After some algebra, we find that $$ \begin{align} \text{d}r^2 &= \left(1 - \frac{r_\text{s}}{4\bar{r}}\right)^2\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^2\text{d}\bar{r}^2,\\ \left(1 - \frac{r_\text{s}}{r}\right) &= \left(1 - \frac{r_\text{s}}{4\bar{r}}\right)^2\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^{-2}, \end{align} $$ so that the Schwarzschild metric can be written in the form $$ \text{d}s^2 = \left(1 - \frac{r_\text{s}}{4\bar{r}}\right)^{2}\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^{-2}c^2\text{d}t^2 - \left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^{4}\left(\text{d}\bar{r}^2 + \bar{r}^2\text{d}\Omega^2\right). $$ Now, in a local frame around a coordinate $(t,r,\theta,\varphi)$, we can treat $r$ in the coefficients as constant, so that $$ \begin{align} \left(1 - \frac{r_\text{s}}{4\bar{r}}\right)^{2}\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^{-2} &\approx \alpha^2,\\ \left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^{4} &\approx \beta^2, \end{align} $$ with $\alpha$ and $\beta$ constants, thus $$ \text{d}s^2 \approx c^2\alpha^2 \text{d}t^2 - \beta^2\left(\text{d}\bar{r}^2 + \bar{r}^2\text{d}\Omega^2\right). $$ Finally, with the coordinate transformation $$ \begin{align} \tilde{t}&=\alpha\, t,\\ \tilde{r}&=\beta\, \bar{r}, \end{align} $$ we obtain the familiar Minkowski metric $$ \text{d}s^2 \approx c^2\text{d}\tilde{t}{}^2 - \left(\text{d}\tilde{r}^2 + \tilde{r}^2\text{d}\Omega^2\right). $$ This transformation can be performed at any coordinate $(t,r,\theta,\varphi)$ along the path, so the local metric is always of Minkowski form.

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Took me ages to work through the algebra, but I then ground to a halt when you said $r$ (presumably you meant $\bar{r}$ ) can be treated as a constant in the new metric coefficients? What's the justification for that? Also, what's the problem with my guess at subsituting $ar=-GM/r$ into the metric? Does that only work for a weak gravitational field? Also, it's still not obvious how the above transformation is linked to something freely falling. Thanks. – Peter4075 Jul 13 '13 at 12:53
@Peter4075 $r$ can be approximated by a constant because the inertial reference frame is local, i.e. it is only valid in the infinitesimal neighbourhood of the specific coordinate $(t,r,\theta,\varphi)$. The inertial frame is not global: each coordinate belongs to a different local inertial frame. A constant $r$ implies that $\bar{r}$ is also constant. Your substitution $ar=-GM/r$ is only valid in the Newtonian weak field limit, when $M$ is small (and the entire metric approaches a global inertial frame). It doesn't work for strong gravitational fields. – Pulsar Jul 13 '13 at 13:25
But if we treat $r$ as a constant why do we need to go to all the trouble of introducing $\bar{r}$ ? Why not just say $\left(1-\frac{2GM}{c^{2}r}\right)=a$ , $\left(1-\frac{2GM}{c^{2}r}\right)^{-1}=b$ and $r^{2}=c$ , where $a$, $b$ and $c$ are constants, and get to the Minkowski metric that way? I'm probably missing the obvious here. – Peter4075 Jul 13 '13 at 14:15
@Peter4075 When I say "treat $r$ in the coefficients as constant", I really mean "consider a region small enough such that the change in $(1-r_\text{s}/r)$ is negligible". The transformation to $\bar{r}$ then allows to write the metric directly in the standard form, in a mathematically rigorous way. In general metrics, the approach would be to write $\text{d}s^2=g_{\mu\nu}(\mathbf{x})\text{d}x^\mu\text{d}x^\nu$ in cartesian form, approximate the coefficients by constants $g_{\mu\nu}$, and find the eigenvectors to write the metric into the standard $\text{diag}(1,-1,-1,-1)$ form. – Pulsar Jul 13 '13 at 14:47
@Peter4075 To clarify: I didn't mean "treat the coordinate $r$ constant" (because then $\text{d}r$ would be zero and we wouldn't have a region of space). What I meant is: treat the coefficients as constant. The $r^2$ in $r^2\text{d}\Omega^2$ isn't really a coefficient in the same way, but rather a part of the Jacobian of the spherical coordinate system. – Pulsar Jul 13 '13 at 14:57

Locally, that is for a fixed point $r,\theta,\phi,t$ you could always find global coordinates, such as the metrics will be Minkowski at this point (but not at other points). At the fixed point, the frame will be a inertial frame (but not at the other points). Of corse, you could use this flat metrics as approximate metrics in the immediate neighborhood of the fixed point.

But if you want global coordinates describing a freely falling observer, it is not possible. If it was possible, this would mean that the Schwartzschild metrics would be a flat metric, and this is obviously false.

An interesting case appears when you are using approximate the metrics near the horizon ($r\sim 2MG$), and in a small angular region ($\theta \sim 0$)

An example is the Rindler metrics:

$$ds^2= \rho^2~d\omega ^2 - d\rho^2 - dX^2 - dY^2$$

with : $$\omega = \frac{t}{4MG},~\rho \sim 2 \sqrt{2MG(r - 2MG)}, ~X \sim 2MG \theta \cos \phi, ~Y \sim 2MG \theta \sin \phi$$

Setting : $$T = \rho ~ sh\omega, Z = \rho ~ ch\omega$$

We get : $$ds^2=dT ^2 - dZ^2 - dX^2 - dY^2$$

This proves, that, near the horizon ($r\sim 2MG$), and for a large black hole ($\theta \sim 0$), the metrics is almost quasi flat.

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