As an Amazon Associate, I earn from qualifying purchases. For more practice problems like this, see Schaum’s Outline of College Physics, 12th Edition.

1. What is being asked?

You are given the standard Schwarzschild metric (which describes the spacetime around a spherically symmetric, uncharged mass) in standard Schwarzschild coordinates. You are also shown the coordinate transformation used to rewrite this metric into isotropic coordinates, where the spatial part of the metric looks conformally flat (resembling spherical coordinates in flat Euclidean space multiplied by a conformal factor).

The question is: Where does the transformation equation for $r$ come from? How do we actually derive or discover the relation between the standard radial coordinate $r$ and the isotropic radial coordinate $r’$?

2. Step-by-Step Derivation

To find where the transformation comes from, we need to start with the spatial part of the Schwarzschild metric and demand that it takes the isotropic form: \(ds_{\text{spatial}}^2 = \left(1 - \frac{2GM}{r}\right)^{-1} dr^2 + r^2 (d\theta^2 + \sin^2\theta d\phi^2)\) (Note: In your prompt, $2m$ or $2GM$ represents the Schwarzschild radius, often set to $2M$ in geometric units where $G=1$).

We want to find a new radial coordinate, let’s call it $r’$, such that the spatial metric becomes conformally flat: \(ds_{\text{spatial}}^2 = \Omega(r')^2 \left[ dr'^2 + r'^2 (d\theta^2 + \sin^2\theta d\phi^2) \right]\) where $\Omega(r’)$ is some conformal factor (a function of $r’$ only).

Step 1: Compare the angular parts

Look at the angular part ($d\theta^2 + \sin^2\theta d\phi^2$) in both metrics. In the standard metric, it is multiplied by $r^2$. In the isotropic metric, it is multiplied by $\Omega(r’)^2 r’^2$.

By equating these two terms, we immediately establish a relation between $r$, $r’$, and the conformal factor $\Omega(r’)$: \(r^2 = \Omega(r')^2 r'^2 \implies r = r' \Omega(r')\)

Step 2: Transform the radial differential $dr$

Now we need to relate the differentials $dr$ and $dr’$. Using our relation $r = r’\Omega(r’)$, we take the derivative: \(dr = \frac{d}{dr'} \big( r' \Omega(r') \big) dr' = \left( \Omega(r') + r' \frac{d\Omega}{dr'} \right) dr'\)

Step 3: Substitute into the radial part of the metric

The standard spatial metric component for $dr^2$ is: \(g_{rr} dr^2 = \left(1 - \frac{2M}{r}\right)^{-1} dr^2\)

Substitute our expression for $dr$ into this term: \(g_{rr} dr^2 = \left(1 - \frac{2M}{r}\right)^{-1} \left( \Omega + r'\frac{d\Omega}{dr'} \right)^2 dr'^2\)

Step 4: Demand the isotropic form

For the metric to be in isotropic coordinates, the coefficient of $dr’^2$ must equal the conformal factor $\Omega(r’)^2$ (because the isotropic spatial metric is $\Omega^2(dr’^2 + r’^2d\Omega^2) = \Omega^2 dr’^2 + \Omega^2 r’^2(d\theta^2 + \sin^2\theta d\phi^2)$).

Therefore, we set: \(\left(1 - \frac{2M}{r}\right)^{-1} \left( \Omega + r'\frac{d\Omega}{dr'} \right)^2 = \Omega^2\)

Take the square root of both sides: \(\left(1 - \frac{2M}{r}\right)^{-1/2} \left( \Omega + r'\frac{d\Omega}{dr'} \right) = \Omega\)

Step 5: Solve the differential equation for $\Omega(r’)$

Rearrange the equation to separate variables or integrate: \(\left(1 - \frac{2M}{r}\right)^{-1/2} \left( 1 + \frac{r'}{\Omega}\frac{d\Omega}{dr'} \right) = 1\)

Recall from Step 1 that $r = r’\Omega$, which means $\frac{r}{r’} = \Omega$. Substitute this back in: \(\left(1 - \frac{2M}{r}\right)^{-1/2} \left( 1 + \frac{r'}{\Omega}\frac{d\Omega}{dr'} \right) = 1\)

Actually, it is much easier to work directly with $r$ and $r’$. Let’s rewrite $\left(1 - \frac{2M}{r}\right)^{-1/2}$ as $\frac{dr}{dr’\Omega}$: From $\left(1 - \frac{2M}{r}\right)^{-1} dr^2 = \Omega^2 dr’^2$, we take the square root directly: \(\frac{dr}{\sqrt{1 - \frac{2M}{r}}} = \Omega \, dr'\)

Since $\Omega = \frac{r}{r’}$, we substitute that in: \(\frac{dr}{\sqrt{1 - \frac{2M}{r}}} = \frac{r}{r'} dr'\)

Rearrange terms to group $r$ on one side and $r’$ on the other: \(\frac{dr}{r \sqrt{1 - \frac{2M}{r}}} = \frac{dr'}{r'}\)

Step 6: Integrate both sides

Now, integrate both sides of the equation: \(\int \frac{dr}{r \sqrt{1 - \frac{2M}{r}}} = \int \frac{dr'}{r'}\)

  • Right side: $\int \frac{dr’}{r’} = \ln(r’) + C_1 = \ln\left(\frac{r’}{C}\right)$
  • Left side: Use the substitution $u = \sqrt{1 - \frac{2M}{r}}$ (or standard integral tables) to find that: \(\int \frac{dr}{r \sqrt{1 - \frac{2M}{r}}} = \ln\left( \frac{\sqrt{1 - 2M/r} - 1}{\sqrt{1 - 2M/r} + 1} \right) + \text{constant}\)

Equating the integrals (and setting the integration constant such that $r \to r’$ at spatial infinity, where $C = M/4$): \(\ln\left( \frac{\sqrt{1 - 2M/r} - 1}{\sqrt{1 - 2M/r} + 1} \right) = \ln\left( \frac{4r'}{M} \right)\)

Exponentiate both sides: \(\frac{\sqrt{1 - 2M/r} - 1}{\sqrt{1 - 2M/r} + 1} = \frac{4r'}{M}\)

Solving this algebraic equation for $r$ in terms of $r’$ yields: \(r = r'\left(1 + \frac{M}{2r'}\right)^2\)

This differential equation requirement is precisely where the coordinate transformation comes from.

3. Final Answer

The coordinate transformation $r = r’\left(1+\frac{M}{2r’}\right)^2$ is not chosen at random; it is derived by demanding that the spatial part of the Schwarzschild metric becomes conformally flat (i.e., isotropic), meaning the cross-terms vanish and the spatial metric takes the form $ds^2 = \Omega(r’)^2 [dr’^2 + r’^2 d\Omega^2]$. Solving the differential equation $\frac{dr}{\sqrt{1 - 2M/r}} = \frac{r}{r’} dr’$ forces this exact relationship between $r$ and $r’$.

4. Common Mistakes

  • Assuming a linear relation: Students often guess a simple scaling like $r = k r’$ or $r = r’ + M$. Isotropic coordinates require a non-linear inverse-square relation to completely eliminate the off-diagonal metric components and flatten the spatial volume element.
  • Confusing $M$ and $2M$ definitions: Different authors define the mass parameter differently ($m$, $GM$, $2M$, or $r_s$ for Schwarzschild radius). Be very careful tracking factors of 2 when integrating the differential equation.
  • Forgetting the conformal factor: Trying to change coordinates by only substituting $r(r’)$ into the $dr^2$ term while forgetting that the angular pre-factor ($r^2$) must also transform ($\Omega^2 r’^2$).

Original question: Schwarzschild metric in Isotropic coordinates on Physics Stack Exchange, licensed CC BY-SA.