Forcing spacetime straight with a moving perfectly rigid grid

We can force the spatial dimensions of spacetime to straighten up with a perfectly rigid object. But we were not able to prove that it would also remove deformation from the time dimension.

A static spacetime has a well-defined concept of simultaneity

Suppose that we have a static spacetime whose spatial metric is euclidean. Suppose that clocks run slower close to the origin of spatial coordinates, but far away, the space is the Minkowski space.

Since the spacetime is static, we have a well-defined concept of simultaneity in it. We can map the time of any spacetime event X to the time of a far-away static observer A by letting a light signal travel from A to X and back. A maps X to

       (t_0 + t_1) / 2,

where t_0 is the signal departure time in A's clock and t_1 is the arrival time.

We can define A's time as the global time. In the Minkowski space area, it is also the clock time of all other static observers, but close to the origin, static observers think that the global time runs faster than their own clocks.

A fast moving perfectly rigid grid

Let us assume that we have a static spacetime. Let us assume that the spatial metric is euclidean throughout the spacetime, but time may run slower in the area T where -1 < x < 1 and -1 < y < 1. Outside that area, the spacetime is strictly Minkowski.

We define a global time coordinate by a clock of a Minkowski observer, and global spatial coordinates through the euclidean metric.

          |        |
    D  --------------
          |        |   
 b(D)--------------  ----> v
          |        |

Let us have a perfectly rigid grid moving to the direction of the x axis. We assume that its speed measured in the Minkowski area is a constant v, and that in the Minkowski area at each global time t, the grid is perfectly rectangular and it is aligned along the directions of the global x and y axes.

Let us look at the movement of the grid bar B whose y coordinate is 0 and compare it to the bar B_2 whose y coordinate is 2.

Let us paint a dot D in B and a brother dot b(D) in B_2, so that their x coordinates in the internal coordinated of the grid are the same.

Let us denote by A the area -1 < x < 1.

The dot D enters A at the same global time as its brother b(D), and exits at the same global time, because the grid is perfectly rigid and rectangular outside the special area T.

The average speed, measured in the global coordinates, of D and b(D), during the journey through the area A is the same. D can move at the most at the local speed of light, measured in the global coordinates.

If we let v approach the speed of light c in the Minkowski space, then we know that the average speed of light in the special area T must be at least as high.

This still leaves open that at some spots, the speed of light might be very slow.

If the speed of the grid is time-independent

If the speed of the grid in the area T only depends on the position, v(x), and not on the time, can we show that length contraction forces the speed of light to the same as the Minkowski speed, throughout the area?

Let us do a little perturbation calculation.

Suppose that it takes a time 2 - d for D to go from x = -1 to 0, and a time 2 + d to x =1.

The average speed is 0.5. Let the speed of light be a constant 1 in the area. The inverse length contraction is

       1 / sqrt(1 - 1 / (2 - d)^2)
    + 1 / sqrt(1 - 1 / (2 + d)^2).

How does this compare to the constant speed inverse contraction

       2 / sqrt(1 - 1/4)?

It depends on the second derivative of

       1 / sqrt(1 - 1 / t^2) = f(t)^-1

at t = 2. The derivative of the denominator f(t),

       f'(t) = (1 - 1 / t^2)^-0.5 * t^-3

is positive.

The second derivative is

       f''(t) = (...)^-0.5 * -3 * t^-4
                  -0.5 * (...)^-1.5 * 2 * t^-3 * t^-3.

The second derivative is negative.

The derivative of f(t)^-1 is

       - f(t)^-2 * f'(t).

The second derivative,

       2 f(t)^-3 * f'(t)^2 - f(t)^-2 * f''(t)

is positive. We see that "perturbing" the constant speed tends to increase the inverse length contraction.

Finding the minimal inverse length contraction when the average speed through the area T is set, and the speed v(x) only depends on x, is a problem of variational calculus. If the speed of light is constant in T, then the optimum might be at a constant speed v(x) throughout the area.

The rigid grid forces the inverse length contraction to be the same in the bars B and B_2, in the area A. If the speed of light is the same throughout the area, v(x) = v gives the minimal inverse length contraction for B. It is the same as for B_2.

If the speed of light is lower at some spot of T, then the inverse length contraction grows for all v(x). Then the inverse length contraction of B is necessarily larger than B_2, which contradicts our assumption about rigidity.

We proved that the speed of light everywhere in T must be the same as in the Minkowski space, but our proof hangs on proving the variational calculus result above.

If we allow v(t, x) depend on the time t, too, then the variational problem is harder. Some strange, fractal-like function might beat the constant speed in optimality. We need to check literature about special relativity and if there are any variational calculus results.

In newtonian mechanics, there is no length contraction, and it is easier to prove that a perfectly rigid grid stays rectangular at each global time moment t.

Does a perfectly rigid small object raise the Schwarzschild solution energy - a perturbation approach

Let us again look at the rubber sheet model of general relativity.

Let us have a small, flat, lightweight object which is relatively rigid.

We have a heavy weight embedded into the rubber sheet. It makes a pit to the sheet with its weight.

If we embed the small flat object to the sheet, will it tend to move closer to the heavy weight or farther away?

The flat object makes a "perturbation" to the system. It slightly deforms the geometry of the sheet.

Let the flat object initially be very far away from the heavy weight. The flat object is "relaxed" - its own deformation energy D_f = 0.

The rubber sheet shape close to the heavy weight has minimized the deformation energy D plus the potential energy V of the weight.

When we move the flat object closer, it tends to increase D + V, because it changes the shape of the system slightly away from the previous local minimum of D + V.

The deformation energy D_f of the flat object itself grows from zero to some small value.

Is it possible that moving the flat object closer could decrease the energy of the complete system? Yes, if the weight of the flat object is large enough. The potential energy V_f will decrease.

But if the flat object is lightweight, it will move away.

The next question is if a similar perturbation argument shows that a mass under the Schwarzshild solution tends to repel a lightweight, relatively stiff small object.

What restrictions does a perfectly rigid object place on the metric of time?

We need to do a detailed study about what is the metric inside a perfectly rigid object in various cases.

A simple case is a rigid half-sphere whose round border is far away in the Minkowski space. Modifying any of the 3D spatial distances within the object would require an infinite energy. The Einstein-Hilbert action would become infinite through any such modification.

We conclude that the metric of the three spatial dimension must be perfectly flat inside the object. The 3D spatial geometry is euclidean.

What about the 4D geometry which involves time? Is it possible that the time dimension is distorted?

An analogous problem in three dimensions is the case where we know that the round border is in a normal global 3D euclidean geometry where the dimensions are x, y, and z. The middle of the object may have a deformed geometry.

We have a foliation of the object where the round border of each folio f is in the plane z = f. That, is, they are horizontal at the round border but may have a deformation in the middle.

The folios themselves have a flat euclidean 2D geometry.

Is it possible that the folios are deformed in the middle? Yes it is. We cannot prove that making 2 dimensions euclidean forces the 3rd dimension to be euclidean.

In the Schwarzschild solution, both time and the radial coordinate are deformed. But there may be solutions in other cases where just the time is deformed.

Actually, we may define a metric where the 3 spatial dimensions are euclidean but time flows slower for smaller r. Then we can calculate the tensor on the left side of the Einstein equation. If we can construct a system whose stress-energy tensor is equal to this, then we have an example of a euclidean 3D geometry but a deformation of the time dimension.

Minimizing the rubber deformation energy

If we have a rubber object under external stresses caused by weights in it, springs, or whatever, it tries to minimize the energy it has to spend on deforming itself plus the energy in the external stresses.

For example, a rubber sheet will bend under a weight, so that some potential energy of the weight is released, at the cost of an increased deformation energy in rubber.

If we enforce further restrictions on the rubber object by preventing its deformation in certain areas to certain directions, the energy of the system will in most cases increase.

If we move a perfectly rigid object close to a spherical mass, then we restrict the metric which spacetime can assume close to the mass. The deformation energy of spacetime will probably increase.

Can we reduce some energy by moving the rigid object there? The rigid object is weightless. We cannot reduce its potential energy.

We assume that the rigid object originally was not under any stresses. Its deformation energy is zero, and will stay zero because it is perfectly rigid.

This argument suggests that a mass will indeed repel a perfectly rigid weightless object, but we need to study this in more detail.

A perfectly rigid object causes repulsive gravity: an anti-gravity vehicle is possible

If the rubber sheet model of general relativity is accurate enough, then a weightless perfectly rigid object repels masses.

We assume that the rigid object was cast in outer space, under the Minkowski geometry. When we bring it close to a mass, where the geometry is Schwarzschild, the object refuses to obey the Schwarzschild geometry. There will be negative and positive pressure within the object. The pressure straightens up the geometry within the object.

It is like bringing a long straight steel bar embedded into a rubber sheet close to a metal ball whose weight has pushed the membrane down. It is obvious that the metal ball will roll farther when the steel bar comes closer. The deformation energy of the sheet is less when the ball is farther away.

In Newtonian gravity, this kind of a repulsion does not exist.

If we cast a perfectly rigid object under a Schwarzschild geometry, then it will refuse to adapt to the Minkowski geometry if moved away. Does it distort the geometry also in the neigborhood of the object or just inside the object? There is energy in the deformation of the rubber sheet. Does that energy cause gravitation?

The rubber model says that the object does change deformation also outside the object. It is like embedding a bent steel bar into the rubber sheet. If there are several bent steel bars, there will be various attractive and repulsive forces between parts of the bars.

We find that the rubber model predicts a rich spectrum of phenomena, while Newtonian gravity is always attractive.

In theory, we could make "anti-gravity" vehicles which would be able to float in the gravitational field of Earth.

An elastic rubber sheet analogy of gravity and other forces

The electric force pushes or pulls in the plane of the elastic rubber sheet

Let us have an elastic rubber sheet. We can model the pressurized vessel thought experiment by embedding a steel ring into the rubber and embedding small steel disks in the area which is surrounded by the ring. Steel is so strong that we can think of steel objects as perfectly rigid.

As we keep embedding small steel disks into the circular area, the rubber has to bulge and stretch to accommodate more.

Instead of steel disks we could embed positive electric charges. Their repulsion causes the rubber to bend and stretch within the circular area.

We see that the electric force is a direct force between charges and its direction is in the plane of the rubber sheet.

More precisely, we should model the electromagnetic field as embedded in the rubber sheet. The field pushes or pulls on charges. Then the electric force is not a direct force between the charges, but a force between the field and the charges.

Gravity is an indirect consequence of a force which pulls perpendicular to the plane of the elastic rubber sheet

Putting weights, for example, steel balls, on a horizontal rubber sheet is a well known analogy for the gravitational force. An outside force F (in this case the real gravity, not the modeled gravity) pulls them to a direction perpendicular to the sheet. The balls want to roll together because by joining their forces they can stretch the sheet more, and can settle into a lower position.

The "attraction" between the balls is an artifact caused by the interplay between the sheet and the perpendicular force which pulls the balls down. There is no "real" direct attractive force between the balls.

In this model, the electric force and the gravitational attraction are very different forces. This may explain why we cannot find a sensible definition for the energy density of the gravitational field - it is because there is no "gravitational force" in the same sense as there is an electric force.

However, we probably can find a sensible formula for the positive energy stored in the deformation of the 3D space. The total energy is the deformation energy plus the potential energy in the field of the outside force F. We can set the potential of F such that all energies are always positive. We get rid of awkward negative energies.

Cosmological models: the de Sitter space and dark energy

Let us model the expanding universe with the usual expanding rubber balloon model.

Dark energy can be explained as a positive energy of empty space, which in turn means negative pressure. The 3D space wants to contract, to turn into a negative curvature saddle shape, to reduce its energy content. In general relativity, this paradoxically leads to an accelerated expansion of the balloon.

Our planar rubber membrane model above, if generalized directly to a rubber balloon, predicts that a negative pressure would make the balloon to contract. We need to improve our model so that it explains the behavior of cosmological models.

Maybe the effect of negative pressure is not that much to contract the rubber membrane, but to make the curvature of the membrane negative. Locally, that would contract the volume enclosed into a ball of a fixed radius.

Positive pressure tries to increase the curvature of the rubber membrane. Locally, that will make the volume of a fixed radius ball bigger.

We see that locally, the effect of a positive/negative pressure is consistent with our planar rubber membrane model. It is like having an infinite balloon.

What to do to extend the model to balloons of a finite size?

In a cosmological balloon model, energy is minimized only locally, not globally

The cosmological balloon is huge and its parts get information only at the speed of light from other parts. It is possible that when each individual part tries to fall into a lower energy state, the balloon as a whole will develop into a higher energy state.

Positive pressure tends to bulge the balloon locally, that is, to increase its positive curvature. The global effect is that the balloon contracts to make the curvature bigger. A global observer outside the universe will notice that the balloon is traveling towards a higher energy state, but no part of the balloon is aware of that. Minimization of global energy might require faster-than-light communication.

Similarly, a global negative pressure in the balloon causes that each individual part wants to reduce its curvature, which in turn leads to the balloon becoming bigger, and the total energy of the whole universe keeps increasing. In a hypothetical inflation scheme in the Big Bang, the energy of the universe would grow phenomenally fast.

Physics in the Minkowski space seem to conserve energy, and try to divide energy evenly among various degrees of freedom. There is no need for faster-than-light communication to implement this. But if we allow varying geometries of the universe, the maybe it is not possible to conserve energy.

Since every concrete, everyday, rubber sheet or balloon model, which we build, lives in the Minkowski space, it conserves energy in the global view. These models are not very good at modeling the behavior of a cosmos where energy is not conserved.

How much positive energy is there in deformation of space?

Our spherical vessel thought experiment from May 23, 2019 can be used to calculate a ballpark value for the positive deformation energy of 3D space.

When we remove the mass, the space deformation in the vessel is produced by pressure.

Let us modify the experiment in such a way that the mass is originally in a thin shell of radius r. The mass is then lowered down in small amounts, so that at the end we have a spherical mass of a radius r, and of a constant density. That is, we build from a thin shell a solid sphere of a constant density. We can collect a "binding energy" E when we lower the pieces of the mass down.

The volume of the sphere of a radius r is then slightly larger than the corresponding sphere in the Minkowski geometry. We then put the rigid vessel around the sphere and fill it with weightless incompressible fluid.

After that we start lifting the mass gradually back to its original position in the thin shell. We may assume that the geometry of space inside the vessel stays roughly constant through the process, because the volume of the vessel stays constant.

We have to do a work E' in lifting the mass. A very rough estimate is that E' = 2E. Then the energy E'' of the pressured vessel is roughly the same as E, the "binding energy" of the original spherical mass.

How do we interpret this? While we lowered the mass, we were able to harvest an amount E of the potential energy. At the same time, an equal amount E of potential energy flowed into the deformation of space. When we lift the mass back up, we have to pay back the energy 2E.

Now we have a rough guess of what is the energy of a deformation of 3D space. It is of the same order of magnitude as the gravitational binding energy for a mass which produces a similar deformation.

Most of the energy in a normal mass is in the mass itself. The deformation of space around it carries a very small amount of energy unless we are dealing with a black hole.

This is not much different from an electric charge. If we assume that all the mass-energy of an electron lies in its electric field, then almost all mass is contained within a few classical electron radii, where the classical radius is 2.8 * 10^-15 meters.

If we have a spherical shell made of metal, and which contains some reasonable amount of charge, then the energy of the electric field outside the shell is much less than the mass of the shell. The energy in that field is at a macroscopic distance from the shell, in contrast to the field of a single electron.

Harvesting kinetic energy of a mass through a gravitational wave antenna

Any elastic solid object acts as a gravitational wave antenna. It resists a change in the geometry of the 3D space inside it. A change in geometry will, in general, produce vibrations.

If we have a spherical mass moving by, it distorts the geometry of space with its Schwarzschild metric. It is like a "gravitational wave" which moves at a slow speed.

The Einstein-Hilbert action

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

The action is

      S = the integral over the whole spacetime

             (1 / (2κ) R + L_M) sqrt(-g) d^4x,

where R is the scalar Ricci curvature of the metric and L_M is the lagrangian density of matter and other fields. The system tries to find a minimum of the action. The first term with R might be interpreted as some kind of energy of the deformation of space.

In the Schwarzschild solution, S is zero outside the gravitating mass. If we model curved space with a rubber membrane, then obviously the deformation does contain positive energy also outside the mass, since the membrane is deformed there. The first term above cannot be the deformation energy density. It can be the correct total energy over the whole space, though.

If a spherical mass moves by us, we can harvest energy from the outskirts of the gravitational field through the deformation it causes in a flexible object. How do we describe that with the Einstein-Hilbert action? The energy L_M increases in the flexible object in the outskirts of the field. It has to be balanced by a change in the kinetic energy of the mass. It is better to keep the spherical mass static. The flexible object flies by and starts to vibrate. The energy came from the kinetic energy of the flexible object. The object did not "harvest" any energy from the deformation but converted some of its own kinetic energy into vibrations.

Pushing an electric charge and pushing a mass - what is the difference in production of waves?

Pushing an electric charge

We have conjectured that the reason why a linearly accelerated electric charge radiates is that its electric field is a "flexible solid object", and the field carries a positive energy density E^2.

The pusher will feel that he is pushing a flexible object. If he wants to give 1 newton second of momentum to the charge, he has to push over a longer distance than he would need in the case of a totally rigid object. He has to do some extra work which does not go to the kinetic energy of the charge. This extra work is radiated away as electromagnetic radiation.

When studying electromagnetism, we can work in a Minkowski space with no gravity. We can assume that the metric stays the same, and we have an infinitely rigid rod with which to push. The extra energy cannot go to the deformation of the rod, as it is infinitely rigid.

The pusher concludes that some of the work he did was lost: it did not go to the kinetic energy and it did not go to the rod. It did not go to the static electromagnetic field of the charge. The energy had to go to the global electromagnetic field.

We have not yet calculated what kind of a flexible object the electric field has to be, to explain the radiative dissipation of energy. Can we assume that its energy density is E^2 and the flexibility comes from the retardation of the field? The energy of the field is infinite if integrated to r = 0. Should we stop at r = classic electron radius if the charge is one electron?

Pushing a mass

The energy content of a gravitational field is negative, if one tries to derive it from the same principles as in the case of an electric field. The concept of an object of mass m carrying a negative energy flexible field is perplexing. Would the inertial mass of the object first appear greater than m to the pusher? He would need to push a shorter distance to convey 1 newton second of momentum than in the case of a totally rigid object? We would have a perpetuum mobile, if the kinetic energy of the mass would increase more than the work done by the pusher.

A flexible positive energy field loses energy as radiation, if the object is pushed. A negative energy field would make the object to gain energy from empty space when pushed.

A negative energy field is a bad idea. Do the pseudotensors of Landau and Lifshitz and others make sense, as they seem to assign a negative energy to the field?

UPDATE May 29, 2019: a perfectly rigid rod seems to have a gravitational repulsion with the mass. We need to think again what actually happens if we try to push with a perfectly rigid rod. See our blog posts on May 28, 2019.

In the previous blog post we showed that if 3D space stretches to accommodate a longer rod when we do the pushing with a perfectly rigid rod, then the pusher does some extra work compared to the case of a fixed metric, and that energy might be the source of gravitational waves.

Does the Schwarzschild exterior metric carry positive energy?

If deformations of 3D space mean positive energy stored in space, one may ask if, for example, the 3D space deformation around a spherical mass carries positive energy.

The local observer finds the stress-energy tensor T zero around the spherical mass. He thinks that the energy content of space is zero there. But it is possible that a global observer would assign a positive energy content to space.

Then the far-away observer might see flexibility in the push from two sources: from the flexibility of the positive energy gravitational field, and the flexibility of 3D space under the infinitely rigid rod. We need to calculate what would be the contributions of these two effects.