UPDATE May 21, 2024: The form of the lagrangian LM has to be
kinetic energy - potential energy
and the variation δg has to be restricted to a finite area of space. Then the variation of the metric of time g₀₀ produces a correct result.
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Wikipedia tells us that James W. York first realized that a boundary term is required. It was probably published in his 1971 paper.
Could it be that the boundary term rescues the Einstein-Hilbert action?
The boundary term
The Wikipedia page contains complicated instructions about how to calculate the boundary term.
The boundary term is the last two terms in the formula above, and its is denoted SGHY,0.
Wikipedia seems to claim that if the boundary term is added, then the following formula holds:
Restrictions on the variation δg
Restriction 1 for δg. Finiteness. Let us restrict ourselves to spherically symmetric variations which only update g in a finite volume of space.
Restriction 2 for δg. Weak energy condition. Let us ban any variation which makes the Ricci scalar R negative anywhere.
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UPDATE May 21, 2024: Restriction 2 is not required, and, actually, cannot be allowed. In a flat metric in empty space, a variation of the metric will produce zones where R < 0.
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Spacetime integration volume in the Einstein-Hilbert action
We work in an asymptotically Minkowski space. We integrate the Einstein-Hilbert action for some spatial coordinate ball
r < r₀,
and for some coordinate time interval T:
t₀ < t < t₁.
Our spacetime manifold has boundaries at those coordinates. To avoid the use of a boundary term, we do not vary the metric g at all at the boundaries, but keep g fixed there.
We assume that the boundary term only depends on what is at the boundary – not on what is in the interior.
Varying the metric of time inside a spherical mass
Let us recapitulate this example from May 10, 2024. We have a uniform sphere M of a mass density ρ. The metric g is the Schwarzschild interior and exterior metric. The metric g is almost flat.
We set κ = 1. The Ricci tensor in spherical coordinates is approximately
R =
1/2 ρ 0 0 0
0 1/2 ρ 0 0
0 0 1/2 ρ r² 0
0 0 0 1/2 ρ r² sin²(θ).
The Ricci scalar:
R = ρ.
These together yield the stress-energy tensor:
T = R - 1/2 R g
=
ρ 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0.
The variation δg. We assume that g₀₀ is very close to -1. We increase g₀₀ by
0 ≤ b(r) << 1
in some short segment of r. The function b(r) must be smooth enough, so that the Ricci scalar stays > 0 inside M.
Let the spatial volume defined by that segment be U and the average value of b(r) in that volume U be
B.
Calculating the change in S.
Integral of R / 2. Wikipedia says that the Ricci scalar R changes locally by
dR = dg^μν * Rμν
= dg⁰⁰ * R₀₀
= -b(r) * 1/2 ρ.
This changes the integral of R / 2 by
-B / 2 * 1/2 ρ U T.
The volume element sqrt(-det(g)) shrinks locally by a ratio b(r) / 2. This changes the integral of R / 2 by:
ρ * -B / 4 * U T.
In total, the integral of R / 2 changes by
-1/2 ρ B U T.
Integral of LM. The mass in the volume U does not change. But the volume element shrinks because time runs slower. This changes the integral of the mass-energy in LM by
ρ * -B / 2 * U T
= -1/2 ρ B U T.
The total change in the action S is
-ρ B U T.
This does not make sense. The change should be zero because we started from a solution of the Einstein field equations.
Conclusions
The Gibbons-Hawking-York boundary term does not help. The Einstein-Hilbert action still produces only nonsensical results.
If there is no matter content in space, then the Einstein-Hilbert action may be correct. All the problems stem from matter. Calculating the energy of the (newtonian) gravity field cannot succeed from mass density alone. But if the mass density is zero, then the Einstein-Hilbert lagrangian might work?
Let us consider electromagnetic waves. They are schematically of the form:
E ~ sin(ω (t - x / c)),
B ~ cos(ω (t - x / c)).
A "curvature" corresponds to the second derivative. We can as well calculate the energy of a wave from second derivatives as from first derivatives, because the second derivatives are nice sine or cosine functions. This may explain why the Ricci scalar is a suitable lagrangian density for gravitational waves!
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