Quantum Entanglement and Teleportation Do Not Work the Way You Think

ABSTRACT:

Quantum entanglement can be explained using decks of cards. This paper shows how these decks of cards violate Bell's inequality, and that "hidden operators" determine a quantum measurement outcome. If these hidden operators could be known prior to measurement, one could predict the outcomes of quantum events. The herein thought experiment demonstrates that no faster-than-light communication takes place between entangled pairs. Further, when teleportation is scrutinized, there is no evidence that information passes through a wormhole.

According to John Stewart Bell, the following thought experiment should not violate his famous inequality: Start with two decks of cards joined together along the red-dotted line:

Because they are joined together, when the decks are randomly shuffled, they correlate (or anti-correlate). Now, divide the two decks along the same red-dotted line. Send one to Alice and the other to Bob. On the face of each card is a |0> or |1>. Neither Alice nor Bob know what card they will each deal from the top of their respective decks. It is as if the decks are in a state of superposition. However, if Alice deals a |0>, she knows which card Bob has dealt. Note there is no communication going on between the decks. They can be any distance apart and Alice will instantaneously know which card Bob has dealt.

Now, if Alice could just take a peek at the cards hidden in her deck before she deals them, she would be able to predict which card she will deal next with 100% certainty.

So, does this thought experiment demonstrate, by analogy, quantum entanglement? There is a test. Let's have Alice and Bob send each of their dealt cards to three polarizers: A, B and C. There is a chance each card will pass through one or more of the polarizers, or none at all. Alice and Bob keep track of the results and each create a data table matching the one below:

Since their datasets are identical, let's focus on Alice's. If a card passed through polarizer A, Alice records a 1, otherwise she records a 0. Same goes for polarizers B and C. At the right of the table, Alice checks if a card passed through A but not B (A!B). If so, she records a 1; otherwise, she records a 0. Then she checks if a card passed through B but not C (B!C) and records the result, and finally, she checks if a card passed through A but not C (A!C). She then adds up the columns. The final result is 2 + 2 is greater than or equal to 2. This is consistent with Bell's inequality:

Thus, according to Bell, the above thought experiment fails to accurately describe quantum entanglement. But what if Bell is wrong? Let's explore this possibility. Equation 2 below is another way to express Bell's inequality. It is expressed in terms of probabilities, sines and cosines, and, most importantly, it is a function of the polarizers' angles. Equations 3, 4, and 5 show that the data at the data table's first row would not be possible if the polarization angles were anything but zero or ninety degrees. This implies that the cards were oriented either vertically or horizontally (90 degrees or 0 degrees) and the polarizer slits were also vertical or horizontal (90 degrees or 0 degrees).

What if the polarization angles were not just zero or ninety degrees? Would Bell's inequality still be valid or would it be violated? Let's modify the above thought experiment. Suppose the cards are made of elastic material that enables them to squeeze through any opening; although, this is not guaranteed--there is a probability a card will squeeze through a polarizer that has an arbitrary polarization angle. Let's see if the cards can violate Bell's inequality if the polarization angles are altered.

Let's assume Alice's deck and Bob's deck are anti-correlated. If Alice deals a |0>, Bob deals a |1>. Below are the probabilities that their cards will pass through polarizers A, B and C:

On the left side of each equation above, the first term is the probablity that a card will pass through a given polarizer; the second term is the probability that a card will fail to pass through. Note that the probabilities add up to one as they should. Also note that the polarization angles are represented by theta-A, theta-B and theta-C. Using the above information, we can create the following version of Bell's inequality:

The first term is the probability of Alice's card passing through A times the probability of Bob's card not passing through B. The second term is the probability of Alice's card passing through B times the probability of Bob's card not passing through C. Finally, the term on the right of the inequality is the probability of Alice's card passing through A times the probability of Bob's card not passing through C. Using equation 6 above, we can calculate what polarization angles are needed to violate Bell's inequality:

At 9 above, we see that if theta-B is less than the term to its right, Bell's inequality will in fact be violated. Of course, theta-B is an arbitrary label. Let's swap it with theta-A:

At 11 above, we assign new angles to polarizers A, B and C: 0 degrees, 22.5 degrees and 45 degrees, respectively. When we plug these angles into inequality 6 and inequality 10, we find that our two decks of cards do in fact violate Bell's inequality. This means that the above thought experiment is an accurate analogy of the workings of quantum entanglement. It also implies that Einstein, Podolsky and Rosen (EPR) had an idea that may be valid after all: "hidden variables." This idea is not so far-fetched when you consider the fact that quantum mechanics has hidden operators. Here are a couple of examples:

The hidden operators are labeled in red. They are the NOT (X) and Identity (I) operators. When given a state of superposition, they secretly conspire with the Hadamard operator (H) to determine the outcome of a measurement. The result seems random to us, since we can't examine these hidden operators ahead of any measurement. Even if we could, we would know the order of the cards in the deck, so to speak, but we still wouldn't know how they got that way. We would be forced to fall back on, "They were randomly shuffled." Nevertheless, Alice could predict what her next card is going to be. She could even know ahead of time if the card will make it through a polarizer, since she might also know the "hidden probability operators."

Even if Alice can never know any hidden operators, she knows which card Bob dealt even if he is light-years away, and, the cards violate Bell's inequality. It appears the two decks of cards are truly entangled--is there a wormhole connecting them? If not, how does quantum teleportation work? There is a five-step teleportation process that apparently shows that Alice can instantaneously send a qubit to Bob, allegedly through a wormhole. Let's examine this process.

Alice wants to send Bob the following qubit (greek letter chi):

Teleportation step 1: Alice and Bob share an entangled pair of particles:

Find the tensor product of chi and the entangled pair:

Note that chi, Alice, and Bob's zeros and ones are colored black, blue and red, respectively. This enables us to track them. Now, we can use the table below to represent the original state of the system. The table has n units of ones and zeros in each column. The top portion has (alpha^2)n units, the bottom portion has (Beta^2)n units.

Teleportation step 2: Alice applies a CNOT gate:

Let's update the table:

Notice that Alice and Bob are no longer entangled! There is no guarantee that their ones and zeros will match. Technically this implies the wormhole is now closed or non-existent, since wormholes allegedly depend on entanglement. Yet, somehow Alice still manages to send the qubit to Bob? Let's continue. Step 3: Alice applies a Hadamard gate:

Time to update the table:

Step 4: Alice measures her pair (which includes chi's one or zero, and her own). Looking at the table above, notice that no changes were ever made to Bob's ones and zeros. Bob's final state is identical to his original state; yet, for example, when chi and Alice have a |00>, there is an alpha^2 probability Bob will have a |0> and a beta^2 probability Bob will have a |1>. Equation 22 confirms this. It's as if Alice really teleported the chi qubit to Bob, but Bob has no clue Alice tried to teleport anything. That's why step 5 is necessary. Alice must send Bob a no-faster-than-light signal through normal channels to let him know about her new disentangled measurements and what gates he must apply. But wouldn't it be faster if Alice uses the wormhole for step 5? What wormhole?

It should be reasonably obvious at this point that entanglement exists without the aid of wormholes or faster-than-light communication. Teleportation seems to work if both Bob and Alice assume Alice's ones and zeros have not changed; therefore, when they compare notes, it seems like Bob's ones and zeros take on new probabilities. This reminds me of relativity, where the train passes the station, or, is it the station that passes the train?

Acknowledgements:

Special thanks to ChatGPT4.

References:

1. Einstein, A., Poldolsky, B., Rosen, N. 03/25/1935. Can Quantum-Mechanical Description of Reality Be Considered Complete. Physical Review, Volume 47.

2. Leonard Susskind | "ER = EPR" or "What's Behind the Horizons of Black Holes?"". Archived from the original on 2021-12-11 – via www.youtube.com.

3. Einstein, A. Rosen, N. 07/01/1935. The Particle Problem in the Theory of General Relativity. Physical Review. 48.

4. McMahon, David. 2008. Quantum Computing Explained. John Wiley and Sons Inc.

5. Fadelli, Ingrid. 09/21/2023. A model probing the connection between entangled particles and wormholes in general relativity. Phys.org

6. Bell's Theorem. Wikipedia, ChatGPT4, and various online sources.

Why Black Holes Don't Need Singularities

ABSTRACT:

According to R.P. Kerr, a black hole need not contain a singularity. Such an assertion prompted this author to explore alternate black-hole models where a singularity is unnecessary. The models presented here show that a sufficiently large black hole could contain a universe; that an average black hole could contain a neutron star. This paper also shows why micro-black holes are untenable and why a typical black hole has a mass of at least approximately three solar masses. As an additional bonus, the models presented conserve quantum information inside the black holes, and, are consistent with black-hole entropy and temperature.

Alice lives in a universe similar to ours. It has an isotropic mass density--its total mass is mostly proportionate to its total volume as well as its Schwarzschild radius. Her universe also has a cosmological horizon which limits how far out Alice can make observations. Her universe is expanding and she observes galaxies receding away from her and they eventually disappear when they cross the cosmological horizon. Below is the relevant mathematics that describe her universe:

As her universe's volume grows, its mass grows proportionately. Equation 4 above shows how the universe's radius r grows as mass m increases. Equation 5 does likewise for the universe's Schwarzschild radius. However, note that the Schwarzschild radius grows exponentially faster than r. At some point in time, the Schwarzschild radius will be greater than her universe's radius. The diagram below assumes such a point in time:

The inner red circle represents Alice's cosmological horizon. The blue circle represents the physical extent of her universe. The black circle is the extent of the Schwarzschild radius. Beyond the black circle lives Bob. When Bob looks at Alice's universe, he sees a black hole. It has gravity consistent with equation 6 above. Alice, on the other hand, believes she lives in a fairly normal expanding universe. Its expansion velocity v is consistent with equation 7.

Bob believes that the black hole has a singularity where there is infinite gravity and infinite mass density. It is where the laws of physics break down. He knows this is true because he watches the Discovery Channel. By contrast, R. P. Kerr (see reference below) suggests that black holes can exist without singularities. He also argues that no one has conclusively proven that black holes must have singularities. I will argue that the above thought experiment, involving Alice and Bob, supports his assertions.

Caveat: The above thought experiment may only prove that extremely large black holes can exist without a singularity. What about smaller black holes? According to the Hawking-Penrose theorem, geodesic paths must meet at a single point. The theorem is based on General Relativity. However, the General Relativity theory preceded quantum mechanics, so its solutions don't take into account quantum mechanics. So perhaps the same can be said about the Hawking-Penrose theorem:

Equation 8 above demonstrates that infinite energy (E) is required to have a singularity where distance x equals zero. Since black holes don't generally have infinite energy, we can infer that the stuff inside must take up some space. The question is ... how much space?

If energy E at equation 8 equals the Planck energy, then x equals the Planck length, but then so does the Schwarzschild radius. To have a black hole we need energy that is equal to or greater than the Planck energy. It seems if the particles invovled are squeezed into the smallest space, x will be less than or equal to the Planck length. Unfortunately, the particles can't be fermions, since fermions must comply with the Pauli Exclusion Principle -- two or more of the same type of fermion can't occupy the same position in space. A single fermion has a physical extent, according to equation 8, where x easily exceeds the Planck length. Multiple fermions take up even more space.

Unlike fermions, bosons don't have to comply with the Pauli Exclusion Principle. It is possible for multiple photons, for example, to occupy the same position in space. However, if a lot of photon energy is concentrated in such a small space they will produce fermion pairs which must take up more space thanks to the Exclusion Principle. If the goal is to pack the most particles in the smallest space possible, gluons seem like the best option. The strong force is weaker when they are closer together, so there is less likelyhood for fermion-pair production. Unfortunately, gluons are hopelessly wedded to quarks due to confinement. Separating gluons from quarks increases the strong force attraction between them (and increases quark-antiquark pairs) which makes the possibility of such separation untenable. Thus, if we want to create a black hole, we are essentially stuck with space-hogging fermions, so it is unlikely we will be creating or observing micro black holes.

The mass of a typical natural black hole is a minimum of around three solar masses or approximately 6E30 kilograms. This is not surprising when you take into account the previous paragraph. The stuff inside a black hole must consist of fermions which take up a lot of space (unless the Pauli Exclusion Principle is violated).

If we use equations 4 and 5 above and use the mass density of, say, a neutron's mass density in place of the universe's mass density, and have m represent mass in general, we find that when we plug into m at least 6E30 kilograms, we get a radius r that is less than or equal to the Schwarzschild radius.

Why use a neutron's mass density? Because the core of a collapsing star has a plasma consisting of electrons and protons that undergo electron capture:

The above Feynman diagram shows how each electron and proton yields a neutron and neutrino. The neutrinos escape, so that leaves the neutrons behind to form a neutron star. The star's neutrons have neutron degeneracy pressure which arises from the Pauli Exclusion Principle. This pressure resists further collapse.

Now, imagine, if you will, a neutron star with a mass that exceeds three solar masses. Its radius is less than its Schwarzschild radius. To an outside observer named Bob, it is indistinguishable from a black hole. To an inside observer named Alice, it is just a neutron star, nothing exotic or mysterious. If we do an accounting of trapped and escaped particles, etc., quantum information is conserved.

One can calculate its temperature and entropy in the usual manner, since neither of these is a function of how much space the stuff inside occupies. The surface area is a function of mass or the Schwarzschild radius, not the radius r--so r doesn't have to be a singularity or zero distance.

References:

1. Kerr, R.P.. 12/05/2023. Do Black Holes Have Singularities? University of Canterbury.

2. Hawking, S.W., Penrose, R. 04/30/1969. The Singularities of Gravitational Collapse and Cosmology.

3. Pecina-Cruz, Jose, N. On the Collapse of Neutron Stars.

Hawking Radiation Doesn't Work the Way You Think

ABSTRACT:

Hawking radiation has not been directly observed. Maybe it exists, maybe it doesn't. Even it it exists, some very fundamental physical laws prevent black hole evaporation. At the quantum level, Hawking radiation can be turned on its head. It could just as easily add mass to a black hole.

Hawking radiation has not been observed for a very good reason: it does not work the way you think. The hypothesis seems sound at first blush: Two particles pop into existence. One outside the black-hole horizon, and the other trapped inside the black hole. The outside particle escapes, and, can be deemed positive energy, since it adds energy to the outside universe. The trapped particle can be deemed negative energy since, according to Hawking, will reduce the black hole's mass. Note that energy is always conserved, since positive energy and negative energy mathematically cancel each other. The bottom line is black holes allegedly evaporate due to Hawking radiation. I say "allegedly" because there is more to the story. The Hawking-radiation hypothesis is incomplete. Let's do a more complete thought experiment and see what happens.

Imagine a star that is virtually all matter, with next to no anti-matter. The star collapes into a black hole. Its mass is still composed of virtually all matter. The stuff that falls into this black hole is virtually all matter. In the black-hole diagram below, we represent a particle of this matter with the Greek letter mu preceded by a plus sign. Ellipses before and after the plus mu's represent multiple particles that may have been crushed into a singularity.

The above diagram represents the starting mass and the state prior to the appearance of a particle (plus mu) and an antiparticle (minus mu). The next state below is where a particle-antiparticle pair appears. The antiparticle escapes the black hole's gravity, but the particle is trapped.

In the next diagram the trapped particle has no anti-particle to interact with, so it adds mass to the black hole! This particle can be deemed the positive energy. The escaped anti-particle is then deemed negative energy and has the potential to interact with any particle it encounters. Such interaction will reduce the mass of the universe outside the black hole.

Of course there is only a 50% probability the black hole will gain mass and the remaining universe will lose mass. Below we see that there is a 50% probability that the particle will escape and the antiparticle is trapped.

The antiparticle has no problem finding particles to interact with:

The black hole loses the mass it previously gained:

The escaped particle and the escaped anti-particle may annihilate each other, or, if they are too far apart, will interact with other particles and antiparticles.

The space outside the black hole returns to nothing and the black hole returns to its starting mass:

The above thought experiment can also be performed with anti-matter black holes. The main problem with Hawking's hypothesis is it has the following implicit assumption: That all black holes have fairly equal amounts of matter and anti-matter. One might ponder whether a star that precedes a black hole can have fairly equal amounts of matter and anti-matter and still exist. Assuming the answer is a resounding no, then black holes don't evaporate via Hawking radiation.

For the sake of argument, let's assume Hawking was right. There is Hawking radiation and it causes black holes to evaporate. Why should black holes have all the fun? Imagine a particle-antiparticle pair appearing above the earth's surface. One escapes earth's gravity, the other does not. When they first appeared, they each had velocity v which is less than light speed. Velocity v was an escape velocity for one but not the other--the other being too close to earth's center of mass. If Hawking was right, the trapped particle should reduce the earth's mass. Over time the earth will completely evaporate. Thus, if Hawking was correct, all planets, stars, etc. should evaporate. The counter-argument is no such evaporation has been observed.

The equations below further demonstrate why black holes, in particular, refuse to evaporate:

Since light can't escape a black hole, a black hole's emissivity is zero. Even if it has an emissivity of one, power (P) according to the Stefan-Boltzmann equation above, is less than zero. This implies there is more radiation entering a black hole than randiation escaping. The minimum mass required to make a black hole is approximately three solar masses. So much mass causes the black hole's temperature to be less than its surrounding environment: deep space. The second law of thermodynamics would be violated if the net thermal transfer favors black-hole evaporation. Black hole entropy increases when a black hole's mass increases:

On the flip side, a compelling argument in favor of black-hole evaporation is the following thought experiment: Imagine a photon-antiphoton pair. Photons and antiphotons are indistinguishable from each other. So if the photon is trapped, it could behave like an antiphoton and annihilate matter inside the black hole, reducing the black-hole's mass. However, the escaped anti-photon can also behave like an antiphoton. If the escaped antiphoton finds another photon first, it becomes the negative-energy particle and reduces the energy of the universe outside the black hole. The trapped photon (or antiphoton) will add energy or mass to the black hole.

Photon-antiphoton Hawking radiation is more likely to cause the black-hole to lose mass if the black-hole's surrounding environment is empty space with a lower temperature. Albeit, this is an ideal and unrealistic condition. The cosmic microwave background raises the temperature of the surrounding environment to approximately 2.73 Kelvin, well above the temperature of the typical black hole. Thus, the following scenario is consistent with equations 1 through 6 above: The escaped photon is more likely to find another photon to interact with. When it does, the two photons vanish. The trapped photon adds mass to the black hole.

References:

1. Hossenfelder, Sabine (23 August 2019). "How do black holes destroy information and why is that a problem?". Back ReAction. Retrieved 23 November 2019.

2. Hawking, Stephen (1 August 1975). "Particle Creation by Black Holes" (PDF). Commun. Math. Phys. 43 (3): 199–220.

3. Susskind, Leonard (2008-07-07). The Black Hole War: My Battle with Stephen Hawking to Make the World Safe for Quantum Mechanics. Little, Brown and Company.

4. Black hole information paradox. Wikipedia.

5. Mathur, Samir D. 03/21/2021. The Elastic Vacuum. Gravity Research Foundation.

6. Chaisson, Eric. Astronomy Today. Englewood, NJ: Prentice Hall, 1993: 503

7. Severino, Paul. THe Black Hole Information Paradox: A Quantum Information Perspective. 03/24/2020

8. Wilkins, Alex. Ilands Poking Out of Black Holes May Solve the Information Paradox. 01/11/2024. UC Berkeley Physics.

Using Quantum Physics to Find the Best Model for Gravity, Gravitational Waves, and the Vacuum

ABSTRACT: According to Einstein's theories of relativity, nothing is faster than light; yet, observations made by Newton, Laplace and Van Flandern led them to believe gravitational information is much much faster than light, virtually instantaneous. To thicken the plot further, LIGO observed gravitational waves propagating within the light-speed limit. Then there's the vacuum energy problem where the vacuum seems to have up to infinite energy! By making use of quantum physics and changing an initial assumption about the vacuum energy, it is possible to connect the dots between quantum physics and General Relativity. By exposing a fundamental flaw in the rubber-sheet model for curved spacetime, it is possible to create a superior model that reconciles the speed of gravitational waves with the illusion of faster-than-light gravitational information.

Quantum physics is probabilistic as opposed to deterministic. Given a vacuum that is composed of numerous (omega) energies it only makes sense to multiply each energy by a probability. The alternative is to simply add all the energies and get up to infinity! Assuming each energy has a wave function, each probability can be determined by squaring each wave function. Each energy is represented by the Hubble energy (Hubble's parameter * Planck's constant) multiplied by n. At equation 1 below, we calculate the vacuum mass density (rho). At equation 2 we determine the cosmological constant:

Now, let's introduce a particle with mass m (or it could be massless: m = E/c^2). It could be located anywhere and everywhere. Its location (x,y,z coordinates) is uncertain at best. We could add m multiple times to cover all its possible locations, but that would lead to a rediculously big number. Or, as we did with the vacuum, we could multiply each m location by a probability (see equations 3 and 4) and that will give us the expectation value for m which is really just m. Thus, multiple m's at multiple locations don't amount to more than just m. The velocity of m is also uncertain, but we can calculate its expected value at equation 5. At 6 we determine the expectation value for the wavelength of m.

At equation 7 we see how multiple locations of m impact its gravity:

The gravity of m is also at multiple locations along with m, but probabilties cut it all down to size. The size being the left side of equation 7: the gravity of the expectation value: m.

At equation 8 below, we redefine m as one or more particles (expected values). At 9 and 10, we calculate the final velocity and the final wavelength, respectively.

At 10 we have the De Broglie wavelength formula. At 11 we below can see that if mass m or final velocity v changes, the final wavelength lambda must instantaneously change to keep Planck's constant a constant. After a few algebraic steps, we derive the Schwarzshild radius at 16.

Note that any change of mass m causes an instantaneous change of its wavelength (lambda prime). This means that when a black hole's singularity mass changes, its Schwarzschild radius instantaneously changes! Equation 17 below confirms this. The light-speed constant c on the left side is not a constant unless mass m and Schwarzschild radius changes are synchronized. Thus, we don't have to wonder how the singularity sends information out as far as the Schwarzschild radius, assuming such information is limited to light speed. It doesn't need to. Every mass simply has, and is defined by, a corresponding wavelength and Schwarzschild radius.

But wait! It gets even better! The diagram below shows the total radius r (in black), the Schwarzschild radius (in red) and the remaining distance (in blue). Let's assume the circle below contains the volume (V) of the entire universe (or any volume you like). If mass m owns the volume as far out as the Schwarzschild radius, then the vacuum's claim along the total distance r is reduced. The rest of the universe owns only the volume along the remaining distance.

One can infer that the remaining distance (in blue) must change instantaneously in response to a change in the Schwarzschild radius (in red) which in turn changes instantaneously to a change in the average wavelength which responds instantaneously to a change in mass m.

Equation 18 shows that proper time, out to radius r, is also reduced and is proportional to the square root of the remaining distance over r. The reader may recognize a variation of the Lorentz factor on the left side. One may also infer that proper time is reduced instantaneously given its dependency on the remaining distance.

Using the diagram above, we can set up equation 19 below. From there we can navigate to Einstein's field equations at 21. At 22, we can further verify the instantaneous relationship between matter, spacetime, and gravity. Again, the constant c is not constant unless spacetime curvature responds instantaneously to a change in the stress-energy tensor.

So far we have shown how matter interacts with spacetime. We should now take a look at how vacuum mass interacts with spacetime. All masses project a Schwarzshild radius regardless of how concentrated or diffuse they are. In each diagram below, the gray area represents the mass concentration. Note that each diagram has a mass of m and the same Schwarzschild radius.

Vacuum mass, like the star and black hole, also curves spacetime. The cosmological constant is that spacetime curvature caused by vacuum mass density. We add this to the field equations:

So a question arises: Why does vacuum mass density and its curvature cause the universe to expand? Below, term A implies term B, and, term C implies term D. Term B shows how matter (E) accelerates. When distance r is larger, the rate of acceleration is less. When r is smaller, the acceleration rate increases. This is because matter (E) is fixed. Contrast this with term D. At term D, the opposite happens: Acceleration increases as r increases and vice versa because vacuum mass is not fixed. It is proportional to volume. Both B and D contribute to spacetime curvature, but are seemingly opposite forces.

At equation 25 we set a scalar version of the Einstein tensor equal to the sum of the curvature caused by mass m and the curavature caused by vacuum mass. From there we derive equations 30 and 31 which show the tug of war between an expanding universe and gravity.

We can simulate the net acceleration rate at equation 31 with an accelerating rocket (see diagram A below). If we throw a shot put across, it will appear to fall along a curved path. Throwing the shot put gives it kinetic energy which may be lost and converted to gravitational waves (GWs, purple curved lines). At diagram B we have a fixed container with a magnetic field. Equations 31 and 32 demonstrate the time-delay difference between gravity and electromagnetism. Imagine t1 is the time it takes to initiate the rocket engine and the electromagnet. The magnetic field takes an additional r/c seconds to develop, since photons must propagate from the floor of B to the shot put (distance r) at speed c. After t1 + r/c seconds have passed, the shot put falls and takes time t2 to hit the floor. By contrast, the shot put at A immediately falls after t1 seconds.

Thus gravity's total time t is t1 + t2 seconds (equation 32). Electromagnetism's total time t is t1 + r/c + t2 (equation 33). Also note that the gravitational waves (GWs) do not cause the shot put to fall, but rather, it is the shot put's lost kinetic energy that causes the GWs.

Observations confirm that the GW strain (h) is consistent with 35 below and not 34. At 34 we have the curvature of the complete energy of the source; whereas, at 35, we just have the curvature of kinetic energy of the source. Notice at 36 and 37 electric and magnetic waves are proportionate and correlate with their respective fields. By contrast, GWs do not correlate with the full gravitational field.

The accelerating-rocket thought experiment above seems like a good approximation of gravity and gravitational waves; however, it seems to contradict the famous rubber-sheet model of gravity. Imagine placing the shot put on a rubber sheet. The shot put will depress the rubber sheet. Such depression, however, does not happen instantaneously. The depression curve takes time to form. As it's forming, one can imagine GWs propagating outward from the center of mass. But what happens if the shot put is moving fast (v > 0) and not at rest (v = 0)? One can imagine it not depressing the rubber sheet:

Imagine the earth is covered with a rubber sheet and the shot put has enough velocity v to orbit. It won't fall towards earth's center, so it won't depress the rubber sheet. We model this fact with equation 39:

One might erroneously conclude that if a mass moves fast enough, it won't curve spacetime! This could not be further from the truth. The truth is velocity enhances the curvature of space time:

Thus the rubber-sheet model has a fundamental flaw. The accelerating-rocket model is superior. It creates the illusion that gravity's speed is faster than light. This is consistent with observations made by Newton, Laplace and Van Flandern. I say "illusion" because nothing in the accelerating-rocket thought experiment exceeds the speed of light.

Given all the forgoing information, we can set up a timeline model for curved spacetime and gravitational waves (GW):

Equations 41 through 44 take into account the instantaneous interplay between matter and spacetime along with gravitational waves that don't exceed the light-speed limit. Equations 45 through 47 below confirm that 41 through 44 are correct; otherwise, the constant c would not be constant if the curvature of spacetime had to wait for gravitational waves or gravitons to propagate.

Acknowledgements:

Amber Strunk. Education and Outreach Lead. LIGO Hanford Observatory.

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