We earlier defined that one coulomb is equivalent to
1.16 * 10¹⁰ kg
of mass. The gravitational force between two such masses is equal to the Coulomb force between two 1 coulomb charges.
An electric dipole source: where is the electromagnetic wave "detached" from the static electric field?
Let us have a one coulomb charge oscillating harmonically along a vertical line whose length is 2 meters. The cycle time is 2 π seconds. The vertical position of the charge is
sin(t),
where t is the time. The acceleration is
a = -sin(t).
The average of a² is 1/2. According to the Larmor formula (see the next section), the radiated power is
P = 1.1 * 10⁻¹⁶ W.
The energy density at the distance 1 meter would be
D = P / (4 π r² c)
= 3 * 10⁻²⁶ J/m³.
We have
1/2 D = 1/2 ε₀ E²,
where E is the electric field strength. At the distance 1 meter, E is
5 * 10⁻⁸ V/m,
calculated from the average E². The peak value of E is
7 * 10⁻⁸ V/m,
on the average over the whole sphere where r = 1. The peak value on the equator might be 20% larger:
8.4 * 10⁻⁸ V/m.
The electric field falls off as 1 / r. At what distance does it equal the vertical component of the static electric field of the 1 coulomb charge?
The peak value of the vertical component of the static electric field at a distance r on the equator is
1 / (4 π ε₀) * 1 / r³.
We can solve that
r = 3.3 * 10⁸ m,
or 1.1 radians of the wave. In radio technology, the detachment is assumed to happen at the distance of 1 radian.
The radiated electromagnetic power of two dipole sources, and their destructive interference
The two black holes were each m = 30 solar masses. Let us assume that they orbit at a distance 350 km from each other, measured between the centers in Minkowski coordinates.
The acceleration is
a = G m / r²
= 10¹¹ m/s².
The speed is
a = v² / (1/2 r)
<=>
v = 1.2 * 10⁸ m/s
= 0.4 c.
The two objects each are a dipole source of radiation. Destructive interference reduces the power output, but how much?
We use the Larmor formula to calculate the radiated power of each dipole source. We hope that the error from a relativistic speed is not too large.
The equivalent electric charge of each black hole is
q = m / (1.16 * 10¹⁰) coulombs
= 5 * 10²¹ C.
The power of the electric binary system would be
P = 1.2 * 10⁵⁰ W
without destructive interference from the two dipole sources.
An electromagnetic wave is "detached" from the local field at a distance of roughly 1 radian.
The length of the orbit is 2200 km, during which time light propagates 5500 km. One radian is 900 km.
When one of the dipole sources is 350 km farther than the other, its field at the distance of one radian is maybe only 70% of the near source. We may assume that the residual field is
~ 1/4
of the field of one source.
The power output is then
~ 1/2 * 1/16
of two dipole sources. That is
4 * 10⁴⁸ W.
Literature tells us that the power output of the merging binary black hole reached its peak
3.6 * 10⁴⁹ W
when the distance was 350 km.
Our extremely crude calculation indicates that a naive electromagnetic model gives only 1/10 of the real power output.
According to Robert C. Hilborn (2017), the accurate calculated ratio is 1/16.
Calculating the power harvested from the spatial metric perturbation
Our observer is 900 km from the center of the binary system.
Let us assume that the metric perturbation is detached from the local metric at the distance of one radian.
•
900 km
|----------------------------------|
● 350 km ●
black holes
•
observers
The Schwarzschild radius of the merged system would be 200 km. The stretching of the radial metric at a distance of 900 km is
200 / 900 = 22%.
How much is the vertical metric stretched in the diagram, between the two observers?
If the angle from the center of the system to an observer is 45 degrees from horizontal, then the vertical metric between the two observers is stretched by some 4%.
We assume that the vertical metric is only half as stretched when the black holes have orbited 90 degrees further. The strething, or strain, oscillates 2%.
We use the pressure system, which we outlined in the previous blog post, to harvest the energy of the vertical spatial metric perturbation in the diagram.
The pressure will be vertical. The observers are 5 times farther from the center than the black holes. To cancel the quadrupole moment of the black holes, we need
3 * 2 m / 5²
= 0.25 m
= M
worth of negative pressure * volume at the distance 900 km.
We can harvest 2% of M c² per a half orbit. That is 3 * 10⁴⁶ J. The time for a half orbit is t = 3 ms. The power is
10⁴⁹ W.
Our pressure system in the diagram harvests the energy that would propagate to the left and to the right. If we put a similar system at an 90 degree angle (up and down in the diagram), we get double the power. We conclude that an extremely crude calculation gives
2 * 10⁴⁹ W
as the power output of gravitational waves. That is reasonably close to the value 3.6 * 10⁴⁹ W given by general relativity.
Question. What is the role of the perturbations in the metric of time? How much energy we could harvest from them?
Calculating the stretched vertical metric from the electromagnetic model
In the diagram, the electric quadrupole radiation has the electric field pointing up or down, far away from the system.
In the Minkowski & newtonian model, we believe that in newtonian gravity, a potential difference causes stretching of the spatial metric in the direction of the gradient of the potential.
Let us calculate an approximation for the potential, and the associated vertical stretching at the distance of r = 900 km.
The power output of the analogous electric quadrupole would be
P = 2.3 * 10⁴⁸ W.
That corresponds to an energy density
D = P / (4 π r² c)
= 8 * 10²⁶ J/m³.
We have
1/2 D = 1/2 ε₀ E²,
where E is the electric field strength. Then the average
E = 10¹⁹ V/m,
and the peak value is
E = 1.4 * 10¹⁹ V/m,
The potential U over 900 km is
U = 1.4 * 10²⁵ V.
The potential for a 1 coulomb charge is 1.4 * 10²⁵ J, or 1.4% of the mass-energy of an equivalent 1.16 * 10¹⁰ kg mass.
In the Schwarzschild metric, a potential of -1.4% * m c² makes the radial metric to stretch by 1.4%.
The calculation shows that the potential of the newtonian gravitational wave might explain the stretching of the metric.
Calculating the stretched vertical metric from general relativity
LIGO measured that the stretching, or strain, at the distance of Earth was 2 * 10⁻²¹. The distance was 1.2 billion light years, or 1.2 * 10²⁵ m. The stretching goes as
1 / r
on the distance. We can conclude that the stretching was 2.6% at the distance of 900 km.
Conclusions
If we have a wave which causes stretching of the spatial metric, then the energy content of the wave seems to be larger than in the analogous electromagnetic wave.
Our calculations assume several things which we have to analyze further:
1. The gravitational wave is "detached" from the local field at the distance of 1 radian. This is a sensible assumption if we assume that the gravitational wave is analogous to an electromagnetic wave, albeit contains more energy.
2. We assumed that negative pressure is equivalent to negative mass like in the Komar mass formula. That is, if we have pressure -p in the direction of the y axis in a volume V, then
m = -1/3 p V / c²
is the equivalent negative mass. Why would this be true in our Minkowski & newtonian model?
3. We ignored the stretching of the metric of time. Does it allow us to harvest even more energy?
4. Our calculations are extremely crude. The accurate numbers might be double or 1/2 of our calculated values.
How is it possible that a gravitational wave can contain more energy than an analogous electromagnetic wave, even when the "field strength" of the waves is the same?
The direct reason seems to be that the gravitational wave causes stretching of the spatial metric.
The fundamental reason might be that gravity acts universally on all mass-energy. Or could it be that the attractive nature of gravity dictates the energy content of a wave to be larger? We have to analyze this in detail.
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