UPDATE August 9, 2024: We forgot that the rate of time is sqrt(|g₀₀|), not |g₀₀|. After correcting the error, we obtain twice the energy density calculated by Chris Hirata (2019). If we assume that the Ricci scalar R = 0 in the "matter part" of the history H, then our calculation agrees with Hirata. This might be a reasonable assumption, if we omit the attractive force of mass-energy altogether.
We also added a section before Conclusions, where we note that the linearized action does not really admit any creation of a gravitational wave because the infinitesimal variation of slowing down time in the "wave part" of the history H changes the value of the action. Maybe we should ban such variations? Then we come closer to an ordinary field theory with canonical coordinates.
----
For the past three weeks we have been trying to figure out why this method gives the correct energy density.
The lagrangian density of the "linearized" Einstein-Hilbert action
Users G. Smith and DanielC (2020) on the Physics Stack Exchange give the following lagrangian density for the metric (from Sean Carroll's book Spacetime and Geometry) perturbation of h in linearized gravity, as well as its coupling to matter (from Richard Feynmann's Lectures on Gravity):
We still add the ordinary matter lagrangian density to the lagrangian:
LM,
which can be of the usual form:
T - V,
where T is the "kinetic energy" and V is the "potential energy".
The lagrangian density and the action for "linearized gravity" is an approximation of the Einstein-Hilbert action S, but the terms are restricted to those which are at most second order in the metric perturbation h.
Let us assume that a history H, where a sine gravitational wave is created, is a stationary point of the linearized Einstein-Hilbert action
Literature says that a sine gravitational wave satisfies the linearized Einstein field equations. Let us believe the literature.
We assume that a gravitational wave was born from matter according to the linearized Einstein-Hilbert action given in the preceding section.
The history H is then a stationary point of the linearized action. We do not know if such a history H exists. In many cases, proving that an action has a stationary point at all, is too difficult. Interacting systems in most cases are nonlinear. Proving anything for nonlinear system is hard.
Linearized Einstein field equations: how to handle static sources?
The course notes (PHYS691) on Robert H. Gowdy's web page give the following formula for the linearized Einstein equations:
in the gauge:
This looks like the Wikipedia equation in the harmonic gauge.
The equation for h is obviously a wave equation. How does a wave equation react to a static mass distribution?
M weight
●
|-----------------------------------|
wall tense string wall
If we put a weight M on a long tense string, the string will take a shape where it hangs ever lower as we make the string longer. This does not look good. A wave equation has problems handling anything but "changing" sources.
The coupling of the gravitational wave should only be to temporal changes in the stress-energy tensor T?
We should put boundary conditions to the system, like the two walls in the diagram, and accept that the equation responds in a strange way to a static source?
The wave equation is about second derivatives. It does not care if the string hangs low, if the hanging string is straight.
M weight
|---------__●__---------|
tense drum skin
In two or more spatial dimensions, the hanging is not as extreme as for a string. If we have a tense drum skin, it can support a static weight beautifully.
If we assume that the drum skin is static under a weight M, then we have an equation:
which for T = 0 says that ∇² of each component of h is zero. It is a reasonable equation for a drum skin outside a static weight M. The forces from the curvature and tension in the skin must cancel each other.
Using the Noether time variation to a drum skin: energy is conserved
The linearized lagrangian looks like a very typical one for wave phenomena.
Let us imagine that we let a weight M oscillate back and forth on a drum skin, and a part of the kinetic energy of M is converted to energy of waves. A Noether time variation should reveal the energy content of a created wave.
The energies associated with M are:
1. the negative potential energy of M in the gravity field of Earth, relative to M being at the height of the drum frame;
2. the kinetic energy of M;
3. the energy of the stretching of the drum skin under M.
The virial theorem states that the positive energy in item 3 is 1/2 of the potential energy lost by M when we place M on the skin and it falls lower as the skin stretches under it.
The energies associated with waves in the skin are:
A. the energy of the stretching of the skin;
B. the kinetic energy of the skin oscillating up and down.
The energy of M is "coupled" to the waves in the drum skin through the lagrangian. The details of the coupling do not matter: Noether's theorem makes sure that energy is conserved in all cases.
start state end state
M M
●---> v ● /\/\/\ wave
static
Let us assume that initially M possesses some kinetic energy. At the end state, all that kinetic energy is in the waves of the drum skin. Let us make "time run faster" by dt at the start state and "slower" by dt in the end state. We will calculate the variations of the action.
We can ignore the energies associated with a static M sitting on the skin, because what we lose at the start we will gain in the end.
Speeding up by dt at the start. Let us speed up the system at the start, so that what it does in 1 second is performed in
1 second - dt.
The lagrangian density is of the type
T - V.
The kinetic energy T = 1/2 M v² grows by a factor
1 + 2 dt,
but the integration time interval is reduced by a factor
1 - dt.
The action integral grows by
T dt.
Slowing down by dt at the end. The wave contains a certain amount of kinetic energy, say, Tw, and potential energy from the stretching of the skin, Vw. The system takes 1 second + dt to perform what it would otherwise do in 1 second. The action integral is reduced by
Tw dt
for the kinetic part, and by
Vw dt
for the potential part.
We assumed that the history is a stationary point of the action. The sum of the infinitesimal variations must be zero. We obtain:
T = Tw + Vw,
as expected. Energy was conserved.
If we denote by h(t) the vertical displacement of a tense string, then the kinetic energy density of a wave is
~ (dh(t) / dt)²
and the potential energy from stretching is
~ (dh(t) / dx)².
This looks a lot like the energy density of a gravitational wave in Chris Hirata's text (2019):
We found an explanation for the Chris Hirata formula. But we still have to find the reason why the energy density can also be obtained from the Ricci tensor component R₀₀.
Using the fact that the linearized action is an approximation of the full Einstein-Hilbert action
If we were using the full Einstein-Hilbert action, then we could use the formula:
to calculate the change in the value of the action if we vary g₀₀, in order to perform the Noether time variation. Maybe the change is the same also for the linearized action?
history H
start state end state
matter M wave
•••••••••• /\/\/\/\
density ρ Ricci scalar R
Ricci scalar κ ρ
----> t
Let us have a history H where some matter is converted into the energy of a gravitational wave. The history H is a stationary point of the linearized Einstein-Hilbert action.
Assumption. Let us perform a Noether time variation to the history H. We assume that the change to the value of the linearized Einstein-Hilbert action is approximately the same as the change to the value of the full Einstein-Hilbert action.
According to the calculation of Chris Hirata (2019), the change is ~ (dh / dt)². The linearized Einstein-Hilbert action is constructed in the way that it should include terms of this magnitude. Thus, the assumption should be true.
1. Let us "speed up time" at the start of the history, by dt. At the start of the history, we have thin static matter of a density ρ. The Ricci scalar R there is κ ρ. The Einstein-Hilbert "lagrangian density"
R / (2 κ) + LM
is
ρ / 2 - ρ = -ρ / 2.
The full Einstein-Hilbert action changes by
ΔS₁ = ∫ -ρ / 2 dV * -dt
V
= M / 2 * dt,
where the integral is over the whole space spatial volume.
2. Let us slow down the history in the gravitational wave part, so that 1 second takes instead
1 second + dt.
The value of g₀₀ is approximately -1, and g⁰⁰ = 1 / g₀₀. If we set g₀₀ = -1 - 2 dt, then g⁰⁰ becomes -1 + 2 dt. The change in R is
R₀₀ * 2 dt.
The change in the Einstein-Hilbert "lagrangian density"
R / (2 κ) + LM
is then
R₀₀ dt / κ.
Since R = 0 originally, adding dt to the time does not affect the action integral of R. The change to the action integral comes from the change in the "lagrangian density". The full Einstein-Hilbert action integral changes by
ΔS₂ = ∫ R₀₀ * 1 / κ dV * dt,
V
where the integral is over the whole space spatial volume.
The sum of the changes to the linearized Einstein-Hilbert action is approximately
ΔS₁ + ΔS₂,
which must be zero because the infinitesimal variation cannot change the value of the action. This implies that the mass-energy density of the gravitational wave is approximately
-2 R₀₀ / κ,
just like the mass-energy density of the matter at the start was ρ.
This is twice the result of Chris Hirata (2019).
But in our derivation we did not refer to a "background metric" at all. The fact that R₀₀ reveals the energy density of the wave comes from R₀₀'s property of describing how much slowing down time by increasing |g₀₀| affects the value of R.
If we assume that the initial "matter phase" has R = 0, then our calculation agrees with that of Hirata.
It may be reasonable to omit the R in the matter phase, since we do not have the attractive metric for the energy of the wave either.
Why is the energy density of a gravitational quadrupole wave 16 times the analogous electromagnetic wave?
The reason obviously is that a gravitational wave couples to matter through a 4 × 4 matrix:
while the coupling for an electromagnetic wave is simpler. Let us have two oscillating masses with a spring attachment:
<-- -->
● /\/\/\/\/\ ●
M M
-----> x
A gravitational wave changes distances in the x direction. It disturbs the oscillation of the two masses. We should some day calculate how much power can a wave absorb from the oscillting masses. The figure should come out 16-fold, compared to an analogous quadrupole wave produced by two oscillating electric charges.
The energy density of an emitted wave is larger if the coupling from the system of masses to a wave is stronger.
What if we superpose the metric of an electromagnetic wave, such that R₀₀ becomes zero in the gravitational wave?
The mass-energy in a gravitational wave most probably should cause a gravitational attraction. It is natural to superpose the metric perturbation which describes an electromagnetic wave of the same energy density. Then R₀₀ becomes zero, and our reasoning in the preceding section fails. Why does it fail?
The obvious reason is that when we add the gravity field of the energy of the wave, then the history no longer is a stationary point of the linearized action. The linearized action is not aware of the mass-energy in the wave, and does not know that it should attract other masses.
-----__ __-----
\/\/\/\/ drum skin depressed
by the weight of a wave
The analogue is the drum skin model which we introduced above. The model does not know that the energy in a wave also has a weight, and the weight should cause a depression into the drum skin.
If we want to fix the lagrangian for the drum skin, the easiest way is to add explicitly into the lagrangian the weight of the energy in the stretching of the skin and in the kinetic energy of the skin. They are usually not included because the weight of these energies is minuscule.
Literature (Butcher, Hobson, Lasenby 2009) claims that adding the "self-interaction of a graviton" (weight) to a gravitational wave inevitably leads to the full Einstein field equations. That is a strange claim, since the full Einstein field equations do not have a solution for the creation of a gravitational wave, as we showed on August 4, 2024. On the other hand, adding the weight of a wave to the drum skin lagrangian most probably does not spoil waves in the drum skin.
In a reasonable lagrangian, the Noether time variation must yield results which show that energy is conserved. A lagrangian based on the integral of the Ricci scalar R cannot work if it forces R to be zero in space free of matter.
The Einstein-Hilbert action is successful in reproducing the Schwarzschild metric, which we empirically know to be correct for weak fields. In dynamic systems, the action fails miserably. In this blog we try to find an action which could also handle dynamic systems.
The calculation actually shows that the history is NOT allowed in linear gravity
If we add an infinitesimal dt to the flow of time in the wave part of the history H, the action changes. This proves H is not a stationary point of the linearized action, and is not allowed in linearized gravity.
Using the linearized action as a physical model fails miserably.
Conclusions
If we assume that the linearized Einstein-Hilbert action with the obvious coupling to matter "correctly" describes the birth of gravitational waves, then we can do a standard Noether time variation, and obtain an energy density of a sine gravitational wave.
The calculation gives twice the value as in the 2019 text of Chris Hirata. In our calculation we do not need to refer to obscure concepts, like the background metric somehow created by the wave.
The wave should attract matter. We will next look at ways to correct the Einstein-Hilbert action, such that it could handle gravitational waves.
In our blog we have suggested that the altered "metric" of spacetime is a mundane side effect of the inertia caused by an (almost) newtonian gravity field. If that is the case, then a metric perturbation is not the right way to describe a gravitational wave. The true description should be some kind of a "mass polarization wave", like inside a massive medium where gravity causes polarization of masses in the medium.
In this blog we have suggested that an "invisible mass polarization" in empty space can be implemented by assuming invisible mass flows, in a fashion similar to how we could describe an electromagnetic wave through invisible electric currents in empty space.
○ /\/\/\/\/\ ○ invisible quadrupole
/ interaction through
/ "metric of space"
/
<-- -->
● /\/\/\/\/\ ● quadrupole
M M
-----> x
The idea is that the oscillating quadrupole in the diagram above interacts with similar invisible quadrupoles in empty space. The interaction is quite strong because it is through an altered metric of spatial distances. This explains why the energy density of a gravitational wave is large.
In this model, a gravitational wave carries energy which is visible in the lagrangian. Adding a gravitational attraction to a wave is simple. The role of the metric is in the interactions. The metric is not "fundamental" in the lagrangian.
The idea resembles our "teleportation" model of waves, where a wave is a method to "teleport" the source close to an absorbing antenna. The polarization is a way to accomplish the teleportation.
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