On April 22, 24, and 26, 2024 we presented examples where a shear stress inside a cylinder does not allow any solution for the Einstein field equations. Let us analyze why there is no solution.
We start from the Einstein-Hilbert action. The state and the development of a physical system should minimize or maximize the action integral. Above, S is the action, R is the Ricci scalar, LM is the lagrangian density of other fields besides gravity, and g is the determinant of the metric g = gμν, det(gμν).
In a cylinder with shear, we were looking for a static solution. There is no kinetic energy in the lagrangian LM. The task is to minimize the potential energy of the system.
On November 5, 2024, the lack of a solution was due to the fact that a "global stationary point" of the action is not a "local stationary point". That is, one cannot find an extremal value of the action through a simple rule which extremizes the action locally at every point. The Einstein field equations assume that we can do the local extremization.
Finding a stationary point is much easier if it can be done locally. That is the advantage of having field equations which are local. However, in this blog we have suggested that the concept of a "field", or a "metric" cannot capture every aspect of the interaction between particles.
Is the strange metric of a thin disk due to the local nature of the Einstein field equations?
In our May 4, 2024 blog post, we wondered if the metric which general relativity gives for a thin disk is correct.
In the Minkowski & newtonian model, the energy shipping happens over a short distance, from the thin disk to the test mass m. But general relativity seems to believe that the field of a thin disk is actually created by a large spherical shell. The metric is like Minkowski & newtonian would predict for a large spherical shell.
The Einstein field equations are local. They cannot recognize what is the source of the field. Minkowski & newtonian does consider the source of the field.
Finding the minimal "potential energy" in the Einstein-Hilbert action S
We are looking for a static solution which minimizes the Einstein-Hilbert action S. There is no kinetic energy. The Ricci scalar R is the "potential energy" density of the gravity field. The matter lagrangian LM is the potential energy density of other fields.
We can save in the potential energy in LM by making
|g₀₀|
smaller, that is, by slowing down time. That makes
sqrt( -det( gμν ) )
in the action integral smaller. The price that we pay is that R becomes larger and increases the value of the action integral.
How do we know that the action integral S has a minimum at all?
It could happen that S can have any value larger than some constant C, but no solution has
S = C.
Another possibility is that the values of S have no lower bound C at all.
If we allow singularities, then S might not have a lower bound. Let us assume that pressure prevents the system from falling into a singularity.
If the possible configurations of the system form a closed set in the space of functions, and we assume that the S is continuous, then it is guaranteed that S does attain the smallest value at some point. This is a basic result of topology.
It looks like that with certain restrictions we can make sure that S has the smallest value at certain point.
The Einstein field equations are derived from the Einstein-Hilbert action by varying the metric g. A variation means that a component, say, g₁₁(x) is changed a little by adding another "infinitesimal" function h(x) to g₁₁(x), and
|h(x)|
is very small at every point x of the spacetime.
The metric g minimizes the action S if the value S cannot be made smaller by adding an infinitesimal function to each component of g.
Note that minimizing the action S here assumes that the coordinate locations of each particle in LM stay constant.
What about minimizing the action S by letting the particles to change their coordinate locations? The global minimum of S has to depend on that, too. It cannot be depend on just the metric g.
In our cylinder examples in April 2024 we tried to make sure that the system already is at a global minimum of LM. The forces on each volume element in the 3D space must add to zero, and the torque must be zero.
The next step is to set the metric g. The metric only differs very little from the flat metric. Can we assume that the position of each particle is constant in the coordinates and that the pressure and the shear are unchanged when g is perturbed a little from the flat metric?
The relative value of the mass density ρ(x), the pressure p(x), and the shear s(x) can only change very little with a small perturbation of the spatial metric. It is a good approximaton to assume that they do not change when g is varied.
The wall-to-wall carpet wrinkle example and the variation of the Einstein-Hilbert action
On November 5, 2023 we analyzed how a changing pressure "breaks" the Einstein field equations.
The metric is initially such that it does satisfy the Einstein field equations. Then the pressure inside the spherical mass changes, which would force the metric to change throughout space. But the metric cannot change faster than light. As the result, there will be a zone around empty space where R is not zero: the Einstein equations are not satisfied.
The Wikipedia article derives the Einstein field equations through variation.
The variation procedure in Wikipedia is based on the book by Sean M. Carroll (2004).
Why the Wikipedia variation method does not work in the case of a changing pressure?
The variation method claims that the Ricci tensor R must be zero in empty space. Locally, we can make R zero by varying the metric. But then the zone of a non-zero R just moves to a new location. It does not help. The variation method fails to take into account this displacement of the R ≠ 0 zone?
That is, the variation method does not calculate everything which changes with the infinitesimal variation ∂g^μν of the metric?
Or the method implicitly assumes that the metric which we vary satisfies the Einstein equations everywhere before the time t coordinate at which we vary the metric?
Wikipedia says:
"By Stokes' theorem, this only yields a boundary term when integrated. The boundary term is in general non-zero, because the integrand depends not only on [the variation of the metric] ∂g^μν, but also on its partial derivatives"
"However when the variation of the metric ∂g^μν vanishes in a neighbourhood of the boundary or when there is no boundary, this term does not contribute to the variation of the action. Thus, we can forget about this term and simply obtain
at events not in the closure of the boundary."
Let us analyze what happens when we try to use the formula above to iron out the "wrinkle" between the different metrics if we (magically) change the mass of a spherically symmetric body.
The metric at r > K + 1 is the Schwarzschild metric for a central mass M, and the metric at r < K is the Schwarzschild metric for
M' < M.
The transition layer of the metric is at K < r < K + 1. We can satisfy the Einstein field equations if we imagine that there is a spherical shell of a uniform density and a mass
M - M'
in the transition layer.
-------------------------------------> r
K K + 1
● |..................|
M' M - M'
central mass transition layer (shell)
We use the (- + + +) metric signature. The flat metric in spherical coordinates is
η =
-1 0 0 0
0 1 0 0
0 0 r² 0
0 0 0 r² sin²(θ).
Assuming the imagined uniform density mass, the Ricci tensor R in the transition layer looks like that for a uniform mass density:
R ~
1 0 0 0
0 1 0 0
0 0 r² 0
0 0 0 r² sin²(θ).
The Ricci scalar R is ~ 2 in the layer, and zero outside it.
At r > K + 1, we have
g⁰⁰ = 1 / g₀₀
= -1 - ε M / r,
where ε is a very small positive constant. Also, we have
g¹¹ = 1 / g₁₁
= 1 - ε M / r.
At r < K, we have
g⁰⁰ = -1 - ε M' / r,
g¹¹ = 1 - ε M' / r.
In the transition layer, the metric g transitions smoothly between these two formulae.
Ironing out the wrinkle by making the transition layer thinner. We try to iron out the wrinkle by making g⁰⁰ and g¹¹ to obey the upper formulae (those with M) at an outer layer:
K + 1 - δ < r < K + 1,
where δ is a small positive constant. Then the values of g⁰⁰ and g¹¹ decrease a little bit in that layer. The Wikipedia formula
claims that (the integral over the entire spacetime of) the Ricci scalar R then decreases a little bit, and there are no other changes. That claim is false. Ironing out the Ricci scalar R in that outer layer clearly increases the integral of R by the same amount in the remaining transition layer.
If we reset g in the outer layer, then the transition becomes steeper in the remaining transition layer and the integral of R does not change over the full spacetime.
Ironing out the wrinkle by reducing M. If we would do the ironing by reducing M everywhere, then in the outer layer the values of g⁰⁰ and g¹¹ would grow a little bit, and also for all r > K + 1. The Wikipedia formula claims that the integral of the Ricci scalar increases, while it, in reality, decreases.
There is a "sign error" (?) in Wikipedia, but that is not the main error. The main error is that the integral of R should not change if we use the first ironing method described above.
Analysis of the Wikipedia variation
Variation of the Ricci scalar R. The Ricci scalar R for a Ricci tensor R is defined as
R = ∑ Rij g^ij,
i, j
where 0 ≤ i ≤ 3 and 0 ≤ j ≤ 3.
The Wikipedia formula
calculates the variation of R, assuming that R stays constant and only the gμν change. This explains the "sign error" in the "ironing out by reducing M" case. The Wikipedia formula ignores most of the changes in R!
In a flat metric, R = 0. A non-zero R is "mostly born" from the curvature which perturbations of the various components of the metric gij cause. The role of multiplying by g^ij is very minor. But the Wikipedia formula claims that the multiplication plays the major role!
Did we misunderstand? Let us check how Wikipedia arrives at the Einstein field equations.
Variation of the determinant sqrt(-det(g)). Wikipedia says that
is the variation. Let us check if the formula is reasonable. Let us assume that the metric g is diagonal and that the flat metric is only slightly perturbed.
Let us assume that g₁₁ is initially 1. We decrease g₁₁ to the value 0.99. The absolute value of the determinant det(g) decreases by 1%, and the value of the square root by 0.5%. The value of g¹¹ increases by 0.01. The formula above says that the square root decreases by 0.5%. That is correct. The variation formula is reasonable.
The Einstein field equations. These come directly from the two variations above:
where
The final result is the familiar Einstein field equations, but the variation of R was very strange.
Above we have links to the 1915 and 1916 papers by David Hilbert, Die Grundlagen der Physik. Do they contain the variation procedure?
On page 404 of the first paper, David Hilbert writes that by using the "designation" introduced above to the variation derivation one gets the following form of the equation:
On page 405 David Hilbert writes that the last equation follows simply, without a calculation, since "Rμν is the only second-order tensor besides gμν, which can be built from g^μν and its first and second derivatives." Hilbert uses the letter K instead of R.
Let us look at the variation procedure in other papers. Does anyone appeal to the fact that R is a second-order tensor?
Ok, we solved the mystery. The variation of the metric, g^μν, probably has to be local. We cannot insert a new mass m anywhere, because the integral over the metric perturbation ~ m / r over the whole space would be infinite. We cannot increase the integral of R with local changes.
The variation ∂g^μν produces a very small change to the integral of R. The variation itself is very small, and the Ricci curvatures Rμν are very small. It is small squared – almost negligible.
We can continue our analysis of the "wrinkle", now that we understand the derivation of the Einstein field equations from the Einstein-Hilbert action.
More analysis of the wrinkle in the carpet – or a pressure change in a spherical mass
The Einstein equations imply conservation of mass-energy of a spherically symmetric system. This is Birkhoff's theorem. But the Einstein-Hilbert action is very tolerant to physical phenomena. It cannot dictate any such law. Let us check how the action reacts to a mass-energy change.
Let us imagine that we have a spherical mass M and the Schwarzschild interior and exterior metrics. A magic trick makes M to disappear suddenly. What does the action say about the development of the system?
Suddenly, we can reduce the "potential energy" of the action by letting the metric g become flat at the former location of M. The Ricci scalar R is reset to zero there, maybe instantaneously.
The speed of light prevents resetting the metric to flat in the entire space instantaneously. We may imagine that a "transition layer" escapes from the former location of M, and the metric in the layer is like it would be with the Einstein equations if there were the mass M in the layer.
The Einstein equations then fail because there actually is no mass in the transition layer.
The variation method above proved that the Einstein equations must hold in the system. But they do not hold. What spoils the variation?
The variation procedure above assumes that there was enough time for the system to settle down to the lowest possible value for the action. If the mass M suddenly changes, then that would require faster-than-light communication.
If the conservation law for a charge is broken, then the variation argument does not hold. The Einstein field equations are not satisfied.
This does not explain why the Einstein field equations fail for a cylinder containing shear. Let us try to find out.
If third derivatives of g affect LM, then the Einstein equations may find a wrong solution
In this example we allow also very strong fields. Let us have a large spherically symmetric mass M embedded inside an extremely rigid thick shell of matter N.
M N
● ■■■■■■■■■
-----------------------> r
Let us assume that we have found a metric g which, when varied only locally with changes in a very small environment U of an arbitrary spacetime point x, keeps the Einstein-Hilbert action S value unchanged.
Let us then consider a variation δg of the metric in the entire volume enclosing both M and N. We may reduce the Ricci scalar R in the entire M, in order to reduce the stretching of N. The elastic energy of N is greatly reduced.
Is there any reason why δg would keep the action S unchanged? If the value of the action S is linear in the sum of small local variations, then we can present δg as a sum of small local variations, and then we know that also δg keeps S unchanged. But, in general, is the action S linear in such small local variations?
Let us assume that the exotic material in N strongly resists adding a varying Ricci scalar R in the volume occupied by N, but falls into a lower energy state if a uniform negative Ricci scalar value is added to the volume of N.
Let us consider local variations δg which add a "positive mass density" at some location, and a "negative" mass density in the volume surrounding that location. The masses cancel each one out. That is, we add a positive Ricci scalar at the location, and a negative Ricci scalar around it.
Now, small local variations increase the energy of N. But a variation enclosing the entire N can reduce the energy of N. Since the Ricci scalar depends on the second derivatives of the metric, the potential energy in the lagrangian LM of our exotic material N depends on the third derivatives of the metric.
The Einstein field equations may claim that the exotic material N is in the lowest energy state, while it is not.
This is probably a known result: if we allow LM to depend on the third derivatives of the metric, then the Einstein field equations do not work. However, such exotic materials are rare in nature.
Question. The stress-energy tensor only measures the variation of dLM / dgij. Could it be that if LM depends on the first or second derivatives of g, that already is enough to confuse the Einstein field equations?
Is it possible to restrict LM in such a way that it does not depend on the third derivatives of g?
Probably not. An exotic material can be built from innocuous parts, each of which does not depend on the third derivatives. But the combination of the parts does depend!
We can build a mechanical device which measures the third derivatives of g, and puts a potential energy V(g''') into a spring according to what the measurement told. Then LM does depend on the third derivatives.
How to define the stress-energy tensor?
The stress-energy tensor T is defined by looking at how the integral of the lagrangian LM reacts to a variation of each gμν.
But what is such a variation like? We increase the value of gμν in a small environment U of a spacetime point x, and measure how much the integral of LM changes?
What if the change in the integral depends on the precise form of the function which we use to increase, say, g₁₁? For example, the precise form of dg₁₁ / dr may affect how much the integral changes.
In that case, the "pressure" component T₁₁ is not well defined.
What if an increase of g₁₁ and g₂₂ has an interaction? The components T₁₁ and T₂₂ do not tell the whole story then.
We have to ban metric variations which introduce a negative Ricci scalar R into empty space
If we do not ban a negative Ricci scalar R, then nothing prevents us from decreasing the Einstein-Hilbert action value without a limit. Let us start from empty space. We simply add negative curvature, as if we would have a negative mass -M sitting in space (though we do not add the mass -M itself to the system – only the metric).
This restricts allowed metric variations δg dramatically.
Let us then consider the following example. We have a uniform density spherically symmetric mass M in otherwise empty space. The metric consists of the Schwarzschild internal and external solutions.
The contribution to the Einstein-Hilbert action consists mostly of the mass density ρ inside, and the corresponding Ricci scalar R = ρ.
The action integral is somewhat reduced because the spacetime volume element sqrt(-det(g)) is somewhat shrunk inside M because of the slowing of time. The mass M in LM contributes less in the integral because it is at a lower potential.
Let us add into the metric g a metric perturbation which would happen if we would move some mass from the outer layer of M to the central part of M. We do not move any mass, we just change the metric.
How does the Einstein-Hilbert action react to this? Does its value grow?
Building a complex dynamic system: does there exist a lowest energy state?
Let us build a complex mechanical machine with springs and rods. The machine is built to form a loop where an "input", say, an impulse hit with a hammer, at a certain location X goes around the loop, and at a later time affects what happens at X.
X -->
|\_=>/\|||\_=>/\||
|\\ \/\
|\\ \/\
|\_<=/\|||\_<=/\||
<-- mechanical effects
circle to this direction
complex machine with
springs and rods
How easy is it to find a "steady state" for this machine?
Probably extremely hard, and a steady state might not exist at all. The system may have an infinite space of states and never returns to the same state again.
What about a static configuration where the machine has the lowest possible energy?
Finding the state may be very hard. Is it guaranteed that such a state exists at all?
If we allow an infinite number of parts in the machine, it might be that it does not possess a lowest energy state. Let the machine consist of all natural numbers and a finite set N which contains n natural numbers. If the energy of the machine is
1 / n,
then there does not exist the lowest energy state.
In the case of the cylinder with shear, we are looking for the lowest value of the Einstein-Hilbert action, i.e., the lowest "potential energy". How do we know that such a state exists?
Conclusions
Let us close this very long blog post. We were not able to arrive at the shear problem yet. We have to study the variational calculus in general relativity first.
What kind of metric variations δg are allowed?
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