2
$\begingroup$

I drop a test-particle in the Schwarzschild geometry, at an initial radial coordinate $r_0 > a \equiv 2 G M$ ($a$ is the Schwarzschild "radius"), with initial velocity $v_0 < 1$ relative to the local stationary observer (I'm using units such that $c \equiv 1$). To simplify the analysis, the initial velocity is perpendicular to the radial coordinate (i.e. in the azimutal $\varphi$ direction). The free particle will follow a timelike geodesic of the spacetime geometry.

There are two constants of motion: $\mathcal{K} = (1 - a/r) \, \dot{t}$ (energy per unit mass) and $\mathcal{J} = r^2 \, \dot{\varphi}$ (angular momentum per unit mass). The dot represents the derivative on the particle's proper time. We could prove the following two relations (the same constants of motion, evaluated at the starting point and formulated as functions of the initial conditions): \begin{align} \mathcal{K}(r_0, v_0) &= \sqrt{\frac{1 - a / r_0}{1 - v_0^2}}, \tag{1} \\[2ex] \mathcal{J}(r_0, v_0) &= \frac{r_0 \, v_0}{\sqrt{1 - v_0^2}}. \tag{2} \end{align} The Schwarzschild metric gives the following equation, from the normalized four-velocity evaluated at the starting point (where $\dot{r} = 0$): \begin{equation} \mathcal{K}^2 = \dot{r}^2 + \Bigl( 1 - \frac{a}{r} \Bigr) \Bigl( 1 + \frac{\mathcal{J}^2}{r^2} \Bigr) \bigg|_{r \,=\, r_0, ~ \dot{r} \,=\, 0}. \tag{3} \end{equation} Since $\mathcal{K}$ is a constant of motion, this equation let us find the minimal and maximal radial coordinates (where $\dot{r} = 0$), for any confined trajectory (typically some kind of elliptical orbit, with a strong relativistic perihelion advance. I'm not interested in the capture trajectories). Since $\dot{r}^2 = 0$ at the turning points, we could recast (3) as this: \begin{equation*} \mathcal{K}^2 \, r^3 - (r - a)(r^2 + \mathcal{J}^2) = 0. \tag{4} \end{equation*} We could substitute (1) and (2) into this third order equation. With some tedious algebric manipulations, we get the following equivalent equation: \begin{equation} (r - r_0) \Bigl( (\, v_0^2 - a / r_0) \, r^2 + (r_0 - a) \, v_0^2 \: r - a \, v_0^2 \: r_0 \Bigr) = 0. \tag{5} \end{equation} If $v_0 \, (r_0 + a) \ge 2 a$ and $v_0^2 \: r_0 < a$, this equation admits three real and positive solutions: the obviously trivial $r = r_0$ solution (the starting point, since the initial velocity is already perpendicular to the radial coordinate), and two distinct real non-trivial solutions: \begin{equation} r_{\pm} = v_0 \, r_0 \: \frac{v_0 \, (r_0 - a) \pm \sqrt{v_0^2 \, (r_0 + a)^2 - 4 \, a^2}}{2 (a - v_0^2 \: r_0)}. \tag{6} \end{equation} Using Mathematica, I checked that (6) are indeed two solutions of equations (3)-(4).

Define $v_{\text{c}}$ as the velocity that gives a perfectly circular motion. Then it is clear that $r_{\text{max}} = r_0$ if $v_0 < v_{\text{c}}$, and $r_{\text{min}} = r_0$ if $v_0 > v_{\text{c}}$. One of the two solutions (6) gives $r_{\text{min}}$ if $v_0 < v_{\text{c}}$, or $r_{\text{max}}$ if $v_0 > v_{\text{c}}$.

But then, why is there another mathematical solution for the extremum of $r$? Why equation (3) is third order? I should only get $r_{\text{min},\, \text{max}} = r_0$ and $r_{\text{max},\, \text{min}}$ as the second extremum point. I don't understand the third extremum.

As far as I know (I may be wrong), the rosace-like trajectories don't have a third extremum point, such that $\dot{r} = 0$ (somewhere between $r_{\text{min}}$ and $r_{\text{max}}$). So what is the meaning of the third solution given by (6)?

$\endgroup$
3
  • $\begingroup$ Have you tried plotting the roots as a function of $v_0$? I suspect there's some algebra error, because looking at (3) as an effective potential equation, it's easy to see that it has one negative solution. $\endgroup$
    – Javier
    Nov 20 '20 at 1:07
  • $\begingroup$ @Javier, how do you see a negative solution? For the exterior trajectories, we have $r > a > 0$. I don't see any algebra errors in my equations, especially since they have the proper non-relativistic limits and other special cases. Also, (6) depends on both initial conditions, not just $v_0$. $\endgroup$
    – Cham
    Nov 20 '20 at 1:24
  • $\begingroup$ @Javier, using Mathematica to do the painful algebra, I just rechecked that (6) are indeed two solutions of equation (3), using (1) and (2) as constants. The third solution is the trivial one: $r = r_0$. It's easy to check that (6) are both real and positive when the conditions after (5) are satisfied (which give rosace-like trajectories). $\endgroup$
    – Cham
    Nov 20 '20 at 1:59
2
$\begingroup$

For simplicity let's set $a=1$, and let's choose $r_0 = 10$ and $v_0 = 0.25$, which satisfy the required conditions. The constants are $\mathcal{K} = 0.979$ and $\mathcal{J} = 2.582$, and the three radial turning points are

$$\{1.208, 10, 13.792\}.$$

If we define the effective potential as the right hand side of equation (3), with $\dot{r}=0$, its plot looks something like this:

effective potential

The horizontal line is $\mathcal{K}^2$; you can barely see the three crossings, corresponding to the three turning points.

Clearly, the largest two are the boundaries of the trajectory we want. The third (and smallest) turning point is actually the maximum radius of a different trajectory, one that falls into the event horizon. It shows up here because it has the same conserved quantities as our original trajectory.

Mathematically, the formulas giving $\mathcal{K}$ and $\mathcal{J}$ in terms of $r_0$ and $v_0$ are not bijective: different pairs $(r_0,v_0)$ can have the same conserved quantities. You can check that taking $r_0 = 1.208$ and $v_0 = 0.906$ gives the same conserved quantities as taking $r_0 = 10$ and $v_0 = 0.25$.

This doesn't happen in the Newtonian problem because there the centrifugal potential dominates as $r \to 0$, so that every nonzero angular momentum has a corresponding minimum radius (the third root is still there, but it's negative). But the gravity of black hole is stronger, and as you can see in the plot above, it brings the potential back down to zero. For every angular momentum, no matter how large, if you place the particle sufficiently close to the black hole, it will get captured. The centrifugal potential still provides a barrier, though: the particle cannot cross from one region to the other.

$\endgroup$
1
  • $\begingroup$ Wow! I've checked your values and everything appears to be fine. Thanks a lot, I think it's clear. I 'll mark your answer in a day or two, in case someone else find another nice way to explain this, but I doubt it! ;-) $\endgroup$
    – Cham
    Nov 20 '20 at 3:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.