Tuesday, February 20, 2024

The spin of the graviton is the same as that of the photon

How do we define the spin or the helicity of a particle?

One way to define it is that a particle whose spin is n can carry

        -n, -n + 1, ..., n - 1, n

times ħ of angular momentum L relative to some axis.

A circularly polarized photon carries either -ħ or ħ of angular momentum relative to its polarization axis. A linearly polarized photon carries zero angular momentum.

Is there any difference in how much angular momentum can a single photon or a single graviton carry?

Apparently, no.


A lopsided quadrupole


Let us have a heavy mass M attached with a rod to a very light mass m. The system rotates around their common center of mass.



                         center
                        of mass
            M     R                      r                 m
              ●----------×-----------------------------•
              |
              v  ω


The center of mass is marked with ×. The distance of M from it is R and the distance of m from it is r. The angular velocity is ω.

An example of such a system is the hydrogen atom where M is the proton and m is the electron. We may imagine that the electron has a pretty high principal quantum number n and orbits the proton relatively far from it.

The system clearly sends out gravitational waves. That can cause the system to decay to a lower energy state. In the case of the hydrogen atom, the system loses a certain quantum of energy E and a certain angular momentum L relative to the center of mass of the system.

Let us use the Bohr model for a hydrogen atom. The small mass m decays to a state where the principal quantum number n is one less. We assume that n > 0 is large. Let

       f  =  ω / (2 π).

Then the energy loss of the system is 

       E  =  h f,

and the loss of the orbital angular momentum of m is

       ħ  =  h / (2 π),

where h is Planck's constant.

The kinetic energy and the angular momentum of the large mass M is negligible in the system.

Let us add to M and m electric charges Q and q of the same sign, such that

       Q / q  = M / m.

Then the system sends out an electromagnetic quadrupole wave which is analogous to the gravitational wave.


A symmetric quadrupole


            M                         M
             ●----------×----------●
             |
             v  ω


Let us have a quadrupole where M and m are equal. Then the system is essentially two dipoles, both M, rotating around their center. A jump to a lower energy quantum state requires that both masses M lose one ħ of angular momentum L around the center.

The graviton in this case is assumed to carry 

       2 ħ

of angular momentum away.

Note that since the system is symmetric, the field returns to its original state already after a rotation through an angle π. One may then imagine that the frequency is 2 f, and the quantum of energy is

       2 h f.

That is, the imagined quantum is two dipole quanta "glued" together.

We did not refer to any special property of gravity. The quadrupole could as well be electric where the two charges have the same sign.


Can we "split" a quadrupole quantum?


What does it mean that a quadrupole quantum has the energy 2 h f and the angular momentum 2 ħ?

Can we split it into two halves?

Let us imagine that inside a classical quadrupole wave there is a classical dipole antenna which is tuned to the frequency f. Obviously, the antenna cannot absorb much energy from the quadrupole wave.

However, if we tune the dipole to the frequency 2 f, then it obviously can absorb energy in units of 2 h f, and angular momentum in units of ħ. However, we cannot really say that it "splits" a quadrupole quantum. Rather, the dipole absorbs a full quantum and then emits a low-energy quantum with the excess ħ of angular momentum.

We may imagine an antenna with a "gearbox" such that it converts a frequency 2 f to a frequency f, and stores the harvested energy to an oscillator whose frequency is f. Such a system might be able to "split" quadrupole quanta.


We can generate pure dipole gravitational waves


The best known gravitational waves come from mergers of black holes. They are quadrupole waves.

However, it is easy to generate dipole waves, too. Simply construct a dipole wave in empty space, and let masses M absorb the energy in that wave. Run time backwards. Now you have masses M sending a pure dipole wave.


What literature says about the spin 2



Lubos Motl (2013) writes that a photon's angular momentum in the z direction, jz, cannot be zero because of "gauge symmetries". This is a strange claim. If we have an electric charge oscillating in the x direction at a frequency f, it should be able to decay to a lower state and give up an energy quantum h f. It will not give up any angular momentum relative to the z axis because it does not have any. That is, we should have a linearly polarized photon where the polarization is to the x direction.

Motl writes that because of diffeomorphisms, a graviton only can have

       jz  =  +- 2 ħ.

Again, this is strange because we can, e.g., have two masses M attached with a rubber band, and oscillating linearly to the x direction. They should be able to radiate gravitons which have jz zero.


Conclusions


In this blog we have speculated that the photon is not really the property of the electromagnetic wave, but describes the decay of an antenna to a lower energy state. That is, the quantum is determined by the antenna which sends it.

We did not find any fundamental difference between electromagnetism and gravity. It makes sense to claim that both the photon and the graviton have the same spin, or helicity, 1.

Monday, February 19, 2024

16-fold energy density of a gravitational wave

We should find out why general relativity calculates the energy density of a gravitational wave correctly, and why the energy density is exactly 16 times the energy density of the analogous electromagnetic wave.


The output of the full nonlinear Einstein equations from a metric perturbation h


Let us have a perturbation h of the flat Minkowski metric η. The Einstein tensor calculates a stress-energy tensor T from the metric

      g  =  η + h:






Above, R is the Ricci scalar curvature, and we assume that the cosmological constant Λ is zero. The constant

       κ  =  8 π G / c⁴  =  2.08 * 10⁻⁴³  *  1 / N,

where G is the gravitational constant.

The stress-energy tensor T tells us what kind of a matter content in space would create the metric perturbation h to an otherwise empty Minkowski space.


The Einstein-Hilbert action for a spherical mass


The Einstein equations are derived from the Einstein-Hilbert action. For a lighweight spherical mass they produce the same field as the Coulomb field is for the analogous electric charge. Why is this?









In the action, the Ricci scalar R can be understood as the energy density of the gravity field. It is "potential energy". Adding Ricci curvature allows us to drop the mass-energy to a lower potential, i.e., we save in the "potential energy" of the mass-energy in the matter lagrangian LM. The price that we pay is that the potential energy in R grows.

We can probably derive the Coulomb field using a similar procedure. If we have a "bare" charge Q, we can drop it to a lower electric potential by creating an electric field around it. The price we pay is the energy that is required to create that electric field.


The energy of a wave in a rubber sheet model: "sound" waves transfer a lot of energy


Two years ago in this blog we calculated that the large energy density of a gravitational wave is due to the fact that it can stretch spatial distances.

Let us use a rubber sheet model of gravity. Let us have two mass systems A and B relatively close to each other. We try to transfer energy from A to B as efficiently as we can.


                A           B
                ●            ●
                ●            ●

               rubber sheet


The systems A and B could be so close to each other that the energy transfer is mostly through (moving) static fields, and there is not much of a "wave" aspect.

In the rubber sheet model, we are able to transfer energy in waves which stretch the rubber sheet. Is this analogous to general relativity?

Maybe we can transfer much more energy in the "sound" waves (horizontal stretching) in the rubber sheet than in transverse waves (up and down movement of the sheet). This analogy would explain why the energy density of a gravitational wave is much larger than that of the analogous electromagnetic wave. Fundamentally, it is due to more degrees of freedom in the stretching rubber. Each degree of freedom opens a new "channel" to transfer energy.

The factor 16 might come from the 16 components of a tensor. If an electromagnetic wave uses only one channel to transfer energy, and a gravitational wave uses 16 channels, then the transfer speed could be 16-fold. A wave which stretches a rubber sheet in one direction involves many more distortions: there is shear involved, too.


Conservation of "energy" in the Einstein-Hilbert action


The action probably implies various conservation laws. One such law is conservation of "energy", where we mean an analogue of energy for an isolated system which sits in an asymptotic Minkowski space.

The ADM formalism proves that there actually is such a law, if we define the "energy" of a system S by the metric which it creates far away. The metric is close to a Schwarzschild metric for a certain mass M. The energy of the system can be defined as M c².


                   S₁                S₂
                    ● ~~~~~~> ●
                             E


If we have a subsystems S₁ and S₂, and S₁ sends the energy E to S₂ through gravitational waves, then, obviously, the waves must generate far away the metric which corresponds to the mass-energy E. Otherwise, conservation of energy would be temporarily broken.

However, we are not happy about this reasoning because we must use the powerful result of the ADM formalism. We should find a simpler explanation.


Energy shipping in a Coulomb-like field and the Ricci curvature: an optical model of gravity


In this blog we have found grounds to claim that the apparent "metric" in the Schwarzschild solution is simply various inertia effects of the gravity field. The true underlying metric is Minkowski.

Let us concentrate on how the metric bends light. The bending is due to the local time running slowly close to a mass, and also due to spatial distances stretching.

We probably can deduce from that how the metric affects orbits of massive particles.

Hypothesis 1. Energy shipping in a force field whose potential is of the form ~ 1 / r, cannot "focus" or "defocus" light outside the charge. This implies that the Ricci curvature of the simulated "metric" is zero outside matter.


In gravity, we are mainly interested in the fields generated by mass-energy and pressure. We will forget about shear stresses for now.

The newtonian potential for a static mass M is ~ 1 / r. If we have a spherically symmetric mass M, then our hypothesis implies that the Ricci curvature is zero outside M.

The "potential" associated with pressure inside matter seems to be ~ 1 / r, too.

What about a mass M which is not symmetric? Maybe the "focusing" power can be "summed" from each mass element dM? Let us assume that the fields are weak.

Hypothesis 2. The "focusing" for light can be "summed" from each mass element dM of a larger mass distribution M, if the fields are weak.


Hypothesis 2 needs some tuning. We probably can sum the perturbations to the metric of time, but the perturbation of the spatial metric has to be deduced from energy shipping. Is there a reason why the Ricci curvature would still be zero in the surrounding empty space if we have two point masses?

Hypothesis 3. For a static mass distribution M, the Einstein field equations calculate the Ricci curvature right. They do it right both for the mass-energy density ρ and for the pressure p inside matter. The curvature is zero outside M, and the equations determine the correct non-zero value inside the mass distribution M.


Here we have an optical model for gravity. Each mass element dM creates a "lens", which makes rays of light to bend around dM just as described by the Schwarzschild metric. Summing the optics of these lenses is not straightforward, though.


              __________
            /___\___/___\


Above we have focusing lenses, which when summed, form a straight glass window. "Summing" of lenses is not simple.

The Einstein field equations "sum" the Ricci curvature caused by each mass element dM. Could this be the correct way to sum lenses? 


            focusing

                  /  \
                /  ●  \  dM
               |         |
               |         |

          rays of light
      

In this case, we would concentrate on the focusing power of each mass element dM when rays of light pass it from different sides, and afterwards adjust the metric outside the mass distribution M in such a way that the Ricci curvature is zero there.


Conclusions


Let us close this blog post. We are not ready to study the 16-fold problem yet. First we have to figure out the exact relationship between our own inertia concept and the Ricci curvature of general relativity.

Friday, February 16, 2024

Gravitational waves: breaking the speed of light?

Andrzej Trautman and others proved in the 1950s that the full, nonlinear, Einstein equations do admit wave-like solutions.


In the link, Donald Salisbury (2019) tells us about these early stages.


Chris Hirata's derivation of the energy density



Chris Hirata (?) (2019) presents a simple way to derive the energy density carried by a gravitational wave. First construct a wave packet solution

       g  =  η + h

for the linearized Einstein equations. There, h is a small perturbation of the flat Minkowski metric η.


If the metric g = η + h were a solution of the full, nonlinear, Einstein equations in the vacuum, then in







we would have the stress-energy tensor T on the right zero. But η + h is not a solution. We obtain a non-zero value for T. The component -T₀₀ supposedly tells us the energy density of the gravitational wave described by h.

Does this make sense? Can we modify g slightly, in such a way that g' would be an approximate solution for the nonlinear Einstein equations? If yes, then why would the improved solution g' carry the energy in -T₀₀?

Let us make a metric perturbation k such that

       η + k

is a solution for the Einstein equation for the above stress-energy tensor T. Then

       g'  =  η + h - k

is an approximate solution for the full nonlinear Einstein equations in the vacuum (if such a solution exists). We assumed here that the Einstein equations are approximately linear.

We may assume that -k contains the "background metric" associated with the gravitational wave packet. That background metric corresponds to the stress-energy tensor -T.

Let us then assume that the large-scale metric of an isolated system S embedded in asymptotically Minkowski space cannot change. That is, the gravitating mass-energy is conserved.

If a subsystem of S creates a wave packet h and loses the energy E from its stress-energy tensor, then the metric -k far away must be equal to the metric created far away by E. Otherwise, the large-scale metric would change. It makes a lot of sense to treat -T₀₀ as the energy density of the wave packet h.


Does the energy density prevent faster-than-light communication inside a gravitational wave?


In this blog we have claimed that a gravitational wave h could break the speed of light limit c in the underlying Minkowski space. That is because if h temporarily shortens the distance between observers A and B (or speeds up the local metric of time), then A and B could exchange information surprisingly quickly.

Now we understand that the energy contained in the wave h slows down the local speed of light, because of the large-scale metric in -k above.

Is -k able to slow down light enough, so that we cannot get superluminal communication?







Apparently not. In the Hirata lecture (2019), the energy density of h is proportional to the square of the time derivative of h. If we make the amplitude of h very small, then its energy content is negligible, and the large-scale metric does not slow the local speed of light enough.

Did we just prove that gravitational waves clash with special relativity? If yes, then the Einstein equations describe gravitational waves incorrectly!

No. We forgot to take into account the gravity field of the mass M which is the source of the gravitational wave. The mass M slows down the local speed of light by

       ~ 1 / r,

where r is the distance from M. If h would increase the local speed of light by ~ 1 / r, the speed can still stay below the global Minkowski speed of light.

Let us move a small mass M back and worth. Its gravity field lags behind: this is the origin of a gravitational wave. The stretching of the radial Schwarzschild metric around M is

       1 + 1/2 rₛ / r,

where r is the distance from M and rₛ is the Schwarzschild radius. The slowdown of time in the Schwarzschild metric is

       1 - 1/2 rₛ / r.

The lagging gravity field creates a wave which "transports" a part of this metric perturbation away, as well as distorts it.

The amplitude of the wave goes as ~ 1 / r. The wave at some large distance r might shorten a local tangential distance by a factor

       1 - 1/4 rₛ / r,

but since the local time runs slower by a factor

      1 - 1/2 rₛ / r,

the local speed of light is still slower than in the surrounding Minkowski space. Faster-light-communication in Minkowski space is not enabled.

Our reasoning indicates that gravitational waves without a source mass might break the light speed limit c in Minkowski space. But are such waves possible?

Let us imagine a localized wave packet which starts from a certain position X far away from any masses. The packet travels a long distance, spreads, and gets absorbed by various masses M. If we play time backwards, we have those masses M producing a tight wave packet at a distant location X. Is the gravity of the masses M strong enough at the faraway location, so that it prevents faster-than-light signals?


Generating a gravitational wave with rods and pressure


We can harvest energy very efficiently from a gravitational wave if we utilize a rigid rod between two masses M. The wave alters the distance of the masses, and we can extract a lot of energy from the pressure inside the rod. Just let segments of the rod slide past each other with a lot of friction. Energy from the gravitational wave is turned into heat.

This rod system might be lightweight enough, so that we are able to break the speed of light at the location X of the previous section?

Most probably yes. The gravitational wave passes past the rod very fast. We do not even need any masses at the ends of the rod, because the rod does not have much time to move. If the rod is almost infinitely rigid, it can harvest a considerable amount of energy from the gravitational wave, while being very lightweight. Such a material does not exist in nature, but in principle it could exist.

Let us assume that we have a lightweight rod system which is able to harvest almost all energy out of a gravitational wave packet. If the play the process backwards in time, our rod system creates the wave packet.

As the wave packet passes two observers A and B, the distance between A and B, the packet probably makes the local speed of light faster than c in the background Minkowski metric. This open ports for a time machine, and the paradoxes which arise from it.

We probably have here a contradiction in general relativity. What is the origin of the contradiction? It is in the assumption that we can describe gravity with a metric. We have earlier in this blog claimed that a metric most probably cannot capture all features of an interaction.

In our own Minkowski & newtonian gravity model, a photon can only gain more inertia in an interaction. An interaction can only make the photon to move slower. Thus, our own model does not suffer from the problems described above.


Conclusions


Earlier in this blog we have argued that a gravitational wave in general relativity can speed up light. That leads to a time machine and paradoxes in special relativity. We cannot accept such behavior.

In this blog post we analyzed the process in more detail. To create a rogue gravitational wave, one must use pressure in a very rigid rod. Moving masses probably cannot do the job.

We have here more evidence that the concept of a metric is flawed. It cannot describe gravity.

But general relativity does predict correctly that the energy density of a gravitational wave is 16X the analogous electromagnetic wave. We have to find out why.

Wednesday, February 14, 2024

Biot-Savart: magnetic gravity; energy density of gravitational waves

Let us try to figure out what implications does our analysis of Biot-Savart have for magnetic gravity. Is it reasonable to assume that the gravity field between m and M works much like the electric field between opposite charges q and Q?

Our analysis of the electric field is based on the energy which is freed or absorbed when we move q closer to Q or away from Q. This energy is completely analogous for weak fields between m and M. We should see analogous phenomena.


The electromagnetic / gravity field of a rotating disk


Ever since our post on August 29, 2023, we have been confused about the gravity field of a mass flow. A typical example of a mass flow is a rotating disk.


                <-- ω
                _____
             /            \
            |                |  +
              \______/
                    +

                    ^  V
                    |
                     • q = e- 


Let us consider a rotating electrically positively charged disk. A negative test charge q approaches it.

The disk is uniformly charged. Let us superpose a static disk onto it, such that the static disk is uniformly charged and cancels the positive charge of the rotating disk in the laboratory frame.

We now essentially have in the disk configuration many wire elements dl creating a magnetic field B. We can calculate the magnetic field of the rotating disk using the Biot-Savart law.

The electric field of the positively charged rotating disk is the opposite of the electric field of the static negatively charged disk.


The 16X power of a radiating gravity quadrupole: gravity is NOT analogous to electromagnetism


If the magnetic effect of a (weak) gravity field is totally analogous to electromagnetism, then we above have a recipe for calculating the gravitomagnetic field of a moving mass.

But the Kerr metric and various other literature claims that the gravitomagnetic effect of is 4X compared to the analogous electromagnetic effect.

Also, we know from binary pulsars that a gravity quadrupole radiates 16X the energy of the analogous electric quadrupole. Is this compatible with the claim that the gravitomagnetic field is completely analogous to electromagnetism?

On December 29, 2021 we tried to calculate the energy content of a gravitational wave by considering its positive and negative pressure effects on a solid body. Our analysis suggested that the large energy density of a gravitational wave comes from its ability to stretch spatial distances.

This is a property which does not exist in an electromagnetic wave. In this respect gravity is not analogous to electromagnetism. If that is the case, can we define magnetic gravity in a reasonable way at all?


Harvesting energy from a wave


If we have any kind of a wave, we can harvest energy from it by putting a system of charges, let us call that S, into it and letting it move those charges. If we put a force which makes the charges resist the movement, we can harvest energy. The movement of the charges must in this case "cancel" part of the wave. By calculating the cancellation effect we can deduce the energy density of the wave.

However, the cancellation effect is not straightforward to determine because, e.g., the electromagnetic wave is not "linear" on the charges generating it. Let D be a dipole antenna generating a wave. If we let another dipole antenna D' harvest energy from the wave, then D' can oscillate almost in sync with the incoming wave, but D' does not contribute to the amplitude of the wave – on the contrary.

Let us consider mechanical waves in a tense string. We can "couple" masses m to the string with springs. The interaction force F between the wave and the mass m does not reveal much about the string.


Conclusions


The energy density of a gravitational wave is a very important concept if we want to understand magnetic gravity. Let us write a new blog post about it.


Stephen M. Barnett (2013) studies the analogy between gravitational waves and electromagnetic waves.


In the link, Chris Hirata (?) (2018) derives the energy density by first forming a wave solution in linearized Einstein equations, and then calculating the error term that the wave solution has in full, nonlinear, Einstein equations. The error term can be interpreted as a "stress-energy tensor" T. The mass-energy density T₀₀ is then taken as the energy density of the gravitational wave.

Why does this work?