GEM: unified field math for the 4 known forces (in technical detail)

25 Mar 2009

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Wed, 03/25/2009 - 08:23

doug

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Where there is new math, there is the opportunity for new physics. Where there is new physics, there is a specific set of experiments to test the new mathematical foundation. Work with strings satisfies the first criteria but not the second.

The new math foundation of the GEM hypothesis to unified the four known forces emerges from 4 properties well established for real and complex numbers: a robust definition of a derivative, multiplication that commutes, a general visualization method, and many connections throughout group theory. Using the standard tool set of field theory (Lagrange densities, the Euler-Lagrange equations, solutions to covariant field equations), I can show that the dynamic exponential metric (the Rosen metric in the literature) with a constant potential is a solution to the covariant field equations. The exponential metric makes exactly the same prediction as the Schwarzschild metric at first order parameterized post-Newtonian (PPN) accuracy, 1.75 arcseconds of bending for light around the Sun. At second order PPN accuracy, the exponential metric predicts 0.8 microarcseconds more bending than general relativity. The current accuracy of light bending experiments is 100 microarcseconds. Effects such as the quadrapole moment and spinning of the Sun could have effects of 1 microarcseconds. Significant effort will be required to confirm or reject the exponential metric on experimental grounds.

We have yet to detect gravity waves. Gravity waves would need to be measured along 6 axes to determine the polarization. Clifford Will made clear in his Living Review article on experimental tests of GR that the polarity of the wave needs to be transverse for GR. For this unified field theory, the transverse modes of emission of spin 1 particles are electromagnetism, leaving the scalar and longitudinal modes for the spin 2 graviton. Physicists are working to detect gravity waves and are aware of the polarization requirement of GR, so I wish the experimentalist good luck.

The Maxwell equations are invariant under a gauge transformation as they must be for photons to travel at the speed of light. My equation for gravity must also be invariant under a gauge transformation so that gravitons travel at c. When gravity and EM are unified, the entire action must be invariant under a gauge transformation for the force mediating particles, yet the field strength tensors for gravity and EM should depend on gauge so that massive particles can interact with the force particles. I will show how this is done without using the Higgs mechanism. The GEM hypothesis make the null prediction that the LHC will not detect a Higgs particle, or any other experiment

The rest of this post does not lead to further tests. It tries to supply consistent reasons for why some issues in the foundation of physics seam odd. When we take standard tools of 1 and 2D math to 3D + time, then physics should have these properties. I will take what appears to be a long math detour before returning solidly to physics, writing out the action for EM, the electroweak and strong forces, and gravity.

Spacetime derivatives

The new math of calculus led to the constellation of physics done by Newton. The limit definition of Leibniz can be applied to real and complex numbers on the real and complex manifolds R¹ and C¹ respectively. Mathematicians have shown that calculus requires a mathematical field, which means a manifold had the group operations of addition, and modulo the additive identify of zero, the group operation of multiplication.

According to a theorem by Frobenius, the next finite dimensional field are known as the quaternions (think of them as complex numbers where the imaginary unit becomes a 3-vector (i, j, k)). Quaternions commute with all scalars and all quaternions that happen to point in the same or opposite direction. Unless there are special conditions, two quaternions will not commute. This is why the Leibniz limit definition fails for quaternions: writing the differential element on the left is different from writing it on the right. There are people who work with right- or left-hand derivatives, but such efforts are unable to do the basics of analysis, such as to show a polynomial such as f(q)=q² is analytic in q. The power of complex analysis comes from working on the manifold C¹, not R². We can do calculus with quaternions on R⁴, but there would be new math with a robust definition of a derivative on the quaternion manifold H¹ (since Q is already taken by the rationals, H is used for the quaternions in honor of Hamilton).

I have proposed a 2 limit definition for a quaternion derivative on one dimensional quaternion manifold. First one lets the imaginary 3-vector go to zero, $(q - q^*) \rightarrow 0$ . What remains is the scalar differential which commutes with all other quaternions. Now let the scalar go to zero. With this two limit definition, writing the differential element on the left is exactly the same as writing it on the right. One in effect gets a directional derivative along the real axis which is well understood by mathematicians.

If the order of the limits are reversed, then one faces the problem again of not knowing which direction the differential element vanished in the limit. This time, what we can know precisely is the norm of the derivative. While not well known, mathematicians have studied norms of derivatives, the average amount of change a system can undergo using multiple paths.

Consider a quaternion to be an event in the manifold of spacetime, the scalar being time, the 3-vector being space. Let us try to describe how a pattern of events in spacetime changes, the calculus of spacetime. To a physicist, if the amount of change in space between events is smaller than the change in time measured in the same units, then the events have a timelike separation. These events can be ordered in time and we can use the directional derivative along the time axis. One event can cause another event to happen. This is the domain of classical physics.

If the amount of change in space between events is greater than the change in time, then the events have a spacelike separation. There can be no causal link between events. The best we can do mathematically is calculate the norm of the derivative. This is what happens in the domain of quantum mechanics.

If we take Newton's calculus and apply it correctly to Einstein's spacetime, we have a mathematical reason for the clear differences between classical and quantum mechanics.

Spacetime Visualizations

No one can visualize 4 spatial dimensions. In Hansen's book, "Visualizing Quaternions", he uses the unit quaternions to reduce the math to 3 spatial dimensions which is manageable. The problem is that most quaternions do not happen to have a norm of 1. A more general approach would view the 4 degrees of freedom inherent in quaternions as 3 spatial dimensions and one for time, Minkowski spacetime. Generate a collection of quaternions using a function, sort those by time. For a 10 second animation with 10 frames per second, or 100 frames total, determine which frame each event belongs in. Use 3D ray tracing software to draw each event (the shadows reduce 3 spatial dimensions to 2 and are an aid to figuring out what is happening).

I found the animation by itself was not enough. The animation is complemented by three complex planes: t-x, t-y, and t-z. The graph is static, although there is a scan line showing where time is for the three complex planes and the animation. The complex planes are familiar. Making math dynamic is odd. The one animation that looks obvious involves the simplest math act: repeated addition. That creates a series of events that travel at a constant velocity in spactime, an inertial observer in the physics lexicon. Interesting that the simplest math act animated - addition - is at the core of special relativity.

A unit circle in a complex plane is easy to visualize. Animated, there suddenly appears one dot that immediately splits in two. The high speed pair creation slows as the events reach their maximal separation, then the two dots rush and annihilate each other. Note my choice of verbs - creation and annihilation - come from the lexicon on second quantization. A normalized quaternion has the symmetry of U(1), a symmetry central to understanding EM.

Because this is a general tool, we can visualize things that have never been seen before. An example relevant to physics the group SU(2), the symmetry of the weak force. Algebraically, a way to represent the unit quaternions uses the exponential of the imaginary 3-vector. The norm of this is always equal to one. If the exponential is tiny, then the animation will be a small sphere around the spacetime point (1, 0, 0, 0), the first term of its Taylor series. That is just a point at time = 1 at the spatial origin for the observer, who always is at zero. It is fun to note that the location of the observer at zero is also the point which much be excluded to make quaternions a group under multiplication. The animation of SU(2) starts at 8 points, the exponential of all permutations of $(0, \pm1, \pm1, \pm1)$ . These all have the same value of time less than one. They grow on the unit sphere into each other, then contract down to (1, 0, 0, 0).

At this time, while I can say the animation is a faithful representation of the group SU(2), the picture is odd and I don't understand the link to the weak force itself. There still is plenty of mystery in the Universe!

Spacetime Multiplication that Commutes

Quaternions are defined as the 4D division algebra where multiplication does not commute. That was the representation developed by everyone who came up with quaternions independently, the first three being Gauss, Hamilton, and Rodgrigues. In my efforts to write the action for gravity, it became clear that I would need a 4D division algebra where elements commuted.

The Hamilton representation of quaternions do not commute because of the cross product. The negative signs flip if the order of multiplication is reversed. If one defines a even representation of the cross product, where all the signs are positive, then changing the order of multiplication does not change a thing. Let us write out the Hamilton representation as a 4x4 matrix:

$Hamilton-rep(t, x, y, z) =\[ \left[ \begin{array}{cccc}e_0 t & -e_1 x & -e_2 y & -e_3 z\\ e_1 x & e_0 t & -e_3 z & e_2 y \\ e_2 y & e_3 z & e_0 t & -e_1 x \\ e_3 z & -e_2 y & e_1 x & e_0 t \end{array} \right] \quad eq ~1 \]$

The even representation drops all minus signs:

$Even-rep(t, x, y, z) =\[ \left[ \begin{array}{cccc}e_0 t & e_1 x & e_2 y & e_3 z\\e_1 x & e_0 t & e_3 z & e_2 y \\e_2 y & e_3 z & e_0 t & e_1 x \\e_3 z & e_2 y & e_1 x & e_0 t \end{array} \right] \quad eq ~2 \]$

Since this is a matrix, there will be times when this has an inverse. To ensure that there is [i]always[/i] an inverse, the divisor that appears in the inverse must never be zero. This is precisely what happens with the Hamilton representation, whose inverse has the norm as the divisor, t² + x² + y² + z². The set of numbers that must be excluded is a point set, (0, 0, 0, 0). To guarantee an inverse for the even representation, zero must also be excluded. I observed something which should be of interest to mathematical physicists: Eigenvalues equal to zero must be excluded to make the even representation matrix invertible. Recall the role of Eigenvalues. These are specific values that a matrix can take, a solution to a particular problem in quantum mechanics. What the tools of quantum mechanics provide is a means of calculating the odds of a system being at a range of specific Eigenvalues. The Eigenvalues are the observables, and we can determine the odds of being at a particular Eigenvalue.

The even representation may never be at an eigenvalue equal to zero. I honestly do not appreciate all the implications of this observation. This topic is open for more work.

Spacetime Group Theory

Group theory is central to particle physics. A way to write both U(1) and SU(2) as quaternions has been introduced. The product of these two is the electroweak symmetry, U(1)xSU(2). This group can be represented using the quaternion product: $\frac{A}{|A|} exp(A~-A^*)$ . The group U(1) is Abelian, and the normalize quaternion will commute with its own exponential.

The group SU(3) underlies the strong force. Its Lie algebra su(3) has eight generators. If we were to multiply one element of U(1)xSU(2) by another, we would get a third element with electroweak symmetry since that is how group theory works. The trick is to change the multiplication table. That can be done by taking the conjugate of the first term in a product:

$(\frac{A}{|A|} exp(A~-A^*))^* \frac{B}{|B|} exp(B~-B^*) \quad eq ~3$

This has a norm of one, it depends on eight independent components, and it is not electroweak symmetry. This representation of SU(3) is remarkable from a physics perspective. It is known by mathematicians that the groups U(1) and SU(2) are subgroups of SU(3), yet it is interesting to see what a direct role they play in forming SU(3). One of the more mysterious aspects of the standard model is containment, where only SU(3) has the property that the farther away partices get, the stronger it becomes. This is why isolated quarks have never been seen. Yet there is no reason whay SU(3) should have that property and not U(1) or SU(2). Now we have a reason why: there is no SU(3) symmetry without a pair of U(1)xSU(2)'s.

The deepest mystery of the standard model is why the groups U(1), SU(2), and SU(3), and not some other combination of groups, dictate the symmetries of the known forces. Now we have a reason. If we want to calculate the norm of how two particles interact using quaternions consistently throughout the calculation, then A[sup]*[/sup] B when normalized to 1 has the symmetries U(1), SU(2) and SU(3).

We know from experiment what gravity is: a force that changes what a length is depending on where you are in a spacetime manifold. Comparing different places in the manifold comes up with differences if there is a gravitational source. The product $(\frac{A}{|A|} exp(A~-A^*))^* \frac{B}{|B|} exp(B~-B^*)$ will always have a norm of 1, yet if we compare two of these at different locations in a spacetime manifold, we need a means for continuously changing distance. That is what the group Diff(M) does, the group of all smooth coordinate transformations on a differential manifold.

The Lie algebras for the standard model have 1 + 3 + 8 = 12 generators. Super symmetric proposals involve larger groups, such as SU(5) and SU(10). Garrett Lisi has strode out to an even larger group, the largest exceptional Lie group known as the "monster", E8. This proposal does something harder: makes due with less. There are (1 + 3)^* + 1 + 3 = 8 generators that together have U(1), SU(2) and SU(3) symmetry. Whether this smaller effort can connect to the vast number of details known about particle physics remains to be seen.

GEM Unified Actions

At long last, we return to physics. Before doing anything new, let us write out the action for EM using quaternions.

$S_{EM} = scalar(\int \sqrt{-g} d^4 x ( - J \frac{A}{|A|} ~-~ \frac{1}{4}(\nabla \frac{A}{|A|} ~-~ (\nabla \frac{A}{|A|})^*)(\frac{A}{|A|} \nabla ~-~ (\frac{A}{|A|} \nabla)^*)) \quad eq ~4$

For those unfamiliar with these arts, an action S is the spacetime integral of a Lagrange density, all the ways energy can be exchange per unit of space, not spacetime. This integral will have an arbitrary value depending on how long a time one sets for the integration. The trick is to use the calculus of variations, where one varies something in the integral, then looks for those things that can change without altering the value of the integral no matter what the duration. That variable quantity is thus conserved by the action. When time can be varied freely, then we get conservation of energy for the action. When space can be varied, we get conservation of linear momentum. When varying angles makes no difference to the action, angular momentum is conserved. If varying the phase makes no difference to the integral, then electric charge is conserved. The action of EM revels the conservation of energy, linear momentum, angular momentum, and electric charge, once you learn to look at it in the proper way.

To make the action apply for flat or curved spacetime, the square root of the metric must be included with the volume element. In curved spacetime, volume elements change, but the square root of the metric accounts for such a variation.

The reason to normalize the potential is that a normalized quaternion is a way to represent the group U(1). That group is the symmetry underlying electromagnetism in the standard model.

The two subtraction are there to ensure the action is invariant under a gauge transformation. That is essential so that photons may travel at the speed of light c.

The scalar part of the current coupling term is exactly what one gets by contracting the negative of 2 4-vectors in spacetime: $-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3$ . This is invariant under a Lorentz transformation (note: I use the numerical subscripts 1, 2, 3 instead of the more comfortable x, y, z to indicate this definition applies no matter what the choice of coordinates happen to be).

Since few readers will have experience parsing expressions using quaternions, let me break down the field strength tensor term, $\nabla \frac{A}{|A|}$ . Div, grad, curl and all that were terms coined by Hamilton to describe the calculus of quaternions in this 4D context. Gibbs skipped the quaternion history to reduce vector analysis to 3D. The simple way to remember the rules for a quaternion written as (scalar, 3-Vector) is (first - last, inside + outside + cross). For a differential operator:

$(\frac{\partial}{\partial t}, \nabla)(\phi, A) = (\frac{\partial \phi}{\partial t} ~-~ \nabla \cdot A, \frac{\partial A}{\partial t} ~+~ \nable \phi ~+~ \nabla \times A) \quad eq ~5$

Div, grad, curl and two time derivatives represents the complete picture of how a 4-derivative can change in spacetime.

Much study is needed to understand something call the connection, a general tool for characterizing curvature. I use the same assumptions as those used in GR, namely that the connection is metric compatible and torsion-free. These assumptions lead to the Christoffel symbol of the second kind as a measure of how the metric changes. Note that the curl is like an exterior derivative, so the antisymmetric nature of the curl cannot have any information about the Christoffel which is symmetric.

The difference between a quaternion and its conjugate, q ~-~ q^* , ensures the scalar is zero. In the field strength tensor term, no matter what one chooses for $\frac{\partial \phi}{\partial t} ~-~ \nabla \cdot A$ , it will not be in the final result, so you are free to choose what you like. This is known as gauge freedom. Gauge freedom is a requirement for forces whose particles travel at the speed of light. In terms of the electromagnetic field terms,

$(\nabla \frac{A}{|A|} ~-~ (\nabla \frac{A}{|A|})^*) =(0, -E ~+~ B)$
$(\frac{A}{|A|} \nabla ~-~ (\frac{A}{|A|} \nabla)^*)) = (0, -E ~-~ B) \quad eq ~6-7$

Multiply these together:

$(0, -E ~+~ B)(0, -E ~-~ B) = (B.B ~-~ E.E, 2 E \times B) \quad eq ~8$

The scalar is the standard Maxwell Lagrangian, while the 3-vector is the Poynting vector. None of this is new, but it is cute that one gets two critical bits of EM theory from one product. Use the Euler-Lagrange equations to derive the source Maxwell equations:

$\rho = \nabla \cdot E$

$J = \nabla \otimes B ~-~ \frac{\partial E}{\partial t} \quad eq ~9-10$

Nothing new so far. By altering the potentials, actions with electroweak and the electroweak-strong symmetry can be written:

$\begin{align*}S_{EW} &= scalar(\int \sqrt{-g} d^4 x ( - J \frac{A}{|A|} exp(A ~-~ A^*) \\ &-~ \frac{1}{4}(\nabla \frac{A}{|A|} exp(A ~-~ A^*) ~-~ (\nabla \frac{A}{|A|} exp(A ~-~ A^*))^*)(\frac{A}{|A|} exp(A ~-~ A^*) \nabla ~-~ (\frac{A}{|A|} exp(A ~-~ A^*) \nabla)^*)) \end{align*}$

$\begin{align*}S_{EWS} &= scalar(\int \sqrt{-g} d^4 x ( - J (\frac{A}{|A|} exp(A ~-~ A^*))^* \frac{B}{|B|} exp(B ~-~ B^*) \\ &-~ \frac{1}{4}(\nabla (\frac{A}{|A|} exp(A ~-~ A^*))^* \frac{B}{|B|} exp(B ~-~ B^*) -~ (\nabla (\frac{A}{|A|} exp(A ~-~ A^*))^* \frac{B}{|B|} exp(B ~-~ B^*) )^*) \\ &*((\frac{A}{|A|} exp(A ~-~ A^*))^* \frac{B}{|B|} exp(B ~-~ B^*) \nabla ~-~ ((\frac{A}{|A|} exp(A ~-~ A^*))^* \frac{B}{|B|} exp(B ~-~ B^*) \nabla)^*)) \quad eq ~11-12 \end{align*}$

These actions have the symmetries of U(1)xSU(2) and U(1)xSU(2), SU(3), respectively, so they will conserve electroweak and strong charge. This many terms can be intimidating, at least it is for me. At one level, it is about systematically working with symmetries of unity for spacetime. The simplest thing was U(1), which uses 1 degree of freedom. The electroweak uses 4 degrees of freedom. There are always 4 terms in the field strength tensor. Two of them ensure that the field strength tensor is gauge-free (the subtraction of the conjugate). The reversal of the location for the differential operator is the second trick to make a Lorentz invariant scalar (look at the EM example in equation 4 to confirm).

I do not know how to use them to make a calculation and see if they are consistent with our current approach. The standard model is a detailed proposal, so making an action that works with the fine grain data would require considerable effort. In the usual approach using covariant derivatives, the potential and group are included in the derivative independently of each other.

An action for gravity is generated using the same terms in the same places. What changes is the rule for multiplication, from the Hamilton representation of quaternions, to what I am calling the even representation, or in the literature, hypercomplex numbers. Note again that nothing changes for addition and subtraction which works like a standard 4-vector. I will use the symbol $\boxtimes$ for the even representation product.

$S_{G} = scalar(\int \sqrt{-g} d^4 x ( - J \boxtimes \frac{A^*}{|A|} ~-~ \frac{1}{4}(\nabla^* \boxtimes \frac{A}{|A|} ~-~ (\nabla^* \boxtimes \frac{A}{|A|})^*))\boxtimes (\frac{A^*}{|A|} \boxtimes \nabla ~-~ (\frac{A^*}{|A|} \boxtimes \nabla)^*))) ) \quad eq ~13$

The additional conjugate is required so that the product of $- J \boxtimes \frac{A^*}{|A|}$ is invariant under a Lorentz transformation. The requirement for a conjugation operator means that associativity does not hold ( $A^* \boxtimes (B \boxtimes C) \neq (A \boxtimes B)^* \boxtimes C$ . I do not understand the implications of the observation, but it does not play a role for the action. Since terms involving $\frac{\partial \phi}{\partial t} ~-~ \nabla \cdot A$ are again subtracted away, the S_G action is invariant under a gauge transformation. Gauge freedom is essential since experimentalist have demonstrated gravity works at the speed of light.

An even smaller number of people have worked with hypercomplex numbers. Fortunately the rules are simpler: everything is positive.

$(\nabla^* \boxtimes \frac{A}{|A|} ~-~ (\nabla^* \boxtimes \frac{A}{|A|})^*) =(0, e ~-~ b)$
$(\frac{A^*}{|A|} \boxtimes \nabla ~-~ (\frac{A^*}{|A|} \boxtimes \nabla)^*)) = (0, -e ~-~ b) \quad eq ~14-15$

where:

$e = +\frac{\partial A}{\partial t} ~-~ \nabla \phi$
$b = (\frac{\partial A_2}{\partial x_3} + \frac{\partial A_3}{\partial x_2}, \frac{\partial A_3}{\partial x_1} + \frac{\partial A_1}{\partial x_3}, \frac{\partial A_1}{\partial x_2} + \frac{\partial A_2}{\partial x_1},) \equiv (\nabla \otimes A) \quad eq ~16-17$

Take the product of these two:

$(0, e ~-~ b) \boxtimes (0, -e ~-~ b) = (b^2 ~-~ e^2, b \otimes b - e \otimes e) \quad eq ~18$

The scalar is the Lagrange density for this proposal. Apply the Euler-Lagrange equation to the scalar to make the field equations which should look vaguely familiar:

$\rho = - \nabla \cdot e$

$J = \nabla \otimes b ~-~ \frac{\partial e}{\partial t} \quad eq ~19-20$

How do these differ from Gauss' and Ampere's laws of EM? In the static case, for EM, Gauss' law becomes $\rho = - \nabla^2 \phi$ . Applying a theorem developed by Gauss, one can show that the difference in sign indicates that like charges will repel for such a stratic law. For the action S_G, the scalar equation is $\rho = \nabla^2 \phi$ , so like charges will attract, as is well known for gravity. A core problem with Newton's gravity equation is that should the mass source change, then those changes must propagate instantaneously. With this proposal, the $\nable \cdot \frac{\partial A}{\partial t}$ component provides a means for the field to adjust in time. Just like the Maxwell equations, the speed of propagation is the speed of light.

What are solutions to these field equations? Let us focus on the simple one, $\rho = \nabla^2 \phi$ . It is essential to note that $\nabla^2$ is not a LaPlacian operator, but is 2 covariant derivatives acting sequentially.

$\rho = \nabla^2 \phi = \frac{\partial^2 \phi}{\partial t^2} - \partial^2 \phi + \frac{\partial}{\partial t} \Gamma^0_{\;\omega 0}A^{\omega} - \partial^i \Gamma^0_{\;\omega i}A^{\omega} \quad eq ~21$

This equation doesn't have a solution because both the potential could vary or the metric - and thus the connection could vary. Where there is a symmetry, there is a conserved charge. In this case since one is free to vary the metric so long as changes in the potential take its place, that means mass is the conserved quantity. What we can do is decide to fix either the metric or potential, then determine what corresponding potential potential or metric solves the eq. 21.

One static solution was known to La Place, a charge/R potential that leads to an inverse square law. That arises if the connection does not vary in time or space.

If $\Gamma$ is for a constant Minkowski metric, then $A = (\frac{q - \sqrt{G} m}{R}, 0, 0, 0)$
$\rho = -\partial^2 \frac{q - \sqrt{G} m}{R} \quad eq ~22$
There is another solution that would not have been apparent to that old master. There is a non-trivial solution if the potential is constant. In that case, all the change would have to come from the curvature of spacetime. It the static case, that would be the last term of equation 21:

$\rho = - \partial^i \Gamma^0_{\;\omega i}A^{\omega}$

We are looking for the divergence of a Christoffel symbol to be equal to the mass charge density.