The Komar mass and the Einstein-Hilbert action for a sphere of incompressible fluid

Let us have a static sphere of incompressible fluid.

According to two sources on the Internet, the Ricci scalar is

       R = 8π (ϱ - 3p)

in the Schwarzschild interior solution. There p is the positive pressure and ϱ is the total mass-energy density, including heat and other microscopic - not macroscopically translational - movement in the material.

The Einstein-Hilbert action is essentially a volume integral on

          [1/2 R + ϱ] * redshift
       = [3/2 ϱ - 3/2 p] * redshift

where the "redshift factor" is 1 in outer space and < 1 close to the mass. We use Planck units, so that G and c are 1.

The Komar mass of the system is a volume integral on

       [ϱ + 3p] * redshift.

These formulas are very different. Can we reconcile them, so that the Einstein-Hilbert action would work in a sensible way?

One is always allowed to multiply a lagrangian by a constant, and that does not change the physics.

If we keep the mass M of the sphere constant and increase the radius r, then the pressure goes as r^-4 and the mass density goes as r^-3. Making the fluid very lightweight, we can make p essentially zero. Then the relevant formulas are

       ϱ

and

      3/2 ϱ.

We see that with a scale factor of 2/3 we can make the Einstein-Hilbert action to agree with the Komar mass for very lightweight material.

But that does not help with pressure. The Komar mass has +3p while Einstein-Hilbert has -p.

The Komar mass calculates the total energy of the system.

In our earlier blog post we raised the possibility that the Einstein-Hilbert lagrangian might contain kinetic energy T, even for a static solution. Then its value

      V - T

would differ from the total energy of the system V + T. But if we have a macroscopically static system whose particles have random or other non-translational motion, then their kinetic energy must be counted in the potential energy V of the system. The differentiation of energy into V and T is used when we study the collective motion of the system. Then T is its translational kinetic energy,

       T = 1/2 V v^2

at slow speeds v.

We do not know the microscopic origin of mass. It could be totally kinetic energy. However, we count such energy in V and not in T in a lagrangian. Likewise, if the energy in spacetime deformation does not have definite translational movement, its possible microscopic kinetic energy must be counted in V.

It seems to be impossible to reconcile the Komar mass and the Einstein-Hilbert action. Such action cannot calculate correctly the motion of a system under an external force?

We are assuming here that special relativity is correct and energy and momentum are conserved.

It is possible that even if the Einstein-Hilbert action miscalculates the total (potential) energy V of the sphere, it may get the inertial mass equal to the Komar mass. Then there would be no contradiction.

Writing a lagrangian in curved spacetime

Let us study forces and accelerations in a gravitational potential well.

We have a problem: in the ordinary lagrangian L_M in the action, how do we define the velocity of a particle, so that we can write its kinetic energy? In the Minkowski space, we can use global coordinates, but such coordinates are not available in general relativity.

Suppose that we have a 100 kg thin cloud of mass. Let us compare it to a 100 kg dense sphere where time runs roughly 10% slower than in outer space.

An observer in the thin cloud can use global Minkowski coordinates to determine his speed.

What does an observer within the dense 100 kg mass do? Maybe he is carrying an accelerometer. He assumes that he is at first static, and measures his speed by integrating what the accelerometer says.

Or, maybe there is a very strong rope which hangs from a global observer in the Minkowski space. The local observer can determine his speed by holding the rope tense and measuring his speed against the rope.

Static equilibrium

If we just study static systems, then we may be able to skip the velocity measurement problem. We may define the velocity as zero in a static system with a static metric.

The lagrangian or hamiltonian should then calculate only the amount of potential energy.

Suppose that we have an area in space where our lagrangian underestimates the energy which is contained therein. Is it possible that the lagrangian still can calculate, in some sense, "right" the equilibrium of the system?

Even if the lagrangian has wrong values for the energy, for example, wrong Hooke's constants for springs, it will keep the calculated energy constant. We have to analyze what the variation of the action really calculates.

A thought experiment: lower a new shell of fluid very slowly on the sphere

Let us have an incompressible fluid sphere and a rigid spherical shell around it. We attach pulley systems to the shell, so that we can lower a new shell of fluid on the surface of the fluid sphere.

We design the pulleys in a way that they allow a very slow lowering process, and the pulleys harvest all the energy which gravity releases in the process.

Let us model the process with the Einstein-Hilbert action. Most of the time the system is in an equilibrium. We can calculate the forces on different parts and they cancel each out.

We assume that the action was scaled, as explained above, so that it agrees with the Komar integral on ϱ over the fluid sphere.

We move the configuration an infinitesimal distance ds at a time. The energy needed is proportional to ds^2 and can be made arbitrarily small.

Let the pulley system lower the fluid and after that we pull the ropes and other gear up to the starting position.

Since the system was essentially static all the time and we used the Einstein-Hilbert action to develop its state, the integral of the Einstein-Hilbert action has to be the same in the starting configuration and in the end configuration.

What are the changes in the Komar masses in the process? There is a difference in how the Komar integral and the action integral handle pressure.

How large is the pressure in the rigid sphere and the pulleys at the start of the process and at the end of the process?

If we assume euclidean geometry, it turns out that the process moves an amount P of positive pressure from the rigid shell to the fluid sphere. The Komar mass stays constant and Birkhoff's theorem is satisfied.

The difference in the Komar mass integral and the action integral is how pressure contributes. The action integral is almost agnostic to pressure while it has a significant contribution to the Komar mass.

In this thought experiment, both the action integral and the Komar mass probably stay constant. There is no contradiction. The distribution of the action integral between the rigid shell and the fluid sphere is different from the Komar mass, but in the end that does not cause any contradiction.

The pressurized vessel

https://arxiv.org/abs/gr-qc/0510041

Ehlers, Ozsvath, Schücking, and Shang (2005) have calculated the result of enclosing a sphere of incompressible liquid inside a membrane with surface tension. Their conclusion is that general relativity and the Komar mass are consistent.

The pressurized vessel experiment is also called Tolman's paradox. If we do not take into account the negative pressure in the vessel wall, then Birkhoff's theorem is broken in the experiment - that is the "paradox".

Let us check their calculations.

The authors write that ϱ, r_0, and σ determine M, and M does not depend on the pressure in the membrane.

But can the pressure vary? The definition of ϱ depends on R, and R depends on r_0 and M. If ϱ and r_0 are fixed, then M is fixed, by definition.

There is an odd consequence of the interior Schwarzschild metric. The spatial metric apparently does not vary if we add more fluid and grow the radius.

Let us consider a sphere of fluid at the center. Let its radius in global coordinates be a fixed s. If we add more fluid on top of s, the pressure grows within s, but the spatial metric stays the same. This is counter-intuitive. We cannot force more fluid within s by increasing pressure. The metric of time does vary when pressure is increased. Like the pressure would be diverted to deform the time and not deform the space.

We have a problem in using the results of the Ehlers et al. paper: the membrane is of an infinitesimal thickness and there is no clear formula for the metric and the Ricci scalar inside such a wall. How to calculate the Einstein-Hilbert action?

C.W. Misner and P. Putnam, Phys. Rev. vol. 116, 1045 (1959).

Misner and Putnam have written another resolution of Tolman's paradox. Let us check their paper.

The anti-gravity device material would allow faster-than-light travel

The construction of an anti-gravity device requires material which can stand an enormous negative pressure and be so light that the positive gravity of its mass does not win the anti-gravity.

Positive gravity causes light to travel slower from the point of view of an external observer. Space appears optically dense. Gravitational lensing is a result of this effect.

Conversely, negative gravity would make light travel faster than light from the point of view of an external observer. Then we would have all the paradoxes of causal loops.

We were able to derive a minimum mass for a force field in an earlier blog post. The minimum mass was derived from momentum conservation.

Now we may have found a stronger limit: the mass of a force field must be high enough to offset the negative gravity produced by pressure on the material. That is, it must not make faster-than-light travel possible.

The Einstein-Hilbert action may be broken - how has this been ignored for 103 years?

UPDATE June 28, 2019: the lagrangian in the Einstein-Hilbert action may, after all, calculate the total energy correctly. The solution may be that the integral on R contains kinetic energy besides potential energy. The kinetic energy may be the kinetic energy of an Einstein cluster, if we assume no other forces than gravity (rest mass still exists).

The slowing down of time in low gravitational potential may be explained by this kinetic energy (or, velocity) which a static observer on the surface has relative to an observer who is not orbiting. This would connect special and general relativity.

The positive energy content in the universe, the dark energy, causes negative pressure. It is logical that positive pressure would be associated with a negative energy content of space inside a body of mass.

We will continue studying the pressurized vessel thought experiment.

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If we believe in conservation of energy, the hamiltonian or the lagrangian of a theory has to conserve energy.

It has been known for a long time that the lagrangian in the Einstein-Hilbert action miscalculates the energy of spacetime deformation. Several pseudotensors were developed to determine the correct energy content of a system. The Landau-Lifshitz and Einstein pseudotensors are the best known ones.

Also, formulas like the ADM mass and Komar mass were introduced. They calculate the correct gravitational and inertial mass of a system in an asymptotic Minkowski space.

To determine the equilibrium of a system we need a hamiltonian which determines the forces correctly, and that requires that the hamiltonian must calculate the correct potential energy for each subsystem.

A typical lagrangian is T - V, where T is the kinetic energy and V is the potential energy in various subsystems. If there are no external forces, V contains the rest mass mc^2 and the energies in the deformation of springs, etc., in the system.

In a static system, there is no kinetic energy, and the potential energy formula of the lagrangian, with the sign flipped, is the hamiltonian. It calculates the total energy of the system.

We cannot find the correct equilibrium if the hamiltonian has an error in the energy of some subsystem. Just this is the case with the Einstein-Hilbert action.
        ____
       /        \______
     |             __###-----/\/\/\/\/\
       \_____/
   balloon    piston   spring

Let us imagine a system where we have a rubber balloon and a piston which pushes air to inflate the balloon. There is a spring which pushes the piston. We can use a hamiltonian to determine the equilibrium for the piston position. But if our hamiltonian has a grossly wrong value for the balloon deformation energy, the result will be incorrect.

In general relativity, we can convert mass-energy into spacetime deformation energy with the exact same system as in the diagram above. Pressure makes the space to curve more, and we can pump more air into the balloon than in newtonian mechanics.

Since the hamiltonian which we get from the Einstein-Hilbert action has an error in the deformation energy, it will give a wrong answer for the equilibrium.

The pressure at the center of a neutron star generates up to 1/4 of the gravitational field there, the mass only 3/4. The error in the Einstein-Hilbert action may show up when we calculate the interior solution for a neutron star.


      +       electric force       +
      ● <--------------------------> O
  dense mass    low-density mass

Let us study a dynamic system where a dense mass interacts with a low-density mass through an electric force.

If we use the lagrangian in the Einstein-Hilbert action, it miscalculates the energies of the two masses and their acceleration. We get wrong paths for the masses, and conservation of momentum is broken.

How is it possible that this problem with the Einstein-Hilbert action has been ignored since the year 1916?

If we would be constructing a bridge and in its strength calculations we would use grossly erroneus formulas for the deformation energy, engineers would immediately recognize that there is a serious problem.

It looks like people have been concentrating on the Einstein field equation and have forgotten about the hamiltonian or the lagrangian. The Einstein field equation is derived by variating the Einstein-Hilbert action. If the action is incorrect, then the field equation is, too.

However, the metric outside the gravitating body may be right when derived from the erroneous Einstein-Hilbert action. Currently, we have no means for measuring the metric inside a large gravitating body. An error in the metric there would go undetected.

The Ricci tensor has to be zero outside a gravitating body?

The Ricci tensor is identically zero outside a gravitating body in Einstein's theory. If we shoot a small spherical dust cloud from a point in spacetime, its volume will very precisely be the same in the Minkowski space as in a laboratory falling freely close to a gravitating body. That means that the Ricci tensor has all the components zero. The form of the cloud will stretch, however, because of tidal forces.

For the corrected Einstein-Hilbert action, the Ricci tensor should probably be zero, too.

In newtonian gravity, the dust cloud volume behaves like in an empty Minkowski space. We thus know that the Ricci tensor outside the mass for the corrected general relativity must be zero or almost zero if the mass is not a neutron star or a black hole. It is a reasonable hypothesis that it is exactly zero.

At this point, we have not yet calculated the Ricci tensor for various rubber models of gravity. We did show in an earlier blog post that a rubber model has the spatial metric close to the Schwarzschild one. We did not calculate the temporal metric yet.

How to calculate the correct spacetime  deformation energy?

For a rubber sheet model we can write a lagrangian where the energies are right. A big difference to the Einstein-Hilbert action is that variation of one component g_ij of the metric will cause a huge or infinite deformation energy unless we vary also other components of the metric. If you want to increase the distance of two pencil-marked points on the x axis of a rubber sheet, you may press a pit in the rubber between the points. But that pit stretches also distances in the y direction.

In the Einstein-Hilbert action, the deformation energy is negative - it is the integral on R, and R is usually positive. In the lagrangian, mc^2 and other potential energies appear negative.

In the Einstein-Hilbert action we probably can vary an individual component g_ij of the metric without causing a huge decrease in the lagrangian value.

It remains to be seen if we can write a lagrangian where the variation is as simple as in the Einstein-Hilbert action.

The Einstein-Hilbert action may estimate the energy of the system approximately right if the pressure inside the mass M is negligible. The integral E on R may in this case be the deformation energy. The binding energy is -E and an energy 2E was "freed" from M through the redshift.

A correct Einstein-Hilbert lagrangian might be something like

       -V(curvature) + T_curv + L_M

where V(curvature) is the spacetime deformation energy and T_curv is the kinetic energy of spacetime curvature. L_M is the usual lagrangian T - V for particles and fields, except the gravitational field. We assume "minimal couplings" between the curvature and other fields.

Clocks slow down within the mass and make the integral of the above lagrangian over spacetime smaller, and effectively release mass-energy from the mass. It is the redshift factor.

In Einstein's gravity, for a static solution, the spacetime outside the mass has a zero Ricci tensor and a zero stress-energy tensor. In rubber sheet models, stretched rubber contains energy throughout the asymptotic Minkowski space. Should this deformation energy gravitate? Can we mine the deformation energy far away from a mass and convert it to conventional mass-energy? If the rubber sheet has assumed its energy minimum, mining the deformation energy would require us to cut the rubber. It might be that the energy does not produce more gravitation.

The role of Birkhoff's theorem

Birkhoff's theorem states that the outside metric of a spherically symmetric system is always a static Schwarzschild geometry. This holds even if we are converting mass-energy to pressure and spacetime deformation inside the spherical system.

The theorem is a principle of energy conservation. An outside observer measures the gravitational mass of the system constant even though the form of energy within the system may change.

If the Einstein-Hilbert action miscalculates the total energy of the system, why does energy conservation still hold in Birkhoff's theorem?

Does general relativity conserve momentum?

https://en.wikipedia.org/wiki/ADM_formalism

ADM have proved that in an asymptotically Minkowski spacetime, the ADM mass and momentum are conserved.

Let us check how ADM handle a few examples which seem to defy conservation of momentum.

An almost massless rigid vessel with pressure

Let us make a vessel of almost massless, extremely rigid material. Let us push almost massless extremely rigid spheres into the vessel.

When the vessel is full in the euclidean geometry, we keep pushing more spheres in. The space inside the vessel stretches under the pressure and we can keep inserting more spheres.

Since we did a lot of work, the ADM or Komar mass of the system is positive.

Let us then pull the vessel and each object inside it with massless tethers. Since each object which we pull is almost massless, we can make the system to accelerate very fast.

The system seems to have an almost zero inertial mass though it contains significant mass-energy. This breaks conservation of momentum as well as several other laws.

Two very massive weights close to each other

https://meta-phys-thoughts.blogspot.com/2019/05/does-constantly-accelerated-mass.html

In an earlier blog post we showed that a 1 kilogram weight on the surface of a neutron star seems to have < 1 kg of gravitational mass, but > 1 kg inertial mass if pulled with a very rigid tether from far away.

Suppose that we have two very heavy weights close to each other. For example, it could be two neutron stars. If we first pull one with an almost rigid tether and then the other, is the combined inertial mass more than the combined masses of the weights, even though the gravitational mass is less?

If we can manipulate the inertial mass of an object, then conservation of momentum breaks.

Pulling a massive object with a tether

Suppose that an almost rigid tether is attached to a neutron star. A far-away observer starts to pull on the tether. What does an observer on the surface measure?

The Einstein-Hilbert action and the pressured vessel

Probably the easiest way to analyze these problems is to study how we can minimize the integral in the Einstein-Hilbert action.

https://arxiv.org/pdf/1010.5557

Hans C. Ohanian has a paper about the gravitational and inertial mass in general relativity.

Let us think about the vessel experiment which we described above, based on the action integral.

We may assume that the vessel and the spheres in it are electrically charged. We use an electric field to accelerate the system.
       
           ______
          | + +    |
          |_____|                 -----> E

     charged pressured    electric field
     vessel

Let us first think of the experiment ignoring gravitation. The action is basically the time integral over the kinetic energy T minus the potential energy V:

       S = ∫ T - V dt.
          time

If we now assume that the vessel and its content are made of massless material, the kinetic energy T = 0, and the vessel would accelerate infinitely fast, unless the gravitational field of the pressure somehow restricts the acceleration.

The integrand in the Einstein-Hilbert action is
 
       1 / (2κ) R + L_M.

In our example, L_M is -V. The role of the kinetic energy T must be played by the increased integral over the Ricci curvature R.

Hypothesis: in a static system, the integral over R can be considered the "energy of spacetime deformation". If the pit in the spacetime moves, then the integral over R is larger, and the growth in the action integral over R can be considered the "kinetic energy of the spacetime deformation."

The action integral for fields and R

The relativistic action of a moving point particle is a time integral of

        m / γ(v),

where v is the speed of the particle and

      γ(v) = 1 / sqrt(1 - v^2 / c^2).

But if we have a moving force field, how do we measure its speed? How do we measure the speed of a moving pit in the metric of spacetime? What is the "speed" of the R term in the action integral?

Let us think of the action integral in a 1+1-dimensional Minkowski space, where we do not use any corrective term γ(v) in the integration.

If we have a cloud of matter moving at a slow speed v, then its contribution is

          -mc^2 sqrt(1 - v^2 / c^2)
       = -mc^2 + 1/2 m v^2

times the global elapsed time in the integration slice.

How do we measure the speed of R, so that we can define the action integral? If R is caused by a static system of particles, then we can use the speed of those particles. What about dynamic systems?

A sign error for pressure in the Wikipedia stress-energy tensor article?

https://en.wikipedia.org/wiki/Stress–energy_tensor

There may be a sign error in Wikipedia about the pressure in the stress-energy tensor.

Let us work in global cartesian coordinates for t and spatial dimensions. The metric g_ij tells us the distances in local coordinates.

Let us keep a physical system, springs, particles, whatever, in fixed positions in global coordinates. In the Wikipedia article about the Einstein-Hilbert action we vary the inverse metric g^ij and study its effect on the action integral. If we did not make a sign error, then in the usual case, T_00 is the mass-energy density and T_11, T_22, T_33 are -P, where P is the pressure.

The Wikipedia link gives T^11 = P, and the formula there gives T_11 = g_11 * g_11 * T^11 = P, if the diagonal elements in the metric are zero.

The R component in the action is negative energy?

The classical newtonian action is an integral on

        T - V

where V contains the mass mc^2 as well as various potential energies, for example, the deformation energy of a spring.  Note that energies appear with a negative sign in tye formula.

Let us assume that there are no external forces on the system.

T is the kinetic energy of the mass-energy in V.

But how do we match this to the Einstein-Hilbert action? The lagrangian L_M contains the above ingredients. The term R can be thought as negative energy which reduces the mass-energy in L_M. However, the (negative) kinetic energy term for R is missing.

R is the Ricci scalar which is coordinate-independent, and should not change when we move to a moving frame. Length contraction reduces the integral on R. The reduction can be thought of as negative kinetic energy. But contraction can only reduce R by an amount less than R. On the other hand, the negative kinetic energy can be arbitrarily high at speeds approaching c.

There is an additional problem: since R is a nonlinear result of the positive mass-energies in the system, we cannot determine what is the "speed" of R, if the system is not static in some frame.

SOLUTION: the relativistic lagrangian of mass m is

        mc^2 / γ(v),

where v is its velocity and

       γ(v) = 1 / sqrt(1 - v^2 / c^2).

It turns out that the length contraction of the integral on R just produces the right formula if we consider the integral as negative energy.

But does the action give the right inertial mass? Suppose that the integral on

        R - mc^2

is almost zero but negative. Classically, that would mean that the system has a very low inertial mass. However, such a system does contain a large energy in general relativity. Conservation of momentum is in danger if we use the Einstein-Hilbert action.

The problem goes away if we assume that R is negative? We have been assuming that R is positive inside the mass. In literature, a positive curvature means a convex surface, like a sphere.

Could we in the lagrangian L_M flip the sign? But then the lagrangian probably would not work in finding the right R for M.

It looks like the Einstein-Hilbert action works in finding the right static equilibrium for a system. But it does not work for a moving system because M - R is not the right total energy for the system. We may have a system where M = 0 and R is large. The action claims that the total energy is negative though it is positive.

Hypothesis: a lagrangian can only work if it calculates the total energy of each isolated subsystem right. If we have a body of mass in an asymptotic Minkowski space, then its total gravitational and inertial mass is the ADM mass. Since the Einstein-Hilbert action does not calculate the ADM mass, it cannot work: it will not calculate the right path if two systems interact in an asymptotic Minkowski space.

This is at odds with the fact that many people have proved conservation of momentum in such a case. We need to check the proofs.

The fact that the Einstein-Hilbert action does not calculate the total mass-energy of a system right, has been known from the start. People have defined various pseudotensors in an attempt to determine the total mass-energy of the system.

If we think of a rubber sheet model, we believe that its lagrangian does calculate the mass-energies right. Maybe general relativity should be made more like a rubber sheet?

How to fix the Einstein-Hilbert action?

When a mass M falls into a gravitational pit, it, in a sense, loses some energy because there is a redshift if a particle wants to send its energy to a far-away observer.

Suppose that the mass is M and the redshift is 1%. But the binding energy E is not M * 1%, but rather half of that. The deformation of spacetime probably took the other half of M * 1%.

We may guess that the deformation energy of spacetime is an integral over the Ricci scalar R times some coefficient. The Ricci scalar is the simplest Riemann curvature measure.

The way to fix the Einstein-Hilbert action is to add there a potential term which corrects the total energy to equal the ADM mass or the Komar mass. Pressure does contribute to the total energy of the system, not just in the deformation energy of the matter.

Maybe we should count R as positive energy? Then the simple variation method in Wikipedia, where we are allowed to vary g^00 independently of g^11 does not work. The rubber model suggests that slowing time down by making the pit deeper necessarily causes changes in the metric in a wide area. If we want to vary the metric of the rubber sheet, we cannot change the component g^00, etc., independently of the other components.

The error in the Frauchiger and Renner paper "Quantum theory cannot consistently describe the use of itself"

UPDATE  June 17, 2019: We corrected the spelling in the headline of this page and made a new blog page.

Note that the paper of Frauchiger and Renner is also logically inconsistent. If the various observers in the experiment do correct quantum mechanical reasoning, then they all will inevitably draw the same conclusions as an external observer. Frauchiger and Renner let the observers do incorrect reasoning and then derive a contradiction - which is no surprise.

---

https://arxiv.org/abs/1604.07422

https://www.nature.com/articles/s41467-018-05739-8

Daniela Frauchiger and Renato Renner claim that quantum theory cannot "consistently describe the use of itself". The paper has appeared in Nature Communications and it was a cover story in New Scientist in March 2019.

In short, the error in the paper is that the authors fail to recognize that a measurement changes the state of a quantum system.

The experiment

      W  external observer does
      weak measurements M1 and M2
  _____________________________
 |  F1   r = heads or r = tails
 |                                          
 |  F2   z = -1/2 or z = +1/2  
 |_____________________________|
      isolated laboratory

Observers F1 and F2 are in a perfectly isolated laboratory. F1 uses a quantum system to generate a random variable r in a pure state where the probability of r = heads is 1/3 and r = tails is 2/3.

Then she measures the variable. If she got r = heads, she prepares an electron spin z = -1/2. If she got r = tails she prepares the electron in a pure state where the probability of z = +1/2 and z = -1/2 is 1/2 each.

F2 then measures the electron spin.

We assume one external observer W(igner). W makes a weak measurement M1 on the macroscopic observer F1 on the basis

       "r = heads"  -  "r = tails".

The quotes mean that W scans the memory of F1. Her memory stored the result of her measurement. By the "weak" measurement we mean that W does not measure the value of "r =..." straight away but only measures something which gives a little hint on the value.

If that measurement yields a non-zero result, we say W got "ok" from M1.

W makes another weak measurement M2 on the other macroscopic observer F2 on the basis

       "z = +1/2"  -  "z = -1/2".

One can calculate that W will 1/12 of the times get "ok" from both M1 and M2.

Try to derive a contradiction

Let us now try to derive a contradiction.

Assume that W got "ok" in M1.

Then if W would measure precisely the macroscopic observer F2, we know that W would get "z = +1/2". That is very simple to calculate. If W would after that measure F1 precisely, he would trivially get "r = tails".

If F1 measured r = tails, then she prepared the spin in a way that W cannot get "ok" from M2, that is trivial. We have an apparent contradiction, because F1 "can deduce" that "ok" from M1 implies not "ok" from M2. But we showed that sometimes M1 and M2 both yield "ok". A contradiction.

The error in the reasoning is that W never measured F2 and got "z = +1/2". A measurement changes the state of the system. Indeed, if W would measure F1 without measuring F2 first, he would sometimes get "r = heads".

In quantum mechanics, if we know that after a measurement M the system will be in the state S, that does not imply that it is in the state S before the measurement. In classical physics this is different.

Another aspect is if the incarnation of F1 who measures r = tails can assume that she is the only incarnation. In our previous blog post we stressed that no one can assume that he personally is able to collapse the wave function and cut off branches of Many Worlds. The incarnation of F1 who measures r = tails in a branch cannot assume that the other branch does not exist.

Scott Aaronson made a similar analysis in his blog:

https://www.scottaaronson.com/blog/?p=3975

Aaronson quotes Asher Peres:
"Unperformed measurements have no results."

The Copenhagen interpretation and any wave function "collapse" is inconsistent with quantum mechanics

UPDATE June 17, 2019: Our blog post on June 17, 2019 contains an example of what happens if an observer (F1 in this case) assumes that she can collapse the wave function. F1 makes wrong deductions about quantum mechanics.

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In this text, we work in the Many Worlds interpretation of quantum mechanics.

In quantum mechanics, we should not let the wave function "collapse" at many places and occasions. A collapse cuts off information from the complete wave function of the universe. It is like a dramatic pruning of the network of Many Worlds, a pruning which will spoil the predictions of quantum mechanics.

The Einstein-Podolsky-Rosen experiment

       A   <-------- antiparallel spins  ------> B
         

                                      C

Let us consider the standard Einstein-Podolsky-Rosen thought experiment. Hidden variables which would specify the direction of the spins in the two particles would, in effect, mean that the wave function has collapsed immediately after the preparation of the system.

Bell's inequality shows that if we assume such a collapse, then we have to assume faster-than-light communication between A and B to make the results match the predictions of quantum mechanics.

A collapse reduces a complex-valued wave function to a real value. It is no wonder that with real values it is hard to reproduce the interference pattern of the original complex-valued wave functions.

Let A and B communicate their measurement results to a third observer C, who will make an interference pattern from the signals S_i he received from A and B. If we assume a collapse of the wave function at A and B, we cut off the necessary information which C needs to form the interference pattern. The interference pattern is very faint: the interference happens only between those signals S1, S2, for which the "state" s(A, B) after sending S1 is equal to the state s(A, B) after sending S2.

In practical setups, C would not observe any interference because A and B are macroscopic objects. The state s(A, B) will never occur again after sending a signal S_i.

The standard quantum mechanical measurement

           quantum system
                         |
    measurement apparatus
                         |
                     gauge
                         |
                        C

The standard quantum mechanical experiment involves a physical system, a measurement apparatus, and one researcher C who looks at the gauge of the measurement apparatus and makes the wave function to collapse.

Wigner's friend

We may ask if having one observer C who makes the wave function to collapse is still too many? The network of the branches of the Many Worlds is cut when C looks at the gauge. Is it so that the branches we cut off are no longer needed in calculating the probability amplitudes?

               quantum system
                             |
        measurement apparatus
                             |
                         gauge
                             |
                            C
                             |
                             |
                            v
                            D

Let us have another human D. Let D form an interference pattern from the measurement results which C communicates to him. In standard quantum mechanics, the signals which D receives from C are complex-valued wave functions and D can form an interference pattern from them. If C would cause a wave function collapse when he looks at the gauge after a measurement C made, then there would be no interference pattern from the signals which D receives from C.

The argument above shows that NO observer can ever cause a collapse of a wave function. All the information in the complete network of the Many Worlds has to be preserved, to reproduce the predictions of standard quantum mechanics.

In the famous Schrödinger's cat thought experiment, the cat cannot make the wave function to collapse, because the experiment would then produce results which are not consistent with quantum mechanics. Since any scientist, including myself, can be "Schrödinger's cat" to another scientist, no one can make the wave function to collapse.

The view of a conscious subject

The above is at odds with our personal experience as conscious subjects: we definitely observe the gauge of an experiment to take a real value. We do not observe a complex-valued probability amplitude. How can we reconcile this with the fact that nothing is ever pruned from the network of Many Worlds?

In our thought experiment, let C come to D and ask to see the results of the experiment. D shows the list of individual measurement results.

In rare cases, the state of C, s(C) was identical after he sent signals S1 and S2. The signals S1 and S2 formed an interference pattern. C cannot reliably remember what he observed in S1 and S2. If he did, there would be no interference. In this thought experiment, the interference of different branches of Many Worlds creeps in because C does not have a 100% reliable memory.

How does a conscious subject end up in a certain branch of Many Worlds? David Bohm's model solves that in the nonrelativistic (the Schrödinger equation) case through hidden variables. The hidden variables would determine where the subject sails in the network of Many Worlds.

We are not sure if anyone has been able to solve the technical problems when quantum field theory is incorporated to Bohm's model. John Bell worried that Lorentz covariance cannot be satisfied in Bohm's model. If the number of particles varies, how can one specify hidden variables for them?

Conclusions

Bell's inequality is one demonstration of the fact that we cannot prune any parts of the Many Worlds network. No collapse is consistent with standard quantum mechanics. The Copenhagen interpretation works in practice, but theoretically it is inconsistent with quantum mechanics.

A rubber sheet model for the Schwarzschild solution

UPDATE July 9, 2019: Note that the solution below is for a rubber sheet which has two spatial dimensions. A model for a 1+3D gravity should have a "sheet" with 3 spatial dimensions.

---

Suppose that we have a horizontal stressed large thin rubber sheet, such that the tension is uniform throughout its area. We paint polar coordinates on its surface, and also paint inside the thin sheet vertical coordinates.

The painted coordinates will be the global coordinates of a global observer.

The horizontal coordinates correspond to spatial coordinates in the Schwarzschild solution, and the vertical coordinate is the time.

Note that these are not the global coordinates of the standard Schwarzschild solution. Besides r, also the coordinate normal to r will stretch in the rubber. In the standard solution, r is the physical radial distance seen by an outside observer. In the rubber global coordinates it will not be the physical distance.

The rubber will stretch in such a way that its volume stays constant.

Suppose that the rubber sheet is very slippery. We put many small light circular steel disks on the rubber. The disks slide together and cause a depression in the rubber.

Is the solution for the rubber sheet the Schwarzschild solution?

Negative pressure and the Weak Energy Condition

We can model mass and the pressure caused by its gravitation simply by putting small steel disks on the rubber.

We can model positive pressure: take a steel ring and try to fit many small steel disks within. The rubber will bulge so that more disks fit in the area within the ring.

A shear stress on a square of rubber is a negative pressure along the diagonal which wants to grow and a positive pressure along the other diagonal.

How can we model negative pressure? In a static setup, a negative pressure is usually accompanied by positive pressure nearby.

The Weak Energy Condition seems to require that the positive mass density always "wins" a negative pressure. In the rubber model, that might mean that if we have a spring pulling some objects together, the mass of the spring will cause the rubber sheet to bulge downward, which means a positive Ricci scalar curvature.

Our anti-gravity device works by causing a large negative pressure in the grid close to a Schwarzschild mass. The Weak Energy Condition may require that the mass of the device must be so big that weight of the mass wins the anti-gravity generated by the negative pressure.

However, there is no proof that the Weak Energy Condition must hold. It is a conjecture.

Positive pressure tends to bulge the rubber sheet and make it thinner. A negative pressure should make it thicker. A thicker sheet means that time flows faster than in the Minkowski space far away.

If we have a steel ring and negative pressure within it, could we use springs to pull more of the rubber sheet inside the steel ring, to make rubber thicker there?

A positive mass stretches the rubber and makes it thinner. A positive pressure works much in the same way. A negative mass and a negative pressure should have the opposite effect.

We have not detected negative mass in nature, but negative pressure certainly exists.

A 2-dimensional rubber sheet is not rigid enough?

Suppose that we do not pre-stress the sheet but attach it to a large circular frame without tension.

Imagine a long equilateral triangle made of the rubber sheet. Assume that we pull with a force F from the sharp tip of the triangle. The negative pressure is then proportional to 1 / x, where x is the distance from the tip. The triangle will stretch an integral of 1 / x from, say 1, to infinity. The integral function is ln(x) and the integral is infinite.

If we pull a 3D rubber cone from the tip, the negative pressure is F / x^2, and the integral function is -1/2 F / x. The integral is finite for an infinitely long cone. The stretching is linearly proportional to F.

Problems with the thin 2-dimensional rubber sheet

If we use a pre-stressed sheet which is attached to a circular frame, then the sheet will probably behave like a harmonic spring if we press it with a finger at the center. Let the pressing small force be F.

Then the vertical depression z of the rubber is

       F = k z.

Let the rubber now form an angle α(r) relative to horizontal, where r is the radial distance from the center. The increase in the radius, measured along the new rubber surface is

        ∫ from 0 to r       [1 - cos(α(r))] dr
        = ∫ from 0 to r    [α(r)^2 / 2] dr

Suppose that we double the depression z at the center. The solution z(r) probably is such that the angle α(r) doubles and α(r)^2 is 4-fold. We see that the stretching of the original rubber metric is 4-fold for a 2-fold load F. But in the Schwarzschild solution, the metric stretches linearly with M.

If we do not pre-stress the rubber before attaching it to the frame, then for a very flexible rubber sheet and a strong force F,

        F = k z^3,

since the stretching is roughly ~z^2 and the angle α(r) ~ z. The metric stretches ~ F^(2/3).

A circular elastic rubber membrane without pre-stressing, under a load from a weight in the middle

           __

         (__( dN  ________

           dR              R

What happens with a very light load F is a harder question. Let us have a roughly square piece of rubber of a size dR × dN at some distance R from the center. More precisely, the piece is a sector of an annular ring.

The rubber is probably curved vertically downward along R, but curved upward along the normal N of R.

Let P be a point on the rubber, originally at a distance r from the center vertical axis. Let R(r) be the new distance measured straight from the center vertical axis, and S(r) the distance measured along the rubber surface. Let z be the vertical displacement of the stretched membrane.

Let N be a local normal axis to R at P. Let (R, N) define a local cartesian coordinate system. There is tension radially as well as normal to it. To keep the infinitesimal square of rubber vertically stationary, the force pulling the rubber down must be equal to the force pulling it up. From the partial second derivatives of z we get:

(1)       F_R d^2 z(R, N) / dR^2
       = -F_N d^2 z(R, N) / dN^2,

where F_R is the radial negative pressure on the square, and F_N is the negative pressure along N. We have assumed that the thickness of the rubber is the same on each side of the square.

The negative pressure F_R is proportional to the radial stretching (strain) of the rubber:

       F_R = k (dS(r) / dr - 1),

and the negative pressure F_N is from the stretching of the circle of an original radius r:

       F_N = k (R(r) - r) / r.

Let us look from the Internet if someone has solved the equations. A brief search does not return anything. If there is a uniform pressure on the membrane, then this is the well-known soap bubble film problem, and the solution is a part of a spherical surface.

We can write

       dS^2 = dR^2 + dz^2.

If dz / dR is small, then

       dS = dR (1 + 1/2 (dz / dR)^2).

We still need to calculate what is d^2 z(R, N)/ dN^2:

              (R + ΔR)^2 = R^2 + N^2,

       2 R ΔR + ΔR^2 = N^2

                          ΔR = N^2 / (2R),

if N is small. We have

       z(R, N) = z(R + N^2 / (2R))
                    = z(g(N)).

We use the formula for the second derivative of a function composition:

          z''(g(N)) * (g'(N))^2 + z'(g(N)) * g''(N)
       = z''(g(N)) * (N / R)^2 + z'(g(N)) * 1 / R.

When N = 0, we have

       d^2 z(R, N) / dN^2 = dz(R) / dR * 1 / R.

The differential equation (1) becomes:

       (R' - 1) (1 + 1/2 (z')^2) z'' = (1 / R - 1 / r) z'.

If z' is small, then we have

(2)   (R' - 1) z'' = (1 / R - 1 / r) z'.

---

We still need another equation to solve R as a function of r. Our differential equation (1) above was about vertical forces. What about the radial force? The radial negative pressure at the square side whose distance is R, is

       (R'(r) - 1) [1 + 1/2 z'(R)^2].

The radial negative pressure at the square side at R + ΔR is

       (R'(r + Δr) - 1) [1 + 1/2 z'(R + ΔR)^2].

In this calculation we cannot assume that the thickness of the rubber is the same at R and R + ΔR. Also, the rubber "square" must be treated as a sector of an annular ring. If there is no stress, then the area of the sector at R is

       ΔN T,

where T is the rubber thickness. The area at R + ΔR is

       (1 + ΔR / R ) ΔN T.

How does pressure affect these? Let us think of a practical example. We have a rubber disk whose radius is 100. There is a weight M in the middle, its radius is 1. The stretching of the rubber is 1% close to M. If cos(α) = 0.99, then α = 0.14. That would correspond to a mild slant close to M.

Let us consider a sector at R = 2. Let ΔR be 0.2. The stretching may make the rubber thickness T to be 1% lower at R = 2, and the thickness T is roughly 0.9% lower at R = 2.2. On the other hand, ΔR / R is 10%. This example shows that we can neglect the changes in rubber thickness within the sector. The dominant thing is that the area of the sector is smaller closer to the center.

In our example, z'(R) may be 0.14, and 1/2 z'(R)^2 = 0.01. Thus:

       1 + 1/2 z'(R)^2 = 1 + 1/2 z'(R + ΔR)^2

almost exactly. We then have that the radial force at R is proportional to

        R'(r) - 1

and at R + ΔR it is

       [1 + ΔR / R(r)] (R'(r + Δr) - 1).

We have ΔR = R'(r) Δr. Thus

       R'(r) - 1 = [1 + R'(r) / R(r) Δr]
                         * (R'(r) + R''(r) Δr - 1)
                     =   R'(r)
                       + R''(r) Δr
                       - 1
                       + R'(r)^2 / R(r) Δr
                       + R'(r) R''(r) / R(r) Δr^2
                       - R'(r) / R(r) Δr
   <=>
(3)           0 = R''(r) R(r) + R'(r)^2 - R'(r)
   <=>
          R''(r) = R'(r) [1 - R'(r)] / R(r),  

where we let Δr approach zero. If r_f is the radius of the frame where we attached the rubber, then

       R(r_f) = r_f.

If r_w is the radius of the weight, we have

        R(r_w) > r_w,

because the rubber has stretched horizontally under the weight. If there is no weight, then R(r) = r should be the only solution.

Let R(r) = r + f(r). The equation (3) becomes

       0 = f''(r) (r + f(r)) + (f'(r) + 1)^2 - f'(r) - 1
          = f''(r) (r + f(r)) + f'(r)^2 + f'(r).

If f(r) is very small and also f'(r)^2 is very small, then we have:

       0 = r f''(r) + f'(r),

a linear differential equation. Let us denote f'(r) = y. Then:

       0 = r y' + y.

Let us try y = C / r:

       0 = -C r / r^2 + C / r.

It holds for any C. Then

       f(r) = C ln(r) + C_1
   <=>
       R(r) = r + C ln(r) + C_1

If there is no weight in the middle, then, for example, R(1) = 1 and R(100) = 100. Then C_1 = 0 and C = 0. The solution is reasonable.

From R(r_f) = r_f we get

       0 = C ln(r_f) + C_1.

If R(r_w) = r_w', then

      r_w' - r_w = C ln(r_w) + C_1.

We can solve C, C_1:

       r_w' - r_w = C (ln(r_w) - ln(r_f))
   <=>
       C     = (r_w' - r_w) / (ln(r_w) - ln(r_f))
       C_1 =  -(r_w' - r_w) * ln(r_f) / (ln(r_w) - ln(r_f)).

---

Let us calculate an example. We set r_w = 1 and r_f = 100. Then

       C     = -(r_w' - 1) / ln(100)
       C_1 = r_w' - 1
       R(r) = r + (1 - r_w') ln(r) / ln(100) + r_w' - 1.

If r_w' = 1.1, then

       R(r) = r - 0.022 ln(r) + 0.1.

---

If we have drawn coordinates on r in an unstretched rubber, then

       R'(r) = 1 - C / r

and the new local radial metric is:

       ds^2 = (1 - C / r)^2 dr^2.

If C / r is small, the local radial metric is

       ds^2 = (1 - 2C / r) dr^2.

In the Schwarzschild solution, the local radial metric is

       ds^2 = (1 - r_s / r) dr^2,

but the Schwarzschild global coordinate r is not the same as our rubber r, but rather it is our R(r).

If R(r) is very close to r, then the Schwarzschild metric, indeed, agrees with our rubber metric, if 2C = r_s.

---

The equation (2) can be used to solve z(R):

       (R' - 1) z'' = (1 / R - 1 / r) z'
   <=>
       C z'' / r = (r - R) / (R r) z'
                   = -(C ln(r) + C_1) / r^2 * z',

where we assumed R is approximately the same as r. Let us denote y = z'. Then

       C / r y' = -(C ln(r) + C_1) / r^2 * y
   <=>
       0 = (C ln(r) + C_1) / r * y + C y'
   <=>
       0 = (ln(r) + C_1 / C) / r * y + y'
   <=>
       y' = - (ln(r) + K) / r * y,

where K is a constant.

What is the minimum possible mass-energy of a force field?

Our anti-gravity device cannot work if there is a minimum for the mass-energy of a force field. A very rigid body would need a force field where the force between two particles has a very sharp and deep pit in the potential energy at a certain distance.

Imagine a force whose minimum happens if particles x and y are at a distance 3 *10^8 meters. The particles themselves are very light, say 1 gram each, but you need a gigajoule of energy to change their distance 1 millimeter.

The particle A is in your hand and the particle B is close to the Moon. It takes one second for B to know that you have started pushing on A. You spent 1 GJ to push A just 1 millimeter in 0.1 seconds. The force is 10^12 newtons and the impulse is 10^11 newton seconds.

What happens when B will know that you have spent that much energy and impulse on A? The kinetic energy would imply that A & B start moving at 10^6 m/s, while the impulse would imply a speed of 5 * 10^13 m/s, much faster than light.

These figures do not make sense. A fix to the discrepancy is that the force field of A and B carries an enormous mass-energy. If pushing A 1 millimeter increases the local momentum and the energy of the field by the above numbers, things start making sense.

The minimum possible mass for a harmonic oscillator spring

       A /\/\/\____________/\/\/\ B
                           d

Let us have two particles, A and B, at a distance d from each other. Suppose that the force field acts like a harmonic oscillator spring, pushing the particles to an exact distance d. The rest mass of both A and B, including the mass of their force field, is m. The mass-energy of the force field is included in m. We calculate in Newtonian mechanics.

The force on the particle A is

       F = -kx,

where x is the displacement of A from the lowest potential distance.

Let us push A to the right with a force

       F_p = F_0 + kx

for T seconds, where T < 2d/c. We assume that since B will not know about the pushing until d/c seconds later, the field of B will remain static at A during the push. The particle A will accelerate to the right:

       a = F_0 / m,

and cover a distance

       s = 1/2 F_0 / m * T^2.

The push force as a function of time is

       F_p(t) = F_0 + k * 1/2 F_0 / m * t^2

The push impulse is the integral of the above over T seconds:

       p = F_0 T + k * 1/2 F_0 / m * T^3 / 3
           = F_0 T [1 + k T^2 / (6m)].

At x = 0, the push force is

       F_p = F_0,

and at x = s it is

       F_p = F_0 + k * 1/2 F_0 / m * T^2.

The work W done by the pushing force is the average of the two forces above, times s:

       W = 1/2 [2F_0 + k * 1/2 F_0 / m * T^2]
                * 1/2 F_0 / m * T^2
            = 1 / (4m) * F_0^2 T^2
               * [2 + k T^2 / (2m)].

The kinetic energy of the whole system A & B from the push impulse p is

       E = p^2 / (4m)
          = 1 / (4m) * F_0^2 T^2
          * [1 + k T^2 / (3m) + k^2 T^4 / (36m^2)].

The kinetic energy E must be less than the total work W done by the force:

       1 + k T^2 / (3m) + k^2 T^4 / (36m^2)
       < 2 + k T^2 / (2m)
   <=> (z = kT^2)
       z^2 / (36m^2)
       < 1 + z / (6m)   
   <=> (y = z / (6m))
       0 < 1 + y - y^2
           = -(y - 1/2)^2 + 5/4
   <=> (ignore negative values of y)
       y < sqrt(5)/2 + 1/2
          = 2.24 / 2 + 1/2
          = 1.62.

We have

        k T^2 / (6m) < 1.62
   <=>
        m > 0.10 k T^2

If we can assume that the field of B stays static at A until the time 2d/c, then we have

        m > 0.40 k d^2/c^2

where m is the mass of a particle, d is the distance of the particles, and c is the speed of light.

A limit on the mass of a space elevator tether

Suppose that we want to lift a payload which weighs F, from the surface of Earth, using a tether attached to the Moon. The length of the tether is L.

We assume that the weight of the tether is negligible.

We let the tether expand a distance e. The strain is then e / L.

Let us assume that the tether consists of Q particles simply chained together under a potential as described above.

Hooke's constant k_0 for the whole tether is

       F = k_0 e
    <=>
       k_0 = F / e

The constant k for an individual pair of particles is k_0 Q. The minimum possible mass for a particle is

       m = 0.40 k_0 Q L^2
               / (Q^2 c^2)

The minimum mass M of the whole tether is Q times the above mass m:

       M = 0.40 k_0 L^2 / c^2
            = 0.40 (F / e) L^2 / c^2.

It does not depend on the number Q of particles.

If the payload is 10,000 N (= 1 metric ton), and we allow an extension e of 10,000 km  in the tether from the Moon to Earth, we get

        M = 0.40 * 0.001 * 1.27^2 kg
             = 0.65 gram

as its minimum possible mass.

https://en.wikipedia.org/wiki/Ultimate_tensile_strength

Graphene has an ultimate tensile strength of 130 gigapascals, and its density is 1,000 kg/m^3. A graphene tether which has a mass of 50 metric tons could extend from the Moon to Earth and could stand the weight of one metric ton of payload.

If a force is mediated by a virtual particle, what is the minimum mass of the force field?

Our calculation above is based on a simple conservation of momentum and energy argument for a force which is mediated at the speed of light.

If we assume that a force is carried by virtual particles of some kind, can we derive bounds for the energy of the force field?

We have conjectured that the electric field of a charge is an elastic body whose mass-energy density is E^2. Electromagnetic waves are born from the vibration of the elastic body.

A limit on the mass of an anti-gravity device

The Schwarzschild radius of Earth is 9 millimeters. The factor

       1 - r_s / r

is roughly 1 - 10^-9 close to the surface of Earth. We conclude that if we want to use a rigid body to force the spatial metric close to Earth to something else than Schwarzschild, the body should stretch much less than 9 millimeters under its own weight.

The gravitational binding energy of Earth is 2.5 * 10^32 J or 3 * 10^15 kg. In an earlier blog post we conjectured that the deformation energy of the spacetime is roughly the same as this number.

If we would have a rigid body several times the size of Earth, we may guess that placing it next to Earth would reduce the deformation energy of the spacetime considerably in the volume which the body occupies.

We may try to calculate the forces on the rigid body from an assumption that the energy would be in the ballpark 10^31 J.

The virial theorem states that the thermal energy of a star is 1/2 of the binding energy. Our conjecture that the deformation energy = the binding energy bears a resemblance to the virial theorem.

The minimal mass for an anti-gravity tether

The tether should stretch much less than 9 millimeters under its own weight. Let us assume that we can hang the tether from a very strong structure at the height of one Earth radius, R = 6,371 km. That is, L = 6,371 km in our formula.

The average gravitational acceleration is the integral

        1 / r^2

from 1 to 2, times g = 9.81 m/s^2. It is 1/2 g.

If the mass of the tether is M, then the force is F = 1/2 Mg. We get

       M > 0.40 [1/2 Mg / e] (L^2 / c^2)

    <=>

       e > 0.40 * 1/2 g L^2 / c^2

          = 0.9 millimeter.

We were able to solve for the extension e > 0.9 mm, but the mass M can be chosen arbitrarily. It looks like an anti-gravity device might be possible.

Suppose that our tether can stand the force 10 N and it stretches 1 millimeter. What is the minimum mass M of the tether?

       M = 0.40 * 10,000 * 4.5 / 10,000 kg
            = 1.8 kg.

The tether can lift a payload of 1 newton.

...WORK IN PROGRESS...

The Einstein-Hilbert action and an anti-gravity device

https://en.wikipedia.org/wiki/Einstein–Hilbert_action

The Einstein-Hilbert action is

       S = ∫ over the whole spacetime
              [1 / (2κ) R + L_M] sqrt(-g) (dx)^4.

There R is the Ricci scalar curvature and L_M is the lagrangian density of matter fields. If the metric tensor has all the off-diagonal elements zero, then g is the product of the diagonal elements.

The metric tensor reveals the mapping of our global coordinates (x_0, ..., x_3) to nature. If we have a map of Earth in the Mercator projection, the metric tensor would reveal how many kilometers in nature corresponds to a centimeter on the paper map to the x or y direction. Everybody knows that the metric varies a lot throughout the Mercator map. The map seriously exaggerates the area of polar countries.

The term sqrt(-g) (dx)^4 is the "volume element". It is the 4-volume of the 4-cube (dx)^4 in our global coordinates when it is mapped into nature. The unit of the volume is a cubic meter times a second.

The integral seems to be infinite. We need to normalize it in a suitable way to treat practical cases.

The energy interpretation of the Einstein-Hilbert action

https://en.wikipedia.org/wiki/Principle_of_least_action

The history of a physical system should be a local minimum of its action. It is the principle of least action.

Suppose that the physical system ends up in a static configuration. Then the action integral is simply the integral over the 3 spatial dimensions times the time elapsed. The action has to be a local minimum, which means that the 3D integral is at its local minimum. Any small change in the configuration of the system would raise the integral over the 3D space.

If we interpret the 3D action integral as potential energy, then we see that the system wants to settle in a minimum potential. For example, if we have a system of attached springs, it will be static if and only if the potential energy of string tensions is at a local minimum.

If we have a static gravitating system, we can interpret the 3D spatial integral of the Einstein-Hilbert action as the "potential energy" of the system.

The Schwarzschild solution

https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric

The Schwarzschild interior solution is a static spacetime with a sphere of incompressible uniform fluid in the middle. The pressure is zero at the surface and grows toward the center.

The system is clearly a static physical system. The external Schwarzschild metric is static, too.

The lagrangian density L_M just contains mass of constant density ϱ kg/m^3. The fluid is incompressible and there is no energy stored into its deformation (compression).

If a curved metric can slow down the flow of time in the mass, then the integral in the Einstein-Hilbert action is reduced for a spacetime slice T_0 < t < T_1 where t is the global time of the static Minkowski observer far away from the mass. A curved metric will increase the integral over

       1 / (2κ) R,

where R is the Ricci scalar curvature.

Since the spacetime does become curved in the Schwarzschild solution, the curved geometry must reduce the integral over the mass more than it increases the integral over R.

In the rubber sheet model, curvature requires energy, but more energy is released from the potential energy of a metal ball on the rubber sheet. We see that the analogy between the rubber sheet and general relativity works in this respect.

In the rubber sheet model, it is the whole deformed area which contains positive deformation energy. In the Einstein-Hilbert action, the deformation energy is calculated from areas where R != 0. In the Schwarzschild solution, R != only inside the mass in the center. The deformation of spacetime outside the mass does not contribute to the integral because R = 0 there. Does this mean that the R term inside the mass accounts also for the deformation energy outside the mass?

One may imagine that for the rubber sheet, the total deformation energy can be determined from the deformation in the area where the sheet is pushed down by a metal ball. But is it so? We need to check if someone has studied rubber sheets.

For the rubber sheet around a metal ball, on a radial line, the second derivative of the elevation of rubber is negative. On a tangential line, the second derivative is positive. For a patch of rubber to stay static, the forces that result from non-zero second derivatives must balance each other. This is similar to R being zero in the Schwarzschild solution around a sphere. R is the average curvature. If R = 0, curvatures on various axes must add to zero.

The analogy between a rubber sheet and general relativity looks strong.

Let us draw x and y axes on a rubber sheet A metal ball resting on the sheet causes both x and y axes to stretch close to the ball. That is, an ant living on the surface of the rubber sheet will measure a longer distance between any points P and P' after the rubber has stretched. The ant uses a rigid ruler.

The volume of rubber stays the same: a z axis embedded inside the rubber will contract as x and y stretch.

https://en.wikipedia.org/wiki/Poisson%27s_ratio

Poisson's ratio for 3-dimensional rubber is almost exactly 0.5: if we stretch 1 % in the x direction, the y and z directions will contract by 0.5 %. The volume stays the same.

In the Schwarzschild solution, a local observer outside the mass measures a temporal distance (t, t') shorter than the global observer, while he measures a radial distance (r, r') longer than the global observer. In the rubber model, the rubber has stretched along the axis r and contracted along the axis t. The rubber has not stretched at directions normal to r.

Suppose that the Schwarzschild mass is a thin shell. Let us extend the external Schwarzschild solution inside the shell.

In the interior Schwarzschild metric, at the center of the mass, the metric of spatial coordinates is the Minkowski metric. The time element is

       4/9 (1 - r_s / r_g) dt^2,

where r_g is the radius of the sphere in global coordinates.

The volume of the rubber inside the shell has decreased: distances in time are less but they are not offset by larger distances along r. Here we have a discrepancy from a realistic rubber model.

An almost rigid body with a low mass close to a Schwarzschild sphere

If we put a rigid body close to the spherical mass in the Schwarzschild solution, then the rigid body restricts the ways spacetime can curve around the spherical mass. It will probably raise the minimal value of the Einstein-Hilbert action which is calculated for the mass M and the scalar curvature R inside and around the mass M.

We need a method to estimate how much the rigid body affects the minimal value of the Einstein-Hilbert action. Then we can calculate the force of anti-gravity on the rigid object.