# Are the Kerr-Schild coordinates for the Schwarzschild metric related to the standard coordinate system for the metric?

In another question, reference is made to the Kerr-Schild coordinate system for the Schwarzschild metric, and there is a question about whether this system is directly related to the standard metric one by an exact coordinate transformation, rather than some sort of linearization of Einstein's equations, because as formulated, it looks very much like some sort of linearization:

$$g_{ab} = \eta_{ab} + C k_{a}k_{b}$$

where $$\eta$$ is the Minkowski metric, $$C = \frac{2M}{r}$$ and (choosing spherical coordinates) $$k = (1, \mp 1, 0, 0)$$

So, what is going on?

• I think it would be more clear in another coordinate, the Eddington–Finkelstein coordinates (u,r,$\theta$, $\phi$), the $k_a$ would just be $(1,0,0,0)$ and $C=\frac{2m}{r}$. It could be seen that KS is just obtained by a coordinate transformation. Commented Aug 15, 2023 at 16:54

## 2 Answers

Yes, it absolutely is an exact coordinate transformation!

Take the expression above, and write it as a line element (also, take $$\epsilon$$ to be either plus or minus 1 to avoid writing $$\pm$$ over and over), and take $$d\Omega^{2} = \left(d\theta^{2} + \sin^{2}\theta\,d\phi^{2}\right)$$:

$$ds^{2} = -\left(1-\frac{2M}{r}\right)dt^{2} + 2dt dr\left( -\frac{2M\epsilon}{r}\right) + \left(1+\frac{2M}{r}\right)dr^{2} + r^{2}d\Omega^{2}$$

By inspection of this expression, we can see that the $$g_{tt}$$ element is already in schwarzschild form, and the angular parts are already correct. We just need the off-diagnonal elements gone, and we need a different $$g_{rr}$$. with all of these constraints, we guess that we cannot do very much, and are pretty much restricted to a coordinate change of the form

$$t = T + f(r)$$

where we expect $$T$$ to be the Schwarzschild time. Substituting this ansatz into our line element, we have:

$$ds^{2} = -\left(1-\frac{2M}{r}\right)dT^{2} + 2dTdr\left(-\frac{2M\epsilon f^{\prime}}{r}\left(1-\frac{2M}{r}\right) -\frac{2M\epsilon}{r}\right) + \left(-f^{\prime 2}\left(1-\frac{2M}{r}\right) -\frac{4M\epsilon f^{\prime}}{r}+ 1+\frac{2M}{r}\right)dr^{2} + r^{2}d\Omega^{2}$$

The diagonal term vanishes if we have:

$$f^{\prime} = -\frac{\epsilon \frac{2M}{r}}{1-\frac{2M}{r}}$$

so, a convenient choice for $$f$$ is $$-2M\epsilon {\rm ln}\left(\frac{r}{2M} -1\right)$$

Finally, all we have to do is check to see whether our expression for $$g_{rr}$$ checks. Substituting the value for $$f^{\prime}$$, we have:

$$g_{rr} = -\left(-\frac{\frac{2M}{r}}{1-\frac{2M}{r}}\right)^{2}\left(1-\frac{2M}{r}\right) -\frac{4M\epsilon}{r}\left(-\frac{\epsilon \frac{2M}{r}}{1-\frac{2M}{r}}\right)+ 1+\frac{2M}{r}$$

The second term is $$-2$$ times the first term, and multiplying the last term by $$\frac{1-2M/r}{1-2M/r}$$ and adding everything together, we find that we do, indeed end up with $$\frac{1}{1-2M/r}$$, and we are done. $$T$$ is indeed the schwarzchild coordinate time.

… looks very much like some sort of linearization … So, what is going on?

In order to understand what is going on a result by Xanthopoulos $$[1]$$ might be useful:

… If $$(M,g_{ab})$$ is an exact vacuum solution of Einstein’s equation, $$l_a$$ a null vector field and if $$l_a l_b$$ satisfies the linearized equation on background $$(M,g_{ab})$$, then $$g_{ab}+l_al_b$$ is an exact vacuum solution.

So, yes $$C k_ak_b$$ solves linearized equations for $$h_{ab}$$ (and the full metric is an exact solution).

1. Xanthopoulos, B. C. (1978). Exact vacuum solutions of Einstein’s equation from linearized solutions. Journal of Mathematical Physics, 19(7), 1607-1609, doi:10.1063/1.523851.