General relativity has problems defining the kinetic energy in matter lagrangians

Let us consider the Einstein-Hilbert action once again.

It contains the ordinary matter lagrangian density L_M. The ordinary matter lagrangian is typically of a form

       L = V - T,

where V is the potential energy and T is the kinetic energy. But how do we determine the kinetic energy T? If we would be working in Minkowski space, then we would measure velocities against fixed inertial coordinates.

                   A                         B

                   ● +                      ● +

                   |---------------------------------> x

                   0       1       2       3

Let us have two electric charges A and B which repel each other. In fixed Minkowski coordinates, a lagrangian density

       L = V - T

calculates their paths right.

Let us then use such coordinates that a small patch of the spatial coordinates moves along both charges: their kinetic energy T stays zero in these coordinates. The coordinates move, but the metric stays as the fixed Minkowski metric. In the diagram, we let the metric to change in such a way that the distance between x = 1 and x = 2 grows.

General relativity allows us to use any coordinates, as long as the spatial coordinates do not move superluminally.

In the new coordinates, the matter lagrangian calculates the paths incorrectly.

How can we fix this? Should we transform the matter lagrangian somehow when we move to new coordinates?

A tensor is a physical parameter which can be "easily" converted between various coordinate systems. The conversion is local. If we only manipulate the metric between x = 1 and x = 2, then a tensor describing locally the charge A or B does not get transformed.

There does not exist a fixed Minkowski coordinate system in general relativity. In our March 17, 2023 blog post we remarked that one cannot use static coordinates at all for a fast moving neutron star because the star moves superluminally relative to static coordinates.

In literature, when working with general relativity, people assume that the sole long-distance force is gravity. Other forces only act locally. Does this save us from problems? One can then use a locally defined, almost Minkowski coordinate system. This helps to alleviate the problem. However, then it should be explicitly stated in textbooks that the coordinates must imitate Minkowski coordinates - one cannot choose the coordinates freely.

Suppose then that A is in faraway space and B in on the surface of a very fast moving neutron star. We cannot use a static coordinate system. When we determine the kinetic energy of B, in which coordinate system we should do it?

In our own Minkowski & newtonian gravity, the coordinate system is always the fixed Minkowski coordinates. Our own gravity model does not suffer from the exact problem described here. However, we do have a problem defining the kinetic energy when we have a complex system of interacting fields. What is the amount of energy and where is it located?

Conclusions

We have not seen in literature any mention about the problems of defining kinetic energy in general relativity. If we cannot define kinetic energy, we cannot define the stress-energy tensor. The problem is theoretically fundamental - though not relevant in practical applications.

Why gravity increases the inertia of the electron but an electric potential does not?

In our previous blog post we realized a profound mystery about interactions. If we put a hydrogen atom into an electric potential (close to large negative or positive charges), then the electron in it carries around a large positive or negative energy. But that does not show up in the hydrogen spectrum. The electron seems to have the same inertia as always. We are pretty sure that a free electron in such conditions possesses a lot more inertia than an electron in empty space. Why the inertia does not show up when the electron is on a stationary orbit in the atom?

On the other hand, that inertia of the electron which is caused by a low gravity potential does show up in the spectrum. The spectral lines are redshifted.

The electron in a hydrogen atom on a stationary orbit

                       ____

                     /        |

                    •  <----                            ● -

                  e-                             large charge

                  in tight

                  circular orbit

One could argue that since the electron cannot radiate away its energy, its electric far field is kind of frozen and cannot add inertia to the electron. But why the same argument does not hold for the gravity field of the electron?

In our April 20, 2023 blog post we argued that a test charge sees the inertia in an electric field beforehand. But it may be that the electric field, after all, does not see in advance the inertia from the energy flow in the electric field.

Since the electron on a stationary orbit does not radiate, its remote field probably is static. If the remote field of the electron in a hydrogen atom is static, then there is no energy flow in the remote field, and in the diagram above, the electron does not feel extra inertia from the large external charge.

The energy density of an electric field is very concrete

We have been arguing that one can locally harvest the energy density of an electric field using a capacitor which cancels the field in a certain volume. In this sense, the energy density of an electric field is very concrete. When we say that energy flows in a electric field, it has a very concrete meaning.

A rubber sheet model of gravity does not work for the hydrogen atom

Suppose that we have two positive electric charges. They repel each other. The repulsion can be explained by the energy density 1/2 ε E² of their combined electric field. One could guess that we get the newtonian gravity field by changing the sign of the field energy to negative. But negative energies behave strangely in classical physics. Maybe it is best to ban negative energies altogether.

In this blog we have introduced various rubber sheet models to describe gravity. In the rubber sheet models, masses lose energy by hanging lower on the rubber sheet. The deformation of the rubber sheet has positive energy. Thus, the energy of the gravity field is always positive. General relativity has a similar idea: to create a low potential one must introduce Ricci curvature. A mass loses energy in a low potential, but the Ricci curvature, in a sense, contains positive energy.

Let us analyze energy shipping in a rubber sheet model of gravity.

        rope           r

        -------

        ____  \     ______                 ___  rubber sheet

               \_•_/           \___●___/

                  ---->

      test mass m     large mass M

If the test mass moves closer to the large mass, both masses can hang lower and potential energy is released. A half of that released energy goes to pulling the test mass m. The test mass gains kinetic energy, or we can harvest that energy if the test mass is attached to a rope.

The second half of the released energy goes to stretching the rubber sheet so that the masses can hang lower.

In this case, the energy flows from the masses m and M to our rope and to a wide area of the rubber sheet.

Let us calculate energy shipments if we lower the test mass dr closer to the large mass. The distance is r. Let us assume that m goes to a potential which is dV lower.

1. The energy shipment m c² dr of the test mass plus the energy in the depression of the rubber sheet under the test mass.

2. Energy dV is shipped roughly a distance ~ r to stretch the rubber sheet.

3. The rope harvests the potential energy dV. A half of that energy probably comes from M, over a distance r.

We conclude that the extra inertia of the test mass m is

       ~ 3/2 dV r / (m c² dr).

In a radial movement of m, there is considerable extra inertia. What is the inertia if we move the test mass m tangentially relative to M?

If the potential of the test mass is -V, then there is extra stretching of the rubber sheet worth the energy V. The inertia of the test mass plus its own depression in the rubber might be m c² - V. If the extra stretching energy V in the rubber moves a shorter distance than m, then it adds < V to the inertia. We conclude that the inertia of m is less than m c² in a tangential movement.

If we then look at the hydrogen atom in the rubber model, the center of mass of the atom does not move. There is very little movement of the rubber sheet far away from the atom, which would mean that there is essentially no extra inertia. The spectrum of the hydrogen atom would not be redshifted in a low gravity potential. It would be blueshifted if the inertia of the electron itself is lower in a low potential.

We conclude that a rubber sheet model of gravity cannot explain the behavior of the hydrogen atom in a low gravity potential. The extra inertia of the electron has to be carried by the point particle itself. The inertia cannot arise from field effects far away.

In general relativity, the metric can be seen a as a field of the large mass M. The extra inertia on the test mass m is imposed on it by the field of M. The extra inertia does not come from an interaction of the fields of M and m.

Analysis of the force and the extra inertia in electromagnetism and gravity

We have to analyze this carefully. We believe that the Coulomb force is imposed on a test charge c directly by the field of a large charge C. Since an electric potential does not affect the spectrum of the hydrogen atom, the Coulomb field, apparently, does not impose extra inertia directly on a test charge.

                   •                                   ●

         test mass m              large mass M

         potential V

But for gravity, the extra inertia is directly imposed on a test mass m. We could imagine that the point mass m carries along with it a "packet" of negative energy V, where V is the potential of m in the field of M.

If we move an amount V of negative energy a distance ds, that is equivalent to moving an amount |V| of positive energy a distance -ds. Thus, the packet of negative energy may add the inertia |V| to m.

There is a problem: in gravitational waves, a gravity field is clearly present. The packet model does not explain the gravity field. Packets are about point particles.

The extra inertia of a test mass m shows that the field of M must be able to absorb momentum and "energy shipping".

A tentative solution to the extra inertia problem

Hypothesis about inertia.

1. The electric field imposes the Coulomb force on a test charge directly, but does not impose extra inertia directly. The extra inertia only comes as a backreaction from energy shipping in the combined electric field of charges.

2. The gravity field imposes the Newton gravity force and the associated extra inertia directly on a test mass. The extra inertia is the absolute value |V| of the negative potential of the test mass, if the potential is not very low.

In general relativity, the metric accomplishes item 2. The hypothesis takes our Minkowski & newtonian gravity model a lot closer to general relativity.

How can a field impose extra inertia? That is easy. We may imagine a field consisting of a grid frame and little robots which manipulate a test mass with their hands. The robots are free to impose a force on the test mass, or simulate extra inertia. Nothing prevents a field from acting in an "intelligent" way like this.

An electron in a stationary orbit cannot radiate gravitational waves. This suggests that its gravity far field is frozen. However, the inertia from the gravity field of an external mass M is present.

The experiment with circularly arranged pendulum clocks: the inertia is the same regardless of the configuration

On March 25, 2023 we suggested arranging synchronized pendulum clocks on the surface of Earth in a circular configuration, such that the pendulums for an expanding and contracting circle. If the extra inertia would be imposed on them by energy flowing in the combined gravity field of the pendulums and Earth, then the inertia of the circular arrangement would be smaller.

Now we realize that the hydrogen atom is such an arrangement. The center of mass in the atom stays at the same place. The gravity field of the system hardly changes at all at remote locations. If the extra inertia would come from the remote field, we would expect essentially no redshift in the spectrum of hydrogen. But the redshift is as expected.

Also, we have been wondering if a collapsing shell of mass feels less inertia than a point mass. It is likely that the inertia is the same.

Conclusions

There is a fundamental difference in the electric field and the gravity field. The electric field does not impose extra inertia on a charge directly, but the extra inertia comes from a backreaction to energy flows in the combined electric field.

The gravity field, on the other hand, imposes the extra inertia directly on a test mass. Indirect effects, like the ordinary tidal effect, may add more extra inertia - a process which most likely is a backreaction and does not happen instantly.

Rubber models of gravity are inadequate because they do not impose the extra inertia on a test mass directly.

We have been claiming in this blog that there is an upper limit on the extra inertia that a mass M can impose on a test mass m - that, is the maximum extra inertia is M. In this we assume that an observer standing on the surface of M measures the inertia of m: M cannot move. Our new insights mean that the extra inertia could be even larger than M. We have to think about this.