Unifying Gravity and EM

[CarlB]

Dr. Crowell's objection reminds me of the objections to Lisi's E8 paper; that one should not be allowed to add together spinors and scalars.

Rather than get involved in the particular objections, I'd like to note that the value of a theory is in its ability to predict reality, not its ability to be mathematically clean in some theoretical sense. What we need is the ability to calculate, not the ability to theorize.

For relativity and quantum field theory, we have absolutely no experimental verification of these theories. What we have instead is good experimental verification of GR calculations and excellent experimental verification of QFT calculations. All theory that lies below the level of calculation is junk DNA that was convenient to frame the calculations, but need not be a part of a newer, more general, theory. The successful calculations, on the other hand, must be retained or replaced with equivalent.

 

[Lawrence B. Crowell]

I have decided to resurrect my little site here I started a year or two ago. I worked up an interesting idea on quantum fields in curved spacetime. This is very simple, only relying upon some basic ideas of geometry in QM and a fibration.

https://www.physicsforums.com/showthr...=115826&page=2

I worked this up in my head as I wrote this, so there might be a boo-boo or two here, but I think the basic idea looks reasonable. In way of boo-boo I posted this notice on my site as well.
Lawrence B. Crowell

 

[Lawrence B. Crowell]

Dr. Crowell's objection reminds me of the objections to Lisi's E8 paper; that one should not be allowed to add together spinors and scalars.

Lisi's paper does add them, but with the application of elements of the Clifford algebra. How Garrett frames spinors and scalars is similar to the application of Grassmannians in supersymmetry. Lisi is a bit glib on this, but it is not fatal to the basic architecture of his paper.

Lawrence B. Crowell

 

[K.J.Healey]

You mean (for a working link) : https://www.physicsforums.com/showthread.php?t=115826&page=2

 

[sweetser]

Three sign changes

Hello Lawrence:

To keep things short and snappy per Lut's request, if this exists:

(\frac{\partial Ay}{\partial z} ~-~ \frac{\partial Az}{\partial y},\frac{\partial Az}{\partial x} ~-~ \frac{\partial Ax}{\partial z},\frac{\partial Ax}{\partial y} ~-~ \frac{\partial Ay}{\partial x})

...and this does not mathematically exist:

(\frac{\partial Ay}{\partial z} ~+~ \frac{\partial Az}{\partial y},\frac{\partial Az}{\partial x} ~+~ \frac{\partial Ax}{\partial z},\frac{\partial Ax}{\partial y} ~+~ \frac{\partial Ay}{\partial x})

then I am more than willing to challenge the status quo of manifold mathematics because it is an error of omission, not a mistake per se.

Doug

 

[Mentz114]

Hello Lut:

To stay rooted in calculations, have you been able to get to field questions for these three Lagrangians:

Maxwell:

(\nabla A - (\nabla A)^*)(- A \nabla - (A \nabla)^*) = (0, -E ~+~ B)(0, E ~+~ B) = (E^2 ~-~ B^2, 2 E \times B) \quad eq 1

Gravity:

(\nabla A2^* ~-~ (\nabla A2^*)^*)(- A2 \nabla^* ~-~ (A2 \nabla^*)^*) = (0, -e ~+~ b)(0, e ~+~ b) = (-e^2 ~+~ b^2, -e \Join e ~+~ b \Join b) \quad eq 2

And GEM:

\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))
= ((-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 \times B ~-~ e \Join e ~+~ b \Join b ~+~ 2 gE ~+~ 2 gb) \quad eq 3

I can do that with pencil and paper and Mathematica. How is your tensor software doing with the challenge?

Doug

I've been playing about with the third one,

E^2 - B^2 -e^2 + b^2
which comes down to

(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})

with the condition that

(\partial^{0}A^{0})^2+(\partial^{1}A^{1})^2+(\partial^{2}A^{2})^2+(\partial^{3}A^{3})^2=0

to get rid of g^2. This is as far as I've got, but I'll get back to it when my schedule allows.

This thread is interesting, please take the time to have a look at it.

https://www.physicsforums.com/showthread.php?t=192422

 

[sweetser]

Check that software

Hello Lut:

It is good to focus on eq 3 which represents the GEM Lagrangian, instead of separately Maxwell and gravity, is a good one to focus on. I don't think it is correct to impose the condition:

(\partial^{0}A^{0})^2+(\partial^{1}A^{1})^2+(\partial^{2}A^{2})^2<br /> +(\partial^{3}A^{3})^2=0 \quad eq 4

What happens algebraically is that a -g² cancels a +g². You need to see exactly the same thing happen with your software.

As a practical programmer type, I appreciate your shortcut to the scalar. The scalar is the same, so should flow through from there the same. The theorist does not agree, because your algebra is working based on a particular gauge constraint - no different from any other gauge theory - while eq 3 has a gauge cancellation. Because eq 3 is based on cancellation, there are infinitely more choices for the gauge that work with eq 3 than for your software that dictates the sum of the squares of the partial derivatives.

Don't toss away this work on

(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) \quad eq 5

That may well be the Lagrangian used for GEM that applies to massive particles. The Lagrangian for the force mediating particles that travel at the speed of light because they are gauge invariant due to cancellation - eq 3 - is not the same as the one for massive particles which must break gauge symmetry in an elegant way, perhaps exactly like eq 5. I have to look at the details of this to see if I like it :-) Lots of things going on, preparing for a few talks, APS talks in New London, Connecticut and St. Louis, Missouri.

Just read the first post in the thread you suggested. My initial reaction - which will be fun to see if anyone else made the argument - is the spin symmetry of the field strength tensor is not consistent with a spin 1 particle needed for like charges to repel.

Doug

 

[Mentz114]

Hi Doug:

OK, I'll find a way to deal with g.

As a practical programmer type, I appreciate your shortcut to the scalar. The scalar is the same, so should flow through from there the same. The theorist does not agree, because your algebra is working based on a particular gauge constraint - no different from any other gauge theory - while eq 3 has a gauge cancellation. Because eq 3 is based on cancellation, there are infinitely more choices for the gauge that work with eq 3 than for your software that dictates the sum of the squares of the partial derivatives.

Sorry, I don't understand a word of that.

I assume it's correct that E^2 - B^2 -e^2 + b^2 is (\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) with the g part removed.

 

[sweetser]

Hello Lut:

The question is why does g make no contribution in the GEM proposal, specifically eq 3 of post 500. There are 4 A's required to generate the squared fields:

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

The first two have curls, the last two have symmetric curls. The gauge g is definitely not set to zero. The g² contributed by the first pair cancels with the second pair. You can tell if your software has faithfully done this because g is not set to zero, but none of the components of g appear in the final result.

In the first pair, it is the order of the differentials that changes. In the second pair, it is which one gets conjugates. The contraction of the asymmetric tensor \nabla_{\mu} A_{\nu} does not do this.

Doug

 

[Mentz114]

Getting rid of g

Doug:

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

F_{(g)}^{\mu\nu} = \[ \left[ \begin{array}{cccc}<br /> 0 &amp; e_x &amp; e_y &amp; e_z\\\<br /> e_x &amp; 0 &amp; b_z &amp; b_y \\\<br /> e_y &amp; b_z &amp; 0 &amp; b_x \\\<br /> e_z &amp; b_y &amp; b_x &amp; 0 \end{array} \right]\]

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

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 &lt;&gt; j,k

b^i = \partial^jA^k + \partial^kA^j, i &lt;&gt; 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}+(\partial^{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.

 

[sweetser]

Managing the Even representation

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/dougsweetser/MaxwellFieldEquations/photo#5171869024067390082

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

 

[Mentz114]

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<br /> +8\partial^xA^3\partial^zA^1<br /> +8\partial^yA^3\partial^zA^2<br /> -8\partial^tA^1\partial^xA^0<br /> -8\partial^tA^2\partial^yA^0<br /> -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.

 

[Mentz114]

Another way to lose g

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

 

[Lawrence B. Crowell]

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

 

[sweetser]

Orthogonal tensors

Hello Lut:

I concur with the postscript statement:

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

 

[sweetser]

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:

\[ \left[ \begin{array}{cccc}<br /> g_t &amp; e_x &amp; e_y &amp; e_z\\\<br /> e_x &amp; g_x &amp; b_z &amp; b_y \\\<br /> e_y &amp; b_z &amp; g_y &amp; b_x \\\<br /> e_z &amp; b_y &amp; b_x &amp; g_z \end{array} \right]\] - <br /> \[ \left[ \begin{array}{cccc}<br /> g_y &amp; -e_x &amp; -e_y &amp; -e_z\\\<br /> -e_x &amp; g_x &amp; b_z &amp; -b_y \\\<br /> -e_y &amp; -b_z &amp; g_y &amp; -b_x \\\<br /> -e_z &amp; -b_y &amp; -b_x &amp; g_z \end{array} \right]\]

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

 

[Mentz114]

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

 

[sweetser]

Inventing math tools

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

 

[Mentz114]

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 realized 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.

 

[sweetser]

Using all modes of emission

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:

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

 

[sweetser]

Everything is antisymmetric (for now)

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

 

[Lawrence B. Crowell]

Gravitation differs from the gauge fields as an interaction determined by an exterior symmetry, rather than an inner symmetry modeled 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

 

[sweetser]

No Lie algebra for a symmetric tensor

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:

http://en.wikipedia.org/wiki/Coleman-Mandula_theorem]In[/PLAIN] 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 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.<br /> <br /> 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.<br /> <br /> Doug<br /> <br /> Big fuzzy picture: If gravity is the force justifying why everything loves and attracts everything else, its algebra must be commuting.

 

[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 modeled by 'hidden variables' is disproved by counter-example. Mathematical theorems always have lots of tight conditions that are often violated in practice.

Lut

 

[Lawrence B. Crowell]

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 modeled 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

 

[sweetser]

Maxwell and GEM derivation on YouTube

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

 

[Mentz114]

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

I can see that the connections are symmetric for geodesics (force-free) motion. But are the generators of the translation group identified with the connections ? The connections are derivatives of the gauge field whose generators commute. I'm busking this, maybe you can set me straight.

Lut

 

[sweetser]

Connections in Standard EM

Hello Lut:

On my latest YouTube talk, I got asked a good question: what would the Maxwell and GEM derivations look like if I included all the connections? I know that because the EM field strength tensor uses an exterior derivative, I could just they that is why they drop out. I wish to see these drop in detail. I will make the same presumptions as happen in GR: the connection is metric compatible and torsion-free, so I can use Christoffel symbols of the second kind. These are not tensors and have 3 indexes. What I hope to learn is how exactly to play with Christoffel symbols in EM, so that they all happen to drop out. Seeing this work for the B field gives us a good place to start:

\nabla \times A = \frac{\partial A^u}{\partial x^v} - \Gamma_{\mu}^{v u} A^{\mu} - \frac{\partial A^v}{\partial x^u} + \Gamma_{\mu}^{u v} A^{\mu}

= \frac{\partial A^u}{\partial x^v} - \frac{\partial A^v}{\partial x^u}\quad eq 1

Because the Christoffel is symmetric in the indexes, the two gammas cancel. I figured a similar thing should happen with the E field, but I must be missing a sign:

E = -\frac{\partial A^u}{\partial t} + \Gamma_{\mu}^{0 u} A^{\mu} -\frac{\partial \phi}{\partial x^u} + \Gamma_{\mu}^{u 0} A^{\mu}\quad eq 2

This time the gammas add. Oops. Where did I misstep?

Thanks,
Doug

 

[Mentz114]

Losing connections

Hi Doug:
I'll try and repeat the calculations above later. The covariant derivatives cancel out in the field tensor.

F_{\mu\nu} = \nabla_{\mu}A_{\nu} - \nabla_{\nu}A_{\mu}

where

\nabla_{\mu}A_{\nu} = \partial_{\mu}A_{\nu} - \Gamma^{k}_{\mu\nu}A_k

and because of the symmetry in the lower indexes the Gamma factors cancel.

As I've pointed out many times, this does not happen with the symmetric field tensor, so you end up with terms containing A^2 in the lagrangian, as well as functions of the metric.

I just don't see how you can mix up a metric theory and a vector potential. Your field equations are four in number and your unknowns are 10 for the metric ( 40 connections) and 4 for the potential. How do you solve that ?

Lut

 

[sweetser]

Mistake in 529

Hello Lut:

It looks like my mistake in Eq. 2 of post #529 was on the first gamma, it should have been negative. For the E field, I need:

E = \nabla_{0}A_{u} - \nabla_{u}A_{0} = -\frac{\partial A_u}{\partial t} - \Gamma^{k}_{0 u}A_k - (\frac{\partial \phi}{\partial x^u} - \Gamma^{k}_{u 0}A_k)


So the minus in the first term happens from the lower script for the A_u which does not change the sign of the gamma. The minus on the second term does flip the switch on the second gamma, making it positive. Not difficult, but I need to get rock solid on how signs work in standard EM first, then out to GEM. Gammas are always negative unless one tosses a minus sign in front.

Doug

 

[Mentz114]

Hi Doug:

with non-zero Christoffel symbols, raising and lowering indices is a nightmare -

A_{\mu} = g_{\mu\nu}A^{\nu} = A_{0}g^{\mu 0}+A_{1}g^{\mu 1}+A_{2}g^{\mu 2}+A_{3}g^{\mu 3}

and this

\Gamma^{ bc}_{a} = g_{ka}g^{bm}g^{cn}\Gamma^{k}_{ mn}

has 64 terms.

Note that most of the GEM calculations I did are with the Minkowski metric.

Lut

 

[sweetser]

3-vector versus 3-vector

Hello:

I am writing code to handle the Even representation of quaternions on the command line so I can make animations using them. In this context, quaternions are always 4 numbers that get piped from one program to another. It is not possible to know if a quaternion being fed into the next program in a chain of these programs was an Even or Hamilton quaternion.

When I got to the nuts and bolts of the programming, I noticed that this issue only matters for binary operations, and then only for that part of the binary operation that has the 2 3-vectors forming a product. In the Hamilton representation, the dot product gets a minus sign, the even dot product stays positive. In the Hamilton representation, the curl is the curl with half the terms positive, half minus. In the Even representation, all the same terms are all positive.

A few issues are missed by using tensors. One is the issue of knowing there is an inverse for both the Hamilton and Even representations of quaternions. I used to justify this on an issue of quantum field theory which is kind of obscure. A more apparent reason might have to do with the observation of the critical importance of group theory in physics. The definition of a group relies on there being both an identity and an inverse for every member of the group. Those are both there for these two representations of quaternions. Tensors don't have to have inverses.

Doug

 

[sweetser]

Complete analysis of vector current coupling

Hello:

I am preparing a talk for an APS meeting in St. Louis. One of the points I intend to make is that Feynman's analysis of the phase of the current coupling term is incomplete. The image really rocks, but I am unsure how to upload images here, so I'll have to get by with LaTeX.

The current coupling term is also known as the source term: -J^{\mu} A_{\mu}. Take the Fourier transform of the potential to get J'/k². Since the indices are the same, this term evaluates to the following scalar:

-J^{\mu} J&#039;_{\mu}/k^2 = (-\rho \rho&#039; ~+~ Ax Ax&#039; ~+~ Ay Ay&#039;~+~ Az Az&#039;)/k^2 \quad eq~1

What Feynman does is consider a specific case of a current moving along the z axis. While it makes the algebra simpler, it made my slide longer, and not as general. Skip all the z-axis specific stuff, and just multiply these two together using the Hamilton representation of quaternions:

-J J&#039;/k^2 = (-\rho \rho&#039; ~+~ Ax Ax&#039; ~+~ Ay Ay&#039;~+~ Az Az&#039;,

-\rho Ax&#039; ~-~ Ax \rho&#039; ~+~ Ay Az&#039;~-~ Az Ay&#039;,
-\rho Ay&#039; ~-~ Ay \rho&#039; ~+~ Az Ax&#039;~-~ Ax Az&#039;,
-\rho Az&#039; ~-~ Az \rho&#039; ~+~ Ax Ay&#039;~-~ Ay Ax&#039;)/k^2 \quad eq~2

The phase is the final 3 lines of eq 2. What Feynman calculated was part of the third line, Ax Ay&#039;~-~ Ay Ax&#039;. These two terms do not add together, so it will take 2 pi to get back to the starting position, the sign of spin 1 symmetry. That is excellent, since these terms are the transverse modes of EM that need spin 1 symmetry so like charges repel.

Now look at the first two terms of the third line, -\rho Az&#039; ~-~ Az \rho&#039;[/itex]. These do work together, and will require only pi rotations to get back to the starting position. This is a property of spin 2 symmetry. Spin 2 particles are needed for systems where like charges attract. A complete analysis of the phase looks good to me.<br /> <br /> Doug

 

[Lawrence B. Crowell]

I can see that the connections are symmetric for geodesics (force-free) motion. But are the generators of the translation group identified with the connections ? The connections are derivatives of the gauge field whose generators commute. I'm busking this, maybe you can set me straight.

Lut

I guess I missed this one. Non-geodesic motion involves some other force on the motion of a particle. Teleparallel connections have a number of meaning as I understand. One is that the geodesic motion is given by a horizontal bundle and the "force" by a vertical bundle. This is Finsler geometry. Another definition, related maybe, is where the motion is parallel or Euclidean-like (latitudinal lines on a globe) and where the force or vertical bundle terms act to sustain this flow. I think this can result in torsional gravity.

Lawrence B. Crowell

 

[CarlB]

New article out that perhaps reads on this subject:

A Spatially-VSL Gravity Model with 1-PN Limit of GRT
Jan Broekaert
In the static field configuration, a spatially-Variable Speed of Light (VSL) scalar gravity model with Lorentz-Poincaré interpretation was shown to reproduce the phenomenology implied by the Schwarzschild metric. In the present development, we effectively cover configurations with source kinematics due to an induced sweep velocity field w. The scalar-vector model now provides a Hamiltonian description for particles and photons in full accordance with the first Post-Newtonian (1-PN) approximation of General Relativity Theory (GRT). This result requires the validity of Poincaré’s Principle of Relativity, i.e. the unobservability of ‘preferred’ frame movement. Poincaré’s principle fixes the amplitude of the sweep velocity field of the moving source, or equivalently the ‘vector potential’ ξ of GRT (e.g.; S. Weinberg, Gravitation and cosmology, [1972]), and provides the correct 1-PN limit of GRT. The implementation of this principle requires acceleration transformations derived from gravitationally modified Lorentz transformations. A comparison with the acceleration transformation in GRT is done. The present scope of the model is limited to weak-field gravitation without retardation and with gravitating test particles. In onclusion the model’s merits in terms of a simpler space, time and gravitation ontology—in terms of a Lorentz-Poincaré-type interpretation—are explained (e.g. for ‘frame dragging’, ‘harmonic coordinate condition’).
http://www.springerlink.com/content/b3758t680gj383j8/

Back on the first of the year, Foundations of Physics (the above journal) picked up a new editor, Gerardus 't Hooft, who writes:

During my first couple of months in this office, it became clear that fundamental questions in physics and philosophy also attract the interest of many laymen physicists.

We receive numerous submissions from people who venture to attack the most basic premises of theories such as Special Relativity, but instead only succeed in displaying a lack of professional insight in how a physical theory is constructed. I suspect that some of these people may have been working somewhere in an attic, deprived from daylight for decades, determined only to reemerge with a Theory of Everything in their hands. Even though they may be very sincere, we have to disappoint such authors. New insights are gained only by intense interactions with professionals all over the globe, and by solidly familiarizing oneself with their findings, and we must make a selection from only those papers whose authors have a solid understanding of the topics they are discussing.

Fortunately they also submit their work, and their clever inventiveness continues to surprise us.
...
I hope to receive your submissions. Acceptation of a paper may not necessarily mean that all referees agree with everything, but rather that the issues put forward by the author were considered to be of sufficient interest to our readership, and the exposition was clear enough that our readers, whom we assume to be competent enough, can judge for themselves.
...
http://www.springer.com/physics/journal/10701

 

[sweetser]

Variable Speed of Light theory

Hello Carl:

The paper cost $32 to download, ouch. I know I have a bias against spatially-Variable Speed of Light theories. On my dinning room wall is a large artwork, The Speed of Light According to Rene Magritte. I think about c as time's relationship to space. Photons travel at the speed of light because time is space. It is hard to see how that statement can vary.

Any reasonable proposal for gravity must agree to the first parameterized Post-Newtonian accuracy, as these folks do. I am a bit surprised they didn't look at the second level of accuracy.

The "scalar-vector model" might have passing similarities to GEM, but I am skeptical.

The attitude of the editor parallels some of the motivations of the Independent Research section of Physics Forums. 't Hooft has a site, How to become a good theoretical physicist, I always recommend to people working alone in the attic.

Should I ever get satisfied with my draft paper, I would consider sending it there. A number of issues need to be addressed first. I need to shift the action from tensor notation to quaternions because the issue of gauge symmetry must be clearly addressed. Second, I still have to get the energy expression in order.

I am emailing from the April APS Meeting in Saint Louis. I heard a talk by someone who is trying to take EM up a level to use the tools similar to GR, basically the opposite direction I am traveling. That talk gave me a good overview of the various approaches to EM as a geometry phenomena. I asked the speaker to send me his slides.

The talks about dark matter and dark energy make me sad. They certainly don't sound like solid physics. In a question and answer section, I confirmed with the speaker that she knew of no large scale systems where Newtonian gravity produces the observed results. That is what I expect since I think an unusual effect of gravity, labeled the relativistic rocket effect, Vc dm/dR Vhat, is never put into their numerical integration systems. I will recall these talks when trying to approach my fears of doing the calculations (they look hard, and I know my limitations too well).

Doug

 

[Mentz114]

Doug:
Jan Broekaert has several papers in the arXiv. Do a search in GR-QC for his name in authors. I'm having a look at one later, for diversion. Wiil report if it's interesting.

Carl:
I enjoyed the quote from t'Hooft.

Lut

[edit : this appears to be the very paper. Save $32 now !]

arXiv:gr-qc/0405015

 

[Lawrence B. Crowell]

New article out that perhaps reads on this subject:

A Spatially-VSL Gravity Model with 1-PN Limit of GRT
Jan Broekaert
In the static field configuration, a spatially-Variable Speed of Light (VSL) scalar gravity model with Lorentz-Poincaré interpretation was shown to reproduce the phenomenology implied by the Schwarzschild metric.

These types of theories are curious, but as Doug pointed out c is just an invariant measure by which space and time are equivalent. The value of c in its particular comes from other units in nature associated with electromagnetism, e, \epsilon_0,~\mu and so forth. If suspect that a variability in c or some energy dependence might come about near the Planck scale, but then again the Planck scale is defined according to c.

Lawrence B. Crowell

 

[CarlB]

These types of theories are curious, but as Doug pointed out c is just an invariant measure by which space and time are equivalent.

This is true only under the assumption that there is no preferred coordinate system. In that sense, the equivalence is logically circular. Any theory of gravity that is built on a flat underlying space has to have a variable speed of light. And in such a theory, the speed of light is well defined and variable.

My favorite such theory is the gauge theory of gravity by the Cambridge geometry group. The math it uses is based on the Clifford algebra of Dirac's gamma matrices (which they call "geometric algebra" following David Hestenes). Their theory is identical to GR (to all orders) outside of the event horizons but is built on a flat (well, Minkowski) underlying space or coordinates. I started a website supporting their theory here that includes links to the papers:
http://www.gaugegravity.com/
I'm planning on redoing the website to make it more informational soon. There are a lot of articles that use these ideas to model electrons and black holes, and I need to link those in.

When you have a flat gravity theory (and from my point of view, Sweetser's theory is flat), the first thing to do is to write down what the theory does for a singular mass, a black hole. This defines a preferred coordinate system, and that system defines the speed of light in the theory.

Since gauge gravity is identical to GR, the gravity you get from it is equivalent to GR written in a particular choice of coordinates. The stationary black hole becomes Painleve coordinates for the Schwarzschild solution. I created a java applet simulation that shows the relation of these to the usual Schwarzschild coordinates here:
http://www.gaugegravity.com/

Now the original name of the applet was "sweet" because I wanted to include Sweetser's gravity as well as Newton, and Einstein's in Schwarzschild and Painleve coordinates. However, I haven't figured out how to actually write down his equations, in the form of the acceleration felt by a test body, and so I haven't modeled it yet. I quit asking, but if he comes up with an equation, I'll type it up into java and add it to the simulation (hint hint).

 

[lba7]

oscillation can respond ?

(translation from french observation)

I think a new theory for source gravity. Maybe the origin of gravity is an oscillation of electrostatic wave. The oscillation of each particle is very very small compared an electron charge. Each particle send a wave, and each wave is accumulate to another wave. The charge of a particle is 0 but the wave is something around 0, one moment + and later - etc. So, each particle send + and - wave around it. It's possible to synchronize all wave because there is a difference between forces if you imagine two sinusoide waves face to face, there is two waves very close and two waves far, the force is with 1/d² so the difference can create a synchronism. This can explain deviation of light and redshift. The formula of perihelion precession of Mercury is the same with electrostatic forces integrate in the formula.

I have create a site for explain (in french for the moment) my ideas: 3w eisog dot com

Maybe this idea is not new, tell me ...

Ludovic

 

[Lawrence B. Crowell]

The breaking of Lorentz symmetry has been proposed by some. I am not sure how I stand on this for certain --- leaning against the idea I suppose. One can work with all sort of pseudotensors and the like with this, though I am not sure if that is what Sweetser is doing.

Gauge gravity, or in a general form with the B-F or Plebanski Lagrangian, recovers Einstein's gravitation at least for weak \alpha^2~\sim~G, though corrections on this for stronger G and higher energy (also I suspect cosmological scales) probably lead to consequences beyond a naive breaking of Lorentz symmetry.

Lawrence B. Crowell

 

[sweetser]

All tensors

Hello Lawrence:

The GEM proposal does not appear to break Lorentz symmetry.

This issue of tensors and GEM remains a concern. We know that \nabla_{\mu} A_{\nu} transforms like a tensor because that is why the machinery of a covariant derivative was built. We also know that the field strength tensor of EM transforms like a tensor, \nabla_{\mu} A_{\nu} ~-~ \nabla_{\nu} A_{\mu}. Correct me if I am wrong, but I believe that the difference of two tensors of the same order is also a tensor. If that is the case, then:

\nabla_{\mu} A_{\nu} - \frac{1}{2}(\nabla_{\mu} A_{\nu} ~-~ \nabla_{\nu} A_{\mu})

= \frac{1}{2}(\nabla_{\mu} A_{\nu} ~+~ \nabla_{\nu} A_{\mu})\quad eq ~1

Is it fair to call this object with the plus sign a tensor? Sure would hope so. Yes all three have different properties, but they all look like they belong in the family of tensors, not pseudotensors.

Doug

 

[CarlB]

Gauge gravity, or in a general form with the B-F or Plebanski Lagrangian, recovers Einstein's gravitation at least for weak \alpha^2~\sim~G, though corrections on this for stronger G and higher energy (also I suspect cosmological scales) probably lead to consequences beyond a naive breaking of Lorentz symmetry.

This is a different gauge gravity than that of the Cambridge geometry group. I guess "gauge" is a cool word to use to describe a theory. The Cambridge version is identical to GR to all orders, at least locally. Where it differs is in that you can't do weird topological things like wormholes. Here's how David Hestenes puts it:

The specialization of GA [i.e. Geometric Algebra] to Minkowski spacetime is called Spacetime Algebra (STA). As explained below, STA clarifies the geometric significance of the Dirac Algebra and thereby extends its range of application to the whole of physics. In particular, STA provides the essential mathematical framework for the new Gauge Theory Gravity (GTG).

The foundations of GTG are fully expounded in a seminal paper, so we can concentrate
on highlights of its unique features. GTG is a gauge theory on Minkowski spacetime, but locally it is equivalent to General Relativity (GR), so it can be regarded as an alternative formulation of GR.8 However, GTG reformulates (or one might say, replaces) Einstein’s vague principles of equivalence and general relativity with sharp gauge principles that have clear physical consequences (Section IV). These gauge principles are more than mere rephrasing of Einstein’s ideas. They lead to intrinsic mathematical methods that simplify modeling and calculation in GR and clarify physical meaning of terms at every stage. In particular, they provide clean separation between gauge transformations and coordinate transformations, thus resolving a point of longstanding confusion in GR. Moreover, GTG simplifies and clarifies the analysis of singularities, for example, in assignment of time direction to a black hole horizon. Finally, since tensors and spinors are fully integrated in STA, GTG unifies classical GR and relativistic quantum mechanics with a common system of gauge principles. Besides facilitating the application of quantum mechanics to astrophysics, this opens up new possibilities for a grand unification of gravitation and electroweak theories, as explained in Section V.
http://modelingnts.la.asu.edu/pdf/procGTG-RQM.pdf

So far, GR has only been verified to 1-PN. I don't think that it will make 3-PN. What I like about Sweetser's gravity is that it seems natural to do as the consequences of a particle theory. The problem I see in doing this is that to end up with a variable speed of light type gravity theory requires that the force of gravity be carried by superluminal particles.

In other words, just as Maxwell's equations end up being a force mediated by a speed of light particle, a theory which produces a variable speed of light needs to be mediated by something that moves at some higher speed. Comments from Doug?

 

[sweetser]

Geometric Algebra and Quaternions

Hello Carl:

Your email inspired me to go make another effort to wrap my head around geometric algebra. David Hestenes will be giving a keynote address at the conference in Brazil. Both Clifford algebras and geometric algebra fans view there tools as more general that quaternions, since quaternions are just Cl(0, 2) or a 0,2-multivector.

There is not much difference between these three approaches, but there is some. I think of the basis vectors differently, due to looking at this with my physics glasses on. A quaternion is described as a 0,2 multivector, which means a scalar for the first part, and three bivectors, e2/\e3, e3/\e1, and e2/\e1 for the 3-vector.

There are two issues the physicists in me objects. First, the main message of special relativity is to bring the scalar in with the 3-vector on equal footing. I look at quaternons at being 4 scalars with 4 basis vectors: (a0 e0, a1 e1, a2 e2, a3 e3). The GA folks like the wedge because it explains the minus signs. I find that claim hollow since is merely shift where the minus comes into the wedge itself. In my view of quaternions, the four parts look like equals.

The second objection comes from my view of gravity. In GEM, gravity can be about the potential, the basis vectors, or a combination of both. When it is about the basis vectors, then I want to be able to shift e0, e1, e2, and e3. For the exponential metric, e0 = exp(-2 GM/c^2 R) and e0 e1 = e0 e2 = e0 e3 = 1. I don't see how I could make similar statements with the GA formalism.

In the cited paper, Hestene explains how to do the Dirac algebra using geometric algebra. The details are different from how I do that with quaternions as triple products (all sixteen possibilities for e_u Q e_v make up the action of the gamma matrices).

My guess is the GA crowd will not be happy with the even representation of quaternions. I will get to find out, since that will lead off my 25 minute talk in Brazil.

The Cambridge geometry group looks like it is trying to recreate general relativity, with some more polite features. The GEM work is in line with a long tradition of directly confronting the dominant idea of the day, saying it is good up to 1-PN, but not 2-PN. One of those binary pulsar guys was at the APS meeting. I asked him if it is correct to take the equations developed for super-weak perihelion precession and apply it to a strong, dynamic system. If he writes back, I'll commmunicate to this thread.

I stay away from "faster than light". Using quaternions for everything is radical enough!

Doug

 

[Lawrence B. Crowell]

This is a different gauge gravity than that of the Cambridge geometry group. I guess "gauge" is a cool word to use to describe a theory. The Cambridge version is identical to GR to all orders, at least locally. Where it differs is in that you can't do weird topological things like wormholes. Here's how David Hestenes puts it:

The specialization of GA [i.e. Geometric Algebra] to Minkowski spacetime is called Spacetime Algebra (STA). As explained below, STA clarifies the geometric significance of the Dirac Algebra and thereby extends its range of application to the whole of physics. In particular, STA provides the essential mathematical framework for the new Gauge Theory Gravity (GTG).

Spinor algebra can reproduce general relativity in a pretty straightforward manner. The Dirac matrices \gamma_\mu define the metric as

<br /> \gamma_\mu\gamma_\nu~=~\frac{1}{4}g_{\mu\nu}<br />

and so if the representation of the spinor basis is local or chart dependent one can pretty easily reproduce GR with the Dirac operator. A gauge invariant form of the Dirac operator would then be

<br /> \partial_\mu(\gamma^\mu\psi)~=~\gamma^\mu D_\mu\psi~=~\gamma^\mu(\partial_\mu~+~A_\mu)\psi,<br />

for A_\mu a gauge term for GR.

For higher Clifford basis elements or with vierbiens extended GR theories can be derived. I suspect these are related in some ways to the string theory for GR which obtains for scales larger than the string length, but has deviations for scales approaching the string length. Of course there are some conformal invariance abuses with the string theoretic approach to GR as a bimetric form of theory. Yet string theory as a form of math-method does have some suggestive elements to it.

It is my sense that GR will survive to very high PN orders. I think deviations from GR obtain on two complementary scales: one scale near the Planck scale or say \le~10^3\sqrt{G\hbar/c^3}, and the other is on the cosmological scale where time translation invariance is "deformed" and there are inequivalent vacua states on scales across the cosmological event horizon. The verdict on this will likely come from gravity wave detection, such as LIGO, and the connection between these measurements with astrophysical events.

Lawrence B. Crowell

 

[CarlB]

Doug, I'm swamped right now with correspondence and I don't have internet connectivity at home so let me postpone writing more about what you've said here until later. But here's a link on another exponential gravitation theory. Can you talk about how this differs from your own?

[Uh, read down to the end of the article]
http://www.insidegnss.com/node/451

Carl

 

[Lawrence B. Crowell]

(translation from french observation)

This can explain deviation of light and redshift. The formula of perihelion precession of Mercury is the same with electrostatic forces integrate in the formula.

I have create a site for explain (in french for the moment) my ideas: 3w eisog dot com

Maybe this idea is not new, tell me ...

Ludovic

What is possible is that the spin quantum Hall effect might result in deviations from the equivalence principle. A circularly polarized photon as the superposition of two polarization states might be split by the effective index of refraction due to a gravity field. This would mean that gravitation has a Berry phase or a topological index associated with the spin of particles. This would be a possible deviation from classical general relativity.

An Einstein lens might provide the way to detect this. A radio telescope that detects photons from a lensed source will for a very narrow band pass filter measure quantum effect for entangled photons in this cosmic beam splitter. In principle the Wheeler Delayed Choice experiment could be performed this way. Also there might be for the two arms of this cosmic beam splitter a spin dependency in how the photons are split.

Lawrence B. Crowell

 

[sweetser]

Making Everything Manifestly Covariant

Hello:

I can finally write about the New England American Physical Society Meeting on Saturday, April 5. Lut can skip this post since it involves the personal stories. The three posts that follow in quick succession will be technical calculations.

My hopes were not high for this local meeting at the Coast Guard Academy. It was held at a military facility, with an official entry gate. People were marching around all dressed to the military nines at 8 am. One positive aspect of the military is they get to wear sharp hats to work, hats that indicate one's station.

The usual cast of characters where at this APS meeting. There were a dozen people in the room, half of whom I was familiar with. The session started off with a familiar fringe guy. As far as I can tell, he claims that everything is in motion. As a skeptic, I attempt to see what notions I need to stretch to see some grain of truth in his near incomprehensible riffs. He claimed that everything was in motion. What I would say is everything moves in spacetime, mostly as time, as things continue to persist. He does make fun of himself a little which helps, otherwise his talks are tragic.

Larry Gold gave his "Let's doubt global warming". It reminds me of Fox News in the US which claims to be unbiased, then makes every effort to selectively sample data needed to support its position.

My talks was titled: "Using Quaternions for Lagrangians in EM and GEM Unified Field Theory." For an audience this tiny and quirky, I could not expect them to know about Lagrangians, or how to use them to derive field equations. I had thought about doing all the equations in LaTeX, but decided to use a fountain pen, good old black ink written by hand, to indicated that despite the complications, this sort of calculation can be done by a person. These sheets of paper packed full of partial differential equations are prized possessions. It looks so hard core. Two of these sheets - a derivation of the Maxwell equations from the Lagrangian, and a derivation of the gravity part of GEM - were at the core of the talk (the rest being a setup for the heavy duty math).

The talk had one technical glitch. My connecting wire is flaky so everything was yellow. Otherwise, the speech went according to plan. My talks often do not elicit questions, but this time someone started asking about the symmetric curl. It was like he was channeling Lawrence. He was wondering why I had all these coordinate dependent equations. After the session broke up, I asked him some more questions, trying to see what he found unsettling.

It certainly is my intension to write coordinate-independent equations that are valid in flat or curved spacetime. When I write a quaternion as "q", it could be in any coordinate system. When I write \nabla q, that is a covariant derivative, the same whether spacetime is flat or curved, independent of coordinates.

When I went to write out the derivative by its component parts, I invariably wrote things in terms of Cartesian coordinates, the rectilinear t, x, y, and z. The progression of symbols for derivatives goes something like this:

\frac{d}{dt} \rightarrow \frac{\partial}{\partial t} \rightarrow \nabla_0 \quad eq~1

Only the last symbol is independent of coordinates, valid in flat or curved spacetime. I had never seen the Maxwell equations derived in a coordinate-free manifestly covariant way, so it was natural for me to just use partial derivatives with respect to t, x, y, and z.

Based on this criticism, I will be changing how I write component expressions. I will only use subscripts 0-3, not t-z. This way, equations that I think of as being manifestly covariant will have the notation required to be manifestly covariant.

For the record, I will repeat the quaternion operator derivations for the Maxwell equations, the gravity part of GEM, and the GEM unified field equations using the covariant notation in the next three posts.

Doug

Stills:
http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations

The talk:

 

[sweetser]

Hello:

In this post I will derive the Maxwell field equations using quaternion operators in a manifestly covariant notation.

Notice that the quaternion differential operator acting on a 4-potential creates a scalar gauge field, the -E and B fields:

\nabla A = (\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~+~ \nabla \times A) = (g, -E ~+~ B)\quad eq ~1

The starting point for the derivation of the Maxwell equations is the Lagrangian which can be viewed as the difference between the scalars of B squared and E squared. This can be achieved by changing the order of the covariant differential operator with respect to the 4-potential, which flips the sign of the curl (B), but not the time derivative of A or gradient of phi which make up E. The scalar gauge field can be subtracted away:

-\frac{1}{8}(\nabla A ~-~ (\nabla A)^*)(A \nabla ~-~ (A \nabla)^*)

=\frac{1}{8}(0, -\nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \times A)(0, \nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \times A)

=\frac{1}{2}(0, E ~-~ B)(0, -E ~-~ B) = \frac{1}{2}(E^2 ~-~ B^2, -2 E \times B)\quad eq ~2

Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms \nabla_0 \phi, \nabla \cdot A because they have been explicitly subtracted away at this early stage. It is also of interest to know the 3-vector part of this expression is the Poynting vector, a conserved quantity in EM.

Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling, -\frac{1}{2 c}(J A + (J A)^*):


\mathcal{L}_{EB} = \frac{1}{2}(-(\nabla_1 \phi)^2 ~-~(\nabla_2 \phi)^2 ~-~(\nabla_3 \phi)^2 ~-~ (\nabla_0 A_1)^2 ~-~ (\nabla_0 A_2)^2 ~-~ (\nabla_0 A_3)^2
~+~ (\nabla_3 A_2)^2 ~+~ (\nabla_2 A_3)^2 ~+~ (\nabla_1 A_3)^2 ~+~ (\nabla_3 A_1)^2 ~+~ (\nabla_2 A_1)^2 ~+~ (\nabla_1 A_2)^2)
~-~ (\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)
-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3

Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable:

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} \phi)}) = -\nabla_1^2 \phi ~-~ \nabla_2^2 \phi ~-~ \nabla_3^2 \phi ~-~ \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho
= \nabla \cdot E - \rho = 0 \quad eq ~4

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_1)}) = -\nabla_0^2 A_1 ~+~ \nabla_3^2 A_1 ~+~ \nabla_2^2 A_1 ~-~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1
= \nabla_0 E_1 - (\nabla \times B)_1 + J_1 = 0 \quad eq ~5

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_2)}) = -\nabla_0^2 A_2 ~+~ \nabla_3^2 A_2 ~+~ \nabla_1^2 A_2 ~-~ \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2
= \nabla_0 E_2 - (\nabla \times B)_2 + J_2 = 0 \quad eq ~6

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_3)}) = -\nabla_0^2 A_3 ~+~ \nabla_2^2 A_3 ~+~ \nabla_1^2 A_3 ~-~ \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3
= \nabla_0 E_3 - (\nabla \times B)_3 + J_3 = 0 \quad eq ~7

This work can be summarized with the Maxwell source equations:

\nabla \cdot E = \rho \quad eq 8

\nabla \times B ~-~ \nabla_0 E = J

This has been a manifestly covariant derivation of the Maxwell equations. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates. Once you get used to keeping track of all these terms, it is kind of fun, honest!

Doug

Stills:
http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations

The talk:

 

[sweetser]

Covariat gravity field equation deivation

Hello:

In this post I will derive the Maxwell field equations using quaternion operators in a manifestly covariant notation.

Notice that the quaternion differential operator acting on a 4-potential creates a scalar gauge field, the -E and B fields:

\nabla A = (\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~+~ \nabla \times A) = (g, -E ~+~ B)\quad eq ~1

The starting point for the derivation of the Maxwell equations is the Lagrangian which can be viewed as the difference between the scalars of B squared and E squared. This can be achieved by changing the order of the covariant differential operator with respect to the 4-potential, which flips the sign of the curl (B), but not the time derivative of A or gradient of phi which make up E. The scalar gauge field can be subtracted away:

-\frac{1}{8}(\nabla A ~-~ (\nabla A)^*)(A \nabla ~-~ (A \nabla)^*)

=\frac{1}{8}(0, -\nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \times A)(0, \nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \times A)

=\frac{1}{2}(0, E ~-~ B)(0, -E ~-~ B) = \frac{1}{2}(E^2 ~-~ B^2, -2 E \times B)\quad eq ~2

Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms \nabla_0 \phi, \nabla \cdot A because they have been explicitly subtracted away at this early stage. It is also of interest to know the 3-vector part of this expression is the Poynting vector, a conserved quantity in EM.

Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling, -\frac{1}{2 c}(J A + (J A)^*):

\mathcal{L}_{EB} = \frac{1}{2}(-(\nabla_1 \phi)^2 ~-~(\nabla_2 \phi)^2 ~-~(\nabla_3 \phi)^2 ~-~ (\nabla_0 A_1)^2 ~-~ (\nabla_0 A_2)^2 ~-~ (\nabla_0 A_3)^2
~+~ (\nabla_3 A_2)^2 ~+~ (\nabla_2 A_3)^2 ~+~ (\nabla_1 A_3)^2 ~+~ (\nabla_3 A_1)^2 ~+~ (\nabla_2 A_1)^2 ~+~ (\nabla_1 A_2)^2)
~-~ (\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)
-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3

Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable:

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} \phi)}) = -\nabla_1^2 \phi ~-~ \nabla_2^2 \phi ~-~ \nabla_3^2 \phi ~-~ \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho
= \nabla \cdot E - \rho = 0 \quad eq ~4

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_1)}) = -\nabla_0^2 A_1 ~+~ \nabla_3^2 A_1 ~+~ \nabla_2^2 A_1 ~-~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1
= \nabla_0 E_1 - (\nabla \times B)_1 + J_1 = 0 \quad eq ~5

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_2)}) = -\nabla_0^2 A_2 ~+~ \nabla_3^2 A_2 ~+~ \nabla_1^2 A_2 ~-~ \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2
= \nabla_0 E_2 - (\nabla \times B)_2 + J_2 = 0 \quad eq ~6

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_3)}) = -\nabla_0^2 A_3 ~+~ \nabla_2^2 A_3 ~+~ \nabla_1^2 A_3 ~-~ \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3
= \nabla_0 E_3 - (\nabla \times B)_3 + J_3 = 0 \quad eq ~7

This work can be summarized with the Maxwell source equations:

\nabla \cdot E = \rho \quad eq ~8

\nabla \times B ~-~ \nabla_0 E = J \quad eq ~9

This has been a manifestly covariant derivation of the Maxwell equations. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates. Once you get used to keeping track of all these terms, it is kind of fun, honest!

Doug

Stills:
http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations

The talk: