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. The local variation of the radial metric g₁₁ does not change the action integral S either, which is correct.
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UPDATE May 15, 2024: We added the section "An analysis of the Wikipedia formula".
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We set κ = 8 π G / c⁴ = 1.
Is LM of the form:
potential energy - kinetic energy,
or:
kinetic energy - potential energy?
The value of R inside a mass M is traditionally taken to be positive. It makes sense to take the potential energy to be positive in LM. The curvature R is then "potential energy" of spacetime. In a rubber sheet model of gravity, it is obvious that a positive R represents positive potential energy.
Reducing distances throughout space must be banned
Let us have a positive and a negative electric charge placed in our coordinate system. Let us assume that the metric g is flat. If we shrink all spatial distances equally, the metric stays flat: the Ricci scalar R is zero everywhere.
But the integral of LM decreases because there is less potential energy. The Einstein-Hilbert action has a lower value then.
This does not make sense. We must require that the metric of space is asymptotically 1 far away.
Is this enough? There can be an immense attractive force between the electric charges. An attractive force constitutes a "negative pressure".
We can shrink the distance between the charges by adding a negative Ricci scalar R between them.
Then we can decrease the integral of R and also the integral of LM in the Einstein-Hilbert action S. This does not make sense.
Increasing the volume of a pressurized vessel: an example of the variation procedure
Let us study a practical variation problem. Let us imagine that we have a spherical vessel where the density of the liquid is ρ. The fields are weak. There is a uniform pressure p throughout the vessel.
The Ricci tensor is approximately
R =
1/2 ρ + 3/2 p 0 0 0
0 1/2 ρ - 1/2 p 0 0
0 0 (1/2 ρ - 1/2 p) r² 0
0 0 0 (1/2 ρ - 1/2 p)
* r² sin²(θ).
The Ricci scalar:
R = ρ - 3 p.
The variation δg. Let us vary the metric by increasing the radial metric g₁₁ at some r. The volume of the vessel grows.
The Wikipedia article says that
Let us assume that g₁₁ is close to 1. We increase g₁₁ by
0 < d << 1
at some short segment of r. Let the spatial volume of the spherical shell defined by that segment be U. Let the integration coordinate time interval be T.
The volume increases d / 2 * U.
Calculating the change in S.
R / 2. Then g¹¹ = 1 / g₁₁ decreases by d. Wikipedia says that the Ricci scalar R changes locally by
δR = δg^μν * Rμν
= δg¹¹ * R₁₁
= -d (1/2 ρ₀ - 1/2 p).
This changes the integral of R / 2 by
-d / 2 * (1/2 ρ₀ - 1/2 p) * U T.
The volume element sqrt(-det(g)) grows locally by a ratio d / 2. This adds to the integral of R / 2:
(ρ₀ - 3 p) * d / 4 * U T.
The notation ρ₀ stresses that it is the density before the stretching operation. We are varying the numerical values of the metric g. It does not matter where those numerical values originally came from. We assume that p remains constant.
In total, the integral of R / 2 changes by
-d / 2 * p U T.
LM. Let us assume that the pressure comes from a repulsive force between particles. The potential energy of the repulsion V changes by
-p d / 2 * U.
The mass of the particles in U does not change. The change in the integral of LM is
p d / 2 * U T.
The total change in the action is 0. This makes sense.
Practical variation problem: varying the metric of time g₀₀ inside a spherical mass M
Let us calculate the most basic variation in general relativity. We should not be allowed to drop a mass M to an arbitrary low gravity potential. We must be punished by a growing Ricci scalar R. In a rubber sheet model of gravity, a weight M cannot fall arbitrarily low. The elastic energy in the stretched rubber sheet must stop it from falling lower.
The Ricci tensor R is the same as in the previous section, but there is no pressure p.
The variation δg. The initial mass density of the sphere is ρ₀. We assume that g₀₀ is very close to -1. We increase g₀₀ by
0 < d << 1
in some short segment of r. Let the spatial volume defined by that segment be U. We integrate S over some coordinate time interval T.
Calculating the change in S.
R / 2. Wikipedia says that the Ricci scalar R changes locally by
δR = δg^μν * Rμν
= δg⁰⁰ * R₀₀
= -d * 1/2 ρ₀.
This changes the integral of R / 2 by
-d / 2 * 1/2 ρ₀ U T.
The volume element sqrt(-det(g)) shrinks locally by a ratio d / 2. This changes the integral of R / 2 by:
ρ₀ * -d / 4 * U T.
In total, the integral of R / 2 changes by
-1/2 ρ₀ d U T.
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
ρ₀ * -d / 2 * U T
= -1/2 ρ₀ d U T.
The total change in the action S is
-ρ₀ d U T.
The result would make sense if the lagrangian LM would have the sign flipped.
The Einstein-Hilbert action works correctly only in empty space?
Note that if ρ and p are zero, then the results of the two preceding sections are reasonable: the small variation δg of the metric does not change the value of S.
There has to be a way to make the variational calculus to work because in a rubber sheet model of gravity it does work. In the rubber sheet, it is not possible to vary g₀₀ and g₁₁ separately. If we press a weight M down on the rubber sheet (increase g₀₀ from -1), that inevitably stretches the rubber: g₁₁ must increase from 1.
Could it be that we must restrict variations to "reasonable" ones, which vary both g₀₀ and g₁₁ simultaneously? How do we define which variation is reasonable?
Varying the Schwarzschild interior and exterior solutions
The variation δg. Let us have a sphere of a mass M of a uniform density ρ and of a coordinate radius K. The mass M is measured from far away.
A "reasonable" metric variation might be one where we replace the Schwarzschild metric g produced by ρ, by the metric g' produced by a slightly larger or lower density ρ':
ρ' ≠ ρ.
Does that variation keep the value of S unchanged?
Note that ρ' varies the metric. We do not vary the mass M. When the spatial metric of r is stretched, that affects also the real, proper mass density ρ. Should we let K to shrink, so that ρ is kept constant? We will see if that makes a difference.
The variation sounds very sensible because it tries to drop the mass M to a lower gravity potential, and thus save on the integral of the LM. The price we pay is a larger integral of R / 2.
Calculating the change in S?
R / 2. The value of R is a constant ρ' within the sphere. Inside the sphere, we use the Schwarzschild interior solution.
where rs = 2 G M / c² and rg is the coordinate radius of the sphere, K. Recall that we set the Einstein constant 8 π G / c⁴ to 1.
Is there a simpler way to calculate?
The contribution of R / 2 to the integral S is
ρ' / 2 * the volume element,
and the contribution of LM is
ρ * the volume element.
Is there a simple reason why the optimum is attained at ρ' = ρ?
A new variation δg. Removing the center of the sphere M in the metric g. Let us calculate a simple case. We modify the metric g as if the mass at the center of the sphere M would be missing.
Let the volume of the central area be
b * the volume of M,
where 0 < b << 1.
R / (2 κ). The Ricci scalar R in SI units is
R = κ ρ = 8 π G / c⁴ * ρ.
The integral of R / (2 κ) over the whole M is
M / 2.
Removing the center reduces the integral of R / 2 by
1/2 b M T,
where T is the integration coordinate time interval.
The metric of time g₀₀ drops closer to -1 throughout M, and the radial metric g₁₁ drops closer to 1. The net effect on the volume element sqrt(-det(g)) is probably zero outside the (small) central area.
LM. The mass M rises to a higher gravity potential. How much higher?
The gravity potential by the mass b M is
-G b M / r,
where r is the radius from the center. The potential of the sphere M, caused by the central mass b M is
K
∫ -G b M / r * ρ * 4 π r² dr
a
≈ -b * G M ρ * 2 π K²,
where K >> a > 0 is the radius of the mass b M, K is the radius of the sphere, and ρ is the density of the sphere.
The formula looks very different from the change in R / (2 κ).
What is the problem? Does the radial metric affect this?
Gradually adjusting the Ricci scalar R of the interior Schwarzschild metric
Let us test a very simple variation of the metric. We use spherical coordinates with a metric signature (- + + +). The fields are weak. The metric g is only slightly perturbed from the flat metric θ.
First, let us prove that the Ricci tensor R looks like this inside a mass of a constant mass density ρ:
R ≈
κ *
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),
where κ is the Einstein gravity constant.
The Ricci scalar R is then
R = κ * (
g⁰⁰ * 1/2 ρ + g¹¹ * 1/2 ρ
+ g²² * 1/2 ρ r² + g³³ * 1/2 ρ r² sin²(θ)
)
= κ * (-1/2 ρ + 1/2 ρ + 1/2 ρ + 1/2 ρ)
= κ ρ.
The stress-energy tensor T is
R - 1/2 R g = T =
κ *
ρ 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0.
We obtained the correct value. The interior Schwarzschild metric is:
We assumed that the Schwarzschild radius rs is very small. The relevant part of the metric becomes:
dτ² = -( 3/2 - 3/4 rs / rg
- 1/2 + 1/2 r² rs / rg)² * dt²
+ 1 / c² * (1 + r² rs / rg) * dr²
= (-1 + 3/4 rs / rg - 1/2 r² rs / rg) * dt²
+ 1 / c² * ( 1 + r² rs / rg ) * dr².
Let us gradually change the metric from flat to the perturbed metric, within the entire sphere. The Ricci curvature for both t and r is within the sphere:
R₀₀ = R₁₁ = κ * 1/2 ρ.
The Wikipedia formula states:
Let us check that we get reasonable values. Let us calculate the value of R if we gradually make the perturbation in g stronger. The value of both R₀₀ and R₁₁ is then, during the process, on the average,
1/2 κ ρ / 2.
The rise of g₀₀ up from -1 adds to R the value:
1/4 κ ρ * ( -3/4 rs / rg + 1/2 r² rs / rg ).
The rise of g₁₁ up from 1 / c² adds to R the value:
1/4 κ ρ c² * -r² rs / rg.
Let us have r = 0. The calculation claims that
1/4 κ ρ * -3/4 * 2 G M / c² * 1 / rg
= κ ρ
<=>
1 = -3/8 G M / c² * 1 / rg.
The formula makes no sense.
The notation δR / δg^μν is not too clear. If δg^μν varies the metric g in a large volume, say, inside the entire spherical mass M, then then the equation might mean the integral over M.
Lifting the underlying mass density ρ of the metric a little bit at a time, in a small spatial volume at a time. But why would we be banned from varying the metric in just a small volume at a time? A very small modification of the local metric preserves R₀₀ and R₁₁ almost the same everywhere. We can change the metric in small parts, and we get the entire metric inside M closer to the target.
More precisely, the initial flat metric corresponds to ρ = 0 in the entire volume of M. Then we gradually let ρ grow to its final value. We do that in many rounds. At the end of each round, ρ is constant inside the entire M.
The Ricci tensor R is locally determined by ρ only. If we increase ρ locally a little, the change in the local Ricci scalar R should obey the Wikipedia formula. The local value of R should change as the formula claims.
For example, at the center, r = 0, we should finally have the strange value calculated above.
This still does not make sense.
"user195583" complains in the link that he cannot get the same result from varying the Einstein-Hilbert action as he gets from the Einstein field equations.
An analysis of the Wikipedia formula
Let us restrict ourselves to metrics inside a spherical mass whose density is ρ(t, x), and initially uniform, and the metric g is almost flat. The metric signature is (- + + +). We set κ = 1. Then, initially:
R(t, r) = ρ(t, r),
R₀₀(t, r) = 1/2 ρ(t, r),
R₁₁(t, r) = 1/2 ρ(t, r).
Let us increase the density ρ(t, r) locally in some volume U of the sphere. Then g₀₀(t, r) slightly increases from ≈ -1, and, at least if the volume U is at the center of the sphere, g₁₁(t, r) slightly increases from ≈ 1.
The value of g⁰⁰(t, r) = 1 / g₀₀(t, r) slightly declines and g¹¹(t, r) = 1 / g₁₁(t, r) slightly declines.
The Wikipedia formula above claims that
dR(t, r) = -R₀₀(t, x) dg₀₀(t, r)
- R₁₁(t, x) dg₁₁(t, r)
<=>
dρ(t, r) = -1/2 ρ(t, x) dg₀₀(t, r)
-1/2 ρ(t, x) dg₁₁(t, r).
The sign is wrong. The density ρ(t, r) increases, but on the right we have a negative number. Let us fix the sign error, and look at the center of the sphere (r = 0), where g₁₁(t, 0) = 1 always.
The perturbation of g₀₀(t, 0) at the center of the sphere up from -1 is linear in ρ, if the entire sphere has a constant density ρ:
g₀₀(t, 0) = -1 + C ρ,
for some constant C > 0, because the newtonian gravity potential V is linear in ρ. Let us increase the density by a constant dρ throughout the sphere.
We get:
dρ(t, 0) = 1/2 ρ(t, 0) * C dρ(t, 0)
<=>
1 = C / 2 ρ(t, 0)
<=>
ρ(t, 0) = 2 / C.
The result is nonsensical. It claims that ρ(t, 0) must be constant.
A toy rubber sheet model – it does not work well
pit
-----___ ___----- rubber sheet
•••••• h
M
circle of mass
The depth of the pit is h. The surface of the rubber sheet acts roughly like a harmonic spring. The elastic energy of the pit is
~ h².
The total energy of the system is
~ (h² + M) * (1 - h).
This looks like the Einstein-Hilbert action if
R / 2 ~ h²,
and
sqrt(-det(g)) ~ (1 - h).
Are these reasonable?
Then h is the metric perturbation, i.e.,
g₀₀ = -1 + h.
The angle of the rubber sheet α at the edge of the circle must be
α ~ M,
for the tension of the sheet to keep M from falling lower.
How much does the rubber sheet stretch horizontally?
The angle α(r), where r is the distance from the center of the mass M, is
α(r) ~ M / r
~ h'(r),
to keep the rubber sheet circle of a radius r from falling lower.
The curvature radius r of the rubber sheet under the circle is then
r ~ 1 / α
~ 1 / M
~ 1 / h.
The curvature Q of the rubber sheet under the mass circle is
Q ~ 1 / r²
~ h².
We guess that Q is zero outside the mass circle.
The energy of the system can be written roughly as:
~ (Q + M) * (1 - h).
It looks like the Einstein-Hilbert action. We have a very crude mechanical model for the Einstein-Hilbert action.
The rubber sheet model may help us in understanding why the Einstein-Hilbert action does not seem to work properly if solely the metric of time, g₀₀, or the radial metric g₁₁ is varied.
However, g₀₀ and g₁₁ in this model do not look like those of the external Schwarzschild solution, maybe because our rubber model has fewer dimensions:
dg₀₀ / dr ~ 1 / r,
g₁₁(r) - 1 ~ 1 / r².
No one has found a rubber model which would replicate general relativity well.
Our own Minkowski & newtonian model says that gravity is not about the spacetime geometry at all. There is no "spacetime substance" which is being deformed by mass, pressure, and shear. Consequently, there probably does not exist a rubber model which would replicate gravity well.
Conclusions
We are not able to get any calculation right: all indicate that the Einstein-Hilbert action is not at the minimum in the interior Schwarzschild metric. Also, the formula
brings incorrect results.
This is perplexing. When we started to calculate Christoffel symbols in March 2024, we immediately got at least something right, even though most of our calculations were erroneous.
In the literature, we have not seen anyone actually doing variations to the metric g and calculating directly the effect on R / (2 κ) and LM. We have to check if anyone has written about this.
In general relativity, the Einstein field equations are the primary subject of study. The action is something which is added later. David Hilbert in his 1915 paper introduces the action. He claims that the Ricci tensor R is the only tensor which can be formed from second derivatives of g, and that implies that the action is correct. He does not try to calculate variations.
Maybe we have misunderstood something. We will write more blog posts about this mystery.
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