Unifying Gravity and EM

[sweetser]

4-momentum charge

Hello:

In my research, I focus on equations, not the words to describe said equations. The reason is that I work intermittently in isolation. There is not a small group of folks working full time on rank 1 unified field theories, so there is no one to chat with about the work.

By making videos for the Stand-Up Physicist - the educational arm of my research efforts - I do have to think about word choices. In creating a show about 4-potentials, I had to give $J^{\mu}_{m}$ a name. I think of this as the same darn thing as the electric current density $J^{\mu}_{q}$ except that the electric charge $q$ has been replaced by a mass charge $\sqrt{G} m$. I have been calling it the mass charge density by reason of analogy.

No one works with a mass charge density, so maybe that is a poor choice of words. The electric charge density is the electric charge in motion. A mass in motion is a 4-momentum. I should call $J^{\mu}_{m}$ the 4-momentum charge density. I include the word "charge" because the units for this tensor are not the same as a 4-momentum. Units matter. The appearance of $G$ indicates the expression is connected to gravity. There is also a $c$, which implies the expression is relativistic.

There might be a deeper message with the shift to "momentum charge". In the current way of thinking, energy-momentum is the most important characteristic of a system. The link to gravity is indirect. Energy-momentum gets added into the stress-energy tensor that curves spacetime according to general relativity. Treated instead as a charge, 4-momentum charge is directly connected to gravity and inertia. A momentum charge is always going to effect what happens with gravity, which is how it should be.

I am off to buy groceries, but I will be playing with the idea that all 4-momentums should be rewritten as 4-momentum charges. It is easy enough to change the units, but it is the implications that are fun to consider while shopping.

doug
TheStandUpPhysicst.com

 

[sweetser]

Hello:

I did get the groceries. Here was the thought of the day I had while trying to choose matching potatoes.

Newton's force law, $\vec{F}=m \vec{A}$, has no need for universal constants, $c, G, h$. It remains a very useful law today for doing work in mechanical engineering and architecture. The enormous advance in the study of natural philosophy inspired directly from Newton was the use of units to quantify observation of Nature. Energy would never be momentum, because the first had units of $M L^{2}/T^{2}$, while the second had units of $M L/T$.

With the advent of special relativity, the line between energy and momentum blurred, because by changing inertial reference frames, the amount of energy and momentum can also change so long as the mass of the system remains invariant. Energy and momentum now have a relationship with each other.

The separation between gravity and EM is clear from the GEM Lagrangian: gravity lives in the irreducible symmetric rank 2 tensor, and EM lives in the irreducible antisymmetric field strength tensor. With GEM, one can calculate directly how much the gravitational attraction takes away from electrical repulsion with $J^{\mu}_{q} - J^{\mu}_{m}$. Using universal constants to match units is a basic step in unifying classically separate forces of Nature.

doug

 

[Norman Albers]

I just uncorked a two-page study where I equate the Schwarzschild metric terms to a dielectic permittivity. This says we can look at GR as a dielectric 'thickening' in a Euclidean space! Very simple, but works beautifully! I don't mean to interrupt, but this is so germain to what you are doing, and I need your help.

 

[sweetser]

Hello Norman:
There is no way this could be germain to the topic at hand: the GEM model makes predictions that are different from GR, and a rank 2 theory is very different from a rank 1 theory. You might want to try and start your own thread at Independent Research Forums, so long as your work meets the criteria.
doug

 

[Don J]

Hello:

In my research, I focus on equations, not the words to describe said equations. The reason is that I work intermittently in isolation. There is not a small group of folks working full time on rank 1 unified field theories, so there is no one to chat with about the work.

Can you explain for us poor laymans and -relativistic-(Einstein sense of the word)what is the ULTIMATE goal of your theory?
Do you think your theory may ultimately replace GR?
For now your theory seem only a "curiosity".
My post is not intended to be harsh!

 

[sweetser]

Hello Don:

Newton's theory of gravity is a good theory for how gravity works. It is still in use today. We know it has a technical flaw, that a change in a mas density must change everything instantaneously.

GR is a better theory, one that supersedes Newton's theory. If one starts from Newton's law and sets out to make it consistent with special relativity, then one ends up with Einstein's field equations. The vacuum field equations are non-linear, something that happens with the weak and the strong force, but not EM. All efforts to quantize GR proper have failed.

The GEM field theory tries to be better than GR. Like GR, one can be certain the theory gets along fine with special relativity, or in fancy works, the field equations are manifestly Lorentz covariant (meaning the form of the equations is not going to change by changing inertial observers). The field equations can be quantized because it is done in graduate-level quantum field theory books on EM field theory (two of the modes for a spin 1 field must be made virtual, but I propose those modes are a very real spin 2 field doing the work of gravity).

If the GEM model is correct, it will supersede GR because it can be quantized and it naturally unifies gravity and EM.

Nature it this harsh one, in that is unreasonable to expect in a lifetime to find a theory consistent with current tests, different from more stringent tests, yet is mathematically sound. To me, the GEM model is as curious as one can ever hope to find.

doug

 

[Rade]

To Sweetser: I have a question from another thread where I received no help. What I look for are the equations that would predict the gravity force between the three quarks (uud) within the proton, then for the anti-proton (u-bar, u-bar, d-bar). Can your GEM model help me here ? I understand this force is not thought to be important--but I care not, I just look for the equations. Plus, do we really know gravity at scale less than 10^-15 meter, perhaps it increases as we move to scale of quark interactions ? I would then use these equations to derive gravity force equations for neutron and anti-neutron, then for higher order isotope interactions, both matter and antimatter (deuteron, helium-3, triton, etc).

 

[sweetser]

Hello Rade:

The short answer is I do not know how to do any calulations with the strong force.

For classical gravity and EM, the source is the electrical charge density minus the 4-momentum charge density. On the other side of the equation, two covariant derivatives act on a 4-potential. The solution can either involve a dynamic metric, a dynamic potential, or some combination of the two. I have done calculations of the field equations for an electron and a neutron, using either the potential exclusively or the neutron exclusive. Those calculations form my proof that GEM has Diff(M) symmetry because I use the same manifold, but different metrics (one Minkowski, one the exponential metric at the start of this forum).

As far as I know, there is no classical way to express the strong force. This is a warning flag for me due to the correspondence principle, which I think should always work both ways. Warning: this warning flag is not many other people's warning flag :-)

The question looks like it must be addressed using quantum mechanics. I have a clear picture of how to do a scattering calculation for an electron. This is a nonlinear equation where one uses Feynman diagrams to make approximations because EM is weak, and gravity is far weaker still. I don't know what the current record is for Feynman diagrams (8?). That calculation required a supercomputer because there are so many permutations. Adding another vertex becomes even harder.

Cool! I just had an idea. The contributions to a scattering calculation depend on the coupling constants. Pretend we have a beam of electrons being fired at a proton. For the Feynman diagram with one vertex, the coupling constant would be
$$\alpha = \frac{\mu_0 c e^2}{2 h} = 4 \pi 10^{- 6} NA^{- 2} 3 \times 10^8 ms^{- 1} 1.6 \times 10^{- 19} / (2 6.63 \times 10^{- 34} Js) = 0.00727$$
(note: I've spent the last hour trying to spot my tex error, but have failed, sorry!)
Now consider the Feynman diagrams with two vertices. The coupling would be
$\alpha^{2}$. There is a simple pattern here: for the number $n$ vertices, the coupling constant is $\alpha^{n}$.

What is the gravitational coupling? With EM, there is one size for electrical charge. For gravity, there are many values for the gravitational charge, so many gravitational alphas. For a proton and electron interaction, the coupling would be:
$$\alpha_{GEM, p - e} = \frac{\mu_0 c G m_p m_e}{2 h} = 4 \pi 10^{- 6} NA^{- 2} 3 \times 10^8 ms^{- 1} 6.67 \times 10^{- 11} m^3 kg s^{- 2} 1.67 \times 10^{- 27} kg 9.11 \times 10^{- 31} kg / (2 6.63 \times 10^{- 34} Js) = 2.55 \times 10^{- 32}$$

Now we need to calculate the number of vertices $n$, such that this gravitational effect will equal that of EM.
$$\alpha^n = \alpha_{\tmop{GEM}, p - e}$$
$$n \log \alpha = \log \alpha \text{_{GEM,p-e}}$$
$$n = \frac{\log \alpha \text{_{GEM,p-e}}}{\log \alpha} = 14.8$$
According to GEM, quantum gravity effects will come into play for Feynman scattering calculations where one has 15 vertices. Even with all the steady improvement in computers, I doubt we will be able to make the calculations in a hundred years time.

***
Sorry for the digression, but it is rare when I start a morning with a calculation. GEM does not provide a reason why gravity is so weak relative to EM or the strong force, the subject of your comment. The difference is so vast they should be treated separately.

doug

 

[Rade]

...According to GEM, quantum gravity effects will come into play for Feynman scattering calculations where one has 15 vertices. Even with all the steady improvement in computers, I doubt we will be able to make the calculations in a hundred years time.

Thank you for your time with my question. Now, your very interesting comment above about quantum gravity effects emerging when there are 15 interacting vertices leads me to another question--how to get 15 quarks interacting to form 15 vertices via quantum gravity to form a stable system--perhaps with added dimensions not now considered. Thus, suppose a thought experiment where we describe the quarks within nucleons to interact (via quantum gravity) in "bags", that is, the proton has 1 bag with 3 quarks, so the neutron. Now, we define a matter Helium-3 isotope, which is macroscopically [PNP], as microscopically 9 matter quarks in 3 bags interacting. Then we attempt to couple this matter isotope [NPN] with an antimatter deuterium isotope [N-bar, P-bar], which has 6 antiquarks in 2 antibags. As you can see, the interaction of this type of asymmetrical matter-antimatter bags of quarks results in a possible quantum gravity solution in which there are a total of 15 interacting quarks, 9 matter and 6 antimatter (can we call these vertices vis-a-vis Feynman??), which should then allow for a mathematical expression using your GEM model ? Taken to a logical (but perhaps not realistic) conclusion, the above would suggest that interactions of 2 and 3 nucleon isotopes (at the level of quarks), both matter and antimatter, form the "fundamental" (lowest energy) physical system in which quantum gravity effects can be experimentally documented, perhaps by your GEM model. Does any of this make any sense to you ? Thanks for your time.

 

[Lawrence B. Crowell]

The 't Hoft Veltman running parameter

g^2(q) = g^2( mu)[1 + (g^2(mu)/12\pi)log(-q^2/mu^2)(2n_f - 11N)]

can handle these problems as it is. For high enough transverse momenta one can do QCD perfectly well. I am not sure how GEM fits into this matter.

The real outstanding question is whether this will address the gauge heirarchy problem. The GEM theory, with all its symmetrical terms and the like is likely to have problems within any attempted exact renormalization group scheme. Without calculation I can see it is likely that GEM will produce all sorts of renormalization flows which are not gauge invariant (eg will involve gauge connections), which will make renormalization problematic.

On my part with Information Preservation in Quantum Gravity I will be soon making an installment which involves regularization schemes in Ricci flows and foliations of three spaces in cosmologies.

Lawrence B. Crowell

 

[sweetser]

Hello Rade:

When I say I understand something in physics, it means I know how to relate it directly to a physics equation or calculation. In my educational efforts at TheStandUpPhysicist.com, I have made videos which have too much math to reach a wide audience, but for me, physics is the equations so they must always be directly referenced. When I claim that gravity and EM are unified, I really am referring to Jq-Jm=Box^2 A, where the Box is a covariant 4-derivative, a standard derivative minus a connection. For me, the link must be that tight.

Let me now detail what I understand about the calculation I made. I took a year of graduate-level quantum mechanics over a decade ago, so key details are no longer with me. When a photon interacts with an electron, the math to characterize it is non-linear. This makes finding a solution darn tricky. Feynman diagrams are a visual aid to making a calculation. For a low energy system, just one or two vertices are needed to get a great approximation. One can add more terms, but due to a growing number of permutations, it becomes necessary to program the process. If one goes out to as many as eight vertices, a supercomputer struggles with the volume of math passing through silicon.

What do these calculations do? Usually it has to do with calculating scattering: one fires an electron at a proton, the electron scatters, and the odds of it scattering say at 14 degrees off the center line is shown by the calculation. The calculation I did was suppose to show that no matter what happens in our lifetime, or our children's lifetime, or their children's lifetime, no one will be able to see how much the EM scattering is changed by gravity scattering. We understand the size of the calculation, and why it is beyond reach.

What goes into the source term of the GEM field equations, Jq-Jm? Anything with an electric current (Jq) or a mass current (Jm). The mass of an electron is equal to the mass of a positron, its anti-particle partner, so the total contribution to the Jm term of an electron and a positron is 2*511 MeV times the 4-velocity. The same goes for quarks, antiquarks, leptons, anti-leptons. All 4-momentum charges go into Jm, no exceptions granted.

The calculation I did was to compare gravity to EM. The number that came out of that process was 14.8, kind of close to 15. At this time, I have no sense - no calculation - which explains why the difference should be as big as it is. It is an observation. It looks to me like you took the 15, and did an exercise in numerology - where combination adds up to 15? The actual number is 14.8, which makes any connection to 15 not relevant. Plus I see zero connection between the calculation I did (relative forces of gravity and EM) and quarks/antiquarks.

doug

 

[sweetser]

Why? Gravity is the least one can do

Hello Don:

Here is a big advantage of the GEM approach a layman can appreciate: the ontology (a fancy why of saying the "why" of gravity and EM). Let's first look at the why behind Newton's law of gravity and general relativity.

There is no "why" for Newton's law of gravity. That law was constructed and shown to be relevant by connection to Kepler's three laws of planetary motion. In theory, gravity could have been a 1/R or 1/R^3 law with the correct adjustment of G (L^2/M T^2 or L^4/M T^2 respectively). An inverse square law has some nice mathematical properties, but that is not a reason why the law should be the product of two masses over the square of the distance.

The situation for general relativity is no better. The best know justification for Einstein's field equations is the following from John Archibald Wheeler:

Spacetime tells mass how to move and mass tells spacetime how to curve.

There is some debate about whether this is an example of circular reasoning. I don't see how or why mass and spacetime should have this sort of dialog, so I am not happy with this perspective.

With GEM, start with an empty Universe. The distance between a pair of events in such a barren place would be governed by the utterly flat Minkowski metric. Why that metric? Calculating a distance is an act of multiplication, and multiplication of a 4-vector as a numerical field is governed by quaternions. Most laymen have never heard of these 4D numbers, and in my informal survey, only about half of physicists recognize the name (even fewer have worked with them). They are the 4D version of 2D complex numbers, which is how the minus sign shows up between time squared and space squared.

Now introduce a mass. The world is not very different. The question is, how does one implement the small amount the Universe has changed because of the mass? No, better, how does one implement the smallest possible way the Universe has changed because of the mass? For systems where one thing barely effects another, there is a universal answer: a simple harmonic oscillator. A less fancy way to say it would be to reference a slinky: let a slinky hang down, give it the smallest of nudges, and it will wobble for a good while. The weaker the connection, the smaller the wobble, but it also lasts longer because not much energy is involved.

Spacetime is nearly flat. It is not perfectly flat, so the smallest step away from perfectly flat involves a 4D slinky. The 4D wave equation, $J_q-J_m=\Box^2 A$ can describe a 4D simple harmonic oscillator. Just because it has the word "simple" in the label does not make it easy to understand. Lots of issues still have me mystified. But the point is to find the math that is all about almost doing nothing. That is exactly what the 4D wave equation with its simple harmonic oscillator give you. The Earth is wobbling around the Sun, been doing that for 4 billion years, and it doesn't take a lot of energy, but it does require the presence of an enormous amount of mass ($2\times10^{30} kg$). Right now, me, and everything around me is trying to wobble around the center of the Earth. Because we get in each other's way, we are in a stalemate. If I ever get a path clear from all the clutter in my way, like at the top of a diving board, I begin my wobble through the center of the Earth to my antipode and back, a trip that would last about an hour and a half. Of course my freedom lasts mere seconds, and back into the traffic jam of mass charges I go.

If you do research, you know the edge of where you do not know. I still wonder how it is I collect the information about the $6\times10^{24}kg$ that makes up the Earth. It is the implementation details that I don't understand. Yet I feel good that gravity is the minimal response spacetime can have due to another mass.

doug

 

[sweetser]

Calculating scattering probabilies

Hello Lawrence:

I've never work with the 't Hoft Veltman running parameter, so I cannot comment on the issue. Nor was I successful with renormalization calculation in my two term quantum field theory class. I doubt I will ever get the training required to answer these technical questions.

I do keep my eye out for a quick and dirty way of doing a calculation, letting those folks with far more refined skills to worry about details. We know for darn sure exactly how to do the calculation of a scattering of an electron off of a proton. We know exactly what the coupling constant is, $\alpha$. We know exactly what the propagator is for a spin 1 field.

This part of the calculation is standard. Nothing about it changes one bit. As stated before, I cannot do the calculation myself at the current time. It is the sort of thing taught to graduate students. The scattering of an electron off a proton is considered "easy", and relatively speaking, it is.

There are only two parts that change: the coupling constant is no longer alpha, but instead one puts in $G m_1 m_2$ where the square of electrical charge used to be in fine structure constant. Secondly, the propagator is for a spin 2 field. Weinberg wrote three articles on the form of such a propagator. I did not understand those articles, so I cannot jot it down and use it in a calculation. Selector rules may change as a consequence, but I don't recall their role either.

If I was handed 6 pages of calculations showing how to calculate the scattering of an electron due to a proton, I could use that as my basis to figuring out the scatter due to gravity. Changing the coupling constant would be the trivial part. Dropping in the correct propagator is probably beyond my reach, but I could spot where the work would have to be focused.

Without doing the calculation, I am more hopeful because I understand with precision where the difference are between a calculation we know how to do (EM scattering using a spin 1 particle) and one we hope to do (gravitational scattering using a spin 2 particle).

doug

 

[Lawrence B. Crowell]

Doug,

The fine structure constant e^2/hbar-c ~1/137 has the unique feature of being dimensionless. The term GMm has units [g cm^3/s^2]. While "dimensionless" in naturalized units it still has a scale dependency. If we set the masses as M = n*m_p, for m_p the Planck unit of mass m_p = sqrt(c/G hbar), and n an integer, then

GMm = (nn')c/hbar.

hbar = h/2pi for h = oint p*dq is unitless in naturalized units and c is also set to unity in naturalized units. However, c requires the use of rulers and clocks to measure and hbar involves the measure of wavelengths (rulers) and momenta (rulers & clocks). Thus alpha and GMm as coupling constants are fundamentally different.

The problem as I see it is that the symmetric structure of GEM leads to a problem with the effective action. For S = S_eff + S',
[tex]
S' = {{\delta S}\over{\delta A}}R_{\Lambda}{{\delta S}\over{\delta A}}
[/itex]
this latter term is not going to satisfy gauge invariant requirements for Wilson/Polchinski renormalization "flow." Here $R_{\Lambda}$ is the regularization cut-off function. This is the central problem here. These renormalization group equations are found from the scale structure of regularization cut-offs /\, stemming from Feynman.

Lawrence B. Crowell

 

[sweetser]

Alpha must be dimensionless

Hello Lawrence:

The fine structure constant is dimensionless. I was not clear on the unit system I was using. I was using cgs, where Coulomb's law is:
$$F=Q Q'/R^2 \hat{R}$$
In cgs, the units for Q Q' are the same as GMm. If one uses Gaussian units, then Coulomb's laws is:
$$F=\frac{1}{4 \pi \epsilon_0} Q Q'/R^2 \hat{R}$$
In Gaussian units, the switch would be:
$$\frac{1}{4 \pi \epsilon_0} Q Q' \rightarrow G M m$$
Let's just check that this works.
$$\alpha = k_c e^2/\hbar c$$
My claim is that for cgs units, the $k_c$ is 1, so a straight swap of GMm should work:
$$\alpha_{GEM} = G M m/\hbar c = (L^3 M^{-1} T^{-2}) M M /((M L^2 T^{-1}) (L/T^{-1}))$$
That is dimensionless. There is no need to use a Planck mass.

Viewed this way, do you agree that the fine structure constant of EM and $\alpha_{GEM}$ could play an identical role? It is too bad my LaTeX was mangled in the previous post (and I cannot edit it now) because I had $\alpha_{GEM}$ defined there.

doug

 

[sweetser]

Perturbation theory should be OK for GEM

Hello Lawrence:

I did some browsing on the web to find out about the Wilson/Polchinski renormalization flow because I was not familiar with it (probably like other readers of this thread). Here is a quote from the start of a 72 page review article (http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Ahep-th%2F0002034):

By "exact renormalization group equation (ERGE)", we mean the continuous (i.e. not discrete) realization of the Wilson renormalization group (RG) transformation of the action in which no approximation is made and also no expansion is involved with respect to some small parameter of the action. Its formulation - under a differential form - is known since the early seventies. However, due to its complexity (an integro-differential equating), its study calls for the use of approximation (an/or truncation) methods. For a long time it was natural to use a perturbative approach (based on the existence of a small parameter like the famous $\epsilon$-expansion for example). But, the standard perturbative field theory turned out to be more efficient and, in addition the defenders of the nonperturbative approach have turned toward the discrete formulation of the RG due to the problem of the "stiff" differential equations...This is why it is only since the middle of the eighties that substantial studies have been carried out via:
* the truncation procedures in the scaling field method
* the explicit consideration of the local potential approximation and of the derivative expansion
* an appealing us, for theoreticians, of ERGE
In the nineties there has been a rapid growth of studies in all directions, accounting for scalar (or vector) fields, spinor, gauge fields, finite temperature, supersymmetry, gravity, etc...

In the classes I took, we only worked with a perturbation approach in EM. It is a good way to go for a small coupling constant like that for EM. This is not a good way to go for the strong force.

For gravity, perturbation would be an awesome way to go. Let's calculate the gravity coupling for an electron and a proton:
$$\alpha_{GEM}=G m_e m_p /\hbar c = 6.67 \times 10^{-11} m^3 kg^{-1} s^{-2} 9.1 \times 10^{-31} kg 1.67 \times 10^{-27}kg/1.62 \times 10^{-34} kg m^2 s^{-1} 3.00 \times 10^8 m/s = 2.02 \times 10^{-42}$$
The size of the coupling constant is important. For much work in quantum field theory like that with the strong force, I could see why the Wilson/Polchinski renormalization method would be important. It is hard to believe that an $alpha_{GEM}^2$ will ever be important in a physical calculation! To be clear, I am not contesting the difficulties of using the Wilson/Polchinski renormalization method that Lawrence believes are there. I am saying this issue is not a "deal breaker" because first order perturbation theory will be sufficient for the calculations because the gravitational coupling constant is so very small.

doug

 

[vebrown]

Hi Sweetser; great work; hope it works out ! I have pondered this myself, but from a different direction. Working from the postulates in the Lorentz Ether Theory that the seat of the Electromagnetic field is the empty space, and that the final irreducible constituent of all physical reality is the electromagnetic field, it turned out like this.

Edit: I just visited your web site; very interesting. You have the graviton as a spin two particle; I always thought it was spin one, maybe I'm wrong. But isn't there a problem with spin two particles moving at the speed of light.

 

[sweetser]

Gravitons

Hello Vebrown:

I clicked through the page cited, but saw neither a field equation nor Lagrange density. For me, that is the "starting height" a proposal must clear before it is worth studying (it is nice to have a clearly defined criteria, so there is nothing personal here, just analytical). I wish you good luck in your studies none-the-less, because it took me several years to go from the GEM field equations to the Lagrange density.

You have the graviton as a spin two particle; I always thought it was spin one, maybe I'm wrong. But isn't there a problem with spin two particles moving at the speed of light.

If a particle has zero mass, then it will travel at the speed of light. As for the spin of the particle needed, let me quote from Brian Hatfield's introduction to "Feynman Lectures on Gravitation":

When the potential is computed in both (even and odd integral spin) cases and the appropriate limits are taken, we find that when the exchanged particle carries odd integer spin, like charges repel and opposite charges attract, just as in the example of electrodynamics. On the other hand, when the exchanged particle carries even integer spin, the potential is universally attractive (like charges and opposite charges attract). Hence, the spin of the graviton must be 0, 2, 4, ...

To be completely truthful, I do not know the details of that calculation. I do know it is widely accepted that the graviton is a massless spin 2 particle that travels at the speed of light. Because we have data that says photons couple to gravity, a spin 0 graviton would not work (I again am paraphrasing Hatfield without completely understanding him). The smallest even integer particle to do the work of gravity is spin 2.

I am not a great student of the particle physics literature, but so far have yet to hear what physicists think is the connection between the graviton which mediates the force of gravity, and the Higgs particle, which adds inertial mass to the standard model. The deep insight of general relativity is that gravitational and inertial masses are precisely the same thing. Therefore the relationship between the graviton and the Higgs must be as tight as can be constructed. Thing is, the Higgs is a spin 0 particle.

How could one make an unbreakable connection between a spin 0 and a spin 2 particle? My thought was the symmetric rank 2 field strength tensor naturally is a source for a spin 2 field. That is completely standard logic, nothing at all radical. What I realized is a rank 2 tensor will always have associated with it the trace of that tensor. The trace would form a rank 0 tensor, and that rank0 tensor could be represented by a spin 0 field. The fact that a rank 2 symmetric tensor always has a rank 0 trace is the reason that gravitational and inertial mass are the same. Cool!

doug

 

[Don J]

Hello Don:

Here is a big advantage of the GEM approach a layman can appreciate: the ontology (a fancy why of saying the "why" of gravity and EM). Let's first look at the why behind Newton's law of gravity and general relativity.

There is no "why" for Newton's law of gravity. That law was constructed and shown to be relevant by connection to Kepler's three laws of planetary motion. In theory, gravity could have been a 1/R or 1/R^3 law with the correct adjustment of G (L^2/M T^2 or L^4/M T^2 respectively). An inverse square law has some nice mathematical properties, but that is not a reason why the law should be the product of two masses over the square of the distance.

The situation for general relativity is no better. doug

Hi sweetser
Your theory must doing something -utile (in the mathematically sense)- that GR cannot do.
Do you have some examples (tests)?
Is there something in the Universe who need to be -explained - with a rank one theory?
If i remember You qualified GR and other -relativistic based theory- of rank two theory

That specific point about rank one and rank two theory needs a clear comparison about its application in the real world (The universe)?

Thanks!

 

[sweetser]

Tests for the GEM proposal

Hello Don:

The tests for this proposal were detailed at the start of this thread (point 2 of that initial post). I try and do new things, but appear to repeat two specific tests: light should bend around the Sun 11.5 microarcsecond versus 10.8 mircoarcseconds predicted by GR when second order PPN measurements are made, and a gravitational wave will not be a transverse wave as predicted by GR.

The vacuum GEM equations are linear, whereas the ones for GR are nonlinear. In theory, and definitely not in practice, one could measure the strength of a test mass, then add another equal test mass, and show either the gravitational field was exactly doubled, or there was a nonlinear increase. Although I have not done the calculation, I believe such a direct measurement will be impossible to do for the foreseeable future because it would require absurdly sensitive measurements.

GR is a rank 2 field theory because it has two Greek letters:
$$R_{\mu \nu} - g_{\mu \nu} R = 8 \pi T_{\mu \nu}$$
GEM is a rank 1 field theory because it has only one:
$$J^{\mu}_q - J^{\mu}_m = \nabla^2 A^{\mu}$$
In GR, gravity couples to the stress energy tensor, $T_{\mu \nu}$. In GEM, gravity couples to the 4-momentum current density, $J^{\mu}_m$. These are tensors of different rank, so we cannot compare them directly. We can look at what goes into each. I believe the only difference is that the energy of a gravity field goes into the stress-energy of GR, but it is excluded in GEM. The effect of gravity is very weak, and the energy of a gravity field is very small, so spotting the contribution of a very small thing to a very weak effect probably will not be possible to measure in the real world. It is very significant for the math guys. In GEM, because the gravity field does not gravitate, it will be much simpler to quantize.

doug

 

[CarlB]

Doug, could you take a look at this paper by Kris Krogh and tell me how it differs from your theory?

Equations of Motion in a Quantum-mechanical Theory of Gravity
Kris Krogh
An earlier paper [1] presented a gravity theory based on the optics of de Broglie waves rather than curved space-time. While the universe's geometry is flat, it agrees with the standard tests of general relativity. A second paper [2] showed that, unlike general relativity, it agrees with Doppler tracking signals from the Pioneer 10 and 11 space probes. There a gravitational acceleration equation plays an important role, accounting for the relative motions of Earth and the probes. Here it's shown that equation also describes Mercury's orbit.
http://www.arxiv.org/abs/astro-ph/0508290

He comes at the problem from a different direction, but he also talks about a relationship with E&M. So I'm wondering if his equations of motion are the same as yours. I figure that I can simulate his with no great effort as they are described clearly enough that a simpleton like me can figure them out.

I've had some success with the gravity simulation program recently, well I should say success in figuring out the equations of motion for GR, and have now updated it to include this. I'm working on Painleve coordinates now.

Carl

 

[Don J]

Doug wrote
The effect of gravity is very weak, and the energy of a gravity field is very small, so spotting the contribution of a very small thing to a very weak effect probably will not be possible to measure in the real world. It is very significant for the math guys. In GEM, because the gravity field does not gravitate, it will be much simpler to quantize.

I think we will have an answer in the coming year...
Gravity Probe B mission, testing Einstein's theory of gravity, completes first year in space
http://www.physorg.com/news4062.html

 

[sweetser]

Orbits in GEM (darn similar to GR)

Hello Carl:

The GEM theory and the one one put forth by Kris Krogh are very different. It all flows from the Lagrangian. We start from exactly the same starting place, the classical EM Lagrangian (OK, he works with forces, not fields, so he has the inertia and current coupling term, but not the field strength contraction term, $F^{\mu \nu}$). What he then does is slap an exponential factor to the side of it. I have issues with that. The particle involved in the force of gravity has got to be different from the photon, because the photon helps like charges repel, but for gravity, like charges attract. Curve spacetime as you like, but one has to account for the particles that will be carrying out the program.

This is one reason I stress the importance of being able to write out a Lagrangian: since a Lagrangian has all the energy interactions that can happen inside a box, looking at it - after some practice - can tell you how the proposal works. There is only one current that binds to the potential, and that is the EM current. In my GEM proposal, there are two currents, one for EM that is made a little bit smaller due to the inertial mass current. It is a flaw of mine how much I like that idea: like electrical charges repel each other because of EM, but they repel a little less due to their own inertia.

So at the level of the Lagrangian, my work does not get along with Krogh's. Krogh has thought through the equations for orbits. I did work in detail on the issue, to get the precession of the perihelion of Mercury. That is a tough calculation! The prediction of GEM is exactly the same as that for GR at this level of accuracy.

Even though it will make this post long, I thought I would include the entire thing here, because you are unlikely to ever see all the steps for the calculation spelled out.

The Precession of the Perihelion of Mercury Based on a Solution to the GEM Field Equations

1. Start with a solution to the GEM field equations for a spherically symmetric, non-rotating, uncharged mass with the gauge choice of a constant 4-potential. The solution is the following metric written in spherical coordinates.

$$d \tau^2 = e^{- 2 \frac{G M}{c^2 R}} d t^2 - \frac{1}{c^2} e^{2 \frac{G M}{c^2 R}} d R^2 - ( \frac{R}{c})^2 (d \theta^2 + \sin^2 \theta d \phi^2)$$

2. Simplify by working in the plane of rotation, so $d \theta = 0$ and [tex]\sin^2 \theta = 1[/itex]:

$$d \tau^2 = e^{- 2 \frac{G M}{c^2 R}} d t^2 - \frac{1}{c^2} e^{2 \frac{G M}{c^2 R}} d R^2 - ( \frac{R}{c})^2 d \phi^2$$

3. Divide by $d \tau^2$ and multiply through by $e^{- 2 \frac{G M}{c^2 R}}$:

$$e^{- 2 \frac{G M}{c^2 R}} = e^{- 4 \frac{G M}{c^2 R}} ( \frac{d t}{d \tau})^2 - \frac{1}{c^2} ( \frac{d R}{d \tau})_{}^2 - \text{} e^{- 2 \frac{G M}{c^2 R}} ( \frac{R}{c})^2 ( \frac{d \phi}{d \tau})^2$$

4. Since the exponent for our solar system is small, use the Taylor series expansion, $e^{- 2 \frac{G M}{c^2 R}} \approx (1 - 2 \frac{G M}{c^2 R})$. View the -4 term as a -2 term squared:

$$(1 - 2 \frac{G M}{c^2 R}) = (1 - 2 \frac{G M}{c^2 R})^2 ( \frac{d t}{d \tau})^2 - \frac{1}{c^2} ( \frac{d R}{d \tau})^2 - \text{} (1 - 2 \frac{G M}{c^2 R}) ( \frac{R}{c})^2 ( \frac{d \phi}{d \tau})^2$$

Note: from here on out, there is nothing different from GEM compared to GR.

5. Define the Killing vectors for this metric. Notice that the expression in 4 is not a function of either time $t$ or angle $\phi$. This means there is a conserved quantity associated with a change in time (energy $E$) and a change in angle (angular momentum $L$). Come back at a later time, and the expression stays the same. Spin around a few degrees, and the metric stays the same. The general structure of a Killing vector is this:

(conserved thing) = (Killing vector)$\cdot$(velocity vector)

Here are the two Killing vectors for 4:

$$\frac{E}{m c^2} = K_t V_t = (1 - 2 \frac{G M}{c^2 R}, 0, 0, 0) \cdot ( \frac{d t}{d \tau}, 0, 0, 0) = (1 - 2 \frac{G M}{c^2 R}) \frac{d t}{d \tau}$$

$$\frac{L}{m c} = K_{\phi} V_{\phi} = (0, 0, 0, \frac{R}{c}) \cdot (0, 0, 0, R \frac{d \phi}{d \tau}) = \frac{R^2}{c} \frac{d \phi}{d \tau}$$

6. Calculate the squares of the energy and angular momentum:

$$( \frac{E}{m c^2})^2 = (1 - 2 \frac{G M}{c^2 R})^2 ( \frac{d t}{d \tau})^2$$

$$( \frac{L}{m c})^2 = \frac{R^4}{c^2} ( \frac{d \phi}{d \tau})^2$$

7. Plug 6 into 4:

$$(1 - 2 \frac{G M}{c^2 R}) = ( \frac{E}{m c^2})^2 - \frac{1}{c^2} ( \frac{d R}{d \tau})^2 - \text{} (1 - 2 \frac{G M}{c^2 R}) ( \frac{L}{m c R})^2$$

What has happened? We have introduced two constant quantities, the energy $E$ and the angular momentum $L$. We still have the factor, $(1 - 2 \frac{G M}{c^2 R})$.

8. Prepare for a change of variable, to $U = \frac{1}{R}$:

$$\frac{d R}{d \tau} = \frac{d R}{d \phi} \frac{d \phi}{d \tau} = (- \frac{1}{U^2} \frac{d U}{d \phi}) ( \frac{L U^2}{m}) = - \frac{L}{m} \frac{d U}{d \phi}$$

9. Plug 8 into 7:

$$(1 - 2 \frac{G M}{c^2} U) = ( \frac{E}{m c^2})^2 - \frac{L^2}{m^2 c^2} ( \frac{d U}{d \phi})^2 - \text{} (1 - 2 \frac{G M}{c^2} U) ( \frac{L U}{m c})^2$$

10. Multiply through by $U$ and bring all the terms to one side:

$$0 = ( \frac{E}{m c^2})^2 - \frac{L^2}{m^2 c^2} ( \frac{d U}{d \phi})^2 - \text{} 1 - \frac{L^2 U^2}{m^2 c^2} + 2 \frac{G M}{c^2} U + 2 \frac{G M L^2 U^3}{m^2 c^2}$$

11. Take the derivative of 10 with respect to $\phi$:

$$0 = - 2 \frac{L^2}{m^2 c^2} \frac{d U}{d \phi} \frac{d^2 U}{d \phi^2} - 2 \frac{L^2 U}{m^2 c^2} \frac{d U}{d \phi} + 2 \frac{G M}{c^2} \frac{d U}{d \phi} + 6 \frac{G M L^2 U^2}{m^2 c^2} \frac{d U}{d \phi}$$

12 Divide all the terms by $- 2 \frac{L^2}{m^2 c^2} \frac{d U}{d \phi}$:

$$0 = \frac{d^2 U}{d \phi^2} + U - \frac{G M m^2}{L^2} - 3 \frac{G M}{c^2} U^2$$

The first three terms are classical Newtonian gravitational physics (implied by the lack of a factor of $c$). The fourth is the correction required by GEM.

13. Write out the Newtonian equation:

$$0 = \frac{d^2 U}{d \phi^2} + U - \frac{G M m^2}{L^2}$$

14. Solve the Newtonian equation. This is a slight variation on an equation with the cosine solution. We must account for the eccentricity of the circle and the constant factor. Guess a solution:

$$U = \frac{G M m^2}{L^2} (1 + \epsilon \cos (\phi - \phi_0))$$

$$\frac{d U}{d \phi} = - \frac{G M m^2}{L^2} \epsilon \sin (\phi - \phi_0)$$

$$\frac{d^2 U}{d \phi^2} = - \frac{G M m^2}{L^2} \epsilon \cos (\phi - \phi_0)$$

$$\frac{d^2 U}{d \phi^2} + U - \frac{G M m^2}{L^2} = - \frac{G M m^2}{L^2} \epsilon \cos (\phi - \phi_0) + \frac{G M m^2}{L^2} (1 + \epsilon \cos (\phi - \phi_0)) - \frac{G M m^2}{L^2} = 0$$
OK!

15. At the perihelion, $\cos (\phi - \phi_0) = 1$ and $R = 1 / U = a (1 - \epsilon)$. Plug into $U$ found in 14:

$$\frac{1}{a (1 - \epsilon^2)} = \frac{G M m^2}{L^2}$$

16. Use the Newtonian solution found in 14 as a start for the correction term, $U^2$:

$$U^2 = \frac{G^2 M^2 m^4}{L^4} (1 + 2 \epsilon \cos (\phi - \phi_0) + \epsilon^2 \cos^2 (\phi - \phi_0))$$

17. Plug this into 12:

$$0 = \frac{d^2 U}{d \phi^2} + U - \frac{G M m^2}{L^2} - 3 \frac{G M}{c^2} \frac{G^2 M^2 m^4}{L^4} (1 + 2 \epsilon \cos (\phi - \phi_0) + \epsilon^2 \cos^2 (\phi - \phi_0))$$

18. Keep only the second $U^2$ correction term. The factor of $\frac{G^3 M^3}{c^2}$ will make the $U^2$ correction tiny. Only if a term is ``on resonance'' - in effect pushing the swing at the same time as the main solution for $U$ - can a term eventually make a contribution to the orbit:

$$0 = \frac{d^2 U}{d \phi^2} + U - \frac{G M m^2}{L^2} - 6 \frac{G M}{c^2} \frac{G^2 M^2 m^4}{L^4} \epsilon \cos (\phi - \phi_0)$$

19. Guess a solution. It must be composed of the previous solution in 14, plus a way to drop the additional cosine term:

$$U = \frac{G M m^2}{L^2} (1 + \epsilon \cos (\phi - \phi_0) + 3 \frac{G^2 M^2 m^2}{c^2 L^2} \phi \sin (\phi - \phi_0))$$

$$\frac{d U}{d \phi} = - \frac{G M m^2}{L^2} (\epsilon \sin (\phi - \phi_0) + 3 \frac{G^2 M^2 m^2}{c^2 L^2} \phi \cos (\phi - \phi_0) + 3 \frac{G^2 M^2 m^2}{c^2 L^2} \sin (\phi - \phi_0)_{})$$

$$\frac{d^2 U}{d \phi^2} = - \frac{G M m^2}{L^2} (\epsilon \cos (\phi - \phi_0) - 3 \frac{G^2 M^2 m^2}{c^2 L^2} \sin (\phi - \phi_0) + 6 \frac{G^2 M^2 m^2}{c^2 L^2} \cos (\phi - \phi_0))$$

The extra terms will drop. OK!

20. Bring the correction term into the cosine. Because $\frac{G^2 M^2 m^2}{c^2 L^2}$ is so small, only the first term of $\phi \sin (\phi - \phi_0)$ will make a contribution to cosine. Rewrite $U$ in 19:

$$U = \frac{G M m^2}{L^2} (1 + \epsilon \cos (\phi - \phi_0 - 3 \frac{G^2 M^2 m^2}{c^2 L^2} \phi))$$

21. Calculate the ratio of the advance of $\phi$ in one $2 \pi$ rotation between the Newtonian solution, and
the GEM correction:

$$\Delta \phi = 2 \pi \frac{1}{1 - 3 \frac{G^2 M^2 m^2}{c^2 L^2}} \approx 2 \pi (1 + 3 \frac{G^2 M^2 m^2}{c^2 L^2})$$

22. Plug the relation in 15, $\frac{1}{a (1 - \epsilon^2)} = \frac{G M m^2}{L^2}$, into 21.

$$\phi_{\tmop{advance}} = 6 \pi \frac{G M}{a (1 - \epsilon^2) c^2}$$

23. Collect the relevant numbers:

$$G = 6.67 \times 10^{- 11} \mathrm{\mathtt{\mathrm{m^3 / \tmop{kg} s^2}}}$$
$$M = 1.99 \times 10^{30} \tmop{kg}$$
$$c = 3.00 \times 10^8 m / s$$
$$a = 5.79 \times 10^{10} m$$
$$\epsilon = 0.206$$
$$( \frac{1 \tmop{revolution}}{88.0 \tmop{days}}) ( \frac{365 \tmop{days}}{\tmop{year}}) ( \frac{100 \tmop{years}}{\tmop{century}}) = 415 \frac{\tmop{rev} .}{\tmop{century}}$$
$$( \frac{180^{\circ}}{\pi \tmop{radians}}) ( \frac{60'}{^{\circ}}) ( \frac{60''}{'}) = 2.0610^5 \frac{''}{\tmop{radians}}$$

24. Plug values in 23 into 22:

$$6 \pi \tmop{rad} . \frac{6.67 \times 10^{- 11} \mathrm{\mathtt{\mathrm{\frac{m^3}{\tmop{kg} s^2}}}} 1.99 \times 10^{30} \tmop{kg}}{5.79 \times 10^{10} m (1 - 0.206^2) (3.00 \times 10^8 \frac{m}{s})^2} 415 \frac{\tmop{rev} .}{\tmop{century}} 2.06 \times 10^5 \frac{''}{\tmop{rad} .} = 42.9 \frac{''}{\tmop{century}}$$

 

[CarlB]

So am I right in thinking that if I do not make the approximation: $e^{- 2 \frac{G M}{c^2 R}} \approx (1 - 2 \frac{G M}{c^2 R}),$ and solve for the equations of motion in the usual manner I will get the correct equations of motion for the GEM theory?

Also Doug, I was beginning to worry about you. I would have called you up but I lost your phone number when they reduced the number of personal messages that you can keep on PF. Please send me your latest so I can pester you when you next take a break.

Carl

 

[sweetser]

yep! [Note, that "yep!" was going to be my entire reply, but I learned there is a 10 character minimum, so this is just filler.]

doug

 

[sweetser]

A proposed numerical calculation

Hello Carl:

Let's see if I can define for you the numerical integral I need to do to test my proposal for solving the rotation profile of thing galaxies.

Background:
At zeroth order, Newton's law of gravity, the Schwarzschild metric, and the exponential metric of GEM are identical. This is the level of accuracy that applies to weak gravitational systems, such as the rotation profile of thin disk galaxies.

At first order, Newtonian gravity is off from measurements by a factor of two. GR and GEM are still identical in their predictions. Note that there effects are subtle, requiring extremely accurate measurements. I've done the calculation for the velocity profile of a thin disk galaxy using Newton's theory or GR, and the results look identical.

At second order, the exponential metric has 12% more bending than Schwarzschild. With all the improvements that have been made in measurements over the years, we are still two orders of magnitude away from detecting anything this subtle. The difference between GEM and GR, real as it is, numerically does not matter for a huge, weak field thing like a galaxy.

When I derived the exponential metric from a relativistic 4-force law, I had this term for the 4-force:
$$\frac{d m V^{\mu}}{d \tau}$$
Two important things to notice. First, force as a change in momentum is a product, governed by the product rule:
$$\frac{d m V^{\mu}}{d \tau}=m \frac{d V^{\mu}}{d \tau} + V^{\mu} \frac{d m}{d \tau}$$
This is the way calculus works, there is no doubt about it. A change in momentum can be due to a change in velocity, the mA of Newton's second law, or a change of momentum could be due to a rocket effect, changing the mass.

The second observation is the change is with respect to the spacetime interval tau. Due to the relationship between the Greek and Roman alphabet, I expected this transition from relativistic to classical physics:
$$d \tau \rightarrow d t$$
I am certain that is what happens to generate the mA term. A change in an interval appears in classical physics as a change in time. There is another possibility: that a change in a spacetime interval appears as a change in space classically:
$$d \tau \rightarrow d R/c$$
With a thin spiral galaxy, the amount of matter changes with respect to the radius of the galaxy, a change which can be described only by the term $\frac{d m }{d R}$. In many thin disk galaxies, the amount of visible matter drops off exponentially with the radius. That is a serious dm/dR function caused by gravity that must be transparent in a calculation.

For Newtonian gravity, there will be three parts: Newton's law of gravity, the centrifugal force, and the acceleration:
$$- \frac{G M_a m_p}{R^2} = m_p \frac{V^2}{R} + m_p \frac{d V}{d t}$$
where $M_a$ is the active mass, basically the sum of all gravitational masses, and $m_p$ is the passive mass, the mass that is going around and accelerating. I can imagine - but have not done - calculating the velocity profile of a galaxy using this equation plus one other, the mass per area. What would be done is to divide the radius R into say 100 chunks, and to divide the circle into say 100 angles. Start with a passive mass at the center, where the velocity is zero. Calculate the mass for all the active masses, and plug it into the above expression. That will generate an acceleration, from zero to a new velocity. Repeat the calculation for the new passive mass at the next step out along the radius traveling at this new velocity V. Repeat for all positions of the passive gravitational mass along the radius. Plot the resulting velocities with respect to the radius.

If this calculation is done correctly, there should be a steep acceleration, a peak, and then continual drop in the velocity with respect to the radius.

The proposal:
For the galaxy NGC3198, we do the classical Newtonian gravity velocity profile calculation as described above. The validity of the calculation can be checked against the numbers for that galaxy (a peak velocity around 150,000 m/s, the velocity drops off after that). Then repeat the calculation with one more term:
$$- \frac{G M_a m_p}{R^2} = m_p \frac{V^2}{R} + m_p \frac{d V}{d t} + V c \frac{d m_p}{d R}$$
If the velocity profile reaches the peak, and remains flat after that, then we have something of significance. If the velocity profile still drops off, then it means this product rule does not explain the velocity profile of thin disk galaxies, and the hypothesis is wrong.

If you understood the problem description, I'll collect the number for that galaxy, and let us both give an independent try at the calculation. Anyone else reading this thread who likes doing a real numerical integral can also join in.

doug

 

[jhmar]

sweetser

I hope no one found the preceding explanation completely satisfying. That is what it feels like to wonder how things really work. There comes a point where what you know runs out.

Your mathematics are way beyond me but are they necessary? Fahr and Heyl published in Astrophysical Notes and on:
http:// arXiv:astro-ph/0606448 v1 19 Jun 2006
a paper on gravity, showing the relationship between mass and radius.

I recently had a particle physics paper rejected that made a similar proposal in that it suggested a relationship between mass, radius, and force. After reading Fahr and Heyl I was able to extend my proposal to show that it produced the same constant (for all particles) as that found by Fahr and Heyl (for gravity) and a revised paper is now under review.

It seems to me that the main difference in our approach is that you use actions to formulate a theory while Fahr and Heyl (and myself) use entities; the advantage of the entity approach to the problem is that what you know does not run out.

I would very like your opinion on the Fahr and Heyl paper.

 

[sweetser]

Cosmological consequences

Hello Jhmar:

If my proposal is correct, then the Schwarzschild radius plays no role in the description of Nature. Oops, that is kind of obnoxious to a large body of work, oh well! In advanced books on GR, they will also explain how that radius is more a consequence of the choice of coordinates than anything else. What is independent of the choice of coordinates is that there is a point singularity. At the point singularity, many terms to describe a system become undefined.

In detail, I do not know how the GEM proposal is going to deal with areas of very high mass density. I do know it will be different. My sense of it is that there will be a singular at tau=0, but that is not a point singularity, it is the surface that all light travels on. That will mean that folks skilled in math can do things, that the description of the physical system will remain defined, unlike the case for a point singularity.

I also believe cosmology is missing a vital classical term, what I have been calling the relativistic rocket effect. Briefly this says that gravity accelerates mass (m A R_hat) and gravity also determines the location in space of moving masses (V c dm/dR V_hat). The presence of the factor of "c" and small spatial scales makes this term insignificant for our solar system. For the rotation profile of galaxies or the big bang, this term may become dominant. That is pure speculation at this time because I haven't figured out how to crunch numbers using this term.

doug

 

[jhmar]

doug

Thanks for a detailed reply which will take some time to digest. But at the risk of making a fool of myself I will make a few comments.

We are approaching the problem of explaining the universe from two completely different points of view; you take the GR approach, I take the classical particle physics approach.

Take for example "there will be a singular at tau=0". Now I would agree that a vacuum field can be reduced to a zero point but the mass (matter, anti-force, Higgs field? call it what you will) cannot be reduced to a point. That is why Fahr and Heyl and myself (although I add 'myself' here with some trepidation) say the constant is 1. The field can be 'zeroed' the field content cannot, but as the content of a single field is a fixed quantity there is a mass,force and radius formula that produces the 1 constant. This is as true of a gravity field as it is of an electron field. Gravity and the other forces can be 'unified' when each is considered in its particle form. That is to say that all force carrying particles have the same structural formula (graviton, lepton and quark are different states of a single fundamental particle), and a universal gravity field, like all formations where particles are contained in a single bonding field; is just another composite particle.

john

 

[sweetser]

Hello John:

It sounds like my efforts to communicate the GEM proposal have fallen short. My proposal is in direct, technical conflict with general relativity. GR is a rank 2 field theory that is generated by varying the action (just the Ricci scalar) with respect to the metric. GEM is a rank 1 field theory generated by varying the action (the contraction of a rank 2 asymmetric tensor) with respect to the potential. There are two testable differences cited in the first post (and I'll try not to repeat myself about them again).

The conservative me says one cannot claim to have a field theory until the action is defined. For a long time, I too did not have the action, and was darn uncertain I would ever find it. It was something like a year and a half before I did find it because at the time I had no practical experience with Lagrangians. That is quite common in the human population!

Most of your points leave me confused. Classical field theory is about Lagrangians, actions, field equations, and solutions to those field equations, all of which are part of the GEM proposal. I know it takes a good long time to understand that volume of detail in this thread, but it is there.

I don't feel I understand the case where tau=0. I would just point out that only particles with a mass of zero will live on that surface.

Elsewhere in the thread I've discussed how the proposal breaks U(1) symmetry due to mass, so there may be no need for the Higgs field.

In the GEM proposal, the stuff of gravity is in the symmetric, rank 2, irreducible tensor, while the stuff of EM is in the antisymmetric, rank 2, irreducible tensor. The graviton cannot do the job of a proton, particularly since like charges attract for the former but repel for the latter.

One must be ultra careful in unified field theory saying what is similar and what is different.

doug

ps. Learning physics is difficult stuff, so I don't consider you foolish at all.

 

[jhmar]

doug

Classical field theory is about Lagrangians, actions, field equations,

My lack of a decent education is the cause of our communication problem as demonstrated your statement. I thought Classical particle physics was about what particles arerather than what particles do.

I thought that only when we know what determines mass, force and radius (physical size), can we understand the cause and outcome of actions. Realising that this was a far simpler problem than the study of actions I set about finding a formula that gives the basic structure of all elementary particles; only to find that there is only one elementary particle that can be transformed into an infinite number of states (all with the same content).

I could only make a crude estimate for the graviton state and my constant is related to force (it is still valid). So I was over the moon to find Fahr and Heyl's paper and quickly realized that by a variation of my formula, my table could be extended to produce the same value (1) constant.

The difference between states is determined by the wave structure. The data given by The Particle Data Group is compared to the theoretical data and falls within the margins of error given by PDG in all but two borderline cases (about 150 are given).

PDG reject many experiments and average the remainder, I make no rejections and do not average. Each experiment is considered to find a specific particle state.

Sorry, have to go, cannot stop to edit reply, hope it is readable

john

 

[Sariaht]

Hello John:

It sounds like my efforts to communicate the GEM proposal have fallen short. My proposal is in direct, technical conflict with general relativity. GR is a rank 2 field theory that is generated by varying the action (just the Ricci scalar) with respect to the metric. GEM is a rank 1 field theory generated by varying the action (the contraction of a rank 2 asymmetric tensor) with respect to the potential. There are two testable differences cited in the first post (and I'll try not to repeat myself about them again).

The conservative me says one cannot claim to have a field theory until the action is defined. For a long time, I too did not have the action, and was darn uncertain I would ever find it. It was something like a year and a half before I did find it because at the time I had no practical experience with Lagrangians. That is quite common in the human population!

Most of your points leave me confused. Classical field theory is about Lagrangians, actions, field equations, and solutions to those field equations, all of which are part of the GEM proposal. I know it takes a good long time to understand that volume of detail in this thread, but it is there.

I don't feel I understand the case where tau=0. I would just point out that only particles with a mass of zero will live on that surface.

Elsewhere in the thread I've discussed how the proposal breaks U(1) symmetry due to mass, so there may be no need for the Higgs field.

In the GEM proposal, the stuff of gravity is in the symmetric, rank 2, irreducible tensor, while the stuff of EM is in the antisymmetric, rank 2, irreducible tensor. The graviton cannot do the job of a proton, particularly since like charges attract for the former but repel for the latter.

One must be ultra careful in unified field theory saying what is similar and what is different.

doug

ps. Learning physics is difficult stuff, so I don't consider you foolish at all.

Hey, I recognise that face! You've been on tv! Coool! What are you working with?

And if i asked you right now, what momentum per time causes gravity, what would you respond?

 

[jhmar]

doug

I cause much confusion by using the wrong terms (lack of training). That aside, I have submitted my work and await the moderator's decision due in about three weeks.
Basically I believe I have demostrated a relationship between wave length and mass, but must now wait and see if I have managed to convince those with a greater knowledge of this subject.
Please keep returning to this forum where I will notify you of the outcome when received; a positive outcome will, of course, appear as a separate forum.
regards
John

 

[sweetser]

The force law

Hello Sariaht:

"The Stand-Up Physicist" is award winning TV seen by almost no one! The show represents ultra-narrow casting. The show appeared on BNN-TV in Boston at 11:30 on a Tuesday night, and no doubt got killed by the likes of Jay Leno & David Letterman. The town of Auburn outside of Worcester gets several doses a week because the guy helping with the post-production work tries to fill the airways. The third place it shows in the town of Acton where I live, four times on the weekend, but I have no idea what time (they schedule things poorly here).

There are all kinds of video festivals out there, particularly if the video has gay content (perhaps as an excuse to get together and party). So far, I found only one festival that had an education category. I entered the episode, "Why Quantum Mechanics is Weird" in the Berkeley Video and Film Festival, and it won Best of Festival in the Education Category. Sounds good, but it was second place to the Grand Prize winner, a film on polar bears. I went up against a polar bear and got mauled.

After reading an article on digital video recording on Slashdot and having discussions with a friend who does professional video work, I spent about $4k to get a 3-CCD digital camera, a great Sennheiser boom microphone, 1500 watts of lights, a tripod, background set holder, and a green screen.

Why bother? Here's my situation: Mathematica and I feel confident that we have found exactly what Einstein was looking for: a unified field theory for gravity, and a reason why causality is different for quantum mechanics versus classical mechanics. In my world view, I don't know if I could construct a more unlikely thesis. It is reasonable for people not to believe it is true. I have no idea how to get through that social barrier. It could be that I am just wrong. That in so many ways is the easiest. Sincere nerd is a moron. I take that seriously because I have made technical mistakes along the way, from the wrong sign for the charge coupling term, to misstatements about the field strength tensor. My honest appraisal is that the foundation still feels solid, and more work can be done.

If and only if the unified field theory is correct, then it is my responsibility to communicate it. I will continue to put my money there in diverse ways. I have a web site, there is the self-printed book, I go to conferences three or four times a year (never as an invited guest, and that limits the audience to those who also were not invited). I make deliberate efforts to toss the idea up to leading researchers in physics, but those folks are busy, busy, busy. There are many more people with more time, but less training, that are interested in the possibility of a unified field theory. That is the group the show is target at. The show is too technical for a general audience. I work through most of the mathematical detail, which is unusual.

The show is just me yakking in front of a camera, very minimalist. I make sure the equations I am thinking about at the moment are on the screen, because physics is math. I do the post-production work at Auburn Community Access Television, in Auburn, MA (a fifty minute drive I have done many times). I use Final Cut Pro, version 5, on a Mac. Sixteen shows are complete, another 8-10 to write, shoot, and produce. The half hour shows can be downloaded from the web (no idea about the stats). The DVDs are for sale internationally, and as of today, the global sales are up to 1 DVD to a friend of mine.

That is the TV story.

> "what momentum per time causes gravity"?

I am going to work to translated this... The units for momentum per time are the units of force. In general relativity, one cannot write an equation for the force of gravity like you can for EM. Instead, there are geodesic paths that objects follow, paths of no force.

GEM is modeled on EM. The answer to what is the force equation of EM is the Lorentz force:
$$\frac{d m V^{\mu}}{d \tau} = q_e V_{\nu} (\nabla^{\mu} A^{\nu} - \nabla^{\nu} A^{\mu}) = \gamma q_e (\vec{\beta} . \vec{E}, \vec{E} + \vec{\beta} \times \vec{B})$$
This is the relativistic force law. At low speed, $\beta\rightarrow 0, \gamma \rightarrow 1$, so the law becomes Coulomb's force law, or:
$$\vec{F} = q_e \vec{E}$$
For my approach, one key change is the field strength tensor is the symmetric partner to the antisymmetric field strength tensor of EM. This has an important consequence: like charges attract. The charge for gravity is independent from EM and must have an opposite sign. This should be enough information to form the gravitational Lorentz force equation:
$$\frac{d m V^{\mu}}{d \tau} = -q_m V_{\nu} (\nabla^{\mu} A^{\nu} + \nabla^{\nu} A^{\mu}) = -\gamma q_m (g_0 + \vec{\beta} . \vec{e}, \vec{g} + \vec{e} + \vec{\beta} \times \vec{b})$$
where small e and small b are the symmetric analogues to the E and B of EM, and g is the four parts along the diagonal of the symmetric field strength tensor. In the classical physics realm, the only term that survives is the small e:
$$\vec{F} = - q_e \vec{e}$$
Newton's law of gravity, bingo. This is darn similar to the case for EM, but the sign is different, which is essential.

Another way to look at this is that I have found the right way to pluck out Newton's law form a relativistic force law from gravity. That wasn't how it was done, but it could be viewed that way.

Hope all had a good thanksgiving (if in the US),
doug

 

[CarlB]

Doug,

Lubos Motl wrote a paper that pretty much states my objections to the requirement that Lagrangian or Hamiltonian formalism be at the foundations of physices here:
http://motls.blogspot.com/2006/12/conventions-vs-physics.html