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.

Does a constantly accelerated mass produce gravitational waves?

If it does, then the accelerator must feel a force which resists the acceleration. If the waves transport energy at the speed of light, the waves can only carry a small amount of momentum away.

Where does the extra momentum go?

We can use, for example, a very long spring to accelerate the mass. The other end of the spring is attached to another mass far away. We assume that we can ignore the field of that other mass, at least for a moment.

Note that if we rotate the mass, then waves which transport energy at the speed of light, can transport a considerable amount of angular momentum if the waves "start their journey" at a great distance r away from the rotating mass. We can keep pumping angular momentum to the field and light-speed waves can take that angular momentum away.

For the electric field, we have suggested that the extra momentum goes to pulling the mass-energy E^2 of the outer parts of the electric field up to the pace of the acceleration. This is completely analogous to pushing a flexible solid object from the middle.

In the case of gravity, there is no concept of an energy density of the field. There can be positive energy in the field, as shown by gravitational waves. A binary neutron star keeps losing its angular momentum to gravitational waves. This means that the waves are carrying mass-energy out of the system.

If we use a newtonian approximation, then the field energy of a gravitational field appears to have a negative energy. Or does it? Suppose that we assign the whole rest mass of an object to its field energy, to lift the field energy positive everywhere. The combined field would contain the newtonian negative energy field plus the rest mass field. The extra momentum could be put to bringing the combined field up to the pace of the acceleration.

We have not found a way to dump the extra momentum to a field of negative energy.

Suppose that we have a static shell of mass, such that it is very close to its Schwarzschild radius. In that case, the negative energy of the newtonian field almost offsets the positive energy of the rest mass field.

General relativity does not have a concept of gravitational field energy. The sources of the stress energy tensor are rest mass, momentum, pressure, and shear stress. None of these is connected to the gravitational field.

The talk about momentum conservation and field energy may make sense in a newtonian approximation of general relativity.

When a binary neutron star loses its angular momentum and energy to gravitational waves, maybe we can interpret that the waves are in the combined field of rest mass and the negative energy newtonian field?

We can think that the whole mass-energy of an electron resides in its electric field. Similarly, we might think that the whole mass-energy of a 1 kg weight resides in its combined field.

Spacetime as a flexible plane

https://arxiv.org/abs/1603.07655
The Mechanics of Spacetime - a Solid Mechanics Perspective on the Theory of General Relativity (2018)

T. G. Tenev and M. F. Horstemeyer have an analogous model of general relativity, where spacetime is a flexible surface which is a flat plane if it is the Minkowski space. Let us check what their model says about the field energy of the gravitational field.

The flexible plane has an elastic modulus Y and its thickness is L.

Gravitational waves are transverse waves of the plane.

The Landau-Lifshitz pseudotensor

https://en.wikipedia.org/wiki/Stress–energy–momentum_pseudotensor

The Landau-Lifshitz pseudotensor seems to satisfy the properties which we would expect from the "energy" of a gravitational field.

Let us check what it says about conservation of momentum if a linearly accelerated mass produces gravitational waves.

For a static, spherically symmetric mass, the L-L pseudotensor must be negative, since the total energy of the system is the stress-energy tensor T plus the L-L pseudotensor. The total energy is less than T.

This does not help us. We have the problem where to put the extra momentum when we start pushing the matter, and we have not found a way to put it to a field of negative energy.

Pushing with a rod from far away

For an electric charge, its inertial mass feels less at the start of the push. How could we make the inertial mass of arbitrary matter feel less at the start of the push? Suppose that we are far away, and start pushing with a long rod. We may assume that the rod is attached to the mass. It is kind of a handle, by which a remote observer can move the mass.

The push create pressure in the rod, which in turn starts to attract the pushed mass.  Could this attractive force absorb the surplus momentum from the pushed mass?

Suppose that after a while, the global observer is pushing with a force F. The mass feels a force F' from the rod. If the global observer pushes 1 meter, the observer on the surface of the mass observes the rod to move the same 1 meter. The force F' has to be greater than F, because if the surface observer would send  the work done by F' as light to the global observer, there would be a redshift.

Suppose that the outside observer applies a 1 newton force to the rod for 1 second. Since gravitational waves can take a negligible amount of momentum away, the 1 newton second momentum must be absorbed by the system mass & rod & their gravitational field.

Exactly as in the case of an electric charge, the extra energy which goes to gravitational waves can be explained if the push takes a longer distance than it would with a totally rigid object.

Transforming momentum in a low gravitational potential to a high gravitational potential

How do we compare a momentum in a low gravitational potential to a momentum in the surrounding Minkowski space? Suppose that we have a weight inside a spherical mass shell and the weight is attached to a long rod which extends far away to the Minkowski space.

The far-away observer uses the rod to move (slowly) the weight.

Let the gravitational potential be V, where V = 1 corresponds to the Minkowski space, and V = 0 would correspond to a black hole horizon. The redshift + 1 is 1/V.

Suppose that the far-away observer applies a force of 1 newton to the rod for 1 second. The rod moves 1 meter, which corresponds to 1 joule of work. He may interpret that he was pushing a weight of 0.5 kg. The final speed of the weight is 2 m/s. Its momentum is 1 newton second and the kinetic energy is p^2/(2m) = 1 joule.

The local observer inside the shell measures that the weight moved 1 meter and it took V seconds. Its final speed is 1/V * 2 m/s. If he measures the weight as 0.5 kg, then its momentum is 1/V newton second and its kinetic energy is 1/V^2 joules. This does not match the fact that if the local observer sends the energy back to the far-away observer, he will receive 1/V joules. We have a perpetuum mobile.

The solution for the discrepancy is that the weight which the local observer measures is only V * 0.5 kg. Then

       1/2 mv^2 = 1/2 * V * 0.5 * 2^2 * 1/V^2 J
                        = 1/V J.

The two observers agree on the length of 1 meter and they agree on the momentum of the weight in newton seconds.

The far-away observer thinks that the inertia of the weight is 1/V times of what the local observer measures. He also measures a time which is 1/V times the time measured by the local observer.

The far-away observer thinks that the energy and the force are V times the energy, and the force which the local observer measures.

Note that if the far-away observer pushes the whole gravitational system, then he will see the inertia of a weight less than the local observer.

Pressure inside a vessel makes the 3D space to stretch

Suppose that we have a spherical vessel with a very strong wall. We put there liquid and apply a very large pressure. Then there is positive pressure inside the vessel and a negative pressure in its wall. The pressure inside the vessel acts as a source in the stress-energy tensor. The geometry inside the vessel probably is not flat.

It would be logical if the space inside the vessel, that is, the 3 spatial dimensions, would stretch slightly to ease the pressure.

Birkhoff's theorem says that the geometry outside a closed spherically symmetric system is a fixed Schwarzschild geometry and cannot change through any process inside the closed system.

If the liquid is incompressible, then we can increase the pressure greatly with just a little bit of locally stored energy.

It has to be that the negative pressure in the vessel wall exactly cancels the geometry change inside the vessel, so that an outside observer sees no change in the geometry.

This analysis suggests that we can consider the 3-dimensional space, at least in slowly changing setups, as flexible, such that it stretches to reduce an increase in pressure.

Consider the following thought experiment. We build a spherical mass by lowering down thin spherical shells of matter. We collect the potential energy E which we gain in the process.

After completion, we enclose this mass with an infinitely rigid weightless spherical shell and fill the shell with incompressible weightless fluid. The metric allows us to put there more fluid than we could in a Minkowski geometry.

We then remove the mass gradually, by lifting spherical shells of mass up. We assume that we have some magic equipment which can move the mass through the infinitely rigid wall. We have to use an energy E'.

By Birkhoff's theorem, the metric outside the wall is a Schwarzschild metric. The circumference of the wall measured in global Schwarzschild coordinates is the same as in local coordinates. The circumference stays the same, as the material of the wall is infinitely rigid.

Once we have moved all mass away, we have a vessel which contains more incompressible liquid than it could in a Minkowski space. There has to be pressure inside which modifies the metric, so that the volume can fit in. The pressure can do work E'' if we let the fluid come out of the vessel.

To avoid a perpetuum mobile, E' has to be equal to E + E''. In the lift phase, there is an extra gravitational field which makes the work to do the lifting E' larger than the energy E we gained from the lowering.

Anyway, we proved that pressure can make more incompressible liquid to fit inside an infinitely rigid vessel. Pressure makes the 3D space to stretch. An incompressible liquid is not incompressible in the global view of an observer just looking at the walls of the vessel, even though it is incompressible for a local observer. This is because the 3D space itself can stretch.

When pushing with a rod, the energy for the gravitational wave comes from the stretching of the 3D space?

Suppose that we have an infinitely rigid rod. That is, the local observer will always measure the distance of atoms in the rod exactly the same, even if the rod is under heavy stress. When a far-away observer pushes a mass with the rod, there is pressure inside the rod. If the 3D space stretches to ease the pressure, the far-away pusher will interpret that the rod is elastic, even though local observers think it is perfectly rigid.

If the pusher pushes for 1 second with a force of 1 newton, he notices that the force will do work over a longer distance than if he would be pushing a totally rigid system. He has to do more work than he would need in the case of a totally rigid system. This extra work does not go to the kinetic energy of the mass but in the deformation of the rod, which itself is caused by the deformation of space. The pusher has pushed some energy into the deformation of space. This energy is probably the energy which is then radiated away as gravitational waves.

Since the rod is infinitely rigid, we cannot store elastic energy into its deformation. All the extra energy went into deforming space.

What happens when a charge falls through a black hole to a white hole?

We now believe that Gauss's law holds for the electric field E measured by local observers, at least in a static gravitational field.

We need to check if there is a proof that Maxwell's equations can be satisfied also under a dynamic gravitational field.

Gauss's law does not hold for a global observer far away, who measures the field strength by the energy he can receive if a test charge is moved a vector ds. The energy is transmitted to the global observer as light, and that light suffers a redshift if it has to climb from a gravitational potential well. The global observer sees the field E weaker than the local observer. The global observer may interpret this that the electric field has polarized space. Some electric lines of force end at opposite charges which polarization has brought there.

The global observer probably sees the zone close to the horizon of a black hole as an equipotential surface, and that the electric lines of force all end before reaching the horizon. For him, the space enclosed by the horizon is an equipotential area where the electric field is zero.

On the other hand, local observers see the lines of force continue unbroken to the horizon. If an observer jumps into the black hole, then the no drama hypothesis implies that he will see electric lines of force continue smoothly also behind the horizon.

Electric field lines which enter a black hole and come out of a white hole

Let us have a positive electric charge close to a black hole.

We may ask what happens if a black hole is connected through a wormhole to a white hole. What kind of an electric field E does an observer measure if he jumps in, during his journey into another universe?

Let us have two observers who jump. Observer A jumps from the side of a positive charge, and B from the opposite side.

A will probably see himself fall in the direction of the field line throughout his journey and then pop up from a white hole.

B will probably see himself falling in the opposite direction of a field line.

Maybe the white hole will appear to be an electric dipole, when A and B observe it in the universe where they fell? The white hole seems to have an induced positive charge on the side of A and an opposite negative charge on the side of B. The universe on the white hole side will contain a dipole field which is extending at the speed of light into the universe.

For a global observer on the black hole side, there seems to be induced negative charge close to the horizon on the side where A jumped, and induced positive charge on the side of B.

An electric charge which falls through to a white hole

Suppose that A throws a positive charge to a black hole and A jumps in after the charge. B jumps in from the other side.

Both A and B will probably observe themselves falling against the direction of electric field lines.

If the field lines do not break for a local observer, then the field lines probably continue unbroken out of the white hole up to the charge which already popped out of the white hole. Then a global observer on the white hole side probably will see an induced negative charge around the white hole horizon. The field lines which start from the positive charge end up at the negative charge close to the horizon.

But how much negative charge does the global observer see in the white hole?

What is the electric field like close to a white hole?

A global observer sees the electric field weak close to a black hole, because the energy that a local observer may get by moving a test charge, is transmitted as light to the global observer, and there is a redshift in the light.

If there is a local observer close to a white hole and he sends light to a global observer, then there is a blueshift. The global observer will see the electric field E stronger than a local observer.

Suppose that we have a positive charge which did not pop out of the white hole but has come to the universe some other way.

A black hole attracts field lines. It is logical that a white hole repels them. Let us assume that local observers see every line avoid the white hole. They may interpret this that behind the horizon there are induced charges which prevent field lines from entering. It would be like what happens to magnetic field lines at a superconductor.

The local observers see E very weak close to the white hole horizon. Maybe a global observer sees it as moderate strength?

The global observer must be able to define a global electric potential. If the local observer sees a very weak field E close to the horizon, then the global one might see a potential which looks reasonable.

Imagine a "neutron star" which is made of negative mass. It would repel most of the electric field lines, but allow a few of them pass through. To enforce energy conservation and avoid a perpetuum mobile, the global observer must see the field strength E roughly the same inside the star and around it. Then the local observer must see E very weak inside the star. This example suggests that the local observer, indeed, sees E weak close to the white hole horizon.

But now we have a dilemma. If a positive charge pops out of the white hole, how can its field lines go inside the hole, as seen by a local observer? A local observer is supposed to see a moderate electric field E all the way as he goes down the wormhole and comes out. A moderate field E for a local observer at the horizon is an infinite field for the global observer.

An electric field does polarize the space around a gravitating object - in the global view

In this blog we have several times mentioned that the electric force seems to be weaker in a low gravitational potential - from the point of view of an observer far away.

When a local observer measures the field, he may get a result that moving a 1 coulomb test charge for 1 meter horizontally consumes 1 joule of his energy. But when he transmits that 1 joule as light to the distant observer, there is a redshift, and the distant observer measures the energy less than 1 joule.

For a static electric field, there should exist a static global potential function to ensure the conservation of energy. The natural way is to define that potential as the energy which the distant observer must use to move a test charge to a specific location in space.

If we define the electric field as the gradient of that global potential, we get the electric field from the distant observer's perspective, and he sees less lines of force than a local observer in a low gravitational potential. Thus, the global observer sees that Gauss's law does not hold in a gravitational field. This assumes that the local observers see Gauss's law holding in their local measurements.

The global observer may interpret this that the electric field has electrically polarized space in a low gravitational potential. Now we see the analogy to an optically dense medium. An electric field causes polarization in the charges of the medium. We may interpret the gravitational lensing effect in the same way as how ordinary lenses work: the speed of light is slowed down by polarizable charges in the medium.

A paradox in an accelerated flexible object and the horizon?

Suppose that we accelerate a charge for some time at a constant acceleration. Part of its infinite electric field falls behind the apparent horizon of the accelerated frame. How does the field behind the horizon "tell" to the accelerating charge that it should send some momentum to keep the field accelerating?

In the Minkowski coordinates, the far behind parts cannot send a light signal to the accelerating charge as long as the charge keeps accelerating. Does this pose a problem?

https://physics.stackexchange.com/questions/440877/why-do-accelerating-bodies-have-an-event-horizon

The apparent horizon is at a distance c^2/a from the accelerating particle, as seen by a static Minkowski observer. For an acceleration a = 10 m/s^2 that is one light year.

If we have any large mechanical, interacting system, then parts of the system may fall behind the apparent horizon of an accelerating part of the system. Thus, this is quite a general question of special relativity. Is that a problem that some far parts of the system cannot send signals to the accelerating part?

If the accelerating object is pulling behind it an inelastic rope, an observer hanging on the rope close to the apparent horizon will feel an enormous acceleration. The rope will break before it reaches the horizon. But an electric field cannot break? What happens to it?

Let us look at the problem in the static Minkowski frame. Let us start accelerating the charge at 10 m/s^2. A year after that, the field one light year away will learn that the charge has started moving. It at first moves at a moderate pace. The electric field of the charge will start deforming in a smooth manner. There are no singularities or infinities at the horizon, from the point of view of the Minkowski observer.

If there were another observer hanging from an inelastic rope which is attached to the charge, that observer would have an infinite acceleration at the horizon, and he would observe something singular or infinite at the horizon. This is like the black hole horizon: it looks like a singularity for a static observer, while for a freely falling observer there is no drama. The natural way to look at the field is the static observer, not the infinitely accelerating one.

If we can find a solution of Maxwell's equations in the Minkowski space, then an apparent horizon poses no problem. The accelerating charge does not need to "know" what happens behind the horizon. It is the global solution of the field equations which has all the necessary information.

The apparent horizon may hide lots of things from the accelerating observer. Earth might fall behind the horizon. There is no problem or contradiction in this if we can find a global solution which respects what is in the incoming light cone of a spacetime point.

In summary, there is no paradox.

A static charge and an accelerating observer in a Minkowski space

We may imagine that the charge has painted the electric lines of force to the entire Minkowski space. The accelerating observer can see the painted lines. What does he see?

Nothing special. His view is like an inertial observer who moves at the same speed as the accelerating observer. If his speed is close to light, he will see the painted space contracted in the direction of his velocity. The field in most cases appears to be almost normal to his velocity vector.

Suppose that our observer is pulling behind him an inelastic rope. Suppose that we have another observer hanging on the rope, close to the apparent horizon of the first observer. This second observer has an enormous acceleration relative to a static Minkowski observer. The second observer moves almost at the speed of light. For him, the field will look almost "singular", but that is just because his own motion is so extreme. The field looks very smooth for a static observer.

We see that the "singular" conditions which an accelerating observer close to the horizon encounters, are just an artifact of the "singular" movement of the observer.

An accelerating observer sees light coming from close to his apparent horizon as enormously red-shifted. He cannot see past the horizon. His inability to see is not because there is something special at the horizon but because of his movement. If there is another observer next to him who moves at the same speed, but does not accelerate, that other observer does not see any apparent horizon. He sees a huge redshift in light coming from the horizon, but he can see also behind the apparent horizon of the accelerating observer.

Does a charged particle fall as fast as a neutral particle?

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

Let us try to solve the paradox. Fritz Rohrlich (1965) claims that a charged particle and a neutral particle fall as fast in a gravitational field. Is that true?

Let us model the charged particle and its electric field with an elastic extended object. The object extends far away, outside the gravitational field. Now it is obvious that the charged particle will not fall as fast as a neutral particle. The parts of the electric field outside the gravitational field will slow down the fall.

Does this break the equivalence principle? No. The system in a falling laboratory is not isolated from its environment. It interacts with the electric field which extends to the infinity.

The interaction with the outer parts of the electric field is small, but it is not small compared to the interaction when we oscillate the charge to create long electromagnetic waves. Thus, we must consider the interaction in our analysis - we cannot ignore it.

Does a constantly accelerated charge radiate?

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

Now that we have a better understanding of Maxwell's equations, we can study this classic problem again.

A charge in a constant acceleration in a Minkowski space

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

Edward M. Purcell's approach fits this case. In the "real" physical world there do not exist eternally accelerated charges. The acceleration has started some time in the past and will end at some time in the future.

The diagram at top right of the link, and also within the text of the link, has the growing circle which contains the end result of the acceleration. Then there is a "transition area", and outside is the field before any acceleration started.

For a long smooth acceleration, the transition area is very wide.

How does Purcell accomplish to connect the lines of force inside the circle and outside the circle?

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

He assumes that there is one "arc" (half a wavelength) of an electromagnetic wave traveling outward at the speed of light. That half-wave pulls the electric lines of force to the right direction, so that they can connect to the lines outside the circle. The magnetic field lines are like in the diagram at the top of the Wikipedia link.

The Purcell approach looks at the system in the frame of a static Minkowski observer.

When the charge is being accelerated, the static observer interprets the electromagnetic field as having a single arc of a forming electromagnetic half-wave. The arc grows in length at the speed of light as long as the acceleration continues. The Poynting vector E × B which the static observer sees, shows energy flowing outward from the charge at the speed of light. We can say that the static observer sees "radiation" flowing out.

When the acceleration ends, the half-wave starts moving outward at the speed of light. Then it is clearly an "electromagnetic wave" moving outward, and everyone would agree that the charge has radiated.

Any inertial observer in the Minkowski space can fill the entire space with static spatial coordinates. By static in this case, we mean that the distances between the intersections of the grid do not change in time. Such a coordinate grid is good for analyzing energy flow.

However, an observer who co-accelerates along with the charge, cannot define such a global static grid which would move along him, because for him, part of the Minkowski space is behind a horizon. He can, however, define local static coordinates. In his coordinates, the electric field of the charge looks static. The field looks static also for an inertial observer who just at this moment t moves along with the charge. The inertial observer does not see any magnetic field close to him. He interprets that the charge is not radiating close to him. The Poynting vector E × B is zero.

Thus, the accelerated charge does radiate from the point of the view of a global inertial Minkowski observer, but in a local partial coordinate system of a co-accelerating observer (or an inertial observer briefly moving along) it does not radiate.

https://www.sciencedirect.com/science/article/pii/0003491680903607

David G. Boulware (1980) writes about the significance of a horizon in the accelerating frame.

Radiation reaction

Does the radiation of an accelerated charge produce a force which slows it down?

https://en.wikipedia.org/wiki/Abraham–Lorentz_force

The Abraham-Lorentz force is proportional to the time derivative of the acceleration. In a constant acceleration, there would be no Abraham-Lorentz force.

We have previously noted that the Abraham-Lorentz formula tries to derive the radiation reaction from incomplete information: it only contains the time derivative of the acceleration of the charge, and does not contain full information of the complete electromagnetic field around the charge.

Let us look at our analogy of a flexible solid object being accelerated. We start pushing the object from the center, and the central part starts pulling static outer parts of the object up to the pace. When the central part pulls on a small outer part, the central part loses some of its kinetic energy. Some of that energy goes to the kinetic energy of the small part. But momentum conservation dictates that some energy is left over. That energy goes to the vibrations of the object, if we assume that the object is perfectly elastic.

The force which slows down the central part is due to the momentum flow to the small part. In that sense, there is no "radiation reaction". The radiation gains its energy from the leftover kinetic energy, when the small part slows down the central part.

In the case of a flexible object, some push energy is lost to vibrations of the object. The "radiation reaction" in this case might also be defined as the energy difference relative to pushing a rigid object of the same mass. In that sense, there is a radiation reaction.

In an inelastic collision, some kinetic energy is lost to deformations. We might define that energy as the radiation reaction.

Suppose that we push a rigid object with a 1 newton force a distance of 1 meter. We give 1 joule of kinetic energy to the object.

Suppose that we push a same mass flexible object in the same way. The push happens faster because the object is "soft": we give less momentum to the object. Only part of the 1 joule I gave ends up being kinetic energy, the rest is spent on vibrations.

The flexibility of the object initially lowers the inertia which we feel when we start the push. If we push for 1 second with a 1 newton force, we need to push a longer distance than we would with a same mass rigid object. That extra distance is the source of the radiation energy. The inertial mass of an electron varies with time during the push.

Is there some sense in saying that a "radiation reaction produced a force" which slowed down the flexible object? All the momentum which our push gave to the object did go to building up its momentum. In this sense, there is no radiation reaction.

But energy-wise, part of the push energy did go to building up vibrations.

If we push a rigid object on a floor, then friction eats up both some momentum and some energy.

Conclusion: if we look at the momentum of the flexible object, there is no radiation reaction. All the push momentum goes to the momentum of the object. But energy-wise, there is a radiation reaction. Vibrations eat up part of the push energy.

A supported charge in a gravitational field

A supported observer sees the electric field static, except far away where the signal from the current setup has not reached yet. He sees the Poynting vector as zero in the local area.

If we think of flexible object supported in a gravitational field, the upward push at the center of the object and the weight of the outer parts have deformed the object. Far away, gravity is weak and the deformation will be small.

The deformation can be considered as a "wave frozen in time". If the forces on the object would disappear, the deformation would start moving mostly outwards at the speed of light, with some reflection of the wave back.

So, does a supported charge radiate? No, its radiation wave is "frozen in time".

What about a freely falling observer? Locally, he sees the configuration like an inertial observer saw an accelerating charge in a Minkowski space. The Poynting vector says that there is energy flow from the charge outward. Where does this energy end up? Freely falling coordinates are not static, and it is not clear if one can meaningfully analyze energy flows in them.

An "almost" freely falling charge in a gravitational field

The charge will not fall freely because it has to pull behind the parts of the electric field which are outside the gravitational field. A freely falling observer will see the charge accelerate upwards. He will see an energy flow. Can we call that radiation? In the local frame, it looks like the accelerated charge Minkowski case which we treated above. We can call that radiation.

Also a supported observer sees the falling charge pull on the outer parts of its electric field, that is, he sees that the charge is not falling freely. He sees radiation.

Does an accelerating object see Unruh radiation?

We have in this blog long stressed that the existence of Unruh radiation would break the conservation of momentum. But can we harmonize them with our new ideas?

The accelerating object is carrying an electric charge as a sensor to the thermal photons it is supposed to see. The conservation of energy requires that these photons should get their energy from the kinetic energy of the electron or from the force which is pushing the electron.

We saw that the flexibility of the electron's electric field allowed it to move momentum to the trailing parts of its field and free some kinetic energy which it radiates away as electromagnetic radiation. Can Unruh use the same mechanism?

An outside agent is able to extract energy from the accelerating object without extracting momentum. The idea is to make use of the changing speed of the object. The agent first helps in acceleration for 1 second with a 1 newton force, and then resists for 1 second with a 1 newton force. The object moves faster in the resist phase and the agent can pick up more energy than he gave. Could this be the mechanism of Unruh radiation?

Indeed, we can imagine the following process with virtual photons: the accelerated electron sends a virtual photon which "borrows" some momentum and energy from the vacuum. The virtual photon helps in accelerating the electron. Later, when the electron is moving faster, it absorbs the virtual photon. Some kinetic energy is left over which the electron radiates away.

The above process does not give out any radiation in the accelerated frame. Otherwise, a statically supported electron on Earth would radiate. Is this a contradiction?

We saw in the previous section that in the inertial frame there is an energy flow from the electron outwards but in the accelerated frame there is no energy flow. If we believe the energy flow consists of "photons" of some kind, then the inertial observer sees the electron radiate but the accelerated observer not. The photons cannot be ordinary full electromagnetic waves because then they would be visible to the accelerated observer.

This reminds us of hypothetical Hawking radiation which is not visible to the freely falling observer.

Actually, the virtual photon cannot extract energy in the way that we described. The electron is in a bound state if it is supported by an electric field. The virtual photon may excite the electron, but the virtual photon probably gives very little - if any - momentum to the electron. When the photon is reabsorbed to the electron, there is no leftover kinetic energy. It looks like the inertial observer does not see any radiation in "ordinary photons". He does see a Poynting vector energy flow in the field, though.

Unruh claims that the accelerating observer can see the electron jump to a higher energy state. That does not happen in our example. It is not Unruh radiation.

If there exists Unruh radiation, then an equivalence principle is broken as a supported electron on Earth does not see it. Could it be that some global phenomenon of the field in the Minkowski space would produce Unruh radiation? That global thing would not be present in the gravitational field of Earth.

In the Minkowski case, the electron keeps pulling the outside parts of its field up to the pace. There must be some kind of a self-force which slows down the electron. But also in the gravitational case, the electron must keep pulling the mass-energy of the field up. Also in that case there has to be some self-force.

If Unruh radiation is real, there has to be a derivation of it also in an inertial Minkowski frame. Since quantum mechanics is formulated in inertial frames, the analysis in an inertial frame is much more trustworthy than the controversial field quantization in an accelerating frame. 

Electromagnetic radiation - finally explained

We had the paradox that a linear acceleration of a charge would not cause any electromagnetic waves, that is, there is no energy flow from the system to the infinity, but a periodic motion of the charge does carry energy to the infinity.

How does the system know when to start transmitting energy to the infinity, if the linear acceleration is at the start of a periodic motion?

When we accelerate the charge, the outer electric field does not yet know of the acceleration, and the outer electric field does contain some mass-energy.

An analogue for the charge and its electric field might be a flexible solid object which is not perfectly elastic. Any deformation causes some friction among the molecules. The friction generates heat - and the heat is radiated out in electromagnetic waves.

We have been claiming that a linear acceleration of a charge cannot produce electromagnetic waves. We were wrong. If the acceleration varies, then the electric field changes its form, and some energy may be lost in electromagnetic waves.

Alternatively, if the static electric field is a perfectly elastic solid object, then a varying acceleration of the charge can make it oscillate (along with the magnetic field). The energy of oscillation is electromagnetic waves.

We need to check if a simple push of a charge might allow the energy loss given by the Larmor formula. Maybe we can save Gauss's law in that case?

If a charge is static in a gravitational field, then nobody believes it can radiate. That suggests that a charge under a constant linear acceleration does not radiate. Is that compatible with Gauss's law?

When we start accelerating a charge linearly, at the start there may be some friction or oscillation in the electric field when it gets deformed. If we continue with a constant acceleration, no new energy is radiated?

If we move the charge in a periodic motion, its electric field is constantly being deformed, and it gives out constant radiation.

https://journals.aps.org/prd/abstract/10.1103/PhysRevD.80.024031

https://arxiv.org/abs/0905.2391

The paper by Gralla, Harte, and Wald from 2009 claims to be the first rigorous derivation of the electromagnetic self-force. We need to compare our ideas to their results.

When a charge is pushed, initially only a part of the electric field moves with it. To bring the rest of the field up to the pace, the charge has to give up some of its own momentum. We could say that the field exerts a self-force on the charge.

This is also the mechanism how the charge can radiate some of its kinetic energy away, even though the radiation carries a negligible amount of momentum. The extra momentum goes to the rest of the static electric field which is brought up to the pace.

Our previous claim that a linear acceleration cannot cause radiation was based on an erroneous thought that the charge and its electric field are rigid and move in unison.

An electric field under a gravitational field

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

https://www.sciencedirect.com/science/article/pii/0003491657900490

Charles Misner and John Wheeler tried to combine Maxwell's equations and a curved spacetime. Their 1957 paper Classical physics as geometry is reprinted in Wheeler's 1962 book Geometrodynamics.

Let us find out how they handled a static electric field under a gravitational field.

The authors suggest that a static charge is really lines of force oozing out of a very small wormhole. They state that lines of force never end and Gauss's law holds.

But how do they treat the case where an electron falls through a big wormhole to a white hole?

A low gravitational potential causes charge polarization?

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

The refractive index of a medium is

      sqrt(relative permittivity

              × relative permeability).

There is polarization of charges in the medium under an electric field.

What about a low gravitational potential? On light it acts like a medium of a high refractive index. Is there polarization of charges?

If there is polarization, then Gauss's law does not hold.

An accelerating charged plane and an observer in the Minkowski space

NOTE May 19, 2019: the analysis on this page is incorrect. We must assume Gauss's law.

----

Our goal is to find out how the electric field of a charge behaves under a gravitational field.

We assume that the electric field is known in the case where the charge is static, or moves at a constant speed in the Minkowski space. Assuming that the Coulomb potential is 1/r for a static charge, Lorentz covariance dictates how the electric field must look like for a charge moving at a constant speed.

The problem becomes more complicated if the charge or the observer is in an accelerating motion.

Let the charge first be static. Then it is pushed rapidly, so that the speed is v. We suggested in our blog post a few days ago that the electric field should be formed by combining the old Coulomb field to the new Coulomb field and that Gauss's law does not hold.

The new Coulomb field is present in the circle of a radius ct around the original position of the charge.

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

It is the familiar diagram top right on the linked page.

A charged plane which is suddenly pushed away from the observer

        observer         push to speed v  --->

                              |

        o                    |

                              |

                              |

        <-- E

We have a static infinite plane with a uniform charge density per area. The electric field is a uniform E_0 at the observer. The plane is suddenly pushed to a relativistic speed v.

The observer will continue to see most of the plane at its original position because the speed of light is finite. But soon the observer will see a disk-like circular area of the plane opposite to him farther away than the rest of the plane. Furthermore, the observer will see the electric field of the circular disk flattened by length contraction.

The observer will initially see the electric field decrease, but it will start growing back towards E_0 as he sees the disk grow bigger and bigger.

How low can the electric field go? If v is very close to c, then the electric field of the disk is flattened extremely much.  Obviously, we can make the electric field E transiently as close to zero as we like.

An observer is suddenly pushed away from a static plane

In this case, the observer at a constant speed v will see the electric field E just as big as a static observer.

http://www.feynmanlectures.caltech.edu/II_26.html

See Table 26-2 in the link. The Lorentz transformation of the electric field E in the direction of v is the identity mapping.

Both the observer and the plane are pushed to the same direction

           --> v

           |

           |

           |            o --> v

           |

           |

                       E -->

Suppose that the initial distance of the static plane from the static observer is 1 meter. Let us quickly push both to a relativistic speed v to the right. We want to keep the distance in the moving frame as 1 meter. To accomplish that, we first need to push the plane, then wait for a little while, and then push the observer. We need to wait because there is length contraction in the frame moving at a constant speed v relative to the original static frame. Without the wait, the distance would be bigger than 1 meter in the v frame.

The observer will see most of the plane at its original position and see that he is moving away from the plane. But soon he will see a circular disk of the plane close and static relative to him, at the the 1 meter distance from him. He will see the electric field E larger than the original field E_0.

         |
         |\
         |   \
         |      \  E -->
         |        o observer
         |      / the cone is 120 degrees wide
         |    /
         | /
         |

Half of the electric field E of a static plane at a static observer is generated by charges which are at most at an angle 60 degrees from the normal as seen by the observer.

The observer is behind the plane and both are pushed to the same direction

       --> v     --> v
                   |
      o           |
                   | 
                   |
           <--- E --->
                 
This is an interesting case. Let us assume that the distance of the observer and the plane is 1 meter both before and after the push, in the frame of the observer.

If the observer attains a relativistic speed, he will see most of the plane behind him. He sees a circular disk in front of him at the distance of 1 meter.

Suppose that v is almost the speed of light and the push is at t = 0. We work in the original static frame. When the observer has moved 2 meters to the right, he is receiving a ray of light sent from the plane sqrt(3) meters above him at t = 0. The observer sees a disk of an angular diameter 120 degrees 1 meter ahead of him. The rest of the plane he still sees at the original position behind him. The electric fields of these parts are opposite. He sees a total electric field E = 0.

An accelerating plane

We believe that in the case of smooth acceleration, the above sudden push examples give a qualitative understanding of what happens to the electric field.

A gravitational field

This is a much harder case. In the above example cases we worked in the flat Minkowski space. What happens when space is curved?

Hanni, Ruffini, Cohen, Wald, and Linet used Gauss's law to construct an electric potential around a black hole. But we saw that Gauss's law does not hold under acceleration. What to do?

Why a periodic motion creates electromagnetic waves but a linear motion does not?

We need a classical mechanics analogue for the electromagnetic field, to have a better intuitive understanding of it.

A possible analogue is a tight drum skin whose mass is zero. The skin can still store energy because it stretches and energy is stored in the deformation.

                  finger
                   | |
   _____       U     ______ drum skin
             \______/

Let us press the skin with a finger. What happens if we move the finger at a constant speed along the drum skin?

The valley which the finger creates in the skin moves. If the skin would have mass, then the upward movement of the skin behind the finger would probably create waves. If a mass starts moving up, it is out of its equilibrium position, and the inertia of the mass will keep it oscillating. Waves are born.

But if the mass-energy of the skin is solely in the stretching of the skin, then it might be that no oscillation will happen? When the finger has moved far away, the mass-energy of the skin is very small and there is not much inertia any more.

Note that if the skin has a positive mass, then the speed of waves in it has a preferred coordinate system: the inertial coordinates determined by the drum. The waves cannot be Lorentz-covariant then.

If we accelerate the finger in a linear fashion on the skin, then the big picture does not change very much from the movement at a constant speed. Only close to the finger there is considerable mass-energy which could sustain waves with its inertia.

But if we let the finger do a periodic motion on the skin, then deformation of the skin will spread to all directions as waves. The deformation carries mass-energy, and that mass-energy has inertia, which will sustain waves.

To exist, waves require inertia. And the waves themselves can carry the required inertia with them because they deform the skin and thus carry mass-energy.

We have presented this idea in our blog also before. We should calculate the wave equation for a massless drum skin, and calculate also the solution for a linearly accelerated finger.

There is at least one obvious problem: if we let a finger pump mass-energy to the skin, then that mass-energy defines a preferred coordinate system. That would be against Lorentz-covariance. However, the existence of photons somewhere does necessarily define a preferred coordinate system. Does our model break Lorentz-covariance or not?

Gauss's law for an accelerating charge

NOTE May 19, 2019. The analysis in this blog post is erroneous. We must enforce Gauss's law to make the lines of force continuous. To that end, we must assume an electromagnetic half-wave which will pull the lines into the bottle when they have exited the neck.

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In this blog we have presented the bottle example, where a charge inside a bottle is quickly pulled from the main part of the bottle to the neck.

Let us assume that the electric field of a charge q is determined by Coulomb's law based on the retarded position of the charge. This is not strictly true. We will look at corrections from Lienard-Wiechert potentials.

We can make the form of the bottle such that at all times, the Coulomb electric field of the charge will point out of the surface.

            _________
           /
         /
     ---
       × ×
     ---
         \
           \_________

The old position of the charge is the right × and the new position is the left ×.

Gauss's law says that a charge is the sole source of the electric field, which implies that inside any closed surface, the charge is equal to the field flux out from the surface.

Let the charge be q. Before the pull, the main body has a flux > q/2 out from the bottle. Right after the pull, the main body still has the flux > q/2, but the neck, too, has a flux out > q/2.

Information about the electric field spreads at the speed of light. That is why the main part of the bottle does not yet know that the charge has moved.

The total flux is > q and Gauss's law appears to be temporarily broken. But can the electric field induced by the changing magnetic field save the law?

Maxwell's equations say that a changing magnetic field produces curl into the electric field, and vice versa. But curl does not create a source of the electric field, it creates field lines which are closed loops. Closed loops do not help in restoring Gauss's law.

Let us put our argument another way: we showed that, assuming that the electric field is determined by Coulomb's law, our source of the electric field, q, produces a flux q' > q out of the bottle. The only way to restore Gauss's law is to assume something else (like magnetic effects) creates a flux q' - q into the bottle.

We can restore Gauss's law by assuming that some effect forces the electric field lines to return back to the bottle when they have exited the neck. If the field lines are unbroken at each moment in the global coordinate frame of our experiment, then Gauss's law is restored at least in this frame. What if we move to another frame? Does the transformation of the electric and magnetic fields keep Gauss's law in force?

Lienard-Wiechert potentials

For a dynamically changing system, the energy to move a test charge q from a spacetime point x to the infinity will vary depending on the route. It is hard to define a global potential.

https://en.wikipedia.org/wiki/Liénard–Wiechert_potential

The Lienard-Wiechert potentials describe the scalar and the vector potential of a moving point charge. We will look at what they say about our thought experiment of the bottle.

Momentum conservation is a problem?

We wrote in our April 2018 blog post about the Larmor formula. If we are pushing a charge and accelerating it, the energy to the electromagnetic field or radiation obviously has to come from the kinetic energy of the charge. But then the charge would lose a lot of momentum, which cannot be carried away by radiation or field energy moving at the speed of light.

It looks like a linearly accelerated charge cannot radiate or lose energy to the field. The Larmor formula must fail if we believe in conservation of momentum.

But Edward M. Purcell has presented a derivation of the Larmor formula where he draws the electric field lines as continuous after a sudden push of the charge. He calculates the electric field energy and finds the Larmor formula true.

What is the solution to this contradiction?

1. Conservation of momentum fails?

2. The field lines are not continuous and Gauss's law for the electric field fails?

3. The field energy is not what Purcell calculated?

4. Something else?

If we believe in conservation of momentum, then a linear push does not add any new energy to the field. That suggests that the field lines are not continuous and Gauss's law fails.

People have not been able to measure the radiation of a linearly accelerated charge. We do not know from experiment what Nature does.

Conservation of momentum is a pillar of physics, and the energy E^2 of the electric field manifests in electromagnetic waves. Most probably we have to abandon 246-year-old Gauss's law in Maxwell's equations and replace it with some principle about the Coulomb potential of a charge plus retardation. Electric lines of force are not continuous if a charge is accelerated linearly.

It would be strange if a charge which is static in a gravitational field would radiate. What about the electric field lines? Are they continuous for such a charge?

Purcell's derivation of the Larmor formula - how much energy is there in a disturbance of the electric field?

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

On the linked page, at top right, or under the headline "Radiation as a Consequence of a Cosmic Speed Limit", we see the electric field lines which Edward M. Purcell drew to derive the Larmor formula. The Coulomb potential inside the circle differs considerably from the Coulomb potential outside the circle. In the electric potential, there is a sharp jump at the circle.

To make the electric field sourceless at the circle, Purcell draws transverse field lines which connect the field lines inside the circle to the lines outside the circle.

In principle, one can accelerate a charge from zero to half the speed of light instantaneously. There is no physical limit imposed on acceleration in special relativity. The transition area A between the inside of the circle and the outside can be made as narrow as we want. There are quite a many transverse field lines in Purcell's diagram that are confined in the transition area. We can make the electric field energy E^2 * A in the transition area as large as we want. This shows that in Purcell's diagram, the electric field certainly carries away energy at the speed of light.

But we have showed in our blog that linear acceleration of a charge cannot convert its kinetic energy to field energy which would move at the speed of light, because conservation of momentum would be broken. We conclude that the transverse field lines of Purcell do not exist.

What about the Coulomb potential? It is strange if there would be sharp jumps in the potential. Since the configuration is not static, a test charge can receive energy from the changing electric field. Can we pump energy from the changing electric field through some mechanism? If yes, then we would a have a breach of conservation of energy.

Suppose that we have an electric dipole at a (large) distance r from where the charge originally was. The dipole is strained by a Coulomb field which is proportional to 1/r^2. Let the dipole be bound with a spring for which Hooke's law is F = kx. We then have the displacement x = 1 / (k r^2) and the energy stored in the spring k / (2 k^2 r^4). After some time, the dipole suddenly sees the electric field change its direction and magnitude. Some energy is released into the vibration of the string.

We can place r^2 of such dipoles in a spherical shell around the charge. The combined energy suddenly released in the shell is at most proportional to 1/r^2. This is the well-known result that you cannot efficiently transmit energy over long distances in longitudinal waves of the electromagnetic field.

At least this thought experiment showed that it is not easy to harvest much energy from the disturbance of the electric field. But even a little bit of energy would breach conservation of energy.

Optical gravity works badly

We introduced optical gravity last year as a way to interpret general relativity. Unfortunately, the optical interpretation only works well for static gravitational fields. In astronomy, we witness (almost) static gravitational lenses caused by galaxies and galaxy clusters. Optics is a nice way to interpret what we see.

We conjectured that the optics of a static Schwarzschild black hole would reflect all incoming waves out before the horizon. That is not true. Several authors have calculated the reflection, or scattering, using the wave equation for curved spacetime. The result is that for short wavelengths and close to the horizon, the scattering is negligible.

There is only significant scattering for wavelengths of the Schwarzschild radius or longer.

If the system is dynamically changing, calculating the optics of spacetime is awkward. For example, the exact location of an event horizon depends also on future events. We do not know the optical density of spacetime before we know the future, too.

The model of a "frozen star", where the interior of a horizon is frozen in time, suffers from the same problem in dynamic systems. If we do not know where the horizon is, how does the material inside the horizon know when exactly it should freeze?

It looks like we have to admit that an astronaut can fall through an event horizon.

We still have not found a way to make the Hawking radiation hypothesis work. Thus, we cannot assume that the Hawking evaporation of the black hole will prevent the astronaut from falling through the horizon.

For a freely falling observer, the event horizon looks completely normal, slightly curved spacetime. There is no drama there, no infinite forces or energies. We would need some unknown physical mechanism to stop the extension of the spacetime manifold inside the event horizon. There seems to be no reason why the astronaut cannot keep living and pass the horizon.

A singularity, on the other hand, is very dramatic. We cannot extend the spacetime manifold smoothly past a singularity.

Does a black hole lose the electric charge?

UPDATE May 23, 2019: Gauss's law holds also for moving charges. The analysis below is erroneous.

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https://meta-phys-thoughts.blogspot.com/2018/04/the-error-in-larmor-formula-for.html

In the blog post on April 5, 2018 we pointed out that Gauss's law for the electric field does not hold if the field is not static. That is, the electric field flux through a closed surface may differ from the total electric charge which is inside the surface.

We posted an example where a charge is held inside the bottle close to the neck, and quickly pulled inside the neck. The information of the movement of the charge spreads at the speed of light. If the main body of the bottle is large, its surface will not know too soon that the charge has moved. The electric flux through the main body will stay as it is. But the narrow neck will soon learn that the charge is now inside the neck, and the electric flux through the neck will increase quickly => the electric flux out of the bottle will increase for a moment.

Lines of Force of a Point Charge near a Schwarzschild Black Hole
Richard Squier Hanni and Remo Ruffini
Phys. Rev. D 8, 3259 – Published 15 November 1973

https://journals.aps.org/prd/abstract/10.1103/PhysRevD.8.3259

The authors seem to assume Gauss's law, even though the situation close to the horizon is not very "static". The charge, which is held at a fixed distance from the horizon, is in a rapid acceleration relative to inertial observers close to it.

Any ray of light which falls into the horizon crosses the horizon orthogonally. This means that the horizon is an equipotential surface for the electric field if the field lines follow paths of rays of light.

The authors assume that the total electric flux is zero through any sphere enclosing the horizon but not the point charge. The horizon in their analysis is like a conducting metal sphere, where the point charge induces an opposite charge at the area closest to the point charge. The rest of the sphere then has a charge of the same sign as the point charge. When the point charge is very close to the metal sphere, then the electric field is almost symmetric, emanating from the horizon.

But why should the total electric flux throught the horizon be zero?

Imagine that the geometry of spacetime is that the black hole is connected to a white hole. If an electric field line passes into the black hole, it will probably come out of the white hole. There is no reason why Gauss's law would hold for a surface which just encloses the black hole, but does not enclose the white hole.

If a black hole can devour electric field lines and not return them back out from the horizon, then we get an extra strong "no hair" theorem: a black hole loses all electric charge which falls into it. The field lines will decay as quickly as the radiation emitted by an object falling into the horizon. Soon the electric charge visible to outsiders is very close to 0.

Why the black hole does not devour the gravitational field, too? The gravitational field is really the spacetime manifold and its metric. We can extend the manifold to a future spacetime point x just by using the information in the incoming light cone of x. For the spherically symmetric case, Birkhoff's theorem guarantees that the metric is the static Schwarzschild metric outside the collapsing mass.

Cohen and Wald forgot to transform the potential to the local frame?

UPDATE May 23, 2019: it looks like Cohen and Wald have absorbed the potential transformation formula into the other coefficients of their equation. Their calculations are correct.

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https:/aip.scitation.org/doi/abs/10.1063/1.1665812

Cohen and Wald in equation (4) have denoted the electric potential A_0 by v.

The electric potential is the electric potential energy of a 1 coulomb test charge.

Is the potential defined relative to an observer at infinity? That would be logical. There is not much use for a "local" potential.

It is logical that equation (4) is the view of the local static observer. But then the chance in the global potential energy dv should be given in local energy. We should multiply dv by 1 / sqrt(1 - 2m/r) to get the view of the local observer. If we send 1 joule of energy from the infinity to the local observer, he sees the energy inflated by a blueshift.

If we have an observer on the surface of a neutron star, and he moves 1 coulomb over 1 volt potential in his own frame, he gets 1 joule of energy. But when he transmits the energy to the infinity, the energy is sapped by a redshift. The global potential difference is less than 1 volt.

Hanni and Ruffini in their paper omit the energy conversion just like Cohen and Wald.