# Unifying Gravity and EM

by sweetser
Tags: gravity, unifying
 P: 361 Hello Jhmar: The standard model is really successful, but there are two "weak" points. The first is why should the three particular groups, U(1), SU(2), and SU(3), be so important? There are lots of other possibilities, yet so far no one can provide a reason for these three. The GEM proposal as it is written can explain these three groups: Diff(M)xU(1)xSU(2) - or gravity and the electro weak force. Gravity comes from the two covariant derivatives. One is free to choose how much the 4D wave propagation is due to changes in the potential or in changes that happen as you move around the manifold. If one write the 4-potential as a normalized quaternion, then the quaternion potential can be written as a unit quaternion times itself, like so: $$q = \frac{q}{|q|} exp(q - q^*)$$ SU(2) is the unit quaternions, the exponential part of the expression above. Quaternions do commute with themselves, so: $$\frac{q}{|q|} exp(q - q^*) = exp(q - q^*)\frac{q}{|q|}$$ The normalized quaternion, $\frac{q}{|q|}$, is now behaving just like U(1), a complex number with a norm equal to one. Let's rewrite the 4D wave equation in the very first post to look like it justifies the electroweak part of the standard model: $$J_q - J_m = \square^2 \frac{A}{|A|} exp(A - A*)$$ The box has Diff(M), A/|A| has the U(1) part, and SU(2) is the exponential. This is good, but not good enough because we need to spot SU(3). One thing I could do with this equation is to calculate its norm. That kind of thing happens all the time in quantum mechanics. It is possible that the norm operation would give the equation SU(3) symmetry. That is speculation I don't have the skills to prove. This is a benefit of GEM I don't discuss much due to my lack of self-confidence in group theory: the GEM field equation in and of themselves justify the symmetry seen in the standard model. Another weakness of the standard model has to do with mass. The standard model out of the box makes one simplifying assumption: all particles have zero mass. Of course that is not true. So now the accepted way to introduce mass back into the model is known as the Higgs mechanism. There has to be a scalar Higgs field everywhere in the universe ready to break the symmetry of the vacuum such that all massive particles get their inertial mass. One of the main reasons for a multi-billion dollar bet being made at CERN is to detect the Higgs. The GEM hypothesis rejects the Higgs mechanism, and the Higgs boson. I've had two people comment on my Lagrange density that it does not have U(1) symmetry. This is mostly true. The Lagrangian has U(1) symmetry if the particles are massless. When there is a mass, the mass charge breaks the U(1) symmetry. The symmetry breaking is EXTREMELY slight - one part in 10^16 for an electron. We only define electric charge to ten significant digits, so the symmetry breaking is beyond our ability to directly measure. This is actually very reasonable. Take a pair of electrons, and a pair of protons, put them 1 cm apart, then measure the acceleration, which is the F/m ratio. To ten significant digits, they are the same. All electric charges repel the same amount once the inertial mass is taken into account. Now take the same electrons and protons, but measure F/m to twenty significant digits. The answer is no longer the same. The gravitational mass of the electron is less than that of the proton, and now the difference can be measured. Gravitational mass breaks the electron charge symmetry. Particle physicists are concerned with the scalar Higgs which is suppose to bring inertial mass to the standard model. People who work with inertial gravity are concerned about the spin 2 graviton. Yet there must be some unbreakable link between the particle of inertia (the Higgs) and the particle of gravity (the graviton). The rank 2 symmetric tensor in GEM is the graviton, and its trace which is a scalar field, does the work of the Higgs. It is clear you can never have an inertial gravity field without having a gravitation field. It was an unexpected gift that the 4D wave equation to give me insight into the standard model. doug
 Sci Advisor HW Helper P: 1,204 Doug, I am still very convinced that you're on the right track here. I just haven't had time to work on my simulation. Thanks for the continuing explanations. Carl
 P: 361 Hello Carl: Thanks. I find struggling to explain the proposal is fun. Remember, I really want a solid technical reason to drop this hypothesis. Although I feel confident about having the right symmetry for gravity and the electroweak forces, if I didn't have a proposal for the symmetry of strong force, that would be a reason to reject the proposal. Three out of four is not good enough, since this fantastic four does all the work in the Universe. In the eternal kids game of "Why?", now that the 4D wave equation may justify the symmetry of the standard model, why is the 4D wave equation so central? The Universe has lasted a good long time. How could it be doing all that it does for such an expanse of time? The key is to do almost nothing. Almost nothing is not the same as nothing. The next door neighbor to doing nothing is the simple harmonic oscillator. The 4D wave equation is the equation that describes the fundamental family of simple harmonic oscillators in spacetime. If you are some particle that happens to exist in 4D spacetime, the closest thing to doing nothing is simple harmonic oscillation caused by that other crap in the Universe. It is really amazing that the Maxwell equations are just a partial rewrite of the 4D wave equations. There is also a set of equations for the symmetric expression which is not an exact clone of Maxwell (like the area of study known as gravitoelectrodynamics, which had identity equations that are not part of GEM). Mapping the classical field equations for GEM back to the 4D wave equation requires getting lots of signs correct, but it is an impressive wad of algebra. A detailed description starts here: http://theworld.com/~sweetser/quater...IAP_2/925.html and goes on for 9 slides. It is hard to generate that many partial differential equations that work together unless there is some truth there. This is the kind of thing I checked with Mathematica because there are so many signs that have to be right, no exceptions. doug
 P: 76 There is the fundamental electric charge, and so far, we have only measured integral amounts of that charge. The quark model does have the fractional charges noted by Carl, but those have not been measured in an experiment. Tsui et al measured two dimensional fractional charge. This is described as thin film indicating that the term two dimensional is being used to described a three dimensional layer with one dimension being to thin to observe, rather than a genuine two dimensional layer which would of course be unobservable. The Nobel prize committee and Scientific American describe this discovery as being of fractionally charged electrons. Note that Tsui makes no reference to pos. or neg.; I read somewhere that this is because the experiment indicated a positive charge that could not be explained by Tsui et al. But have lost the ref.. see: http://www.ee.princeton.edu/people/Tsui.php Note that the dashed (all fractions) line is accompanied by the stepped experimental results line with steps of decreasing value. This is the basis of my proposal for particle structure which has been rejected on the grounds that I have used the two dimensional experiment as a base for a three dimensional object (particle). But I would maintain that any theory must take into account the stepped nature of particle structure. It should also being able to explain what mass and charge are and what causes them to exist. Personally I think the interpretation (but not the mathematics) of Tsui work is wrong and I note he does not use the interpretation (i.e. the term fractionally charged particles) himself.
HW Helper
P: 1,204
 Quote by sweetser Remember, I really want a solid technical reason to drop this hypothesis.
http://www.grg18.com/

I find that I'm getting convinced into going. Plenty of time left to think about it.
 P: 361 Hello Carl: Unless I become an invited speaker, it conflicts with a different physics event. I am taking a 4 day class at MIT titled "Relativity, Gravity, and Cosmology". It is part of MIT's Professional Institute, a way to milk alumni cows (actually, any cow willing to fork over \$2k for 4 days can step into the machine). For a full description of the class, go here: http://web.mit.edu/mitpep/pi/courses...y_gravity.html I will be going to the April APS meeting in Jacksonville FL. I think I forgot to post my abstract, so here it is: Title: Geometry + 4-potentials = Unified Field Theory Abstract: Geometry without a potential is like a bed without a lover. The Riemann curvature tensor is only about the change in moving around the manifold, the geometry of the bed sheet. The exterior derivative of the EM field strength tensor is only about changes in the potential, isolated from geometry. In my work, changes in the potential lay on top of changes in the manifold. A covariant 4D wave equation can describe both gravity and light. The metric solution passes weak field tests of gravity and tests of the equivalence principle. At higher resolution, 0.8 microacrseconds more bending of light around the Sun is predicted than GR. Quantize using the Gupta/Bleuler method, but the scalar and longitudinal modes are the spin 2 graviton. The misses found the first line fun. I expect to accomplish zero, but it is great that provocative text comes straight out of the math. At the end of May, beginning of June, there should be EGM 10 at Columbia, but they are still working out the details for that meeting. doug
 P: 361 Hello: The standard model of particle physics defines three out of the four fundamental forces of physics. The key to organizing the particle zoo is a set of three symmetries: U(1) - complex numbers with a norm of 1, SU(2) - unitary quaternions, and SU(3). Gravity has yet to be incorporated into the Standard Model. Two riddles are why Nature chose these three particular symmetries and how can gravity come into the picture. I've been writing software to animate quaternion expressions (URL at the end). An obvious target for animation is SU(2) which can be written as $$SU(2)\rightarrow exp(q - q*)$$ Pick a bunch of random quaternions, plug them in to the exponential, then create an animation. It looks like a slice of a sphere that grows and shrinks. From there it was easy enough to get to U(1)xSU(2), which is known as the electro-weak symmetry that unifies EM with the stuff of radioactive decay, like so: $$U(1)xSU(2)\rightarrow \frac{q}{|q|}exp(q - q*)=exp(q - q*)\frac{q}{|q|}$$ A quaternion commutes with itself, that is why $\frac{q}{|q|}$ can be written on either side. The Abelian property is necessary if this is a valid representation of U(1). It actual is U(1) in the usual sense if y=z=0, but the way it is written, this represents U(1) for arbitrary quaternions. The animation looks like a more complete expanding sphere, but it has a bias, the points the make up the sphere are not evenly distributed. Two thirds of the way there! The question was how to get to SU(3). The Lie algebra that can be used to generate this group has eight elements to it. My thought was to multiply conjugates together, q* q' (q != q' so it has 8 components like the Lie algebra su(3)): $$U(1)xSU(2)xSU(3)\rightarrow (\frac{q}{|q|}exp(q - q*))* \frac{q'}{|q'|}exp(q' - q'*)$$ The result is an evenly distributed collection of points in an expanding/contracting sphere. When you do get if q=q'? The result is a single dot at (1, 0, 0, 0)! The standard model may all be about the symmetry of one in quaternion spacetime. Fun to think about. The final question is where is gravity? It all has to do with 1, and not quite 1. For an arbitrary point in a spacetime manifold, you calculate the norm of q* q', and it comes out to be exactly 1.0. Now you go to a neighboring point, repeat the calculation, and, oops, the norm is a little bit smaller than 1. That's gravity, because gravity is about how measurements change as one moves around the spacetime manifold. The act of multiplication means metrics must be involved. This can be explicitly connected to the GEM field equation: $$U(1)xSU(2)xSU(3)\rightarrow g_{\mu \nu} J*^{\mu} J'^{\nu} = g_{\mu \nu} ((\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})\frac{A^{\mu}}{|A^{\mu}|}exp (A^{\mu}-A^{\mu}*))* (\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})\frac{A'^{\nu}}{|A'^{\nu}|}exp (A'^{\nu}-A'^{\nu}*)$$ Why do this big complicated expression? We are trying to pack all the known forces of Nature into one expression: EM as U(1) represented by $\frac{A^{\mu}}{|A^{\mu}|}$, the weak force as SU(2) represented by $exp (A^{\mu}-A^{\mu}*)$, the strong force as SU(3) represented by q* q', and gravity as Diff(M) represented by the (possibly) dynamic metric $g_{\mu \nu}$. That's a long sentence too! The animation gives a clue: this is how one fills up a volume of spacetime consistently with groups. This equation is the most basic spacetime wave equation. The collision of spacetime waves, group theory, and quaternions explains the origin of the fantastic four forces of Nature. Stunning, if true. doug You can see all the images that make the story here: http://www.theworld.com/~sweetser/qu...ard_model.html Warning: it will take getting use to the 6 graphs. The animation is center top. On the right are three complex planes, ty, tz, and tx. On the left is the superposition of all states the system can be in for the animation, formed by merging everything in the animation. That was inspired by quantum mechanics.
 P: 361 Hello: The newsgroup sci.physics.research was very important to me as a way of learning about physics and physics research. I began reading the posts in 1995, almost as soon as I had my first access to the Internet (using trn, if I recall correctly). As I continued to see just how much I could do with quaternions, the moderators of the newsgroup tired of me. I know one of the key players in SPR, a math guy named John Baez, no longer thought my quest to work extensively was interesting. John was supportive of non-professionals, unless they were weasels like myself (it is an obtuse skill I have, the ability to tick of math guys). It is hugely frustrating to get a post rejected as being "overly speculative", a nice vague definition they can whip out on a whim. For the most part, I have stopped posting there. I did make an expectation recently. This work with the cause of the 4 forces of Nature is just too cool. I was not confident they would accept it, but it got in. No one is discussing it so far, but that is par for the course. Here it is, both playful and technical. Hope you enjoy... [Post to sci.physics.research] Hello: "Where's Waldo" is a cartoon phenomena whose goal is to spot Waldo somewhere in a densely drawn image. Waldo is there, you just have to work to spot him. A simple game that has meant millions for its author. In contrast, the standard model of physics is super serious, dictating three out of the four forces of Nature: EM, the weak force, and the strong force. If you have ever seen the Lagrangian, it is densely drawn, with generators of groups, gamma matrices, binding constants, and wave functions. I do not find it enlightening. A simpler approach to the standard model focuses on the symmetries: U(1)xSU(2)xSU(3), which are related to EM, the weak force, and the strong force respectively. The Lie algebras for these continuous groups have vector spaces with 1, 3, and 8 dimensions, which exactly matches the number of bosons involved in these forces: 1 photon for EM, +/-W, Z for the weak force, and 8 gluons for the strong force. The group U(1) has all the properties of the complex numbers with a norm of 1. The group SU(2) is the unitary quaternions, quaternions with a norm of 1. The continuous group SU(3) is also a special (norm of 1) unitary group. So now the hunt is on for these groups somewhere in EM, the weak force, or the strong force. Although I have worked with EM, I cannot say the same for any equations involving the weak or the strong force. I have read what they do, but in an equation-free way. I cannot start with the weak or the strong force, but need to keep an eye out for them. Let's write out EM in a way Feynman called "a beautiful set of equations!" (Lectures, II, 18-11), the Maxwell equations in the Lorentz gauge: J^u = (1/c d^2/dt^2 - c Del^2) A^u Both J^u and A^u are 4-vectors. Quaternions can also be viewed as 4-vectors. So treat J^u and A^u as quaternions that happen to have indices (it's just a label after all, and in this case it is restricted to run from 0-3). Now that this is a quaternion wave equation, how would we write a unitary quaternion? Take the exponential of a quaternion where the scalar has been dropped: exp(q - q*) is an element of SU(2) There is a problem though, because the J and A have four degrees of freedom, but exp(q-q*) only has 3. We need to plug 1 degree of freedom back in. We could just grab the scalar using (q+q*), but recall the purpose of the exercise: lets go for the Abelian group U(1) since it's Lie algebra also has 1 degree of freedom. Rewrite A^u like so: A^u = A^u/|A| exp(A^u - A*^u) = exp(A^u - A*^u) A^u/|A| A^u commutes with itself, and with the exponential of itself because it points in the same direction so the cross product is zero (that's the non-commuting part). Together, A^u/|A| exp(A^u - A*^u) has electroweak symmetry, U(1)xSU(2)! The Lie algebra su(3) has eight dimensions, twice the number we have in this equation. What would be a reasonable way to "double" this equation, doing some standard operation in quantum mechanics? Recall the notation, which is a* b. Imagine 2 current densities, J^u and J'^v. I'd like to calculate the inner product of the two, but need a metric to do so: g_uv J*^u J'v = g_uv (1/c d^2/dt^2 - c Del^2)(A^u/|A| exp(A^u - A*^u))* A'^u/|A'| exp(A'^u - A'*^u) An initial objection for this expression being a representative of U(1)xSU(2)xSU(3) might go like this. If a quaternion represents U(1)xSU(2) when written like q/|q| exp (q - q*), then the product of two quaternions should be in the same group. That's how groups work! A detail was missed: we are taking the conjugate of one of these quaternions and multiplying it by the other. As John Baez pointed out to me, that means that multiplication is no longer associative, since: (a b)* c != a* (b c) but norm ((a b)* c) = norm (a* (b c)) I happen to call this sort of non-associative multiplication "Euclidean multiplication" because the scalar part of q* q is t^2 + x^2 + y^2 + z^2. The multiplication table will necessarily be different because regular (or what I call Hamiltonian) multiplication is associative. It is important to remember that quaternion multiplication, even Euclidean multiplication, preserves the norm. Therefore the norm of this q* q' will be one. There is still the same identity, (1, 0, 0, 0), and every quaternion will have an inverse under Euclidean multiplication. With eight numbers to plug into q* q' and a norm of 1, I believe q* q' is a way to represent SU(3). There is a bonus to viewing this 4-vector equation as a 4-vector with the properties of an indexed quaternion, which not only can be added, subtracted and multiplied by a scalar, but also multiplied and divided with each other. The bonus is the explicit appearance of the metric g_uv. We have placed no constraints on the metric. It can be whatever, and this wave equation does not change its form. If the metric is the flat Minkowski metric, then we have the same multiplication rules set out by Hamilton for the scalar part. Calculating the inner product of two current densities will use the same equation no matter what the manifold is, whether it is static or dynamic. In terms of group theory, the equation has Diff(M) symmetry which is at the heart of general relativity. This was a fun game of algebra, but Where's Waldo is a _visual_ exercise. It is time to translate this algebraic story into pictures. Quaternions are 4 dimensional, so how do we deal with that? Go to the movies! Imagine generating a thousand quaternions at random: different values of t, x, y, and z. Take all of them, and sort them by time t. Make an animation lasting 10 seconds, at 30 frames per second, or 300 frames total. Figure out the range of time, from earliest to latest, and divide that by 300. Any particular frame will correspond to a range of time values. If a quaternion happens to fall in the range, place it at the appropriate place given the x, y, and z values. The software for analytic animation using quaternions has been written. Since SPR is a text based newsgroup, I'll describe the images (URL at the end if you want to see the results, but they are not trivial to grasp since we are not accustom to seeing functions of spacetime). The easiest one to deal with is SU(2) because all that needs to be done is to generate thousands of quaternions randomly, then calculate exp(q-q*), and plot the result. None of the events have a time less than zero. The norm always has to be equal to 1, so the first points start out at the extremes of +/-x, +/-y, +/-z. The image then forms a ball that shrinks to a small radius, because by that time, most of t^2+x^2+y^2+z^2=1 is the t^2 part. When exp(q-q*) is multiplied by itself normalized, q/|q| exp(q-q*), the result is a sphere that grows and shrinks, but has a decided bias. Most of the points now have a negative time. The animation of (q/|q| exp(q-q*))* (q'/|q'| exp(q'-q'*)) is a completely smooth rendering of spacetime from (-t,0,0,0) to (+t,0,0,0). It doesn't appear like one could devise a smoother way to fill up a volume of spacetime with events. The take home message is simple: the symmetry of taking the inner product of two indexed quaternion 4D wave equations with a norm of one is Diff(M)xU(1)xSU(2)xSU(3). This may be the reason behind the four forces of Nature: gravity, EM, the weak force, and the strong force fill up spacetime. Way to go Waldo! doug http://www.theworld.com/~sweetser/qu...ard_model.html http://quaternions.sf.net/
 Sci Advisor HW Helper P: 1,204 I am amazed that they put this up on SPR. The only sociological guess I can come up with is that they're a bit jealous now that sci.physics.foundations is up and running and is moderated with a lighter hand. You know, the longer I look at this the more difficulty I have understanding it. I liked: "exp(q - q*) is an element of SU(2)". When we get to "Together, A^u/|A| exp(A^u - A*^u) has electroweak symmetry, U(1)xSU(2)!" I start having problems. Now in the above, I'm assuming that you mean A^u to mean a collection of four quaternionic numbers. There are therefore a total of 16 degrees of freedom running around here. I can't seem to agree that A^u/|A| is at all a U(1) symmetry. The way I understand it, it's not a normed quaternion, it's a normed vector of quaternions and must be something rather complicated. Even if it were A^u/|A^u| it would be a pretty complicated beast. When I've seen SU(3) naturally pop out of combinations of things, the things involved showed up in 3s. Three is a very important number for SU(3), my intuition says that it won't show up naturally as a product of U(1)xSU(2) even with a lack of associativity. Maybe you would end up with a representation of the octonions. My guess is that the way to approach this is by computation. Also, the way that degrees of freedom and bosons were adding up made sense when you combined U(1) and SU(2) to get 1+3 = 4 degrees of freedom. But when you add SU(3) to the list you should end up with 12 degrees of freedom, not the eight you have. You could fix this by adding in another q, which gets back to those 3s that are so important to SU(3).
P: 361
Hello Carl:

Sounds like we have miscommunication going on. Hopefully I will clarify.

 Quote by CarlB Now in the above, I'm assuming that you mean A^u to mean a collection of four quaternionic numbers. There are therefore a total of 16 degrees of freedom running around here.
This is not what I meant. There is 1 quaternion 4-potential, $A$, which I happen to write as $A^{\mu}$ so when I calculate the scalar part of two quaternions, say $A^{\mu} B^{\nu}$, it makes sense how to use a metric $g_{\mu \nu}$ to calculate the scalar part of that quaternion product. The usual quaternion product always and exclusively uses the Minkowski metric to calculate the scalar part of the quaternion product. Let me write out in detail what I mean:
$$g_{\mu \nu}A^{\mu}B^{\nu} = (g_{00} A^0 B^0 + g_{11} A^1 B^1 + g_{22} A^2 B^2 + g_{33} A^3 B^3, A^0 B^1 + A^1 B^0 + A^2 B^3 - A^3 B^2, A^0 B^2 + A^2 B^0 + A^3 B^1 - A^1 B^3, A^0 B^3 + A^3 B^0 + A^1 B^2 - A^2 B^1)$$
For the Minkowski metric, $g_{00}=1, g_{11}=-1, g_{22}=-1, g_{33}=-1$, and for this choice of metric, the product will be exactly the same. The ability to use a different metric only comes into play at the last step of the game, where I want to see the Diff(M) symmetry that underlies any metric theory for gravity. For the sake of clarity, I'll drop the mu's in the discussion.

So A the quaternion has 4 degrees of freedom. The way to calculate and write out a unitary quaternion is exp(A-A*). It is the A-A* step that tosses away one of the four degrees of freedom, because it gives back only the 3-vector.

The group U(1) is usually introduced as the unit circle in the complex plane. This is an Abelian group, with one degree of freedom in its Lie algebra. A quaternion A/|A| might have the right norm, but it has 4 degrees of freedom, and is non-Abelian.

Now we consider a particular product: A/|A| exp(A-A*). We agree that the exp part is a fine way to represent SU(2), whose Lie algebra has 3 degrees of freedom. Since we are using the same A throughout, and since a quaternion commutes with itself, then for this product: A/|A| exp(A-A*) = exp(A-A*) A/|A|. A/|A| must point in the same direction as exp(A-A*), so it really only has one degree of freedom.

I am going to stop here, to see if we are on the same page as far as metrics, labels, SU(2) and U(1)xSU(2).

doug
 P: 76 Gravity has yet to be incorporated into the Standard Model. The key to incorporation is to recall that Relativity is a classical theory and that gravity is carried by gravitons. Then by relating classical particle structure to Einstein's equation; classical particle physics is incorporated into classical relativity. The formula is: E(c squared) = m = Linear Force /2radius As 2r = wavelength substitute λ for 2r Ec2 = m = Fl/λ E = energy c = speed of light m = mass Fl = linear force (2.8799296) r = radius Check this and you will find that it gives the Classical electron radius, and (given the wide experimental margin of error) the proton radius (3 elementary particles = 3Fl) and the neutron radius (5 elementary particles [3 quarks 1 electron and 1 neutrino] = 5Fl). This is not to say that QT is wrong, it is simply an overly complicated way of finding part of the answer. The formula I have given matches the three radii found by experiment, the fractional waves found in TFQHE and the mass found by Einstein's equation. I am waiting for my submission to be approved but whilst waiting I have made a major correction to one of the tables. The correct wave fraction and wavelength for the electron is 1/5, λ = 5.635882. The fractional wave sequence is 1 1/2 1/3 1/4 1/5 1/6 etc and all elementary particle radii are proportional to the electron radius. I used the particle data from the 2004 PDG list to show that all particles found by experiment match the predicted data; no rejections, no averaging. All particles are compaction's of a graviton.
 P: 361 Hello jhmar: Good luck in starting a thread. I won't comment on your proposal, as that belongs to a different thread. I will say what my proposal says about particles of the standard model. In standard quantum field theory, a force can have like particle attract if and only if the particles that mediate the force are even spin. A force can have like particles repel if and only if the particles that mediate the force are odd spin. In the GEM unified field theory, the 4D wave equation contains 2 fields, the even spin 2 field for gravitons, and the odd spin 1 field for photons. The standard standard model works for massless particles. The Higgs mechanism was a way to break the symmetry using a scalar field as a false vacuum. The problem with the Higgs is there is no connection to the graviton at the quantum level, so no apparent means of enforcing the equivalence principle. The symmetric rank 2 tensor that generates the gravitons will always have associated with it a scalar field by taking the trace of the graviton field. I propose the trace serves the same function as the Higgs, but there is no need for the false vacuum field. Recall that in this thread, while we respect the success general relativity has had, the GEM model is necessarily in conflict with GR. The graviton in GR is predicted to be a transverse wave, while GEM predicts the longitudinal and scalar modes of emission are the only ways gravitational energy waves can be set through space. doug
 P: 76 doug Thanks for your comments. As a crude definition I like to think that QT tells us what particles do, classical theory tells us what particles are. You seem to deal only with one class of wave, but I think that force and anti-force produce two different waves; one is associated with vacuum force and the other with anti-vacuum force. Each particle has one vacuum wave that by torque actions winds up the anti-vacuum wave into an odd number of waves; the middle anti-vacuum wave lies at 90 degrees to the vacuum wave. This pattern is caused by the requirement that both wave patterns rotate around a central Zero Vacuum Point. My concern with 0 charged particles is again to describe what they are, not to predict there existence, a function that QT already does quite well. QT does not explain why Charged and 0 charge particles behave differently, that's the work of classical theory. Interpretation (the function of classical theory) should not contradict QT except where mathematical proof linked to experiment can be shown as in the case of the allocation of fractional charge to quarks, which quite simply is an assumption to far. Time for me to stop or I'll be here all night, regards john
 P: 361 Hello John: Classical and quantum mechanics are both so precisely defined, they can start from exactly the same place. The GEM 4D wave equation, $J_q - J_m = \square^2 A$, without a single modification, can be used for either classical or quantum mechanics. What changes are the implementation details, which can get very confusing. In a big picture way, the classical approach will lead to expressions where the measurements are numbers. In the quantum approach, operators are the observables which leads to measurements that are averages. What bothered Einstein was why causality for classical physics was different from causality in quantum mechanics. Classical physics is simple: if I punch someone in the nose, you can make a high speed film of the assault and order all the picture frames, one after the other. In quantum mechanics, if a photon punches an electron in the face, we must average all the possible ways a photon could contact an electron to get the right answer, confirmed to every digit of accuracy experimentalists can push the data. The reason for the difference in my opinion has to do with doing 4D calculus correctly with quaternions. Defining a quaternion derivative the simple way by exactly copying the one in any calculus book causes a problem: dividing the differential element on the left is different from dividing on the right. I steal a play from L'Hospital who stole it from Bernoulli, and use a dual limit process, where the pesky 3-vector goes to zero first, then the commuting real part goes to zero. Technically it is called a directional derivative along the real axis. It works just like a real derivative because after the 3-vector goes to zero, it is a real derivative. The important idea for mathematical physics is what happens if the order is now reversed, so that the real goes to zero, then the pesky 3-vector. Writing the differential on the left is different from the right, unless you take the norm of the vector. That is well defined, and always the same for the differential on the left or the right. So one absolutely must always work with norms of derivatives. That is what goes on in quantum mechanics: an amplitude can be calculated, but it is the averages that are observed. The "differential element" sounds like an abstraction, but I think of it as a ratio of changes in space over changes in time (more technically, over changes in the interval). When the change in space is smaller than the change in time, information travels at less than the speed of light, and we have classical causality. When the change in space is larger than the change in time, information travels at less than the speed of light, so all we can measure is average values of what happens. There is a way to visually understand this issue. Take 10 frames of someone getting punched in the nose. In the classical view, these can be ordered in time, and you get a movie out of the event. In the quantum view, one cannot make a movie. Instead all the frames of the movie can be superimposed, so you see all valid states of someone getting punched in the nose. This contains all the information about the system, the proverbial wave function. Making a measurement is the act of picking out a particular frame out of all the possible ones, the collapse of the wave function. About the vacuum... I consider this one of the sadder issues in physics, and no, this is not a dig at John, but at the larger professional community. The pros want the vacuum to do important things. Breaking the vacuum symmetry with the Higgs mechanism is suppose to give particles mass. The false vacuum energy of empty space is suppose to be cranking up the acceleration of the Universe. On the fringe of physics, zero point energy folks want to power everything and its brother with the vacuum. To all these people, my message is clear. A vacuum is empty and can do absolutely nothing. Ever. To the folks at CERN betting billions on the Large Hadron Collider to detect the Higgs, I will go on record to say they are going to fail. For the people working on a non-zero cosmological constant, I will go on record to say Einstein's greatest mistake will remain a mistake. Anyone chasing after zero point energy is on a fools errand. The variation for any measurement, even zero energy is not zero, but that is all about the numbers we must use to make measurements in spacetime, namely quaternions, which are not a completely ordered set. Think of baseball for a moment. We all know they talk about the average number of hits for a baseball player: that statistic is important. For each of those players, there is also a deviation from average, and almost no one knows what those numbers are. It may well be that batters with low variations also have higher averages, but I don't know, it is not a statistic that is discussed much. It is important to feel the difference between an average - the thing that matters most - and the deviation from the average. In measuring energy, the deviation from the average cannot be made arbitrarily small because quaternions (three complex numbers) are being used. Details for that claim are made on quaternions.com, under the uncertainty principle (borrowed directly from a lecture on the impact of complex numbers on quantum mechanics). My three take home messages are these: 1. Doing quaternion calculus in spacetime explains why classical physics uses a directional derivative along the real time line, while quantum mechanics must necessarily use normed derivatives for average measurements. 2. Doing group theory with quaternions explains why there are 4 known forces of Nature via their symmetries Diff(M)xU(1)xSU(2)xSU(3) which are needed to characterize any possible collection of events that appear in a volume of spacetime. 3. The simplest quaternion wave function is a way to unify gravity and EM. This claim can be tested two ways: by measuring light bending to the next level of accuracy, or by measuring the polarity of a gravitational wave. Things have gelled nicely this year (point #2 being brand new). Have a good day. doug
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 Quote by sweetser This is not what I meant. There is 1 quaternion 4-potential, $A$, which I happen to write as $A^{\mu}$ so when I calculate the scalar part of two quaternions, say $A^{\mu} B^{\nu}$, it makes sense how to use a metric $g_{\mu \nu}$ to calculate the scalar part of that quaternion product.
Actually, this was what I thought you meant the first time I read it. It was only on the third or fourth read that I started thinking of it literally.

 Quote by sweetser Since we are using the same A throughout, and since a quaternion commutes with itself, then for this product: A/|A| exp(A-A*) = exp(A-A*) A/|A|.
Of course A commutes with A. Does A also commute with A*? I think it does too, but that is from my intuition with SU(2). Yes, I'm sure it does.

 Quote by sweetser A/|A| must point in the same direction as exp(A-A*), so it really only has one degree of freedom. I am going to stop here, to see if we are on the same page as far as metrics, labels, SU(2) and U(1)xSU(2).
Yes, I am back on track.

Sorry for not replying sooner. I relied on PF to send me an email when you replied but it apparently didn't work.
 P: 53 What do you think about the sentence that EM is just a perturbation of gravity, and the gravity is just a perturbation of the space ? Supposing it is true, can we say that gravity moves at light speed? that is space moves at light speed?
 P: 361 Hello Carl: No problem. I knew I was trying to mix an apple with an orange, and come up with a new drink, as it were. As you know the metric tensor's job is to decrease the rank of a tensor expression by two. I want the metric to be used to calculate the distance, but keep the rest of the quaternion expression in its place. I am certain math nerds will say that is not legal. I kind of would prefer to work with the "square root" of a metric, so that the vector part of the product could be $g_0 t g_1 x + ...$ I'm certain the square root of a metric is not a new idea, but haven't read up on the topic. Here is a stunner: for the GEM metric solution, the eponential metric, if I were to work with square root of the metric, then the vector is invariant because $g_0 g1 = g_0 g_2 = g_0 g_3 = 1$. Special relativity preserves the scalar of a quaternion square product, gravity preserves the 3-vector of the same product. That's a simple invariance principle. doug
 P: 361 Hello Pippo: In my proposal, the wave for gravity and EM both travel at the same speed, c. You can call it the speed of light, or the speed of gravity, and they are both the same. As a practice, I always try to talk about spacetime. I don't think it makes sense to talk about space or spacetime moving. I am not clear about the statement on perturbations, so this is what I normally say. The Universe has survived as long as it has by doing almost nothing to nothing. That is durable! A completely empty Universe would have the Minkowski metric rule the rulers. The problem is there is stuff in the Universe (very little, despite your own experience). So the question becomes how to do almost nothing in a mathematically formal way. Think of a slinky that is held up with one hand. Give it a slight nudge, and it will wobble. Watch closely, and it wobbles for quite some time. This is a simple harmonic oscillator. That's the equation I am using for gravity and light. The first path I used to get to the exponential metric used perturbation theory in a critical step (I'll skip the details, there was a lot of math). doug

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