https://en.wikipedia.org/wiki/Binary_pulsar
Empirically, we know that this is the correct energy density. Binary pulsars proved this in the 1980s and 1990s.
Why does this method give the right energy density?
A gravitational wave must act as a source of gravity
Let us introduce an example configuration which we call a "convertible mass" M:
Let us have a mass M which is almost spherically symmetric. We can send energy radially between different layers of M using gravitational waves. The waves cannot be exactly spherically symmetric, because then they would be longitudinal waves. Presumably, we can make the waves almost spherically symmetric.
It is natural to assume that the almost Schwarzschild gravity field of M far away does not change appreciably in the process.
We conclude that the waves must act as sources of gravity, exactly according to their energy content.
One may imagine the following conversion: the mass M is a spherical vessel filled with photon gas. We use some method to convert the photons into gravitons. It is plausible that the large-scale metric inside M should remain unchanged, except that now there are mini gravitational waves bouncing around.
On June 17, 2024 we heuristically proved that the Ricci tensor component R₀₀ of a gravitational wave must differ from zero, because the wave acts as a source of gravity.
The gravitational wave "derived" from the Einstein equations
On May 21, 2024 we showed, using rogue variations, that the Einstein field equations do not seem to have a solution for any dynamic system at all. Thus, one cannot really "solve" the Einstein equations for a binary system where two masses M₁ and M₂ orbit each other.
The metric g of the quadrupole wave has to be derived in some way, but g is not a solution of the Einstein field equations.
The question is why does this "some way" produce:
1. the right metric g for a wave, and
2. why does the Einstein tensor G then give the correct energy density for g?
The gravitational wave metric g is derived from linearized Einstein field equations.
In the link someone (probably Chris Hirata, 2019) has derived the energy density of a gravitational wave.
We want to solve the metric g, where h is a perturbation and η is the flat metric.
The Einstein tensor G is decomposed into a linear part (1) and nonlinear parts (2), etc.
Then it is guessed that the energy density can be obtained from a nonlinear part (2) of the equation. We want to find out why this works.
The electromagnetic lagrangian
Let us analyze what is "potential energy" and what is "kinetic energy" in the lagrangian density.
If we have a static charge Q, it corresponds to a current j into the "direction of time". The magnetic field B in a static configuration is zero.
We can make a potential pit for Q by introducing an electric field E. We can then put Q into a lower electric potential, but the integral of E² grows.
The lagrangian L(x) seems to be of the type:
potential energy - kinetic energy.
Let us analyze with this lagrangian the process:
1. we make a charge Q to move back and forth, attached to a spring;
2. Q radiates a dipole wave which travels over a large distance;
3. the wave is absorbed by many dipole antennae far away.
The start configuration of the action is the charge Q oscillating back and forth. The end configuration has Q almost stopped and the charges in the receiving antennae oscillating back and forth.
Why does the electromagnetic wave behave in the well known way?
The Einstein-Hilbert action in "canonical coordinates"
The May 21, 2024 rogue variations can be eliminated if we introduce canonical coordinates, against which we measure the kinetic energy of particles. That is, the kinetic energy in the Einstein-Hilbert action matter lagrangian LM is calculated against these coordinates.
Good canonical coordinates might be standard coordinates of the underlying Minkowski space, where the particles move.
The Einstein-Hilbert action should then work much better?
The energy of gravitational waves
In the formula above,
LM = kinetic energy - potential energy.
The formula treats R as "kinetic energy". The wave g does act as a source of gravity, just like an electromagnetic wave does in LM.
What exactly is the reason why we believe that the Einstein-Hilbert action does not allow R ≠ 0 in space empty of ordinary matter? Obviously, it comes from the variational calculus, which is based on the work of Riemann, Christoffel, and Ricci-Curbastro. Could there be some error there, such that the error only is manifested when a wave moves at the speed of light?
The Einstein-Hilbert action gives a reasonable solution for a spherical static mass M. It is the Schwarzschild metric. But for the "kinetic energy" of a gravity field (= wave) we get a strange result.
It might be that the correctness of the action for a static, or an almost static field, makes the wave metric g in literature approximately correct, when it is derived in literature using whatever methods. But the action is not correct for a field moving at the speed of light?
Could it be that the Einstein-Hilbert action actually says that R ≠ 0 inside a gravitational wave?
The carpet analogy: a gravitational wave necessarily breaks the Einstein field equations
Our November 9, 2023 carpet model may cast light on the variation problem.
In a section above we introduced the "convertible mass" M, of which a part can temporarily be converted to an almost exactly spherically symmetric gravitational wave, and then M absorbs the wave again.
If the Ricci curvature R₀₀ inside M would change in the "conversion", then outside M, the Ricci curvature R₀₀ has to differ from zero, too. That is, the Einstein field equations are broken in every attempted solution of the new metric g.
Now it looks natural that the Ricci scalar R of a gravitational wave can be > R. The Einstein equations will be broken anyway.
This is just like the carpet analogy. One can press the carpet to the floor at any chosen location, but cannot press it to the floor everywhere.
If we try to make the Ricci scalar R zero in a zone which contains just gravitational waves and no matter, then, apparently, R will inevitably grow elsewhere. The variation which is used to derive the Einstein field equations, may ignore the fact that R just moves around. Just like if we have a wrinkle in a carpet which is too large: the wrinkle just moves around. Once again, we have to check the details in the variation used at:
It may be that the variation calculation is otherwise correct, except that it does not take into account that faster-than-light signals are not allowed. If they were allowed, then we could instantaneously adjust the metric throughout the entire space, to reflect the mass-energy "lost" in M into gravitational waves.
The metric g in a gravitational wave really is close to the metric which we get from the linearized Einstein equations. Furthermore, the Ricci tensor R inside the wave is not zero, thus breaking the full Einstein field equations.
C. Denson Hill and Pawel Nurowski (2017) seem to claim that gravitational waves must satisfy the full Einstein equations. Above we obtained the opposite result: a real-world gravitational wave must necessarily break the full Einstein field equations.
The history in a gravitational wave would be a stationary point of the Einstein-Hilbert action, in the allowed histories where superluminal signals are banned.
The ADM formalism may have got this right, since it starts from the Einstein-Hilbert action. Has anyone noticed that the ADM formalism breaks the Einstein field equations?
Why does an approximate solution from the linearized Einstein equations give the right Ricci tensor component R₀₀?
Above we reasoned that gravitational waves are not solutions of the full Einstein field equations. But why are they solutions of the linearized equations, and on top of that, give even the right value for R₀₀, such that it tells the energy density right?
A possible solution: we must add to the Einstein-Hilbert action the energy of gravitational waves as mass-energy?
Noether's approximate theorem
Let us have a particle moving in a potential V. Let the path x(t) of the particle be a stationary point of a newtonian action S. We then slightly modify the potential. Let the new potential be V'.
The new lagrangian density is
kinetic energy 1/2 m x'(t)² - potential energy in V'.
The old path x(t) is an almost stationary point of the action, in the sense that varying x(t) by some fixed amount only slightly changes the value of the action S.
The total energy
kinetic energy + potential energy
is almost conserved in the path x(t).
Noether's approximate theorem. If a history is an almost stationary point of the action S, in the sense that changing the history by some fixed amount only changes the value of S very little, then energy is almost conserved in that history.
A model with a vibrating string
waves waves
------ ~~~~ ------ ~~~ ------- string
| |
v v
gravity pulls on energy
Let us consider a vibrating string in newtonian mechanics, where newtonian gravity also pulls on the kinetic and elastic energy of the vibration.
If we ignore newtonian gravity, then the string is governed by a linear wave equation. This might be analogous to the linearized Einstein equations.
If we solve the vibration using the linear wave equation, we get a solution which approximately solves the behavior of the string. This is analogous to solving a gravitational wave g from linearized Einstein equations.
Let us then apply the full equation of the system to the approximate solution. The full equation is derived through variational calculus, and the full equation calculates how much the action S of a history changes when the history is varied. In this case, the variation we are interested in is if the string is pulled slightly lower at the locations where there is wave energy.
The full equation calculates how much the potential energy of the system is reduced if we pull the string 1 meter lower. We see that the value of the full equation measures the energy of a wave!
Similarly, the component T₀₀ from the full Einstein equation might calculate the energy of a gravitational wave g.
We still have to check the details. This may explain why the full Einstein equations produce the energy density of a gravitational wave g.
The Chris Hirata (2019) calculation of R₀₀
The metric g in the TT gauge makes the spatial metric of x and y coordinates to oscillate according to the perturbation h₊:
The average R₀₀ over many wavelengths is:
The calculation claims that R₀₀ then has a value which, on the average, is negative. In this blog we have remarked that R₀₀ measures how initially static test masses become denser as time flows. But why would a wave which only manipulates spatial distances, in an oscillating fashion, cause any change in the density of test masses in the long run?
Could it be that a wave packet manipulates the density of the test masses? But why would it decrease the density of the test masses, rather than increase?
Let us use the geodesic equation to determine the orbit of a test mass which is initially at some fixed coordinates x, y, z, and only "moves" through time, t.
Above, we have μ any of 1, 2, or 3. Let μ be 1. The rightmost term is zero because dx¹ / dt is zero. The first term on the right side of the equation is zero because Γ¹₀₀ is zero, as calculated by Hirata.
Thus, test masses stay at their initial coordinate positions x, y, z. In the metric, geodesics going to the time direction are not "focused" or "defocused".
We probably found a solution to the mystery:
R₀₀ really measures the focusing of test masses which start from the same point and move to every direction at very slow speeds. It is not enough to study test masses which start static.
Let the test masses start to every direction at a very slow coordinate speed v at a moment when the spatial metric is flat. The spatial metric inside a gravitational wave undulates. Then the test masses travel about half of the time at a coordinate speed which is
v / (1 + ε),
and a half of the time at
v / (1 - ε).
The average coordinate speed is larger than v.
This might mean that the average R₀₀ < 0 inside the wave.
Then we can try to "compensate" the negative average value of R₀₀ by adding another metric perturbation h', which has R₀₀ > 0, on the average.
The faraway metric in h' will act like mass pulling test masses toward the wave packet. The wave appears to have mass-energy.
Maybe we should make the compensation h' to move along the gravitational wave? Then we can probably make R₀₀ zero or almost zero inside the gravitational wave. Outside the wave packet, h' pulls test masses toward the packet.
However, can this really work? If the wave packet pulls test masses toward itself, is it possible that R₀₀ = 0 everywhere?
In two dimensions, if initially parallel geodesic lines going to the t direction approach each other in the proper distance in the x direction, then certainly R₀₀ > 0?
What about t and two spatial dimensions x, y?
Does a gravitational wave have R₀₀ = 0 in some coordinate system?
Initially static test masses never move away from their initial spatial coordinates x, y, z. Thus the coordinate system which we use is "freely falling". Above we argued, and the literature says that R₀₀ > 0 there. The wave is not a solution of the full Einstein field equations.
Somewhere someone claimed that we can choose coordinates where R = 0. We do not see how that could be possible.
Equivalence of two notions of Ricci curvature
If initially static test masses approach each other, in this blog we have claimed that then R₀₀ > 0 somewhere.
This is equivalent to saying that initially parallel lines approach each other.
But could it be that using the algebraic definition of R this is not always true?
Let us determine curvatures using vector parallel transports. Let us assume that we just have the coordinates t and x.
We divide a square 0 < t < 1 and 0 < x < 1 into small subsquares. Is it so that the rotation of a transported vector is obtained by summing the rotations for each subsquare? That seems to be the case.
Let us have a gravitational wave g which squeezes and stretches alternately the x and y directions. We put initially static test masses at various coordinates x, y, z. The masses remain at the same coordinates.
Let us parallel transport a vector pointing to the z direction around a subsquare whose sides are in the t and x directions.
If the wave packet is localized and the surrounding space has a flat metric, then no vector is rotated at all in a parallel transport. This means that all components of the Riemann tensor are zero, on the average, inside the wave packet. Then all components of R are zero, on the average.
Then R₀₀ = 0, on the average, in a wave packet, contrary to what the calculation of Chris Hirata (2019) claims.
Does the difference come from different weights that we give to volumes inside a wave packet?
The averaging by dividing a cubic volume into subcubes is the logical way to do the averaging, of course.
The Gauss-Bonnet theorem says that we obtain the geodesic curvature of the edge of a 2-dimensional region by integrating the gaussian curvature K inside the region.
Does the analogous theorem hold for Riemann curvature?
Conclusions
Let us close this long blog post. We were not yet able to determine why R₀₀ averaged over a gravitational wave happens to give the correct energy density of the wave.
We will next look at Riemann curvature. Are the formulae with Christoffel symbols correct? Chris Hirata (2019) gets an average R₀₀ < 0 over a gravitational wave, while the Gauss-Bonnet theorem suggests that the average of R₀₀ over a wave packet should be exactly zero. Real-world gravitational waves are wave packets.
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