How to derive the energy density of a gravitational wave?

UPDATE January 9, 2022: The reasoning that we can "balance" the shipping of energy using negative pressure in a simple way is incorrect.

The negative pressure should be positioned at the same location as a positive mass for this to work. In a gravitational wave, the location of "positive mass" in ambiguous.

A massless very rigid wall does not stop or reflect a gravitational wave. The wall stretches less than the stretching of the metric in empty space. The energy is transmitted as elastic energy of the wall. The wall is like a high refractive index material for light: the electric field inside the material is less, but the energy is transmitted.

To reflect a gravitational wave, we need a massive rigid wall, or pairs of large masses where the pair is connected with a very rigid rod. The movement of the masses changes the quadrupole moment of the pair. The changing quadrupole moment causes destructive interference to the passing wave => some energy is reflected back.

----

Our newtonian model says that a gravitational wave makes distances appear longer, never shorter. The stretching is due to increased inertia of a test mass m.

                  test mass m

                           • ->

           ● <-------- r --------> ●

   black hole           black hole

           <----------------------- energy of gravity

In the diagram, the black holes pull the test mass to opposite directions. If we let the test mass to move a short distance to the right, then energy is shipped from the black hole on the right to the left one.

The inertia of the test mass increases because when we move the test mass a short distance, we also ship energy over a long distance.

The obvious way to cancel the shipping of energy is to put negative pressure close to both black holes. When the test mass moves right, it expands the spatial volume on the right. It has to do work to expand the volume. Thus, energy is shipped from the left to the right.

Hypothesis. If we balance the shipping of energy to the left and to the right, then the extra inertia is canceled. The spatial metric becomes 1 again.

   field of black hole 1     field of black hole 2

                    g                                      g

   2 r <-----------------------  •  -----------------------> 2 r

                                  test mass

In the previous blog post we were able to explain the stretched metric inside a gravitational wave when we guessed that both fields ship energy over the distance 2 r. The field strength of both fields is g.

Why is the energy shipped over 2 r, and not over r?

The reason might be that if we harvest energy from the electric field of a wave, we must harvest the corresponding energy from the magnetic field, too. That might explain the extra factor 2.

The gravity of a volume of pressure far away from a test mass; the Komar mass

Let us calculate the force that the negative pressure -p in a volume V exerts on the test mass m. We denote the distance of V from m by R.

                  m <-------- R --------> O

                                                 V, -p

                            pressure in the direction of R

                  --------------------------------> x

The test mass causes the metric in the direction of R to stretch by a factor

       1  +  1/2 r_m / R

        = 1  +  G / c² * m / R

where r_m is the Schwarzschild radius of m.

The energy of the negative pressure grows by

       W = p V G / c² * m / R

            = G m * (p V / c²) / R

            = G m M / R.

We see that W is just like the gravitational potential for two objects: one of mass m, and the other of mass

       M = p V / c².

If we have pressure into all three orthogonal directions, then the potential at the center of V is three times the potential of pressure just to one direction. That looks like the Komar mass of uniform pressure p in a volume V:

        3 p V / c².

Calculating the effect of harvesting energy through negative pressure

We want to extract energy from negative pressure which the stretching of the metric causes. The extraction process attenuates the gravitational wave - but how much?

                                            s

                                            -->

            

            O <------ 2 r ------>  •  <------ 2 r ------> O

          V, -p                        m                         V, -p

Let us put pockets V of negative pressure -p on both sides of the test mass m, at the distance 2 r from it. The pressure is only in the direction of g, not to orthogonal directions.

Let m move a short distance s to the right. The energy displacement to the left, by the two fields g, is

       C = 4 r * m g s.

Let the stretching of the metric be 1 + d. Then

       d = C / (m c² * s)

          = 1 / c² * 4 r g

<=>

       g = c² / 4 * d / r.

The negative pressure -p cancels a part of the energy displacement C because when we move m to the right, then the left pocket V contracts and the right pocket V expands. Energy is shipped over a distance 4 r to the right.

Let us denote by f the "field" of each pocket V of negative pressure -p.

The field strength is

       f = G / c² * p V / (4 r²).

Let us use the fields f to cancel a part of the fields g. Let us cancel, say, 1% close to the pressure system.

       f = 1% * g

<=>

       p V = 4% * c² / G * r² g.

Let the maximum stretching of the metric be

       1 + d

along the x axis. The pressure system harvests the energy

       E = 2 d p V

          = 8% * c² / G * r² g * 1 / c² * 4 r g.

          = 32% * 1 / G * r³ g²

The "size" of our negative pressure system is 4 r. It causes destructive interference within a certain zone. How large is the zone?

https://en.wikipedia.org/wiki/Antenna_aperture

The cross section of a typical isotropic receiving antenna in radio technology is 

       ~ λ² / (4 π).

However, our pressure system does not try to collect the maximum energy, but only 2%. The cross section might be much larger, up to λ².

Let us put (λ / 2) / (4 r) pressure systems in a "series". The system removes 2% of the wave energy over the volume λ³.

Note that the claim that a "series" works this way is yet another guess.

We have

       E * (λ / 2) / (4 r) = 2% * D  λ³,

where D is the energy density of the gravitational wave. Then

       D = 50/8 E / r * 1 / λ²

           = 1 / (2 G) * r² g² * ω² / (π² c²)

           = 1 / (2 π²) * 1 / (G c²) * ω² r g² 

           = 1 / (2 π²) * 1 / (G c²) * ω² r²

               * c⁴ / 16 * d² / r²

           = 1 / (32 π²) * c² / G * ω² d²

           = 1 / (8 π²) * c² / G * ω² H²,

where d = 2 H, and H is the maximum value of h+.

https://www.tat.physik.uni-tuebingen.de/~kokkotas/Teaching/NS.BH.GW_files/GW_Physics.pdf

Kostas D. Kokkotas (2002) gives the above formula for t₀₀^GW, which is the energy density in the stress-energy pseudotensor.

The formula looks similar to our formula for the energy density D, except for a constant factor. Our calculation was extremely crude. It is pure luck that the constant factor is close to our value in D.

Do the fields of the two black holes really exist separately?

We have been explaining the increase in the inertia of the test mass by claiming that the fields of the two black holes in the gravitational wave "exist" individually, and pull the test mass m to opposite directions.

However, it might be that the increase of inertia, after all, is due to the sum of the fields. The sum is the quadrupole field. The field is quite weak, and complicated in form. The field of the test mass m does strengthen and weaken the quadrupole field at faraway locations. If we move the test mass, energy is shipped over great distances in the quadrupole field. This shipping might be enough to explain the increase in inertia.

We have to study this. The separate existence of fields is an ugly assumption.

                                          s

                                          -->

                  ●                     •                    ●

              mass 1        test mass          mass 2

In the case of static fields, in the diagram we have a typical configuration. Moving the test mass the distance s ships energy from the field of the mass 2 to the field of the mass 1. We do not need to assume the separate existence of the fields for the two masses. Moving the test mass ships energy from the field behind the mass 2 to the field behind the mass 1.

It might be that the tug-of-war on the test mass m is correctly described by either

1. each mass i (i = 1 or i = 2) pulling the test mass with the newtonian gravity force, and storing the energy in the mass i itself, or

2. calculating the energy of the sum of fields of the masses 1 and 2, and calculating how moving the test mass m ships energy around in the field.

A wave is a greatly stretched "image" of the above configuration. If the test mass is at a distance R from the source, energy may be shipped over a very great distance, ~ 2 R. The quadrupole field may be strong enough to explain the increase in inertia.

In the above sections, we assumed that the fields ship energy over a distance 2 r, where r is the separation of the black holes. The distance scale 2 r is strangely short, since the fields at a distance of R span a length ~ 2 R. Maybe 2 R is the relevant distance scale, and the fields are weaker because of the destructive interference in a quadrupole field.

Conclusions

We were able to derive a formula for the energy density of a gravitational wave. The formula looks a lot like the corresponding formula of general relativity. Is this a coincidence? We do not think so. The energy of a gravitational wave seems to be in the stretching of the transverse metric.

To explain the transverse metric we had to make a major assumption: the fields of the black holes exist separately and ship energy over a distance 2 r, where r is the separation of the black holes. Alternatively, the fields do not exist separately, and energy is shipped over great distances in the quadrupole field.

In the calculation of the field energy we had to make another major assumption: a "series" of receiving antennas of gravitational waves behaves like antennas in radio technology.

The latter assumption might be wrong. If we ship field energy to opposite directions, why would these shipments cancel the extra inertia which is caused by shipping energy over large distances? How does Nature know how to optimize shipments so that canceling of extra inertia happens?

The calculations above were extremely crude. We should refine them. Also, we could calculate the effect of harvesting energy with two test masses connected with a rigid rod.

We believe that the extra inertia happens also inside an electromagnetic wave for a test charge. This implies that we are studying general field theory, rather than just gravity.

The gravitational + polarization wave from newtonian gravity

UPDATE January 6, 2022: We updated the text concerning whether we should count the effect of a dipole twice or once. The guess is S = 2 r or 4 r depending on that.

UPDATE December 27, 2021: We realized that g₁₁ - g₂₂ varies by 4 h+, and updated the text accordingly.

----

We now understand well enough the origin of the stretching of the spatial metric, and can calculate the amplitude of the + polarization wave from a binary black hole. We calculate using the newtonian gravity force and the electromagnetic analogue.

The electric field of a dipole wave

https://scholar.valpo.edu/cgi/viewcontent.cgi%3Ffilename%3D29%26article%3D1000%26context%3Dengineering_oer%26type%3Dadditional

From the link we find the electric field strength in a wave generated by a dipole source:

      ^

      |   a       

      |         angle θ

     ● Q <-------------- R ----------------> •  observer

Let us denote the r in the formula by our own symbol R.

We assume that a charge Q accelerates upward and the acceleration is a. An observer is at a distance R. The angle between a and R is θ.

The electric field E is transverse: it is normal  to R. Since

       μ₀ = 1 / (c² ε₀),

we have

       E = 1 / (4 π ε₀) * Q a / (c² R) * sin(θ).

The analogous formula for the newtonian gravity force field is

       g = G m a / (c² R) * sin(θ).

The newtonian field in a wave generated by a binary black hole

If the components of a binary black hole do not orbit at a relativistic speed, then the binary creates two dipole waves whose field strengths almost completely cancel each other in destructive interference. The dipoles have a 180 degree phase shift.

However, it looks like the stretching of the spatial metric is not canceled. On the contrary, we have to sum the stretching for the two dipoles.

                        z

                        ^

                        |       •  observer

                        |      /

                        |    /  R

                        |  /

             ● <---------> ●

                       r

         binary black hole

Let us have two black holes, each of a mass M. They move in a circular orbit and their separation is r.

The centripetal acceleration is

       a = G M / r².

We have

       g = G² M² / (r² c²) * 1 / R * sin(θ)

as the field strength of a single dipole. We can take θ = 90 degrees, because in the  movement, the projection of each black hole on a normal of R is a harmonic oscillation whose amplitude is r.

The stretching of the spatial metric

Let us move the test mass m to the direction of a short distance s. The field g increases the inertia of a test mass m by a factor

         (m s +  m g s / c² * S) / m s

         = 1 + g / c² * S,

where S is the (long) distance over which field energy is shipped.

Let us guess S = 2 r.

The increase in the inertia is by a factor

       1  +  2 / R * G² M² / (r c⁴),

and the stretching of the spatial metric has the same factor. The stretching is in the direction of g.

For two dipoles, the stretching d is 

       1 + d  =  1  +  2 / R * 2 G² M² / (r c⁴)?

Or should we only count the effect once? If only once, then we should set S = 4 r.

                        

                        ^

                        | θ_i  •  observer

                        |      /

                        |    /  R

                        |  /

             ● <---------> ●

                       r

         binary black hole

https://arxiv.org/abs/1710.04635

The metric perturbation h+ is

             0         0         0         0

            

             0         h+      0         0

             0         0        -h+      0

             0         0         0         0

Let us compare our result to the metric given by Robert C. Hilborn (2017), above.

Note that the stretching d is 1/2 of the perturbation of the metric because, e.g., g₁₁ is the square of the stretching in the x direction.

Let us denote the maximum value of h+ by H. The value g₁₁ - g₂₂ varies by 4 H. In our own formula, d = 2 H.

We conclude that our result agrees with that of Hilborn in the case θ_i = 90 degrees.

What about θ_i = 0 degrees?

Let us in the diagram have the x axis pointing out of the screen. Our observer measures the metric horizontally, which is his x direction, and vertically, which is his y direction.

When θ_i = 90 degrees, there is stretching in the metric in the x direction, but none in the y direction. Let the stretching be d. The difference

       g₁₁ - g₂₂

varies between 0 and 2 d.

If θ_i = 0 degrees, there is stretching in the x metric as well as in the y metric. The difference g₁₁ - g₂₂ varies between -2 d and 2 d.

Robert C. Hilborn uses a traceless gauge. The relevant figure is then g₁₁ - g₂₂, and it has a double amplitude in the case θ_i = 0 compared to θ_i = 90 degrees.

We conclude that our formula agrees with his.

What about an arbitrary θ_i?

The field g to the y axis direction is proportional to cos(θ_i). 

Now we have to make another guess: the distance s over which the field energy is shipped in the general case is r cos(θ_i).

After this second guess, our formula agrees with that of Robert C. Hilborn.

Conclusions

We were able to reproduce the amplitude of the + polarization gravitational wave from the newtonian field and the electromagnetic analogue.

We had to assume that we can sum the metric perturbations for each dipole source of the wave. For this, we have to assume that the field of each mass exists separately: the fields cannot be summed and treated as a one whole in the "pool field" of a spatial volume V.

Also, we had to make two guesses about the distance over which the field energy is shipped when we move a test mass. In the future, we must give good grounds for those guesses.

Our result is very encouraging. Our result also shows that the "true" metric of space is always stretched, never contracted. That prevents superluminal communication.

Gravitational energy travels from the central mass: that explains the radial stretching in the Schwarzschild metric

We have been claiming that the radial Schwarzschild metric of a spherical mass M is stretched because moving a test mass m radially ships energy from the gravity field of M over a long distance.

             ● <------------  r  -------------> •

            M                                         m

What is that long distance? The obvious candidate is that the energy is shipped from the center of M to our small test mass.

Let us calculate the extra inertia which m gains for a radial movement. The force of gravity is

       F = G m M / r².

The work that we gain when we move the test mass radially a distance s toward the central mass is

       W = s G m M / r².

That work corresponds to a mass W / c² and that mass is shipped over a distance r. The mass displacement is

       d' = s G m M / c² * 1 / r.

The mass displacement of the test mass m over a distance s is

       d = s m.

The extra inertia in a radial movement is

       d' / d = G M / c² * 1 / r

                 = 1/2 r_s / r,

where r_s is the Schwarzschild radius of M.

https://en.wikipedia.org/wiki/Schwarzschild_metric

The radial Schwarzschild metric is

       1 / sqrt(1 - r_s / r)

       = 1  +  1/2 r_s / r,

when r_s is small.

Everything travels to the radial direction a little bit slower, including light, because the inertia is slightly larger than to the horizontal direction. We take as an axiom:

If light travels to a certain direction by a factor f < 1 slower, then distances in that direction appear to be stretched by a factor 1 / f > 1.

The stretching of the spatial metric and the newtonian field of a gravitational wave

Robert C. Hilborn (2017) gives the formulae for the polarizations + and × of a gravitational wave from a binary black hole (equation (83) in the paper):

https://arxiv.org/abs/1710.0463

                      z

                      ^

                      |    

                      |   θ_i    •  observer

                      |          /

                      |        /

                      |      /  R

                      |    /

                      |  /

           ●       r       ●  black holes

                      |

         -----------------------------------> x

           

Let the black holes orbit in the xy plane around the z axis. The distance of the black holes from each other is r, and the distance of the observer is R. The angular velocity of the orbit is ω. The angle of R from the z axis is θ_i. The global time of the observer is t, and t_R is the retarded time at the observer, that is,

       t_R = t - R / c.

The stretching of the spatial metric is due to increased inertia to one direction; the sum rule of inertia

In this blog we claim that the stretching of the spatial metric is due to the inertia of a test mass being larger to one direction than the other. The extra inertia is a consequence of energy being moved over large distances in the combined newtonian gravity field of the test mass and other masses around it.

                 ●                    <-- • -->

              mass             test mass

If we have just one mass, then the inertia of a test mass is larger to the radial direction, because energy from the field far away is shipped over a large distance to the test mass when we move the test mass radially.

What if we have several masses around the test mass?

Then we obviously have to sum the extra inertia due to each mass. Even if the combined newtonian gravity force on the test mass would be zero, one still has to ship energy of the gravity field from one faraway location to another.

Robert C. Hilborn mentions at equation (83), which we copied above, that the electromagnetic analogue of the binary black hole would have no radiation to the direction of the z axis, though the gravitational wave has the largest amplitude to that direction.

             analogue field of black hole 1

                                 --------->

                                 <---------

             analogue field of black hole 2

The electromagnetic analogue field of the wave on the z axis is zero because the field vectors point to opposite directions for each black hole.

Linearized Einstein equations calculate the sum of the metric perturbations for each black hole. One can add the stretching of the metric due to the black hole 1, and due to the black hole 2.

We must assume that the field of each mass exists individually?

Let the observer be at the z axis far away from the binary black hole. Because of the symmetry, the combined newtonian gravity field is very weak in the gravitational wave generated by the black holes, just like we argued in the previous section for the analogue electromagnetic wave.

If we want to explain the stretching of the spatial metric at the z axis at an observer far away, we cannot do it based on the sum of the newtonian fields in the vicinity of the observer. We must assume the separate existence of the field for each black hole. We can then calculate the stretching of the metric for each black hole, and sum the values.

In this blog we have been sceptical of the existence of "the electromagnetic field" in a spatial volume V; that is, sceptical of the existence of a "pool field" which would hold the fields of all charges, as well as hold all electromagnetic waves.

A photon seems to be born in the field of an individual accelerated charge. If there exists "the electromagnetic field" of a spatial volume V, when does the photon leave the field of the charge and enter the "pool field"?

Inertia in electromagnetics

If we move a test charge in an electric field, we expect to observe all the same effects about inertia as we observe in gravity.

We probably must sum the extra inertia due to each charge.

The problems that we face in determining the extra inertia inside an electromagnetic wave are probably identical to gravity.

We need to develop a framework for determining inertia in various fields.

The classical potential associated with the magnetic vector potential A: Aharonov-Bohm effect is classical

In the previous blog post we had the configuration where an electron is passing a tube of magnetic flux from the outside. Let us calculate if we can explain the vector potential A by the effect that the magnetic field of the electron has on the energy of the magnetic field B in the tube.

                   B

             ×    ×    ×               ^

             ×    ×    ×               |   v

             ×    ×    ×               e-

                             r

https://en.wikipedia.org/wiki/Magnetic_vector_potential

Suppose that B is constant over an area S. The magnetic flux is

       Φ = S B,

and the vector potential at a distance r is

       A = Φ / (2 π r).

https://en.wikipedia.org/wiki/Hamiltonian_mechanics

The hamiltonian H is:

Suppose that r is large. Then q A² is negligible. The contribution of A to the hamiltonian is

       W = -1 / m * m v * q A

            = -q v A

            = -1 / (2 π) * v q S B / r.

Is the value equal to the energy which the magnetic field of the electron adds to the field B?

The energy density of the field B is

       1/2 * 1 / μ₀ * B².

The magnetic field of the electron at the distance r is

       B' = 1 / c² * v × E

           = 1 / c² * v * 1 / (4 π ε₀) * q / r²

           = μ₀ / (4 π) * v q /  r².

If |B'| is much smaller than |B|, then B' changes the energy density by

       -1 / μ₀ * B B'.

In how big a volume does the change of energy density take place? The volume might be 2 r S.

The energy change in the magnetic field in the tube is

      W' = -1 / (2 π) * v q S B / r

We see that W = W'.

We get an intuitive definition for the vector potential (at least in this case):

The "potential"

       -q v A

is the energy which the magnetic field of a moving small charge q adds to the magnetic field B which generates the vector potential A.

That is, if we are moving the charge at a velocity v, we must do that much work against the magnetic field B to bring q to a location where the vector potential is A.

The Aharonov-Bohm effect

The Aharonov-Bohm effect is due to the classical "potential" -q v A. The effect is not quantum mechanical. Of course, the interference pattern on the screen is quantum mechanical but that is not the essence of the effect. The essence is that a magnetic field B, which the electron does not touch, still affects the path of the electron.

https://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect

The Wikipedia article claims that the effect proves that the vector potential A is "physical" in quantum mechanics.

But we showed that quantum mechanics is not relevant here. The hamiltonian of a classical charge does depend on A, and consequently, A does affect the path of a classical charge just like a quantum particle.

The excerpt above is from Wikipedia. Let us comment on them.

1. We could claim that the magnetic field B is "physical", as well as the field of the electron. There is no need to claim that the vector potential A is "physical". 

2. The same holds for action principles. There is no need to claim that they are "fundamental".

3. The field of the electron spans the whole universe. To find out how the electron moves, we cannot restrict ourselves to the immediate vicinity of the electron.

We cannot reduce complex electric phenomena to the local field

Suppose that we have an electric point charge. We move an infinitesimal test charge in its field.

The inertia of the test charge depends on the amount of energy that the test charge displaces in the combined field of the two charges.

Since the field of a single point charge is so simple, a spherically symmetric field, it may appear that we can determine the inertia based on just the value of the local electric field vector at the test charge. However, that does not work in the general case.

Consider a system of two charges. If we move a test charge, its inertia depends on the energy that is displaced in the individual fields of the two charges. Clearly, we cannot know the displacement from the local field vector at the test charge. We must know the field from a large volume.

Our Minkowski & newtonian model explains the metric of spacetime from the inertia which a test mass has in the newtonian gravity field. It seems to work ok for the Schwarzschild metric around a spherical mass.

Just like in the case of the electric field, we cannot determine the inertia from the local gravitational field at the test mass. We must know the field in a large volume.

Now we face a question both for electromagnetism and for gravity: how do we determine the inertia of a test charge or mass inside a wave? In the case of gravity, the inertia determines the metric. In the case of electromagnetism, the inertia is a phenomenon which has not been measured yet.

The inertia may be additive if we have several charges around the test charge.

Gravitational waves in the linearized Einstein equations are additive.

However, if the local field does not determine the inertia, how do we determine the inertia inside a wave? It looks like that we have to assume an individual field for each charge. That is, the field of each charge extends to infinity, and we cannot reduce a set of fields to the sum of the fields. Then there would not exist such a thing as "the electromagnetic field" in a spatial volume. It would always be a collection of separate fields for each charge in the universe.

The notion of a separate field for each charge reminds us again of the Wheeler-Feynman absorber theory. A photon is always associated with the field of a single elementary charge.

The Aharonov-Bohm effect and the classical limit

Question. Is the Aharonov-Bohm effect a consequence of the "electromagnetic fields cannot be summed" principle? Even though the magnetic field outside a solenoid is negligible, the zero field still affects the phase of an electron.

If the answer to the Question is yes, then the Aharonov-Bohm effect is classical - not a quantum effect.

         e- ● -------->  O  solenoid

When an electron approaches a solenoid, it acquires inertia relative to the electrons on the near side of the solenoid, as well as inertia relative to electrons on the far side. The combined result is that the electron acquires angular momentum and tends to start rotating along the electrons in the solenoid. It is like frame dragging in general relativity.

However, the Aharonov-Bohm effect depends on the sign of the charge carriers in the solenoid. Frame dragging probably is independent of the sign.

The Aharonov-Bohm effect is a consequence of the "minimal coupling" in the Dirac equation. Frame dragging seems to be a multipole effect, and is probably different from Aharonov-Bohm.

The Aharonov-Bohm effect may be due to the following very simple classical effect.

                       B

                 ×    ×    ×                 ^

                 ×    ×    ×             ×  |  •

                 ×    ×    ×                 e-

The crosses at B denote a magnetic field whose vector points inside the screen. The cross and the dot close to the electron denote its magnetic field.

If the electron flies past the field on the right, then the summed magnetic field of the electron and B has more energy than if the electron flies past on the left. Thus, there is a higher potential for the different paths. We have to calculate if the potential difference explains the Aharonov-Bohm effect.

The diagram above is yet another example where one cannot determine the effect of an electromagnetic field from its local value at the electron. One must consider a large volume.

People sometimes claim that classically, the electron in the diagram feels no force and moves along a straight path, while in quantum mechanics there is a phase shift which is due to the nonzero vector potential.

Let us think about the classical limit. If we increase the charge and the mass by some factor C, then in quantum mechanics, the path of the electron stays qualitatively the same. In the case of a tunneling perticle, the path is drastically changed by increasing the required energy, but for a particle under a moderate force field there is little change.

We conclude that at the classical limit, the charge cannot move along a straight path. A classical particle must feel the vector potential just like a quantum particle does.

Gauge changes in a gravitational wave do not genuinely make the metric of time -1

Let us have a binary black hole. In literature, people first derive the perturbed metric of a gravitational wave in standard coordinates, which we assume are Minkowski coordinates.

https://www.kceta.kit.edu/grk1694/img/01_Theory.pdf

(Kostas D. Kokkotas, 2016)

The metric of time g₀₀ undulates in the solution, and there are also other nonzero values on the row 0 and the column 0 of the metric tensor g.

Then people do a change of coordinates:

       x -> x + ξ(x).

A suitable function ξ makes the metric tensor simpler, and time becomes orthogonal to spatial coordinates:

               -1            0             0             0

        

                0            1 + h+    h×           0

                0            h×          1 - h+     0

                0            0            0              1

Let us call the above nice metric g+×. It is the transverse traceless (TT) gauge.

The (small) value h+ is the amplitude of the + polarized gravitational wave, and h× is the amplitude of the × polarized wave.

The coordinate change ξ is called a gauge transformation. It does not affect any physical phenomena.

However, the change of coordinates assumes that we have the entire Minkowski space filled with a plane wave, and that the space does not contain any matter.

Suppose that the gravitational wave encounters an infinitely rigid wall, inside which we use the standard Minkowski coordinates and the standard Minkowski metric. If we try to extend the coordinate transformation inside the wall, the metric no longer is orthogonal inside the wall. That is not what we want.

When the wave hits the wall, we can measure the "true" metric of the wave from the pressures it induces inside the wall. The "true" metric is not affected by a coordinate transformation.

We can make gravitational waves look like the TT metric g+× above, but the "true" metric is revealed when the wave hits obstacles. We cannot glue the metric g+× in a nice way into the "natural" metric which we would like to use inside the obstacle.

https://arxiv.org/abs/1310.2871

Michael J. Koop and Lee Samuel Finn (2013) stress that we are interested in the results of actual physical measurements of gravitational waves. Measurements are gauge and coordinate independent, while the metric depends on the choice of coordinates. They prefer an approach where the geometry of spacetime is specified through the Riemann curvature.

The electromagnetic analogue

Suppose that we have an electromagnetic wave and a small test charge. We can calculate the motion of the test charge from the electric field of the wave.

However, if the test charge is very big, then the electromagnetic wave cannot move it much because that would require more energy and momentum than is available in the wave.

One might use comoving coordinates of small test charges and claim that these are "good" coordinates to use. But the "true" coordinates of space are revealed by putting into space huge charges which the wave does not have enough muscle to move.

A coordinate transformation transforms the gravitational wave, but what does it do to the "interface" to matter?

     ●  ●            )    )    )    )    )    )    )            •   •   •

     binary                    wave                           matter

     black hole

Suppose that we have calculated the wave in some convenient standard coordinates which we also use with a matter obstacle.

Suppose then that we change the coordinates in the wave zone and make the metric of the wave to look nice. We may obtain the TT metric g+×, for example.

But if the "interface" becomes complicated to the coordinates and the metric which we use inside the matter obstacle, then it is not a good idea to change coordinates.

The interface to the binary black hole may become complicated, too.

Are gravitational waves produced by a binary black hole purely spatial and transverse?

The above question appears on the Internet. Most people will answer yes.

However, the answer depends on the coordinates which we use. For an infinite plane wave produced by a binary black hole, if the wave propagates in otherwise empty space, then there exists a TT choice of coordinates which makes the metric of time -1, and time orthogonal to spatial dimensions.

We do not know if such a choice of coordinates exists for a realistic, burst wave produced by a binary black hole merger, even if that burst wave propagates in empty space.

In general, the question if a wave is in the metric of time or in the metric of space depends on the coordinates. If we measure the time signal from a remote clock to undulate in the frequency, we may alternatively interpret that as a result of a change of distance, or a change in the rate of the clock.

Conclusions

When we change coordinates to normalize the metric of time to -1 inside a gravitational wave, and make time orthogonal to spatial dimensions, it is sleight of hand which assumes an infinite plane wave.

What matters is if we can conveniently use the new coordinates to calculate the response of matter to gravitational waves. We need to check if that really is the case with the g+× metric.

Also the change of coordinates may be hard to extend in a nice way to empty space outside the gravitational wave. Outside observers would not see the waves having the nice structure indicated by the g+× metric.

We can explain the Schwarzschild metric with frame dragging

The gravitational wave from a binary black hole does have an undulating metric of time g₀₀ in the standard coordinates in which we calculate. But there is also undulating frame dragging because the wave is born from rapidly moving large masses.

The metric looks like the following:

             g₀₀        g₀₁        g₀₂        g₀₃

             g₁₀         .

             g₂₀                      .

             g₃₀                                     .

There is undulation on the row 0 and the column 0 in the matrix, as well as in g₁₁ and other purely spatial elements of the metric.

In literature it is shown that we can change the coordinates in such a way that g₀₀ becomes -1 and other elements on the row 0 and the column 0 become 0:

             -1            0            0           0

              0             .

              0                          .

              0                                         .

Thus, with a change of coordinates we show that there really was no undulation in the metric of time. But why we did have undulation when we were using our original coordinates?

The reason seems to be frame dragging. Frame dragging makes the time and space axes in our original coordinates non-orthogonal. That shows up, e.g., as a nonzero value of g₀₁. When we change the time coordinate to the proper time, so that g₀₀ becomes -1, we skew the coordinate axis of time relative to the spatial coordinate axes. It happens that this skewing makes the axes orthogonal again.

Frame dragging

It is as if the undulation in the metric of time in the original coordinates were due to frame dragging. In frame dragging, the "natural" inertial frame starts to move along with a large moving mass. If we want a clock to stay static relative to faraway observers, we have to exert a force on the clock. Otherwise, the clock would move along with the inertial frame.

Since we exert a force on the clock, it starts to move relative to the inertial frame and ticks slower.

If the black holes in the binary system move at a relativistuc speed, then gravitational waves become prominent. Also frame dragging is significant at such speeds.

Question. How does a faraway observer see the frame dragging that is included in a gravitational wave? The local observer inside the wave sees no undulation in the metric on the row 0 and the column 0. But can an outside observer see the undulation?

If a gravitational wave meets a very heavy object, then, obviously, frame dragging cannot drag that heavy object along. Then the distorted metric of time probably is revealed.

Time dilation and the stretching of the radial metric in the Schwarzschild solution can be explained with frame dragging

Let us drop a test mass from very far and let it fall toward a spherical mass.

Our freely falling test mass defines the "natural" inertial frame. An observer A who is static close to the mass is moving relative to the natural inertial frame. The clock of the observer ticks slower and the ruler that he uses to measure radial distances has become shorter.

We can explain the change in the clock rate and the length contraction of the ruler by the relative velocity v of the observer A relative to the natural inertial frame.

We can claim that the spherical mass "drags", or attracts, the natural inertial frame. Since the observer A is moving relative to that natural inertial frame, his clock appears to tick slower and the ruler is contracted.

This explains why the distortion of the metric in the Schwarzschild solution looks like the corresponding distortion for a rapidly moving object in special relativity. Time is slowed down by a factor γ and distances have grown by the factor 1 / γ in the direction of the movement.

There is a flaw, though, in our explanation. A clock attached to our freely falling test mass does not tick any faster than the clock of the observer A. The reason might be that the process of falling down is one-way. When a gravitational wave tugs on a test mass, the process is periodic. A clock floating freely in space will tick faster than a clock which we force to stay static in our initial coordinates.

A wave is "falling" in a highly distorted force field

If we wave an electric charge in our hand up and down, its Coulomb field produces electromagnetic waves far away. Far away, the electric field alternately points up and down.

We may claim that an opposite test charge far away is "falling" in a highly distorted Coulomb electric field of the charge in our hand. The test charge will mostly move up and down, but it will eventually reach the charge in our hand.

Our model may explain why some phenomena occur both in a free fall in a force field, and also if we let a test charge float freely inside a wave of the force field.

Conclusions

We found an explanation for the claim of literature that gravitational waves are purely spatial and that the metric of time does not undulate. Next we must find out how outside observers see the waves. In a black hole merger, the final phase lasts less than a second and produces a burst of gravitational waves. How do observers who are outside the burst see the passing of the burst? Can they measure it with precision clocks?

We were able to link the Schwarzschild metric to special relativity. That may help us in analyzing gravitational waves.

Large energy of gravitational waves vs. electromagnetic waves

Robert C. Hilborn (2017) calculated that gravitational waves hold 16 times the energy content of analogous electromagnetic waves.

The calculations in our previous blog post suggest that the reason for the high energy density is that gravitational waves perturb the metric of space. We are able to extract a lot of energy using rigid rods.

Why do gravitational waves change the metric of space?

Larger inertia in the radial direction

Our Minkowski & newtonian model explains the stretching of the spatial metric by the fact that the inertia of any object is larger when it moves radially in a gravity potential well. The inertia is smaller if the object moves horizontally.

The speed of light is slower in the radial direction of the potential well. All objects move slower in the radial direction if they make a 90 degree turn from a movement in the horizontal direction.

We introduced the axiom that a slower speed of light to some direction necessarily involves the squeezing of all force fields in that direction. Every measuring rod becomes shorter in that direction. One may speculate that this is due to a slower speed of light.

Thus, everything moves slower or is shorter in the radial direction. That is equivalent to a stretched metric.

Why is the inertia larger in the radial direction? We conjectured that the object draws energy from the gravity field, or puts energy into it. The energy apparently comes from a distant location. Shipping mass-energy over a long distance involves inertia.

From what distant location does the energy come? If gravity would only have finite reach and have homogeneous strength, we were able to explain the inertia by assuming that the energy comes from the "surface" of the gravity field.

If we have a test mass hanging from a rope which at the "surface" is attached to a spring to keep the system in balance, then moving the test mass lower transfers energy from the test mass to the spring system at the surface. We would expect inertia to be larger when energy moves a long distance over the rope.

The inertia in the horizontal direction has increased because the test mass ships with it negative energy of the gravity field. But the inertia in the radial direction is even larger because besides shipping the existing negative energy, the test mass draws more negative energy from a distant location ("surface"), or returns negative energy back to the surface.

What exactly is required to harvest a lot of energy from a gravitational wave?

To harvest energy from a gravitational wave, we need rigid rods whose effective length gets contracted inside the wave.

It is enough that the gravity field squeezes some other force field. That probably happens if gravity is coupled to the quantum of that other force field. The coupling allows gravity to tamper with the other force field.

In this reasoning we did not use the attractive nature of gravity at all.

The electromagnetic analogue of the first LIGO observation

Let us calculatete the radiated power of the electric analogue of the first observed black hole merger, as well as other estimates of the power.

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.

https://en.wikipedia.org/wiki/Larmor_formula

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.

https://arxiv.org/abs/1710.04635

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

https://en.wikipedia.org/wiki/First_observation_of_gravitational_waves

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.

The energy of a gravitational wave - reflection from an infinitely rigid wall

We have struggled to find a way to determine the energy of gravitational waves. In literature, people estimate their impact on the background geometry, and deduce the mass-energy from that. We want to find a more direct way to calculate the energy.

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.

https://en.wikipedia.org/wiki/Quadrupole_formula

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.

https://www.tat.physik.uni-tuebingen.de/~kokkotas/Teaching/NS.BH.GW_files/GW_Physics.pdf

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.