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## Unifying Gravity and EM

Doug:

The easiest way is just to ignore it, so we start by defining the field tensors,

$$F_{(g)}^{\mu\nu} = $\left[ \begin{array}{cccc} 0 & e_x & e_y & e_z\\\ e_x & 0 & b_z & b_y \\\ e_y & b_z & 0 & b_x \\\ e_z & b_y & b_x & 0 \end{array} \right]$$$

$$F_{(em)}^{\mu\nu} = $\left[ \begin{array}{cccc} 0 & -E_x & -E_y & -E_z\\\ E_x & 0 & B_z & -B_y \\\ E_y & -B_z & 0 & B_x \\\ E_z & B_y & -B_x & 0 \end{array} \right]$$$

$$L_{(g)} = F_{(g)}^{\mu\nu}F_{(g)}_{\mu\nu} = b^2 - e^2$$

$$L_{(em)} = F_{(em)}^{\mu\nu}F_{(em)}_{\mu\nu} = B^2 - E^2$$

and with the usual definitions of E,B, e and b -

$$E^i = \partial^0A^i - \partial^iA^0$$

$$e^i = \partial^0A^i + \partial^iA^0$$

$$B^i = \partial^jA^k - \partial^kA^j, i <> j,k$$

$$b^i = \partial^jA^k + \partial^kA^j, i <> j,k$$

from which I finally get this
$$B^2 - E^2 + b^2 - e^2 =$$
$$(\partial^{x}A^y)^{2}+(\partial^{x}A^z)^{2}+(\partial^{y}A^x)^{2}+(\par tial^{y}A^z)^{2}+(\partial^{z}A^x)^{2}+(\partial^{z}A^y)^{2}-(\partial^{t}A^x)^{2}-(\partial^{t}A^y)^{2}-(\partial^{t}A^z)^{2}-(\partial^{x}A^t)^{2}-(\partial^{y}A^t)^{2}-(\partial^{z}A^t)^{2}$$

The next step is to apply Euler-Lagrange, which doesn't seem to lead anywhere. Where did I go wrong ? ( Apart from losing some factors of 4 and butchering the notation !).

[later] I seem to be getting

$$\Box^2A^{\mu} = 0$$

Too tired to continue right now. I've got a feeling I've made a meal of something simple.
 Recognitions: Gold Member Hello Lut: You were only suppose to get the cross terms :-) I scanned in my hand-drawn derivations of the field equations here: http://picasaweb.google.com/dougswee...69024067390082 The step that may be tripping up your program is multiplying the even quaternion representation that is (0, b + e)(0, b - e) should give the scalar b2 - e2, not the negative of this. My bet is B2 - E2 - b2 + e2 would work in your software as is. Doug
 Blog Entries: 4 Recognitions: Gold Member Doug: Yes, I changed a sign and now I get $$B^2 - E^2 - b^2 + e^2 =$$ $$8\partial^xA^2\partial^yA^1 +8\partial^xA^3\partial^zA^1 +8\partial^yA^3\partial^zA^2 -8\partial^tA^1\partial^xA^0 -8\partial^tA^2\partial^yA^0 -8\partial^tA^3\partial^zA^0$$ (apologies for mixed notation, but I'm sure you get the drift) In agreement ( up to a sign) with your doodle. I won't repeat the algebra to get the field equations. I suppose the only point of this is that one can start with the traceless field tensors and go from there to get the same formulation as your funny quaternions. Much easier for the lay ( non-quat.) person to grasp. Lut PS : before I closed down the calculator I did this - $$g_{ik}F_{(g)}^{mk}g_{mn}F_{(em)}^{ni}$$. It comes to zero, zip, nothing. So the field tensors are orthogonal in a way.
 Blog Entries: 4 Recognitions: Gold Member Doug: there is a neat way to lose g. The symmetric field tensor has a dual, in analogy with EM theory, defined thus - $$\tilde{F_g}_{mn} = s_{mnij}F_g^{mn}$$ where s is the totally symmetric pseudo-tensor equal to |e|. The dual has no diagonal terms and $$\tilde{F_g}_{mn}\tilde{F_g}^{mn} = -8(b^2 - e^2)$$ So, putting the square of the dual in the Lagrangian density does the job. The symbol s also allows the symmetric curl of a vector to be defined as $$s_{ijk}\partial^{j}E^{k}$$ Lut

 Quote by Mentz114 Doug: [later] I seem to be getting $$\Box^2A^{\mu} = 0$$ Too tired to continue right now. I've got a feeling I've made a meal of something simple.
Which is the Lorentz gauge. If you put the g in there and let this be something "dynamical" then this inserts group dependent structure into the theory. The elliptic complex determines the bundle curvatures $\Omega^{2}(ad(g))$ which give fields "mod-g," for here g means group. The gauge condition is set to constrain the appropriate variables.

Lawrence B. Crowell

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Hello Lut:

I concur with the postscript statement:

 Quote by Mentz114 $g_{ik}F_{(g)}^{mk}g_{mn}F_{(em)}^{ni}$ It comes to zero, zip, nothing. So the field tensors are orthogonal in a way.
When I thought about GEM in terms of tensors, the symmetric one was the average amount of change in the 4-potential in 4 spatial dimensions, while the EM tensor is the deviation from the average amount of change. My recent shift to quaternions puts a stress on the underlying math tools, the Hamilton versus Even representations. At this time I don't understand all the links between these two perspectives, but it is nice having multiple perspectives. Thanks for adding to my views via this calculation.

Doug
 Recognitions: Gold Member Hello Lut: I am not clear on post #514. $F_{(g)}^{\mu\nu}$ has nothing down its diagonal - where the terms of a gauge live - so its dual also has this property. That appears to be starting too far down the page as it were. Here is what I am doing with quaternions, written as your tensors: $$$\left[ \begin{array}{cccc} g_t & e_x & e_y & e_z\\\ e_x & g_x & b_z & b_y \\\ e_y & b_z & g_y & b_x \\\ e_z & b_y & b_x & g_z \end{array} \right]$ - $\left[ \begin{array}{cccc} g_y & -e_x & -e_y & -e_z\\\ -e_x & g_x & b_z & -b_y \\\ -e_y & -b_z & g_y & -b_x \\\ -e_z & -b_y & -b_x & g_z \end{array} \right]$$$ The second tensor is the conjugate of a symmetric tensor. It shows operationally why there is a significant difference between $F_{(g)}$ and $F_{(em)}$, where both define the conjugate as flipping the signs of the later three parts, but $F_{(em)}$ can do so by taking the transpose of the matrix representation. That does not work for $F_{(g)}$. Nice. Doug
 Blog Entries: 4 Recognitions: Gold Member Hi Doug, in post #514, I'm defining the dual of F_g using a pseudo-tensor, in analogy with the EM case. The pseudo-tensor s is something I just invented, because we go from EM/anti-symmetric -> gravity/symmetric. This dual is traceless and gives the right energy. It's marginally less dodgy than inventing a new quaternion algebra ( maybe). Lut
 Recognitions: Gold Member Hello Lut: I thought you might be inventing things. This is a good sign. It dovetails with Lawrence's complaint about my work not fitting into the proud tradition of differential geometry. I think I referred to the wrong post. Post #514 is an extension of #511. It was #511 where you wrote out F(g), saying one should just ignore the gauge. It would be great if you could write the software to make it so. I don't count setting to zero a good answer, or just ignoring it is good practice. I want to catch the math responsible for gauge symmetry. That is what is most interesting. Doug
 Blog Entries: 4 Recognitions: Gold Member Hi Doug: I understand your unwillingness to ignore, or set to zero the trace of the grav. field tensor. We can play at finding mathematical ways to dispose of it, but this might not throw light on the physical justification. I'm not sure why you call it 'the gauge'. Anyhow, I'll try and justify my use of the 'symmetric dual' for the field tensor. In EM theory, one gets the dual using the Levi-Civita antisymmetric pseuso-tensor like this, $$\tilde{F}_{pq} = \epsilon_{mnpq}F^{mn}$$..............(1) furthermore $$F_{mn}F^{mn} = -\tilde{F_{mn}}\tilde{F^{mn}} = -2(E^2 - B^2)$$.....(2) So, to carry on the analogy between the EM theory, with anti-symmetric F and curl, to the gravitation case with a symmetric F and symmetric curl, we make a dual using a symmetric equivalent of the L-C pseudotensor, which we call 's' ( for 'symmetric'). $$\tilde{F}_{pq} = s_{mnpq}F^{mn}$$............(3) from which it follows $$F_{mn}F^{mn} - 2g^2 = -\tilde{F_{mn}}\tilde{F^{mn}} = -2(e^2 - b^2)$$....(4) because $$\tilde{F}_{pq}$$ is traceless. While I was typing the above, I realised that your field Lagrangian can now be very succinctly written as $$(s_{\mu\nu\rho\sigma}\partial^{\mu}A^{\nu})^2 - (\epsilon_{\mu\nu\rho\sigma}\partial^{\mu}A^{\nu})^2$$........................(5) this has no g terms as required, with the two pseudo-tensors doing the work of the odd and even quaternions. I still think GEM is wrong but the Lagrangian is tidier ! Lut PS the software worked faultlessly and did all the donkey work.

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Hello Lut:

Good work! I particularly like equation (5) and will make a mental note of it. Many group theory pros would object, saying the tensor is reducible, and reducible tensors cannot be used to represent fundamental forces. This issue has been raised indirectly in that thread you recommended reading in the first reply:
 Quote by blechman The "trouble" with gauge theories is that you are including unphysical degrees of freedom (such as the longitudinally polarized photon) and you have to make sure that they don't appear in the final calculations, and this complicates matters when you have to calculate things. However, despite the headaches, this can be done.
What I have argue since post #1 is the longitudinal mode of emission is a spin 2 graviton traveling at the speed of light doing the work of gravity. It is the scalar mode of emission that causes more serious problems since for a spin 1 field it implies negative probabilities.

People familiar with EM quantum field theory know these problems, and impose a constraint on quantizing EM such that the scalar and longitudinal modes are always virtual. Grad students grinding through these calculations often complain: it looks like a hack. Yet the hack needs to be there for a spin 1 field, and eventually they shut up and accept it.

The hack looks like a golden opportunity to put two modes of emission to work for gravity. Those modes will not harm EM theory in any way. A spin 2 scalar field would not have the indefinite metric problem, no negative probability problem.

No one else has picked up on the omission in Feynman's analysis of the current coupling term. I think that represents the greatest barrier to a rank 1 field theory.

Someone who bought a DVD from me thought I was within spitting distance of a unified field theory. I like that characterization slightly crude as it is.

Doug
 Recognitions: Gold Member Hello: I thought I'd share a small epiphany I had since it sheds a bit of light on the conflict I have had with Lawrence. The best field theory we have is the Maxwell field equations for EM. The second rank field strength tensor is: $$F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} \quad eq 1$$ It gives me confidence to know this is the foundation of my GEM proposal. Yet it also was the basis for the weak, the strong force, and Einstein's work on EM. For EM, the weak, and the strong forces, what changes is what group gets plugged into the machinery, U(1), SU(2), and SU(3) for EM, the weak, and strong forces respectively. Gravity is a bit different, but not so different. I don't know the details, but I recall reading how Einstein looked to EM for guidance on how to proceed. Take a peak at the Riemann curvature tensor: $$R^{\rho}{}_{\sigma \mu \nu}=\partial_{\mu} \Gamma^{\rho}{}_{\sigma \nu}-\partial_{\nu} \Gamma^{\rho}{}_{\sigma \mu}+\Gamma^{\rho}{}_{\mu \lambda}\Gamma^{\lambda}{}_{\nu \sigma}-\Gamma^{\rho}{}_{\nu \lambda}\Gamma^{\lambda}{}_{\mu \sigma} \quad eq 2$$ We have a subtraction thing going on again. This is a much more complicated animal, but we only need two contractions (the Ricci tensor minus the Ricci scalar), for the Einstein field equations. Now I can see the resistance to tossing in a "+" for the minus in eq 1. That equation is good enough for EM, the weak, and the strong force, which covers 3/4 known forces of Nature. We have GR which has withstood every subtle test, while failing all large scale ones (velocity profiles of galaxies, the big bang, and our current acceleration). I can be at peace with that resistance. I don't think the objection to a symmetric field strength tensor will stand the test of time, but it is rational to not embrace it. Doug
 Gravitation differs from the gauge fields as an interaction determined by an exterior symmetry, rather than an inner symmetry modelled as a vector space on a principal bundle. The bundle structure is then a fibration of the symmetries of the space (spacetime) in an atlas of local tangent spaces. This is all spelled out in the Coleman-Mandula theorem, which gives a mathematical reason for the structure of interactions which can possibly exist. It does not tell us their explicit form, but as a "no-go theorem" of sorts it tells us what is not permitted. What you say above about "gravity being different" is frankly a bit on the silly side of things. Lawrence B. Crowell

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Hello Lawrence:

Thank you for clarifying the difference between an inner symmetry and an exterior symmetry. I do lack the experience to speak the jargon of what is known. I thought I had admitted I was speaking imprecisely.

So I go read up about the "Coleman-Mandula theorem, which gives a mathematical reason for the structure of interactions which can possibly exist [under the assumptions used in the theorem]." I hope you do not object to the clause, which is appears banal. Plucking out a one-line summary:
 Quote by http://en.wikipedia.org/wiki/Coleman-Mandula_theorem In other words, every quantum field theory satisfying certain technical assumptions about its S-matrix that has non-trivial interactions can only have a symmetry Lie algebra which is always a direct product of the Poincare group and an internal group if there is a mass gap: no mixing between these two is possible.
A Lie algebra has the property that it anti-commutes. If one makes a strategic decision to walk in a new direction with a field strength tensor $\partial_{\mu} A_{\nu} + \partial_{\nu} A_{\mu}[/tex] which is a different combination of exactly the same players in [itex]F_{\mu \nu}$ (when the symmetric field tensor trace is zero). One of the first observations is that any generating algebra for the group would have to commute, ergo any conclusions of the Coleman-Mandula theorem do not apply.

Sorry Lut, this is the same non-debate Lawrence and I have had before. I had to find out where the imposed domination of anti-commutivity was written.

Doug

Big fuzzy picture: If gravity is the force justifying why everything loves and attracts everything else, its algebra must be commuting.
 Blog Entries: 4 Recognitions: Gold Member Hi Doug: You needn't apologise, it is always diverting to read your and LBC's posts, as long as they are short and to the point. The gauge group of teleparallel gravity, the translation group, has a commuting Lie algebra. I don't believe all mathematical theorems necessarily apply in physics, except in so far as they prevent inconsistencies. For example, von Neumann's 'proof' that QM cannot be modelled by 'hidden variables' is disproved by counter-example. Mathematical theorems always have lots of tight conditions that are often violated in practice. Lut

 Quote by Mentz114 Hi Doug: You needn't apologise, it is always diverting to read your and LBC's posts, as long as they are short and to the point. The gauge group of teleparallel gravity, the translation group, has a commuting Lie algebra. I don't believe all mathematical theorems necessarily apply in physics, except in so far as they prevent inconsistencies. For example, von Neumann's 'proof' that QM cannot be modelled by 'hidden variables' is disproved by counter-example. Mathematical theorems always have lots of tight conditions that are often violated in practice. Lut
I am not trying to say that the Coleman-Mandula theorem is proven in physics, where after all in science nothing is ever proven as such. It is just that the CM theorem is a fairly strong aspect of theoretical physics.

Teleparallel connections are commutative for special cases of nongeodesic motion. Christopher Columbus sailed to the new World on a nearly teleparallel path.

Lawrence B. Crowell
 Recognitions: Gold Member Hello: I will be going to the New England APS/AAPT Meeting in New London, Connecticut on Saturday to give the talk "Using Quaternions for Lagrangians in EM and GEM." At just under 15 minutes, you will be able to see the Maxwell equations derived from its Lagrangian. Rarely will you see this much detail about this central part of physics. GEM is a small variation on the theme, where you can see the 12 of 24 terms flip signs, generating a relativistic cousin of Newton's field equations for gravity. Enjoy. Doug http://youtube.com/watch?v=P9TUqUXGgpE

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