- #1

JimWhoKnew

- 33

- 11

- Homework Statement
- Must a body at rest in Boyer-Lindquist coordinates move (or not) in Doran coordinates?

- Relevant Equations
- see below

I read this question in the second edition of "Exploring Black Holes" by Wheeler, Taylor & Bertschinger. The book can be freely accessed and downloaded from the author's website.

Chapter 17 deals with Kerr's solution on the equatorial slice ## \theta = 0 ##.

In Doran coordinates, the 2+1 metric is given by $$ d\tau ^2 = dT^2 - [(\frac{r^2}{r^2 + a^2})^{1/2} dr + (\frac{2 M}{r})^{1/2} (dT - a d\Phi)]^2 - (r^2 + a^2) d\Phi ^2 $$ (Eq. 5 p. 17-4)

By the transformation (Eqs. 113-114 p. 17-36) $$ dT = dt + \frac{R \beta}{r H^2} dr $$ $$ d\Phi = d\phi + \frac{\omega R}{r H^2 \beta} dr $$ where ## r ## is the same in both coordinate systems, one gets the metric in Boyer-Lindquist coordinates (Eq. 112 p. 17-35) $$ d\tau ^2 = (1-\frac{2 M}{r})dt^2 + \frac{4 M a}{r}dt d\phi - \frac{dr^2}{H^2} - R^2 d\phi ^2 $$ with $$ R^2 \equiv r^2 + a^2 + \frac{ 2 M a^2}{r} $$ $$ H^2 \equiv \frac{1}{r^2} (r^2 - 2 M r + a^2) $$ $$ \omega \equiv \frac{2 M a}{r R^2} $$ $$ \beta \equiv (\frac{2 M}{r})^{1/2} (\frac{r^2 + a^2}{R^2})^{1/2} $$ (p. 17-27)

(the calculation is straight forward but lengthy; I have carried it out and got the desired result).

Now, on page 17-38 the question is asked:

Show that a stone at rest in Boyer-Lindquist global coordinates (## dr = d\phi = 0 ##) is not at rest in Doran global coordinates; in particular, ## d\Phi \neq 0 ## for that stone.

My problem is that from the transformation relations (presented above), ## dr = d\phi = 0 ## in B-L, trivially transforms into ## dr = d\Phi = 0 ## in Doran, and vice versa. What am I overlooking? How can this (apparent?) contradiction be resolved?

Chapter 17 deals with Kerr's solution on the equatorial slice ## \theta = 0 ##.

In Doran coordinates, the 2+1 metric is given by $$ d\tau ^2 = dT^2 - [(\frac{r^2}{r^2 + a^2})^{1/2} dr + (\frac{2 M}{r})^{1/2} (dT - a d\Phi)]^2 - (r^2 + a^2) d\Phi ^2 $$ (Eq. 5 p. 17-4)

By the transformation (Eqs. 113-114 p. 17-36) $$ dT = dt + \frac{R \beta}{r H^2} dr $$ $$ d\Phi = d\phi + \frac{\omega R}{r H^2 \beta} dr $$ where ## r ## is the same in both coordinate systems, one gets the metric in Boyer-Lindquist coordinates (Eq. 112 p. 17-35) $$ d\tau ^2 = (1-\frac{2 M}{r})dt^2 + \frac{4 M a}{r}dt d\phi - \frac{dr^2}{H^2} - R^2 d\phi ^2 $$ with $$ R^2 \equiv r^2 + a^2 + \frac{ 2 M a^2}{r} $$ $$ H^2 \equiv \frac{1}{r^2} (r^2 - 2 M r + a^2) $$ $$ \omega \equiv \frac{2 M a}{r R^2} $$ $$ \beta \equiv (\frac{2 M}{r})^{1/2} (\frac{r^2 + a^2}{R^2})^{1/2} $$ (p. 17-27)

(the calculation is straight forward but lengthy; I have carried it out and got the desired result).

Now, on page 17-38 the question is asked:

**BL-7. Not “at rest” in both global coordinates**Show that a stone at rest in Boyer-Lindquist global coordinates (## dr = d\phi = 0 ##) is not at rest in Doran global coordinates; in particular, ## d\Phi \neq 0 ## for that stone.

My problem is that from the transformation relations (presented above), ## dr = d\phi = 0 ## in B-L, trivially transforms into ## dr = d\Phi = 0 ## in Doran, and vice versa. What am I overlooking? How can this (apparent?) contradiction be resolved?