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I am working with the Kerr metric (suppose it is $g_{\mu \nu}$) where I want to understand how the Teukolsky master equation is derived using the Newman-Penrose formalism. The original article can be found here. Given the Kerr metric, he considers a tetrad of four vectors given by $ l^{\mu}, n^{\mu}, m^{\mu} \text{ and } \bar{m^\mu}$ where the bar represents complex conjugation. Consider the explicit form of $l^\mu$ in the coordinate basis (as I understand it).

$$ l^{\mu} = [(r^2 + a^2)/\Delta, 1, 0, a/\Delta] $$

where $\Delta = r^2 - 2Mr + a^2$. These are supposed to form null vectors and hence when I transform them to the tetrad basis, I should have a null tetrad set. Taking the conditions on l,n,m one can show that the metric to lower tetrad indices can be given by:

$$\eta _{ab} = \left[\matrix{0 &1 &0 &0 \\ 1 &0 &0 & 0 \\ 0 &0 & 0&-1 \\ 0&0 &-1 &0} \right]$$

For the rest of the question I take greek indices to mean coordinate indices and roman indices to mean tetrad indices. I want to find out the components of the vector l in the tetrad basis. I guess this can be given as $$ l^a = E_\mu ^{ \text{ }a}\text{ }l^{\mu} $$ where $E_{\mu}^{\text{ } a}$ represents the transformation from the coordinate to the tetrad basis.

  1. How do I find out this $E_\mu ^{ \text{ }a}$?

I know using the knowledge of the metric, I can write $$\eta _{ab} = E_{a}^{\text{ }\mu} E_{b}^{\text{ }\nu} g_{\mu \nu} $$

and since I know both $\eta_{ab}$ and $g_{\mu \nu}$ I can in principle, find the transformation matrix (at least modulo coordinate transformations). However, I am unable to find out this matrix.

  1. I want to explicitly show that $l_a l^a = 0$ (tetrad basis inner product - lowering is done using $\eta$)

This is one of the conditions for choosing the Newman Penrose tetrad vectors. I have also tried to compute $l_\mu l^{\mu}$ (coordinate basis inner product - lowering done using $g$) but I am not getting this to be zero. Shouldn't this also be zero, since the norm of the vector must be invariant?

EDIT: Had a typo in one of the equations.

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  • $\begingroup$ I don't understand the equation $l_a=E_\mu^a l^\mu$. The indices don't match. And you have a tetrad vector on both sides! $\endgroup$
    – MBN
    Commented Dec 18, 2021 at 19:12
  • $\begingroup$ @MBN Yeah the index was a typo sorry. Yes I am considering the same vector in the coordinate basis and in the tetrad basis. I mentioned that roman indices like $a,b,c$ stand for tetrad indices and greek indices like $\mu, \nu$ stand for coordinate indices. I want to find the components of the same vector in both bases. $\endgroup$
    – newtothis
    Commented Dec 19, 2021 at 5:10
  • $\begingroup$ So $E^a_{\mu}$ represents transformation from complex orthonormal null tetrads to "usual" tetrad, right? Which means $E^{\mu}_aE^{\nu}_bg_{\mu\nu}=\eta_{ab}$ is just the Minkowski metric.The expression for this transformation matrix for any axis symmetric space time is given in chapter 2 of "The mathematical theory of black holes" by S. Chandrashekhar. $\endgroup$
    – paul230_x
    Commented Dec 19, 2021 at 6:28
  • $\begingroup$ In this case, $\eta_{ab}$ will not be Minkowski, since I am not choosing the local frame to be Minkowski (although it is still constant). The normalization and the orthogonality properties from the definition of the NP tetrads ensure that the "tetrad" metric is the one given in my question. So $E_{\mu}^{\text{ }a}$ represents this transformation, and not to Minkowski. $\endgroup$
    – newtothis
    Commented Dec 19, 2021 at 6:50
  • $\begingroup$ In the book by Chandrasekhar as well, $\eta_{ab}$ is given in this way, but he does not explicitly derive (or mention) $E_{\mu}^{\text{ }a}$ $\endgroup$
    – newtothis
    Commented Dec 19, 2021 at 6:51

1 Answer 1

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I think you are slightly confused about the overall picture. Let's review it again:

All is about the construction of suitable frames. By "frame", I mean a set of four basis vectors at each point of spacetime. Given a coordinate system over the spacetime, we already have one: the "coordinate basis" given by partial derivatives $\partial_\mu$ along each coordinate (e.g. $\{\partial_t,\partial_r,\partial_\theta,\partial_\phi \}$). However, the problem with this frame is that it is not orthonormal in a curved spacetime, since $\partial_\mu\cdot \partial_\nu=g_{\mu\nu}$. For various reasons, we prefer to work with orthonormal frames. Therefore we transform from the coordinate basis $\partial_\mu$ to another set of four basis vectors $e_{(a)}=e_{(a)}{}^\mu\partial_\mu$ for $a=\{0,1,2,3\}$ such that \begin{align} e_{(a)}\cdot e_{(b)}=\eta_{(a)(b)} \end{align} I put parantheses around "$a,b$" indices to stress that they are "label indices" to be distinguished from coordinate indices $\mu$. For a Rimannian geometry it is natural to take $\eta_{(a)(b)}=\text{diag}(1,1,1,1)$, which implies that the four vectors are orthonormal, i.e. of unit norm and orthogonal to each other. For a Lorentzian spacetime, we instead typically consider $\eta_{(a)(b)}=\text{diag}(-1,1,1,1)$ so that $e_{(0)}$ is timelike while the other three are spacelike.

However, in some situations (including black hole perturbation theory), we prefer to work with null basis vectors, i.e. $|e_{(0)}|^2=e_{(0)}\cdot e_{(0)}=0$ and similarly for the other three basis vectors. To form a complete basis, then one typically takes $e_{(0)}\cdot e_{(1)}=-1, e_{(2)}\cdot e_{(3)}=1$ and the rest of the inner products to be zero. It is also conventional to give these basis vectors separate names: $e_{(0)}^\mu=l^\mu,e_{(1)}^\mu=n^\mu,e_{(2)}^\mu=m^\mu,e_{(3)}^\mu=\overline{m}^{\mu}$, where overline means complex conjugation. Therefore the matrix $\eta_{(a)(b)}$ looks like what you wrote above [except that you have replaced (0,1) with (2,3)]. In Boyer-Lindquist coordinates for Kerr spacetime, a choice for the basis vectors is \begin{aligned} l^{\mu} &=\frac{1}{\Delta}\left(r^{2}+a^{2}, \Delta, 0, a\right), \\ q^{\mu} &=\frac{1}{2 \rho^{2}}\left(r^{2}+a^{2},-\Delta, 0, a\right), \\ m^{\mu} &=\frac{1}{\sqrt{2}} \frac{1}{r+i a \cos \theta}\left(i a \sin \theta, 0,1, \frac{i}{\sin \theta}\right), \end{aligned} and $e_{(a)}{}^\mu$ is a 4$\times$4 matrix whose columns are the basis vectors, i.e. $e_{(a)}{}^\mu=\Big(l^\mu\,|\,n^\mu\,|\,m^{\mu}\,|\,\overline{m}^{\mu}\Big)$.

The inverse of $e_{(a)}{}^\mu$ is instead a set of four one-forms $e^{(a)}=e^{(a)}{}_\mu dx^\mu$ whose components are given by $$e^{(a)}{}_\mu=\eta^{(a)(b)}g_{\mu\nu}e_{(b)}{}^\nu$$

From this we see that as a column matrix $e^{(a)}{}_\mu=\Big(-n_\mu\,|\,-l_\mu\,|\,\overline{m}_{\mu}\,|\,{m}_{\mu}\Big)$

Now it is very clear that $e^{(a)}{}_\mu l^\mu=(-n_\mu l^\mu, -l_\mu l^\mu,\overline{m}_\mu l^\mu,m_\mu l^\mu,)=(1,0,0,0)$. So as you see $l^{(a)}=e^{(a)}{}_\mu l^\mu$ the vector $l$ is the zeroth basis vector and hence contains no particularly useful information. Also obviously $l^{(a)}l_{(a)}=\eta_{(a)(b)}l^{(a)}l^{(b)}=\eta_{(0)(0)}=0$. this is another way to say that $l^\mu$ is by construction a null vector.

I hope this helps.

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  • $\begingroup$ Thank you so much! $\endgroup$
    – newtothis
    Commented Dec 29, 2021 at 6:48

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