Unifying Gravity and EM

[sweetser]

The quantum equivalence principle

Hello:

Let's start with things researchers agree on:

1. A spin 0 quantum field is necessary to give fundamental particles like electrons and quarks inertial mass.

2. A spin 2 quantum field is required to mediate the effects of gravity.

3. The equivalence principle is a classical doctrine confirmed by experiments that the inertial mass is equal to the gravitational mass.

4. The correspondence principle shows that classical laws arise from the aggregate of quantum events.

In my limited exposure, folks who are searching for the Higgs particle do not bring up gravitons. Likewise, the gravity wave detector crowd appears unconcerned with the Higgs. Yet the equivalence principle must arise from an absolutely unwavering link between the quantum source for rests mass, a scalar field, and the quantum source for gravitational mass, a spin 2 field. The two fields can never "get out of step" with each other, or the equivalence principle will be violated, which it is not.

In the GEM proposal, the link between the scalar field and the spin-2 field is direct and rock solid: the spin 2 field for gravitational mass is the symmetric field strength tensor $\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}$, and the spin 0 field for inertial mass is the trace of the same symmetric tensor, $tr(g_{\mu \nu}(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}))$. Einstein was the first to argue the elegant case for the classical equivalence of gravitational and inertial mass. I am pleased the GEM proposal sings the same song on the quantum level.

doug

 

[Lawrence B. Crowell]

nonassociativity, octonions & field theory

The nonassociative product
$$(e_ie_j)e_k~-~e_i(e_je_k)~=~\upsilon_{ijkl}e_l$$
can be defined where $\upsilon_{ijkl}$ is defined without reference to octonions. However, this system will not satisfy a closed algebra unless it defines the Cayley four-form. One can of course restrict things to this rather myopic view.
\\
I am not sure what you mean by the Hamilton product. It appears that you are referencing a Jordan product in some form. The Hamilton product to me is
$$i^2~=~j^2~=~k^2~=~ijk~=~-1.$$
and a general quaternion is defined by
$$H~=~a\sigma_x~+~b\sigma_y~+~c\sigma_z~+~d{\bf 1}$$
In four dimensions the Pauli matrices are generalized to Dirac matrices.
\\
At any rate, nonassociativity without reference to the octonions means that there is no closed multiplication table defined by the structure constant $\upsilon{ijkl}$, of which there are 480 possible representations. I could go into the cyclic group and $PSL(2,~7)$ Hurzewitz discrete group theory behind this, but that gets too far into maths.
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Gauge theory has some sort of algebraic structure, eg. $SU(n)$, and the like. One of course defines gauge connections according to representations of this group. However, for there curvatures on the principle bundle are found by $F~=~dA~+~A\wedge A$, where the tensor components of the two-form $F$ in the case of an abelian field (EM) is the antisymmetric field strength tensor containing the electric and magnetic field components. The antisymmetry is imposed not by any group structure, but rather from a much more basic theory of chains and cycles on any manifold.
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Physics was not built from the ground up, but rather from a domain of experience that existed at the time. Faraday’s description of changing magnetic fields, currents and the like were based on observational experience of the time. Of course now we can easily see that this all stems from $F~=~dA$, and things appear elementary. Neither Faraday nor Maxwell had any notion of differential forms or integrating on chains in n-dimensional manifolds. Maxwell indeed had things according to spinning ether vortices and the like as his “description.” In part because of this his original theory was actually quaternionic. Now in our modern age we can see that these types of theories have antisymmetric structure because of these basic “facts,” and hold for any gauge theory --- even gravity.
\\
Lawrence B. Crowell

 

[sweetser]

Non-associative quaternion multiplication

Hello Lawrence:

This is the definition for a Hamilton quaternion product I am using:
$$i^2~=~j^2~=~k^2~=~ijk~=~-1.$$
While this is true:

The nonassociative product.
$$(e_ie_j)e_k~-~e_i(e_je_k)~=~\upsilon_{ijkl}e_l$$
can be defined where $\upsilon_{ijkl}$ is defined without reference to octonions.

it is not relevant because I am using a conjugate operator, or more technically an anti-involutive automorphism. The conjugate flips the sign of the 3-vector.

What I call a Euclidean product has a closed multiplication table:
$$i^{*} i= j^{*} j = k^{*} k= (i^{*} j)^{*} k = 1$$
The standard definition of a non-associative product you provide does not use a conjugate operator as I do. Because the automorphism keeps the quaternions within the quaternion manifold, the multiplication table is closed: $i^{*} i = 1$. Let's see this in action, how always taking the conjugate of the first quaternion in the binary quaternion multiplication makes the triple product non-associative.
$$a (b c) = a^{*} (b^{*} c) = a^{*} b^{*} c$$
$$(a b) c = (a^{*} b)^{*} c = a b^{*} c$$
These are not the same. The norm of these two are same, but they point in different directions.

Our current gauge theories for gravity and EM are both spectacular and inadequate. Both match experimental data extraordinarily well. Yet we cannot quantize EM without choosing a gauge. Even if a gauge is chosen, gravity cannot be quantized.

Let me reestablish how I am working from the "domain of experience". Gravity and EM as currently formulated work in 4 dimensions, as does GEM. It is odd that simple statement is at odds with the majority of work on gravity today. The group Diff(M) was used by Einstein to make his equations covariant, and it plays a role in GEM. EM symmetry must be broken by a scalar field, as it is in GEM by the trace of the field strength tensor. The 4D EM wave equation can be quantized by the Gupta-Bleuler method, but two of the modes of emission must be eliminated by a supplementary condition. GEM uses the same creation and annihilation operators, but it also has a spin 2 field that will not have the same technical issues the spin 1 scalar mode has. And in keeping with the most important of all scientific traditions, should we refine the level of accuracy of measuring light around the Sun, GEM predicts 0.7 microacrsecond more than GR, so unlike all the work in string theory, I am willing to put every chip I own on the line and let the data decide.

How things get written matters. You write the field strength tensor as $F = dA$. Nothing can be simpler than that. Sure, it has all the differential forms stuff behind it, but since that is the starting point, I can appreciate that there appears nothing simpler. I write it differently, also in an established way, as a $F=\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}$. In my notation, I can explain even to my girlfriend that the F thingie is not a complete story because of the act of subtraction: there are 4 parts in the triangle, 4 parts in the A, sixteen parts total, but the study of light needs only 6 of the 16. What I cannot explain to her is why the brightest folks on the planet do not share my feelings of obligation to work with the other 10 terms.

dougHistorical note: Maxwell was familiar with quaternions because Hamilton trumpeted their cause (too loudly). Scalar, vector, div, grad, curl were inventions of Hamilton in his quest to understand quaternions. In Maxwell's first edition of his treatise, he used "pure quaternions" which is one with a zero for the scalar, effectively as a stand-in for 3-vectors. He did nothing at all fancy with quaternion algebra. By the third edition, he only hoped that quaternions would play a role in describing EM. Many folks have figured out how to toss in an extra imaginary number and get the Maxwell equations. Peter Jack was the first person to do so with real quaternion (1996?) and I did the same independently a year later. There is a rumor on the Internet that some 200 quaternion equations were deleted from the first to the third edition. From my viewing of an on-line version of the first treatise, that is an inaccurate description: the quaternions were used like any 3-vector, and he did not write the homogeneous or source equations as an exercise in quaternion algebra.

 

[Lawrence B. Crowell]

Below is a multiplication table of octonions that I devised for fermions. It was derived with respect to black hole physics and Bogoliubov transformations --- but that is another story. These fields are the operators for ingoing and outgoing modes into a black hole. These fields include their conjugations. I hope this shows up right!
\\
$$\left(\begin({\array}{}* & * & e_1 & e_2 & b_k & e_4 & b_k^\dagger & b_{-k} & b_{-k}^\dagger \\ *& * & *& * & * & * & * & * & *\\ e_1 & &-1 & e_4 & b_{-k}^\dagger & -e_2 & b_{-k} & -b_k^\dagger & -b_k \\ e_2 & & -e_4 & -1 & b_k^\dagger & e_1 & -b_k & b_{-k}^\dagger & b_{-k} \\ b_k & & -b_{-k}^\dagger & -b_k^\dagger & 0 & b_{-k} & e_2 &-e_4 & e_1 \\ e_4 & & e_2 & -e_1 & -b_{-k} & -1 & b_{-k}^\dagger & b_k & -b_k^\dagger \\ b_k^\dagger & & -b_{-k} & b_k & -e_2 & -b_{-k}^\dagger & 0 & e_1 & e_4 \\ b_{-k} & & b_k^\dagger & -b_{-k}^\dagger & e_4 & -b_k & -e_1 & 0 & e_2\\ b_{-k}^\dagger & & b_k & b_{-k} & -e_1 & b_k^\dagger & -e_4 & -e_2 & 0}\end{array}\right)$$
\\
You say that (ab)c and a(bc) have the same norm but differ in their direction. That is standard with octionions. You might notice this if you examine this table.
\\
A couple of other points. A choice of gauge amounts to lifting the base manifold up the fibre. Think of the base manifold as a sheet of paper. Then perpendicular to each point is a line coming up, which for the case of triviality defines the tensor product of the vector space in the line with the manifold. Now to compute the action of this internal vector space one needs to move from fibre to fibre. But how does one do this? One takes the base manifold (the sheet of paper) and lifts a copy of it so it cuts through the fibres. One can lift the copy paper any way one wants, but once done things are computed with that choice. This is what a gauge choice is. One needs to define a bundle section in some way. It does not matter how this is done, but it must be done and used consistently. Well, 't Hoft and Veltmann did a gauge choice change in mid-stream with QCD to get renormalizability, but one has to be very careful about that! At any rate a gauge choice is nothing more than a way of fixing a frame, akin to defining a reference frame in relativity (or Newtonian mechanics for that matter) in order to do calculations.
\\
As for why only 6 of the possible terms end up as real is because the antisymmetry imposes a structure on the field. Anyway, we only have 3 E-fields and 3 B-fields. Imposing a structure on the theory involves the elimination of some of the possible tensor terms. The antisymmetry of gauge field 2-forms does that automatically.
\\
Lawrence B. Crowell

 

[Creator]

In the GEM proposal, the link between the scalar field and the spin-2 field is direct and rock solid: the spin 2 field for gravitational mass is the symmetric field strength tensor $\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}$, and the spin 0 field for inertial mass is the trace of the same symmetric tensor, $tr(g_{\mu \nu}(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}))$.
doug

Hello Doug;
In QFT isn't mass obtained, at least some fraction of it, by radiative corrections? How does that relate here ?

 

[sweetser]

Radiative corrections

Hello Creator:

The short answer is I do not know how to incorporate radiative corrections into the proposal.

I do know that for compound particles such as protons and neutrons, most of the mass is from the movement of gluons and quarks. Any radiative corrections would be just another source of energy to add into the picture. I am unable to calculate what percentage of the proton mass comes from the Higgs mechanism. I'm sure some folks on the planet know how to do that calculation, but I do not. The ice is very thin where I am skating!

doug

 

[sweetser]

Hello Lawrence:

Unfortunately, your table for the Octonians did not work out :-( I would love to see it. If you come back to the forum, there should be an edit button at the bottom if you have logged in. The first two lines look something like this:

\left(\begin({\array}{}* & * & e_1 & e_2 & b_k & e_4 & b_k^\dagger & b_{-k} &
b_{-k}^\dagger \\ *& * & *& * & * & * & * & * & *\\

So the "*"'s need to be replaced with what goes in there. A second effort for this multiplication table would be appreciated. Note that after the save button is hit and the page appears, the new graphic does NOT appear. The equation is a graphic which the computer keeps in cache. One must hit the reload button while holding down the control key to see the improvements.

doug, whose LaTeX here is NEVER done perfectly the first time.

 

[Lawrence B. Crowell]

I have an inordinate difficulty with the LaTeX here, which I do not have otherwise. Anyway, I have taken a dvi page with this table and prduced an image file. I am going to try to attach that to this post.

Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

Thanks for the dvi page. I am a bit confused about one aspect of this table. Octonions are a non-associative, non-commutative division algebra. The zeros that appear in the table would create a problem for the division algebra. I think those should be -1's. For completeness, the identity should be included. Here is a table from the web without zeroes: http://www.geocities.com/zerodivisor/obasis.html

doug

 

[Lawrence B. Crowell]

octonion table

The zeros are meant to convey fermionic content with b^2 = 0. The anticommutators of the Fermionic fields are "treated" as commutators in order to obtain associators

[a, b, c] = (ab)c - a(bc).

I suppose I should have considered the anti-associator

{a, b, c} = (ab)c + a(bc)

instead to avoid this matter. However, the basic import is that the xero reflect a topological content of fermions. Since b^2 = 0 this is equivalent to d^2 = 0, and that the fields are defined in ker(b)/im(b). This topological element is what skirts the problem with division algebras. I think this extends to the sedenions, but that is another issue.

Maybe this was not the best example for this, for in most cases there is none of this topological implication underlying it. I attach a more familiar octonion multiplication table.

 

[Lawrence B. Crowell]

Follow on with octonions

This might be a bit outside of the GEM theory, but I figured I would try to clarify a couple of things.
\\
Let $b_q~\rightarrow~\phi_q b_q$ for $q~=~\pm k$ and $\phi_q$ a scalar field that obeys
$$[\phi_q,~\phi_{q^\prime}^\dagger]~=~f(q,~q^\prime),$$
$$[\phi_q,~\phi_{q^\prime}]~=~g(q,~q^\prime)$$
Then the commutator of $\phi b_q$ is
$$[\phi_q b_q~,\phi_{q^\prime}b_{q^\prime}]~=~[\phi_q,~\phi_{q^\prime}]b_q b_{q^\prime}$$
$$=~1/2/{b_q,~b_{q^\prime}/}g(q,~q^\prime)~=~e_i,$$
for the appropriate i on the table. In this way one can treat fermions as nonassociative
//
The division algebra is one that want e_ie_j = e_k =/=0. If e_k is zero without either e_i or e_j being zero then there is said to be no algebra. At the level of octonions it is said that this is the final algebra.
//
The whole process of construction from reals, complexes, quaterions and octonions involves a pairing of each other. The simplest of course is the complex plane where $z~=~(x,~y)$ and the defined multiplication
$$c*z~=~(a,~b)*(x,~y)~=~(ax~-~by)~+~i(bx~+~ay)$$
$$(a,~b)*(x,~y)~=~(ax~-~by,~bx~+~ay)$$
The same goes for the quaternions, they are a pairing of complex numbers. Octonions are then in turn a pairing of quaternions. At each level one loses ordering, commutivity and finally associativity. This also reflects the so called Cayley numbers and the multiplication rule with pairing defines what is called the Cayley-Dickson algebra.
\\
The octonions are pairs of quaternions. Consider the octonion $O~=~(A,~B)$ and $O^\prime~=~(X,~Y)$. The multiplication of these two is then
$$O\cdot O^\prime~=~(AB~+~e^{i\phi} Z^\dagger B,~BX^\dagger~+~AP).$$
where the argument is usually taken as $\phi~=~\pi/2$. For a system of quaternions $\sigma_i$ and $\bf 1$ this defines four additional elements $e_i$.
\\
Now for quaternionic valued operators as fields which satisfy the BRST quantization condition $Q^2~=~0$, where $\psi~\in~ker(Q)/im(Q)$ gives the field as purely topological. In other words $\psi~\ne~Q\chi$. In this way it is possible to have the square of an element in the octonions being zero without it in a strict sense being unalgebraic. Of course in the octonions of operators e_je_j =/=0 for i =/=j.
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What is this good for? I might go on about this in greater detail in another post, but the Dirac operator $\Gamma^a\partial_a$ has the same topological information as the field. Further, in general the Dirac matrices (the quaterions) may have a representation which depends upon the chart on the base manifold. Thus a more general Dirac operator is $\partial_a\Gamma^a\_$. In this way the quaternions are operators.
\\
In the case of sedenions one has $e_ie_j~\ne~0$ and algebra is lost. There are eight octonions within it that are "islands" of algebra, but "outside" of them appears to be algebraic "chaos." However, I think that by Bott periodicity there is structure there and I think it is involved with some sort of topology. The sedenions define $S^7\times S^7\times G_2$, which probably constrains this topology.
\\
Lawrence B. Crowell

 

[sweetser]

A symmetry, not a field

Hello:

I spend time wondering why I have trouble communicating this proposal. The action looks like the simplest one that can be made! I have fun finding tension between what we now know and the GEM proposal. In this post I will outline an example.

Why is energy conserved in EM? Take a look at the action:
$$S_{EM}=\int \sqrt{-g} d^{4} x (-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))$$
Vary time:
$$\delta S_{EM}=\int \sqrt{-g} d^{4} x (-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu})) \delta t$$
There is no time in the action, so it is not going to be changed in the slightest: integrate over a nanosecond or a century, the integral will be the same under a variation in time. That makes the action symmetric under time transformations. Where there is a symmetry, there is a conserved quantity. In this case, that is energy, $E_{cons.}=m \frac{\partial t}{\partial \tau}$. The definition of energy conservation says that there is no change in the derivative of time with respect to the interval $\tau$.

Why is linear momentum conserved in EM? Look at the action above. No distance R appears in it, so like time, varying R will keep the variation in the integral at an extremum, and the conserved quantity is momentum, $P = m \frac{\partial R}{\partial \tau}$. The same story.

We know what mass is from special relativity, the difference between the square of energy and the square of momentum. Since energy and momentum both arise from a symmetry in the action, mass which is calculated from this two must also arise exclusively as a symmetry in the action.

That is not how it works for the Hilbert action of general relativity. Instead of hunting for a symmetry, the metric is treated as a field and varied.

In the GEM proposal, one looks at the definition of a covariant derivative to spot a symmetry. One can vary changes in the potential or changes in the metric (the connection). So long as the changes offset each other, there would be no difference in the integral of the action with those changes. There must be a conserved quantity. Since the metric is changing, it is reasonable to propose mass is conserved. Logical consistency appears to favor the GEM proposal.

Convincing someone who has made the investment in understanding general relativity at the nuts and bolds level may be beyond my reach, but I still like the view from my chair: it's drop dead gorgeous.

doug

 

[Lawrence B. Crowell]

There is something rather mysterious here. You say there is no time in the action for the Lagrangian [/itex]{\cal L}~=~1/4 F_{ab}F^{ab}$. Agreed there is no explicit function of time, but one does have elements$F_{0j}~=~-\partial A^j/\partial t~-~\nabla_jA^0$which are the electric field components. A variation with respect to time is going to pick out$\partial_0F^{0j}$terms of the form$\partial E/\partial t$, and$\partial_0F^{ij}$of the form$\partial B/\partial t$. These are just the time derivative parts of the Maxwell equations. In fact your$\delta S_{EM}$is wrong, for the variation with time is not going to involve just a multiplication by$\delta t[itex], but partial derivatives of this with time.
//
Lawrence B. Crowell

 

[Lawrence B. Crowell]

retry: symmetries & fields

There is something rather mysterious here. You say there is no time in the the Lagrangian ${\cal L}~=~1/4 F_{ab}F^{ab}$. Agreed there is no explicit function of time, but one does have elements $F_{0j}~=~-\partial A^j/\partial t~-~\nabla_jA^0$ which are the electric field components. A variation with respect to time is going to pick out $\partial_0F^{0j}$ terms of the form $\partial E/\partial t$, and $\partial_0F^{ij}$ of the form $\partial B/\partial t$. These are just the time derivative parts of the Maxwell equations. In fact your $\delta S_{EM}$ is wrong, for the variation with time is not going to involve just a multiplication by $\delta t$, but partial derivatives of this with time.
//
Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

There are partial derivatives with respect to time, but nothing that is just time. A delta t is not a t. One can imagine a Lagrangian where the effects dissapate after a certain amount of time. That is not what happens with the classical EM Lagrangian. If one comes back in 10 years, the delta t's will still be the same. That is the source of energy conservation as I see it.

I believe my $S_{EM}$ is standard, although most people write it in flat spacetime.

Looks like your first itex bracket should not have a /.
doug

 

[Lawrence B. Crowell]

Conservation of energy

The variation of the action results in the Euler-Lagrange equations, which are the equations of motion. Conservation of energy for fields comes from the momentum-energy tensor
$$T^{ab}~=~\partial{\cal L}/\partial g_{ab}~-~g^{ab}{\cal L}$$
with the continuity condition $\partial_cT^{ab}~=~0$. The variation of the action gives dynamical equations, or $F~=~ma$ stuff, and the momentum-energy tensor gives the conservation laws a'la Noether's theorem.
\\
Lawrence B. Crowell

 

[sweetser]

Equations of motion and forces

Hello Lawrence:

My equation about $\delta S_{EM}$ above is wrong. Partial derivatives with respect to t are required.

I can see that my description of the issue was informal, but not unconventional. Here is a quote from http://en.wikipedia.org/wiki/Noether's_theorem:

The word "symmetry" in the previous paragraph really means the covariance of the form that a physical law takes with respect to a one-dimensional Lie group of transformations which satisfies certain technical criteria. The conservation law of a physical quantity is usually expressed as a continuity equation.

The most important examples of the theorem are the following:

* the energy is conserved iff the physical laws are invariant under time translations (if their form does not depend on time)
* the momentum is conserved iff the physical laws are invariant under spatial translations (if the laws do not depend on the position)
* the angular momentum is conserved iff the physical laws are invariant under rotations (if the laws do not care about the orientation); if only some rotations are allowed, only the corresponding components of the angular momentum vector are conserved

A Noether charge is a physical quantity conserved as an effect of a continuous symmetry of the underlying system.

The form of the GEM and EM Lagrangians do not depend on time, so energy is conserved.

Landau and Lif****z gave me a really great lesson on this topic. I knew that once the Lagrangian is set, then everything can flow from that, be it equations of motion or dynamical equations or stress tensors. But how are these related? Let's start with the GEM Lagrangian:
$$\mathcal{L}_{GEM}=\frac{-\rho_{m}}{\gamma}-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu} -\frac{1}{4c^{2}}(\nabla^{\mu} A^{\nu}-\nabla^{\nu} A^{\mu})(\nabla_{\mu} A_{\nu}-\nabla_{\nu} A_{\mu}) - \frac{1}{4c^{2}}(\nabla^{\mu} A^{\nu}+\nabla^{\nu} A^{\mu})(\nabla_{\mu} A_{\nu}+\nabla_{\nu} A_{\mu})$$
Vary A and its derivative, keeping the velocity field fixed, and that generates the equations of motion, the stuff of the Maxwell equations. The second, third, and fourth terms come into play.

Now start with the same Lagrange density, but keep A and its derivative fixed, and vary the velocity V, which appears in the terms $\frac{-\rho_{m}}{\gamma}$ and in the current densities $J_{q}^{\mu}-J_{m}^{\mu}$. That will generate the Lorentz force. One can also get to the Lorentz force through the stress tensor. Only the first and second terms come into play.

One important thing to notice is the role played by the second term, the so-called charge coupling term. In EM, the equations of motion indicated like charges repel, which is the same message as arises from the Lorentz force equation. By having the opposite sign for the gravitational charge in the second term, like charges attract for the field and force equations.

doug

 

[Chronos]

Pardon my dumb question sweetser, but is this a quantum treatment?

 

[sweetser]

Possible to do quantum calculations

Hello Chronos:

I hope to give a sophisticated answer to your dumb question. Bur first, the short answer: their are concrete technical reasons to hope the GEM field equations can be quantized, but due to my own limitations, I have not done any quantum calculations, such as a scattering cross section of an electron fired at a proton. A well-behaved, finite calculation of a scattering cross section would demonstrate the proposal is consistent with quantum mechanics.

How does one know if a theory can or cannot be quantized? There are fancy answers, but I prefer simple ones (but not too simple). Zeroes for observables of a classical theory are bad. In a classical field theory, an observable such as energy or momentum is a number. To make the same theory a quantum theory, an operator must be found for that observable that acts on the wave function. The operator gets plugged into a commutator that should be some multiple of Planck's constant. If a classical observable is zero, the quantum operator is zero, and the commutator will be zero, not a multiple of Planck's constant. Failure.

That is exactly what happens to the classical EM Lagrange density:
$$\mathcal{L}_{EM}=-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))$$
What is the energy/momentum 4-vector? One thing not appreciated widely enough is the answer to this question is automatic: all it takes is a calculation, no thought involved, a no-brainer. One takes the derivative of this with respect to the time derivative of the potential A, and out comes the canonical energy momentum. I don't know if you can buy a book that does this in detail. It is not a hard calculation, but it is painful to do the LaTeX. I like to see all the details of an easy calculation, so here it is. First, this is the EM Lagrangian written out in all its component parts:
$$\mathcal{L}_{EM}= - \frac{1}{2} ( ( - ( \frac{\partial \phi}{\partial x} )^2 - ( \frac{\partial \phi}{\partial y} )^2 - ( \frac{\partial \phi}{\partial z} )^2 \\ - ( \frac{\partial A_x}{c \partial t} )^2 + ( \frac{\partial A_x}{\partial y} )^2 + ( \frac{\partial A_x}{\partial z} )^2 \\ - ( \frac{\partial A_y}{c \partial t} )^2 + ( \frac{\partial A_y}{\partial x} )^2 + ( \frac{\partial A_y}{\partial z} )^2 \\ - ( \frac{\partial A_z}{c \partial t} )^2 + ( \frac{\partial A_z}{\partial x} )^2 + ( \frac{\partial A_z}{\partial y} )^2 \\ - 2 \frac{\partial A_x}{c \partial t} \frac{\partial \phi}{\partial x} - 2 \frac{\partial A_y}{c \partial t} \frac{\partial \phi}{\partial y} - 2 \frac{\partial A_z}{c \partial t} \frac{\partial \phi}{\partial z} \\ - 2 \frac{\partial A_y}{\partial z} \frac{\partial A_z}{\partial y} - 2 \frac{\partial A_z}{\partial x} \frac{\partial A_x}{\partial z} - 2 \frac{\partial A_x}{\partial y} \frac{\partial A_y}{\partial x} )$$
Now take the derivative of the Lagrangian with respect to 4 things: $\frac{\partial \phi}{\partial t}, \frac{\partial A_x}{\partial t}, \frac{\partial A_y}{\partial t}, \frac{\partial A_z}{\partial t}$. Looks painful. Start with the first one, $\frac{\partial \phi}{\partial t}$, and you will notice there is no such term in the classical EM Lagrangian. Even if one's calculus is rusty, if a variable is not there, the derivative with respect to there's no there there is zero. This classical EM Lagrangian cannot be quantized for that reason alone.

So what do people do? They tack in a term to cover up this problem, and surround the patch job with discussions of gauge theory. The sport is called fixing the gauge. There are all kinds of super sophisticated things to say about this topic. I prefer the simple question: why is there this problem? The answer is also simple: there was a subtraction step in the Lagrangian, so of course you are missing something. All the GEM proposal does is keep all the parts of the 4-derivative of a 4-potential together so nothing is missing. Let's look at the GEM Lagrangian written out explicitly:
$$\mathcal{L}_{GEM}=- \frac{1}{2} ( ( \frac{\partial \phi}{c \partial t} )^2 - ( \frac{\partial \phi}{\partial x} )^2 - ( \frac{\partial \phi}{\partial y} )^2 - ( \frac{\partial \phi}{\partial z} )^2 \\ - ( \frac{\partial A_x}{c \partial t} )^2 + ( \frac{\partial A_x}{\partial x} )^2 + ( \frac{\partial A_x}{\partial y} )^2 + ( \frac{\partial A_x}{\partial z} )^2 \\ - ( \frac{\partial A_y}{c \partial t} )^2 + ( \frac{\partial A_y}{\partial x} )^2 + ( \frac{\partial A_y}{\partial y} )^2 + ( \frac{\partial A_y}{\partial z} )^2 \\ - ( \frac{\partial A_z}{c \partial t} )^2 + ( \frac{\partial A_z}{\partial x} )^2 + ( \frac{\partial A_z}{\partial y} )^2 + ( \frac{\partial A_z}{\partial z} )^2 )$$
That certainly looks complete. Calculate the canonical momentum:
$$\pi^{\mu} = h \sqrt{G} \frac{\partial \mathfrak{L}}{\partial ( \frac{\partial A^{\mu}}{c \partial t} )} = h \sqrt{G} ( - \frac{\partial \phi}{c \partial t}, \frac{\partial A_x}{c \partial t}, \frac{\partial A_y}{c \partial t}, \frac{\partial A_z}{c \partial t} )$$
Nothing is zero, so quantizing the theory is possible. [For fun, notice the units required to get mass*length^2/time^2.]

Being skeptical of my own skills, the only time I get confident is when I see that others have already done a nearly identical thing. The field equations written in the first post have been quantized, the energy and momentum turned into operators. The problem is that this was done for a massless spin 1 field of EM only. There are 4 modes of emission, but only 2 of them could be described by a massless particle (something about constraints on polarization because the particle is travel ling a the speed of light). The scalar mode is non-sense, allowing for negative probability. So Gupta makes the ad hoc supplementary condition whose entire purpose is to eliminate the scalar mode of emission as well as the longitudinal one.

In the GEM approach, the same operators for energy and momentum used by Gupta and Bleuler are used. Instead of just a spin 1 field doing only EM, GEM has two fields, a spin 2 for gravity, a spin 1 for EM.

The fact that I can open a graduate level quantum field theory book, go to the section on the Gupta/Bleuler quantization method, and point to the operators for energy and momentum, says the theory can be quantized. The next step is to do a calculation, for example the scattering of an electron off of a proton. It is known how to do this for EM. This is where Feynman diagrams come in. There would be two changes to go from EM scattering to gravity scattering: the coupling constants and the propagator. The couple constants are easy: $e^2 -> G m_{e} m_{p}$. If you plug in numbers, the coupling goes from weenie to extra-super-wimpy: $2.56 \times 10^{-38} C^2 -> 1.01 \times 10^{-67} C^2$. The propagator, the squiggly line between two nodes of a Feynman diagram, must change from a spin 1 to a spin 2 propagator. Weinberg wrote papers on this topic in the early 60's. I Xeroxed them, and knew I would never understand the contents. I gave those papers to a friend with a recent Ph. D. in cold matter physics from MIT over his Christmas break, and he could not decipher the information. For someone actively doing scattering calculations in quantum field theory, this calculation is probably simpler than what they are getting paid to investigate: it is a variation on a calculation they are trained to do, and variations are easier to perform than making up something new. I have no idea how to reach a person who could do that work, and at this time have no authority to compel them to crank through it.

doug

 

[Chronos]

Thanks for the thoughtful reply, doug. I didn't phrase the question very eloquently, but you went right to the heart of the matter. Quantization is a difficult prospect in the best of circumstances. It's not pretty, much less elegant, and that rubs me wrong. I've always been intrigued, and am curious what you think of this paper:

That strange procedure called quantisation
http://www.arxiv.org/abs/quant-ph/0304202

 

[sweetser]

Preparing for a talk

Hello Folks:

I have been preparing for a 15 minute talk at the 9th Eastern Gravity Meeting happening this Friday and Saturday at MIT. I am slated for noon, March 25. The title is "Unifying Gravity and EM, a Riddle You Can Solve". The slides are available, and the pdf has additional text approximating the main points I will say. Please feel free to check out the slides and give feedback.

I am also moving into a new home which takes mountains of time and effort, hence my delay in replying to Chronos.

doug

The slides:
http://theworld.com/~sweetser/quaternions/talks/riddle/riddle.html
[note: The page uses css & javascript, use the click, arrows, pageup and pagedown to advance]

The slides + comments:
http://theworld.com/~sweetser/quaternions/ps/riddle.pdf

 

[sweetser]

Response to talk

Hello:

The 9th Eastern Gravity Meeting at MIT was full of folks who work with Einstein's field equations for their daily bread. Some are working on trying to make gravity wave detectors, others work with computers trying to develop models of spiraling black holes that crash into each other. There was one string theory guy who still does not have any way to test the proposal.

My observation of the meeting in general was there was very little communication between different workers. It was only clusters of folks who happened to be working on related problems that could bring up decent questions. That may just be the nature physics: it's too technical today to be a generalist.

I can say with some confidence that no one trained in physics today works with a vector proposal for gravity anywhere in their education. It is either Newton's scalar theory or Einstein's rank 2 theory. I don't consider the two paragraphs a piece in MTW, or papers by Gupta and Thirring to count (I am embarrassed by the logical weaknesses involved in those short comments).

There was one work on general relativity - I cannot remember which one - that after introducing the covariant derivative, pointed out that one could take the divergence of the connection:
$$\partial_{\mu}\Gamma_{\sigma}{}^{\mu\nu}A^{\sigma}$$
The author then quickly pointed out that there is no need to ever do this sort of calculation because the term does not transform like a tensor. If we want second order derivatives of the metric that transform like a tensor, then we MUST use the Riemann curvature tensor. The logic is that simple and straightforward.

What's wrong with this logic? It is an omission. If you put the divergence of a connection in the right place, it transforms like a tensor:
$$\partial_{\mu}\partial^{\mu}A^{\nu}- \partial_{\mu}\Gamma_{\sigma}{}^{\mu\nu}A^{\sigma}$$
If the potential happens to be constant, the result is the divergence of the connection, the thing that was dismissed as silly to bother to calculate. This has second derivatives of the metric in a term far simpler that the Riemann curvature tensor.

My sense of the project is this: I will not be able to communicate this proposal until by some chance combination I find someone else who actually does that calculation for the Rosen metric. If you don't go through the exercise, you don't see it. That was the [non]reaction I got. Oh sure, I can be a bit more entertaining, the graphics are easier to read than the average presentation, but there was no effective communication about the core idea of the talk. I don't think there is anything too unusual about that.

doug

 

[Lawrence B. Crowell]

talk & tensors

To be honest the objection appears valid. Consider this term
$$\partial_a\Gamma_b^{ac}A^b$$
then the gravity connection transforms by U(1) by
$$A^b~=~U^{-1}A_b'U~+~U^{-1}\partial_b U$$
the partial on the gravity connection will transform homogeneously by local Lorentz transformations. However, this whole term does not transform homogeneously under the gauge action
$$\partial_a\Gamma_b^{ac}A^b~=~\partial_a\Gamma_b^{ac}(U^{-1}A_b'U~+~U^{-1}\partial_b U)$$
The second order term $\partial\partial A$ (indices suppressed) does not cure this, since this will transform homogenously under the gauge action and thus contains no inhomogenous terms which cancel out the inhomogeneous term $U^{-1}\partial_b U$. So this is not a tensor. Thus the objection is valid.

Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

We both agree that $\partial_a\Gamma_b^{ac}A^b$ does not transform like a tensor. That is a non-issue.

I will always start instead with this:
$$\partial^a A^c-\Gamma_b^{ac}A^b$$
This is a tensor. This is the definition of a covariant derivative I am starting with. I choose to work with a potential $A^b$ that is constant, so the derivatives of $A^b$ are zero. Then I act on it with a partial derivative, $\nabla_a$ (Note: I used the triangle there because it too has a partial derivative and a connection). The second derivative of the potential does not get stapled in at the end. Instead I start only with valid tensors, and act on those tensors with tensor operators.

U(1) is an approximate symmetry of this proposal, not an exact symmetry. For the special case of massless particles, the U(1) symmetry is exact. Mass breaks U(1) symmetry extremely weakly (1 part in 10^16 for an electron). You are correct that the connection breaks the U(1) symmetry, but that is not a constraint on the proposal. If one only works with tensors throughout, the result must be a valid tensor expression.

Breaking symmetry with in this way is a good thing. It means that the Higgs particle and Higgs mechanism are unnecessary. In the standard model, the mass of an electron has to be included via the Higgs and the spontaneous symmetry breaking of a false vacuum. I use something we already associate with mass - the metric, and the changes in the metric or connection - to break very slightly U(1) symmetry with something physically relevant.

doug

 

[sweetser]

Hello:

I think I am going to be a bit more assertive in the coming months about what I may have accomplished. Unifying gravity and EM in a way that can be quantized should get people's attention. In practice, delivering that line is not enough. This confuses me, because whenever someone makes that claim to me, being a good skeptic, I investigate, and a little digging usually determines there is no math there (and thus for me, no content).

Prof. Clifford Will is going to MIT to give a talk. He is a fellow who is an expert on experimental tests of general relativity. He wrote a great review article on the topic in living reviews (long paper, can be found via Google). Someone suggested I read it because it was an exhaustive survey of approaches to gravity. I read it carefully, and like everyone else, he did not mention the possibility of a vector field theory for gravity. He did cite vector-scalar and vector-tensor theories, but no plain old vector. See, Newton's field theory for gravity is the simplest scalar theory that can be written. It is still a darn good theory, used in all rockets except those that carry atomic clocks. The simple scalar theory has a technical flaw: it does not respect the speed of light limit.

A better theory is the simplest rank 2 field theory, that is general relativity. It too has a technical flaw: it cannot be quantized.

The only thing between the simplest rank 0 and the simplest rank 2 field theory is the simplest rank 1 field theory. It should be a big, honking red warning light that someone like Will has this sort of sin of omission. So I will be going to his talk and seeing if I can point this out to him. The odds are very bad - it is like pointing out a shade of green to someone who is color blind. He has zero experience with a vector field theory for gravity, $Jq^{\mu}-Jm^{\mu}=\nabla^{2}A^{\mu}$.

doug

 

[nrqed]

Hello:

I think I am going to be a bit more assertive in the coming months about what I may have accomplished. Unifying gravity and EM in a way that can be quantized should get people's attention. In practice, delivering that line is not enough. This confuses me, because whenever someone makes that claim to me, being a good skeptic, I investigate, and a little digging usually determines there is no math there (and thus for me, no content).

Prof. Clifford Will is going to MIT to give a talk. He is a fellow who is an expert on experimental tests of general relativity. He wrote a great review article on the topic in living reviews (long paper, can be found via Google). Someone suggested I read it because it was an exhaustive survey of approaches to gravity. I read it carefully, and like everyone else, he did not mention the possibility of a vector field theory for gravity. He did cite vector-scalar and vector-tensor theories, but no plain old vector. See, Newton's field theory for gravity is the simplest scalar theory that can be written. It is still a darn good theory, used in all rockets except those that carry atomic clocks. The simple scalar theory has a technical flaw: it does not respect the speed of light limit.

A better theory is the simplest rank 2 field theory, that is general relativity. It too has a technical flaw: it cannot be quantized.

The only thing between the simplest rank 0 and the simplest rank 2 field theory is the simplest rank 1 field theory. It should be a big, honking red warning light that someone like Will has this sort of sin of omission. So I will be going to his talk and seeing if I can point this out to him. The odds are very bad - it is like pointing out a shade of green to someone who is color blind. He has zero experience with a vector field theory for gravity, $Jq^{\mu}-Jm^{\mu}=\nabla^{2}A^{\mu}$.

doug

Good luck Doug.

I am just starting to read this long and fascinating thread so I will read more before starting to ask questions that I am sure have already been asked by others. Of course, the first comment is that the force between like charges interacting through a vector field repel so it seems as if a gravitational force based on such a field would lead to gravitational repulsion. But I am sure you have debated that with others so I will read on.

Regards

 

[sweetser]

Hello:

There are two papers in the literature that also claim that a vector field equation must have like charges attract just like EM. That result comes from copy EM too closely! My replies to the question are in posts 23 and 33.

The key is one perfectly placed minus sign in the Lagrange density, which then appears in the subsequent field equations and in the Lorentz force laws. The charge coupling term for EM has the same sign as the rank 2 field strength tensor contraction. That is what generates like electrical charges that repel in Gauss's law. The same sign for the coupling term and the inertial mass term lead to the Lorentz force law where like electric charges repel.

By changing the sign of the charge coupling term, the couping and field strength tensor terms have opposite signs, and like mass charges attract. The different sign for the coupling term and the mass term lead to a Lorentz force law where like mass charges attract.

The signs are actually required due to the asymmetric field strength tensor. See, that gets split into two, an irreducible symmetric field strength tensor, and an irreducible antisymmetric field strength tensor. The symmetric rank 2 field strength tensor will be represented by spin 2 particles, and those must attract (read that in Hatfield's introduction to the Feynman lectures on gravity, go there if you want a good explanation). The antisymmetric field strength tensor will be represented by spin 1 particles, and like charges will repel.

Sorry for my delay in replying, the site forgot to send me email. This is an issue my proposal must get right, or it is not worth bothering anyone.

doug

 

[sweetser]

Interaction with Prof. Clifford Will

Hello:

In this note, I will describe my interaction with Prof. Clifford Will of the University of Washington, St. Louis, an authority on experimental tests of gravity, that happened during his visit to MIT Thursday, May 18, 2006.

I am conflicted about my own body of work. On one hand, I consider the equations to be as gorgeous as the come. I've taught my mailman the field equations which he remembers to this day with the mnemonic, "Always give 2 Brownies to Jim", the reverse of $J=Box^2 A$. The exponential metric is prettier that the algebraic fragment of the Schwarzschild metric in the Schwarzschild coordinates, and completely trumps the Schwarzschild metric in isotropic coordinates with its 4th power terms (most folks studying GR probably aren't even familiar with it, but it is the form used in experimental tests). I can see the symmetries in the GEM Lagrangian that lead to energy, momentum, and mass conservation. As long as I keep focused on the equations, I can feel joy in their elegance.

On the other hand, I am familiar with every misstep made along the way, even in this forum. I am aware of my own inadequacies as a messenger, without any degree in physics, just the ad hoc collection of graduate-level courses audited from MIT, Harvard, and BU (Boston may be the best town to audit courses). It feels like a cruel cosmic joke that someone with such limited skills in math and physics should have found this cache of rocket fuel. As I age and see those limited skills shrink, so does my confidence.

For about a week, I thought about the questions I would like to ask Will. That was fun. Not fun was the self-doubt which can derail the task. I decided to wear the Turquoise Einstein t-shirt featured at the top of quaternions.com for two reasons: I like the shirt enough that it boosts my self confidence, and it has the Always Give 2 Brownies to Jim equation which might be useful in a technical discussion.

Will was to be at an MIT physics lounge at 3:45, with the talk in 10-250, a large hall, at 4:15. I showed up right on time. Will looks much like Ted Turner, with white hair and mustache. He was seated on a couch, the room sparsely occupied by small clusters of graduate students talking about grading undergrads. I went straight for the cookies, wondering when I should say hello. Rather than have a long debate, I decided to get it over with sooner rather than later.

Eye contact made, I told Will that he was the center of a small debate on the Internet (sci.physics.research several years ago). I was investigating a simple approach to gravity in 4D. A fellow suggested I read Will's Living Reviews article on tests of gravity because it covered EVERY respectable approach to gravity, bar none. I read it. It was a very good article (that brought a smile). The paper cited vector/scalar theories, and vector/tensor theories, but no simple vector-only theories. Newton's law of gravity is the simplest rank 0 field theory one can construct. It is a darn good theory, still used in the guidance systems of most rockets. If those rockets carry atomic clocks, we realize two technical limitations of Newton's law: that the speed of gravity is wrong (no speeds are infinite) and gravity bends spacetime more than Newton's law predicts. A better theory is the simplest rank 2 field theory that can be constructed to explain how gravity works - Einstein's field equations for general relativity. It really is a darn simple approach, with the Ricci scalar sitting alone in the Hilbert action. The speed of light is respected. All weak field tests are passed. All strong field tests are passed. The rank 2 field theory also has a technical problem: it cannot be quantized. The brightest folks in physics have all tried to no avail.

Between the simplest rank 0 field theory and the simplest rank 2 field theory is the simplest rank 1 field theory. This formal possibility was not discussed in this otherwise exemplary review article. I asked him what he thought about that omission.

He replied that there was no need to disprove a vector theory because we already had proof that a metric theory was required. All the tests of the equivalence principle were effectively tests that gravity must be a metric theory.

I nodded. I was aware that for my own efforts to succeed, I would have to precisely deflate this position. I may do that in a subsequent post here, but that was not my purpose in the discussion with Will. Rather, I had a rare chance to talk technically with an expert, and wanted to find out his perspective on several issues. I had it, so it was time to move on.

I pointed to my t-shirt, saying this was the 4-vector field equation I happened to study. Most physicists if asked would think it was the Maxwell equation in the Lorenz gauge, where like charges repel, and thus not applicable to gravity. If the metric had a +2 signature instead of -2, then like charges would attract. The key to understanding the proposal is that the box^2 is not a D'Alembertian operator (a scalar operator consisting only of the second time derivative minus a Laplacian operator). Instead the box^2 represents two covariant derivatives. A covariant derivative has a normal derivative minus the connection, so the equation reads: normal derivative minus a connection, applied to a normal derivative minus a connection, applied to a 4-potential. One could have a differential equation that was the divergence of the connection of the potential. I claimed that the Rosen metric (the one at the start of this forum) solves that very differential equation.

He shrugged his shoulders. That was too dense to follow, and he didn't. I offered to send him a copy of my paper. He told me he was extremely busy, and there was no way he would have the time to look at it. I took him at his word, and promised him I would not email him. I am not going to beg for people to look at elegant equations.

He asked if the theory was a full one. I told him I had a Lagrange density and the field equations from the action, along with solutions to the field equations that were physically relevant.

An objection I was sure people would raise is that the proposal has linear vacuum field equations. Like EM, once particles start interacting, things become nonlinear. My impression was that physicists believe that the vacuum field equations had to be nonlinear, that non-linearity was a way to decide if a proposal was mature or just a toy.

Will said there was no experimental data on the topic. After a bit of going back and forth, he said essentially that the collection of good candidate theories for gravity happened to all be nonlinear, so it was reasonable given that observation for there to be a sense that nonlinear field theory was the correct approach. The Schwarzschild metric was linear in one coordinate system, and nonlinear in another, so linearity is coordinate dependent, and of limited value.

I asked if there were any plans to do gravity tests to second order PPN accuracy. My proposal is identical at first order PPN accuracy, but about 12% different at seccond order PPN accuracy. He said those experiments are very demanding. There were nothing coming soon. He wondered if I had shown my proposal worked for the precession of the perihelion of Mercury. I claimed that it did, a four page calculation (available at quaternions.com). I also asserted that the lowest mode of gravity wave emission should be a quadrupole because the theory conserves momentum, and there is no other field to store momentum.

He nodded along, but was not engaged, so I thanked him for his time, shook his hand, and went back for another cookie. He eventually got off the couch and chatted with a few professors. I realized that had I been delayed, there would have been no way to have had such a long and detailed private discussion. He left at 4:05 to go to the lecture hall.

The talk itself was good. He covered tests of the equivalence principle, solar tests (weak field), binary pulsars (strong field), and gravity waves. He was one of the folks who developed the PPN system. For general relativity, $\gamma=1, \beta=1$, all 8 others are equal to zero. At the end of the lecture, I asked if there were any other theories where $\gamma=1, \beta=1$, all 8 others are equal to zero. [I was interested because this is a property of my own proposal.] He said Rosen, of Einstein, Poldolski, Rosen fame, came up with such a theory. It works for some tests, but for binary pulsars, the theory predicts the frequency should increase instead of the decrease observed. It is vital to check a proposal in all regions.

Someone asked a question on MOND, the Modification of Newtonian Dynamics that is consistent with some data on the velocity profile of disk galaxies (as well as other astronomical data sets). He said that someone had figured out a Lagrange density for MOND, he had seen it, and it was UGLY. I found that funny. The strength of the GEM proposal is that the Lagrangian is fit and trim. Can it match all the data collected in eighty years? Not in a brief private discussion.

I did feel it was a good day for the GEM proposal. I certainly did not convert Will: that would require a formal demonstration that the proposal works for all weak field tests, all strong field tests, all tests of the equivalence principle, and gravity waves as a starting point. On an informal level, it was not dismissed out of hand. It felt like there was stuff worth talking about. I was glad I had attended and asked as many questions as I did.

doug

 

[sweetser]

Metric OR Potential Theory

Hello:

One of the things that gives me confidence is when my points of contention with standard theory become subtle instead of confrontational. Too many folks brighter than myself have thought about these issues before. It is unreasonable to expect them to be wrong. It is natural for a blind spot to remain unseen.

Clifford Will has argued that gravity must be explained using a metric theory. He is not the first one to state that position. I've seen it espoused in Misner, Thorne, and Wheeler (the beginning of chapter 40 to be more precise). I agree, gravity must be explained with a metric theory.

Here is where the subtle issues start. The question not asked is how many ways are there to implement a metric theory? The standard approach is to work with the Riemann curvature tensor. If you are not familiar with this rank 4 monster, it is essentially the difference between two curved paths. It measures how much the curvature of spacetime is changing. I call it a monster because one time I tried to write it out in terms of metrics and it was a futile exercise, there were just too many terms. [More detail than necessary: the Riemann curvature tensor is the difference between two derivatives of a Christoffel symbol, and the difference of two products of two Christoffel symbols, and a Christoffel symbol itself has three derivatives of a metric that get contracted with another metric, so the details of the Riemann curvature tensor can be overwhelming.]

There is an implied message behind the pitch for a metric theory: that the approach cannot use a potential. I forget who did the calculation (someone between Newton and Einstein), but he showed that gravity will bend light. Einstein also did the calculation in 1911. The answer was exactly half right. It did make exactly the same prediction for bending the time term of the metric as does general relativity, the g_00 term of the Schwarzschild metric, which gets a little smaller than one. The error is that Newton's theory leaves the space part unchanged. Experimental tests show the g_11 term is a little greater than one.

What the data unambiguously proves is that a scalar potential theory cannot explain light bending around the Sun. The coefficient in front of the dt part of the metric is less than one, whereas the coefficient in front of the dR part of the metric is greater than one. A one parameter potential can not be both greater and lesser than one. Proof complete.

Let's shine light on the blind spot. A potential that has more than one parameter could be consistent with experimental data. If one used 100 parameters, it would be trivial to match any data set. The question is what is the smallest number of parameters needed? It has been established that one is too small. The next thing to try after a scalar tensor potential is a vector tensor potential. In spacetime, that would have 4 parameters. There is sufficient freedom with 4 parameters to match the experimental results.

Another subtle issue is that people if pressed would view this discussion as a metric theory versus a 4-potential theory. It cannot be that, since I have already agreed the theory must be a metric theory - up to a diffeomorphism. That is a fancy way to say I am proposing a metric theory and/or a 4-potential theory. Some might say that is a slimy political move, but I believe that is a beautiful symmetry slight of hand. It's magic! Remember, general relativity is magic too, saying gravity is exclusively about spacetime geometry, no force needed. Incorporating diffeomorphisms correctly to a vector theory allows me to say gravity is all about spacetime geometry or potential, or any combination of geometry and potential you choose. The GEM proposal is not only a unification of gravity and EM, but the union of Newton's exclusively potential theory with Einstein exclusively metric theory. That rocks!

This work keeps getting sexier than string theory :-)
doug

 

[sweetser]

Visualizing GEM

Hello:

More of our brain is devoted to visual information processing than any other, so it is vital to have a visualization of the GEM proposal. It can be found in this link:

http://TheWorld.com/~sweetser/quaternions/gravity/em2gem/figure_1.jpg

The image is a collage of two figures most people are familiar with. The Newtonian potential looks a little like a ladle, and particles hang out at the bottom of the dip. The lines on the graph paper in the background for the x and y coordinates are always as straight as can be drawn.

For Einstein's explanation of gravity, artists use a rubber sheet, with the mass stretching the sheet. Now a "straight line" - one that stays within the warped lines - looks curved to us from afar.

What the GEM theory proposes is that either or a combination of both images is correct. Newton's potential theory for gravity does not work because it can only bend one way, and spacetime needs at least four parameters to describe its bending as is possible with a 4-potential versus a scalar potential theory.

Newton's theory gets the light bending around the Sun half right. Specifically, it gets the bending of the time portion correct. If one used a metric that was flat for the time part ($g_{00} = 1$), but was bent the appropriate amount in the space part ($g_{11} = 1 + 2 GM/c^{2}R$), then the combination of a Newtonian potential with a curved-only-for-R metric would be consistent with experimental tests of gravity to first order PPN accuracy. The curved-only-for-R metric would not be a solution to Einstein's field equations which explain gravity as exclusively arising from spacetime curvature. It is unfortunate, but the GEM theory says that exclusive metric theories like general relativity are too restrictive. The potential well/rubber sheet symmetry must be embraced to unify gravity and light.

doug

 

[sweetser]

Pitching the program

Hello:

People have suggested I submit a paper to a physics journal. I have set a specific goal to achieve before I go through that process. I would like to find someone who understands general relativity and the Maxwell equations at the level of the action, have that person read my draft paper, and tell me if s/he thinks it is valid. Based on the ensuing give and take, I would either submit the paper, or post to my website (and here) why the proposal was flawed. In this post I will show my latest effort to find someone with the skills required to review my work on a technical level.

A few posts ago on this thread, I told how I had a productive discussion with Clifford Will, but he was too busy to read my work. The Internet was developed by physicists at CERN for physicists working together over the globe. It is my sense that in measurable ways, the pace of life for theoretical physicists is more frantic than most other professions. I expect doors to be closed not due to aloofness - Will was both approachable and polite - but because there is too much stuff crammed in the room to open the door.

Well-known physicists at MIT, my alma mater, are frantic. I've seen it up close when I worked at the bench for Prof. Richard Young. It is part of my ethics not to bother a busy scientist unless I meet one of two criteria: either I have a well-formed question or I have a result. Being a skeptic, it is important to me that at the very least Mathematica has checked the algebra contained in the draft (Mathematica did not alway approve of my efforts, but that is a different story). Professors that are less well-known are probably even more frantic, trying to establish a name.

What follows is my last email to a less well-know physicist, which references an email to a well-known physicist. The short story is the less-well known prof. has not written back, and the well-known prof. documented why he cannot read my draft. I am not mad or frustrated because I understand why they behave as they do. I remain persistent.

doug

--------
From: Doug <dougsweetser@gmail.com>
To: slloyd
Date: Jun 8, 2006 7:22 AM
Subject: Fwd: The work of the stand-up physicist

Hello Seth:

Dave Pritchard gave me your name (email exchange at the end). I am an
amateur physicist who has a Lagrangian for a unified field theory.
There have been a few times in the history of science when an amateur
- after a few pitches by mail - has mad a contribution. What is not
part of the lore is the number of appeals of limited value. I hope
the details in my note to Dave and the attached pdf file show it is
possible my work has some of the claimed significance.

I hope you get a chance to review the work. Thinking about the
action, the field equations, or the solutions to the field equations
gives me a quiet sense of joy (unusual for such abstractions).

Have a good day despite the rain,
doug

---------- Forwarded message ----------
From: Dave Pritchard
Date: Jun 7, 2006 12:30 PM
Subject: Re: The work of the stand-up physicist
To: sweetser@alum.mit.edu
Cc: Sarah Smith

Doug,

Right now I have about 6 papers of which I'm a coauthor (often senior
authro)to read through and correct - I can't take on another (I spent
the weekend dusting off 3 overdue referee reports ahd nave a big
proposal to review, and another one to do also). So adding to this list
is impossible.

Your ideas sound like something theorists might have tried.

[I'll reply here since it would have been impolite to reply to Dave.
The action would certainly have been tried back in the 19th century.
This was before the notion of a diffeomorphism was developed, a key to
riddle. Clifford Will wrote a Living Review article on the modern
view, and despite being over a hundred pages and quite thorough, it
never brought up the simplest vector theory, as a resonable proposal
between the simplest scalar theory, Newton, and the simplest rank 2
theory, GR. Will recently visited MIT, and I asked him why his paper
had this omission. He said that there is more than enough data to
indicate that gravity must be a metric theory. A potential theory
will only generate half the bending of light that has been measured.
He, like Pritchard, said he was too busy to view my work. Gravity
must be explained via a metric theory, I agree on that point. An open
but overlooked question is how to implement a metric theory. All
efforts to date have relied on the Riemann curvature tensor in some
way or another. I work from a simpler starting point, the connection
as it appears naturally in a covariant derivative. It is true that a
scalar potential theory of gravity can only get the smaller than one
g_00 term correct (the g_11 term of the Schwarzschild metric is
greater than one, and a single parameter cannot do both). It would be
imprecise to presume that a 4-potential theory would be insufficient
to achieve the results seen in experiment.]

[...back to Dave]

I think you might try to talk to someone in the field who's not so famous (Seth Lloyd at MIT has been working on GR and QM at MIT).

I wish you luck, and am sorry I can't help.

Dave

David E. Pritchard 617/253-6812
Associate Director, Research Laboratory of Electronics
Cecil and Ida Green Professor of Physics
Room 26-241
MIT
Cambridge, MA 02139
617/253-4876 fax
http://www.rle.mit.edu/pritchard
Education Research: http://relate.mit.edu



Doug wrote:
> Hello Dave:
>
> I was having lunch at Mary Chung's with Sarah Smith. I was lamenting
> my plight in theoretical physics, and she suggested I drop you a note.
>
> It would be great if gravity and EM could play nicely with each other.
> That is not the case today, with general relativity standing outside,
> refusing all efforts to be quantized.
>
> Mathematica and I think we have a solution: a Lagrange density, the
> field equations generated by varying the action, physically relevant
> solutions to those field equations, the force law, a dynamic metric,
> an appreciation of the group theory perspective for the proposal, and
> two experimental tests to distinguish the approach from general
> relativity. That's a long list. My problem is that I am an amateur
> physicist, no funding from the government or industry. I know it
> takes an hour and a half to go from the classical EM Lagrangian to the
> Maxwell equations. In my approach, I toss in another current and a
> symmetric field strength tensor to do the work of gravity right next
> to EM. The vacuum field equations for gravity are every bit as linear
> as EM, and so quantization is just like EM, but with a spin 2 field so
> like charges can attract.
>
> Of course I should just publish, but a man has to know his
> limitations. I am an amateur. I do not have a Ph.D., a masters, or
> even an undergraduate degree in physics. In the 90's, I sat in on one
> graduate physics class a semester. I worked at a lab bench, and know
> how to shut up and get more data. Writing a technical paper in
> physics is not a craft I have mastered. I have attached to this email
> a pdf draft of my best effort to date. It is highly probably that I
> have a garbled line or two in the text that I am unable to spot. That
> is all that is needed for a paper to be rightly rejected.
>
> I know you are not an expert in the study of gravity. If an amateur
> and a symbolic math program can figure it out, so can you. It is a
> variation on the Maxwell equations, so it requires just as much work
> as the Maxwell equations to understand. You probably know from
> teaching that most folks can get Newton's scalar law of gravity.
> Electromagnetism is far more difficult to teach. No one bothers to
> teach general relativity to undergraduates (almost no one at least,
> Edwin Taylor is giving it a try). My work requires an amount of
> effort between Maxwell and GR, which is to say it takes considerable
> work.
>
> In some ways, the thesis is all about doing no work. That is what
> life in spacetime is like, flat as a pancake. Only there is a little
> bit of other stuff around, so everyone does the very least they can,
> which would be simple harmonic oscillation. Do a SHO in 4D, and there
> are two transverse modes for light, longitudinal and scalar modes for
> gravity. The rest is details.
>
> There are a lot of details. At APS meetings I get all of 12 minutes
> to explain five hours of differential equations. You want to write a
> novel and put it in a fortune cookie format? I have a web site, but
> who could possibly click through that much math? I am the kind of
> fellow who prefers a creative solution no matter what the cost.
> People will listen to someone telling stories, to someone teaching
> with a passion. So I decided to create a community access TV show,
> "The Stand-Up Physicist", and it has a companion web site,
> www.thestandupphysicist.com. The show serves as an outlet for me.
> These equations for unifying gravity and EM are drop-dead gorgeous.
> It all works in 4D, no ten or eleven dimension BS needed. It makes
> sense physically and mathematically. If experimentalists could
> measure the bending of light three orders of magnitude better than
> done today, they might see the 0.8 microarcseconds more bending my
> theory predicts than general relativity. It is funny and a little
> tragic that someone with my limited skill set has uncovered these gems
> (my one strong suit is creativity, and that can be measured by
> painting, piano, French pastries, swing dancing, and recumbent bike I
> designed and ride).
>
> I hope you can scan the paper and see if it looks logically coherent,
> something that cannot be faked. Like "The Old Man and the Sea", I am
> trying to land a fish that I know is too big for me to handle by
> myself. I am willing to talk about it at anytime at your convenience
> (I work for a software company in Waltham, live in Acton, and like to
> eat at Mary Chung's).
>
> Have a good day, elegance governs the heavens.
> doug

I attached a file which can be viewed either as a pdf or in HTML
http://TheWorld.com/~sweetser/quaternions/ps/em2gem.pdf
http://TheWorld.com/~sweetser/quaternions/gravity/em2gem/em2gem.html

Thanks for reading,
dougd

 

[CarlB]

Doug,

Could there be any chance that this new result has anything to do with an interaction between gravity and E&M that is consistent with what you are doing?

New Experimental Results on the Lower Limits of Local Lorentz Invariance
Fabio Cardone1, 2, 3, Roberto Mignani4, 5, 6 and Renato Scrimaglio1

(1) Istituto per lo Studio dei Materiali Nanostrutturati (ISNM-CNR), via dei Taurini 19, 00185 Roma, Italy
(2) Istituto di Radiologia, Facoltà di Medicina, Università di Roma “La Sapienza”, Roma, Italy
(3) I.N.D.A.M.—G.N.F.M., Sesto Fiorentino, Italy
(4) Dipartimento di Fisica “E. Amaldi”, Università di Roma “Roma Tre”, Via della Vasca Navale, 84, 00146 Roma, Italy
(5) Sezione di Roma III, INFN, Roma, Italy
(6) Dipartimento di Fisica, Università dell’Aquila, Via Vetoio, 67010 Coppito, L’Aquila, Italy

Received: 7 October 2004 Published online: 9 May 2006

An experiment aimed at detecting a DC voltage across a conductor induced by the steady magnetic field of a coil, carried out in 1998, provided a positive (although preliminary) evidence for such an effect, which might be interpreted as a breakdown of local Lorentz invariance. We repeated in 1999 the same experiment with a different experimental apparatus and a sensitivity improved by two orders of magnitude. The results obtained are discussed here in detail. They confirm the findings of the previous experiment, and show, among the others, that the effect is independent of the direction of the current. A possible interpretation of the results is given in terms of a geometric description of the gravitational and the electromagnetic interactions by means of phenomenological, energy-dependent metrics.

Foundations of Physics, Volume 36, Issue 2, Feb 2006, Pages 263 - 290, DOI 10.1007/s10701-005-9014-z, URL:
http://dx.doi.org/10.1007/s10701-005-9014-z

I still haven't forgotten about putting together a program to compare your gravitation with Einstein and Newton. But I've been too busy to get it written. When it's done, it will be written in Java. As a test to see if you can compile my variety of Java, try downloading the java source code for my Sudoku solver, which also has a link to where you can get the free Java development program from Borland:
http://www.brannenworks.com/SU/

Carl

 

[sweetser]

Invariant and Covariants in GEM theory

Hello Carl:

Thanks for the reference to the paper. Based only on the abstract, I was not able to justify the $30 cost for downloading the pdf, so I cannot comment on the specific content.

Here's a skeptical (in the postive sense of the word) view. In the 1800s, people thought that electric and magnetic phenomena should somehow be linked. When Oelmsted found that link, experimentalist got excited and did a barrage of work. Experimentalists are like that today. On the positive side, there was a frenzy to find ever higher temperature superconductors. On the negative side, there was a squall of work to detect a 5th force and to see the signs of cold fusion.

I have not picked up a buzz concerning a local violation of Lorentz symmetry. The experiment was first done in 1998, then apparently repeated by the same folks in 1999, and published some 5 years later. I believe the journal "Foundations of Physics" has a reputation for publishing work of questionable long term value.

What does the phrase "violate local Lorentz symmetry" mean? I'm not sure, but will discuss this symmetry as best as I understand it, and how it relates to my work.

Empty spacetime is governed by the Minkowski metric. Square the spacetime difference between any two events, and all inertial observers agree on the value of the interval (Lorentz invariance), and the all disagree about time and space measurements in a way all observers understand is related to their relativistic velocities (Lorentz covariance).

A complete answer to an observation always included both the invariant and the covariant quantities. This pairing is often forgotten. For example, people will point out that the speed of light is an invariant, and forget to mention that the frequency and wavelength of light are covariant, changing by a relativistic doppler equation. Many popular sciences sources will talk about rulers and clocks disagreeing (Lorentz covariance), without mentioning the invariant interval.

So I have an ecletic rule: I must always talk about the invariant and the covariant measurements in order to be complete. Once I have a rule, I try not to break it ever, so let's see how this goes...

In spacetime with mass in GEM theory, either one continues to use a flat spacetime metric with a dynamic potential, or one uses a dynmaic metric with a constant potential, or some combination of the two (in GR, it must exclusively be the metric that changes). There are only an infinite number of choices to be made! That's diffeomorphic symmetry for you. Let's choose to work with the boring potential, so everything is due a dynamic metric. Another way to say it is that the metric is different depending on where you are. We know how it changes. So the interval is now a covariant quantity, not an invariant one, because the interval changes in a way we understand.

That raises the question: what is the invariant quantity? I know exactly what it is, but I don't know its name! If you recall the metric that appeared in the first post of this thread, the dt term had a exponential with a negative exponent, while the dx, dy, dz all had exponentials with a positive exponent. It turns out that the products dt dx, dt dy, and dt dz are invariant under the introduction of mass into empty spacetime!

I don't know what to call dt dx, dt dy, or dt dz, so for now I'll make up a name: a 3-rope. Inertial observsers conserve the interval, and folks in a Universe with mass conserve the 3-rope. Is there any connection between an interval and a 3-rope?

This is where the story gets downright unbelievable. They have a simple, direct connection. Write an interval as a quaternion (if you are unfamiliar with quaternions,visit my website devoted to the topic, quaternions.com, to learn about the next division algebra after the real and complex numbers, which as 4 parts that can be added, subtracted, mutliplied or divided). Here is an interval written as a quaternion:
[tex]\xi=(dt, dx, dy, dz).[/itex]
Now square the interval quaternion:
$$\xi^2=(dt^2 - dx^2 - dy^2 -dz^2, 2 dt dx, 2 dt dy, 2 dt dz).$$
In special relativity, it is the first term of this square that is invariant, while the 3-rope is covariant because we know exactly how it changes. Now toss in a mass and for the GEM theory, the first term is covariant while the 3-rope is invariant. If one wanted to be more precise, I would have to add a caveat that all the theory could claim is that there exists a choice of coordinates such that the 3-rope is invariant (I think there is also a caveat like this for special relativity, that pathological coordinates are pathological and mess nice statements up).

If anyone reading this knows the offical name for the 3-rope symmetry, it would be a big help to me. If I knew the name of the beast, then I could read up on it.To get back to Carl's question, the electric and magnetic fields live inside the same irreducible field strength tensor, the antisymmetric tensor $\nabla^{\mu}A^{\nu} -\nabla^{\nu}A^{\mu}$. The reason there are all those interactions between E and M is because they live inside the same irreducible tensor. The fields for gravity live in a different irreducible tensor, the symmetric $\nabla^{\mu}A^{\nu} +\nabla^{\nu}A^{\mu}$. I identified three gravitational fields, and those should be able to mix with each other as happens for E with B. The gravity and EM fields cannot mingle so directly according to GEM theory. To be completely honest, I still am unclear about their relationship. Both are caused by the same 4-potential, something that cannot be measured directly. What can be measured is the change in the 4-potential, and that change falls into these two irreducible tensors.

The bottom line at this point looks like gravity and EM should not mix together in an obvious way according to GEM theory, and the 3-rope symmetry in the context of quaternions is a fun way to wonder what is going on.

doug

 

[CarlB]

Here is an interval written as a quaternion:
[tex]\xi=(dt, dx, dy, dz).[/itex]
Now square the interval quaternion:
$$\xi^2=(dt^2 - dx^2 - dy^2 -dz^2, 2 dt dx, 2 dt dy, 2 dt dz).$$
In special relativity, it is the first term of this square that is invariant, while the 3-rope is covariant because we know exactly how it changes.

Since my background is in elementary particles, I'd prefer that you go with the flat space and explain it again from that point of view and would appreciate that.

But your equation with quaternions is interesting from a particle point of view. Let's see how you can map your way of doing bizness into the Pauli algebra. If

$$\xi = idt + \sigma_xdx +\sigma_ydy +\sigma_zdz$$

then

$$\xi^2 = (-dt^2 + dx^2 + dy^2 + dz^2) + 2i\sigma_x(dt\;dx) +2i\sigma_y(dt\;dy) +2i\sigma_z(dt\;dz)$$

and all the other cross terms cancel because of anticommutation of the Pauli sigma matrices, giving a result very similar to your own. I suppose I should multiply through by i.

Now as it turns out, the above use of the Pauli algebra is what I was pushing before I decided to abandon "Euclidean relativity" in favor of submitting to Einstein's forms. This was not because it was wrong, but because I got tired of being the nail that stuck out and kept getting hammered down. I was using an algebra where the natural differential operator $$\nabla$$ is defined as:

$$\nabla = \hat{t}\partial_t + \hat{x}\partial_x + \hat{y}\partial_y + \hat{z}\partial_z + \hat{s}\partial_s$$

where $$\hat{t}^2 = -1$$ and all the other hats square to +1, and the t hat commutes with everything while the x,y,z and s hats anticommute with each other. The "s" coordinate is for proper time in a short cyclic coordinate. If you go to massive particles, the s coordinate goes away, so ignore it.

This was written up in my original paper classifying the fermions. See the comment on page 5, "For those manifolds that do not explicitly include time, an extra commuting operator (...) accounting for momentum versus position must be included and this increases the value of k by one." http://brannenworks.com/a_fer.pdf see

What this boils down to is that if you want to model spacetime and you insist that time NOT be a geometric part of the manifold (as time is considered geometric by Minkowski), then you have to have an extra degree of freedom to distinguish between traveling backwards and forwards in time. That is, vectors in space-time can give directions forwards and backwards in time. Vectors in space can not, and thus need an extra degree of freedom. This you can accomplish either by having double sets of canonical variables, i.e. keeping track of Ps and Qs in the traditional classical mechanics manner, or you can accomplish this extra degree of freedom by adding an extra "notation" coordinate to your geometry. As a notation coordinate, it must commute with everything else.

There are also some simple ways of embedding the quaternions into the Dirac algebra that might be less unsettling. The key is that you have to arrange for the time component to commute with the spatial components, while the spatial components anticommute. But every example of this will be analogous to the above Pauli example. That is, the algebra is completely defined by these commutation rules, along with the rules telling what the squares need to be.

Carl

 

[Don J]

Since my background is in elementary particles, I'd prefer that you go with the flat space and explain it again from that point of view and would appreciate that.

Hello Carl
I think this paper can complete your research ...because you work in that field access to this rare paper should be granted to you.
Your opinion will be appreciated
http://prola.aps.org/abstract/PR/v105/i2/p735_1