Schwarzschild in Cartesian: Tricks for Transformation

[epovo]

TL;DR Summary

What is the appropriate coordinate change?

According to Schutz, the line element for large r in Schwarzschild is
$$ ds^2 \approx - ( 1 - \frac {2M} {r}) dt^2 + (1 + \frac {2M} {r}) dr^2 + r^2 d\Omega^2 $$
and one can find coordinates (x, y, z) such that this becomes
$$ ds^2 \approx - ( 1 - \frac {2M} {R}) dt^2 + (1 + \frac {2M} {R}) (dx^2+dy^2+dz^2) $$
where $R \equiv (x^2 + y^2 + z^2)^{1/2}$

This makes sense, as it is the same as the weak field metric with $\phi = -M/r$ (Newtonian gravitational field).
However, if I try to go from the first to the second using the usual spherical to Cartesian transformation, I get something different. Am I missing a trick here?

 

[vanhees71]

I don't know, why you look at the approximation. You can introduce another radial coordinate such that you get an isotropic form. See

https://en.wikipedia.org/wiki/Schwarzschild_metric#Alternative_coordinates

 

[PeroK]

However, if I try to go from the first to the second using the usual spherical to Cartesian transformation, I get something different. Am I missing a trick here?

I recall I had this same confusion when I was learning GR!

 

[Vanadium 50]

Why by the pendulous yarbles of Zeus would you want to take a system with spherical symmetry and work it in rectangular coordinates?

 

[epovo]

Why by the pendulous yarbles of Zeus would you want to take a system with spherical symmetry and work it in rectangular coordinates?

I don't have any special interest, I was just following Schutz and intrigued by his phrasing "One can find coordinates in which..." rather than the more natural "In Cartesian we have...". So just applied the transformation and I got an extra term that refuses to go away:

$$ ds^2 \approx - ( 1 - \frac {2M} {R}) dt^2 + (1 + \frac {2M} {R}) (dx^2+dy^2+dz^2) -2RM d\Omega^2 $$

 

[PeroK]

I don't have any special interest, I was just following Schutz and intrigued by his phrasing "One can find coordinates in which..." rather than the more natural "In Cartesian we have...". So just applied the transformation and I got an extra term that refuses to go away:

$$ ds^2 \approx - ( 1 - \frac {2M} {R}) dt^2 + (1 + \frac {2M} {R}) (dx^2+dy^2+dz^2) -2RM d\Omega^2 $$

I think the problem is that $r \ne R$. You need to take $R$ as some function of $r$. This will be buried in my notes somewhere.

 

[epovo]

I think the problem is that $r \ne R$. You need to take $R$ as some function of $r$. This will be buried in my notes somewhere.

Well, Schutz says explicitly that $R \equiv (x^2 + y^2 + z^2)^{1/2}$

 

[PeroK]

Well, Schutz says explicitly that $R \equiv (x^2 + y^2 + z^2)^{1/2}$

First, all that means is what it says. That the metric involves coordinates $t, x, y, z$ with $R$ defined as above to simplify the terms in the denominators.

Second, there is no sense in which $r^2 = x^2 + y^2 + z^2$.

 

[epovo]

But I thought $r^2 = x^2 + y^2 + z^2$ is part of the coordinate transformation!

 

[PeroK]

But I thought $r^2 = x^2 + y^2 + z^2$ is part of the coordinate transformation!

That was your chosen coordinate transformation. It was the wrong one. Which is what I'm trying to point out!

 

[epovo]

That was your chosen coordinate transformation. It was the wrong one. Which is what I'm trying to point out!

I see! But then, which is the right one, that would give me the result I am after???

 

[PeroK]

I see! But then, which is the right one, that would give me the result I am after???

That's the question. What I would do is look at $x, y, z$ with their usual spherical definitions and compare them to $r$ and $R$. I think that's how I figured it out.

 

[epovo]

That's the question. What I would do is look at $x, y, z$ with their usual spherical definitions and compare them to $r$ and $R$. I think that's how I figured it out.

Thank you for the tip. I will try some more

 

[ergospherical]

Why by the pendulous yarbles of Zeus would you want to take a system with spherical symmetry and work it in rectangular coordinates?

A fair question, but in this case probably to linearise the scwarzschild metric (taking only first order in 2M/r) to identify the components of the linearised metric, $h_{00}$ = 2$\phi$/$c^2$ and $h_{ij}$ = 2$\phi$ $\delta_{ij}$/$c^2$. (As opposed to doing it by solving the linearised field equations).

 

[Vanadium 50]

Note that you still have x, y and z's buried in the Ω.

 

[PeterDonis]

But I thought $r^2 = x^2 + y^2 + z^2$ is part of the coordinate transformation!

No. $R = x^2 + y^2 + z^2$ is part of the coordinate transformation. But $R$ (capital R) is not the same as $r$ (lower case r). The latter is the "areal radius", i.e., $r = \sqrt{A / 4 \pi}$, where $A$ is the surface area of a 2-sphere at $r$. In flat spacetime, we would indeed have that $r = R$; but you are not asking about flat spacetime.

 

[vanhees71]

Let's do the transformation explicitly. We start with the usual "Schwarzschild coordinates":
$$\mathrm{d} s^2=\left (1-\frac{2m}{r} \right) \mathrm{d} t^2 -\frac{1}{1-2m/r)} \mathrm{d} r^2 + r^2 \underbrace{(\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2)}_{\mathrm{d} \Omega^2},$$
where $m=GM/c^2=r_{\text{S}}/2$.We want to introduce a new radial coordinate, $R$, such that the spatial part is
$$\propto \mathrm{d} \vec{x}^2 = \mathrm{d} R^2 + R^2 (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2).$$
The spatial part of the line element is
$$\mathrm{d} \sigma^2 = \frac{1}{1-2m/r} \mathrm{d} r^2 + r^2 \mathrm{d}\Omega^2 = \frac{1}{2-2m/r} r^{\prime 2} \mathrm{d} R^2 + \frac{r^2}{R^2} R^2 \mathrm{d} \Omega^2.$$
The prime denotes the derivative wrt. the new radial coordinate, $R$.

To get "isotropic coordinates" we thus want
$$\frac{r^{\prime 2}}{1-2m/r} = \frac{r^2}{R^2}.$$
This ODE is easily solved by separation, leading to [EDIT: Corrected typo, the factor of 2 on the left-hand side in view of #18]
$$\frac{2R}{m} =\frac{r-m}{m} + \sqrt{\frac{r(r-2m)}{m^2}}$$
or
$$r=R [1+m/(2R)]^2.$$
Plugging all this into the line element indeed leads to what's given in Wikipedia
$$\mathrm{d} s^2 = \left (\frac{1-m/(2R)}{1+m/(2R)} \right )^2 \mathrm{d} t^2 - [1-m/(2R)]^4 (\mathrm{d} R^2 +R^2 \mathrm{d} \vartheta^2 + R^2 \sin^2 \vartheta \mathrm{d} \varphi^2],$$
i.e., introducing the usual "Cartesian coordinates"
$$\vec{x}=R \begin{pmatrix}\cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta \end{pmatrix},$$
i.e.,
$$\mathrm{d} s^2= \left (\frac{1-m/(2R)}{1+m/(2R)} \right )^2 \mathrm{d} t^2 - [1-m/(2R)]^4 \mathrm{d} \vec{x}^2.$$
Note that the Schwarzschild Radius is $r_S=2m$.

The advantage of these coordinates is the same advantage that have Cartesian coordinates over spherical coordinates, i.e., the invariance under the full rotation group $\vec{x} \rightarrow \hat{R} \vec{x}$ with $\hat{R} \in \mathrm{SO}(3)$ is manifest, while in spherical coordinates it's not, because you distinguish the polar axis as a "preferred direction".

 

[epovo]

Let's do the transformation explicitly. We start with the usual "Schwarzschild coordinates":
$$\mathrm{d} s^2=\left (1-\frac{2m}{r} \right) \mathrm{d} t^2 -\frac{1}{1-2m/r)} \mathrm{d} r^2 + r^2 \underbrace{(\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2)}_{\mathrm{d} \Omega^2},$$
where $m=GM/c^2=r_{\text{S}}/2$.We want to introduce a new radial coordinate, $R$, such that the spatial part is
$$\propto \mathrm{d} \vec{x}^2 = \mathrm{d} R^2 + R^2 (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2).$$
The spatial part of the line element is
$$\mathrm{d} \sigma^2 = \frac{1}{1-2m/r} \mathrm{d} r^2 + r^2 \mathrm{d}\Omega^2 = \frac{1}{2-2m/r} r^{\prime 2} \mathrm{d} R^2 + \frac{r^2}{R^2} R^2 \mathrm{d} \Omega^2.$$
The prime denotes the derivative wrt. the new radial coordinate, $R$.

To get "isotropic coordinates" we thus want
$$\frac{r^{\prime 2}}{1-2m/r} = \frac{r^2}{R^2}.$$
This ODE is easily solved by separation, leading to
$$\frac{R}{m} =\frac{r-m}{m} + \sqrt{\frac{r(r-2m)}{m^2}}$$
or
$$r=R [1+m/(2R)]^2.$$
Plugging all this into the line element indeed leads to what's given in Wikipedia
$$\mathrm{d} s^2 = \left (\frac{1-m/(2R)}{1+m/(2R)} \right )^2 \mathrm{d} t^2 - [1-m/(2R)]^4 (\mathrm{d} R^2 +R^2 \mathrm{d} \vartheta^2 + R^2 \sin^2 \vartheta \mathrm{d} \varphi^2],$$
i.e., introducing the usual "Cartesian coordinates"
$$\vec{x}=R \begin{pmatrix}\cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta \end{pmatrix},$$
i.e.,
$$\mathrm{d} s^2= \left (\frac{1-m/(2R)}{1+m/(2R)} \right )^2 \mathrm{d} t^2 - [1-m/(2R)]^4 \mathrm{d} \vec{x}^2.$$
Note that the Schwarzschild Radius is $r_S=2m$.

The advantage of these coordinates is the same advantage that have Cartesian coordinates over spherical coordinates, i.e., the invariance under the full rotation group $\vec{x} \rightarrow \hat{R} \vec{x}$ with $\hat{R} \in \mathrm{SO}(3)$ is manifest, while in spherical coordinates it's not, because you distinguish the polar axis as a "preferred direction".

Thank you, @vanhees71.
I would have never figured this out myself.
I have followed your derivation and only found that the initial conditions of the ODE demand a factor of 2 here:
$$ 2\frac{R}{m} =\frac{r-m}{m} + \sqrt{\frac{r(r-2m)}{m^2}}$$
That is from demanding that for large r, $R \approx r$
otherwise it doesn't work.

 

[vanhees71]

You are right. That was a typo. I'll correct it in the OP.