https://en.wikipedia.org/wiki/Ergosphere
Optical gravity explains in a simple way the ergosphere of a rotating black hole. If we have an optically dense sphere which is rotating very fast, then the local speed of light at a point inside the sphere may be lower than the rotation speed of that point measured in the global coordinates. A fast spinning glass sphere will take even light to rotate with itself in the carousel.
https://en.wikipedia.org/wiki/Penrose_process
One can extract the rotation energy of a black hole through the Penrose process. From where does the energy come from in the process? Does the event horizon of the black hole shrink as energy is extracted from it?
In an extremal black hole, the horizon in the global coordinates is rotating at the speed of light. Suppose that we have an object that passes by the black hole at a speed less than light. Then its interaction with any mass in the part of the horizon close to the object will obviously slow down the rotation of the black hole and transfer angular momentum to the object. The interaction of the object with the far side of the horizon will similarly slow down the rotation of the black hole, and move angular momentum to the object.
If we think frame dragging as a carousel, then the carousel in an extremal black hole rotates at the global speed of light, also to an observer who is far away. The carousel does not slow down when we go further, even though the effect of the frame dragging becomes weaker.
Monday, May 21, 2018
Monday, May 14, 2018
Frame dragging explains the movement of a black hole in optical gravity
Our blog post on April 26, 2018 left open the problem how black holes can orbit in optical gravity if the local speed of light is zero.
http://meta-phys-thoughts.blogspot.fi/2018/04/how-do-black-holes-orbit-and-merge-in.html
The frame-dragging phenomenon of a black hole offers a solution. In optical gravity, a moving mass drags the optically dense medium along with it.
https://en.wikipedia.org/wiki/Frame-dragging
Close to the horizon of a black hole, the local speed of light as seen by a faraway observer can be arbitrarily slow. But the black hole can still rotate fast or move linearly fast relative to the distant observer. If the distant observer lowers a rope close to the horizon, the local speed of the rope relative to the black hole cannot exceed the local speed of light. The rope must be dragged along with the horizon. Close to the horizon, everything must move in the direction of the global movement of the horizon.
Similarly, if we have a glass sphere of a very high optical density, a ray of light that enters the ball must move along the ball.
What does frame-dragging say about the merging of two black holes? It is like two spheres of very high optical density colliding. The spheres are infinitely rigid because the local speed of light is essentially zero inside them. As soon as the horizons of the two black holes touch, their relative movement should stop. The result should be a dumbbell-shaped horizon.
The merged system will lose substantial energy in gravitational waves as the dumbbell keeps rotating. Does the horizon shrink because of the lost energy? It has to, because eventually the rotation will slow to a crawl and the combined ADM mass of the dumbbell is much lower than the sum of the masses of the original black holes.
As the horizon shrinks, parts of the old black holes will "melt". The local speed of light is greater than zero again in those parts. Thus, those parts can fall towards the new, smaller horizon.
What is the dynamics of the dragged space like in the collision of two black holes? In the center-of-mass frame the horizons of the black holes will appear squeezed in the direction of the movement. If the movement suddenly stops, what will the immediate shape of the horizon be?
What if we have a merger of two heavy neutron stars? The local speed of light is slow. The collision process has to happen like in a slow-motion film. The relative kinetic energy of the neutron stars will slowly be transformed into heat in their matter. Since the local speed of light is slow, the collision of the neutron stars is like two spheres of very viscous liquid colliding. The relative speed of the spheres can be high but the deformation of the spheres in the collision is necessarily slow.
The merger of two black holes colliding head-on should be like the merger of two infinitely viscous heavy neutron stars. Apparently, the relative kinetic energy of the matter in the black holes will remain kinetic energy after the collision, but since the local speed of light is essentially zero, there is no movement. The kinetic energy is kind of frozen in place.
The dumbbell will keep its shape after the contact of the horizons. If not much energy is lost in gravitational waves, the horizon will not shrink much. We are left with a dumbbell-shaped nonrotating black hole.
The creation of a dumbbell black hole can be viewed as a collision process that slows down infinitely at the first contact of the horizons.
What about a mini black hole that falls into a big black hole? Can it become infinitely squeezed at the big horizon? The length contraction is disturbed by the field of the mini black hole.
Our discussion above casts some light on the problem how a gravitational field can reflect photons. Where does the extra momentum go when a photon suddenly turns back in its path? The "viscosity" of the gravitational field may explain this. Since the gravitational field appears viscous to a faraway observer, it can absorb momentum from the photon. A black hole appears infinitely rigid and can easily absorb the extra momentum.
http://meta-phys-thoughts.blogspot.fi/2018/04/how-do-black-holes-orbit-and-merge-in.html
The frame-dragging phenomenon of a black hole offers a solution. In optical gravity, a moving mass drags the optically dense medium along with it.
https://en.wikipedia.org/wiki/Frame-dragging
Close to the horizon of a black hole, the local speed of light as seen by a faraway observer can be arbitrarily slow. But the black hole can still rotate fast or move linearly fast relative to the distant observer. If the distant observer lowers a rope close to the horizon, the local speed of the rope relative to the black hole cannot exceed the local speed of light. The rope must be dragged along with the horizon. Close to the horizon, everything must move in the direction of the global movement of the horizon.
Similarly, if we have a glass sphere of a very high optical density, a ray of light that enters the ball must move along the ball.
What does frame-dragging say about the merging of two black holes? It is like two spheres of very high optical density colliding. The spheres are infinitely rigid because the local speed of light is essentially zero inside them. As soon as the horizons of the two black holes touch, their relative movement should stop. The result should be a dumbbell-shaped horizon.
The merged system will lose substantial energy in gravitational waves as the dumbbell keeps rotating. Does the horizon shrink because of the lost energy? It has to, because eventually the rotation will slow to a crawl and the combined ADM mass of the dumbbell is much lower than the sum of the masses of the original black holes.
As the horizon shrinks, parts of the old black holes will "melt". The local speed of light is greater than zero again in those parts. Thus, those parts can fall towards the new, smaller horizon.
What is the dynamics of the dragged space like in the collision of two black holes? In the center-of-mass frame the horizons of the black holes will appear squeezed in the direction of the movement. If the movement suddenly stops, what will the immediate shape of the horizon be?
What if we have a merger of two heavy neutron stars? The local speed of light is slow. The collision process has to happen like in a slow-motion film. The relative kinetic energy of the neutron stars will slowly be transformed into heat in their matter. Since the local speed of light is slow, the collision of the neutron stars is like two spheres of very viscous liquid colliding. The relative speed of the spheres can be high but the deformation of the spheres in the collision is necessarily slow.
The merger of two black holes colliding head-on should be like the merger of two infinitely viscous heavy neutron stars. Apparently, the relative kinetic energy of the matter in the black holes will remain kinetic energy after the collision, but since the local speed of light is essentially zero, there is no movement. The kinetic energy is kind of frozen in place.
The dumbbell will keep its shape after the contact of the horizons. If not much energy is lost in gravitational waves, the horizon will not shrink much. We are left with a dumbbell-shaped nonrotating black hole.
The creation of a dumbbell black hole can be viewed as a collision process that slows down infinitely at the first contact of the horizons.
What about a mini black hole that falls into a big black hole? Can it become infinitely squeezed at the big horizon? The length contraction is disturbed by the field of the mini black hole.
Conservation of momentum when the horizon reflects light
Our discussion above casts some light on the problem how a gravitational field can reflect photons. Where does the extra momentum go when a photon suddenly turns back in its path? The "viscosity" of the gravitational field may explain this. Since the gravitational field appears viscous to a faraway observer, it can absorb momentum from the photon. A black hole appears infinitely rigid and can easily absorb the extra momentum.
Saturday, May 12, 2018
Why do closed loop Feynman integrals diverge?
We will now tackle the main problem that has been vexing quantum field theorists for the past 85 years - divergent integrals and renormalization.
Feynman, R. P. (1949) The theory of positrons. Physical Review, 76 (6). pp. 749-759.
https://authors.library.caltech.edu/3520/
Feynman, R. P. (1949) Space-time approach to quantum electrodynamics. Physical Review, 76 (6). pp. 769-789.
https://authors.library.caltech.edu/3523/
Feynman's second 1949 paper considers the path of two electrons in a scattering experiment.
The starting point is no interaction. The two electrons trace their path as free particles according to the Dirac equation. The free particle case is easy to handle: the wave function is the direct product of two free particle wave functions.
Then Feynman adds the repulsive Coulomb force that only is switched on for a short time dt. The Hamiltonian is modified for the interval dt, which creates a correction term to the original wave function. The term is small. That is why this is called the perturbative method.
By integrating over all dt we get a correction term for the interaction for the whole time interval. The correction term is now bigger, but we still use the, perhaps misleading, name perturbation.
An electric field invokes vacuum polarization. The electric field is slightly distorted because it creates a disturbance in the Dirac field. In the Feynman formulas, the disturbance is modeled by a term which can be interpreted as the creation and annihilation of a virtual electron-positron pair.
Unfortunately, the term diverges because it allows an arbitrary value for the momentum p of the virtual pair. If we limit p to some sensible interval, we get finite values and can do extremely accurate calculations of the real physical process. This is called renormalization.
The goal of this blog post is to understand intuitively why the divergence happens and what it means.
If we modify a wave function, its integrated probability over all possible outcomes should stay as 1.
Why the Feynman loop term does not respect this rule? If the term becomes infinite, then it breaks the rule maximally.
It looks like there is some error in the Feynman formula in the first place. Maybe the probability amplitude associated with large values of p is exaggerated?
If we take the Feynman diagram literally, the probabilities associated with each path of real and virtual particles should add up to 1.
Let us consider a simpler case: the double slit experiment.
We trace all possible paths for the photon. One half of the paths go through slit A and the rest through slit B. The interference pattern on the screen adds up the complex number probability amplitudes. A constructive interference and a destructive interference alternate. The alternation makes the integral of probabilities over the whole screen as 1.
If we would use some approximation method in the calculation of the interference pattern, the probability integral might become infinite if we either inflate the amplitudes in some path or exaggerate greatly the constructive interference. Which is the case in the Feynman loop integrals?
Feynman integrals involve an "infrared" divergence in the "self-energy" graph, where an electron sends a virtual photon to itself.
~~~~~ virtual photon
/ \
e- --------->-------------
For a free electron, hypothetical virtual self-energy photons are not relevant. We can describe a free electron with the Dirac equation.
But when an electron is accelerated, then self-energy has an impact. What is this self-energy? Our discussion of the Larmor formula offers a clue:
http://meta-phys-thoughts.blogspot.fi/2018/04/the-error-in-larmor-formula-for.html
Radiation is produced by the interaction of charges. The charges together cause a disturbance to the electromagnetic field. Far away from charges the field is under an essentially linear wave equation and its disturbance has a Fourier decomposition. Far away, we can talk about electromagnetic radiation and photons.
But close to the charges, the system is nonlinear. The interaction between the field and the charges is complex. Momentum and energy can be transferred to the field and back to the charges in a complex way. This explains why an electron can move momentum and energy to the field and receive it back later.
Thus, the self-energy is nothing mystical but a regular phenomenon of classical electrodynamics. In classical electrodynamics the amount of momentum transferred from the electron to the field is not arbitrarily large, as the Feynman formulas postulate, but restricted by the geometry of the situation.
Open problem 1. Is there a reason why Feynman formulas should allow an arbitrary momentum for the self-energy virtual photon?
The following analogue might clarify a nonlinear scattering process. If we have two people strirring water with paddles that are close to each other, then the interaction between paddles is mediated mostly by water currents. But far away, the movement of water can be approximated by waves. The interaction to a far away paddle would be mediated by waves. "Virtual photons" are the water currents while real photons are waves.
The name "virtual photon" is misleading. A better name would be a direct force impulse. A force impulse can be between charges or between a charge and the electromagnetic field.
We have a hypothesis that electromagnetic waves are just polarization of virtual electron-positron pairs. In this framework, a direct force impulse is always between electrons or positrons, virtual or real. The electromagnetic field is just an abstraction.
Let us then try to decipher where and why arbitrarily large momentums enter the Feynman calculations. Feynman uses Green's functions to approximate solutions to an inhomogeneous linear differential equation. Inhomogeneous means that we have a linear system which is disturbed by a "source" function. An example is a string which is plucked.
A Green's function describes an infinitely sharp "point impulse" to a wave equation and the diffusion of the impulse afterwards. The point impulse is a Dirac delta function which starts to develop in time and diffuse according to the wave equation.
A Dirac delta function can be built by adding an infinite number of plane waves where all waves have a constructive interference at time t = 0 at position x = 0. The waves have arbitrary momentum p. This is obviously the source of arbitrary momentums in the Feynman formulas.
The question is whether allowing arbitrary momentums makes sense in the nonlinear system we are studying.
In The theory of positrons, section 6, Feynman says that the Fourier transform of K_+(2, 1) is of the form given in his equation (31). The integral in (31) contains (p - m)^-1, which introduces two poles in the integral, at E and -E in the notation of Feynman.
The poles are the first symptom of diverging integrals in Feynman rules. Feynman handles the poles by adding an infinitesimal imaginary mass to m, which moves the poles out of the real axis.
Let us try to decipher what is the origin of these poles.
The notation p - m is not clear. The p is apparently a 4 x 4 matrix built by summing gamma matrices for each of the 4 components of the momentum of the electron. The fourth component is apparently the mass-energy "moving" forward in time.
But what is m? Do we multiply the fourth gamma matrix with the rest mass of the electron to obtain m?
What is the physical intuition why the Fourier decomposition of K_+(2, 1) should contain an infinite component when p - m happens to be a zero matrix?
Why handling the poles by adding an infinitesimal imaginary part to m produces sensible results?
The Dirac equation with a potential is
(i∇ - m) ψ = A ψ.
Let the potential A = 0. We can "disturb" the equation with a "source" pulse
(i∇ - m) ψ = δ(x - x_1)
at a spacetime point x_1.
Feynman uses a Green's function to "create" a particle at point x_1. He lets a source to disturb the homogeneous Dirac equation with a Dirac delta -shaped pulse. Then he tracks the time development of the wave function ψ. The Dirac equation contains the first time derivative of ψ.
This is not analogous to hitting a string with an infinitely narrow hammer. The hammer will give a momentum to an infinitely short segment of the string in an infinitely short time. The wave equation of a string contains the second time derivative of the vertical displacement of the string. The source in such an equation means the momentum of the string segment.
Electromagnetic waves are analogous to a string, but the Schrödinger equation is not.
We want to set ψ to some initial value at t = 0, then calculate its development forward in time.
If we want to do the calculation by summing the time development of small "spike" functions, we should set ψ = spike and calculate forward. We set a small spike as the initial value the wave function by setting a Dirac delta function (a "big" spike) as the source of our linear equation.
Basically, we can write
dψ / dt = Lψ(x, t) + source(x, t),
where source is a Dirac delta function in t and x, and zero elsewhere.
Thus, a short timestep forward:
ψ = ψ + Lψ dt + source dt.
When we loop the above algorithm for the very short time interval that source differs from 0, the equation above is dominated by the source and we can let it set the desired value as the initial value of the wave function.
Using the source method makes sense when some other field "disturbs" our ψ for a very short time in a small spatial zone and the disturbance can be modeled as a source function in our equation. But why would we use the source method to solve a simple initial value problem?
In Feynman diagrams, the most important thing is the vertexes where a field disturbs another field and creates particles to the other field. In those vertexes, it makes sense to use a Green's function, because it is a disturbance, that is, a source that creates particles.
Is it ok to use a Dirac delta function as the source, or should we use smoother functions?
A Dirac field might be seen as describing a point particle, the quantum of the field that has a sharp point location. The particle then disturbs the electromagnetic field and creates quantums.
In bremsstrahlung, the photons are created in the interaction of two charges. Does it make sense to put them in a Feynman diagrams as originating from a single charge?
What is the role of virtual particles in the Lagrangian and Feynman diagrams? The Feynman integrals are done in the "energy-momentum space". How do we describe a virtual particle there versus a real particle?
Feynman, R. P. (1949) The theory of positrons. Physical Review, 76 (6). pp. 749-759.
https://authors.library.caltech.edu/3520/
Feynman, R. P. (1949) Space-time approach to quantum electrodynamics. Physical Review, 76 (6). pp. 769-789.
https://authors.library.caltech.edu/3523/
Feynman's second 1949 paper considers the path of two electrons in a scattering experiment.
The starting point is no interaction. The two electrons trace their path as free particles according to the Dirac equation. The free particle case is easy to handle: the wave function is the direct product of two free particle wave functions.
Then Feynman adds the repulsive Coulomb force that only is switched on for a short time dt. The Hamiltonian is modified for the interval dt, which creates a correction term to the original wave function. The term is small. That is why this is called the perturbative method.
By integrating over all dt we get a correction term for the interaction for the whole time interval. The correction term is now bigger, but we still use the, perhaps misleading, name perturbation.
An electric field invokes vacuum polarization. The electric field is slightly distorted because it creates a disturbance in the Dirac field. In the Feynman formulas, the disturbance is modeled by a term which can be interpreted as the creation and annihilation of a virtual electron-positron pair.
Unfortunately, the term diverges because it allows an arbitrary value for the momentum p of the virtual pair. If we limit p to some sensible interval, we get finite values and can do extremely accurate calculations of the real physical process. This is called renormalization.
The goal of this blog post is to understand intuitively why the divergence happens and what it means.
If we modify a wave function, its integrated probability over all possible outcomes should stay as 1.
Why the Feynman loop term does not respect this rule? If the term becomes infinite, then it breaks the rule maximally.
It looks like there is some error in the Feynman formula in the first place. Maybe the probability amplitude associated with large values of p is exaggerated?
If we take the Feynman diagram literally, the probabilities associated with each path of real and virtual particles should add up to 1.
Let us consider a simpler case: the double slit experiment.
We trace all possible paths for the photon. One half of the paths go through slit A and the rest through slit B. The interference pattern on the screen adds up the complex number probability amplitudes. A constructive interference and a destructive interference alternate. The alternation makes the integral of probabilities over the whole screen as 1.
If we would use some approximation method in the calculation of the interference pattern, the probability integral might become infinite if we either inflate the amplitudes in some path or exaggerate greatly the constructive interference. Which is the case in the Feynman loop integrals?
Self-energy of an electron
Feynman integrals involve an "infrared" divergence in the "self-energy" graph, where an electron sends a virtual photon to itself.
~~~~~ virtual photon
/ \
e- --------->-------------
For a free electron, hypothetical virtual self-energy photons are not relevant. We can describe a free electron with the Dirac equation.
But when an electron is accelerated, then self-energy has an impact. What is this self-energy? Our discussion of the Larmor formula offers a clue:
http://meta-phys-thoughts.blogspot.fi/2018/04/the-error-in-larmor-formula-for.html
Radiation is produced by the interaction of charges. The charges together cause a disturbance to the electromagnetic field. Far away from charges the field is under an essentially linear wave equation and its disturbance has a Fourier decomposition. Far away, we can talk about electromagnetic radiation and photons.
But close to the charges, the system is nonlinear. The interaction between the field and the charges is complex. Momentum and energy can be transferred to the field and back to the charges in a complex way. This explains why an electron can move momentum and energy to the field and receive it back later.
Thus, the self-energy is nothing mystical but a regular phenomenon of classical electrodynamics. In classical electrodynamics the amount of momentum transferred from the electron to the field is not arbitrarily large, as the Feynman formulas postulate, but restricted by the geometry of the situation.
Open problem 1. Is there a reason why Feynman formulas should allow an arbitrary momentum for the self-energy virtual photon?
The following analogue might clarify a nonlinear scattering process. If we have two people strirring water with paddles that are close to each other, then the interaction between paddles is mediated mostly by water currents. But far away, the movement of water can be approximated by waves. The interaction to a far away paddle would be mediated by waves. "Virtual photons" are the water currents while real photons are waves.
The name "virtual photon" is misleading. A better name would be a direct force impulse. A force impulse can be between charges or between a charge and the electromagnetic field.
We have a hypothesis that electromagnetic waves are just polarization of virtual electron-positron pairs. In this framework, a direct force impulse is always between electrons or positrons, virtual or real. The electromagnetic field is just an abstraction.
Let us then try to decipher where and why arbitrarily large momentums enter the Feynman calculations. Feynman uses Green's functions to approximate solutions to an inhomogeneous linear differential equation. Inhomogeneous means that we have a linear system which is disturbed by a "source" function. An example is a string which is plucked.
A Green's function describes an infinitely sharp "point impulse" to a wave equation and the diffusion of the impulse afterwards. The point impulse is a Dirac delta function which starts to develop in time and diffuse according to the wave equation.
A Dirac delta function can be built by adding an infinite number of plane waves where all waves have a constructive interference at time t = 0 at position x = 0. The waves have arbitrary momentum p. This is obviously the source of arbitrary momentums in the Feynman formulas.
The question is whether allowing arbitrary momentums makes sense in the nonlinear system we are studying.
The poles in the energy-momentum representation
In The theory of positrons, section 6, Feynman says that the Fourier transform of K_+(2, 1) is of the form given in his equation (31). The integral in (31) contains (p - m)^-1, which introduces two poles in the integral, at E and -E in the notation of Feynman.
The poles are the first symptom of diverging integrals in Feynman rules. Feynman handles the poles by adding an infinitesimal imaginary mass to m, which moves the poles out of the real axis.
Let us try to decipher what is the origin of these poles.
The notation p - m is not clear. The p is apparently a 4 x 4 matrix built by summing gamma matrices for each of the 4 components of the momentum of the electron. The fourth component is apparently the mass-energy "moving" forward in time.
But what is m? Do we multiply the fourth gamma matrix with the rest mass of the electron to obtain m?
What is the physical intuition why the Fourier decomposition of K_+(2, 1) should contain an infinite component when p - m happens to be a zero matrix?
Why handling the poles by adding an infinitesimal imaginary part to m produces sensible results?
The Dirac equation with a potential is
(i∇ - m) ψ = A ψ.
Let the potential A = 0. We can "disturb" the equation with a "source" pulse
(i∇ - m) ψ = δ(x - x_1)
at a spacetime point x_1.
Feynman uses a Green's function to "create" a particle at point x_1. He lets a source to disturb the homogeneous Dirac equation with a Dirac delta -shaped pulse. Then he tracks the time development of the wave function ψ. The Dirac equation contains the first time derivative of ψ.
This is not analogous to hitting a string with an infinitely narrow hammer. The hammer will give a momentum to an infinitely short segment of the string in an infinitely short time. The wave equation of a string contains the second time derivative of the vertical displacement of the string. The source in such an equation means the momentum of the string segment.
Electromagnetic waves are analogous to a string, but the Schrödinger equation is not.
We want to set ψ to some initial value at t = 0, then calculate its development forward in time.
If we want to do the calculation by summing the time development of small "spike" functions, we should set ψ = spike and calculate forward. We set a small spike as the initial value the wave function by setting a Dirac delta function (a "big" spike) as the source of our linear equation.
Basically, we can write
dψ / dt = Lψ(x, t) + source(x, t),
where source is a Dirac delta function in t and x, and zero elsewhere.
Thus, a short timestep forward:
ψ = ψ + Lψ dt + source dt.
When we loop the above algorithm for the very short time interval that source differs from 0, the equation above is dominated by the source and we can let it set the desired value as the initial value of the wave function.
Using the source method makes sense when some other field "disturbs" our ψ for a very short time in a small spatial zone and the disturbance can be modeled as a source function in our equation. But why would we use the source method to solve a simple initial value problem?
In Feynman diagrams, the most important thing is the vertexes where a field disturbs another field and creates particles to the other field. In those vertexes, it makes sense to use a Green's function, because it is a disturbance, that is, a source that creates particles.
Is it ok to use a Dirac delta function as the source, or should we use smoother functions?
A Dirac field might be seen as describing a point particle, the quantum of the field that has a sharp point location. The particle then disturbs the electromagnetic field and creates quantums.
In bremsstrahlung, the photons are created in the interaction of two charges. Does it make sense to put them in a Feynman diagrams as originating from a single charge?
What is the role of virtual particles in the Lagrangian and Feynman diagrams? The Feynman integrals are done in the "energy-momentum space". How do we describe a virtual particle there versus a real particle?
Thursday, May 10, 2018
Is a Schwinger process possible for a gravitational field?
In the Schwinger process, a strong static electric field creates electron-positron pairs. If the electric field exceeds 10^18 V/m, the Schwinger process becomes relevant.
In the Hamiltonian thinking, the Schwinger process happens because it lowers the energy of fields and frees heat. Heat has a large entropy. Therefore, the process happens to one direction.
Problem 1. What about a strong gravitational field? Can we find a process which frees energy as heat?
If a virtual electron has zero or negative energy, it can gain energy by moving in the right direction in an electric field and become a real electron. In a gravitational field, a virtual particle cannot do the same, because as soon as its mass-energy becomes zero, it will no longer gain energy by moving in a gravitational field.
In Feynman diagrams we require that our process must eventually produce only real particles.
Thought experiment 2. What about the metaphorical Hawking process where a negative energy particle A enters a black hole and a positive energy particle B escapes?
The positive energy particle must have momentum upward to be able to escape. Is there any way that we could get rid of the negative energy particle and its downward momentum? If we would have its antiparticle A' coming up with the right energy and momentum, then we could let them annihilate and we would be left with zero energy and momentum close to the forming horizon. But we do not have such upward-traveling particles at the horizon. Furthermore, such annihilation would lower the entropy, which means the process is extremely unlikely to happen. Also, the process is better described as the upward shooting particle A' escaping on its own. There is no need to invoke A and B.
The metaphorical Hawking process is banned by Feynman diagram type of thinking, or has an infinitesimal probability to happen.
Thought experiment 3. What about a process where the positive energy particle drops to the horizon and the negative energy particle escapes? Again, the negative energy particle cannot gain enough energy from the gravitational push and cannot become a real particle.
Let us then forget about Feynman diagrams and think in terms of fields and their energy, momentum, and all conserved quantities. That is, use the Hamiltonian way of thinking.
Thought experiment 4. If we are collapsing a photon sphere to form a black hole, then we can satisfy conservation laws by letting two photons to reflect upward on the opposite sides of the forming black hole. But this is better understood as reflection rather than Hawking radiation or the Schwinger process. Our optical gravity predicts such process.
Let a dark matter shell collapse to form a black hole. We can satisfy conservation laws by letting two photons to materialize and escape from the opposite sides of the forming black hole, and by deducting the energy for photons from the kinetic energy of dark matter particles.
Is this process possible? Suppose that we have two dark matter particles approaching each other in empty space. We could satisfy conservation laws by deducting their kinetic energy and letting that energy to escape as two photons. Is such a process possible?
Let us think about the Hamiltonian. We have some fields and a function which assigns an energy to a field configuration. The fields themselves have energy and the fields have interactions which add to the energy.
If there is no direct or indirect interaction between two fields, then it is obvious that the fields cannot exchange energy.
The gravitational field does have an interaction with all fields. Can it transfer energy from any field to another field?
If we have two field configurations with the same energy, momentum, electric charge, etc., is it always possible that one configuration evolves to the other? Or are there other conserved quantities that ban such an evolution? Intuitively, most such evolutions are banned or have just an infinitesimal probability of happening. For example, the electron-positron annihilation process is extremely unlikely to happen in the exact same way in the opposite direction, though the annihilation happens very easily. The opposite process would require fine-tuned photons meeting just in the right way.
If we have a time-varying electromagnetic field, it will always produce a time-varying gravitational field. Actually, such electromagnetic field is always accompanied with a non-zero gravitational field, since mass-energy always produces a gravitational field.
What about a time-varying gravitational field? If the electromagnetic field has zero energy, can a time-varying gravitational field transfer energy to it?
If we have a totally smooth surface of water and a breeze of wind, ripples will appear to the water. The ripples get their energy from the breeze. The breeze amplifies tiny fluctuations in the surface of water.
An electromagnetic ripple can gain substantial energy from a gravitational field by moving closer to a black hole horizon. But from where does the ripple get its original energy? If it borrows it from a virtual photon, how do we get rid of the virtual photon? If the ripple originally has some energy of its own, we can describe the process by the blue shift of a photon moving down a gravitational field. Furthermore, if we measure the energy of the photon from the point of view of a global Schwarzschild observer, the photon really does not gain any more energy because it is offset by a lower gravitational potential. The ripple changes the gravitational field very little as observed from far away.
The above is in contrast with the Schwinger effect. A strong electric field around a big charge can tear electric charge from "empty" space and the charge will change the electric field in a significant way.
What if the electromagnetic ripple gets energy from a gravitational "elevator" process where it first goes down a gravitational potential, say, close to a neutron star and then the neutron star is suddenly taken apart so that the ripple moves to a higher gravitational potential without using its own energy? If we would have such elevators in nature, they could greatly amplify small electromagnetic ripples. Maybe they could create electromagnetic ripples from space also in the case where the electromagnetic field originally has zero energy? But again, how do we get rid of the negative energy ripple from which our ripple has to borrow energy?
Classically, a zero energy electromagnetic field cannot gain any energy from a time-varying gravitational field because there is no "handle" in it to which the gravitational field could get a grip.
In quantum mechanics, we can never know that a field has exactly zero energy. We must assume that there are small disturbances around. But the author of this blog has not been able to find a way to transfer much energy to those disturbances from a time-varying gravitational field.
Conjecture 5. A time-varying gravitational field cannot in natural circumstances transfer much energy to a zero or almost zero energy electromagnetic field. There is no Schwinger process for a gravitational field.
In the Hamiltonian thinking, the Schwinger process happens because it lowers the energy of fields and frees heat. Heat has a large entropy. Therefore, the process happens to one direction.
Problem 1. What about a strong gravitational field? Can we find a process which frees energy as heat?
A Feynman diagram point of view
If a virtual electron has zero or negative energy, it can gain energy by moving in the right direction in an electric field and become a real electron. In a gravitational field, a virtual particle cannot do the same, because as soon as its mass-energy becomes zero, it will no longer gain energy by moving in a gravitational field.
In Feynman diagrams we require that our process must eventually produce only real particles.
Thought experiment 2. What about the metaphorical Hawking process where a negative energy particle A enters a black hole and a positive energy particle B escapes?
The positive energy particle must have momentum upward to be able to escape. Is there any way that we could get rid of the negative energy particle and its downward momentum? If we would have its antiparticle A' coming up with the right energy and momentum, then we could let them annihilate and we would be left with zero energy and momentum close to the forming horizon. But we do not have such upward-traveling particles at the horizon. Furthermore, such annihilation would lower the entropy, which means the process is extremely unlikely to happen. Also, the process is better described as the upward shooting particle A' escaping on its own. There is no need to invoke A and B.
The metaphorical Hawking process is banned by Feynman diagram type of thinking, or has an infinitesimal probability to happen.
Thought experiment 3. What about a process where the positive energy particle drops to the horizon and the negative energy particle escapes? Again, the negative energy particle cannot gain enough energy from the gravitational push and cannot become a real particle.
A Hamiltonian point of view
Let us then forget about Feynman diagrams and think in terms of fields and their energy, momentum, and all conserved quantities. That is, use the Hamiltonian way of thinking.
Thought experiment 4. If we are collapsing a photon sphere to form a black hole, then we can satisfy conservation laws by letting two photons to reflect upward on the opposite sides of the forming black hole. But this is better understood as reflection rather than Hawking radiation or the Schwinger process. Our optical gravity predicts such process.
Let a dark matter shell collapse to form a black hole. We can satisfy conservation laws by letting two photons to materialize and escape from the opposite sides of the forming black hole, and by deducting the energy for photons from the kinetic energy of dark matter particles.
Is this process possible? Suppose that we have two dark matter particles approaching each other in empty space. We could satisfy conservation laws by deducting their kinetic energy and letting that energy to escape as two photons. Is such a process possible?
Let us think about the Hamiltonian. We have some fields and a function which assigns an energy to a field configuration. The fields themselves have energy and the fields have interactions which add to the energy.
If there is no direct or indirect interaction between two fields, then it is obvious that the fields cannot exchange energy.
The gravitational field does have an interaction with all fields. Can it transfer energy from any field to another field?
If we have two field configurations with the same energy, momentum, electric charge, etc., is it always possible that one configuration evolves to the other? Or are there other conserved quantities that ban such an evolution? Intuitively, most such evolutions are banned or have just an infinitesimal probability of happening. For example, the electron-positron annihilation process is extremely unlikely to happen in the exact same way in the opposite direction, though the annihilation happens very easily. The opposite process would require fine-tuned photons meeting just in the right way.
If we have a time-varying electromagnetic field, it will always produce a time-varying gravitational field. Actually, such electromagnetic field is always accompanied with a non-zero gravitational field, since mass-energy always produces a gravitational field.
What about a time-varying gravitational field? If the electromagnetic field has zero energy, can a time-varying gravitational field transfer energy to it?
If we have a totally smooth surface of water and a breeze of wind, ripples will appear to the water. The ripples get their energy from the breeze. The breeze amplifies tiny fluctuations in the surface of water.
An electromagnetic ripple can gain substantial energy from a gravitational field by moving closer to a black hole horizon. But from where does the ripple get its original energy? If it borrows it from a virtual photon, how do we get rid of the virtual photon? If the ripple originally has some energy of its own, we can describe the process by the blue shift of a photon moving down a gravitational field. Furthermore, if we measure the energy of the photon from the point of view of a global Schwarzschild observer, the photon really does not gain any more energy because it is offset by a lower gravitational potential. The ripple changes the gravitational field very little as observed from far away.
The above is in contrast with the Schwinger effect. A strong electric field around a big charge can tear electric charge from "empty" space and the charge will change the electric field in a significant way.
What if the electromagnetic ripple gets energy from a gravitational "elevator" process where it first goes down a gravitational potential, say, close to a neutron star and then the neutron star is suddenly taken apart so that the ripple moves to a higher gravitational potential without using its own energy? If we would have such elevators in nature, they could greatly amplify small electromagnetic ripples. Maybe they could create electromagnetic ripples from space also in the case where the electromagnetic field originally has zero energy? But again, how do we get rid of the negative energy ripple from which our ripple has to borrow energy?
Classically, a zero energy electromagnetic field cannot gain any energy from a time-varying gravitational field because there is no "handle" in it to which the gravitational field could get a grip.
In quantum mechanics, we can never know that a field has exactly zero energy. We must assume that there are small disturbances around. But the author of this blog has not been able to find a way to transfer much energy to those disturbances from a time-varying gravitational field.
Conjecture 5. A time-varying gravitational field cannot in natural circumstances transfer much energy to a zero or almost zero energy electromagnetic field. There is no Schwinger process for a gravitational field.
Collision of gravitons
Let us return to Feynman diagrams and think about a scattering experiment.
A collision of electrons and positrons always produces photons. A collision of high energy photons can create electron-positron pairs:
https://en.m.wikipedia.org/wiki/Two-photon_physics
A collision of electrons and positrons always produces photons. A collision of high energy photons can create electron-positron pairs:
https://en.m.wikipedia.org/wiki/Two-photon_physics
What about a collision of gravitons? Can it produce photons or electron-positron pairs?
Since gravitons themselves carry mass-energy, the collision of gravitons will produce more gravitons.
Pair production from photons looks like this:
/ e-
^
photon / ~~~~~~ photons
~~~~~~
|
^
|
~~~~~~
photon \ ~~~~~~ photons
^
\ e+
Time flows to the right. The wave equation is in a way symmetric with respect to time and position. We can interpret the scattering in various ways. One such interpretation is that a positron travels backward in time and meets a transient electric potential. The positron cannot enter the zone with a potential but scatters to the opposite direction in time. The time reflection reverses the sign of the frequency of the wave function of the positron: it becomes an electron.
If we replace the photons in the diagram with gravitons, there is a prominent difference from the original diagram: the gravitational charge is positive for both the electron and the positron.
If we would replace the e+ in the diagram with e-, that would be a better analogue for the graviton scattering. But we cannot do such replacement because that would break conservation of charge.
If it were possible to create an electron-positron pair + some other particles from a collision of just gravitons, then by time symmetry, it would be possible to run the reaction backwards. The reaction run backwards would mean that we could with some probability > 0 convert all electromagnetic energy that we feed to the process to gravitons. Is such a process possible?
Conjecture 6. In a Feynman diagram, if a photon, electron, or positron enters the diagram, then the scattering amplitude for a process where a sizeable portion of its energy is converted to gravitons is zero or infinitesimal.
As a simple example, let us think about a diagram where just two photons enter and they are totally converted to two graviton.
photons electron-positron
loop gravitons
~~~~~~~~ O --------------------
~~~~~~~~ --------------------
We need 2 particles to make the total spin 0 in the diagram.
A brief look at literature reveals that calculating scattering amplitudes in a simplified quantum gravity is difficult and at two loops the formulas diverge. There is not much hope of proving our conjecture anytime soon.
Since gravitons themselves carry mass-energy, the collision of gravitons will produce more gravitons.
Pair production from photons looks like this:
/ e-
^
photon / ~~~~~~ photons
~~~~~~
|
^
|
~~~~~~
photon \ ~~~~~~ photons
^
\ e+
Time flows to the right. The wave equation is in a way symmetric with respect to time and position. We can interpret the scattering in various ways. One such interpretation is that a positron travels backward in time and meets a transient electric potential. The positron cannot enter the zone with a potential but scatters to the opposite direction in time. The time reflection reverses the sign of the frequency of the wave function of the positron: it becomes an electron.
If we replace the photons in the diagram with gravitons, there is a prominent difference from the original diagram: the gravitational charge is positive for both the electron and the positron.
If we would replace the e+ in the diagram with e-, that would be a better analogue for the graviton scattering. But we cannot do such replacement because that would break conservation of charge.
If it were possible to create an electron-positron pair + some other particles from a collision of just gravitons, then by time symmetry, it would be possible to run the reaction backwards. The reaction run backwards would mean that we could with some probability > 0 convert all electromagnetic energy that we feed to the process to gravitons. Is such a process possible?
Conjecture 6. In a Feynman diagram, if a photon, electron, or positron enters the diagram, then the scattering amplitude for a process where a sizeable portion of its energy is converted to gravitons is zero or infinitesimal.
As a simple example, let us think about a diagram where just two photons enter and they are totally converted to two graviton.
photons electron-positron
loop gravitons
~~~~~~~~ O --------------------
~~~~~~~~ --------------------
We need 2 particles to make the total spin 0 in the diagram.
A brief look at literature reveals that calculating scattering amplitudes in a simplified quantum gravity is difficult and at two loops the formulas diverge. There is not much hope of proving our conjecture anytime soon.
Wednesday, May 9, 2018
No paradox of reflected waves in a uniform gravitational field
Our discussion in the previous blog post raised the following question:
Paradox 1. Static observers in a homogeneous uniform gravitational field will see some of an incoming wave packet to reflect back, but freely falling observers see it to propagate freely in a flat Minkowski space.
Solution. There is no such thing as a uniform gravitational field in the sense that we would need for the paradox!
If we try to define such a field, let us put there two static observers connected with a tight rope. Both observers will feel a constant acceleration g.
But it is a well-known fact that in a Minkowski space, a tight rope set between rockets that both have an acceleration g will break. QED.
Optical gravity claims that to a global observer in the Schwarzschild solution, a gravitational potential well will appear as an optically dense zone.
Suppose that we have a string or a rope under tension. Its weight per meter is analogous to the optical density of a zone in space.
Suppose the string is in the left-right direction and that the weight of the string increases as we go right.
. • ● ■ weights
|---------------------------------------------------|
string under tension
The setup is equivalent to a string of uniform weight to which we attach individual weights whose mass increases as we go right.
Suppose that we have a simple "bump" wave packet in the string proceeding right. The bump is a gaussian wave packet which extends infinitely to the left and the right. The diagram below does not show this infinite extension.
___ ---->
______/ \______
Some of the energy in the bump will be spent on moving the attached weights. Each weight will act as a new source of waves:
● weight
___ original wave ---->
______ / \_____
<--- \___/ left-moving new wave
\___/ --> right-moving new wave
In the diagram above, we have greatly exaggerated the new waves. The inertia of the weight resists the up and down movement of the string. Therefore, the effect of a weight is roughly equivalent to:
1. at the position of the weight, first pull the string down with a finger so that you exert an impulse p down;
2. then pull it up and exert a impulse 2p up;
3. pull down again and exert an impulse p down.
Our pulling procedure will make new waves to proceed to both the left and the right.
The new waves will deform the original wave. How do they deform it?
___ original wave
_______________/ \_______
____ ______ ________
\___/ \___/ new wave 1
________ ___ _________
\___/ \___/ new wave 2
In the diagram above, we have drawn two examples of new waves. We have drawn both the left-moving and the right-moving new wave. When we sum these waves, we see that there is kind of a "sonic boom" progressing along with the original wave.
If the weight of the string increases to the right, then the sonic boom wave will be very steep on its right side, but much less steep on the left side.
The sum of the original gaussian wave plus the sonic boom wave will appear to be lower amplitude on the left side than the original wave.
The Fresnel formula can be used to calculate the reflection if there is a sharp change in the optical density. If we have a smooth change, maybe we can use the Fresnel formula if we divide the path into parts of one wavelength?
If we have two steps in the optical density within one wavelength, maybe the reflection should be calculated by adding up the steps and using the Fresnel formula?
Paradox 1. Static observers in a homogeneous uniform gravitational field will see some of an incoming wave packet to reflect back, but freely falling observers see it to propagate freely in a flat Minkowski space.
Solution. There is no such thing as a uniform gravitational field in the sense that we would need for the paradox!
If we try to define such a field, let us put there two static observers connected with a tight rope. Both observers will feel a constant acceleration g.
But it is a well-known fact that in a Minkowski space, a tight rope set between rockets that both have an acceleration g will break. QED.
Optical gravity claims that to a global observer in the Schwarzschild solution, a gravitational potential well will appear as an optically dense zone.
Suppose that we have a string or a rope under tension. Its weight per meter is analogous to the optical density of a zone in space.
Suppose the string is in the left-right direction and that the weight of the string increases as we go right.
. • ● ■ weights
|---------------------------------------------------|
string under tension
The setup is equivalent to a string of uniform weight to which we attach individual weights whose mass increases as we go right.
Suppose that we have a simple "bump" wave packet in the string proceeding right. The bump is a gaussian wave packet which extends infinitely to the left and the right. The diagram below does not show this infinite extension.
___ ---->
______/ \______
Some of the energy in the bump will be spent on moving the attached weights. Each weight will act as a new source of waves:
● weight
___ original wave ---->
______ / \_____
<--- \___/ left-moving new wave
\___/ --> right-moving new wave
In the diagram above, we have greatly exaggerated the new waves. The inertia of the weight resists the up and down movement of the string. Therefore, the effect of a weight is roughly equivalent to:
1. at the position of the weight, first pull the string down with a finger so that you exert an impulse p down;
2. then pull it up and exert a impulse 2p up;
3. pull down again and exert an impulse p down.
Our pulling procedure will make new waves to proceed to both the left and the right.
The new waves will deform the original wave. How do they deform it?
___ original wave
_______________/ \_______
____ ______ ________
\___/ \___/ new wave 1
________ ___ _________
\___/ \___/ new wave 2
In the diagram above, we have drawn two examples of new waves. We have drawn both the left-moving and the right-moving new wave. When we sum these waves, we see that there is kind of a "sonic boom" progressing along with the original wave.
If the weight of the string increases to the right, then the sonic boom wave will be very steep on its right side, but much less steep on the left side.
The sum of the original gaussian wave plus the sonic boom wave will appear to be lower amplitude on the left side than the original wave.
The Fresnel formula can be used to calculate the reflection if there is a sharp change in the optical density. If we have a smooth change, maybe we can use the Fresnel formula if we divide the path into parts of one wavelength?
If we have two steps in the optical density within one wavelength, maybe the reflection should be calculated by adding up the steps and using the Fresnel formula?
Friday, May 4, 2018
Erik Verlinde's entropic gravity and optical gravity
Erik Verlinde has suggested that gravity is not a "fundamental force", but guided by entropy.
Erik P. Verlinde
On the Origin of Gravity and the Laws of Newton
https://arxiv.org/abs/1001.0785
Recall the optical coordinates that we introduced in our blog post:
Erik P. Verlinde
On the Origin of Gravity and the Laws of Newton
https://arxiv.org/abs/1001.0785
Recall the optical coordinates that we introduced in our blog post:
In a sense, from the point of view of an electromagnetic wave, the space close to the horizon is much bigger than the R^3 Euclidean space far away.
We could try to explain gravitation as an entropic force. A wave will tend to spread more to the direction where the space looks bigger to it, in some sense, there are more degrees of freedom available for it in that direction.
In the free particle solution of the nonrelativistic Schrödinger equation, the wavenumber k for a massive particle is proportional to its momentum p. If we have a potential well, the de Broglie wavelength for the particle grows shorter when the particle falls into the well, because the momentum of the particle grows there.
It seems to be a general rule that waves in various settings tend to steer into zones where the wavelength becomes shorter, or the speed of waves is slower. We say that those zones have a lower potential, because they attract waves. We interpret this behavior as due to an attractive force.
When a wave enters a zone where its wavelength grows shorter, the form of the wave tends to become more complex. Can we say that the "entropy" of the wave has increased? Can we measure the entropy of a wave by its Fourier decomposition?
Gravitation as a force seems to be no special with waves, except that gravitation affects all known mass-energy. If gravitation is an "entropic" force, then all forces are "entropic" forces?
In Wikipedia, the formula for the entropic force is derived from the "degrees of freedom" in the border surface (holographic screen) of a zone in space. If mass-energy is contained in the zone, then we can calculate the "temperature" of that mass-energy of the screen.
The gravitational acceleration is set such that the temperature of the screen matches the Unruh temperature of the acceleration. The author of this blog does not know an answer to the following:
Open problem 1. Why the Unruh temperature of the gravitational acceleration should match the temperature of the screen. What is the physical intuition behind that?
Waves tend to steer into the direction of a shorter wavelength
Our optical view of gravity might offer an intuitive explanation why the correct acceleration for gravitation can be obtained from seemingly entropic concepts.
If we have a static observer in a gravitational field, then in the optical coordinates, the space appears stretched on the side of the the lower gravitational potential, that is, the wavelengh of light that we use as the measuring stick is shorter there.
For a static observer, there are more "degrees of freedom" when we go downhill in the potential, because the space appears stretched there and thus bigger. Waves tend to turn to the direction where there are more degrees of freedom available. Here we consider both light-speed waves and also the nonrelativistic wave function of a massive particle.
The static observer sees waves accelerating downhill in the potential. On the other hand, a freely falling observer sees the geometry of the space more flat. The degrees of freedom appear to be symmetric on each side of the freely falling observer. Therefore, he will not observe acceleration of waves.
In the above way, the degrees of freedom in a zone in space is connected to the acceleration of a wave downhill in the potential. But is it a fruitful point of view to claim that the degrees of freedom is the primary thing and the force, or acceleration, is secondary?
Also, rather than referring to degrees of freedom, it is much simpler to say that waves tend to steer to the direction of a shorter wavelength. From the point of the view of waves, the geometry of the space is described with a measuring stick of one wavelength. In that geometry, the waves or rays tend to move in a straight line. But from a static observer's view, that straight line steers downhill in the gravitational potential.
Unruh temperature of reflected waves in a gravitational field?
Open problem 1 above raises the question of how we can associate a temperature to acceleration, or to a gravitational field.
How can we associate a temperature to a specific value of acceleration? The Unruh temperature is one such way. Suppose that a static observer H is in a gravitational field and the gravitational acceleration is g at his location. A static source far away sends light waves toward H.
In the vicinity of H, the form of light waves of a very short wavelength is not much distorted by the gravitational field. Short length waves do experience blueshift when we move downwards, but since that happens under a great number of wavelengths, an individual wave can keep its form quite well.
If we have a very long wavelength wave, its electric field E will appear almost constant in the vicinity of H. We could say that the also a long wave keeps its form well in the vicinity of H, even though the waveform is very much distorted at longer distances. As an extreme case, we may have a static electric field E in the gravitational field. The electric field will have some constant value at each location. There is no reflection, as there is no wave oscillation.
If we have a medium wavelength wave, then its form is considerably distorted in the vicinity of H. We can associate a temperature with the "most distorted" wavelength. Is this the Unruh temperature for acceleration g?
If we have a rocket with acceleration g in flat Minkowski space, then the Unruh temperature tells us the wavelength of the light for which the acceleration, in some sense, distorts the waveform most.
Recall that the amount of "negative frequencies" in the Fourier decomposition of the wave determines how much Unruh radiation we are supposed to observe for a specific acceleration g.
We showed in previous blog posts that Unruh radiation most probably does not exist. What is the interpretation of the Unruh temperature then?
A static gravitational field certainly scatters or reflects back some of an incoming planar wave, because the waveform gets distorted and destructive interference cannot possibly cancel all reflected waves that are dictated by the Huygens principle.
Waves sent by a freely falling source in a gravitational field
Above we assumed that the source of waves is static and far away from the gravitating mass. Waves sent by the source appear as fixed frequency waves for each static observer in the field. For observers further down, the frequency appears higher. The observers observe a blueshift of the waves.
In the Unruh setup, we have an accelerating rocket and we study waves that are sent by an inertial source in the Minkowski space.
To mimic the Unruh setup under a gravitational field, the source of the waves actually should be freely falling in the field, and the rocket is replaced by a static observer. This is equivalent to having a freely falling laboratory where we have a source of waves, and an observer who is static relative to the gravitating mass. Inside the laboratory, the observer appears to accelerate upwards.
The source of waves in this case is like an accelerating ambulance coming downhill the gravitational potential. Its siren sounds as a chirp in the ears of the static observer.
How do such waves behave in the gravitational field?
Waves sent by an accelerating source in a Minkowski space
If we have a static observer in a gravitational field and he receives planar waves from a static source far away, that is analogous to having an accelerating rocket send planar waves to another accelerating rocket behind it. The observer sits in the second rocket.
Static observers in the gravitational field will see some of the waves reflect back. Similarly, observers in the accelerating rockets will see some waves starting to propagate to the direction of the acceleration.
Waves in a time-varying gravitational field
Open problem 3. Assume that we build a wave packet from purely positive frequencies and let it pass through a time-varying gravitational field. The waveform, as calculated by the canonical transformation (see http://meta-phys-thoughts.blogspot.fi/2018/04/does-unruh-radiation-exist.html, Definition 6) may afterwards contain negative frequencies. What is the role of these negative frquencies?
Can a time-varying gravitational field produce photons?
Our critique of Hawking radiation contains a conjecture that a time-varying gravitational field cannot produce photons if there is no ordinary matter present. Ordinary matter is coupled to the electromagnetic field, and can produce photons when the gravitational field pushes its particles around.
If a time-varying gravitational field is a gravitational wave, or it is produced by hypothetical dark matter, then photons cannot be produced, according to our conjecture.
This is at odds with the fact that a time-varying electromagnetic fields, or photons, create a time-varying gravitational field. If there is a coupling to that direction, why not to the other direction?
If we have a small random quantum fluctuation in the electromagnetic field, can a time-varying gravitational field make it to grow to real photons? Can the gravitational field give enough energy?
Let us study an analogous situation. The Dirac electron-positron field is coupled to the electromagnetic field. A time-varying Dirac field will always create a time-varying electromagnetic field.
In the Schwinger process, an electric field will create electron-positron pairs. That is, even a static electric field creates a permanent disturbance of the Dirac field, if the potential difference in the electric field is at least 1.022 MV.
In the coupling of the Dirac field and the electromagnetic field, there is an asymmetry: any small disturbance in the Dirac field always produces photons, but only a relatively large 1.022 MV potential difference in the electric field creates a permanent disturbance to the Dirac field. The quantum of the Dirac field has an energy 511 keV, while a photon can have any energy.
We have a conjecture that electromagnetic waves really are waves in the polarization of virtual electron-positron pairs. There is really just one field, the electron-positron field, whose temporary polarization manifests as photons, and permanent polarization as electrons and positrons. A big difference in the electric potential can turn a virtual electron-positron pair into a real one.
What about the gravitational field?
Conjecture 1. Gravitational waves really are polarization of virtual pairs, one having a positive mass-energy and the other having the opposite negative mass-energy.
Can a strong static gravitational field create pairs of photons?
Suppose that a quantum fluctuation creates a transient pair of wavelength b virtual photons. One photon hass mass-energy E and the other -E. The E photon travels down the gravitational potential. Let its wavelength get shorter by some ratio R. It gains energy from gravitational pull, so that it has energy R * E. If the photon did not have enough energy to be real at the start, it cannot gain enough energy to become real as it travels down.
The -E photon will travel up (?). It gains positive energy from the gravitational push. Its mass-energy will approach zero but can never become zero. Thus, creation of photons by this process seems to be prohibited by quantum field theory.
Can a static gravitational field create electron-positron pairs?
Suppose we have a virtual electron e- with mass-energy E and a positron e+ with mass-energy -E.
The electron will travel downward and gain momentum and mass-energy from the gravitational pull. If it gains the 511 keV, do we have then a real electron?
The positron has the same problem as the photon: its negative mass-energy will diminish, but can never reach zero. It cannot become a real particle.
Looks like a static gravitational field cannot create pairs of particles. The difference to the Schwinger process is that the mass-energy of the negative energy particle can never become positive.
If a time-varying gravitational field is a gravitational wave, or it is produced by hypothetical dark matter, then photons cannot be produced, according to our conjecture.
This is at odds with the fact that a time-varying electromagnetic fields, or photons, create a time-varying gravitational field. If there is a coupling to that direction, why not to the other direction?
If we have a small random quantum fluctuation in the electromagnetic field, can a time-varying gravitational field make it to grow to real photons? Can the gravitational field give enough energy?
Let us study an analogous situation. The Dirac electron-positron field is coupled to the electromagnetic field. A time-varying Dirac field will always create a time-varying electromagnetic field.
In the Schwinger process, an electric field will create electron-positron pairs. That is, even a static electric field creates a permanent disturbance of the Dirac field, if the potential difference in the electric field is at least 1.022 MV.
In the coupling of the Dirac field and the electromagnetic field, there is an asymmetry: any small disturbance in the Dirac field always produces photons, but only a relatively large 1.022 MV potential difference in the electric field creates a permanent disturbance to the Dirac field. The quantum of the Dirac field has an energy 511 keV, while a photon can have any energy.
We have a conjecture that electromagnetic waves really are waves in the polarization of virtual electron-positron pairs. There is really just one field, the electron-positron field, whose temporary polarization manifests as photons, and permanent polarization as electrons and positrons. A big difference in the electric potential can turn a virtual electron-positron pair into a real one.
What about the gravitational field?
Conjecture 1. Gravitational waves really are polarization of virtual pairs, one having a positive mass-energy and the other having the opposite negative mass-energy.
Can a strong static gravitational field create pairs of photons?
Suppose that a quantum fluctuation creates a transient pair of wavelength b virtual photons. One photon hass mass-energy E and the other -E. The E photon travels down the gravitational potential. Let its wavelength get shorter by some ratio R. It gains energy from gravitational pull, so that it has energy R * E. If the photon did not have enough energy to be real at the start, it cannot gain enough energy to become real as it travels down.
The -E photon will travel up (?). It gains positive energy from the gravitational push. Its mass-energy will approach zero but can never become zero. Thus, creation of photons by this process seems to be prohibited by quantum field theory.
Can a static gravitational field create electron-positron pairs?
Suppose we have a virtual electron e- with mass-energy E and a positron e+ with mass-energy -E.
The electron will travel downward and gain momentum and mass-energy from the gravitational pull. If it gains the 511 keV, do we have then a real electron?
The positron has the same problem as the photon: its negative mass-energy will diminish, but can never reach zero. It cannot become a real particle.
Looks like a static gravitational field cannot create pairs of particles. The difference to the Schwinger process is that the mass-energy of the negative energy particle can never become positive.
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