Extrinsic curvature of Kerr-Schild using ADM equations

[ergospherical]

Homework Statement

Exercise 2.33 of Shapiro & Baumgarte. In Kerr-Schild coords, $ds^2 = (\eta_{ab} + 2H l_a l_b) dx^a dx^b$ where the null vector $l_a$ has Cartesian components $l_t = 1$ and $l_i = x^i/r$ and $H := M/r$. Identify the lapse, shift, spatial metric and show that the extrinsic curvature is$$K_{ij} = \frac{2H\alpha}{r}(\delta_{ij} - (2+H)l_i l_j)$$

Relevant Equations

ADM equations

I can think of a couple of ways to go about determining the extrinsic curvature, but the most direct seems to be straight from the ADM equation for the evolution of the spatial metric,$$\partial_t \gamma_{ij} = \beta^m \partial_m \gamma_{ij} + \gamma_{m(i} \partial_{j)} \beta^m - 2\alpha K_{ij}$$From the form of the metric in coordinates adapted to the 3+1 split, it's easy to write down the lapse $\alpha = 1$, shift $\beta^i = 2Hx^i/r$ and the spatial metric $\gamma_{ij} = \delta_{ij} + 2H x^i x^j/r^2$. I find the following terms:\begin{align*}
\beta^m \partial_m \gamma_{ij} &= -4H^2 l_i l_j \\
\gamma_{mi} \partial_j \beta^m &= \frac{2H}{r}(\delta_{ij} - 2(2+H)l_i l_j)
\end{align*}The spatial metric being time independent means that$$2\alpha K_{ij} = \beta^m \partial_m \gamma_{ij} + \gamma_{m(i} \partial_{j)} \beta^m$$The expressions computed above don't give the quoted result for $K_{ij}$ (as far as I can tell). Can anyone spot my error?

 

[JimWhoKnew]

I can't really address your question without refreshing my memory on ADM, which currently I don't have the time for. But reading it, I (think I) can see some things:

$\gamma_{ij} = \delta_{ij} + 2H x^i x^j/r^2$

Seriously?

$x_i=g_{i \nu}x^\nu$ . Did you remember to account for that in the differentiation $\frac{\partial x_i}{\partial x^j}$ ?

$\beta^m \partial_m \gamma_{ij} = -4H^2 l_i l_j$

It appears as not having the same units as the line below and the textbook's solution.

One more: even if $\alpha = 1$, its appearance once as proportional to $K_{i j}$ and once as inverse, seems odd (superficially).

 

[ergospherical]

Because $l^a$ is a null vector (with respect to both $\eta$ and $g$), then you can check that $l_i = l^i = x^i/r$. So $g_{ij} = \delta_{ij} + 2H l_i l_j = \delta_{ij} + 2H x^i x^j/r^2$, with indices up in the last term...

 

[JimWhoKnew]

I looked it in the books. You are right.

If $\beta$, $H$ and $\gamma_{ij}$ are unitless (as you presented them), then $\beta^m \partial_m \gamma_{ij}$ should be of units 1/length

 

[ergospherical]

Yeah, the dimensions aren't right. There must be a sloppy mistake somewhere in my work -- including possibly some missed raised/lowered indices. I'll look tomorrow!

 

[JimWhoKnew]

You might find table 2.1 on page 50 interesting

 

[ergospherical]

You might find table 2.1 on page 50 interesting

For sure! So I have the completely wrong lapse (and contra-variant shift). They get, $\alpha^2 = (1 + \tfrac{2M}{r})^{-1}$ and $\beta^i = \tfrac{2M}{r} \alpha^2 l^i$. I've checked it's indeed consistent with the general form of the metric in ADM-adapted coordinates, e.g. \begin{align*}
\beta_i = \gamma_{ij} \beta^j &= [\delta_{ij} + \tfrac{2M}{r}l_i l_j] \tfrac{2M}{r} \alpha^2 l^j \\
&= \tfrac{2M}{r} \alpha^2 l_i \left( 1 + \tfrac{2M}{r} l^j l_j \right) \\
&= \tfrac{2M}{r} l_i \\
&= g_{0i}
\end{align*}I guess it's not supposed to be obvious that this is the correct decomposition, which is why they've given it in the table.