Gravitational wave luminosity and energy transfer to masses in vicinity of merging mid-sized black holes

Question

Would the energy transmitted by gravitational waves to stars or planets in the vicinity of a mid-sized black hole merger be enough to disrupt those bodies? If so, how far away would the gravitational wave disruptions be lethal?

Answer 1

Suppose you have two black holes of the same mass $M$ and $m = GM/c^2$. The radius of each black hole is then $r = 2m$, and the horizon area is $A = 4\pi r^2$ $ = 16\pi m^2$. Two constraints are imposed. The first is that the type-D solutions have timelike Killing vectors, which are isometries that conserve mass-energy, and with the merger the gravitational radiation is in an asymptotically flat region where we can again localize mass-energy. So the initial mass $2M$ is the total energy. The entropy of the two black holes is a measure of the information they contain and that too is constant. So the horizon area of the resulting black hole is the sum of the two horizon areas, $A_f = 2A$ $ = 32\pi m^2$, that has $\sqrt{2}M$ the mass of the two initial black holes. Now with mass-energy conservation $$ E_t = 2M = \sqrt{2}M + E_{g-wave} $$ and the mass-energy of the gravitational radiation is $.59M$. That is a lot of mass-energy!

However, this is an idealization. I am assuming the area of the coalesced black hole is the same as the initial black holes. We know however that only about 5% of the mass of two black holes goes into gravitational radiation. This is about $1/6$ times the initial black hole masses. This is because in the near field situation a lot of curvature near the just coalesced black hole goes back into the black hole. So we then expect that maybe $.1M$ in gravitational radiation.

Does this mass-energy in the gravity wave demolish planets? The Einstein field equation is $G_{ab} = (16\pi G/c^4)T_{ab}$, where I am going to as a back of envelope calculation consider the gravitational wave's matter interaction as just its energy density. The $T_{ab}$ then pertains to the interaction of the gravitational wave with a set of masses, and the mass-energy of the gravitational radiation is absorbed by these masses. Let us focus in on the $T^{00} = \rho$ or the mass-energy density. To get this density was consider this mass-energy in the form of a gravity wave in a volume $V = (4\pi/3)r^3$. The $G_{00}$ curvature term is then $$ G_{00} = \frac{16\pi G}{c^4}Mc^2/V = 4.1\times 10^{-43}N^{-1}\times .1M\times 9.0\times 10^{16}m^2/s^2/V, $$ where I am now going to assume $M = 10M_{sol}$ $ = 2\times 10^{31}kg$ $$ G_{00} = 7.4\times 10^{5}m/V $$ Now assume you are $1\times 10^{9}$m away. The curvature is then about $1.8\times 10^{-22}m^{-2}$.

How much gravity would I expect form this? The Riemann curvature for gravitation at the surface of the Earth is $R = GM/c^2r^3$ or $$ R = \frac{6.7\times 10^{-11}Nm^2/kg^2\times 6\times 10^{24}kg}{9\times 10^{16}m^2/s^2\times (6.4\times 10^{6}m)^3} = 1.7\times 10^{-23}m^{-2}. $$ Thus if you were about a million kilometers from the coalescence of two black holes the curvature induced would be comparable to the gravitational curvature here on Earth.

This sounds a bit surprising, for if $.1M$, or about a solar mass, amount of mass energy is generated by the collision of two black holes, then that would seem to imply a huge amount of local violence. It is that coupling term $\frac{16\pi G}{c^4}$ being so small that makes the gravitational effect so small. It is why detecting gravitational radiation from many light years away is so difficult. As a result the energy deposited by gravitational radiation onto a nearby star or stellar system of planets is relatively small. Gravitation is a very weak force, far weaker than the weak interactions, and as a result the graviton is a bit of a boson version of the neutrino. It is hard to catch them, and it took a long time before the LIGO was determined to actually get a gravitational wave.

Answer 2

It is a good question. But it probably is not the most likely way that merging black holes (BHs) affect nearby stars unless this is happening in high density regions of galaxies. Gravitational waves could have a disruptive effect on the Earth and on human bodies but we'd have to be at lunar distances or less away from the BH, and somehow have survived all the pseudostatic tidal effects. Stars would have been disrupted (with some exceptions, see the first link below for a white dwarf getting eaten by a BH at lunar distances), for the most part before/further away.

First, near galactic centers, as galaxies are getting formed and then in their youth, there is a large amount of star creation, while simultaneously the galactic center is forming and growing what eventually becomes a supermassive BH. The accretion of matter into the supermassive BH happens as it devours matter around it. Matter near it without enough angular momentum to escape gets pulled in, and stars or other objects near it will get broken up by the tidal forces of the strong gravity there and go into an accretion disk, which is then accreted into the BH. As that gets accreted jets of microwave and X Rays (and some particles also) will be expelling those. We see those as quasars, and X Ray sources.

So the principal effect of BHs are really their huge tidal forces which disrupt most everything nearby, and can cause a huge amount of energy (and mass) to be radiated. See for instance a close in star being eaten by a BH, in the closest orbit seen so far (about twice the distance from the earth to the moon) https://www.nasa.gov/mission_pages/chandra/news/star-discovered-in-closest-known-orbit-around-likely-black-hole.html

So in most cases merging BH may have some other matter or stars nearby, but much of it may have been absorbed already. But it is possible that there is other matter around, it just depends on how close. If close enough it would have already been swallowed up

So for some numbers, from the CalTech write up on calculating the numbers at http://www.tapir.caltech.edu/~teviet/Waves/gwave.html. It shows the strain equations. Note that the strain is proportional to them released mass, and inversely proportional to the distance away. See the equations for h (the strain) or g' (the tidal force). Both go like 1/r. This is the quantity that determines whether there is tidal disruption or not. It's important because if the radiative effect (in some measurable way), went as 1/$r^n$ with n higher than 1 it would be very much harder to detect. The time varying quadrupole moment, which is the radiative source of a gravitational field, determines, along with the distance away, h. In LIGO h detected was about $10^{-22}$, which for the size of the LIGO legs is about 1/100th the diameter of a proton.

ADDED NOTE PER REQUEST: FROM A COMMENT BY @COUNTTO10 BELOW ADDING A COUPLE OF TERMS CLARIFICATIONS. First on quadrupole moment. Please note that the quadrupole moment is really just one way of determining the mass anisotropies of a configuration. It represents asymmetries beyond dipole moments. Two bodies orbiting each other, together have a quadrupole moment which is varying, have a changing quadrupole moment and will produce ravitational radiation, even if very little. Other anisotropies (anisotropic rotating objects also can have changing quadrupole moments). His reference is good, it is https://en.m.wikipedia.org/wiki/Quadrupole#Gravitational_quadrupole. Still to see more detail on the exact dependence the CalTech reference is ok, and if you google quadrupole gravitational radiation you'll find other more explicit and mathematical treatments.

Second, for the meaning of quasistatic. It is used in this posting to mean simply a static (in our case gravitational) field, except that the source of the field can be changing slowly. The gravitational field from a static body, or BH, is static. If that body moves not too fast, it'll still be the same at your position, effectively. An example is the field of the Sun on the Earth as the Earth moves around it its pseudostatic, just the direction and maybe a little fo the strength changes. Similarly one can use the same idea to the field of the BHs. It is not a radiating field at the lowest approximation.

END OF ADDED NOTE

If you then take those numbers and calculate the strain on an object say our size, say 1 meters (a short person), for a merger emitting 1 solar mass at a distance of the Sun from us it is about

distance to FIRST observed BH/distance earth-sun = 3 Billion ly/100 million kms =(approx.) $10^{15}$

So for the same BHs at the distance from the Sun we'd have strain = $10^{-22} 10^{15} = 10^{-7}$ which not that small, and then for someone 1 meter tall (short guy) the tidal force would stretch and compress us alternatively in two orthogonal directions (see the Caltech link for a figure) by distances of

Stretch and compression distances for a person = $10^{-7}$ meters= 0.1 um

The earth would be stretched compressed by

Stretch and compression of earth diameter = $10^{-7} X 12 X 10^{6}$ meters = approximately 1 meter.

That is about the effect of the Moon on the tides, about 1 meter. Except, the Earth would be stretching and compressing like that at the rate of the wave's frequency, maybe 100 Hz or so. Objects which are pretty small like us might or might not notice the 0.1um effect, but the Earth changing like that 100 times per second I'd guess would have a huge effect.

If the BHs were 10 times closer than the Sun, the effect would be 10 times more, or 10 meters on the Earth, and it might destroy it. If the BHs were at a distance to the Moon, 400,000 kms, the effect would be another 20 times larger, or 200 meters. It'd probably break apart the Earth. It would stretch/compress us as well, by about 20 um. That's still small effect. Whether that kills us or not is a medical or biological question. Maybe stretching and compressing us like that 100 times a second would bad, maybe have the effect on us that a a kitchen grinder has on food and ice.

But note that's just from the gravitational radiation. At those distances the actual tidal force from a solar mass BH at the location of the Moon to us, even if it was not radiating gravitational waves, would be

$20^3$ = 8000 meters (if it was water). It'd break up the Earth, and kill us

(pseudostatic tidal forces go like $1/r^3$)

So, in summary

1)Stars which get too close to a BHs (merging or not) will be disrupted by the pseudostatic tidal force from the BHs. We have pictures of those being accreted into BH (merging or not). Too close can be even some solar distances away

2)Gravitational radiation from merging BHs can have a disruptive effect on nearby stars and bodies, if close enough. From the calculations it looks like solar distances might not be close enough for gravitational waves to disrupt planets or stars or human bodies, but the pseudostatic gravitational tidal force would be able to disrupt stars (and in essence get them absorbed into the BH over time). For gravitational waves to disrupt the Earth it seems that the BHs would have to be at lunar distances or so, order of magnitude. To affect us as human bodies maybe closer in