8
$\begingroup$

Imagine two black holes on the x-axis coming together at the origin (not rotating around each other, just falling towards each other).

In which direction would the most intense gravitational waves be emitted when they collided?

Would they be emitted most strongly along the x-axis, namely in the directions the black holes came from?

Or would they radiate in circles on the yz-plane?

Or would they radiate in spherical shells evenly in all directions?

Or would there be no waves at all, since the black holes are not rotating around each other?

$\endgroup$
  • $\begingroup$ “Most intense” is too vague. Are talking about the size of metric perturbation in various directions? If so, which component? Or are you talking about the angular distribution of the radiated energy flux? Or the radiated momentum flux? $\endgroup$ – G. Smith Mar 7 at 23:21
  • 1
    $\begingroup$ @Smith which direction would carry off the most energy. $\endgroup$ – zooby Mar 7 at 23:43
14
$\begingroup$

In a comment you explained that you are interested in the angular distribution of the power radiated as gravitational waves.

When the two black holes are close together, the analysis is probably messy, requiring numerical relativity. But when the separation of the two black holes is considerably greater than their Schwarzschild radii, the nonrelativistic analytical approach in the 1963 paper "Gravitational Radiation from Point Masses in a Keplerian Orbit" by Peters and Matthews is valid.

The angular power distribution is given by equation (5),

$$\frac{dP}{d\Omega}=\frac{G}{8\pi c^5}\left[\dddot{Q}_{ij}\dddot{Q}_{ij}-2n_i\dddot{Q}_{ij}n_k\dddot{Q}_{kj}-\frac12(\dddot{Q}_{ii})^2+\frac12(n_in_j\dddot{Q}_{ij})^2+\dddot{Q}_{ii}n_jn_k\dddot{Q}_{jk}\right],$$

where $\hat n$ is the unit vector in the direction of radiation and

$$Q_{ij}=\sum_n m^{(n)}x^{(n)}_ix^{(n)}_j$$

is the mass quadrupole tensor of the two masses around their center of mass.

Taking the $z$ (not $x$) axis to be along the direction of motion of the two holes, there is only one nonzero component of $Q_{ij}$, namely $Q_{zz}$, and it will be proportional to the square of the separation. Its third time derivative is irrelevant to the question of the angular distribution.

Since only $Q_{zz}$ is nonzero, one finds that only $n_z=\cos\theta$, where $\theta$ is the usual polar angle from the $z$-axis, enters into the contractions. The result is easily found to be

$$\begin{align} \frac{dP}{d\Omega}&=\frac{G}{8\pi c^5}\left[\dddot{Q}_{zz}\dddot{Q}_{zz}-2n_z\dddot{Q}_{zz}n_z\dddot{Q}_{zz}-\frac12(\dddot{Q}_{zz})^2+\frac12(n_zn_z\dddot{Q}_{zz})^2+\dddot{Q}_{zz}n_zn_z\dddot{Q}_{zz}\right]\\ &=\frac{G}{8\pi c^5}(\dddot{Q}_{zz})^2\left[1-2\cos^2\theta-\frac12+\frac12\cos^2\theta+\cos^2\theta\right]\\ &=\frac{G}{16\pi c^5}(\dddot{Q}_{zz})^2\sin^2\theta. \end{align}$$

So no power is radiated along the line of motion; the radiated power is greatest perpendicular to the line of motion; and the radiation pattern is symmetric relative to that perpendicular plane even when the two black holes have different masses.

Addendum: As @mmeent explains in another answer, my last point about the symmetry relative to the perpendicular plane no longer holds when smaller, higher-order multipole contributions are also considered. The calculation here is a leading-order (quadrupole) calculation.

| cite | improve this answer | |
$\endgroup$
9
$\begingroup$

Much of this question can be answered qualitatively based on basic symmetry arguments without doing any calculation or resorting to any approximation.

The key observation here is a system consisting of two (non-spinning) black holes on a trajectory for a head-on collision is symmetric around the axis of motion. This implies that the resulting gravitational waves must have the same symmetry.

Now, gravitational waves have the interesting property that any axisymmetric (and smooth) configuration of gravitational waves must vanish along the symmetry axis. Ultimately, this is a consequence of the the gravitational waves being described by a spin-2 field. Consequently, there is no gravitational radiation emitted along the symmetry axis.

A secondary consequence is that the polarization of all gravitational waves must be aligned with the symmetry axis.

If both black holes have the same mass, then the system has an additional symmetry (when viewed in the center of mass frame): exchange of the two black holes, or equivalently mirroring of the system across the plane perpendicular to symmetry axis and containing the center of mass. Since the emitted gravitational must satisfy this symmetry, the distribution of gravitational waves must be symmetric around this central plane. In particular, the net linear momentum carried away by the gravitational waves must be zero.

If the masses are unequal the system is asymmetric around the central plane, and in general there will be more gravitational waves emitted in one direction than the other (although still none exactly along the symmetry axis!). The result is a net emission of linear momentum, and consequently the black hole formed from the merger will have some linear momentum in the original center of mass frame.

Of course, to obtain a quantitative result one must actually do some calculation. One can do the weak field approximation as in G. Smith's answer (although do to the quadrupole approximation used this missed the asymmetry for unequal masses). Due to the high degree of symmetry this system has been the subject of some of the earliest investigations in numerical relativity. (E.g. the first equal mass simulation gr-qc/9309016, or the first unequal mass simulation gr-qc/9806031) Alternatively, one can study head on collisions in the small mass ratio limit (gr-qc/9609012).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "and in general there will be more gravitational waves emitted in direction" You're saying that there's no emission in the actual directions of the symmetry axis, so presumably you're saying something along the lines of that if we allow the plane perpendicular to this axis to divide space into two hemispheres, and look at the total x-components of all the emissions for the two hemispheres, they are unequal. I don't know a simple way of saying that, though. $\endgroup$ – Acccumulation Mar 8 at 18:36
  • $\begingroup$ There is a typo : it should read no gravitational radiation emitted along the symmetry axis. Regards $\endgroup$ – chris Mar 8 at 19:23

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.