Black Hole Trajectories

Just for fun, let’s plot some particle trajectories of the Schwarzschild solution of the Einstein Field Equations, a.k.a a black hole! An easy starting point is the geodesic equation, $$\frac{ d^2 x^r}{ds^2} + \Gamma^r_{mn}\frac{dx^m} {ds} \frac{ dx^n} {ds} = 0.$$ Given a point and a direction, the geodesic equation tells how a particle will move through spacetime. This form actually turns out to be a very poor choice since there’s a singularity at the event horizon, but we’ll deal with it. (Mind-expanding fact: as seen by a distant observer, nothing crosses the event horizon in a finite amount of time—finite, like ever—but if it’s you falling into the black hole, you most certainly get to the center in a very short amount of time!) Anyway, to the best of my knowledge, transforming the equations into proper time avoids this mathematical singularity (though the physical weirdness persists), but this form is easy, so we’ll go with it.

With no further ado, here’s the result:

interactive-frame[width=100% height=600](/2015/03/11/black-hole-trajectory-demo/)

Click and drag to move launch point, option-click to change launch direction. Ctrl-click to zoom, shift-click to pan.

The mathematical details

To calculate this, I started with the geodesic equation above. \(x^r\) is the four-dimensional spacetime position, so with some initial conditions, it’s just a differential equation we can integrate. That means we need \(\Gamma^r_{mn}\), which as far as we’re concerned, are just a bunch of coefficients describing the curvature of spacetime. Coefficients in hand, the problem is reduced to a system of ordinary differential equations. Using the spherical space-time coordinate system,

$$\begin{eqnarray}y^0 &:=& x^0 = r \\ y^1 &:=& x^1 = \theta \\ y^2 &:=& x^2 = \phi \\ y^3 &:=& x^3 = t, \end{eqnarray}$$

and choosing some nice \(y\) variables for integration, we’re left with our eight first order, coupled, nonlinear ordinary differential equations:

$$\left\{ \begin{eqnarray} \frac{dy^0}{ds} &=& y^4 \\ \frac{dy^1}{ds} &=& y^5 \\ \frac{dy^2}{ds} &=& y^6 \\ \frac{dy^3}{ds} &=& y^7 \\ \frac{dy^4}{ds} &=& \frac{r_s}{2r(r-r_s)}(y^4)^2 + (r-r_s)(y^5)^2 + \\ && (r-r_s)\sin^2\theta(y^6)^2 + \frac{c^2r_s(r_s-r)}{2r^3}(y^7)^2 \\ \frac{dy^5}{ds} &=& -\frac{2}{r} y^4y^5 + \sin\theta\cos\theta (y^6)^2 \\ \frac{dy^6}{ds} &=& -\frac{2}{r}y^4y^6 - 2cot\theta y^5y^6 \\ \frac{dy^7}{ds} &=& \frac{r_s}{r(r_s-r)}y^4y^7.\end{eqnarray} \right.$$

One last note:

I’m plotting a two-dimensional projection, \((x,y)\), of the four-dimensional space-time coordinates, \((x,y,z,t)\). As long as you think of these coordinates as labels only and agree not to regard distances on the screen as meaningful, there aren’t any serious problems. I really can’t state it much better than my Tensor Calculus book:

“It is most important to note that Riemannian geometry is built up on the concept of the distance between two neighboring points, rather than the concept of finite distance. … Nor do we think of Riemannian space as necessarily embedded in a Euclidian space, as we are tempted to do when we discuss the intrinsic geometry of a sphere.”

So basically, the plot is informative, but distances aren’t what they appear!