Unifying Gravity and EM

[Lawrence B. Crowell]

Hello Lawrence:

And just as seriously, I understand why this holds together so tightly from a logical perspective. I have even learned from you how odd it is to try and tack on something symmetric to this construction, which is not what I am trying to do. Every tight web of logic has an underlying assumption. What underlies this is the assumption that the Riemann curvature tensor is necessary to describe the physical force of gravity. GR does work that way, we have darn great data to say GR is correct. All vetted researchers try to recreate GR in a wider context, or do a technical variation on the rank 2 field theory theme. I hope to show that GR, as good as it is, is not good enough for a unified field theory, it will be a challenge to challenge. I heard no reply to the long standing energy problem which is well known and well ignored today. There is no trivial way around that problem. If the Riemann curvature tensor is not relevant to the way unified fields in Nature work, then the Bianchi identities - a property of the Riemann curvature tensor - are also not important, nor are the bundles built on top of it all.

You have fallen into the trap! There is no energy problem with general relativity, at least if you think about it correctly. What you see as a problem is in fact an astounding fact of cosmology which is vitally important! Here is what you think the problem is, which I will state in fairly precise terms. The deSitter spacetime, which our cosmology appears to be asymptoting towards, as the metric

$$ds^2~=~-dt^2~+~e^{\beta t}(dr^2~+~r^2d\Omega^2)$$

The metric terms are time dependent and it is not possible to find a $k_t$ so that there is a stationary condition ${\cal L}_{K_t}g~=~0$ , for a Lie derivative ${\cal L}_{K_t}~=~\kappa\partial/\partial t$. This Lie derivative is defined according to brackets so that

$${\cal L}_{k_t}g(X,~Y)~=~g([K_t,~X],~Y)~+~g(X,~[K_t,~Y]),$$

where the brackets $[K_t,~X]$ are not zero because the vectors X are functions of time. So there is no involutary system which defines a conservation of energy on the entire spacetime. The basis vectors for the deSitter cosmology are $]X_i~=~exp(\sqrt{\Lambda/3}~t/2)$ and so the above expression gives

$${\cal L}_{K_t}g_{\mu\nu}~=~\sqrt{\Lambda/3}g_{\mu\nu}$$

For the cosmological constant $\Lambda~=~3H^2\Omega/c^2$ this equation is an eigenvalued equation, and the nonvanishing of the Hubble constant is a measure of how $k_t$ fails to be a proper Killing vector.

$${\cal L}_{K_t}g_{\mu\nu}~=~Hg_{\mu\nu}$$

The Hubble constant defines a velocity-distance rule $H~=~{\dot R}/R$, for $R(t)$ a scale factor for the cosmology.

This gives a nonconservation of energy! There is no symmetry in general which defines a conservation of energy in a cosmology. This might for some be a horrible problem --- for me it is an astounding miracle! Now there is still the continuity equation $\nabla_a T^{ab}~=~0$, which is related to Bianchi identities etc, but the energy is the projection of a manifold basis vector on the momentum energy tensor

$$E^a~=~\Big(\int_{V^3}~-~\int_{V'^3}\Big) e_bT^{ab}$$

which is defined in a region of four spacetime bounded by three bounding spatial surfaces. The generalized Stokes' law then tells us this is the same as the differential of the integrand evaluated on the enclosing four spacetime

$$E^a~=~\int_{V^4}d(e_bT^{ab})$$

which can be expressed according to the covariant derivative. This will give a covariant derivative on the basis vector with $de_a~=~{\underline\omega_a}^b e_b$, and the differential one form is a connection form. This is coordinate dependent and so the energy can't be localized. In the above case with cosmology, this is a similar result, but here the energy can't be defined globally and conservation of energy is not a global law.

A cosmology with a nonzero cosmological constant, or what is likely a parameter set into a constant (or approximately constant) value by the inflaton or dilaton in a $spin(4,2)~\sim~su(2,2)$ model, is one where in general there is no unitarily equivalency between states in all regions of the cosmology. Even if the spatial surface is flat the accelerated expansion of the cosmology means that there is no such equivalency, and this comes about because there is no Killing vector K which when it acts on the energy KE = const. Without a Killing vector of this sort it means there is no isometry in the spacetime which maintains a constant energy on all paths in the cosmology. So the unitarity inequivalence of vacua in the earliest universe, where a vacua of unitary states is defined on a region $\sim L_p$ in a superposition of other such vacua on about the same scale, is frozen into the classical cosmology after inflation. In a more general setting the Coleman-Mandula theorem is then a local principle. This gives the maximal set of symmetries of the S-matrix as the $(0,~1/2)\oplus(1/2,~0)$ spinorial Lorentz group for external symmetry, an internal symmetry $[A_i,~A_j]~=~c_{ijk}A_k$, and the discrete CPT symmetry. The "maximal" extension on this is called supersymmetry. A cosmology with a non-zero cosmological constant necessarily means this is a local law, it does not apply globally. This is likely a source for what we call dark energy.

I will leave this at this point. This might sound odd, but this is a tremendous blessing. This is not something physicists should try to bury away, but embrace it. If thought about properly the consequences are astounding.

Lawrence B. Crowell

 

[sweetser]

Local energy

Hello Lawrence:

It is good to read that my powers of prediction are spot on.

Classical theories, such as the Maxwell equation of EM which can also be integrated seamlessly with quantum mechanics, allows one to define energy at a point. That utterly fails for Einstein's GR. Experts in differential geometry claim this bug is a feature. People who know bugs are bugs, and to distrust people claiming bugs are features.

Actually, I was off on tone:

This might sound odd, but this is a tremendous blessing. This is not something physicists should try to bury away, but embrace it. If thought about properly the consequences are astounding.

I would like to explain this conflict in a friendly way. As usual, I learn a few more things about the standard approach to dealing with technical issues from Lawrence's posts. If one understands GR in a non-trivial way as Lawrence does, then the logic of his argument is spot on. There is no misstep along the way. I was aware of the "take home message":

This is coordinate dependent and so the energy can't be localized. In the above case with cosmology, this is a similar result, but here the energy can't be defined globally and conservation of energy is not a global law.

If this is in fact the way that Nature handles energy for gravitational systems, then Lawrence is correct to say the consequences of nonlocal energy are astounding. There might even be links to inflation or dark energy. The opportunity to make a contribution in these areas justifies the excitement reflected in Lawrence's post.

Anyone getting paid a working wage in gravity today would endorse in various tones the logic presented by Lawrence. This is because the logic is consistent, and GR is the only game that pays a working wage. Back in the early days of GR, there were people who thought this was a significant issue (I am not enough of a GR historian to know who took which sides, or how the debate evolved).

Let me start my minority view with the words for a pop tune:

There's nothing I hate more than nothing
Nothing keeps me up at night
I toss & turn over nothing
Nothing could cause a great BIG fight
Hey -- what's the matter?
Don't tell me nothing.

It is my unwavering belief that the vacuum state, that volume of spacetime that is devoid of all events, can accomplish nothing. From the perspective of logic, there are no events to do anything. This apparently conservative belief is radical. It goes against many research themes viable today: the false vacuum need by the Higgs mechanism, perhaps a key to inflation, perhaps a key to dark energy.

This will not cause a great BIG fight. My ultra-conservative view will be summarily dismissed as indicating I do not understand the issue. I recognize that there are measurable effects created by the variation of the energy of the vacuum. These effects, real as they are, cannot be engine that flings entire galaxies apart at a greater rate. The deviation from the average measured amount of energy is not energy.

The one and only true engine of the Universe must be math done right.

The road between a classical field theory to a quantum theory is well defined, even if not discussed often. I read this first in a quantum field theory book by Kaku. The idea is that you take the field equation, which related the potential to the source, and then invert this equation to get the propagator used in Feynman diagram calculations. How straightforward! There are a few hitches, such as gauge theories like the Maxwell equations and GR cannot be inverted until one picks a gauge. This is one reason I am excited that I have expressed the action both EM and GEM using quaternion operators. Those expressions were gauge free because the gauge was explicitly subtracted away when constructing the action. Yet the field equations are necessarily invertible because that is a property of a division algebra. Have I constructed the propagators from my fields and done a few quantum field theory calculations? No, not a one. I would like to do so, but would need technical help, or a pretty significant time block to try and work it out. I do think it holds promise.

Since the 1930s, people far brighter than myself have tried to figure out how to quantize an approach to gravity. One consistent theme holds: despite the current excitement measurable in the day, all efforts have failed. There are people today excited about super symmetry, others are into loops. I think every last one of these sincere people are wrong for the same reason. If energy cannot be localized, then one can find a place where the energy is zero, and will not be able to invert the field equations to get a propagator because dividing by zero is not allowed by ultra-conservatives like myself.

In an if...then statement, if the clause in the if is not true, nothing in the then clause matters. From my lofty station in the Independent Research forum, I am stating clearly that the "If GR is true" clause is false. I have a an alternative: "If GEM is true, then...". A rank 1, linear set of field equations would be a cake walk for those skilled in the quantum field theory arts. It would be so easy, that ease would be a reason people would dismiss it. Ironic, but true.

Nothing you said sounded odd to me. I am taking aim at the foundations which are in the action of any field theory.
Doug

 

[Lawrence B. Crowell]

All actions are Lagrange multipiers on the bare action

$$S_b~=~\int pdq$$

and give a set of constraints required to solve the DE that emerge from the Euler-Lagragne equations. In ADM GR the constraints are $NH~+~N_iH^i~+~\dots$, where the rest are Gauss' law results and the rest on the number of sources. The gauge term is required because constraints on gauge fields $1/4 F^{ab}F_{ab}$ are insufficient. So an additional gauge terms is required in the Lagrangian to constrain four additional variables, say in the case of EM.

As a classical theory of gravitation I think basic GR is correct. This might be a "bias," but I let that be as it is. Some results of this are a bit striking, in particular with respect to cosmology. It also leads to some strange results when you look at quantum fields in curved spacetime. I think these are simply forced upon us, and quantum gravity requires that we generalize or abstract certain canonical aspects of physics. In particular it demands that we quantize gravity, or maybe as I have said to "gravitize the quantum," in ways which manage noncompact group structure for exterior symmetries that exist locally, and to "patch" these together (atlas-chart constructions in diff-geom) in some consistent manner for a global theory.

Now one can say that GR is incorrect, but frankly I think this is in line with some late 19th century prosaic ideas about fixing gravity so it is not a 1/r^2 for but a 1/r^n force for some n = 2 + a small bit. To do new physics right it often requires that certain aspects of current physics be abandoned in order that different aspects of the world are viewed as aspects of a single system. I could go at great lengths about this with general relativity and quantum mechanics as "relationships" between particles or observables. One is geometric and local, the other is nonlocal and does not describe these relationships according to metric geometry. I am not sure it is of value to write about this in great depth here.

As for energy conservation, or the conservation of a component of a momentum energy tensor projected out by a local basis element, it simply turns out that in general this is not conserved. In particular for a cosmology with a time dependent metric it is not possible to define energy conservation within standard approach. Some people say this is a "disaster," but frankly to me it means that the vacuum on a local region is not unitarily equivalent to a vacuum "out there." This requires that we address the problem of what we mean by a vacuum in a QFT, where much of what physicists think of might just be an aspect of local quantization.

At any rate if one adopts your view that "anything goes," as Cole Porter said, with Lagrangians then maybe one can make any theory, even if it has no proper connection to differential geometry. Your symmetric field tensors and the rest have no connection to differential or algebraic geometry (where the latter is required for quantum fields) and so in effect you might be able to blast your way to what you want. Yet at the end of the day you might find, if you have not already, that few people are really paying any attention.

Lawrence B. Crowell

 

[sweetser]

GEM has a n=2 force law

Hello Lawrence:

You certainly are free to believe that GR is correct, and I hope I have shown respect for that practice. As a gambling man, GR is a safest bet on the table. Some of the security in working with strings is that one can pick out GR within its formalism. I prefer to call it a belief to a bias since it gives direction to studies on a topic.

It is my belief that GR is not correct. This is part of what guides my research. This is a minority belief. People who hold minority beliefs get falsely accused of many things. Here is one in your post:

Now one can say that GR is incorrect, but frankly I think this is in line with some late 19th century prosaic ideas about fixing gravity so it is not a 1/r^2 for but a 1/r^n force for some n = 2 + a small bit.

Why is GR consistent with a 1/R² force law? Because $g_{00} = 1 - 2 G M/c^2 R$, take the derivative, and out pops the 1/R² force law. There is absolutely no difference between this line of reasoning and the GEM proposal. If the GEM proposal was an n=2+delta, I would consider that a deal breaker, a solid reason to reject this line of research. It is unfortunate that you thought GEM does not have an n=2 exactly force law.

One difference between Newtonian theory and GR is in a force formulation, there will be a 1/R³ term in GR, whereas there will be no such term in Newton's proposal for gravity. GEM also has exactly the same term. It is the coefficients for the 1/R⁴ that are about 10% different. The difference between GR and GEM is subtle, and according to my knowledge - and a direct question to Clifford Will - there is no experimental data to second order PPN accuracy for weak gravitational systems, and no experiments are being funded at this time to look to that level of precision of static sources.

The GEM proposal could not be a 19th century idea because new math is necessary. Any well-trained physicist who has read the Feynman lectures on gravity would know that the phase of the current coupling term $J^{\mu} A_{\mu}$ would have spin 1 symmetry for the transverse wave. That must be the case for EM, a transverse wave where like charges repel. If one asked about the other parts of that analysis, the scalar and longitudinal parts, one would see it had spin 2 symmetry, the particles needed by gravity where like charges attract, and the flight of photons is changed by gravity. You may have missed that issue, but it is an essential 21th century line of reasoning.

The newest addition is the Even representation of quaternions as a 4D commuting algebra. Nothing is ever completely new: a similar idea is packaged under the name of hypercomplex numbers. That introduces a new imaginary number, the hypercomplex number, but in practice of the Even representation, no new number is needed, just a new representation, based on a real 4x4 matrix. It was shown how this is a division algebra once the Eigen values are excluded. One of the cool things is that for some quaternions, this will exclude any quaternion living on the light cone. Neat, since the Even representation is related to gravity whose particles do not live on the light cone.

As for energy conservation, or the conservation of a component of a momentum energy tensor projected out by a local basis element, it simply turns out that in general this is not conserved. In particular for a cosmology with a time dependent metric it is not possible to define energy conservation within standard approach. Some people say this is a "disaster," but frankly to me it means that the vacuum on a local region is not unitarily equivalent to a vacuum "out there."

If you believe in GR, this is an important and productive issue to think about. I try and be more cold and analytical, not labeling it a "disaster", just a reason why GR, great as it has been, is flawed and must be rejected.

Energy conservation is looking better these days for the GEM proposal. Recall how Lut said I needed gauge invariance to get conservation of energy. I also know it is needed so that the gravitons travel at the speed of light. That was accomplished in post #442 for gravity alone by subtracting it away, and in post #457 for the GEM field equations by a fortuitous cancellation between gravity and EM.

"Anything goes"? I have gone to great effort to construct a theory consistent with experimental tests of weak field gravity to first order PPN accuracy. It differs slightly at second order PPN accuracy for spherically symmetric, static sources. My action is well defined. I have made the necessary link to spin 2 in the current coupling and field strength tensors. Because of my belief that GR is wrong, it eliminates the need to make what you label a "proper connection to differential geometry". In practice, that is always some road to the Riemann curvature tensor or Bianchi identities or some antisymmetric system. There is some antisymmetry in my proposal, just enough to get the job done for EM, no more than that, a rare case of intellectual minimalism. I do use a dash of differential geometry in calculating the Christoffel symbol of the second kind for the Rosen metric, getting a GM/R potential out of the exercise, showing I work with exactly n=2 for the force law.

The only thing I care about are technical arguments about the GEM proposal. I can number the exchanges I have had with Ph.D. level physicists. These have always had a one sided nature: they tell me what I need to work on, I do so, but what I get done does not make it back, they are too busy. Again, this is an observation. I told Alan Guth I had a unified field equation. He said I needed a field theory, with the Lagrangian defined, the field equations derived by varying the action, find solutions, show those solutions are both consistent with current tests and different at higher order. That took more than two years of work, but I did it. It is impossible to explain something long and complicated to the man, he will fall asleep, I have seen it happen :-) The entire issue about spin was created based on comments from Steve Carlip. In your own indirect way, I found the graph for both the Hamilton and Even representation of quaternions based on discussions here.

Yet at the end of the day you might find, if you have not already, that few people are really paying any attention.

There is nothing technical in this comment, it is all social. People get paid enough money to pay for mortgages studying GR and slight variations of GR. At the current time, there are no international meetings on quaternions. They will be on the stage at the International Conference on Clifford Algebras which happens once every three years. This thread is over 40k views which is considerably more than most threads at Physics Forums.

I am patient on resolving technical issues, and indifferent to the social aspects.

Doug

 

[Lawrence B. Crowell]

Hello Lawrence:

Why is GR consistent with a 1/R² force law? Because $g_{00} = 1 - 2 G M/c^2 R$, take the derivative, and out pops the 1/R² force law. There is absolutely no difference between this line of reasoning and the GEM proposal. If the GEM proposal was an n=2+delta, I would consider that a deal breaker, a solid reason to reject this line of research. It is unfortunate that you thought GEM does not have an n=2 exactly force law.

Doug

I indicated this by way of comparison. There is a standard theorem that a central force given by $F~\propto~r^n$ determines closed elliptical orbits for $n~=~\pm 2$. There were proposals to modify Newtonian gravity to account for the precession of Mercury's orbit.

To be honest what you are doing is to impose symmetric terms into a field tensor, which must necessarily be zero, in a way so as to break gauge symmetry. You then figure this is some sort of coup, for now your theory requires no gauge fixing term or gauge condition ---- apparently seen as some auxilliary "baggage" you have eliminated. The "procedures" you go through are comparatively elementary, which would suggest that if these were appropriate that some physicists would have discovered and applied them by now. After all people such as Feynman were quite on the bright side of the intelligence scale. These symmetric field terms you fold into the theory, which are presumed to account for gauge conditions, must either be necessarily zero or if not then you are breaking up gauge symmetry. In the first case this means your theory would in some way reduce to standard YM gauge field theory, or if not then you are saying that field theory is not at all about gauge theory. You did make a statement about principal bundles and the rest being irrelevant. Hermann Weyl was one of the smarter guys to do theoretical physics and proved that EM was a gauge system.

As for quaternions, they are noncommutative and were employed initially by Maxwell in his EM equations to account for the curl-equations. To be honest a lot of what you say about quaternions does not make much sense, or at least what you are calliing quaternions appear to be something else.

Lawrence B. Crowell

 

[sweetser]

Particular omission

Hello Lawrence:

I indicated this by way of comparison.

Fair enough. I felt it necessary to point out that this particular comparison was not accurate on a technical level. You still have your serious reservations, but this is not one of them.

The "procedures" you go through are comparatively elementary, which would suggest that if these were appropriate that some physicists would have discovered and applied them by now. After all people such as Feynman were quite on the bright side of the intelligence scale.

The logical conclusion based on accepting this analysis is that one should not bother to try and do physics, unless one could demonstrate in a measurable way they were smarter than Feynman. One of Feynman's chief characteristics is to challenge anybody, no matter what their station. It is ironic that you place Feynman on an unreachable pedestal.

I do not put Feynman there. He was amazing, and human. Let's challenge Feynman specifically on the completeness of his analysis. Do you have "Feynman Lectures on Gravitation" in your possession? If not, one can look inside at amazon.com to review the pages in question (pages 31-34). He does an analysis of the current coupling term in EM, $J^{\mu} A_{\mu}$. He takes the Fourier transform of the potential to get a current-current interaction. He restricts the analysis to a current moving along the z axis. The goal is to figure out the phase of the transverse current, the Jx and Jy terms (he used J1 and J2 in his lectures). The equation on page 34 is the basis of the statement that for EM, the transverse wave will take a 2$\pi$ radian change in phase to get back to where it started, a property of spin 1 particles. Spin 1 particles mediate a force where like charges repel, so photons can do the work of gravity.

His reasoning is both flawless and incomplete. He does not considered the rho-Jz current coupling (or J3-J4 in the lecture). For that term, the J and J' work together, and because they work together, it will require a change of $\pi$ radians to get back to the start. That is a property of particles with spin 2. Feynman's analysis was not complete, which is different from wrong.

During my first three years at MIT, I played poker against a guy named Rob that went on to win the World Series of Poker. We were roughly the same level, although we both knew Dean was better. Rob kept up and intensified his study of the game, although I did not. I respect the conservative bet that it is unlikely that anyone posting to Independent Physics at Physics Forums has not found something new. Aware of these odds, what I actively look for are areas that have not been explored. I know neither Feynman or Einstein worked with quaternions. I know P.A.M. Dirac was asked if he was interested in a formulation of relativistic quantum field theory with quaternions (in other words, the Dirac equation), and after pausing a really long time as was his way, said he would only be interested if they were the real-valued quaternions (the person asking the question was crushed, since he had worked with complex-valued quaternions). This implies that Dirac did not work with 19th century quaternions.

Rob won that year on the flip of the last card, where is opponent got a flush, but he pulled a full house. In this particular instance, I have documented how Feynman's published analysis was not complete. It is nice it ties in so closely with the content of this thread.

I always get skeptical when I see "quotes" around "words". It usually indicates a breakdown in communication. A differential equation written with a division algebra should always be invertible. This would be a great time to cite such a proof for that assertion, or just do the proof myself, but I know my limitations.

The games with gauges are precise. One needs EM theory as well as gravity to be invariant under a gauge transformation for two reasons. First, both the particles that mediate the forces - the photon and the graviton - travel at the speed of light. Second, according to Lut, it is much easier to demonstrate a gauge theory conserves energy.

So what exactly do I mean by gauge theory in the context of the GEM proposal? Consider this scalar field:

$$g=\frac{\partial \phi}{\partial t} - \frac{\partial Ax}{\partial x} -\frac{\partial Ay}{\partial y} -\frac{\partial Az}{\partial z}$$

Not a one of these terms ends up in the field equations in GEM field equations derived in most 438, 442, or 457. You are free to let $\nabla . A = 0$, known as the Coulomb gauge. You could work in the static gauge, setting $\frac{\partial \phi}{\partial t} = 0$, or the Lorenz gauge, $\frac{\partial \phi}{\partial t} + \frac{\partial Ax}{\partial x} + \frac{\partial Ay}{\partial y} + \frac{\partial Az}{\partial z} = 0$.

Peter Jack was the first person to write operators with real-valued operators to generate the Maxwell field equations. A year later, I did that trick independently. We has the mark of independent researchers: we did not do it the right way, just a way that worked. Post 438 is significant because the derivation is the first to use a quaternion to generate the Lagrange density. Once one gets E² - B², the rest is completely triple grade A standard field theory. And I got the Poynting vector as a freebee. An accident? I wouldn't bet against that one. Elegance is an essential guide in the search for truth.

One of the steps used there is familiar to anyone who decides to play with quaternions, and that is to eliminate the scalar using a conjugate, q - q*. That is what was done on the road to E² - B². We know the Maxwell equations is gauge invariant, and when the Lagrangian is formulated with quaternions, the reason is clear: it got subtracted away. Nice. Do the same exercise with the Even representation of quaternion - an idea from February 2008 - and one gets the field equations which are the natural relativistic form of Newton's law of gravity.

There was no way I could have planned it, but to find the unified field theory, do the exact same as EM and G separately, just skip the q - q* business. Then both Maxwell and G toss in the squared gauge, but with opposite signs, so they drop. An accident? See above comment again.

I also need to formulate GEM in a way that is not invariant under a gauge transformation. This will apply to the multitude of particles that do not travel at the speed of light. I haven't done that yet here, I got distracted defending the virtue of this work, and preparing talks, and living life (broken arm managements, yadda, yadda).

It is clear that the word symmetric is of concern to someone steeped in the technical nature of approaches to GR. As often repeated, I am not doing a variation on GR, I am doing a variation on the Maxwell equations. As you may know, one needs to supply the background metric as part of the mathematical structure of the Maxwell equations to put it to use (people usually use a flat metric, but it is a choice). There is no differential equation to solve that constrains what the metric can be.

What I am doing is a variation on the Maxwell equations, just barely enough to provide a differential equation whose solution is a dynamic metric based on the physical conditions.

It is a fine thing to question how a symmetric component could integrate into the math structure of GR to provide a non-zero result. As a variation on GEM, the concern is silly. Here are the undoubtedly non-zero terms found in the fields of Maxwell:

$$E = -\frac{\partial A}{\partial t} - \nabla \phi$$
$$B = \nabla \times A$$

And from the same soil, here are the two fields I refer to as the symmetric analogues needed for gravity:

$$e = \frac{\partial A}{\partial t} - \nabla \phi - 2 \Gamma_\nu^{\mu 0}A^{\nu}$$
$$b = -\nabla \Join A - 2 \Gamma_\nu^{i j}A^{\nu}$$

where $\Join$ is defined as the symmetric curl, composed of the same terms as the curl, but all the signs are positive.

The fields of E and B are manifestly free to be non-zero. The fields e and b, no matter what labels we attach to them, are also free to be non-zero.

So how are these four different? If one decides to work with a metric compatible, torsion-free connection, then the way a dynamic metric changes will not change a calculation of the fields E and B, but will change e and b.


There are many claims on the Internet about Maxwell and quaternions. The idea of the curl was due to quaternions, as was the gradient, the dot product, the divergence, scalars and vectors. In the first edition, he used pure quaternions, where the scalar is equal to zero. That is a different way to write a 3-vector. The pure quaternions were removed by the third edition. In the introduction, he predicts that someone will someday figure out how to do all the work with quaternions, a point of pride for me, completing a task Maxwell defined.

Maxwell would not have been concerned about the issue of spin 2 symmetry, it was before his time. He certainly wouldn't be concerned with the divergence of the Christoffel of the Rosen metric. Smart guy, but imperfect at seeing into the future, a common problem.

Doug

 

[jillz]

ok; lots of great info, but what is the effect of mass on height??

ok; lots of great info, but what effect does mass have on height??
If all parameters remain constant except the mass varies from 1kg to 10kg, the maximum height remains the same, so what's the effect of mass on height?

 

[Mentz114]

Doug & Lawrence,

this is getting repetitive. You're not going to agree for the simple reason that Doug doesn't see any deep truth in the 4D manifold or the theorems of diff geom or gauge theory ( even while claiming that 'elegance' is a guide to truth ). Where does the 'truth' lie ? You are both convinced that your theoretical structures are 'right', but it's experimental facts that will decide not the mathematics you espouse.

Doug, your epic posts are getting too long. All the biographical detail does not add to the physics, or make your arguments more convincing. Didn't someone say 'brevity is the soul of something' ?

It's time to shut up and calculate !

Lut

 

[sweetser]

Three Lagrangians to evaluate

Hello Lut:

Scientific conflicts are hard to manage. They quickly degenerate to name calling, which Lawrence and I have so far avoided (the personal stories might be one way of dodging that descent, plus I think people unskilled in these arts might find those side notes entertaining). Yes, the underlying difference in opinion stands, and I have no doubt will remain. The focus of exchanges has shifted, and that is informative, at least to me.

To stay rooted in calculations, have you been able to get to field questions for these three Lagrangians:

Maxwell:

$$(\nabla A - (\nabla A)^*)(- A \nabla - (A \nabla)^*) = (0, -E ~+~ B)(0, E ~+~ B) = (E^2 ~-~ B^2, 2 E \times B) \quad eq 1$$

Gravity:

$$(\nabla A2^* ~-~ (\nabla A2^*)^*)(- A2 \nabla^* ~-~ (A2 \nabla^*)^*) = (0, -e ~+~ b)(0, e ~+~ b) = (-e^2 ~+~ b^2, -e \Join e ~+~ b \Join b) \quad eq 2$$

And GEM:

$$\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))$$
$$= ((-g, E ~+~ B)(g, -E ~+~ B) ~+~ (g, e ~+~ b)(g, -e ~+~ b))$$$$=(-g^2 ~+~ E^2 ~-~ B^2 ~+~ g^2 ~-~ e^2 ~+~ b^2, 2 E \times B ~-~ e \Join e ~+~ b \Join b ~+~ 2 gE ~+~ 2 gb) \quad eq 3$$

I can do that with pencil and paper and Mathematica. How is your tensor software doing with the challenge?

Doug

 

[Lawrence B. Crowell]

I think Lut has a point. I do have to say that you are working up equations which have n-chains equated to m-chains for m =/= n. This is in part what your scalar equation stuff does, which leads you to the symmetric curl, as you call it, which does not mathematically exist. Sure, once you spin something like that up from whole cloth you can then do calculations with it. Yet even if those calculations are flawless they are ultimately based on mathematical nonsense.

Yet it is clear I am not going to make you see these points. I indicated something about some recent measurement of the orbits of neutron stars and their agreement with GR in ppN expansion. Ultimately this is where the verdict will lie. At this point it may come down to a choice between a theory well grounded in differential geometry and one with questionable differential geometric content. If you are right then it not only overturns GR, but it means that the foundations of manifold mathematics from Gauss to Riemann and all the way up to Taubs and Atiyah needs to be seriously modified as well.

Lawrence B. Crowell

 

[CarlB]

Dr. Crowell's objection reminds me of the objections to Lisi's E8 paper; that one should not be allowed to add together spinors and scalars.

Rather than get involved in the particular objections, I'd like to note that the value of a theory is in its ability to predict reality, not its ability to be mathematically clean in some theoretical sense. What we need is the ability to calculate, not the ability to theorize.

For relativity and quantum field theory, we have absolutely no experimental verification of these theories. What we have instead is good experimental verification of GR calculations and excellent experimental verification of QFT calculations. All theory that lies below the level of calculation is junk DNA that was convenient to frame the calculations, but need not be a part of a newer, more general, theory. The successful calculations, on the other hand, must be retained or replaced with equivalent.

 

[Lawrence B. Crowell]

I have decided to resurrect my little site here I started a year or two ago. I worked up an interesting idea on quantum fields in curved spacetime. This is very simple, only relying upon some basic ideas of geometry in QM and a fibration.

https://www.physicsforums.com/showthr...=115826&page=2

I worked this up in my head as I wrote this, so there might be a boo-boo or two here, but I think the basic idea looks reasonable. In way of boo-boo I posted this notice on my site as well.
Lawrence B. Crowell

 

[Lawrence B. Crowell]

Dr. Crowell's objection reminds me of the objections to Lisi's E8 paper; that one should not be allowed to add together spinors and scalars.

Lisi's paper does add them, but with the application of elements of the Clifford algebra. How Garrett frames spinors and scalars is similar to the application of Grassmannians in supersymmetry. Lisi is a bit glib on this, but it is not fatal to the basic architecture of his paper.

Lawrence B. Crowell

 

[K.J.Healey]

You mean (for a working link) : https://www.physicsforums.com/showthread.php?t=115826&page=2

 

[sweetser]

Three sign changes

Hello Lawrence:

To keep things short and snappy per Lut's request, if this exists:

$$(\frac{\partial Ay}{\partial z} ~-~ \frac{\partial Az}{\partial y},\frac{\partial Az}{\partial x} ~-~ \frac{\partial Ax}{\partial z},\frac{\partial Ax}{\partial y} ~-~ \frac{\partial Ay}{\partial x})$$

...and this does not mathematically exist:

$$(\frac{\partial Ay}{\partial z} ~+~ \frac{\partial Az}{\partial y},\frac{\partial Az}{\partial x} ~+~ \frac{\partial Ax}{\partial z},\frac{\partial Ax}{\partial y} ~+~ \frac{\partial Ay}{\partial x})$$

then I am more than willing to challenge the status quo of manifold mathematics because it is an error of omission, not a mistake per se.

Doug

 

[Mentz114]

Hello Lut:

To stay rooted in calculations, have you been able to get to field questions for these three Lagrangians:

Maxwell:

$$(\nabla A - (\nabla A)^*)(- A \nabla - (A \nabla)^*) = (0, -E ~+~ B)(0, E ~+~ B) = (E^2 ~-~ B^2, 2 E \times B) \quad eq 1$$

Gravity:

$$(\nabla A2^* ~-~ (\nabla A2^*)^*)(- A2 \nabla^* ~-~ (A2 \nabla^*)^*) = (0, -e ~+~ b)(0, e ~+~ b) = (-e^2 ~+~ b^2, -e \Join e ~+~ b \Join b) \quad eq 2$$

And GEM:

$$\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))$$
$$= ((-g, E ~+~ B)(g, -E ~+~ B) ~+~ (g, e ~+~ b)(g, -e ~+~ b))$$$$=(-g^2 ~+~ E^2 ~-~ B^2 ~+~ g^2 ~-~ e^2 ~+~ b^2, 2 E \times B ~-~ e \Join e ~+~ b \Join b ~+~ 2 gE ~+~ 2 gb) \quad eq 3$$

I can do that with pencil and paper and Mathematica. How is your tensor software doing with the challenge?

Doug

I've been playing about with the third one,

$$E^2 - B^2 -e^2 + b^2$$
which comes down to

$$(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})$$

with the condition that

$$(\partial^{0}A^{0})^2+(\partial^{1}A^{1})^2+(\partial^{2}A^{2})^2+(\partial^{3}A^{3})^2=0$$

to get rid of g^2. This is as far as I've got, but I'll get back to it when my schedule allows.

This thread is interesting, please take the time to have a look at it.

https://www.physicsforums.com/showthread.php?t=192422

 

[sweetser]

Check that software

Hello Lut:

It is good to focus on eq 3 which represents the GEM Lagrangian, instead of separately Maxwell and gravity, is a good one to focus on. I don't think it is correct to impose the condition:

$$(\partial^{0}A^{0})^2+(\partial^{1}A^{1})^2+(\partial^{2}A^{2})^2 +(\partial^{3}A^{3})^2=0 \quad eq 4$$

What happens algebraically is that a -g² cancels a +g². You need to see exactly the same thing happen with your software.

As a practical programmer type, I appreciate your shortcut to the scalar. The scalar is the same, so should flow through from there the same. The theorist does not agree, because your algebra is working based on a particular gauge constraint - no different from any other gauge theory - while eq 3 has a gauge cancellation. Because eq 3 is based on cancellation, there are infinitely more choices for the gauge that work with eq 3 than for your software that dictates the sum of the squares of the partial derivatives.

Don't toss away this work on

$$(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) \quad eq 5$$

That may well be the Lagrangian used for GEM that applies to massive particles. The Lagrangian for the force mediating particles that travel at the speed of light because they are gauge invariant due to cancellation - eq 3 - is not the same as the one for massive particles which must break gauge symmetry in an elegant way, perhaps exactly like eq 5. I have to look at the details of this to see if I like it :-) Lots of things going on, preparing for a few talks, APS talks in New London, Connecticut and St. Louis, Missouri.

Just read the first post in the thread you suggested. My initial reaction - which will be fun to see if anyone else made the argument - is the spin symmetry of the field strength tensor is not consistent with a spin 1 particle needed for like charges to repel.

Doug

 

[Mentz114]

Hi Doug:

OK, I'll find a way to deal with g.

As a practical programmer type, I appreciate your shortcut to the scalar. The scalar is the same, so should flow through from there the same. The theorist does not agree, because your algebra is working based on a particular gauge constraint - no different from any other gauge theory - while eq 3 has a gauge cancellation. Because eq 3 is based on cancellation, there are infinitely more choices for the gauge that work with eq 3 than for your software that dictates the sum of the squares of the partial derivatives.

Sorry, I don't understand a word of that.

I assume it's correct that $$E^2 - B^2 -e^2 + b^2$$ is $$(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})$$ with the g part removed.

 

[sweetser]

Hello Lut:

The question is why does g make no contribution in the GEM proposal, specifically eq 3 of post 500. There are 4 A's required to generate the squared fields:

$$\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))$$

The first two have curls, the last two have symmetric curls. The gauge g is definitely not set to zero. The g² contributed by the first pair cancels with the second pair. You can tell if your software has faithfully done this because g is not set to zero, but none of the components of g appear in the final result.

In the first pair, it is the order of the differentials that changes. In the second pair, it is which one gets conjugates. The contraction of the asymmetric tensor $\nabla_{\mu} A_{\nu}$ does not do this.

Doug

 

[Mentz114]

Getting rid of g

Doug:

The easiest way is just to ignore it, so we start by defining the field tensors,

$$F_{(g)}^{\mu\nu} = \[ \left[ \begin{array}{cccc} 0 & e_x & e_y & e_z\\\ e_x & 0 & b_z & b_y \\\ e_y & b_z & 0 & b_x \\\ e_z & b_y & b_x & 0 \end{array} \right]\]$$

$$F_{(em)}^{\mu\nu} = \[ \left[ \begin{array}{cccc} 0 & -E_x & -E_y & -E_z\\\ E_x & 0 & B_z & -B_y \\\ E_y & -B_z & 0 & B_x \\\ E_z & B_y & -B_x & 0 \end{array} \right]\]$$

$$L_{(g)} = F_{(g)}^{\mu\nu}F_{(g)}_{\mu\nu} = b^2 - e^2$$

$$L_{(em)} = F_{(em)}^{\mu\nu}F_{(em)}_{\mu\nu} = B^2 - E^2$$

and with the usual definitions of E,B, e and b -

$$E^i = \partial^0A^i - \partial^iA^0$$

$$e^i = \partial^0A^i + \partial^iA^0$$

$$B^i = \partial^jA^k - \partial^kA^j, i <> j,k$$

$$b^i = \partial^jA^k + \partial^kA^j, i <> j,k$$

from which I finally get this
$$B^2 - E^2 + b^2 - e^2 =$$
$$(\partial^{x}A^y)^{2}+(\partial^{x}A^z)^{2}+(\partial^{y}A^x)^{2}+(\partial^{y}A^z)^{2}+(\partial^{z}A^x)^{2}+(\partial^{z}A^y)^{2}-(\partial^{t}A^x)^{2}-(\partial^{t}A^y)^{2}-(\partial^{t}A^z)^{2}-(\partial^{x}A^t)^{2}-(\partial^{y}A^t)^{2}-(\partial^{z}A^t)^{2}$$

The next step is to apply Euler-Lagrange, which doesn't seem to lead anywhere. Where did I go wrong ? ( Apart from losing some factors of 4 and butchering the notation !).

[later] I seem to be getting

$$\Box^2A^{\mu} = 0$$

Too tired to continue right now. I've got a feeling I've made a meal of something simple.

 

[sweetser]

Managing the Even representation

Hello Lut:

You were only suppose to get the cross terms :-) I scanned in my hand-drawn derivations of the field equations here:

http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations/photo#5171869024067390082

The step that may be tripping up your program is multiplying the even quaternion representation that is (0, b + e)(0, b - e) should give the scalar b² - e², not the negative of this. My bet is B² - E² - b² + e² would work in your software as is.

Doug

 

[Mentz114]

Doug:

Yes, I changed a sign and now I get

$$B^2 - E^2 - b^2 + e^2 =$$

$$8\partial^xA^2\partial^yA^1 +8\partial^xA^3\partial^zA^1 +8\partial^yA^3\partial^zA^2 -8\partial^tA^1\partial^xA^0 -8\partial^tA^2\partial^yA^0 -8\partial^tA^3\partial^zA^0$$
(apologies for mixed notation, but I'm sure you get the drift)
In agreement ( up to a sign) with your doodle. I won't repeat the algebra to get the field equations.

I suppose the only point of this is that one can start with the traceless field tensors and go from there to get the same formulation as your funny quaternions. Much easier for the lay ( non-quat.) person to grasp.

Lut

PS : before I closed down the calculator I did this -

$$g_{ik}F_{(g)}^{mk}g_{mn}F_{(em)}^{ni}$$. It comes to zero, zip, nothing.

So the field tensors are orthogonal in a way.

 

[Mentz114]

Another way to lose g

Doug:

there is a neat way to lose g. The symmetric field tensor has a dual, in analogy with EM theory, defined thus -

$$\tilde{F_g}_{mn} = s_{mnij}F_g^{mn}$$

where s is the totally symmetric pseudo-tensor equal to |e|. The dual has no diagonal terms and

$$\tilde{F_g}_{mn}\tilde{F_g}^{mn} = -8(b^2 - e^2)$$

So, putting the square of the dual in the Lagrangian density does the job.

The symbol s also allows the symmetric curl of a vector to be defined as $$s_{ijk}\partial^{j}E^{k}$$

Lut

 

[Lawrence B. Crowell]

Doug:


[later] I seem to be getting

$$\Box^2A^{\mu} = 0$$

Too tired to continue right now. I've got a feeling I've made a meal of something simple.

Which is the Lorentz gauge. If you put the g in there and let this be something "dynamical" then this inserts group dependent structure into the theory. The elliptic complex determines the bundle curvatures $\Omega^{2}(ad(g))$ which give fields "mod-g," for here g means group. The gauge condition is set to constrain the appropriate variables.

Lawrence B. Crowell

 

[sweetser]

Orthogonal tensors

Hello Lut:

I concur with the postscript statement:

$g_{ik}F_{(g)}^{mk}g_{mn}F_{(em)}^{ni}$ It comes to zero, zip, nothing. So the field tensors are orthogonal in a way.

When I thought about GEM in terms of tensors, the symmetric one was the average amount of change in the 4-potential in 4 spatial dimensions, while the EM tensor is the deviation from the average amount of change. My recent shift to quaternions puts a stress on the underlying math tools, the Hamilton versus Even representations. At this time I don't understand all the links between these two perspectives, but it is nice having multiple perspectives. Thanks for adding to my views via this calculation.

Doug

 

[sweetser]

Hello Lut:

I am not clear on post #514. $F_{(g)}^{\mu\nu}$ has nothing down its diagonal - where the terms of a gauge live - so its dual also has this property. That appears to be starting too far down the page as it were.

Here is what I am doing with quaternions, written as your tensors:

$$\[ \left[ \begin{array}{cccc} g_t & e_x & e_y & e_z\\\ e_x & g_x & b_z & b_y \\\ e_y & b_z & g_y & b_x \\\ e_z & b_y & b_x & g_z \end{array} \right]\] - \[ \left[ \begin{array}{cccc} g_y & -e_x & -e_y & -e_z\\\ -e_x & g_x & b_z & -b_y \\\ -e_y & -b_z & g_y & -b_x \\\ -e_z & -b_y & -b_x & g_z \end{array} \right]\]$$

The second tensor is the conjugate of a symmetric tensor. It shows operationally why there is a significant difference between $F_{(g)}$ and $F_{(em)}$, where both define the conjugate as flipping the signs of the later three parts, but $F_{(em)}$ can do so by taking the transpose of the matrix representation. That does not work for $F_{(g)}$. Nice.

Doug

 

[Mentz114]

Hi Doug,

in post #514, I'm defining the dual of F_g using a pseudo-tensor, in analogy with the EM case. The pseudo-tensor s is something I just invented, because we go from EM/anti-symmetric -> gravity/symmetric. This dual is traceless and gives the right energy.

It's marginally less dodgy than inventing a new quaternion algebra ( maybe).

Lut

 

[sweetser]

Inventing math tools

Hello Lut:

I thought you might be inventing things. This is a good sign. It dovetails with Lawrence's complaint about my work not fitting into the proud tradition of differential geometry.

I think I referred to the wrong post. Post #514 is an extension of #511. It was #511 where you wrote out F_(g), saying one should just ignore the gauge. It would be great if you could write the software to make it so. I don't count setting to zero a good answer, or just ignoring it is good practice. I want to catch the math responsible for gauge symmetry. That is what is most interesting.

Doug

 

[Mentz114]

Hi Doug:

I understand your unwillingness to ignore, or set to zero the trace of the grav. field tensor. We can play at finding mathematical ways to dispose of it, but this might not throw light on the physical justification. I'm not sure why you call it 'the gauge'.

Anyhow, I'll try and justify my use of the 'symmetric dual' for the field tensor.

In EM theory, one gets the dual using the Levi-Civita antisymmetric pseuso-tensor like this,

$$\tilde{F}_{pq} = \epsilon_{mnpq}F^{mn}$$.....(1)

furthermore

$$F_{mn}F^{mn} = -\tilde{F_{mn}}\tilde{F^{mn}} = -2(E^2 - B^2)$$...(2)

So, to carry on the analogy between the EM theory, with anti-symmetric F and curl, to the gravitation case with a symmetric F and symmetric curl, we make a dual using a symmetric equivalent of the L-C pseudotensor, which we call 's' ( for 'symmetric').

$$\tilde{F}_{pq} = s_{mnpq}F^{mn}$$...(3)

from which it follows

$$F_{mn}F^{mn} - 2g^2 = -\tilde{F_{mn}}\tilde{F^{mn}} = -2(e^2 - b^2)$$...(4)

because $$\tilde{F}_{pq}$$ is traceless.

While I was typing the above, I realized that your field Lagrangian can now be very succinctly written as

$$(s_{\mu\nu\rho\sigma}\partial^{\mu}A^{\nu})^2 - (\epsilon_{\mu\nu\rho\sigma}\partial^{\mu}A^{\nu})^2$$......(5)

this has no g terms as required, with the two pseudo-tensors doing the work of the odd and even quaternions.

I still think GEM is wrong but the Lagrangian is tidier !

Lut

PS the software worked faultlessly and did all the donkey work.

 

[sweetser]

Using all modes of emission

Hello Lut:

Good work! I particularly like equation (5) and will make a mental note of it. Many group theory pros would object, saying the tensor is reducible, and reducible tensors cannot be used to represent fundamental forces. This issue has been raised indirectly in that thread you recommended reading in the first reply:

The "trouble" with gauge theories is that you are including unphysical degrees of freedom (such as the longitudinally polarized photon) and you have to make sure that they don't appear in the final calculations, and this complicates matters when you have to calculate things. However, despite the headaches, this can be done.

What I have argue since post #1 is the longitudinal mode of emission is a spin 2 graviton traveling at the speed of light doing the work of gravity. It is the scalar mode of emission that causes more serious problems since for a spin 1 field it implies negative probabilities.

People familiar with EM quantum field theory know these problems, and impose a constraint on quantizing EM such that the scalar and longitudinal modes are always virtual. Grad students grinding through these calculations often complain: it looks like a hack. Yet the hack needs to be there for a spin 1 field, and eventually they shut up and accept it.

The hack looks like a golden opportunity to put two modes of emission to work for gravity. Those modes will not harm EM theory in any way. A spin 2 scalar field would not have the indefinite metric problem, no negative probability problem.

No one else has picked up on the omission in Feynman's analysis of the current coupling term. I think that represents the greatest barrier to a rank 1 field theory.

Someone who bought a DVD from me thought I was within spitting distance of a unified field theory. I like that characterization slightly crude as it is.

Doug

 

[sweetser]

Everything is antisymmetric (for now)

Hello:

I thought I'd share a small epiphany I had since it sheds a bit of light on the conflict I have had with Lawrence. The best field theory we have is the Maxwell field equations for EM. The second rank field strength tensor is:

$$F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} \quad eq 1$$

It gives me confidence to know this is the foundation of my GEM proposal. Yet it also was the basis for the weak, the strong force, and Einstein's work on EM. For EM, the weak, and the strong forces, what changes is what group gets plugged into the machinery, U(1), SU(2), and SU(3) for EM, the weak, and strong forces respectively.

Gravity is a bit different, but not so different. I don't know the details, but I recall reading how Einstein looked to EM for guidance on how to proceed. Take a peak at the Riemann curvature tensor:

$$R^{\rho}{}_{\sigma \mu \nu}=\partial_{\mu} \Gamma^{\rho}{}_{\sigma \nu}-\partial_{\nu} \Gamma^{\rho}{}_{\sigma \mu}+\Gamma^{\rho}{}_{\mu \lambda}\Gamma^{\lambda}{}_{\nu \sigma}-\Gamma^{\rho}{}_{\nu \lambda}\Gamma^{\lambda}{}_{\mu \sigma} \quad eq 2$$

We have a subtraction thing going on again. This is a much more complicated animal, but we only need two contractions (the Ricci tensor minus the Ricci scalar), for the Einstein field equations.

Now I can see the resistance to tossing in a "+" for the minus in eq 1. That equation is good enough for EM, the weak, and the strong force, which covers 3/4 known forces of Nature. We have GR which has withstood every subtle test, while failing all large scale ones (velocity profiles of galaxies, the big bang, and our current acceleration). I can be at peace with that resistance. I don't think the objection to a symmetric field strength tensor will stand the test of time, but it is rational to not embrace it.

Doug

 

[Lawrence B. Crowell]

Gravitation differs from the gauge fields as an interaction determined by an exterior symmetry, rather than an inner symmetry modeled as a vector space on a principal bundle. The bundle structure is then a fibration of the symmetries of the space (spacetime) in an atlas of local tangent spaces. This is all spelled out in the Coleman-Mandula theorem, which gives a mathematical reason for the structure of interactions which can possibly exist. It does not tell us their explicit form, but as a "no-go theorem" of sorts it tells us what is not permitted. What you say above about "gravity being different" is frankly a bit on the silly side of things.

Lawrence B. Crowell

 

[sweetser]

No Lie algebra for a symmetric tensor

Hello Lawrence:

Thank you for clarifying the difference between an inner symmetry and an exterior symmetry. I do lack the experience to speak the jargon of what is known. I thought I had admitted I was speaking imprecisely.

So I go read up about the "Coleman-Mandula theorem, which gives a mathematical reason for the structure of interactions which can possibly exist [under the assumptions used in the theorem]." I hope you do not object to the clause, which is appears banal. Plucking out a one-line summary:

http://en.wikipedia.org/wiki/Coleman-Mandula_theorem]In[/PLAIN] other words, every quantum field theory satisfying certain technical assumptions about its S-matrix that has non-trivial interactions can only have a symmetry Lie algebra which is always a direct product of the Poincare group and an internal group if there is a mass gap: no mixing between these two is possible.

A Lie algebra has the property that it anti-commutes. If one makes a strategic decision to walk in a new direction with a field strength tensor [itex]\partial_{\mu} A_{\nu} + \partial_{\nu} A_{\mu}[/tex] which is a different combination of exactly the same players in $F_{\mu \nu}$ (when the symmetric field tensor trace is zero). One of the first observations is that any generating algebra for the group would have to commute, ergo any conclusions of the Coleman-Mandula theorem do not apply.

Sorry Lut, this is the same non-debate Lawrence and I have had before. I had to find out where the imposed domination of anti-commutivity was written.

Doug

Big fuzzy picture: If gravity is the force justifying why everything loves and attracts everything else, its algebra must be commuting.

 

[Mentz114]

Hi Doug:

You needn't apologise, it is always diverting to read your and LBC's posts, as long as they are short and to the point.

The gauge group of teleparallel gravity, the translation group, has a commuting Lie algebra.

I don't believe all mathematical theorems necessarily apply in physics, except in so far as they prevent inconsistencies. For example, von Neumann's 'proof' that QM cannot be modeled by 'hidden variables' is disproved by counter-example. Mathematical theorems always have lots of tight conditions that are often violated in practice.

Lut

 

[Lawrence B. Crowell]

Hi Doug:

You needn't apologise, it is always diverting to read your and LBC's posts, as long as they are short and to the point.

The gauge group of teleparallel gravity, the translation group, has a commuting Lie algebra.

I don't believe all mathematical theorems necessarily apply in physics, except in so far as they prevent inconsistencies. For example, von Neumann's 'proof' that QM cannot be modeled by 'hidden variables' is disproved by counter-example. Mathematical theorems always have lots of tight conditions that are often violated in practice.

Lut

I am not trying to say that the Coleman-Mandula theorem is proven in physics, where after all in science nothing is ever proven as such. It is just that the CM theorem is a fairly strong aspect of theoretical physics.

Teleparallel connections are commutative for special cases of nongeodesic motion. Christopher Columbus sailed to the new World on a nearly teleparallel path.

Lawrence B. Crowell