# Jacobi Matrix between Cartesian and Schwarzschild coordinates

Let $$\mathcal P$$ be a photon at position $$\vec x =(x,y,z)$$ with 3-velocity $$\vec v=(v_x,v_y,v_z)$$, where both are given in local Cartesian coordinates. I want to follow the photons geodesic by numerically solving the geodesic equations, which can be written in 3+1 form as,

\begin{align} \frac{dx^i}{d t} &= \frac{1}{p^0}\frac{dx^i}{d\lambda} = \gamma^{ij}\frac{p_j}{p^0}-\beta^i,\\ \frac{dp_i}{d t} &= \frac{1}{p^0}\frac{dp_i}{d\lambda} = -\alpha p^0\partial_i\alpha+p_k\partial_i\beta^k -\frac{1}{2}\partial_i\gamma^{lm}\frac{p_l p_m}{p^0},\\ \frac{dt}{d\lambda} &= p^0 = \frac{1}{\alpha}\sqrt{\gamma^{ij} p_i p_j}. \end{align}

I extract the quantities $$\alpha,\beta^i,\gamma^{ij}$$ from the Schwarzschild metric and use a Runge-Kutta 4 algorithm to integrate the ODE's.

From my given Cartesian starting values $$\vec x$$ and $$\vec v$$ I can calculate the starting values for the differential equations $$x^\bar i$$ and $$p_\bar i$$, where the bared indices stand for Cartesian coordinates.

Question: What are the coordinate transformations and how do I determine the Jacobi matrices $$\Lambda^i_{\ \ \bar i}$$ and $$\Lambda_i^{\ \ \bar i}$$ to transform my Cartesian starting values to Schwarzschild coordinates?

\begin{align} \begin{array}{l} r=r(x,y,z) \\ \theta = \theta(r,y,z)\\ \phi = \phi(x,y,z) \end{array} ,\qquad p_i = \Lambda_i^{\ \ \bar i} p_\bar i \end{align}

On first glance Schwarzschild coordinates look like spherical polar coordinates, but if i transform them accordingly and calculate the norm of my velocity vector with the 3-metric of the Schwarzschild spacetime, the norm is not preserved,

\begin{align} |\vec v| = \sqrt{v_x^2+v_y^2+v_z^2} = 1 \neq \sqrt{\gamma_{ij}v^i v^j} \end{align}

• I don't see how you can just "change coordinates". That only works when the spaces have the same geometry, but Cartesian coordinates are for flat space and Schwarzschild implies curved Mar 12, 2021 at 14:17
• You can always construct a local reference frame around any observer which is locally flat.
– Tom
Mar 12, 2021 at 19:27
• Only to quadratic order. The Christoffle symbols vanish to linear order. physics.stackexchange.com/questions/392521/… Mar 12, 2021 at 23:07

The transformation of a vector from local Cartesian coordinates to Schwarzschild coordinates can be done in two steps.

Step 1: Transform the Cartesian vector to spherical coordinates with the Jacobian,

\begin{align} v^\hat i = \Lambda^\hat i_{\ \ \bar i} v^\bar i. \end{align}

Indices with a bar and hat correspond to Cartesian and spherical coordinates respecitvely.

Step 2: Transform the spherical coordinates to Schwarzschild coordinates. The corresponding Jacobian can be extracted from the transformation law of any tensor which we know in both systems,

\begin{align} \gamma_{ij} = \Lambda^\hat i_{\ \ i}\Lambda^\hat j_{\ \ j} \gamma_{\hat i\hat j}. \end{align}

Here I use the 3-metric which we know in both coordinate frames. Unadorned indices represent tensors in the Schwarzschild spacetime. In general the above equation yields a system of 9 quadratic equations which can be difficult to solve. However, In the case of spherical and Schwarzschild coordinates it simplifies to 3 quadratic equation, due to both 3-metrics being diagonal,

\begin{align} \gamma_{11} &= \Lambda^\hat 1_{\ \ 1}\Lambda^\hat 1_{\ \ 1} \gamma_{\hat 1\hat 1}, \quad \gamma_{22} = \Lambda^\hat 2_{\ \ 2}\Lambda^\hat 2_{\ \ 2} \gamma_{\hat 2\hat 2}, \quad \gamma_{33} = \Lambda^\hat 3_{\ \ 3}\Lambda^\hat 3_{\ \ 3} \gamma_{\hat 3\hat 3},\\ (1-\frac{2M}{r})^{-1} &= \Lambda^\hat 1_{\ \ 1}\Lambda^\hat 1_{\ \ 1} 1, \qquad r^2 = \Lambda^\hat 2_{\ \ 2}\Lambda^\hat 2_{\ \ 2} r^2, \qquad r^2\sin^2(\theta) = \Lambda^\hat 3_{\ \ 3}\Lambda^\hat 3_{\ \ 3} r^2\sin^2(\theta). \end{align}

Thus, we get the following Jacobian,

\begin{align} J = \Lambda^\hat i_{\ \ i} = \left( \begin{array}{ccc} (1-\frac{2M}{r})^{-1/2} & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array} \right). \end{align}

Note that this Jacobian transforms contravariant Schwarzschild tensors to contravariant spherical tensors or covariant spherical tensors to covariant Schwarzschild tensors,

\begin{align} v^\hat i = \Lambda^\hat i_{\ \ i} v^i\qquad v_i = \Lambda^\hat i_{\ \ i} v_\hat i. \end{align}

In order to transform a contravariant spherical tensor to Schwarzschild coordinates we have to use the inverse Jacobian transformation,

\begin{align} v_\hat i = \Lambda_\hat i^{\ \ i} v_i\qquad v^i = \Lambda_\hat i^{\ \ i} v^\hat i. \end{align}

with $$\Lambda_\hat i^{\ \ i} = (\Lambda^\hat i_{\ \ i})^{-1} = J^{-1}$$.