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

[sweetser]

Constructing a link between GEM and a quantum SHO

Hello:

In this post I will start from the GEM unified field equation, and see what needs to be done to get to a quantum simple harmonic oscillator. This will establish a link between the GEM field equations and quantum mechanics.

A 4D homogeneous wave equation is the GEM vacuum equation:

(\frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 \phi, \frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 A) = 0

The 4D inhomogeneous equation is the source GEM equation:

(\frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 \phi, \frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 A) = (\rho_q - \rho_m, J_q - J_m)

We can see that the scalar part of this equation can deal with Gauss' law of EM and Newton's law of gravity in a way consistent with special relativity since this equation in manifestly covariant under a Lorentz transformation (a fancy way of saying we know exactly how everything changes). The waves travel at the speed c. It is critical to note that this equation uses covariant derivatives, the sort with a connection that can work fine on a curved manifold. It is unfortunate that some skilled people see a d'Alemberian operator on a flat Minkowski background (it is there), but do not also see this expression can have a dynamic metric and work in a curved spacetime. I have shown earlier how the Rosen metric solves this 4D inhomogeneous equation for a static point charge.

The next step is to add in a term to convert the wave equation into a simple harmonic oscillator. One needs to subtract the norm of a factor that depends on R times the 4-potential. I'll write out the expression in its smallest parts, so you can notice the similarities:

(\frac{1}{c} \frac{\partial}{\partial t}, c \nabla)^*(\frac{\partial}{\partial t}, \nabla)(\phi, A) - (\epsilon, R)^*(\epsilon, R)(\phi, A)= (\rho_q - \rho_m, J_q - J_m)

Multiply things out.

(\frac{1}{c} \frac{\partial^2 \phi}{\partial t^2} + c \nabla^2 \phi - (\epsilon^2 + R^2) \phi, \frac{1}{c} \frac{\partial^2 A}{\partial t^2} + c \nabla^2 A - (\epsilon^2 + R^2) A) = (\rho_q - \rho_m, J_q - J_m)

To make this simpler, imagine that the potential in no way depends on time. A few terms drop:

(c \nabla^2 \phi - (\epsilon^2 + R^2) \phi, c \nabla^2 A - (\epsilon^2 + R^2) A) = (\rho_q - \rho_m, J_q - J_m)

What we now have looks almost like a quantum simple harmonic oscillator, as written here: quantum SHO. The one difference is that the \epsilon is positive in that expression, and it is a squared negative here. Since that term contains energy, it may end up depending how positive and negative energy are defined.

The quantum simple harmonic oscillator is a particular class of the time-independent Schrödinger equation. I'll put up a quaternion derivation of the Schrödinger equation soon.

doug

 

[sweetser]

Video on Spin 2 Symmetry Available

Hello:

I gave a talk this morning at an APS meeting titled "Seeing Spin 2 Symmetry in GEM, a Unified Field Theory", and already it is up on YouTube



There were seven people in the room, and due to the nature of the meeting, they were all folks who work with atoms, molecules, and lasers, not gravity. I am using Keynote '08 for my presentations these days, and they have made it simpler to record and upload to YouTube. I did that at 11am on Friday night, and now three people have viewed the 15 minute talk, one of them giving it 5 stars. I like the global reach of YouTube! The number of folks that flock to a Saturday 8am meeting in Connecticut was limited exclusively to the other people giving talks and the moderator.

The video contains another presentation of the content in posts #319 and #320, but I got some new insights in preparing for the talk. Let me explain again what the problem is. There is a current coupling term in my vector unified field theory, J^{\mu}A_{\mu}. In Feynman's lecture book on gravity, chapter 3, he shows why a vector coupling term like J^{\mu}A_{\mu} has spin 1 symmetry, a great thing for EM because spin 1 particles dictate that like charges repel. That creates a HUGE problem for a unified field theory because for gravity, like charges attract. I think this is the key calculation why bright folks don't play with something like GEM.

The calculation that shows the current coupling term has spin 1 symmetry is correct as far as its assumptions go. The hidden assumption has to do with the orientation of the current and the 4-potential. You are free to work with a 4-potential pointing in any darn direction you want relative to the current J. The freedom to choose a different directions for the orientation of the potential means there might be different symmetries to the current coupling term. Look at it this way: J^{\mu}A_{\mu} evaluates to a scalar, and we cannot detect A_{\mu} directly (only how it changes), so we must consider different possible angles between the currents. In short, Feynman shows that J^* J' has spin 1 symmetry, and I show that (i J i)^* J' has both spin 1 and spin 2 symmetry. It is a great little calculation, and helps me understand why folks as bright as Feynman never stopped to take a serious look at a rank 1 field theory for gravity. The particle physics is wrong, until you recognized the freedom to orient the coupled currents differently to each other.

doug

 

[Graviman]

I feel humbled corresponding here. Your YouTube video is about the only leading edge physics on the internet that makes any sense! I tried to order your DVDs, but there seems to be something wrong with the PayPal setup.

I have studied Maxwell, Relativity, and Quantum with the Open University. I am intending to teach myself Differential Geometry, and will probably try to understand Dirac too. What else do i need to read to do your GEM principle justice?

Thanks in advance,

Mart

 

[Mentz114]

Hi Doug,

re-the APS talk. I don't know the details of the Feynman calculation, and I'd like to see how the Fourier transforming of the potential leads to the Grassman (?) product between the currents. From that point on, it looks reasonable. But if there's a degree of freedom to rotate one vector independently without affecting the coupling scalar - is it physical ?

I'd like to read what Feynman has to say on gravity - do you have a fuller reference ?

Regards,
Lut

 

[sweetser]

Learning GEM

Hello Mart:

The humbling thing here is how cool Nature is. I work by making every possible mistake along the way, and learning from those mistakes, putting policies in place so I don't repeat a mistake.

Take for example relativity. People say they have studied relativity, they can solve problems, they "get it". All true, but how do you make sure you always work in a way that will be consistent with special relativity and the effects of gravity? [note to careful readers: I did not say general relativity since GEM is in technical conflict with GR, but both approaches must be able to explain gravity as the result of a dynamic metric]. My answer to the question is to always use quaternions which are just 4-vectors that one can also multiply or divide. Some folks might think that silly for problems in classical physics, but I say solving such problems in 4D is great practice: you accept the results, can believe them in a classical context, so when the same problem appears (same diff. eq.), the same approach will solve the relativistic problem. This comment is relevant to GEM, since the central equation for GEM is a 4D simple harmonic oscillator.

To learn the GEM proposal specifically, download and print these PDFs:

http://theworld.com/~sweetser/quaternions/ps/day1.pdf
http://theworld.com/~sweetser/quaternions/ps/day2.pdf
http://theworld.com/~sweetser/quaternions/ps/day3.pdf

Then you need to push a pencil. I do lots of calculations, always with a specific "start", always having 8 or fewer steps. It is important for you to be able to do those yourself, and repeat it until you can do it quickly (that is a sign that the brain does really get it). Redraw the little graphics that accompany the slides. The little graphics represent some of the information contained in the algebra, but the drawing will work a different part of the brain than the algebra, so you will get a wider grip on what is going on with this obtuse stuff. I toss around the phrases symmetric and antisymmetric tensors a bit, but I also have a small doodle that gives me a visualization on the issue. It has black and red squares, could be art, but is physics.

If you understand all the calculations in those notes from 2003, that is the foundation of this proposal. The YouTube lecture on spin 2 is an important technical issue, but could not be a complete explanation of the proposal, which probably would run on the order of five hours (the Maxwell equations are not simple, and the necessary bridges to gravity takes time to develop).

Understanding the GEM proposal take a real block of time and effort. If at any point, you get stuck, I want to know about it.

doug

 

[sweetser]

Feynman's Lecture Notes

Hello Lut:

Here is the reference:

https://www.amazon.com/dp/0813340381/?tag=pfamazon01-20
Feynman Lectures on Gravitation

I liked Brian Hatfield's introduction. I have not read the entire book, instead focusing on chapter 3, particularly pages 34-38. I used to be able to "look inside" on Amazon at those pages, but it is not working for me now.

You bring up a most excellent point, one I have been thinking about. EM is a gauge invariant theory, GEM is not. With EM, you can do a transformation like so:

A \rightarrow A' = A - \nabla \phi

This new A' points in a different direction. That is a "legal move" in EM, not GEM, an allowable gauge transformation. I am making this a bit more specific, by looking at a rotation of A:

A \rightarrow A' = A - \nabla \phi = (i A i)^*

The exercise is to compare the symmetry of the phase of A (spin 1) to A' (spin 1 and spin 2).

For GEM, I am not free to rotate A. Instead, I am free write A like so:

A = 1/2(A + (i A i)^*) + 1/2(A - (i A i)^*)

The 1/2(A + (i A i)*) has spin 1 symmetry, while 1/2(A - (i A i)*) has spin 2 symmetry in the phase. I have the freedom to represent the 4-potential in this way.

doug

 

[Mentz114]

Doug,

that looks promising.
The book is not available at the moment. I'm lost on the notation. Is your i from the vector basis i,j,k ? What is the '*' operator doing ?

Lut

 

[Mentz114]

I've been looking at the Lagrangian and your splitting of the potential. I've alway been worried about the hidden (implied) coupling between EM and grav currents.

If you write

A_g = \frac{1}{2}(A + L(\theta)A)
and
A_{em} = \frac{1}{2}(A - L(\theta)A)

Where L is a 3D spatial rotation, you can split the interaction term
into two
A_g^{\mu}J_{\mu} + A_{em}^{\mu}I_{\mu}, where J and I are the currents.
Now the coupling is explicit, and depends on theta. For me, this takes away one major problem with GEM.

The meaning of theta is a bit obscure, but as theta changes, energy moves between the spin 2 and the spin 1 fields.

[later] I've just realized that this changes the Lagrangian too much. Back to the scratch-pad.

 

[sweetser]

Spliting the Potential

Hello Lut:

The "i" is from the 4-basis vector (e, i, j, k). The way I think about curved spacetime is none of these has the norm of one, although ei=ej=ek=1 (e gets smaller, i,j, and k get larger).

The * is the conjugate operator, so (e, i, j, k)* = (e, -i, -j, -k). On my flight, I plan on playing with two sorts of splits of the potential:

A_g = \frac{1}{2}(A + i A i) = (0, 0, A_y, A_z)
A_{EM} = \frac{1}{2}(A - i A i) = (\phi, A_x, 0, 0)

One thing I like about this is that it has two degrees of freedom for gravity and two degrees of freedom for EM. This is done for the current coupling term, but not the field strength tensor, because that splits the spin 1 and spin 2 fields based on the symmetry of the second rank tensors.

The second possibility was my initial suggestion:

A_g = \frac{1}{2}(A + (i A i)^*) = (\phi, 0, A_y, A_z)
A_{EM} = \frac{1}{2}(A - (i A i)^*) = (0, A_x, 0, 0)

Gravity has too much of the potential, EM to little, so I am skeptical of this one working out.

I had been ignoring the coupling term because it looked too simple to do anything with. Now there may be a way to separate the coupling of like charges that attract from the coupling for like charges that repel. Need to do some work with pencil and paper... And do my day job, and pack.

doug

 

[sweetser]

One Current Density, Two Coupling Terms

Hello:

I realize there is a logical inconsistency in my proposal, and therefore there is a technical problem that must be addressed or the GEM proposal is wrong.

I have discussed how the 4D wave has two transverse modes of emission for EM and two modes for gravity. Yet I have written many times about 2 separate 4-currents, J_q and J_m. Together, those 4-currents have eight degrees of freedom which is too many and thus is wrong. I must make due with one 4-current density.

There are three constraints on the current coupling term.

1. The coupling term must be spilt into two: one for gravity and another for EM since they have different properties in regards to the behavior of like charges.

2. The two coupling terms must have opposite signs, so for EM like charges repel and for gravity like charges attract.

3. The EM coupling term must have a phase with spin 1 symmetry, while the gravity coupling term has spin 2 symmetry.

Nothing like a fun riddle!

To make things easier, consider motion along the z axis. The 4-potential can be split along the the transverse part of the potential, Ax and Ay, from the scalar and longitudinal parts, phi and Az:

A = \frac{1}{2} (A - k A k) + \frac{1}{2} (A + k A k)

A = (0, A_x, A_y, 0) + (\phi, 0, 0, A_z)

If I wanted to generalize this, I could refer to the parallel versus the transverse parts of the potential, A_{tr} and A_{||}. Constraint one solved.

The end result of these games must be the Lorentz invariant contraction of the current and the potential as it appears in the GEM action:

-\rho \phi + J_x A_x + J_y A_y + J_z A_z

Let's get the EM part first. I will be treating the current and potential 4-vectors as quaternions equipped with the usual ijk kind of algebra, but keep only the scalar part since that is all the action needs. Just multiply the current and potential together for the transverse part:

scalar (J \frac{1}{2} (A - k A k)) = - J_x A_x - J_y A_y

Close, but not quite. One needs to include a minus sign, which is great, since it satisfies half of the second constraint. To be explicit:

scalar (- J \frac{1}{2} (A - k A k)) = J_x A_x + J_y A_y

Where there is a scalar, there is a vector, necessary to look at the symmetry of the phase. Fourier transform the 4-potential to a current:

vector (- J \frac{1}{2} (A - k A k)) = J_x J'_y - J_y J'_y

This satisfies half of constraint 3, not a surprise since that is standard EM analysis.

Now on to the gravity part. If we do exactly the same as before, we get exactly the same result:

scalar (J \frac{1}{2} (A + k A k)) = \rho \phi - J_z A_z

These have the wrong signs, and so we would have to toss in a minus sign and fail to meet constraint number 2. This would be good in the sense that I would finally be able to forget the GEM proposal. Yet I am too crafty for such a defeat! This is one scalar product, but not the only one. Let's take what I call the first conjugate, (i q i)^* = (-t, x, -y, -z). This will flip the sign of phi and Az:

scalar (J \frac{1}{2} (i (A + k A k)) i)^*
=scalar((\rho, J_x, J_y, J_z)(-\phi, 0 , 0, -J_z))
=-\rho \phi + J_z A_z

Now we have the correct signs for the scalar with a positive J A contraction. That completes the requirements for constraint number 2. Now look at the vector after the potential has been Fourier transformed into a current:

vector (J \frac{1}{2} (i (A + k A k)) i)^* =
=vector((\rho, J_x, J_y, J_z)(-\rho', 0, 0, -J'_z))
=(0, 0, 0, -\rho J'_z - J_z \rho')

The phase has spin 2 symmetry! These two will subtract together, so will take only pi radians to return.

The rain in London was not heavy enough to prevent my writing this note while waiting for the tickets to Wicked. I hope the math stands up to anyone's analysis.

doug

ps. I did get front row tickets, nice.

 

[Mentz114]

Hi Doug,

I look forward to seeing the new Lagangian. I've been looking at current conservation and have found the energy-momentum tensor of the EM fileld is (from Stephani )-

T^{mn} = g_{ak}F^{am}F^{nk}-\frac{1}{4}g^{mn}F^{pq}F^{ba}g_{ap}g_{bq}
The condition for energy conservation is -

(1) T^{mn}_n = -\frac{1}{c}F^{mn}j_n

The thing on the left is the divergence of T, as you'll recognise. Having found a nice new tensor problem I set it up in the calculator and tried to verify equation 1. The interesting thing is that I get expressions like
T^0 = B_x(\partial_zE_y-\partial_yE_z)
+B_y(\partial_xE_z-\partial_zE_x)
+B_z(\partial_yE_x-\partial_xE_y)
+E_z(\partial_xB_y-\partial_yB_x)
+E_x(\partial_yB_z-\partial_zB_y)
+E_y(\partial_zB_x-\partial_xB_z)

which goes to

T^0 = -B_x\dot{B_x}-B_y\dot{B_y}-B_z\dot{B_z}+E_x(\dot{E_x}+j_x)+E_y(\dot{E_y}+j_y)+E_z(\dot{E_z}+j_z)

In the source free case, this is zero because B and E are orthogonal and J = 0. Yippee. I'm not so sure what emerges if there is a current.

The other terms of T^n look like this -

T^1 = B_x\partial^i\dot{B_i}+E_x\partial^i\dot{E_i}-B_y\dot{E_z}-B_z\dot{E_y}+E_y\dot{B_z}+E_z\dot{B_y}

The first two terms look like divs of curls, but I'm not sure about the other four. They could be parts of a curl of a curl and so zero.

The reason I've shown all this detail is because it shows that the energy conservation conditions are actually Maxwell's equations.

This means that this analogy, and the use of the E-M tensor could show coupling of the current and potential as the equations we get by working out the divergence of the E-M tensor of the gravitational force tensor.

Watch this space.

 

[Mentz114]

Doug:
Some time later : I have realized that the terms I get for T_0 are not zero. Something wrong somewhere, back to the drawing board.

[Later]
For the source free case, with an EM field traveling in the z direction, we can say

T^0 = -B_y\dot{B_y}+E_x\dot{E_x}

which comes to

2kB_0E_0 - \omega (B_0^2+E_0^2) multiplied by a phase,

which may well be 0.

 

[Mentz114]

Hi Doug,
I've been re-reading your last post (#371) and disregarding my dislike of using the same potential and current for gravity and EM, it looks OK.

Back to the energy conservation. I did some pencil work and discovered that the divergence of the EM stress tensor actually hinges on this F^{\mu\nu}_{,\nu}, which is easily seen if you apply a differentiation operator to T^{\mu\nu} above.

If F^{\mu\nu}_{,\nu} = J^{\mu} then T^{\mu\nu}_{,\nu} goes to zero even if there's a current. This is a good thing in my opinion. So F^{\mu\nu}_{,\nu} is an interesting thing to calculate. For the EM anti-symmetric F the result looks like a 4-current. With A^{\mu} = ( \phi, \vec{A} )

J^0 = -\nabla^2\phi + \partial_{t} \vec{\nabla}.\vec{A}

J^1 = \partial_{t}E_x + \partial_{y}B_z + \partial_{z}B_y

Repeating the above with the symmetric field tensor

F^{\mu\nu} = \partial^{\mu}A^{\nu} + \partial^{\nu}A^{\mu}

I get,

F^{\mu\nu}_{,\nu} = \partial^{\mu}(\partial_{k}A^{k}) + \partial^{k}\partial_{k}A^{\mu} = J^{\mu}

[edit] I found the Tex 'Box' operator so is this -
F^{\mu\nu}_{,\nu} = \partial^{\mu}(\Box A) + \Box^2A = J^{\mu}


which looks like your field equations. Or is that an extra term ?

So, provided the conditions above are met, we have energy conservation in the field tensor. I'm not sure if this only works in Minkowski space where covariant differentiation commutes. The gravity is all in the potential, and the field equations tell us how to get the potential from the current. It looks too linear to be a theory of gravity without the coupling term, which maybe will tell us how the current changes the potential, which restores the non-linearity.

I might have a go at constructing a full stress tensor from the gravity part of the Lagrangian, then check the divergence for non-linearity.

 

[sweetser]

Non-linearity

Hello Lut:

It is quite common to think that non-linearity is a requirement for a full theory of gravity. What is going on in my opinion is one of those linked circle of statements one sees in math, where one thing implies another thing implies yet another. They all hang together tough. One can even change the order of the links, and the logic is consistent. If you are familiar with that idea in general, here is how it applies for GR:

1. The field equations are rank 2
2. The gravity field can be a source of gravity
3. The non-interacting vacuum field equations are non-linear
4. No efforts to quantize a GR type of theory of gravity have succeeded

For GEM, there is a different collection:

1. The field equations are rank 1
2. The gravity field cannot be a source of gravity just like in EM
3. The non-interacting vacuum field equations are linear (interactions make the equations non-linear, and must be approximated using thngs like Feynman diagrams)
4. The completely relativistic approach to quantizing EM (known as Gupta/Bleuler) can be applied to GEM, but the scalar and longitudinal modes of emission are spin 2

I asked Clifford Will once if there was direct experimental data about non-linearity, and I think he said no (but worry that my memory read in what it wanted to hear, or at least filtered out whatever technical issues he raised. Damn brain). I know there are people who claim there is data for the non-linearity. What I recall Will saying is that so far, all of our best, most consistent theories for gravity have to date all been non-linear. Time for a change! It is just the non-interacting, vacuum equations that are linear.

doug

 

[CarlB]

For GEM, there is a different collection:

1. The field equations are rank 1
2. The gravity field cannot be a source of gravity just like in EM
3. The non-interacting vacuum field equations are linear (interactions make the equations non-linear, and must be approximated using thngs like Feynman diagrams)
4. The completely relativistic approach to quantizing EM (known as Gupta/Bleuler) can be applied to GEM, but the scalar and longitudinal modes of emission are spin 2

I agree completely with the first three. I think that they make complete sense from a parrticle theory point of view. Number 4 is unfortunately over my head.

By the way, have you given the question of the equations of motion any more thought?

Carl

 

[Mentz114]

Doug:

How did you derive your field equations ? Do you have any remark about the formula in my last post ?

I think that can be written ( now I've found 'Box' )
F^{\mu\nu}_{,\nu} = \partial^{\mu}(\Box A) + \Box^2A^{\mu} = J^{\mu}


Lut

 

[sweetser]

I'll give a try at answering both of these questions over the weekend.

doug

 

[sweetser]

Getting the Field Equations, the Standard Way

Hello Lut:

The field equations were the first things I derived when I wrote the GEM Lagrangian. They were the test that I had the right Lagrangian. What I did was write the Lagrangian in terms of the components:

I used this definition of the field equations:

which if you write out all the components looks like so:

Although there are lots of terms, it is kind of simple, since everything ends up being positive. The field equations are generated by taking the derivative of the Lagrangian with respect to the sixteen possible combinations of taking the 4-derivative of the 4-potential. The result is this:

This looks like

\Box^2A^{\mu} = J^{\mu}

Notice that I am using the complete asymmetric tensor, not the symmetric term in isolation. I'll have to look at what the scalar is from the contraction of the symmetric tensor. That should be a fun calculation! In other words, I have not calculated the field equations just for gravity.

doug

 

[Mentz114]

Field equations

Hi Doug,

thanks for posting all that. I can see you've used Euler-Lagrange with the
A^{\mu}'s as dynamical variables. I reckon the result adds up.

I've been finding out some interesting wrinkles about the EM Lagrangian from Itzykson&Zuber ( Warning; they write \Box for our \Box^2 )

Starting with

L = \frac{1}{4}F^2+j\cdot A

they get field equations

\Box^{2}A^{\mu} - \partial^{\mu}(\partial_{\nu}A^{\nu}) = j^{\mu}

and by applying the Lorentz condition \partial_{\nu}A^{\nu} = 0 the field equations become

\Box^{2} A^{\mu} = j^{\mu}

Now, under the Lorentz condition, only gauge transformations of the type

A^{\mu} \rightarrow A^{\mu} + \partial^{\mu}\phi where \Box\phi = 0 are allowed.

The point of this is to show how they lose the extra term from the field equations by restricting allowed gauge transformations.

Back to GEM. We have,

for EM field : \partial_{\nu} (\partial^{\nu}A^{\mu}-\partial^{\mu}A^{\nu}) = j_{EM}^{\mu}

for Grav field : \partial_{\nu} (\partial^{\nu}A^{\mu}+\partial^{\mu}A^{\nu}) = j_{G}^{\mu}

which when summed gives your original field equations. There is no need to apply the Lorentz condition - it has canceled out.

What can it mean ? I'm off for food, more later.

Lut

 

[sweetser]

Hello Lut:

I haven't really integrated the difference in notation yet, but this looks cool. Although I started out with 2 4-currents, now there is only one. So a bit of simplification is:

For EM:
\partial_{\nu} (\partial^{\nu}A^{\mu}-\partial^{\mu}A^{\nu}) = j_{tr}^{\mu}
For Gravity:
\partial_{\nu} (\partial^{\nu}A^{\mu}+\partial^{\mu}A^{\nu}) = j_{||}^{\mu}
For GEM:
\partial_{\nu} \partial^{\nu}A^{\mu} = j^{\mu}

So far, I have missed the detail concerning the \partial_{\nu} \partial^{\mu}A^{\nu} factor.

doug

 

[sweetser]

The G field equations

Hello Lut:

Great question about the G field equations. The Lagrangian needed for this looks like so:

\mathcal{L}_G = \frac{1}{c} J^{\mu} (A_{\mu} - I_{||} A_{\mu} I_{||})<br /> -\frac{1}{4c^{2}}(\nabla_{\mu}A_{\nu}+\nabla_{\nu}A_{\mu})(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})<br />

Make sure you write out the coupling term in a way that splits out the transverse from the parallel parts. The parallel part is mediated by a spin 2 field where like charges attract.

Write this out in its component parts:

\mathcal{L}_G = -\frac{1}{c} \rho A_0 + \frac{1}{c}J_{||} A_{||}

+ \frac{1}{2c^2}( -(\frac{\partial \theta}{\partial t})^2 + (\frac{\partial A_x}{\partial x})^2 + (\frac{\partial A_y}{\partial y})^2 + (\frac{\partial A_z}{\partial z})^2

- (\frac{\partial A_x}{\partial t})^2 + 2 \frac{\partial A_x}{\partial t} \frac{\partial \theta}{\partial x} - (\frac{\partial \theta}{\partial x})^2

- (\frac{\partial A_y}{\partial t})^2 + 2 \frac{\partial A_y}{\partial t} \frac{\partial \theta}{\partial y} - (\frac{\partial \theta}{\partial y})^2

- (\frac{\partial A_z}{\partial t})^2 + 2 \frac{\partial A_z}{\partial t} \frac{\partial \theta}{\partial z} + (\frac{\partial \theta}{\partial z})^2

+ (\frac{\partial A_y}{\partial x})^2 + 2 \frac{\partial A_x}{\partial y} \frac{\partial A_y}{\partial x} + (\frac{\partial A_y}{\partial x})^2

+ (\frac{\partial A_z}{\partial x})^2 + 2 \frac{\partial A_x}{\partial z} \frac{\partial A_z}{\partial x} + (\frac{\partial A_x}{\partial z})^2

+ (\frac{\partial A_y}{\partial z})^2 + 2 \frac{\partial A_y}{\partial z} \frac{\partial A_z}{\partial y} + (\frac{\partial A_y}{\partial z})^2)

Wow, that's a lot of partial derivatives! Notice the simple pattern: either a term is squared with a 1/2 in front of it, or the term is mix, and has a coefficient of 1. When we start taking the derivatives of these things, the derivative of d \frac{1}{2}x^2/dx = x and d (x y)/dx = y. Now it is time to apply the Euler-Lagrange equation. First do the current, which is the derivative of the Lagrangian with respect to the potential:

\frac{\mathcal{L}_G}{\partial A^{\mu}} = -\frac{1}{c} \rho + \frac{1}{c}J_{||}}

The tougher part is taking the derivative with respect to all the fields. This part takes some practice, but it is cool if you can get it to all work out. Do this one guy at a time:

\frac{\partial}{\partial x^{\mu}}\frac{\mathcal{L}_G}{\partial \frac{\theta}{x^{\mu}}} = -\frac{\partial^2 \theta}{\partial t^2} + \frac{\partial^2 A}{\partial t \partial x} - \frac{\partial^2 \theta}{\partial x^2} + \frac{\partial^2 A}{\partial t \partial y} - \frac{\partial^2 \theta}{\partial y^2} + \frac{\partial^2 A}{\partial t \partial z} - \frac{\partial^2 \theta}{\partial z^2}

There are a few things to notice from this, the first of four field equations. The current density rho has the same negative sign as the second derivative terms of theta. That is good, because it is consistent with like charges attracting each other. We can rewrite pairs of terms in this kind of way:

+ \frac{\partial^2 A}{\partial t \partial x} - \frac{\partial^2 \theta}{\partial x^2} = + \frac{\partial}{\partial x}(\frac{\partial A}{\partial t} - \frac{\partial \theta}{\partial x})

Notice the difference in the sign between these two terms. For the E field, they are both minus signs. For gravity, they have different signs. Time to get the next field equation:

\frac{\partial}{\partial x^{\mu}}\frac{\mathcal{L}_G}{\partial \frac{A_x}{x^{\mu}}} = -\frac{\partial^2 A_x}{\partial t^2} + \frac{\partial^2 \theta}{\partial t \partial x} + \frac{\partial^2 A_x}{\partial x^2} - \frac{\partial^2 A_x}{\partial y^2} - \frac{\partial^2 A_y}{\partial x \partial y} - \frac{\partial^2 A_x}{\partial z^2} - \frac{\partial^2 A_z}{\partial x \partial z}

Here again we can rewrite a pair of these terms like so:

- \frac{\partial^2 A_x}{\partial y^2} - \frac{\partial^2 A_y}{\partial x \partial y} = - \frac{\partial}{\partial y}(\frac{\partial A_x}{\partial y} + \frac{\partial A_y}{\partial x})

This is "curl-like" in the way the different derivatives swap places, but it certainly is not the curl since both have the same sign. There is also the term + \frac{\partial^2 A_x}{\partial x^2} which is probably the "Lorentz gauge canceling" term.

I did this all first with pencil and eraser. It might be a good exercise in 19th century math to see if you can spot mistakes (it is hard to proof in LaTeX format). Kind of amazing to work with this many partial differential equations, and have it make sense. It does look consistent with Lut's statements based on tensor analysis.

doug

 

[Mentz114]

Hi Doug:

on first reading this, I recognise those not curl-like terms. I've calculated currents and field-tensor contractions for potential A^m and A^m+phi^m to get the extra terms, and check if they vanish like divergences. I'm struggling to make sense of the extra terms, but I'll post them later so you can have a go at solving the puzzle, or even checking the calculation ( Mathematica ?).

Later ...

Lut

 

[Mentz114]

Hi Doug,

I have to say that I don't think a current can be represented with 2 degrees of freedom, so I have some doubts about the one-current proposal. I've been thinking about stress tensors instead of currents.

I'm intrigued by the cancellation of the extra terms from the field equations when EM and gravity are included. This could mean something or just be an mathematical accident.

Another thing is that if the potential is augmented thus -

A^{\mu} \rightarrow A^{\mu} + \phi^{\mu} where \phi^{\mu} = \partial^{\mu}\Phi

the field equations \Box^{2}A^{\mu} = j^{\mu} have an extra current

2\partial_{\nu} \partial^{\nu}(\partial_{\mu}\phi^{\mu})

which means we can add any potential as long as the above is zero.

But the important thing is to check the Lagrangian when the gauge transformation is done. After much effort I've ended up with 32 terms which I've emailed to you. It would be good to reduce them, and even to check the calculation, if Mathematica is up to it !

My software takes about 4 minutes to run the script, but with no simplification. Some of the extra terms will be divergences and will go to zero when integrated. The remainder are extra conditions to ensure energy conservation. If they cannot be met physically, the action is not gauge invariant. In EM this is bad thing, but whether it has any significance in the GEM context, I don't know yet.

Check your email...

Lut

 

[sweetser]

Not gauge invariant in the usual sense

Hello Lut:

Looks like we have a few technical disagreements based on your last post. Let's look at them seperately.

> I have to say that I don't think a current can be represented with 2 degrees of freedom, so I have some doubts about the one-current proposal.

Let me say it this way: a 4-current has 4 degrees of freedom, and a current-current interaction has two types of symmetry, both spin 1 and spin 2 symmetry. The two sorts of symmetry seen in the contraction of a current with a potential comes from the ability to arbitrarily point the potential relative to the current density it represents. Part of the potential will be parallel to the current it represents, part of it will be transverse. If you split the current into these two parts, that algebraic act by itself changes nothing. The phase of the parallel (||) and transverse (tr) potentials are:

J A_{||}-&gt;J FT[A_{||}] = J J&#039; = \rho \vec{J&#039;} - \vec{J} \rho&#039;

J A_{tr}-&gt;J FT[A_{tr}] = J J&#039; = \rho \vec{J&#039;} - \vec{J} \rho&#039;

The phase of these two contractions are the same, spin 1 symmetry.

What is needed is another twist of the potential for the transverse current:

J (I_{tr} A_{||} I_{tr})^*-&gt;J FT[(I_{tr} A_{||} I_{tr})^*] = -J J&#039; = - \rho \vec{J&#039;} - \vec{J} \rho&#039;

Notice those minus sign! When I did this work initially, I had to work with 2 4-currents, and put in the sign difference by hand. Now I get to work with 1 4-current with 4-degrees of freedom, and by twisting the orientation of the 4-potential, get to find phases with spin 1 and spin 2 symmetry and two signs.

Onward...

This gauge transformation:

A^{\mu} \rightarrow A^{\mu} + \phi^{\mu} where \phi^{\mu} = \partial^{\mu}\Phi

is not allowed in the GEM proposal. The reason is there are two fields, E for EM, and e for gravity, and this transformation will mess up one or the other. The fields are defined like so:

E = - \frac{\partial A}{\partial t} - \nabla \phi

e = + \frac{\partial A}{\partial t} - \nabla \phi

If I were to work only with E, there is a gauge transformation of the sort you refer to. If I were to work only with e, there is also such a gauge transformation. But these two gauge transformations are not the same.

The title here refers to an unusual way to think about gauge symmetry. Gauge means "to measure". You are exploring the usual use of gauge symmetry, the ability to add in an arbitrary derivative of a scalar field. When I look at the GEM action, and see the covariant derivative, I think of gauge symmetry as the ability to measure the action as the result of a potential in flat spacetime, or as the curvature of the spacetime manifold with a dull potential, or any combination in between.

I will work on the gauge symmetry, to make certain the kind of gauge transformation you suggest is not allowed for the full GEM proposal.

doug

 

[Mentz114]

Hi Doug :

I thought that

e = + \frac{\partial A}{\partial t} + \nabla \phi ?

I assume that A^{\mu} = (\phi, \vec{A}).

I can see your point, I think.

The only reason gauge transformations are of interest is to see if energy is conserved, and I think it is because the gauge current can be eliminated with the field condition.

Regarding the the current business. You can split the 4 degrees of freedom into 2 orthogonal vectors but you have an electrical and a matter current to describe, and only 2 df each. Or do I misunderstand ?

Lut

 

[sweetser]

Hello Lut:

I appreciate the importance of having an explicit way to show that energy is conserved. One reason I am not worried that we will eventually find it is that there is no "t" in the Lagrangian, so one can vary t without changing the action. A symmetry like that means that there is a conserved quantity, and it turns out to be energy.

There is no t in the action of GR either. Energy is conserved in GR, but it is not localized, one needing a volume of spacetime before one can say energy is conserved. The nonlocalization happens because one varies the metric field.

> Regarding the the current business. You can split the 4 degrees of freedom into 2 orthogonal vectors but you have an electrical and a matter current to describe, and only 2 df each. Or do I misunderstand ?

That sounds right. Both the graviton and photon travel at the speed of light, and thus only need 2 degrees of freedom each to describe a mediating particle where like attract (gravity) and like repel (EM).

doug

 

[sweetser]

Field Theory of Gravitation

Hello:

I was given an optional assignment by Lut to read a paper: "Field Theory of Gravitation: Desire and Reality" by Yurij V. Baryshev, arxiv:gr-qc/9912003. It was a fun read, and I will give you the GEM spin on the article.

Baryshev sees two types of theories out there in the literature: field theories with flat metric backgrounds and spacetime curvature theories. Nearly all the work done with gravity since Einstein has been spacetime curvature. The few papers on field theories bow to the powers that be, and say they are effectively the same thing as a geometric approach. Baryshev points out that while this might be true for weak gravity fields, the treatment of black holes will necessarily be quite different. He argues that there will be no black holes in a field theory approach, even though there is a Disney movie about such hellish places in the Universe. Einstein and Eddington would be pleased, but not the folks who churn out black hole papers, or model the collapse of black holes.

In Baryshev's hands, the argument against black holes is simple: the energy is positive definite, and built to stay that way by energy conservation (see section 5.3 for details, since I still don't completely get it).

With my GEM bias, I enjoyed the clarity with which he said there were only two types of proposals in the literature. I view GEM as being a third approach, exactly between a field theory with a flat metric and a dynamic spacetime connection (notice: I avoided the word curvature and the Riemann tensor on purpose, using the connection which is a derivative of the metric instead).

GEM does work as a pure potential theory with a flat metric. According to Baryshev, that is a great thing because it means energy conservation will work as a local effect, which is good because that is the way energy works for all other field theories we understand well. I have calculated the T^{00} of the GEM proposal before, but want to redo the exercise using only quaternions (my idea of challenging fun).

If GEM only worked with a flat metric, folks like Prof. John Baez would dismiss it with a one liner (he did actually, this very complaint of working only with a fixed background metric like all other field theories to date).

Actually, based on my education in GR, the thought of working with a flat metric did not occur to me (strange, but true). I learned special and general relativity from Edwin F. Taylor, the coauthor with Wheeler of "Spacetime Physics". That book was devoted to special relativity. They also wrote another book together, "Exploring Black Holes: Introduction to General Relativity". This book was all about playing with the Schwarzschild metric of GR. They did not derive the Schwarzschild metric. They avoided any technical discussions about connections and Riemann curvature tensors. Instead, the book looked at all the consequence of the metric.

In August of 1999, I had a field equation with a 4-potential in it. From my training, I had to find a way to a metric. I was not sure I could find such a path, but gave myself the time to look for one. I don't emphasis the first road I found often, fearing the potholes. What I did was to work with a force equation, F^{\mu} = J_{\nu} d^{\mu} A^{\nu}. I found a solution to that equation which was a 4-velocity, c^2 = (k_1 c \frac{dt}{d \tau}, \vec{K_2} \frac{d \vec{R}}{d \tau}). I squared this result, using the condition that for flat spacetime I should get a flat Minkowski metric to eliminate the constants k1 and K2. This lead to the exponential metric written in the first post of this long thread.

I derived a metric equation in a way no one else has done. Being skeptical of myself means I was not sure such a derivation was right. To the point of being a bore, I repeat that the exponential metric can be tested against the Schwarzschild metric, using the same experiment that was used to rule GR beats Newton: look at light bending around the Sun. In 1919, that was a big technical challenge. The data was not as clear cut as the public was led to believe. In time, we have solidified the GR victory.

The playing field for GR versus GEM involves a million fold improvement in the light bending measurement. We have a ten thousand fold improvement, so a few orders of magnitude refinement are required. Even so, the difference amounts to 12%. At the microarcsecond level, the rotation of the Sun, and the quadrapole moment also have to be accounted for.

If the only path to the testable exponential metric was the funky chicken walk through the force equation, I would not have the confidence that I am "within spitting distance of a unified field theory". I told the story earlier in this thread of finding a differential equation I had to solve using the Christoffel symbol of the exponential metric. For three weeks I avoided doing the calculation because I had never done a calculation with a Christoffel, and was unsure I could figure it out. The exponential metric did solve the field equation.

In all my reading on gravity, I have never seen another approach that got to a metric in the ways I have. So it goes.

doug

 

[Mentz114]

Hi Doug:

With my GEM bias, I enjoyed the clarity with which he said there were only two types of proposals in the literature. I view GEM as being a third approach, exactly between a field theory with a flat metric and a dynamic spacetime connection (notice: I avoided the word curvature and the Riemann tensor on purpose, using the connection which is a derivative of the metric instead).

You could be right. But one can always transform away the first derivatives of the metric. In GEM you'll have to make the action in the connection go to the potential to preserve an energy density.

I came across this today in a PF thread

As a funny aside, it's a theorem in QFT that the ONLY self-consistent theories with spin-1 bosons are gauge theories,
and the analogous theorems get even stronger as you go to higher spin. For example, the ONLY self-consistent theory of
(massless) spin-2 bosons is Einstein's Gravity!

It is a rather wonderful result (proved by Dick Feynman many years ago, if I remember correctly) that the only way to
write down a self-consistent theory of a *massless, spin-2 * object is Einstein's GR. There is simply no other way to
do it! The proof is quite non-trivial, but it's a famous result. Of course, there are some important assumptions, such
as Lorentz invariance, energy conservation, etc. But given some rather weak assumptions like that, then the only theory
that you can write down with massless spin-2 is GR. There is NOTHING at all "ad-hoc" about Einstein's Field Equations
(I'm assuming that's what "EFE" stands for). High-spin theories (s>1) are VERY constraining.

People are out to modify GR all the time, and that's okay. But almost always, the way they do it is to ADD something,
such as new scalar fields, extra dimensions, etc. But they always end up with Einstein PLUS stuff. You can never get
rid of Einstein entirely. Every model I know of that tries always has deep problems!

Is this discussed in your copy of the Feynman's gravity lectures ?

I'm sure you'll say it's wrong in any case.

[aside]
I have been informed that the potential does not transform as a 4-vector and sorted out the boosting of the EM field tensor and the equivalent operation on the potential.
https://www.physicsforums.com/showthread.php?t=197942

Lut

 

[sweetser]

Where the first derivatives of the metric go

Hello Lut:

Understanding this line is central:

> But one can always transform away the first derivatives of the metric.

I have to agree with this, it is true, but I always have my odd slant, promise! When you transform away the first derivative, where does it go? I have a specific technical answer. In GEM, the first derivative of the metric lives in one and only one place - the covariant derivative:

\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}

All the first derivatives live in the Christoffel symbol of the second kind. In GEM, when you change the first derivative of the metric in any way for any amount, you must make a compensating change in the potential so the derivative of the potential exactly balances it out. The potential and metric are separate, but changes in the potential play with the first derivative of the metric. You are free to play with the first derivative of the metric. You are not free to change the covariant derivative of the potential.

On the aside...
There is a lot of truth in the aside. I've seen it in action, specifically the Rosen bimetric theory that has the exponential metric in it. The problem is that it does add something, a fixed background metric, so that is a place to store energy/momentum, and it predicts dipole moments for gravity waves. We have good data to say the lowest form of gravity wave emission is a quadrapole.

I am not trying to make a theory of EM. I am not trying to make a theory for gravity. I am trying to make a theory of EM and gravity. The two play together in the covariant derivative as discussed about. This is how I avoid the proofs, I claim they do not apply to the "and" situation.

I don't know if Feynman provides a discussion in his book. He is a rank 2 field theory guy, like every other smart person making their money doing research on gravity. I can only follow along before hopping off that train of thought.

Fun discussion you have going on there. The way I read it, both the differential element and the potential do transform like vectors, but each comes with its own technical issue. For the differential, the contravariant tensor, d^u, needs to have minus signs for the 3-vector. For the potential, you need to have the current-coupling term around to keep things honest. So one can either say they do not transform like 4-vectors because they have these extra rules, or they do transform like 4-vectors so long as these rules are applied. People will argue over which is right, even if the net result is identical.

doug

 

[Mentz114]

Hi Doug:

I have to agree with this, it is true, but I always have my odd slant, promise! When you transform away the first derivative, where does it go? I have a specific technical answer. In GEM, the first derivative of the metric lives in one and only one place - the covariant derivative:

\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}

All the first derivatives live in the Christoffel symbol of the second kind. In GEM, when you change the first derivative of the metric in any way for any amount, you must make a compensating change in the potential so the derivative of the potential exactly balances it out. The potential and metric are separate, but changes in the potential play with the first derivative of the metric. You are free to play with the first derivative of the metric. You are not free to change the covariant derivative of the potential.

Yes, you've made this clear in the past, thank you. I'm asking about the next step. Suppose I make a coordinate transformation X -> X' = f(X) which removes the Christoffel symbols - by what rule or process does the gradient of the potential A(X) change to compensate ?

Another thing that bothers me is the fact that if the connections are non-zero then the Lagrangian ( the energy ) depends on the value of A as well as its derivatives. This is new territory, because GEM is no longer a gauge theory in the normal sense. Obviously adding a local or global field Phi changes the total energy.

I sent you a dumb question by email - it's important for my understanding of GEM to know the answer.

Lut