Broken Maxwell EM ?

[Newton]

Don't you know that there are Riemannian manifolds without torsion? Unless your definition is a "personal one". I am not even sure that know what is a Riemannian manifold and what is torsion.

Let me try to teach you something. Riemannian manifold is a manifold equipped with a Riemannian metric. Such a manifold has a natural connection (called Levi-Civita connection), and this connection has no torsion.

On the other hand you can put on any manifold a connection with torsion. Torsion has nothing whatsoever to do whether manifold is Riemannian
or not. Torsion is not property of the metric. It is a property of a connection. Connection and metric can be given as totally independent.

Now, after all that, do you understand why your statement is simply either wrong or just nonsensical?

Please read http://ej.iop.org/links/rru9OOqRG/lqaQGcVJ2xGHMR1xav5vpA/jhep021999024.pdf , you'll see torsion is associated both manifolds and connections.

 

[Newton]

Don't you know that there are manifolds that are both Riemannian and symplectic that are not Kahler? I am getting little bit brutal when I see poor logic :)

This is an excellent discussion - We're building up to the mathematics of SUSY transforms.

Let me give you concrete examples of symplectic and Riemannian manifolds. The state-space (orbits) of solutions to linear PDEs (of less than 3rd order) is a Riemannian manifold. The state-space of solutions to 2nd order irreducible nonlinear PDEs is a symplectic manifold. So the question is: can a linear PDE have the same state space as a nonlinear one? Ans. Only if you treat a linear PDE as a non-linear PDE where some coefficients go to zero. ( Actually the same argument applies to PDEs with constant vs periodic coefficients)

Let me be brutal... Riemannian manifolds have (support) the inner product. Symplectic manifolds have a symplectic form, which means these manifolds can't support an inner product. That is, 2-forms don't support the inner product - well known in algebraic geometry.

The symplectic manifold is torus shaped. The Riemannian manifold has quadratic curvature. There's no diffeomorphism between the two.

Riemannian manifolds have integer dimensions. Symplectic manifolds have fractional dimensions.

Riemannian manifolds do not have knots. All symplectic manifolds have knots.

So the 4-dim space of quarternions, Spin(4) is very,very special. And very, very different from what most physics think it is.

 

[John G]

I'm afraid stochastic flows have no place in physics, all flows are Hamiltonian.

Ark could be right, maybe there's some term definition differences cause come to think of it Tony considers his Feynman Checkerboard to be a stochastic flow and you liked Tony's version of the Checkerboard.

As for Hamiltonian flows:

In other words, 26-dim String Theory can be interpreted as being the theory of movement of a "Bohm Point" in configuration space. ( Since I like to use a Lagrangian formulation using spacetime as opposed to a Hamiltonian spatial formulation with time as an "outside" variable, the configuration space is not particle positions (points) in fermion representation spaces, internal symmetry space, and spatial space but is particle world-lines (strings) in fermion representation spaces, internal symmetry space, and spacetime including both spatial dimensions and time dimension.

So maybe you and Tony also have some interpretation differences for the same bosonic string math.

Tony is correct that spin(6) is associated with gravity, but actually spin(4) x spin(4), Boson interactions lead to spin(6) with a left over scalar (constant acceleration). The scalar is probably near-field gravity, and spin(6) is a fermion related (maybe a Quark) since it has so much energy.

Tony gets spin(6) gravity from the spin(8) bosons [like SU(5) GUT] after dimensional reduction (D3 from D4). The Higgs scalar field/VEV comes from the dilation within Spin(6) and near-field gravity comes from spin(6) via something called the MacDowell-Mansouri mechanism. Spin(6) has rotations/boosts/translations/dilation/special conformal transformations (Tony calls them graviphotons) and they are all bosons for Tony.

Notice that spin(4) is symplectic (has knots), but spin(6) has an octonioin manifold (no knots) and is a soliton. The interaction of spin(6) x spin(6) is well known for solitons to be a phase shift (which is exactly the same as gravity). Interesting isn't it?

Tony I think would consider spin(6) like a Lorentz version of the spin(4) Euclidean quaternion vector and spin(10) like a Lorentz version of the Euclidean spin(8) octonion vector.

Tony gets O(3) aka CP1 solitons for protons and this is related to the special conformal transformations of spin(6) but more directly to Spin(10) I think. He also has solitons for ball lightning and that would be related to the Spin(6) special conformal transformations also.

 

[John G]

So the 4-dim space of quarternions, Spin(4) is very,very special. And very, very different from what most physics think it is.

I can't resist posting for this a quote from Tony that began with a question I asked:

John Gonsowski said, in a post to sci.physics.research:
".. John Baez discusses triality ... [saying] ...
"Triality is a cool symmetry of the infinitesimal rotations
in 8-dimensional space.
It was only last night, however, that I figured out what triality
has to do with 3 dimensions. ... Look at the group of all
permutations of {i,j,k}. This is a little 6-element group which
people usually call S_3, the "symmetric group on 3 letters". ..."
... Back to me [John Gonsowski] instead of John Baez:

Is there any way to relate this linear transformation to D3? ...".

In reply, I said:
The Weyl group of D3 is S3 x (Z2)^2 (of order 6x4 = 24).

The S3 part of the Weyl group of D3 comes from the symmetries
of the 3-Euclidean-dimensional cuboctahedron, which is the
root vector polytope for D3.

The geometric picture that I have in my head is that
the S3 comes from
using reflections through hyperplanes through the origin
to interchange 3 elements of the triangular faces of the cuboctahedron.



Although D3 does not have as obvious a triality of its representations
as does D4 (for which the vector represenation is 8-dimensional,
as are the two half-spinor representations),
here is how I see a similar (but less obvious) version of triality in D3:

D3 has two 4-dimensional half-spinor representations
and
a vector representation that is 6-dimensional.
However,
D3 is the Lie algebra of the conformal group of Minkowski spacetime,
which is 4-dimensional.
Therefore,
although the conventional linear vector representation of D3
is 6-dimensional,
you could (and I do) say that
D3 has a NONlinear 4-dimensional representation due to conformal
transformations,
and
I see a triality among:

two D3 4-dim half-spinor representations
and
one D3 NONlinear conformal (sort of vector-like) 4-dim representation.


(This is related to how I use D3 in my physics model,
after dimensional reduction of spacetime from 8 to 4 dimensions,
but that is a digression that I won't pursue further here.)



For comparison, the D4 Dynkin diagram is,
where v denotes the conventional linear vector representation,
a denotes the adjoint representation,
ad s+ and s- denote the two half-spinor representations:

s+
|
v - a - s-

The dimensionalities of these representations are:

8
|
8 - 28 - 8

The D4 triality is the S3 group of permutations of the outer three
8-dimensional representations of the Dynkin diagram.

-----------------------------------------------

The D3 Dynkin diagram is,
where v denotes the conventional linear vector representation,
a denotes the adjoint representation,
ad s+ and s- denote the two half-spinor representations:

s+
|
v - s-

Note that the adjoint representation "goes away" and is not an
elementary fundamental representation of D3.

The conventional linear dimensionalities of these representations are:

4
|
6 - 4

Note that you can make the 15-dim adjoint representation of D3
by taking the tensor product of 4x4 to make a 16-dim group
that includes the 15-dim adjoint as a subgroup.

By using the conformal non-linear representation c in place of
the conventional linear vector representation,
the D3 diagram becomes

s+
|
c - s-

Using the non-conventional conformal c instead of the vector v,
the dimensionalities of these representations are:

4
|
4 - 4

The D3 triality is the S3 group of permutations of the three
4-dimensional representations of the Dynkin diagram.

Note that the D3 Dynkin diagram can be written equivalently as

4 - 6 - 4 (conventional linear vector)

or

4 - 4 - 4 (conformal)

which shows the isomorphism between D3 and A3.

The A3 viewpoint makes it easier to see that the 4x4 = 16 gives
you 16-dimensional U(4), which is reducible to SU(4)xU(1),
and the 15-dimensional SU(4) is the adjoint of A3 and is
also (by isomorphism) the Spin(6) adjoint of D3.
You could also use signatures such as U(2,2) and Spin(4,2).


The D2 Lie algebra of (1,3) Minkowski physical spacetime and Spin(1,3) = SU(2) x SU(2), inherits from D3 a quaternionic type of triality.

For Cl(1,3) the 2x2 quaternionic matrices have Full Spinors
that are 1x2 quaternion column vectors.
Each Half-Spinor space is one quaternion variable,
which has a 1-2 correspondence with first generation fermions,
and
also corresponds 1-1 with the (1,3) vector space of
physical Minkowski spacetime,
resulting in a quaternionic version of triality
(diluted by the 1-2 nature of the fermion correspondence)
that is related to
the reducibility of the D2 Lie algebra Spin(1,3).

 

[ark]

Let me be brutal... Riemannian manifolds have (support) the inner product. Symplectic manifolds have a symplectic form, which means these manifolds can't support an inner product. That is, 2-forms don't support the inner product - well known in algebraic geometry.

The symplectic manifold is torus shaped.

Now, you are being brutal indeed, brutal with yourself, because you are talking a brutal nonsense.

Let me try to teach you therefore. (I mean if you are teachable).

First: a symplectic manifold CAN support inner product. For instance two-dimensional real plane. This was your first nonsense. If you are teachable (I am not sure if your homotopy group allows teachability....) then, from now on, you will know that a symplectic manifold can support inner product.

Second: Symplectic manifolds are NOT necessarily torus-shaped. Example: two-dimensional real plane. This was your second nonsense. If you are teachable (I am not sure if your cohomology class allows teachability....) then, from now on, you will know that a symplectic manifold is not necessarily torus shaped ... I mean unless you sit on your torus and make it into a pancake :)

 

[ark]

Please read http://ej.iop.org/links/rru9OOqRG/lqaQGcVJ2xGHMR1xav5vpA/jhep021999024.pdf , you'll see torsion is associated both manifolds and connections.

1) The link does not work
2) Please learn from some good textbook what is torsion. But if you have no access to good textbooks, let me quote for you from Kobayashi-Nomizu

p. 120:

We define the torsion form Theta of a linear connection Gamma by:

Theta = D theta

where theta is the canonical form on the bundle of linear frames.

But I understand that you may have aversion to Japanese authors. So, let me quote from Sternberg "Lectures in differential geometry":

The form D omega is called the torsion of the connection. (Note: Sternberg denotes the canonical form by omega)

So, now, perhaps, you will know what torsion is. Of course there are OTHER kinds of torsions - for instance Reidemeister torsion. But each tiem you are using the term "torsion" it is absolutely necessary to DEFINE PRECISELY what you mean. Otherwise you will be creating NOISE.

 

[Newton]

That link worked last night, but here's a better link
http://arxiv.org/PS_cache/hep-th/pdf/9901/9901016.pdf

Actually the paper's quite good irrespective of our torsion discussion!

Let me give you another concrete example, this time for torsion in state-space manifolds. Let's take a spherical gyro and spin it up on an axis (obviously a Riemanniam state-space). Then release the axis constraint... The gyro then tumbles in wild directions. So therefore the Riemannian state-space has natural torsion otherwise it would not tumble. The unconstrained (by the axis of rotation) dynamics has a symplectic state-space (no torsion since the constraining axis was removed).

It would be so much easier if physicists actually understood classical mechanics (Hamitonian flows). Mechanical engineers are well aware of the nonlinear (ie symplectic) nature of spinning shafts.

The above paper explains the relationship (Dirac operators) between manifolds with torsion and without.

 

[Newton]

Let me be brutal... Riemannian manifolds have (support) the inner product. Symplectic manifolds have a symplectic form, which means these manifolds can't support an inner product. That is, 2-forms don't support the inner product - well known in algebraic geometry.

The symplectic manifold is torus shaped.

Now, you are being brutal indeed, brutal with yourself, because you are talking a brutal nonsense.

Let me try to teach you therefore. (I mean if you are teachable).

First: a symplectic manifold CAN support inner product. For instance two-dimensional real plane. This was your first nonsense. If you are teachable (I am not sure if your homotopy group allows teachability....) then, from now on, you will know that a symplectic manifold can support inner product.

Second: Symplectic manifolds are NOT necessarily torus-shaped. Example: two-dimensional real plane. This was your second nonsense. If you are teachable (I am not sure if your cohomology class allows teachability....) then, from now on, you will know that a symplectic manifold is not necessarily torus shaped ... I mean unless you sit on your torus and make it into a pancake :)

Pure nonsense. Do you know what a symplectic form is? :)

Symplectic manifolds are in Banach space - no inner product! Symplectic manifolds result from constraints. In the case of EM, and special relativity, the constraint is the max phase velocity of light, c. In classical mechanics the constraint is conservation of angular momentum in three dimensions (In two dimensions, interestingly enough, there is no constraint). Because of those constaints, the degrees of freedom are fractional so in projections to a Euclidean frame you'll see fractals!

 

[John G]

Riemannian manifolds have (support) the inner product... Symplectic manifolds are in Banach space - no inner product!

What happens when you have a phase space (symplectic) manifold with a Riemannian metric? Doesn't that give you an inner product and make it a Riemannian manifold too?

For example:

http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Aquant-ph%2F9805014

That link worked last night, but here's a better link
http://arxiv.org/PS_cache/hep-th/pdf/9901/9901016.pdf

There is a nice relationship between Atiyah-Singer, Kerr-Newman, torsion, and spin(6). I think all the torsion is coming from a connection isn't it? Here's something from Saul-Paul Sirag via Tony echoing Ark's earlier comment:

"... if a space is curved, it is impossible to compare two distant vectors without some method of parallel transport of vectors throughout the curved space. The amount of curvature is a measure of the mismatch of a vector with a copy of itself which has undergone a complete circuit. ... The parallel transport is provided by a structure which is added to the manifold and is called the connection. In the theory of general relativity, the connection is provided by an object calledthe Christoffel symbol G_ij^k. This is a very compact notation for a set of 40 (= 64 -24) functions on the 4-d spacetime. If the symbol carried two asymmetric lower indices, there would be 64 (= 4 x 4 x 4) functions; but the symmetry of the lower indices reduces the independent functions to 40. The standard Christoffel symbol of general relativity is symmetric in the two lower indicies i,j, and generates a connection called the Levi-Civita connection. However, there are geometries for which an asymmetic Christoffel symbol is employed in addition to the the symmetric Christoffel symbol. The asymmetry is carried by a tensor T called the torsion. We can write:

G_ij^k - G_ji^k = T_ij^k
Thus although the Christoffel symbols are not tensors, their difference is a tensor. In physics, we expect tensors to correspond to measurable quantities. If T is 0, then the torsion is zero, and the symbol must be symmetric. A very special case of parallel transport is called absolute parallelism. While ordinary parallel transport guarantees that the vectors will be rotated only by the curvature along the particular path in the circuit, an absolute parallelism connection guarantees that the vectors will remain unrotated by travel along any circuit that follows vector field flow lines. This implies that there is no curvature for this absolute parallelism connection. However [there] will, in general, be a gap in this circuit caused by a "vertical" motion of the ... moving vector. After making the ciruit, the moving vector and its stay-at-home twin will, end up parallel to each other but separated by this "vertical" gap. This gap is called the torsion. ... The connection structure which provides curvature, is based on the symmetric Christoffel symbol. Thus this connection (called the Levi-Civita connection) has zero torsion. By contrast, the absolute parallelism connection which provides torsion has zero curvature. ... there are good examples of spaces carrying both these connections. These spaces are Lie group manifolds. In fact, later work by Joseph Wolf proved that the only spaces that carry an absolute parallelism (Cartan) connection are Lie groups--with one exception: the seven-sphere S7. ... the only spheres that carry an absolute parallelism are spheres of dimension 1, 3, and 7. And the only spheres that are Lie groups are spheres of dimensions 1 & 3. The Lie group structures of these spheres are called U(1) and SU(2). Moreover, S1 (= U(1)) is the set of all unit complex numbers, while S3 (= SU(2)) is the set of all unit quaternions, and S7 is the set of all unit octonions (or Cayley numbers); it is because octonions are not an associative algebra that S7 fails to be a Lie group; but the octonion structure provides an absolute parallelism on S7. ... it is the left-invariance (or right invariance) of the Lie algebra vector fields the provides absolute parallelism. As Cartan discovered, there are three canonical connections on a Lie group manifold. These three connections are generated by three different actions of the Lie group on itself:

(1) Left action: g --> h g (where g and h are group elements of Lie group G)
(2) Right action: g --> g h [(where g and h are group elements of Lie group G)]

(3) Adjoint action: g --> h^(-1) gh (where h^(-1) is the inverse element of h )

... The set of all ... tangent planes together form a vector bundle called the tangent bundle of the Lie group. For the Lie group G, the symbol for the tangent bundle is TG, and it is simply the direct product of the Lie group G and the Lie algebra g. ... In contrast to the case of an ordinary manifold, which is not a Lie group, we say that TG is a trivial bundle because it is direct product of the base space G with the the fiber g, this implies a global trivialization of the bundle structure; moreover, this global trivialization corresponds to the absolute parallelism afforded by the group action on the group manifold and thus on the parallel transport of vectors of the Lie algebra, as described above. The intimate relationship between the Lie group G and the Lie algebra g, has the consequence that the torsion of G ... is simply the Lie product, [x,y], of g ... for the torsion T of a Lie group manifold we can write:

[ X_i ,Y_j ] = T_ij^k Z^k
... where the componets of T are the structure constants of the Lie algebra; and X, Y, and Zare Lie algebra elements, i.e., left-invariant vector fields on G. For right invariant vector fields the torsion tensor is the would be -T_ij^k. ... In general, the curvature tensor describing the curvature of the Lie group manifold is the Riemann curvature tensor which can be written in terms of the Lie algebra structure constants:

R_i,kl^k = (1/4) C_hi^j C_kl^ h
The Riemann curvature tensor is the tensor generated by the Levi-Civita connection, for which there is zero torsion. Thus on the Lie group manifolds we have two radically different connections: the Cartan asymmetric connection which has torsion but no curvature, and the symmetric Levi-Civita connection which has curvature but no torsion. Given two connections on the same manifold, the difference between the Christoffel symbols is a tensor, called the difference tensor. In the case of these two connections on a Lie group manifold the difference tensor is the contorsion tensor K.

We can write:

K_ij^k = G_ij^k - G_ij^k
where the first Christoffel symbol is the Cartan connection of absolute parallelism (with 64 independent functions); the second ... Christoffel symbol is the Levi-Civita connection (with 40 independent functions ...); and K is the contorsion tensor (with 24 independent functions). The contorsion tensor is also the tensor of Ricci rotation coefficients. ...".

 

[Newton]

What happens when you have a phase space (symplectic) manifold with a Riemannian metric? Doesn't that give you an inner product and make it a Riemannian manifold too?

For example:

http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Aquant-ph%2F9805014

John

That paper is an excellent example of 'junk math'. Riemannian manifolds have 'quadratic curvature' and symplectic manifolds have 'saddle points'. And the Lie algebra for each is quite different.... Riemannian manifolds have a 1st order polynomial Lie algebra and symplectic manifolds are 2nd order (hence the saddel points).

I find it amazing that the obvious differences between Riemannian Manifolds (complex variables) and symplectic manifolds (quarternions) isn't visible in the physics community!

Again, if you see the word stochastic, it's junk science or 'folk' math.

 

[ark]

That link worked last night, but here's a better link
http://arxiv.org/PS_cache/hep-th/pdf/9901/9901016.pdf

Actually the paper's quite good irrespective of our torsion discussion!

The above paper explains the relationship (Dirac operators) between manifolds with torsion and without.

You probably do not understand what the paper is about, otherwise you would not be writing nonsense like

Riemannian manifolds have torsion, and symplectic are torsion free.

You certainly do not know what a Riemannian manifold is. Please, in the future, avoid using terms that you do not undertsand. That creates noise on this Forum.

 

[ark]

Pure nonsense. Do you know what a symplectic form is? :)

Symplectic form is not a nonsense. It is explained in textbooks.

Symplectic manifolds are in Banach space - no inner product!

Not true. Symplectic manifolds do not have to be in a Banach space.

Symplectic manifolds result from constraints.

Some symplectice manifolds result from constraints. Some other live well without constraints.

In the case of EM, and special relativity, the constraint is the max phase velocity of light, c.

This depends on which manifold you are talking about. You are never being clear. You should try to avoid being not clear as your posts create noise on this forum. If this continues I will have to take measures and put a limit to this noise. This is anot a joke.

 

[Newton]

Ark,

I gave perfectly clear examples from classical mechanics of every idea I presented.

You can read Classical Mechanics,Goldstein, 1980 to learn about the properties of symplectic manifolds and why all Hamiltonian systems have this phase space manifold.

I do agree with you that when you deal with only 1 or 2 dimensional models that the phase space manifold of Hamiltonian systems reduces to Riemannian (1 and 2-dim Hamiltonian systems are not symplectic - perhaps that's the confusion with QM, since very little QM is done in 3-dim). But in a 3 dimensional world, the real world, phase space manifolds are always symplectic.

Physicists must learn Classical Mechanics before they attempt Particle Physics, and that is NOT done at unversities. Junk physics it the result!

Operators do not commute on symplectic manifolds, but they do commute on Riemannian manifolds. I can't see how you can think symplectic and Riemannian manifolds can ever be equivalent.

Ark, are you affiliated with Cornell Univ? You seem to have their bias. I'm sure Tony can tell you how they try to shut him up when he presents quality mathematics which differs from the current thinking.

Newton

 

[ark]

Ark,

I gave perfectly clear examples from classical mechanics of every idea I presented.

You can read Classical Mechanics,Goldstein, 1980 to learn about the properties of symplectic manifolds and why all Hamiltonian systems have this phase space manifold.

You see, I was studying classical mechanics, Goldstein and others when I was in my second year of studies, and that was some fourty years ago. So, I do not have to learn about symplectic manifolds, because I know about them (as a matter of fact I today I know much more than you can find in Goldstein). Guillemin and Sternberg's "Symplectic techniques ..." goes much further

I do agree with you that when you deal with only 1 or 2 dimensional models that the phase space manifold of Hamiltonian systems reduces to Riemannian (1 and 2-dim Hamiltonian systems are not symplectic - perhaps that's the confusion with QM, since very little QM is done in 3-dim). But in a 3 dimensional world, the real world, phase space manifolds are always symplectic.

Phase space manifolds are by definition symplectic. Even in one or two dimensional models. From what you wrote I am deducing that you do not even know what is a phase space manifold. And it has nothing to do with quantum mechanics.

Physicists must learn Classical Mechanics before they attempt Particle Physics, and that is NOT done at unversities. Junk physics it the result!

You probably do not know much about universities either. There are different universities, with different programs of studies. Do not forget that the devil is always in the details. Do not forget details. Otherewise you can become the devil.

Operators do not commute on symplectic manifolds, but they do commute on Riemannian manifolds. I can't see how you can think symplectic and Riemannian manifolds can ever be equivalent.

From what you wrote I deduce that you do not even know what an "operator" is. Some operators commute with other operators. Some not. And it has absolutely nothing to do with symplectic manifolds. Again you are creating noise by using terms that you do not understand.

Ark, are you affiliated with Cornell Univ?

I am affiliated with myself.

You seem to have their bias.

My bias is fighting with nonsense in science. My other bias is open-mindness. As a rule I am patient. But because of this first bias, my patience has limits.

I'm sure Tony can tell you how they try to shut him up when he presents quality mathematics which differs from the current thinking.

Newton

Tony's work does not fit the criteria of rigour. That is why his papers are being rejected. But Tony has a number of very interesting ideas, insights, and half-developed theories. The point is that scientific journals do not accept half-developed theories. In my opinion there should be special journals that would accept papers a'la Tony. And, in fact, there are such journals. Unfortunately they are being abused by others who create just noise.

 

[Newton]

You see, I was studying classical mechanics, Goldstein and others when I was in my second year of studies, and that was some fourty years ago. So, I do not have to learn about symplectic manifolds, because I know about them (as a matter of fact I today I know much more than you can find in Goldstein). Guillemin and Sternberg's "Symplectic techniques ..." goes much further

Yes... Guillemin and Sternberg's Symplectic techniques in physics. That book from the early eighties by the MIT and Harvard duo is a real gem. That was my first venture into phase-space optics.

It has the core elements for completing a full Bosonic String theory! As I recall, it was missing Kac-Moody Lie algebra for spin(4) and it didn't go into Riccati equations and superposition on nonlinear manifolds. Witten turned me on to studying symplectic manifolds and cyclotomic fields. Eventually I got into Galois theory because of the parallel between irreducible polynomials and the Lie algebra of irreducible PDEs. Are you following Kac's work at MIT on gauge invariance?

BTW. Bosonic string theory is a slam-dunk. The technology is the same as solving 2nd order irreducible PDEs (mathematical physics isn't there yet, but industry research is). It's true, you can develop the full theory of Bosons based only on the constraint c, the max phase velocity of light.

You are correct about the rigor of Tony's work (although his ideas was incredibly helpful for me in my research), but there are so many far, far worse things being published. I personally get a tremendous laugh when I see papers on measure theory. (It turns out, measure theory only applies to topological spaces, it does not apply to non-topological smooth manifolds - verified experimentally on nonlinear second order systems)

I'd make sure his stuff is published, even if others need to go in and edit to add rigor.