How Can Gravity and Electromagnetism Be Unified Through a Rank 1 Field Theory?

[Lawrence B. Crowell]

Obviously with the 421 kg mass in a box or a unit charge it will be easy to tell the difference. The mass will attract all masses equally, while the charge will attract or repel the same whether that has the same or opposite charge. The proton will have negligable attraction to an uncharged mass.

The NS metric is for a spacetime determined by a sperically symmetric mass with a charge. One could well enough compute the orbit of a charged particle in this spacetime. Yet what I compare is the fall of a neutral mass around a charged black hole and a charge falling into a neutral black hole. What I find is that the "braking" of a charged particle fall by the emission of radiation by purely classical means is equivalent to the geodesic rate of fall for a neutral particle entering a black hole with the equivalent charge. This result illustrates I must be at least on the right track. It is in line with the "no hair" theorem for black holes is that the details of how a black hole reaches a final state by the acquisition of charge and mass by the infall of matter and fields is independent of the details by which that occurred. A black hole will end up with the same final mass M~=~M_0~-~(q/r)^2 indepdendent of which masses brought in what charge.

Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

The thought experiment did not concern itself with the ability to tell the difference between the behavior of electric charge or mass charge. Rather, it was an effort to find a condition where the behavior was indistinguishable.

> One could well enough compute the orbit of a charged particle in this spacetime.

If and only if the charged mass had the same sign as the charge on the test mass would the NS metric work. If th test had the opposite sign, it would fail because the metric must demonstrate attraction.

I appreciate that you looked into the consequences of the NS metric, and that other skilled people have worked with it. But it looks unacceptable to me because it cannot model electrical attraction.

Doug

 

[Lawrence B. Crowell]

Hello:

A basic property of electromagnetism is that like charges repel and different charges attract. That must be a property of any proposal related to EM. The Reissner-Nordstrom can only accept like charges that repel, since it uses Q², which always has the same sign (and opposite the M/R term). That looks like a deadly flaw to me, one that prevents me for doing anything further with the metric.

Doug

The RN metric uses (Q/r)^2 as the self-energy of a charged mass. This can be derived by using the electromagnetic field tensor as a source of the gravitational field. From there the RN metric is a solution to the Einstein field equation. In more advanced settings solutions of this type are BPS black holes, where gauge "charges" are the source of the black hole metric. The charge here is the charge of the black hole. One might then consider the motion of a charged mass in this spacetime, where that charge can have either sign.

This is fairly standard stuff.

Lawrence B. Crowell

 

[Lawrence B. Crowell]

Hello Lawrence:



> One could well enough compute the orbit of a charged particle in this spacetime.

If and only if the charged mass had the same sign as the charge on the test mass would the NS metric work. If th test had the opposite sign, it would fail because the metric must demonstrate attraction.

Doug

I am not sure where you got this idea. You can well enough compute the orbit of a charge with any value or sign in an RN metric.

Lawrence B. Crowell

 

[sweetser]

Coupling to a current, not a stress-energy tensor

Hello Lawrence:

I agree, you have represented standard analysis, which is darn good most of the time. If I see a killer flaw, I will point it out.

Even better is when I realize I was wrong about the killer flaw. I "get" the standard way, and can pinpoint where I drive a different direction. That is what happened with regards to the Reissner-Nordstrom metric.

In general relativity, gravity binds to the stress energy tensor. How much energy will the electric field contribute to the metric? G Q^2/c^4 R^2[/tex]. If the charge is positive, or the charge is negative, it does not matter, the same amount goes in.<br /> <br /> For the GEM proposal, the charge coupling term is -\frac{1}{4 c}(J A + (J A)^* - J^{*1} A^{*2 *3} - (J^{*1} A^{*2 *3})^*) With GEM, the effort is to do both gravity and EM. All these conjugates are required to get the phase to have both spin 1 and spin 2 symmetry. The coupling of gravity is not to energy, but instead to the 4-current density. The energy of the electric field makes zero contribution to gravitational effects.<br /> <br /> What we get in return for giving up the energy connection is a new equivalence principle for particles that attract each other. The attraction might be caused by electric charge or by a mass charge. Attraction looks the same either way. Sure, you could do other tests to tell which one was at work, but one could have videos of a particle being attracted by gravity or by EM that are absolutely identical. That is the bridge between gravity and EM.<br /> <br /> People who work on EM almost never deal with the tools of differential geometry. Calculate the divergence of the Christoffel symbol of the second kind for this metric:<br /> <br /> d \tau^2 = exp(2(\sqrt{G} Q - GM)/c^2 R) dt^2 - exp (-2(\sqrt{G} Q - GM)/c^2 R) dR^2/c^2 - R^2d\Omega^2/c^2<br /> <br /> Since this is for a static, spherically symmetric metric, to be logically consistent with EM, the answer had better be both Gauss&#039; and Newton&#039;s potential theory for EM and gravity,\nabla^2 \frac{\sqrt{G} Q - GM)}{R}, which it is. Lucky? I don&#039;t think so.Doug

 

[sweetser]

Generating the gauge-invariant GEM field equations

Hello:

In this post, I will apply similar, but not quite identical, approaches to generating the field equations for the fields E, B, e, and b.

If one hopes to model particles that travel at the speed of light, that requires that the field theory be invariant under a gauge transformation. This is one of those constraints on finding a solution I have heard, I have accepted, and I don't understand as well I should.

In deriving the Maxwell field equations using quaternion operators, the gauge invariance was achieved by noting all the derivatives that make up a gauge are in the first term of -A \nabla. The first term was subtracted away using the time honored quaternion trick subtracting the conjugate, q - q* = vector(q). The identical method was applied to the symmetric fields e and b. If were were to just add these two results together, there would be no link between the two sets of equations.

This time we will not use the trick of subtracting the conjugate, yet no terms with g appear in the final field equations because the g² terms in one cancel the g² terms of the other in the scalar. There is a g in the 3-vector, but that is not used for generating the field equations.

Start by taking the derivatives of 4-potentials in two ways. For quaternions written in the Hamilton basis, change the order of the differential operator with the potential, which flips the sign of B. For quaternions written in the Even basis, change which term gets conjugated. Calculate:

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

=((-\frac{\partial \phi}{\partial t} ~+~ c \nabla . A, -\frac{\partial Ax}{\partial t} ~-~ \nabla \phi ~+~ \nabla X A)(\frac{\partial \phi}{\partial t} ~-~ c \nabla . A , \frac{\partial Ax}{\partial t} ~+~ \nabla \phi ~+~ \nabla X A)
~+~ (\frac{\partial \phi}{\partial t} - c \nabla . A,\frac{\partial Ax}{\partial t} ~-~ \nabla \phi ~-~ \nabla .X2. A)(\frac{\partial \phi}{\partial t} ~-~ c \nabla . A,-\frac{\partial Ax}{\partial t} ~+~ \nabla \phi ~-~ \nabla .X2. A))

= ((-g, E ~+~ B)(g, -E ~+~ B) ~+~ (g, e ~+~ b)(g, -e ~+~ b))

(-g^2 ~+~ E^2 ~-~ B^2 ~+~ g^2 ~-~ e^2 ~+~ b^2, 2 E X B ~-~ e .X2. e ~+~ b .X2. b ~+~ 2 gE ~+~ 2 gb) \quad eq 1

The g field is not in the scalar due to a cancelation, but is in the 3-vector. The field equations are generated from the scalar, not the 3-vector, so any choice for the gauge g will not effect the field equations I am about to derive.

The current coupling term is complicated by the need to have spin 1 and spin 2 symmetry in the phase. This was worked out earlier in the thread, and here is the solution:

-\frac{1}{4}(J A ~+~ (J A)^* ~-~ J^{*1} A^{*2 *3} ~-~ (J^{*1} A^{*2 *3})^*) = ~-\rho \phi ~+~ Jx Ax ~+~ Jy Ay ~+~ Jz Az \quad eq 2

Write out the Lagrangian by its components, including the current coupling terms:

\mathcal{L}_{BEbe} ~=~ -c^2 \frac{\partial \Ay}{\partial z} \frac{\partial Az}{\partial y} ~-~ c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x} ~-~ c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}
-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t} ~-~ c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t} ~-~ c \frac{\partial \phi }{\partial z} \frac{\partial Az}{\partial t}
-\rho \phi ~+~ Jx Ax ~+~ Jy Ay ~+~ Jz Az \quad eq 3

There are fewer terms than in either the Maxwell Lagrangian or the symmetric field Lagrangian because terms between the two cancel. The fields in the field equations will need to do the same. Calculate the first field equation by taking the derivative of \mathcal{L}_{BEbe} with respect to the 4 derivatives of phi.

\frac{\partial}{\partial x^{\mu}}\left( \frac{\partial \mathcal{L}_{BEbe}}{ \left( \frac{\partial \phi}{\partial x^{\mu} \right)}} \right) ~=~ -c \frac{\partial ^2 Az}{\partial t\partial z} ~-~ c\frac{\partial ^2 Ay}{\partial t\partial y} ~-~ c\frac{\partial ^2 Ax}{\partial t\partial x}\right) ~-~ \rho

=~ \frac{1}{2}\nabla . (E ~-~ e) ~-~ \rho ~=~ 0 \quad eq 4

Nice, the E and e terms work together to isolate the A derivatives. And yet, you can spot Gauss' law for EM where like charges repel. There is a Gauss-like law for like charges that attract. Repeat for the derivative with respect to Ax:

\frac{\partial}{\partial x^{\mu}} \left( \frac{\partial \mathcal{L}_{BEbe}}{ \left( \frac{\partial Ax}{\partial x^{\mu} \right)}} \right) ~=~ -c^2 \frac{\partial ^2 Az}{\partial x\partial z} ~-~ c^2 \frac{\partial ^2 Ay}{\partial x\partial y} ~-~ c\frac{\partial ^2\phi }{\partial t\partial x}\right) ~+~ Jx ~=~ \frac{1}{2}(-\nabla X B ~-~ \nabla .X2. b ~+~ \frac{\partial(Ex ~+~ ex)}{\partial t}) ~+~ Jx ~=~ 0 \quad eq 5

You should be able to spot Ampere's law.

What has been done

There is now a formulation of the GEM proposal that uses quaternions exclusively. The standard quaternion algebra inherited from the nineteenth century needed to be extended in two ways. First the idea of a conjugate (or anti-involutive automorphism in fancy jargon, or thingie that flips the sign of all but one part in simple words) had to be expanded to *1, *2, and *3 to get the phase symmetry right for the current coupling term.

The second advance in quaternion algebra needed is the Even representation of quaternion multiplication. Here the eigenvectors of the representation must be excluded to make the representation an algebraic field. I have consistently said I am more comfortable when I found out someone else has done this before. The multiplication table is known as the Klein four-group. I will have to see if others have noticed what happens when the eigenvectors are excluded.

With these two innovations, the field equations for E, B, e and b have been generated. These field equations are gauge invariant because g may be whatever one chooses since it is canceled out in the process. This is vital since the graviton and proton both travel at the speed of light.

What needs to be done

I need to develop a non-gauge invariant set of field equations for massive particles, where the gauge symmetry is broken by the mass charge. I have something technical to keep me off the streets.

Doug

 

[sweetser]

The goal: a 4D wave equation

Hello:

A gauge-free unified field equation has been constructed using quaternion operators (previous post), a very good thing because the particles that mediate gravity and EM travel at the speed of light. The particles that interacted with these mediators of force - massive and possibly electrically charged particles - do not travel at the speed of light, so the gauge symmetry must be broken.

In a standard approach to EM, gauge symmetry is broken with the Higgs mechanism. This is done by postulating that there is a Higgs field everywhere any particle ever goes. The vacuum state is a false vacuum, actually higher than a nearby state that will give particles like quarks mass without breaking the symmetry needed by EM.

False vacuums remind me of false gods, something akin to the aether that had to be everywhere, no place in the Universe could be without. The Universe is a clumpy place, and it would be a magical coincidence if the Higgs field got to every place it needed to in the right density in order to give every particle the same symmetry breaking experience needed so all protons, neutrons, and electrons have the same mass. These are reasonable skeptical objections, but they get no air time today for a good reason: there is no alternative. One choice makes things simple.

A thought experiment may show that mass charge does break electric charge symmetry. You have a pair of electrons sitting 1 cm apart, and you measure how fast they accelerate away from each other, a_e = F/m_e[/tex]. Repeat the experiment, but for a pair of protons, a_p = F/m_p[/tex]. Let&amp;#039;s say your experimental system is so good you are able to measure the two accelerations to ten significant digits. The value of the two accelerations is identical. This equivalence indicates that electric charge is universal.&lt;br /&gt; &lt;br /&gt; Repeat the experiment without changing anything concerning the setup. There are two electrons over here rushing away from each other, two protons doing the same thing. The one difference is the acceleration is now measured to twenty significant digits. Now the two accelerations will not be the same. The reason is the gravitational mass has a trivial effect that will keep the heavier protons from accelerating as fast as the pair of electrons because the protons attract each other gravitationally more than the electrons. &lt;br /&gt; &lt;br /&gt; Unlike the standard model + Higgs which characterize inertial mass and ignores gravity, the GEM proposal is about gravity and EM working together ever so lightly. We saw that in post 457, eq 1, where the g&lt;sup&gt;2&lt;/sup&gt; term contributed from EM canceled with the g&lt;sup&gt;2&lt;/sup&gt; tossed in by gravity.&lt;br /&gt; &lt;br /&gt; This time, I will combine the method used to generate the Maxwell equations (post 438) with its Even quaternion representation clone (post 442). If you look carefully at the two Lagrangians, you would notice all the cross terms are the same, and they all have a minus sign (but don&amp;#039;t look too close or you will notice I got the sign of one term wrong). This means if we subtract one from the other, all the mixed terms drop, leaving only 12 squared terms.&lt;br /&gt; &lt;br /&gt; When I was using tensors, I found the contraction of the two asymmetric rank 2 tensors, \nabla_{\mu} A_{\nu} \nabla^{\mu} A^{\nu}, had these same twelve terms, along with 4 others. The field equations that come out of the asymmetric field strength tensor contraction are drop dead gorgeous (from page 1 of this long thread):&lt;br /&gt; &lt;br /&gt; J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}&lt;br /&gt; &lt;br /&gt; &lt;blockquote data-attributes=&quot;&quot; data-quote=&quot;Feynman lectures&quot; data-source=&quot;&quot; class=&quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&quot;&gt; &lt;div class=&quot;bbCodeBlock-title&quot;&gt; Feynman lectures said: &lt;/div&gt; &lt;div class=&quot;bbCodeBlock-content&quot;&gt; &lt;div class=&quot;bbCodeBlock-expandContent js-expandContent &quot;&gt; What a beautiful set of equations! They are beautiful, first, because they are nicely separated - with the charge density, goes \phi; with the current, goes &lt;b&gt;A&lt;/b&gt;. Furthermore, although the &lt;div style=&quot;text-align: right&quot;&gt;side looks a little funny - a Laplacian together with a (\partial/\partial t)^2 - when we unfold it we see...it has a nice symmetry in the x, y, z, t - the [c&amp;#039;s are] necessary because, of course, time and space &lt;i&gt;are&lt;/i&gt; different; they have different units.&amp;#8203;&lt;/div&gt; &lt;/div&gt; &lt;/div&gt; &lt;/blockquote&gt;&lt;div style=&quot;text-align: right&quot;&gt;&lt;br /&gt; &lt;br /&gt; To get the context correct, Feynman was extolling the virtues of writing the Maxwell equations in the Lorenz gauge. Earlier in this thread we went over how Feynman showed the current coupling term J^{\mu} A_{\mu} has spin 1 symmetry, necessary for EM where like charges attract. I know Feynman did not analyze the spin of (i J^{\mu} i)^* (k(j A_{\mu} j)^* k)^*. That has spin 2 symmetry, necessary for gravity where like mass charges attract. That&amp;#039;s the strength of this proposal, it keeps getting more subtle.&lt;br /&gt; &lt;br /&gt; I think I will stop here tonight and let the goal sink in...&lt;br /&gt; &lt;br /&gt; Doug&amp;#8203;&lt;/div&gt;

 

[Lawrence B. Crowell]

Hello:

The field equations that come out of the asymmetric field strength tensor contraction are drop dead gorgeous (from page 1 of this long thread):

J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}


To get the context correct, Feynman was extolling the virtues of writing the Maxwell equations in the Lorenz gauge.
Doug

The problem with this wave equation which gives the currents is that it assigns a mass current to an electromagnetic potential. I doubt many are going to cotten on to this idea.

The electromagnetic and gravity fields have their sources because of their group structure. Electromagnetism is an abelian gauge field A\wedge A~=~0[/tex] which has the unitary group U(1). The two roots of this group are the real parts of the circle on the Argand plane C of compex numbers. Those are of course the numbers +1 and -1. Gravitation is the Lorentz group which is the hyperbolic group SO(3,1). This group is similar to SO(4)~=~SO(2)\times SO(2) though the hyperbolic group is noncompact. SO(3,1)~=~SL(2,C)\times Z_2 and the special linear group is SU(2)\times SU(1,1). These two parts define the set of three rotations and three boosts which give the six generators of the Lorentz group. <br /> <br /> Compact groups have nice properties that a set of transformations of the group generator will converge in a Cauchy sequence. Hyperbolic groups are non-compact and so a sequence is likely to go off to asymptopia and not converge. <br /> <br /> If we think of gravity as a gauge-like theory with F~=~dA~+~A\wedge A for nonabelian gauge fields the DE&#039;s for these on the classical level are nonlinear. Yet we can quantize these, but renormalization is a bit complicated. We can well enough quantize a SO(4) theory obtained in euclideanization. But gravity is a strangely different. Why? The gauge group SU(1,1) is hyperbolic. In the Pauli matrix representation we have that \tau_z~=~i\sigma_z. So we form a gauge connection<br /> <br /> &lt;br /&gt; A~=~A^{\pm}\sigma_{\pm}~+~iA^3 \sigma_3&lt;br /&gt;<br /> <br /> and for the group element g~=~exp(ix\tau_3) =~exp(-x\sigma_3)the connection term transforms as<br /> <br /> &lt;br /&gt; A&amp;#039;~=~g^{-1}Ag~+~g^{-1}dg~=~e^{-2x}A^{\pm}\sigma_{\pm}~+~iA^3\sigma_3&lt;br /&gt;<br /> <br /> and for x~\rightarrow~\infty this gives A~\rightarrow~iA^3\sigma_3. Now A^{\pm}\sigma_{\pm} and A^3\sigma_3 have distinct holonomy groups and are thus distinct points (moduli) in the moduli space. But this limit has a curious implication that the field F~=~dA~+~A\wedge A for these two are the same and the moduli are not separable. In other words the moduli space for gravity is not Hausdorff. This is the most serious problem for quantum gravity.<br /> <br /> Because of this it is not possible to construct the connection coefficient for gravitation from an internal gauge connection that is the generator of a compact group. The basic point set topologies for the two are different. The point set topology for gravitation is the Zariski topology, and defines certain algebraic varieties and a sheaf bundle system that is fundamentally different from the more trivial principal bundle system for the unitary U(1) group of electromagnetism. Penrose&#039;s twistor theory exploits this property and in some ways is a fairly honest approach to quantum gravity, even if it has not proven to be terribly workable in more recent time.<br /> <br /> Lawrence B. Crowell

 

[sweetser]

The Dihedral group

Hello Lawrence:

The problem with this wave equation which gives the currents is that it assigns a mass current to an electromagnetic potential. I doubt many are going to cotten on to this idea.

Yeah, people will probably quickly jump to the wrong conclusion. I cannot stop that, nor do I care about such a null reaction.

If you want to look at the group theory behind the GEM proposal, you must look at the Lagrange density, not the field equations. The field equations don't say a word about the spin symmetries, or the symmetries of the field strength tensor, for Maxwell or GEM.

You fortunately are smarter than that, and discuse F. That is in the GEM proposal as - A \nabla + (A \nabla)^* = (0, E + B)=F. The way we write these - you preferring differential geometry, I choosing to work with quaternions - does not have an impact on group theory: it is the unitary group U(1) as it must be to characterize electromagnetism.

The problem as you note is gravity. We agree on that. You then discuss what the group for gravity is, which is kind of surprizing since we don't have a quantum gravity theory yet. How we know what the group theory for the theory we don't have is beyond me. You do know the limitations of current efforts to apply group theory to the problem of gravity. That was a fun challenge to read.

It would appear like you missed a comment made in post #457 about group theory. There I noted that I am using Klein four group, Z_2~X~Z_2, also known as the dihedral group, Dih_2. Since the group is finite, it is compact, a good thing as you have pointed out. This avoids the problem you cited about hyperbolic groups. My proposal is a linear theory for gravity because the spin-coupling is between 2 4-vectors, just like in EM. It is far better to quantize a linear theory than a nonlinear one!

If the technically skilled take a brief glance at my proposal and assume I only use U(1), I can wait for the more reflective to think about the advantages and challenges of using Z_2~X~Z_2 for gravity and U(1) for EM.

Doug

 

[Lawrence B. Crowell]

Hello Lawrence:

It would appear like you missed a comment made in post #457 about group theory. There I noted that I am using Klein four group, Z_2~X~Z_2, also known as the dihedral group, Dih_2. Since the group is finite, it is compact, a good thing as you have pointed out. This avoids the problem you cited about hyperbolic groups. My proposal is a linear theory for gravity because the spin-coupling is between 2 4-vectors, just like in EM. It is far better to quantize a linear theory than a nonlinear one!

If the technically skilled take a brief glance at my proposal and assume I only use U(1), I can wait for the more reflective to think about the advantages and challenges of using Z_2~X~Z_2 for gravity and U(1) for EM.

Doug

A dihedral group, or any polytope is a Coxeter-Weyl A-D-E system define a system of roots. These are the discrete symmetries for the bundle framing of a system. The continuous symmetry are the group generators.

I have not looked at your reference, but the relationship between cyclic groups such as Z_n are

<br /> D_n~=~\{(x_1,~\dots,~x_n)~\in~Z_n: \sum_{i=1}^n x_i = 2Z (even)\}<br />

which puts constraints on the D_n group that can be used for Voronoi cell "glue vectors." But at any rate the simple cyclic group Z_2 corresponds to the }D_2~\simeq~SO(4) or the Lorentzian version SO(3,1). Each of the x_i corresponds to the rotations in an SU(2). The product of these would correspond to a tetrad where one group is for local transformations of affine connections on the base manifold and the other would correspond to an internal bundle.

I could go on a whole lot more about this and how it leads to D_4~=~SO(8) and how SO(7,1) and how the spin(6)~\rightarrow~spin(4,2) embedded in SO(7,1) is the group for conformal gravity. This group has the cyclotomic ring defined by the Galois field GF(4). The Galois field is GF(4)~=~(0,~1,~z,~z^2) with z~=~{1\over 2}(i\sqrt{3}~-~1) with z^2~=~z^*. GF(4) is the Dynkin diagram for the Lie Algebra D_4~=~spin(8). The properties of the basis elements that produce a commutator are

<br /> z^2~=~z~+~I,~z^3~=~I,~{\bar z}~=~z^2,<br />

and defines the hexcode system C_6

None of this gets away from the problem that the holonomy groups for the Euclideanized versions of groups do not contain those in the Lorentzian or hyperbolic case. The Berger classification of groups according to their holonomy groups for loop variables of affine variables does not cover groups of the sort SO(n)\times SL(2,C). This is in spite of the fact that SL(2,C)~=~SU(2)\times SU(1,1) and both SU(2) and S(1,1) have the same Dynkin root system.

Lawrence B. Crowell

 

[sweetser]

The dihedral division algebra

Hello Lawrence:

I am trying to decipher what is good or bad about the specific proposal in this thread. A gentle reader would need a good graduate level class on group theory to appreciate all the relationships you have pointed out. For those readers of this thread further behind than myself, I will point out a few things I have learned so far.

The Klein four-group is abelian, so two elements in the group commute. That is consistent with my definition of the Even representation of quaternion multiplication.

The Klein four-group can be viewed as a subgroup of A_4. It does not have a simple graph. Instead the graph has 4 vertices, where only two of them are connected:

.-.
. .

where the dots are the vertices, the dash a connecting edge.

Since I came to this group from a quaternion, it was interesting to read this on wikipedia:

http://en.wikipedia.org/wiki/Klein_four-group]The[/PLAIN] Klein four-group is the group of components of the group of units of the topological ring of split-complex numbers.

Quaternions in the usual Hamilton representation are 3 complex numbers that share the same real. In the Even representation, the complex numbers are split. Nice.

A cyclic graph is created by taking powers of the group to the n. The graph for the Dih₂ has the identity element in the middle, connected to the three other elements of the group. For the chemists in the audience, it looks like an amino group.

Another element of this puzzle are the finite fields or Galois fields. A way to represent the Klein group is as the set of four elements, {1, 3, 5, 7} modulo 8. So 1*3=3, 3*7 Mod 8=5, 5*7 Mod 8 = 3. The representation is Abelian, since 3*5 = 5*3. The Klein group is isomorphic to GF(4) because there are 4 prime numbers in the representation shown.

So far so good.

One way in which the Even representation of quaternions is different from the Klein group is that the group is modulo the eigen vectors of the 4x4 real matrix representation. This is necessary so that the group is a division algebra, a subject I did not come across in my readings on the topic. The moding out of the eigen vectors might have big implications, because then from one element of the group, there would necessarily be a way to get to any other element. This implies that the graph for Dih₂ Mod (eigen vectors) is simply connected.

Now we come to the objection:

None of this gets away from the problem that the holonomy groups for the Euclideanized versions of groups do not contain those in the Lorentzian or hyperbolic case.

The gentle reader might wonder what a holonomy group is, having never read about it in the funny pages. This is a topic that is central to differential geometry. It is about the relationship between the connection and the manifold. If one moves around the manifold, and does not quite get back to where one started, that is the subject of this topic.

I can now appreciate the problem at hand. If the graph for a group is not simply connected, then one cannot represent the Lorentz group - a picture of flat spacetime - or the small deviations from the Lorentz group needed for gravity to be a dynamic metric theory. If the GEM proposal used the group Dih₂ for gravity, it is reasonable to say it could not represent gravity as a smooth metric theory. Although not sure how to write this, I am using a different group, Dih₂ Mod (eigen vectors). One can travel anywhere on the manifold with the connection, a necessary thing. At this time, I don't know the impact on parallel transport.

Doug

 

[Lawrence B. Crowell]

Doug,

For some reason it took a long time to get on here, so I don't have much time. I have a Schubertiade to atttend to. The Wiki-p page indicates what I said. This is involved with D_2, or its roots. These structures are useful in deriving group structures, and I am actively involved with work along these lines, though more up the ladded to Leecha lattices, Conway groups and up to the Fischer-Griess "monster." Yet at the end this does not directly address the issue of non-compact group structure and holonomies. There is in my thinking a system of projective varieties over these in the form of quantum codes, which give light cone structure. Penrose's twistors are related to this type of structure.

Anyway, I will try to address more directly what it is that you say above later this weekend.

Lawrence B. Crowell

 

[sweetser]

Graphs of quaternion represetations

Hello:

In this post I will propose what graphs describe the Even and Hamilton representations of quaternions.

A graph has a number of vertices and then edges which are made of pairs of vertices. The Hamilton and Even representations both have the same 4 vertices: (e, i, j, k). To keep this possibly related to physics, I like to think that in flat spacetime, the absolute value of each of these is equal to 1, but in the curved spacetime of GEM theory, the absolute value of e is the inverse of the absolute value of i, j, and k.

For the Even representation of quaternions, all the vertices are connected to each other. The graph looks like a box with an X:

.-.
|X|
._.

vertices: (e, i, j, k)
edges: ((e, i), (e, j), (e, k), (i, j), (i, k), (j, k)
name: K₄:6

This is known as a complete graph because every pair of distinct vertices is connected to an edge.

Here was a bit of fun I had mixing my physics in with pure math. I was thinking about the labels for the edges. For the edge (e, i), I thought I would just use i, similarly for j and k. For the edges connect two 3-vectors, it would be the third 3-vector. The problem is that no labels use e. That didn't sound good to me, ignoring e. What if I labeled each edge as i/e, j/e, or k/e? The physics of GEM suggest the absolute value of this will be one in curved or flat spacetime, so the edge is invariant. Nice. If one is at a vertex and wants to get to another vertex, read the label, and form the product, such as e (j/e) = j.

What about the Hamilton representation of quaternions? We still have four vertices. The part of the graph that connects with e is exactly the same. What changes are the edges that connect the 3-vectors. Because the 3-vectors do not commute, the edges must be directional. The edge that connects i to j has a label of -k/e because -i k/e = j. The directional edge from j to i has the label k/e. Here is the graph for the Hamilton representation of quaternions:

.=.
|X()
._.
vertices: (e, i, j, k)
edges: ((e, i), (e, j), (e, k))
directed edges: ((i, j), (j, i), (i, k), (k, i), (j, k), (k, j))
name: K₄:9 ?
Note: the \ of the X should be two directed lines, the limitations of ASCII graphs.

I am not certain if the K classification allows for including directed edges. It is significant that the graphs for the Hamilton and Even representations of quaternions are not the same. Gravity is not the same as EM, in fact, gravity is a little bit simpler (one charge, only attracts), and it is gratifying that the graph for the Even representation is a little simpler than that for the Hamilton representation.

This was a lot of fun, I hope you enjoyed. I have never applied graph theory to anything before in my life.

Doug

 

[Lawrence B. Crowell]

This is in part a test. I am having trouble getting a post sent here.

 

[Lawrence B. Crowell]

What about the Hamilton representation of quaternions? We still have four vertices. The part of the graph that connects with e is exactly the same. What changes are the edges that connect the 3-vectors. Because the 3-vectors do not commute, the edges must be directional. The edge that connects i to j has a label of -k/e because -i k/e = j. The directional edge from j to i has the label k/e. Here is the graph for the Hamilton representation of quaternions:

.=.
|X()
._.
vertices: (e, i, j, k)
edges: ((e, i), (e, j), (e, k))
directed edges: ((i, j), (j, i), (i, k), (k, i), (j, k), (k, j))
name: K₄:9 ?
Note: the \ of the X should be two directed lines, the limitations of ASCII graphs.


Doug[/QUOTE]

It is best to go all the way and consider the 120 icosian of quaterions, and its extension to the 240 cell with the 128 elements which give the 240 roots plus the 8 Cartan center elements.

The basic group system for gauge theory is the heterotic group E_8, which is one of the Heterotic groups. There has been a lot of activity with this, and the Vogan DeCloux group found its system of root representations and possible impacts on elementary particles and gravitation. There is also a business called supersymmetry, which has 2^{N} elements in its representation. For something called N = 8 this has 256 elements. There is then the Clifford algebra, noted as CL(16)~=~CL(8)\times CL(8) which gives a relationship between these 256 elements and the 240 roots of E_8 and its 8 weights (weights are the Cartan centers or eigen-matrices of the roots). The roots are what define the physical states. Now "half" of this CL(8) is defined by

<br /> CL(8)~=~1~+~8~+~28~+~56~+~70~+~56~+~28~+~8~+~1<br />


for spins 2, 3/2, 1, 1/2, 0, -1/2, -1, -3/2, -2. The 248-dim Lie algebra E_8 = 120-dim adjoint Spin(16) + 128-dim half-spinor spin(16) is rank 8, and has 240 root vectors that form the vertices of an 8-dim polytope called the Grosset polytope.

Now the Clifford basis decomposition of the 256 there are the following elements

<br /> CL(8)~=~1~+~8~+~(24+*4*)~+~(24+4+28)~+~(32+3+3+32)~+~(28+4+24)~+~ (24+*4*)~+~8~+~1,<br />

The * * eclosed give the 4+4 = 8 that corresponds to the 8 E_8 Cartan subalgebra elements that are not represented by root vectors (they are the eigen-matrices), and the black non-underlined 1+3+3+1 = 8 correspond to the 8 elements of 256-dim Cl(8) that do not directly correspond elements of 248-dim E8. Now for the spins \pm 1/2 there is a 24 = 8 + 8 + 8, which are the three generations of fermions. These are the long roots associated with the 24, which corresponds to the elements of gauge fields which are the SU(3)xSU(2)xU(1) of the standard model. Now the 8 + 8 + 8 are the short roots which complement the roots of the SU(3), and define a triality condition on the fermionic sector.

This triality condition appears to be a very central aspect of group theory and irreducible representations (ir-rep). I have been doing some calculations on a three fold structure with mutually unbiased bases on Jordan algebras (an ir-rep of E_8 elements with octonionic OP^2 structure), so that the heterotic E_8 is embeded in a three fold structure within the Leech lattice and a general system of modular forms.

But to bring things to a more common level, the E_8 is a rich structure on symmetries with the numbers 2, 3, 5, 8. The two is on the helicity match up above, the 3 is on the triality condition of subgroup ir-reps, the 5 is a symmetry on an icosahedral system "dual" to the triality symmetry, and the 8 the whole system. This naturally embeds in a further three fold system called the Steiner system with the [5, 8, 24] structure for a "code." So the reason there are three families of quarks, three families of leptons (8 + 8 + 8) etc is due to this rich algebraic structure.

Why there are four forces is again an aspect of group decomposition. The Cartan center for the E_8 is related to the other heterotic groups G_2,~F_4,~E_6~ and~ E_7. A basic representation is F_4~=~C_{E_8}(G_2) (the Cartan center weighted on G_2), where G_2~=~SU(3)~+~3~+~{\bar 3}~+~1, which gives the nuclear force plus a hypercharge "1". The F_4 is a D_4 group \sim~SO(7,1) which embeds gravity and the weak interactions plus the 8 + 8 + 8.

Yet this structure operates equally for Euclidean and Lorentzian signatures. There is a loss of information, which is being glossed over. This does not solve the problem of ambiguities in holonomic structure for non-compact groups. In quantum field theory it is a common practice to Euclideanize time by letting time go to t~\rightarrow it. This time turns out to be related to temperature by

<br /> t~=~\frac{\hbar}{kT}<br />

This time is not exactly the same as the Lorentzian time, what we measure on a clock (or might we say what a clock “produces”), but is really a measure of the time where quantum fluctuations may be observed. Sometimes the term quantum fluctuations causes trouble, so it really is more the distance in a Euclidean 4-dim space where an instanton (a tunnelling state etc) with a certain magnitude can appear. As this temperature becomes very small the fluctuation time becomes large and the strength of the fluctuation, if we qualitatively invoke the Heisenberg uncertainty principle

and consider this instanton time t as this uncertainty in time. As the temperature heats up it also means that the fluctuation is stronger or its coupling is made larger and a phase transition will ensue.

This gets into some fascinating stuff! The Lorentzian time implies that the moduli space is not separable. Two moduli, points in the space of “gauge equivalent connections,” are not separable in a Hausdorff point-set topological definition. The topology is Zariski. I can go write more about this if needed, but this gets us into some rather serious stuff. So there is the Lorentzian time and there is this Euclidean time and there is a Wick rotational map between the two. So distinct fluctuation in the Euclidean case which have moduli that are Hausdorff, separable and “nice” correspond to a set of moduli in the Lorentzian case which are not. Physically this means there is some scale invariant physics (again to go into would require a bit of writing work) of phase transitions associated with the correspondence between fluctuations at various pseudo-time scales (or temperatures) and their Lorentzian versions.

This correspondence and the phase transition is a quantum critical point, which have been observed with High temp superconductors and in the physics of Landau electron fluids in metallic crystals in the actinide plus transition range. This also shares some aspects of physics with the Hagedorn temperature of strings.

Lawrence B. Crowell

 

[sweetser]

Not a Clifford algebra

Hello Lawrence:

Thanks for all the information on the group E₈, the largest exceptional Lie group. Garrett Lisi made international news back in November for a unification proposal that used E₈ without resorting to strings.

Unfortunately, I am not able to implement your suggestion. A brief description of icosian make it clear one is dealing with "a non-commutative algebraic structure". That does work of the Hamilton representation of a quaternion. I am using the word quaternion to mean a 4D division algebra. Many people work productively with the assumption that the only implementation of a 4D division algebra is the one Hamilton developed so many years ago (Gauss got there first by the way).

In this thread, I found it necessary to formulate a new representation of a 4D division algebra where the elements commute. To make the Hamilton representation a mathematical field, one only needs to exclude the additive inverse 0 from the quaternions. For the Even representation, there are more quaternions that need to be excluded (0 and the eigen vectors) but it can be done.

The Even representation of quaternions is not a Clifford algebra. Clifford algebras have orthogonal basis vectors, e_i e_j = -e_j e_i. For the Even representation, e_i e_j = e_j e_i. Oops. I am going to go to the 8th International Conference on Clifford Algebras and their Applications to Mathematical Physics at the end of May in Brazil, and claim the generalization of quaternions that Clifford Algebras represent is not enough to unify gravity and light (that will not go over well!).

Let's think about the two graphs again, something I have done all weekend with great amusement. My central thesis is that we have yet to do the math of 4D like Nature practices it. In the graph for the Hamilton and Even representations, both have 4 vertices. Sometimes I think of them as events - a t, x, y, and z - but even better might be a difference between two events - a dt, dx/c, dy/c, dz/c. Recall the labels I used, which with this application would be dx/dt c, or \beta_x.

The Even and Hamilton representations also share three edges, those for (e, i), (e, j), and (e, k). These seven shared elements indicate an overlap between the two representations. So how are they different? When one wants to connect two vertices that are in the 3-vector, there are two choices: use and edge or a directed edge. The Even representation uses an edge, .-., while the Hamilton representation uses two directed edges, .=. (please imagine arrows on both lines in opposing directions). I got so excited by this simple, direct message, that I designed and made a button with the graph for gravity and light last Saturday night, wearing it to a Boston swing dance. Fresh math! If anyone reading these notes would like a similar button, just email me off this thread with your snail mail address, and I will put one in the mail in a few weeks (I should reward those that are doing the work of trying to follow these nerdly riffs).

doug
sweetser@alum.mit.edu

 

[Lawrence B. Crowell]

Hello Lawrence:

Thanks for all the information on the group E₈, the largest exceptional Lie group. Garrett Lisi made international news back in November for a unification proposal that used E₈ without resorting to strings.

It is an framing scheme which puts the particles into the F_4 and G_2 exceptional subgroups of the E_8. It is a neat idea in some ways, but there are few ambiguities, in particular how spin(4,2) fits into the scheme. The heterotic string has E_8\times E_8 for two chiralities and in this one can supersymmetrize the theory. One does not need to have string theory explicitely in here, but the Jordan subalgebra naturally contains stringy structure, such as the 26-dimensional bosonic string.

Unfortunately, I am not able to implement your suggestion. A brief description of icosian make it clear one is dealing with "a non-commutative algebraic structure". That does work of the Hamilton representation of a quaternion. I am using the word quaternion to mean a 4D division algebra. Many people work productively with the assumption that the only implementation of a 4D division algebra is the one Hamilton developed so many years ago (Gauss got there first by the way).

The quaternionic division algebra is noncommutative by definition. The wiki-p site just discusses the Hamilton vertex permutation on the icosian. There is a whole lot more structure to this system.

In this thread, I found it necessary to formulate a new representation of a 4D division algebra where the elements commute. To make the Hamilton representation a mathematical field, one only needs to exclude the additive inverse 0 from the quaternions. For the Even representation, there are more quaternions that need to be excluded (0 and the eigen vectors) but it can be done.

This is a contradition in definition. The Cayley numbers 1, 2, 4, 8 lead to the reals, complexes, quaterions and octonins. Quaternions as a division algebra are noncommutative.

The Even representation of quaternions is not a Clifford algebra. Clifford algebras have orthogonal basis vectors, e_i e_j = -e_j e_i. For the Even representation, e_i e_j = e_j e_i. Oops. I am going to go to the 8th International Conference on Clifford Algebras and their Applications to Mathematical Physics at the end of May in Brazil, and claim the generalization of quaternions that Clifford Algebras represent is not enough to unify gravity and light (that will not go over well!).

This appears related to a graded structure. The even quaterionic structure involves elements e_i~=~\xi_a e^a_i, for \xi^a a Grassmannian element. The quaterions then define supermanifold coordinates with

<br /> y^i~=~x^i~+~{\bar\theta}\sigma^i\theta.<br />

You will then have \{e^a_i,~e^b_j\}~=~\eta^{ab}g_{ij}, which are fermionic correspondences with quaterionic elements. The quaterionic fields are framed within a Clifford basis, such as a connection form

<br /> {\cal A}~=~A~+~\Gamma\cdot\psi<br />

So that in the case of QCD the SU(3) is combined with the 3~+~\bar 3 in G_2. Similarly the fields can be again framed with Grassmannian elements.

This will then extend these elements to a graded algebraic correspondence between bosons and fermions. I have had some communciations with Garrett Lisi and others on just this issue of extending the framing system to a graded Lie algebraic structure.

Lawrence B. Crowell

 

[sweetser]

Representation theory

Hello Lawrence:

The question is whether representation theory can play a role in the description of quaternions. As a warmup exercise, let's write out two different representations of what I have been carefully calling the Hamilton representation of quaternions.

<br /> q1 (t, x, y, z) = \left(\begin{array}{cccc}<br /> t &amp; -x &amp; -y &amp; -z\\<br /> x &amp; t &amp; -z &amp; y\\<br /> y &amp; z &amp; t &amp; -x\\<br /> z &amp; -y &amp; x &amp; t<br /> \end{array}\right)\quad eq 1<br />

<br /> q2 (t, x, y, z) = \left(\begin{array}{cccc}<br /> t &amp; x &amp; y &amp; z\\<br /> -x &amp; t &amp; z &amp; -y\\<br /> -y &amp; -z &amp; t &amp; x\\<br /> -z &amp; y &amp; x &amp; t<br /> \end{array}\right)\quad eq 2<br />

These two 4x4 real matrix representation of quaternions are not identical. Yet if we look at the multiplication table, they turn out to be the same. We would say the real 4x4 matrix representation q1 is isomorphic to the one for q2. These matrices have addition, subtraction, and multiplication. The one step that requires a light amount of work is to get the inverse and thus show the matrix is always invertible. Turns out that is not the case because there is a division by the norm, t² + x² + y² + z². If one excludes zero, then division works.

Representation theory does apply to the definition of quaternions. Working exclusively with isomorphic representations can get dull quickly. Let's find a representation that is not isomorphic to eq 1 (and thus eq 2). That would be the Even representation of quaternions, here written in as a 4x4 real matrix:

<br /> q1 (t, x, y, z) = \left(\begin{array}{cccc}<br /> t &amp; x &amp; y &amp; z\\<br /> x &amp; t &amp; z &amp; y\\<br /> y &amp; z &amp; t &amp; x\\<br /> z &amp; y &amp; x &amp; t<br /> \end{array}\right)\quad eq 3<br />

Again it should be obvious that such a matrix has addition, subtraction, and multiplication well defined for all members. Fire up Mathematica, and you find out, just like eq 1, the matrix is not always invertible. To make it invertible, zero and the eigen vectors of this matrix are omitted.

One of the things that gives me confidence is when I spot errors. This might appear contradictory, but mathematical physics can be so abstract, it can be difficult to spot the right direction. If a question can be asked with enough precision, and shown to be in error, then a correction to the course can be made.

I have said several times that one needs to exclude the Eigen vectors to construct the Even representation of the quaternions. Being self-skeptical, I fired up Mathematica to confirm my statements. Here is the output.

The Hamilton representation:

q1[t_, x_, y_, z_] := \left(<br /> \begin{array}{cccc}<br /> t &amp; -x &amp; -y &amp; -z \\<br /> x &amp; t &amp; -z &amp; y \\<br /> y &amp; z &amp; t &amp; -x \\<br /> z &amp; -y &amp; x &amp; t<br /> \end{array}<br /> \right)\quad eq 4

Simplify[Inverse[q1[t,x,y,z]].\{1,0,0,0\}]
\left\{\frac{t}{t^2+x^2+y^2+z^2},
-\frac{x}{t^2+x^2+y^2+z^2},
-\frac{y}{t^2+x^2+y^2+z^2},
-\frac{z}{t^2+x^2+y^2+z^2}\right\} \quad eq 5

The Even representation:

q3[t_, x_, y_, z_] := \left(<br /> \begin{array}{cccc}<br /> t &amp; x &amp; y &amp; z \\<br /> x &amp; t &amp; z &amp; y \\<br /> y &amp; z &amp; t &amp; x \\<br /> z &amp; y &amp; x &amp; t<br /> \end{array}<br /> \right) \quad eq 6

Calculate the inverse:

Factor[Inverse[q3[t,x,y,z]].\{1,0,0,0\}]
\left\{\frac{t^3-t x^2-t y^2+2 x y z-t z^2}{(t+x-y-z) (t-x+y-z) (t-x-y+z) (t+x+y+z)},
-\frac{t^2 x-x^3+x y^2-2 t y z+x z^2}{(t+x-y-z) (t-x+y-z) (t-x-y+z) (t+x+y+z)},
-\frac{t^2 y+x^2 y-y^3-2 t x z+y z^2}{(t+x-y-z) (t-x+y-z) (t-x-y+z) (t+x+y+z)},
-\frac{-2 t x y+t^2 z+x^2 z+y^2 z-z^3}{(t+x-y-z) (t-x+y-z) (t-x-y+z) (t+x+y+z)}\right\} \quad eq 7

Calculate the Eigenvalues:

Eigenvalues[q3[t,x,y,z]]
\{t+x-y-z,t-x+y-z,t-x-y+z,t+x+y+z\} \quad eq 8

The mistake I made was it is the Eigen values that need to be excluded. In the Hamilton representation, the same value of zero must be excluded 4 times. In the Even representation, zero needs to be excluded, along with three other values. That is not too different, but does indicate the representations are not isomorphic.

I have no problem with the statement that quaternions have traditionally been defined exclusively as a 4D division algebra that is noncommutative. I have read that a thousand times. Yet I do research. Sometimes you need to knock heads. I do so using representation theory in this case. That nice every-element-is-positive matrix is 4D in eq 3, can be added to other elements just like itself, can be subtracted from other elements like itself, can be multiplied to create other elements just like it, and should its Eigen values be excluded, can promise that a inverse necessarily exists.

I might be able to decrease conflict by talking about 4D division algebras, one of which happens to be the noncommuting quaternions, the other being a hypercomplex number. I am not a fan of this approach since it does not pay respect to representation theory. There is nothing "hyper" about the matrix in eq 3: all the elements are positive. No wonder it may turn out to be the math behind the one universally attractive force in Nature, gravity.

Doug

 

[Lawrence B. Crowell]

Hello Lawrence:

The question is whether representation theory can play a role in the description of quaternions. As a warmup exercise, let's write out two different representations of what I have been carefully calling the Hamilton representation of quaternions.

Doug

As this goes on I am beginning to think that you have very different definitions for various things. It is a bit as if you have some alternate mathematics, such as symmetric field tensors and the like.

Of course quaternions or a Clifford basis has a representation. The Dirac matrices are a particular represenation of Cl(3,1) in the (0,~1/2)\oplus (1/2,~0) spinor representation of the Lorentz group. It is not too difficult to impose a local representation theory on the Dirac matrices and define connection terms and curvatures for general relativity in a spinorial form. Then the gravity field terms are equivalent to the Clifford spinors, and their product is just a Clifford bispinor.

Lawrence B. Crowell

 

[sweetser]

New meth means new physics

Hello Lawrence:

It certainly is possible that I am misusing standard technical terms. I point out one such error in my last post, where I should have used Eigen value instead of Eigen vector.

It would be a concern to me if as a general practice I was using "very different definitions of various things". This sort of problem does happen for people who do Independent Research. This happens because those independent people are so isolated, they do not learn correct definitions.

In post 469 I talked about three different representations for quaternions. For two of them, I said they were isomorphic representations. In the examples section of http://en.wikipedia.org/wiki/Group_representation, they show trivially different matrices that end up having the same multiplication table. If every representation was isomorphic, we would not need the word isomorphic, it would be part of what representation theory means. Representations of groups that are not isomorphic are an important branch of representation theory.

I like this comment:

It is a bit as if you have some alternate mathematics, such as symmetric field tensors and the like.

I am a subversive, but I am a bit sensitive to the phrase "alternate mathematics" since it is close to "math that makes no sense". My math is precise enough for Mathematica to understand it, defined in equation 6 of post 469, it found the inverse in equation 7, and the Eigen values in equation 8. One cannot put a vague or BS math into a symbolic math package and have it spit out sensible results.

I have been so precise, so into the standard lexicon of math, that I think I am trying to use the graph K₄:6 as the foundation for gravity, and K₄:9 for EM.

What you have done consistently is try an throw a noncommuting curveball, and I am not swinging at it. Yes, the Hamilton representation of quaternions - the only ones anyone ever learns - is the Clifford algebra CL(0, 2). That is part of my proposal. Not being a Ph.D., I practice the rare art of intellectual minimalism. I can get by with few things, used creatively.

What was Clifford trying to do? He was trying to generalize quaternions, focusing on that strange animal, the cross product. In my work, one cross product is enough. I don't need many varieties of a cross product. What I need is something that is genuinely different from a cross product, hence the two K4 graphs. I don't need more than 4 dimensions. Actually, I think if a proposal works in 5 or more dimensions, it is wrong on the grounds of dimensional analysis alone. I recognize that people in academia may find such a dismissive stance of all work done on Kaluza-Klien and strings harsh, oh well. If people choose to work with the Riemann curvature tensor, the Ricci tensor, the Ricci scalar, or Einstein's field equations, they may do so to describe gravity. The GEM proposal is dedicated to the idea that work with gravity and the tools of Riemann curvature will continue to fail to connect with quantum mechanics as they have since the 1930s (about the time when people thought GR and quantum mechanics had to be united somehow). CL(0, 2) is enough, 4D is enough, the Christoffel is needed but the Riemann curvature tensor is not. Lucky I am not trying to get a grant.

If by "alternate mathematics" you mean I cannot connect to the vast amount of work done on gravity in the past, I can accept that as a compliment. It is a good sign for originality in an area of study that needs a serious slap in the face.

Doug

 

[Mentz114]

Hi Doug:

The competition has been busy. You really should have a look at this

Gravitomagnetism in teleparallel gravity

E. P. Spaniol, V. C. de Andrade‡
Instituto de Fisica, Universidade de Bralsilia
C. P. 04385, 70.919-970
Brasilia DF, Brazil

arXiv : 0802.2697v1

extract -

In the present work, a dfferent approach will be adopted to reexamine
gravitomagnetism. Due to the fundamental character of the geometric structure
underlying gauge theories, the concept of charges and currents and, in particular, the
concept of energy and momentum are much more transparent when considered from
a gauge point of view. Accordingly, we shall consider gravity to be described
as a gauge theory for the translation group, which gives rise to the so-called
teleparallel equivalent of GR. In this scenario we recover all the aspects predicted by
GR and moreover we have all the formal structure of a gauge theory, which is naturally
close to electromagnetism due to its abelian character. Therefore, the concepts of
gravitoelectric and gravitomagnetic fields emerge, as we will see, in the same way as
in the electromagnetic theory, that is, as components of the field strength of the gauge
theory.
The paper is divided as follows: in section 2 we review the fundamentals of
teleparallel gravity; in section 3 the gravitational Maxwell equations are introduced
in their exact form...

You might even meet these guys on your jaunt.

I'd like to hear what Lawrence B. thinks about Teleparallel gravity. It's very appealing that it's a simple gauge theory.

 

[Lawrence B. Crowell]

Hello Lawrence:

It certainly is possible that I am misusing standard technical terms. I point out one such error in my last post, where I should have used Eigen value instead of Eigen vector.

It would be a concern to me if as a general practice I was using "very different definitions of various things". This sort of problem does happen for people who do Independent Research. This happens because those independent people are so isolated, they do not learn correct definitions.

---------

If by "alternate mathematics" you mean I cannot connect to the vast amount of work done on gravity in the past, I can accept that as a compliment. It is a good sign for originality in an area of study that needs a serious slap in the face.

Doug

The problem that I have is that some of what you say simply makes little sense to me. The reasons for anti-symmetric field tensors, for instance, are due to some basic results in differential geometry. The reason for some of these structures are mathematically determined by some very well grounded theorems in mathematics. I say this as someone familiar with the theorems of Uhlenbech, Freedman and Donaldson on the differential topology of gauge theory on four manifolds, and the celebrated Atiyah-Singer index theorem which determines the structure of moduli spaces. There is nothing in the mathematical literature which points to anything which you allude to.

There is a reason why people work with dimensions larger than four, or five in the case of the EM Kaluza-Klein theorem. It is likely that the structure of elementary particles is intimately associated with quantum gravity and the structure of the universe. The Maldacena result on the dual isomorphism between the Anti-deSitter spacetime and the conformal structure of field theories is a clear indication that quantum gravity necessitates a unification with gauge field theory and their fermionic sources (quarks, leptons, Higgs, dilatons etc). And as much as you might not like it this gets one into all the complexity of supersymmetry, some stuff with string theory, loop quantum gravity, maybe twistor theory and ... .

My point about nonholonomic loops and noncompactness indicates that I think there is a major physical (and equivalently mathematical) element which is missing from all of physics out there. I am working on a number of possiblities to address this question. I also lean a bit on subjects such as solid state physics, the theory of quantum gases (boson condensates etc) and quantum liquids, quantum phase transitions and so forth. I am primarily interested in approaching this from a physical basis, and exploiting the mathematics to make it work where necessary.

Also, I and anyone can solve equations on MATHEMATICA. However, if those equations are arrived at by wrong mathematics then the solutions don't mean a whole lot, even if done by computer.

Lawrence B. Crowell

 

[Lawrence B. Crowell]

Hi Doug:

The competition has been busy. You really should have a look at this

Gravitomagnetism in teleparallel gravity

E. P. Spaniol, V. C. de Andrade‡
Instituto de Fisica, Universidade de Bralsilia
C. P. 04385, 70.919-970
Brasilia DF, Brazil

arXiv : 0802.2697v1

extract -


You might even meet these guys on your jaunt.

I'd like to hear what Lawrence B. thinks about Teleparallel gravity. It's very appealing that it's a simple gauge theory.

This has some interesting prospects. The Finsler geometry can be used to intertwine all possible frames. So potentially in a more general setting, potentially useful for quantum gravity, accelerated frames can be brought into a form of equivalence with inertial frames. Then accelerated frames are in a more general setting a form of comoving frame with an underlying isometry.

Lawrence B. Crowell

 

[sweetser]

Missing elements

Hello Lawrence:

I think we should be able to part on good terms.

The problem that I have is that some of what you say simply makes little sense to me. The reasons for anti-symmetric field tensors, for instance, are due to some basic results in differential geometry. The reason for some of these structures are mathematically determined by some very well grounded theorems in mathematics. I say this as someone familiar with the theorems of Uhlenbech, Freedman and Donaldson on the differential topology of gauge theory on four manifolds, and the celebrated Atiyah-Singer index theorem which determines the structure of moduli spaces. There is nothing in the mathematical literature which points to anything which you allude to.

All the physicists eggs are in the anti-symmetric field tensor basket for good technical reasons. It takes an impressive investment to understand the groundwork of differential geometry as you have done.

Important breakthroughs are the sport of the young who have yet to learn the foundations so well. Those youngsters need to be part of a cell that solve enough nagging questions that the establishment has to pay attention.

I am not particularly young, and I do not have a "cell". I am using the Internet to try and construct one via my web assets (this thread on physicsforums.com, quaternions.com, quaternions.sourceforge.net, TheStandUpPhysicist.com, and probably most important, YouTube.com). I can quantify the number of folks which continue to read this thread, and have statistics on my web sites and YouTube (18k downloads).

There is a reason why people work with dimensions larger than four, or five in the case of the EM Kaluza-Klein theorem. It is likely that the structure of elementary particles is intimately associated with quantum gravity and the structure of the universe. The Maldacena result on the dual isomorphism between the Anti-deSitter spacetime and the conformal structure of field theories is a clear indication that quantum gravity necessitates a unification with gauge field theory and their fermionic sources (quarks, leptons, Higgs, dilatons etc). And as much as you might not like it this gets one into all the complexity of supersymmetry, some stuff with string theory, loop quantum gravity, maybe twistor theory and ... .

I know researchers are rational. My objection was technical, and it was not addressed. The units for spacetime are wrong. You start off in Mathematica wrong, everything that follows is wrong. You start off wrong in theoretical work, and everything that follows is wrong. Same harsh logic.

My preferences are not going to change the flow of research money or efforts into work on strings or loop quantum gravity. I am not bitter that physics research happens to be going that way. Nature doesn't care if we get the right answer or the wrong one, and neither do I. I have no fear of anyone. I have no problem telling the brightest physicists on the planet that because the units of higher dimensional spacetime are wrong, what they work on is wrong and will not last the test of time. They will go off on a compatification riff which as my mother, God bless her soul, would say is a bunch of crap. They will get funded, go to conferences, talk about the latest new result, but in time, all my chips are bet on the collapse of extra dimensional spacetime work.

People who do work on gravity do not even acknowledge the risk that their work could be wrong. This is purely a technical issue: if gravity gets united with the rank 1 field theory for EM by also being a rank 1 field theory, then every paper that presumes gravity must be rank 2 is wrong.

My point about nonholonomic loops and noncompactness indicates that I think there is a major physical (and equivalently mathematical) element which is missing from all of physics out there. I am working on a number of possiblities to address this question. I also lean a bit on subjects such as solid state physics, the theory of quantum gases (boson condensates etc) and quantum liquids, quantum phase transitions and so forth. I am primarily interested in approaching this from a physical basis, and exploiting the mathematics to make it work where necessary.

I wish everyone luck in their research. Your point about both group theory and noncompactness effected me in a good way. I was trying to keep up with all your comments on group theory, and was not satisfied with what I could say about group theory as it applied to GEM. One of the big accomplishments I have had is to visualize the groups SU(2) and SU(3) with animations of quaternions. That work did not link to the discussions on the Hamilton and Even representations of quaternions. I recently got a book, "Quaternions, Clifford Algebras and Relativistic Physics" based on the title alone :-) The author Patrick Girard works within the standard limitations people bring to the topic, not realizing the Maxwell equations can be written with real quaternions as has been done here. On page 4, he wrote out the multiplication of the Klein four-group, and it looked identical to what I had posted here for the Even representation of quaternions. Nice.

So I read more about this group. That eventually led to the graph of the dihedral group:

._.
. .

Although I had a group, it did not look right. We start from a flat metric:

d \tau^2 = dt^2 - dR^2/c^2

Take the Newtonian step away from this flat spacetime:

d \tau^2 = (1 - 2 \frac{G M}{c^2 R}) dt^2 - dR^2/c^2

This is Newton's law written as a metric theory. The potential that is consistent with field theory is (1 - 2 \frac{G M}{c^2 R}), not -\frac{G M}{c^2 R} as is often written. Take one more step away to get to the first-order parameterized post Newtonian metric:

d \tau^2 = (1 - 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2) dt^2 - (1 + 2 \frac{G M}{c^2 R}) dR^2/c^2

The Schwarzschild metric a solution to the Einstein field equations, or the exponential metric a solution to the GEM field equations, both have the same Taylor series expansion that match these terms. It is the next terms where GR and GEM part ways.

GR:
d \tau^2 = (1 - 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2 - 3/2 (\frac{G M}{c^2 R})^3) dt^2 - (1 + 2 \frac{G M}{c^2 R} + 3/2 (\frac{G M}{c^2 R})^2) dR^2/c^2

GEM:
d \tau^2 = (1 - 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2 - 4/3 (\frac{G M}{c^2 R})^3) dt^2 - (1 + 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2) dt^2) dR^2/c^2

At second order parameterized post-Newtonian accuracy, the GEM proposal is a testable hypothesis. Nothing this precise has come out of the vast amount of work done in loop quantum gravity or strings.

So we can see this shift from a flat metric, to Newtonian, to weak field, to strong field, with all four terms gently changing. I felt if my graph was this:

._.
. .

then my proposal about the Even representation of quaternions was wrong. I put that key part of my proposal on the firing line. I felt that graph would never be able to do a smooth transition needed for a metric solution (there is also a potential solution to the GEM field equations, and a metric/potential solution to the GEM field equations, but this is a possibility people trained in GR cannot entertain because there is no potential in the Rienmann curvature tensor, it is exclusively about the metric.)

I read up on graph theory, and realized that the graph that described the Even representation was this one:

._.
|X|
._.

This is both compact and the vertices are all connected to each other. This sort of finite group is compact in the formal sense of the word. It is part of standard math, the group K₄:6, but is not part of the literature devoted to gravity. This makes it a candidate for the missing piece of physics.

Also, I and anyone can solve equations on MATHEMATICA. However, if those equations are arrived at by wrong mathematics then the solutions don't mean a whole lot, even if done by computer.

I was trying to figure out what I didn't like about this comment. It was not an attack on me, because we both know from experience that processing equations through Mathematica is an acceptable check of form, but like all computer programs, what one puts in is the most important aspect of what comes out. The first line of your reply indicates you don't get where I am going. You know your land well. The best reply I can think of was written by a folk singer:

Come mothers and fathers
Throughout the land
And don't criticize
What you can't understand
Your sons and your daughters
Are beyond your command
Your old road is
Rapidly agin'.
Please get out of the new one
If you can't lend your hand
For the times they are a-changin'.

If the GEM proposal is correct - or some similar technical variation of a rank 1 field theory to unify gravity and EM - then the work of Uhlenbech, Freedman, Donaldson, Wheeler, Hawking, Feynman, Kaluza, Klein, all string theorist, all loop quantum gravity people, and even the work of Albert on gravity will collapse. I cannot care what your opinion is on that clear yet radical sentence because I know you haven't calculated the Christoffel symbol of the second kind for the Rosen metric, finding some erudite reason to not bother. I am much happier finding wonderful new gems about GEM like the graph theory for the Hamilton and Even representation of quaternions than bonking heads like I have done in this paragraph.

This Mathematica warning is also not germane to the question at hand: can one formulate a 4 dimensional, commuting division algebra? If the answer is yes, that would be interesting because everyone is instructed that the only 4 dimensional division algebra has the property that it is non-commuting. That is math worth talking about, not a banal caution about symbolic math programs.

Just for fun, I will take a different approach on the value of Mathematica. What passes for physics research today is so vague it cannot be translated into a proposal that can be confirmed by machine. A measure of the value of work is the ability to translate it to symbolic code. The Universe is constructed out of parts that do not think, they do. The math used to describe the Universe should be the same way. Sure, the high priests that believe they are above the mundane nuts and bolts of math will look scornfully at such a stance, but that is fine with me. I prefer nuts and bolts, one can build real things with them.

Doug

 

[Lawrence B. Crowell]

Hello Lawrence:

The Schwarzschild metric a solution to the Einstein field equations, or the exponential metric a solution to the GEM field equations, both have the same Taylor series expansion that match these terms. It is the next terms where GR and GEM part ways.

GR:
d \tau^2 = (1 - 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2 - 3/2 (\frac{G M}{c^2 R})^3) dt^2 - (1 + 2 \frac{G M}{c^2 R} + 3/2 (\frac{G M}{c^2 R})^2) dR^2/c^2

GEM:
d \tau^2 = (1 - 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2 - 4/3 (\frac{G M}{c^2 R})^3) dt^2 - (1 + 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2) dt^2) dR^2/c^2

At second order parameterized post-Newtonian accuracy, the GEM proposal is a testable hypothesis. Nothing this precise has come out of the vast amount of work done in loop quantum gravity or strings.

Doug

To be honest at this point I think I can rely upon experimental evidence. Last year the motion of two neutron stars was measured and compared to Parameterized Post Newtonian parameters up to O(1/c^4), or ppN in the standard line element parameters with ds^2 in units of distance. The motion agreed with Einstein's GR up to one part in 10^7 in this order. If your equations were correct over those of GR the deviation to this order, which appears as the O(1/c^6) in your proper time, would have been blatantly apparent.

I hate to say but I think your GEM theory, or at least the metric you claim above, has been falsified.

Lawrence B. Crowell

 

[sweetser]

Neutron star data

Hello Lawrence:

If you have a reference for this work, I would appreciate it. I certainly would cite it in discussions of my work.

If your equations were correct over those of GR the deviation to this order, which appears as the O(1/c⁶) in your proper time, would have been blatantly apparent.

We cannot measure the mass or angular momentum of neutron star number 1 directly, it must be modeled. We cannot measure the mass or angular momentum of neutron star number 2, it too must be modeled. The most we could hope to say is that we can construct a model using the Schwarzschild metric that is consistent with the data we have. I am sure they have done that, which is good news.

I would need to see that they tried to plug in the Rosen or other alternative metrics at the start, which could change the mass and angular momentum descriptions of both neutron stars, and that such a model could not be adjusted to also be consistent with the data they collected. The coefficients of the two metric have the same sign but are 12% different at second order PPN accuracy. There are 4 critical numbers that have to be generated, and that gives one freedom to match a range of alternative metrics. I hope they are not claiming a system with 4 parameters can be used to eliminate alternative metrics.

You, the great skeptic of the value of the Mathematica, I am surprised you would think a system that inherently depends on the construction of two models for two spinning masses would think this data could distinguish alternative metric solutions for gravity. It is vital people do not over-claim what their data shows. They can demonstrate consistency, but I am skeptical about broader claims.

For light bending around the Sun whose mass we know from other effects, doing a measure at second order PPN accuracy is also dependent on models, such as the quadrapole moment of the Sun. That will have an effect in the range of 1 microarcsecond, which is the size of the difference between the Schwarzschild and GEM metrics (10.8 versus 11.5 respectively). The rotation of the Sun also must be accounted for at this level of detail, and that is darn difficult. I would bet the folks with the neutron star data did not put an effort to have a quadrapole moment to each of the stars, yet it would necessarily change the model at the resolution they were investigating.

The only non-model dependent data for higher order effects I know is about gravity waves. The rate of energy loss is consistent with a quadrapole moment. If a system lost energy as a dipole, then the loss would be far greater than what we have seen. This is why the Rosen metric is not a serious competitor. He proposed a fixed background metric as well as a dynamic background metric. An isolated system could store energy in the fixed metric field, which would allow the system to have a dipole moment while still conserving energy. This is why many theories that add on a new field fail: the new field can store energy and allow for dipole emission of gravity waves.

The GEM proposal is simpler than GR in that it uses only one derivative of a connection, not the difference between two derivatives of connections in the Riemann curvature tensor. The lowest order of gravity wave emission will be the water balloon wobble, the quadrapole moment.

If we were to ever measure a gravity wave, there could be a clean way to distinguish GEM from GR. In GR, they copy EM so closely, it is widely claimed that the waves must be transverse. In GEM which has EM integrated, the transverse waves are EM, leaving the longitudinal and scalar modes for gravity. We would need to detect the same gravity wave along 6 axes to to figure out the polarity, but that would be non-model dependent way to confirm or reject the proposal.

I am looking forward to reading the reference. I would not "hate" such a result because I have a professional scientific attitude. I have worked on other research projects that had to be abandon because of the data. The limitations of data must be skeptically assessed. Falsification is tough.

Doug

 

[Lawrence B. Crowell]

Hello Lawrence:

If you have a reference for this work, I would appreciate it. I certainly would cite it in discussions of my work.


Doug

Here is an overview of this type of work from the Cliff Will, a war horse on the experimental verification of GR:

http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.2589v1.pdf

I tried to look up in APJ or in Science an article which discussed a specific neutron pair system. I'll have to look further. Yet the paper by cliff illustrates aspects of what I said. This paper also illustrates why your objections are not terribly relevant.

Lawrence B. Crowell

 

[CarlB]

This might be the place to start looking:
http://en.wikipedia.org/wiki/PSR_J0737-3039

This is probably what you're looking for, or at least should be referenced by it:

Tests of general relativity from timing the double pulsar
http://arxiv.org/abs/astro-ph/0609417

 

[Lawrence B. Crowell]

This might be the plae to start looking:
http://en.wikipedia.org/wiki/PSR_J0737-3039

The references in the Wiki-p site

http://www.physicsweb.org/articles/world/18/3/6/1
http://www.jb.man.ac.uk/news/doublepulsar/

http://www.atnf.csiro.au/research/highlights/2003/manchester/manchester.html

are of help, in particular the last one. The graph in figure 3 I recall appeared in the journal paper. The paper was in Science or APJ. The case was pretty convinceingly made that this was a verification of GR beyond the Hulst result of the late 1970s.

Lawrence B. Crowell

 

[sweetser]

The Will paper

Hello Lawrence:

The cited paper by Clifford Will is not relevant to your assertion that "at least the metric you claim above, has been falsified". What Will does in that paper is try to connect the post-Newtonian math machinery to what folks do in numerical relativity for the final few cycles of collapse of binary systems. There are so many assumptions that go into those models - big ones like the correct field equations for gravity are the Einstein field equations - that zero of this article has to do with the first seven coefficients of the Taylor series expansion for the metric of gravity. He is playing guessing games with eccentricities and M/R errors.

The paper shows a mathematical consistency between PN approaches and numerical general relativity, one nicer than he expected. A fine result, but not germane.

I have asked Clifford Will questions about the GEM proposal twice. The first time was at an Eastern Gravity Meeting where he gave a keynote address. I asked the first question: when will we get data to second order PPN accuracy? He said he did not know, and did not know of any experiments designed specifically to get that data (that is one reason I was taken aback by you referencing a paper by Will and saying this was relevant to the coefficients). I also asked another question, about the polarization of gravity waves, and in post #477 put his reply into my own words: it may not happen.

Will came to MIT to give a talk. There was a coffee session before, and I was sure to go although I don't drink coffee or tea. He looks like Ted Turner. No one was talking to him, so I went up to him and asked him about his Living Review article on experimental tests of GR. I pointed out that in section three, he went through lots of alternatives to GR. The article was pointed out to me by someone who claimed that of course a rank 1 field theory of gravity would be cited in his exhaustive review, but I said it wasn't there. He confirmed that a rank 1 theory was not there. The reason was that no rank 1 theory had been proposed that would pass all the basic tests, from the equivalence principle, to light bending around the Sun, to the precession of the perihelion of Mercury. He also made clear he would not be able to read my efforts to accomplish these goals.

Because there is a solution to the GEM field equations that can be expressed as a metric, it will pass the tests done of the equivalence principle, where all objects no matter what their composition move according to the metric equation.

The light bending calculation was done for GEM in post #351. Light bends a little more than what is predicted by GR, but the differences is only 9.1% according the the calculation in the post (11.69-10.96/10.96). The fact that light can bend at all in the proposal is why I have discussed the coupling term at length in this thread. One has to show spin 2 symmetry in both the coupling term and the field strength tensor term.

The third test is the precession of the perihelion of Mercury, which I did in post #233. It took me along time to get all the details of that one right, 24 steps in all.

Per Will's request, I have not forwarded my draft paper to him. There is a rank 1 theory involving gravity that meets his basic criteria not referenced in his review.

Doug