# How to obtain initial conditions to image Kerr black hole?

I'm reading Gravitational Lensing by Spinning Black Holes in Astrophysics, and in the Movie Interstellar to make a raytracer code to image Kerr black holes. The paper introduces a Fiducial Observer perpendicular to the time foliations of the Kerr black hole with Boyer-Lindquist coordinates and a Camera with it's own reference system. The relation between this reference system is depicted in this image from the paper:

The paper provides in Apendix A.1 a way to transform initial conditions in the camera's reference system to the Fiducial Observer reference system. This is done as follows:

1. Specify the camera’s location $(r_c, \theta_c, \phi_c)$, and its speed $\beta$ and the components $B_r,B_θ,B_φ$ of its direction of motion relative to the FIDO at its location; and specify the ray’s incoming direction $(\theta_{cs}, \phi_{cs})$ on the camera’s local sky.
2. Compute, in the camera’s proper reference frame, the Cartesian components (The upper figure) of the unit vector $N$ that points in the direction of the incoming ray $$N_x = sin \theta_{cs} cos \phi_{cs} , N_y = sin \theta_{cs} sin \phi_{cs} , N_z = cos \theta_{cs}$$.
3. Using the equations for relativistic aberration, compute the direction of motion of the incoming ray, $n_F$ , as measured by the FIDO in Cartesian coordinates aligned with those of the camera: $$n_{Fy}=\frac{-N_y+\beta}{1- \beta N_y}, n_{Fx}=\frac{-\sqrt{1-\beta^2}N_x}{1- \beta N_y}, n_{Fz}=\frac{-\sqrt{1-\beta^2}N_z}{1-\beta N_y}$$ 4.From these, compute the components of nF on the FIDO’s spherical orthonormal basis: $$n_{Fr}=\frac{B_\phi}{\kappa} n_{Fx}+B_r n_{Fy}+\frac{B_r B_\theta}{\kappa} n_{Fz}$$ $$n_{F\theta}= B_\theta n_{Fy}-\kappa n_{Fz}$$ $$n_{F\phi}=-\frac{- B_r}{\kappa} + B_\phi n_{Fy}+\frac{B_\theta B_\phi}{\kappa} n_{Fz}$$

where $\kappa=\sqrt{1-B_\theta^2}=\sqrt{B_r^2+B_\phi^2}$.

The question is how to obtain the equations in the point 4 (these last equations). I do not know how to relate the FIDO's reference system with the camera's one. I know the relation is in the image, but i do not know how to propperly obtain these equations.

Can you provide a detailed calculation of how to arrive to these last equations?

Al the information missing in this question can be found at the paper.

Thank you!

Ultimately, we are just converting between a spherical coordinate system and a Cartesian one. I'll try to stick to the notation of the paper, with one critical change: components of vectors will be denoted with superscripts, while subscripts will be reserved for identifying which vector from a set is being referenced.

We have a vector $\mathbf{n}$ (aka $\mathbf{n}_F$), and we know its components in Cartesian coordinates $n^x,n^y,n^z$. We want its spherical coordinates $n^\hat{r},n^\hat{\theta},n^\hat{\phi}$. The basis vectors for the Cartesian system are $\mathbf{e}_x,\mathbf{e}_y,\mathbf{e}_z$, and for the spherical system are $\mathbf{e}_\hat{r},\mathbf{e}_\hat{\theta},\mathbf{e}_\hat{\phi}$, where the hats remind us that this is an orthonormal basis rather than the equally common orthogonal coordinate basis.1 We need to express one basis in terms of the other, where we can assume we have the components $B^\hat{r},B^\hat{\theta},B^\hat{\phi}$ of the unit vector $\mathbf{B}$ indicating the camera's direction of motion.2 The precise definition of the basis is given in §2.1, point (iii) of the paper.

The procedure can be done in these steps:

1. Define $\mathbf{e}_y = \mathbf{B}$. Then we can trivially write $\mathbf{e}_y$ in our spherical basis: $$\mathbf{e}_y = B^\hat{r} \mathbf{e}_\hat{r} + B^\hat{\theta} \mathbf{e}_\hat{\theta} + B^\hat{\phi} \mathbf{e}_\hat{\phi}. \tag{1}$$

2. We seek the spherical components of $\mathbf{e}_x$. Since we take $\mathbf{e}_x \in \operatorname{span}(\mathbf{e}_\hat{r},\mathbf{e}_\hat{\phi})$, we know we can write $\mathbf{e}_x = e_x^\hat{r} \mathbf{e}_\hat{r} + e_x^\hat{\phi} \mathbf{e}_\hat{\phi}$. We have the orthogonality and normalization constraints \begin{align} 0 & = \mathbf{e}_y \cdot \mathbf{e}_x = B^\hat{r} e_x^\hat{r} + B^\hat{\phi} e_x^\hat{\phi} \\ 1 & = \mathbf{e}_x \cdot \mathbf{e}_x = (e_x^\hat{r})^2 + (e_x^\hat{\phi})^2. \end{align} Solving this system of equations tells us $$\mathbf{e}_x = \frac{B^\hat{\phi}}{\kappa} \mathbf{e}_\hat{r} - \frac{B^\hat{r}}{\kappa} \mathbf{e}_\hat{\phi}, \tag{2}$$ where an arbitrary sign choice was made.

3. Two orthogonality constraints, one normalization constraint, and another sign choice (that determines the handedness relation between the coordinate systems) determine the three spherical components of $\mathbf{e}_z$. The equations are \begin{align} 0 & = \mathbf{e}_y \cdot \mathbf{e}_z = B^\hat{r} e_z^\hat{r} + B^\hat{\theta} e_z^\hat{\theta} + B^\hat{\phi} e_z^\hat{\phi} \\ 0 & = \mathbf{e}_x \cdot \mathbf{e}_z = \frac{B^\hat{\phi}}{\kappa} e_z^\hat{r} - \frac{B^\hat{r}}{\kappa} e_z^\hat{\phi} \\ 1 & = \mathbf{e}_z \cdot \mathbf{e}_z = (e_z^\hat{r})^2 + (e_z^\hat{\theta})^2 + (e_z^\hat{\phi})^2 \end{align} Solving this system yields $$\mathbf{e}_z = \frac{B^\hat{r}B^\hat{\theta}}{\kappa} \mathbf{e}_\hat{r} - \kappa \mathbf{e}_\hat{\theta} + \frac{B^\hat{\theta}B^\hat{\phi}}{\kappa} \mathbf{e}_\hat{\phi}. \tag{3}$$

4. Finally, we just assemble the coefficients from (1), (2), and (3). For example, $$n^\hat{r} = e_x^\hat{r} n^x + e_y^\hat{r} n^y + e_z^\hat{r} n^z = \frac{B^\hat{\phi}}{\kappa} n^x + B^\hat{r} n^y + \frac{B^\hat{r}B^\hat{\theta}}{\kappa} n^z.$$

1By coordinate basis I mean one in which $\mathbf{e}_r$ scales proportional to $r$, so for instance we have $\lVert\mathbf{e}_r\rVert = r$ instead of $\lVert\mathbf{e}_\hat{r}\rVert = 1$. Such a basis would be convenient if we were looking at vectors originating at different points (as indeed is done when actually integrating the null geodesics), but since we are only considering directional unit vectors at a single point (the camera's location), we don't have much to loose by making our basis orthonormal.

2That this is the direction of motion is not relevant for this derivation.

• Such a fantastic answer!! One question before accept the answer: In point 2, You mean $e_x= e_x^r e_r + e_x^\phi e_\phi$ instead of $e_x= e_x^r e_r = e_x^\phi e_\phi$ ? – Dargor Sep 9 '15 at 18:37
• By the way, I'd love if there was a chance to vote multiple times up this answer! – Dargor Sep 9 '15 at 18:40
• Yes, that was a typo. – user10851 Sep 9 '15 at 19:01