Robert C. Hilborn (2017) calculated that the energy is 16-fold compared to a naive electromagnetic analogue.
Reflection of spatial gravitational waves from an infinitely rigid wall
If we have an infinitely rigid wall, then perturbations of the spatial metric cannot proceed inside the wall, because that would require infinite energy. In the rubber membrane analogue, the membrane is at the wall made rigid, for example, by gluing it to a steel plate.
The waves have to be reflected by the wall. What could cause the reflection? The obvious answer is the negative and positive pressures which the wave creates inside the wall.
Let us assume that we have + polarized gravitational wave.
y
^
|
| ^
| | • test mass
| ● ● x | measuring rod
| | • test mass
| v
black holes
We know that negative pressure behaves much like negative mass. If we have a binary black hole, then the stretched phase along the y axis is produced when the masses are in the position as in the diagram above.
The quadrupole moment, or inertial moment, in the y direction changes rapidly in that phase.
One could "cancel" a part of the wave if one could suddenly create negative mass at the location of the cross in the diagram.
If the gravitational wave hits an infinitely rigid wall at the position of the measuring rod, it creates negative pressure inside the wall. The negative pressure probably creates a wave which causes total destructive interference for the wave which would propagate to the right in the diagram. The gravitational wave is reflected from the wall, but shifted in phase 180 degrees. That is exactly like any wave hitting a solid wall.
How much pressure do we need to cancel the wave?
The formula of the Komar mass suggests that if p is uniform pressure to the x-, y-, and z-directions, then a positive pressure is worth
m = p V / c²
of mass, where V is the volume of the pressurized material and c is the speed of light.
If we want to cancel the wave very close to the binary black hole, we obviously have to use negative pressure which amounts very approximately to the mass M of the black holes. What about far away? The oscillation of the metric decays as
1 / r,
where r is the distance. Should we use pressure worth
M,
or
M r,
or
M r² ?
Let us delay the decision until we have calculated some numbers.
Extracting energy from a partially reflected wave
If we have a tense string, we can extract energy from a wave by fixing the string at some location to a structure which is not completely rigid.
___ loose attachment
______/ \______•__________ tense string
wave -->
amplitude A
If the amplitude of the incoming wave is A, and the amplitudes of the reflected and transmitted waves are 1/2 A, then our attachment stole a half of the energy of the wave.
The energy content of a gravitational wave
Let us make a wall from infinitely rigid rods, but we only allow them to attain some specified (positive or negative) maximum pressure p. If the pressure would become larger, we let the rods slide past each other, and harvest the energy that is available in the sliding motion.
In this way we can make a wall which, for an incoming wave of an amplitude A, reflects and transmits waves of an amplitude 1/2 A. From the harvested energy we can calculate the energy of the gravitational wave.
Canceling the wave very close to a relativistic binary black hole
If the black holes move at almost the speed of light, then their metric wave is "detached" from the local geometry very close to the system. Let us calculate with the numbers of the first LIGO observation of a black hole merger. The black holes both had approximately a mass
M = 30 solar masses.
The stretching of the metric very close to the merging system might be ~ 0.1. If the equivalent mass m of our negative pressure volume is
m = 0.1 M,
then we might be able to extract energy worth
2 * 0.1 * 0.1 M = 0.6 solar masses
per cycle.
The LIGO group estimated that the system lost 3 solar masses as gravitational waves.
If we put the pressurized volume farther away, its volume grows as
~ r²
and the stretching goes as
~ 1 / r.
The pressure has to go as
~ 1 / r,
so that the harvested energy stays the same.
This is analogous to a rubber model where the pressure is linearly proportional to the strain, or stretching.
The pressure seems to be linearly proportional to the strain, and the harvestable energy is something like
p V s,
where s is the strain.
In our example, the angular velocity of the binary is close to the maximum possible since the black holes are very close to each other and move at a relativistic speed.
Kostas D. Kokkotas (2002) gives the above formula for the energy density. Why does it depend on the square of the angular velocity ω?
If the gravitational wave would consist of a genuine perturbation of the transverse spatial metric, then our pressure calculation might make the energy per cycle independent of ω. However, the metric which Kokkotas uses is probably the metric of comoving coordinates of test masses.
The test masses can move: if the spatial metric seems to change in the comoving coordinates, that may be due to the test masses moving and not due to a genuine change in the spatial metric.
We need to investigate the dependency on ω.
Conclusions
Now we understand better how much energy can be harvested from gravitational waves. Our discussion above was qualitative. We cannot yet say why the energy density is 16-fold relative to a naive electromagnetic model. That requires further analysis.
We also need to analyze if our Minkowski & newtonian model can account for the stretching of the spatial metric in a gravitational wave.
What happens with waves where the metric of time varies? Can they pass an infinitely rigid wall? Probably not, because we believe that if the metric of time varies, then the metric of space must vary, too. We need to find out what is the role of the metric of time in gravitational waves.
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