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}

  • $\begingroup$ 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 $\endgroup$
    – mike stone
    Mar 12, 2021 at 14:17
  • $\begingroup$ You can always construct a local reference frame around any observer which is locally flat. $\endgroup$
    – Tom
    Mar 12, 2021 at 19:27
  • $\begingroup$ Only to quadratic order. The Christoffle symbols vanish to linear order. physics.stackexchange.com/questions/392521/… $\endgroup$
    – mike stone
    Mar 12, 2021 at 23:07

1 Answer 1


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


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