Simple Kaluza Klein question

[MikeL#]

When searching for KK stuff in forums, I see that when g(5) is applied to L(5 R), this effectively splits the equation of motion into a (4) gravitational part, (4) electromagnetic part and the Plank mass associated with a miniscule x^5. I have no problem accepting that GR can do this and calling it the ‘KK miracle’.

But when g(5) is applied to the simple L(5 u) or L(5 d) we simply get the classical eqn of motion, Klein Gordon and electron mass m. This is associated with a large x^5 approx= proper time tau (i.e. NOT 1/the Plank mass). Is there a name for this second miracle?
Please help me to find out more about it.
Thanks.‘Appendices’
This idea is not new, back in 1978 Franchi and in 1984 Kubo developed 5d QFT using s = x^5. id_s <=> mass of electron etc.

Definitions i,j run 0-3 a,b run 0-3,5 u = d_tau x
L(5 R) involves curvature & GR
L(5 u) = u^a g(5)_ab u^b m /2
L(5 d) = d_a f* g(5)^ab d_b f /2

Where u^a g(5)_ab u^b = u^i g(4)_ij u^j + u^a g(5)_a5 g(5)_5b u^b
and g(5) has -++++ signature, g(5)_55 = 1
and g(5)_j5 = g(5)_5j = A_j q/m (as it turns out by the 2nd KK miracle)

Useful bits from calculations:
A = A(4) ensures g(5)^ab g(5)_bc = KronDelta^a_c
So u^a A_a q/m = 1 => u^5 = (1 - A_a u^a q/m) approx= 1
So x^5 approx= tau >>>>>>>>>>>> the x^5 associated with Plank mass.

 

[MikeL#]

I am just checking that I have set things up properly as this is my first PF question. Having discovered the edit button, I corrected an error in the penultimate line.

But now I am concerned about the dogey looking rating symbol that other messages do not have. Have I comprssed the message too much? I am on a learning curve and realize I should try harder to use Tex for symbols.

 

[MikeL#]

I gather edit only lasts for a limited time, so here goes with another update.
Spotted an advanced option which allows attachment - now deleted.

I would love to know where to find a help page for PF. Thanks for the reply about LaTeX.
Anyway I will gradually apply it while learning how to use it as long as the edit mode lasts & expand my Kaluza Klein question here.

tips so far:
0. Save text in rtf or doc so can edit it when the option to do this on PF goes.
1. the help for LaTeX is given by clicking on $$\LaTeX {}$$. $I use ‘tex' by default to stand out but use ‘itex’ to blend in$ '/tex' is end of 'tex' loop not '\tex'.
2. the \ for single space works & inside 'mbox' . \\ for new page does not work:
3. resulting in equations too long which then distort the wrap-around
4. when save, only some of changes take place. Really need to go right out of PF & come back minutes after the change was made.
5. use advanced option in edit so can back out change - typical scary error is a missing '/tex.' (but it is slow) Definitions:
i,j run 0-3 while a,b run 0-3,5 $$\ \ \ u = d_\tau x \ \ \ \ \ \ F_{ij} = A_{j,i} - A_{i,j} = \partial_i A_j - \partial_j A_i$$
L(5 R) = 5d Lagrangian density involving Ricci & GR curvature) mentioned in passing
L(5 u) = 5d Lagrangian simply in terms of g(5) & u_a

$$u^a g(5)_{ab} u^b = u^i g(4)_{ij} u^j + u^a g(5)_{a5} g(5)_{5b} u^b \ \ \ \ \ g(5)_{55} = 1$$
So g(5) has -++++ signature, and $g(5)_{j5} = g(5)_{5j} = \kappa A_j$

$$g(5)^{ij} = g(4)^{ij}, \ \ \ g(5)^{55} = (1 + g(5)_{5j}g(5)^{j5}), \ \ \ g(5)^{j5} = g(5)^{5j} = \kappa A^j$$
$$\mbox{A = A(4) \neg \noteq \neq A(5) ensures } g(5)^{ab} g(5)_{bc} = \delta^a_c$$[I hope to show it turns out that $$\kappa = q/m$$ by the 2nd Kaluza Klein miracle.]Put on a classical hat and consider planet mass m, charged up to q (say by a Solar flare) in a field A due to a star mass M.

$$L(5 \ u) = u^a g(5)_{ab} u^b m/2 = (u^i g(4)_{ij} u^j + u^a g(5)_{a5} g(5)_{5b} u^b )m/2$$
$$p(5)_a = \partial L(5 \ u)/ \partial u^a = g(5)_{ab} u^b m$$

$$p(5)_i = (g(4)_{ij} u^j + g(5)_{i5} g(5)_{5b} u^b )m = p(4)_i$$
$$\mbox{ \ \ provided \ \ } g(5)_{j5} = \kappa A_j = A_j q/m \mbox{ \ \ and }$$
$$1 = g(5)_{5b} u^b = u^5 + g(5)_{5j} u^j \Longrightarrow u^5 = 1 - g(5)_{5j} u^j$$

So (almost at simple school level M,L,T,Q-dimensional analysis), we get $$u^5 = (1 - A_j u^j q/m)$$ where q and m are the charge & mass of the macroscopic object being considered (not a Plank mass).In the following: assume (5d) for L, p, g_5a and (4d) for the rest.
So for the classical equation of motion via $$\partial L/\partial x^a = d_\tau(p_a)$$ gives
$$d_\tau(u^a) = d_\tau( g(5)^{ab}u_b) = - \Gamma(5)^a_{bc} u^b u^c$$
$$d_\tau(u^i) = - \Gamma(4)^i_{jk} u^j u^k} + \kappa F^i_j u^j$$

But to conform with Maxwell physics, this again forces $$\kappa = q/m$$

If anyone shows any interest, I could show the working in Appendix 1.

In the Newtonian approximation
$$- \Gamma(4)^i_{jk} u^j u^k = - \Gamma(4)^i_{00} u^0 u^0 = g_{00,i} (u^0)^2 /2 = (u^0)^2 \partial_i (MG/r)$$
$$\Longrightarrow m \partial_ t(u^i)} = m (u^0) \partial^i (MG/r)} + q F^i_j \partial_t (x^j)$$
So this ‘classical’ view makes no mention of Plank mass. Now consider going back to school with the Bohr atom, then q/m refers to the electron - not a Plank mass.

Again I could ‘derive’ the Klein Gordon equation by assuming from:

$$L(5 \ d) =( \partial_a \psi ^*) g(5)^{ab} ( \partial_b \psi ) /2$$ where:
$$g(5)^{ij} = g(4)^{ij}, \ \ \ g(5)^{j5} = g(5)^{5j} = A^j q/m, \ \ \ g(5)^{55} = (1 + \kappa ^2 A_j A^j )$$

But to conform with the Klein Gordon equation, this again forces $$\kappa = q/m$$
$$L(5 \ d) = (D_j \psi)^* D^j\psi + m^2 \psi ^*\psi = L(4 \ d)$$

Then from here we can take non-relativistic approximation to ‘derive’ the familiar Schrodinger form.

Yet again we could ‘derive’ classical mechanics by choosing an appropriate wave-function: $$\psi \propto e^{-iS}$$
$$S = \int{p_a dx^a} \mbox{ \ \ \ \ And can show that \ \ } p_5 = m.$$ (Appendix 2 &3)

$$\Rightarrow L = (u_j u^j + 1) m^2 \psi ^*\psi$$ gives a consistent equation of motion
$$(u_j u^j + 1) m^2 \psi = 0.$$
This sketch is rather rushed. If done more carefully (Appendix 2), we can see that L splits into the 1st inertial part (= -1) and a second electromagnetic part (= +1).The moral of the story is that
1. classical kinematics & electro-magnetism can be linked using a vary simple g(5) and q/m.
2. simple connections between QM & CM can be established without resort to feeding in A_j ‘by hand’ into L. I must admit F_ij F^ij/4 still has to be fed in by hand if you want the full blown Lagrangian density. This is not an attempt to put QM/CM on the same footing - but just to show consistency.
3. Plank mass does not appear in these L. I leave that to the ‘Kaluza Klein miracle’ using L(5 R) of GR curvatures mentioned earlier.
4. So could someone tell me why 1 & 2 get away with a 2nd type of KK miracle allowing any old mass with no need for the x^5 having to be curled up.
5. Could a g(6) on similar lines be any help to those looking at Yang Mills etc? Would it not be great to be able to have a baryon, quark or lepton mass rather than a Plank mass as ‘lowest’ mass starting point?Appendices 1,2,3 in progress.

 

[MeJennifer]

Mike, there is no need for "ascii-latex", PF supports $$\LaTeX{}$$.

Just click on it to see the tags for regular $$\LaTeX{}$$ or $\LaTeX{}$ for inline.

 

[MikeL#]

Jennifer, thanks for the tip, site & my first reply - much appreciated.

Appendices 1,2,3 in progress @@@ experiment

[tex] ( gab ) = \left( \begin{array } {cc} gij & gi5 \\ g5j & g55 \end{array} \right) [\tex]

 

[MikeL#]

I would appreciate a reply/comment on the physics (or technical short-cuts or help sites) to see if this question is of any benefit to anyone out there. Thank you.

Overview:
Time was a parameter rather than a dimension before 1905, but it could still have been used to ‘simplify’ Lagrangian & equations of motion by using it as a ‘pseudo-dimension’.
So could $x^5 \approx \tau$ used here be a new ‘pseudo-dimension’ to ‘simplify’ electromagnetism, and save having to put in L both e.m and mass ‘by hand’.
The main motives for asking about this are:
1. I have not seen this obvious version in texts
2. This could provide an alternative to the enormous Plank mass as the ‘lowest’ mass to play with.

As I got timed out on the edit (1440 minutes = 1 day) I will attempt some of the appendices 1,2,3 here. It is slow going as still learning LaTeX and no-one has replied to the physics. More tips:
Not equal is \neq

Scripting convention.
In the following c = 1 and the e.m. field = A_j. For this to appear in a simple metric it must be ‘dimensionless’ in the sense of being a number. So write $$B_j = \tilde A_j = \kappa A_j = g^{(5)}_{5j}$$.

My apologies if this is not the conventional use for $\kappa$ which here will (turn out to) = q/m. And to be honest, I am using B_j to simplify the generation of equations. Likewise I use the (5) superscript when referring to 5d (except for g_5j or g^5j or g_55 where it is obvious), and may omit the (4) for 4d.


$$( g_{ab} ) ^{(5)} = \left(\begin{array}{cc} {g_{ij}} & {g_{i5}} \\ {g_{5j}} & {g_{55}}\end{array}\right) ^{(5)} = \left(\begin{array}{cc} {g_{ij}+ B_i B_j} & {B_i} \\ {B_j} & {1}}\end{array}\right)$$


$$( g^{ab} ) ^{(5)} = \left(\begin{array}{cc} {g^{ij}} & {g^{i5}} \\ {g^{5j}} & {g^{55}}\end{array}\right) ^{(5)} = \left(\begin{array}{cc} {g^{ij } & {-B^i} \\ {-B^j} & {1 + B_k B^k}}\end{array}\right)$$


Appendix 1 - here L is a Lagrangian.

$$2 L(5 \ u) = m u^a g(5)_{ab} u^b = m (u^i g(4)_{ij} u^j + u^a g(5)_{a5} g(5)_{5b} u^b )$$
$$p(5)_a = \partial L(5 \ u)/ \partial u^a = m g(5)_{ab} u^b$$

$$p(5)_i = m (g(4)_{ij} u^j + g(5)_{i5} g(5)_{5b} u^b ) = p(4)_i$$
$$\mbox{ \ \ provided \ \ } g(5)_{j5}= \kappa A_j = A_j q/m \mbox{ \ \ and }$$
$$1 = g(5)_{5b} u^b = u^5 + g(5)_{5j} u^j \Longrightarrow u^5 = 1 - g(5)_{5j} u^j$$

$$So \ \ p(5)_5 = \partial L(5 \ u)/ \partial u^5 = m( g(5)_{5j} u^j + g(5)_{55} u^5) = m g(5)_{5a} u^a = m$$
$$And \ \ L(5 \ u) 2/m = (u^i g(4)_{ij} u^j ) + (u^a g(5)_{a5} g(5)_{5b} u^b )$$
The first part = -1, the second part = + 1
(And $g_{55} = 1 \ \ because \ \ g_{55} g_{55} = g_{55}$).

It is going to take ages to show the following using matrices, so I will add it to Appendix 1b at end.
$$d_\tau(u^i) = d_\tau( g(5)^{ib}u_b) = - \Gamma(5)^i_{bc} u^b u^c = - \Gamma(4)^i_{jk} u^j u^k} + \kappa F^i_j u^j$$


Appendix 2 - here L is a Lagrangian density.

To ‘derive’ classical mechanics from Quantum, choose an appropriate wave-function:
$$\psi \propto e^{-iS} \ \ \ S = \int{p_a dx^a} \ \ where \ \ p_5 = m.$$
This assumption is discussed in Appendix 3. For p_5 = m, see Appendix 1.

$$2 L(5 \ d) =( \partial_a \psi )^* g(5)^{ab} ( \partial_b \psi )$$
$$= \left(\begin{array}{cc} {\partial_i \psi } & {\partial_5 \psi }\end{array}\right) ^* \left(\begin{array}{cc} {g^{ij } & {-B^i} \\ {-B^j} & {1 + B_k B^k}}\end{array}\right) \left(\begin{array}{cc} {\partial_j \psi } \\ {\partial_5 \psi }\end{array}\right)$$

$$= ( \partial_j \psi ^* \partial^j \psi ) - ( \partial_j \psi ^* B^j \partial_5 \psi + \partial_5 \psi ^* B^j \partial_j \psi ) + (1 + B_k B^k ) ( \partial_5 \psi ^* \partial^5 \psi )$$
$$= (D_j \psi)^* D^j\psi + m^2 \psi ^*\psi \ \ where \ \ D_j = (\partial_j + i m B_j)$$

For this to give the Klein Gordon Lagrangian density, we must make $m B_j = m \kappa A_j = q A_j, \ \ so \ \ \kappa = q/m.$ So for the ‘classical wave-function’:

$$i D_j \psi = (i \partial_j - q A_j) \psi = ( p_j - q A_j) \psi = m u_j \psi \ \ so$$
$$2 L = (u_j u^j + 1) m^2 \psi ^*\psi$$
Remember L is a field density here, and $L = L(\psi^* , \psi^*_{,a), \psi , \psi_{,a} ) \ \ with \ no \psi_{,a}$!
So the ‘equation of motion’ comes from differentiating w.r.t $\psi^* \ \ or \ \ \psi \Rightarrow (u_j u^j + 1) = 0$ as expected.


Appendix 3.

The ‘classical wave-function’ should really use WKB approximation for the slowly varying part. A Gaussian wave packet should do. I worked it out years ago extending the elementary 3d idea of expressing omega as omega(k) resulting in packet centred on (x - vt). In 4d, we use (x - u tau) etc.

Getting back on track to $S = \int{2 L(5 u) d \tau } = \int{p_a u^a d \tau } = \int{p_a dx^a}$.

This then allows $\partial_a S \Leftrightarrow p_a \ \ for QM \ \ \Leftrightarrow CM.$


If you don’t like the way I sneaked in a 2 above, (which I don’t) then how about trying:
$$\partial_a S = \int{\partial_a L(5 u) d \tau } = \int{d_{\tau}p_a d \tau } = \int{d p_a } = [ p_a ]$$
But ignoring the lower limit of integration is pretty crass as we are playing with path integration between 2 proper times.
I am sure there are better ways to show the result we want.


Appendix 1b (added later)

$$d_\tau(u^l) = d_\tau( g(5)^{lb}u_b) = - \Gamma(5)^l_{bc} u^b u^c = - g(5)^{la} \Gamma(5)_{abc} u^b u^c$$

$$But \ \ \Gamma(5)_{abc} u^b u^c = ( g_{ab,c} - g_{bc,a}/ 2) ^{(5)} u^b u^c$$
$$= ( g_{ij,k} - g_{jk,i}/ 2) ^{(4)} u^j u^k + [ (B_a B_b)_{,c} - (B_b B_c)_{,a} / 2 ] u^b u^c$$
$$= \Gamma(4)_{ijk} u^j u^k + [ B_m B_{b,c} - B_b ( B_{c,a} - B_{a,c} ) ] u^b u^c / / So$$

$$\Gamma(5)_{ibc} u^b u^c = \Gamma(4)_{ijk} u^j u^k + [ B_m B_{b,c} - B_b \kappa F_{a,c}] u^b u^c$$
$$\Gamma(5)_{5bc} u^b u^c = [B_m B_{b,c}] u^b u^c \ \ ( as \ \ B_5 = g_(55) = 1)$$

Then after a lot of index gymnastics & simplification:

$$d_{\tau} \left(\begin{array}{cc} {u^l} \\ {u^5} \end{array}\right) = \left(\begin{array}{cc} {g^{li} & {-B^l} \\ {-B^i} & {1 + B_k B^k}}\end{array}\right) \left(\begin{array}{cc} {- \Gamma(5)_{ibc} } \\ {- \Gamma(5)_{5bc} }\end{array}\right) u^b u^c$$


$$= \left(\begin{array}{cc} {- \Gamma(4)^l_{jk} u^j u^k + \kappa F^l_k u^k } \\ {- \kappa [ A_{j,k} u^j u^k + A_i (- \Gamma(4)^i_{jk} u^j u^k + \kappa F^i_k u^k) ] }\end{array}\right)$$


Note the expression for $d_\tau(u^5)$ above is entirely consistent with $d_\tau (1 - \kappa A_j u^j)$!

 

[member 11137]

I am not sure that I completely understand your question; this is why I am not sure that I can help you. I like one thing in your posts: the formalism you try to introduce and to work with; namely the matrices and the way to expose clearly the results concerning the 4D part and those concerning only the fifth dimension. For example, the first line of your last matrix in post 6 is the Lorentz Einstein Law in a 4D formulation, so far I can recognize it.

 

[MikeL#]

rephrasing the questions

Danke sehr Blackforest for your encouragement.

To clarify.
In GR the 5d action S decomposes (splits) into a Gravitational & F_ij F^ij/4 electromagnetic part where $\kappa^2 = 16 \pi G, \ \ \ \kappa = 1/PlankMass$. This decomposition or 'derivation' of electromagnetism from geometry is known as the Kaluza Klein miracle.

An example of this is from Albrecht December 2005 paper:
Kaluza Klein theories & Physics of extra dimensions 22 pages
http://www.itp.phys.ethz.ch/proseminar/partphys05/albrecht.pdf
I am using the \kappa of page 7 (not page 6) and set \phi = 1.

Classical mechanics uses the classical L & Gamma (described in post 6) and
this Gamma decomposes as you observed. But here $\kappa = q/m \neq 1/PlankMass$. There is a similar result for 'classical' Quantum Mechanics.


My questions are:
1. Why does this simple Classical description not appear in textbooks?
2. What is the name for this other 'classical' Kaluza Klein miracle?
3. Why are the values of kappa so different?
4. surely someone must notice the benefit - see moral of the story at end of post 3.

Thanks for your patience - Mike.



post script: corrections to Appendix 1b of post 6.

$$d_\tau(u^l) = d_\tau( g(5)^{lb}u_b) = - \Gamma(5)^l_{bc} u^b u^c = - g(5)^{la} \Gamma(5)_{abc} u^b u^c$$

$$But \ \ \Gamma(5)_{abc} u^b u^c = ( g_{ab,c} - g_{bc,a}/ 2) ^{(5)} u^b u^c$$
$$= ( g_{ij,k} - g_{jk,i}/ 2) ^{(4)} u^j u^k + [ (B_a B_b)_{,c} - (B_b B_c)_{,a} / 2 ] u^b u^c$$
$$= \Gamma(4)_{ijk} u^j u^k + [ B_a B_{b,c} - B_b ( B_{c,a} - B_{a,c} ) ] u^b u^c \ \ \ So$$

$$\Gamma(5)_{ibc} u^b u^c = \Gamma(4)_{ijk} u^j u^k + [ B_i B_{b,c} - B_b \kappa F_{i,c}] u^b u^c$$
$$\Gamma(5)_{5bc} u^b u^c = [B_{b,c}] u^b u^c \ \ ( as \ \ B_5 = g_{55} = 1)$$

Then
$$d_{\tau} \left(\begin{array}{cc} {u^l} \\ {u^5} \end{array}\right) = \left(\begin{array}{cc} {g^{li} & {-B^l} \\ {-B^i} & {1 + B_k B^k}}\end{array}\right) \left(\begin{array}{cc} {- \Gamma(5)_{ibc} } \\ {- \Gamma(5)_{5bc} }\end{array}\right) u^b u^c$$

after a lot of index gymnastics & simplification.


$$= \left(\begin{array}{cc} {- \Gamma(4)^l_{jk} u^j u^k + \kappa F^l_k u^k } \\ {- \kappa [ A_{j,k} u^j u^k + A_i (- \Gamma(4)^i_{jk} u^j u^k + \kappa F^i_k u^k) ] }\end{array}\right)$$


Note the expression for $d_\tau(u^5)$ above is entirely consistent with $d_\tau (1 - \kappa A_j u^j)$!

 

[MikeL#]

experiment to see if can update here

Note the expression for $d_\tau(u^5)$ above is entirely consistent with $d_\tau (1 - \kappa A_j u^j)$!

 

[MikeL#]

tips & references

I would like to add some more tips I have learned for new PF users, summarise PF references to the question, comment on nomenclature & finally make some comments about Appendix 2.
The first post here is a summary of the question. Later posts re-express the summaries and expand the equations (as I became more familiar with LaTeX).

More tips for new PF users:
0a. Save text in rtf or doc so can edit it when the option to do this on PF goes (which is 1440 minutes = 1 day after the last edits).
0b. ‘Physics Help and Maths help - Physics Forums’. Click ‘User CP’, then click ‘View all subscribed threads’ to see all the threads to which you have added a post. Here can see what the letter icons mean.
0c. More Relativity Resources (by Chris Hillman - how/where to question, other sites, ref to John Baez etc.) at http://math.ucr.edu/home/baez/RelWWW/HTML/more.html
0d. To reduce the LHS margin display an equation no longer than:
$$m = m = m = m = m = m = m = m = m = m = m = m = m = m = m = m = m$$
otherwise portrait print will be truncated even with zero L & R margins.
1-4. in post #3
5. To proof read LaTeX equations, use: http://rogercortesi.com/eqn/index.php Try out the string in 0d to set the size of the web page window. Then you know if & how your equation will sit on your print.
6. Not equal is \neq

PF references to ‘Re Kaluza-Klein help needed’

October 11, 2006 (8.56, 8.43, 9.15pm) in:
‘Physics Help and Maths help - Physics Forums > Physics > General Physics > sci.physics.research’.

1. https://www.physicsforums.com/showthread.php?t=136063 I discovered this one recently and is the most relevant. Markwh04 uses K where I use \phi and \kappa. The result of not making this distinction is that p_5 = charge instead of mass! In other words he has used the wrong metric as well. See ‘nomenclature’ below. Otherwise he gives me a lot to think about in a language I can understand, does not pull \Gamma out of a hat, and without LaTeX it was easy to copy.

2. https://www.physicsforums.com/showthread.php?t=136010 gar uses A for my (-B) and introduces vielbein formalism.

3. https://www.physicsforums.com/showthread.php?t=136104 charge, length & Yang Mills.


PF references to ‘Re Kaluza-Klein help needed’ of May 11, 2007 (6.17pm)
4. https://www.physicsforums.com/showthread.php?t=136016 Ken Tucker replies to questioner who uses q for my x and h_ for my g_(5). Same problem as in 1 above with wrong metric.
(dq/dt)^5 = h^{5a} (dq/dt)_5 - is wrong, if anything, it should be:
(dq/dt)^5 = h^{5a} (dq/dt)_a - but we don’t know the RHS. So try instead
(dq/dt)^a h_{5a} = 1 which gives (dq/dt)^5 = 1 - (dq/dt)^j h_{5j}
I referred questioner to this thread - but I don’t think the communications in sci.physics.research work too well as nearby it holds pages of repeated threads rather than replies to previous posts.

Special & GR forum: Dec 12-14 2005
5. https://www.physicsforums.com/showthread.php?t=103981 asking why we don’t hear more about KK.
This points to several refs in 23-30 Dec 2005 https://www.physicsforums.com/showthread.php?t=101151

6. In the above post #8 rpbphys (in his words) ‘stumbled across’ http://www.vttoth.com/kaluza.htm where Victor clearly describes the maths on his web page that I have been attempting here. He uses g_44 for my \phi and has same problem as in 1 above with wrong metric. But he does not dwell on p_5, so the problem is hidden and the classical mechanics is nearly reproduced like mine. I prefer the method of 1 - but without the error!

Moral of the story is - do a background search early on to save re-discovering the wheel. But on the other hand re-discovering the wheel can save reproducing the same faults!


Nomenclature.
With hindsight, I could have used ‘G’ or ‘h’ instead of g(5), leaving ‘g’ exclusively for g(4). But as this is not a common convention, I had best leave this idea well alone.

Also for 4d, I should have used the familiar indices $( \alpha, \beta, \gamma, \lambda )$ instead of (i, j, k, l) - but when I started, I did not know about LaTeX and I modeled the question from a reference to ‘Re Kaluza-Klein help needed’ above. Henceforth I will use ($\alpha, \beta$) = 0-3 for 4d, and (a, b) = 0-3,5 for 5d. Unfortunately this makes LaTeX essential, which prevents such easy copying. A way round this would be to attach the Rich text format that I now increment before adding new posts. In this I also correct past errors & amend approach without having to quantise changes into posts.

Scripting convention.
In the following c = 1 = \hbar and the e.m. field = A_j is a 4d concept (NOT 5d). For this to appear in a simple metric it really MUST be converted into a ‘dimensionless’ quantity in the sense of being a number when dealing with Cartesian co-ordinates. So use:
$$B_5 = 1, \ \ B_\beta = \tilde A_\beta = \kappa A_\beta \ ( = g^{(5)}_{5 \beta } \ if \ \phi = 1)$$.

To be honest, I originally used B_j to simplify the generation of equations. But now that I have seen the confusion generated by mentally merging \kappa with \phi, I feel it is even more important to keep using B_j. I use the matrix representation of the metric from Albrecht (December 2005): Kaluza Klein theories & Physics of extra dimensions (page 7 of 22) http://www.itp.phys.ethz.ch/proseminar/partphys05/albrecht.pdf . This was chosen to make the 5d GR Lagrangian density (that contains electromagnetism and Plank mass) self-consistent.

You can mentally set the dilaton field \phi = 1 to reproduce the simplicity of Kaluza’s original idea (before Klein introduced the need to make x^5 so small). I have shown in previous posts that \kappa = q/m = charge/mass. So by writing g_ in terms of A you can see how other people have misused a symbol like g_55 (or g_44 or K) to simultaneously stand for 3 different things, namely $\phi \kappa^2, \ \phi \kappa, \ \phi$. This misuse results in the nonsense of (p_5 = q). We need (p_5 = m) as an essential ingredient with Kalusa’s idea to make Klen-Gordon quantum mechanics (QM) compatible with classical mechanics (CM).

$$( g_{ab} ) ^{(5)} = \left( \begin{array}{cc} {g_{\alpha \beta} & {g_{ \alpha 5}} \\ {g_{5 \beta}} & {g_{55}} \end{array} \right) ^{(5)} = \phi^{-1/3} \left( \begin{array}{cc} {g_{\alpha \beta}+ \phi B_\alpha B_\beta} & {\phi B_\alpha B_5} \\ {\phi B_5 B_\beta } & {\phi B_5 B_5} } \end{array} \right)$$

$$= \phi^{-1/3} \left( \begin{array}{cc} {g_{\alpha \beta}+ \phi \kappa^2 A_\alpha A_\beta} & {\phi \kappa A_\alpha} \\ {\phi \kappa A_\beta } & {\phi } } \end{array} \right)$$


$$( g^{ab} ) ^{(5)} = \left(\begin{array}{cc} {g^{\alpha \beta}} & {g^{\alpha 5}} \\ {g^{5 \beta}} & {g^{55}}\end{array}\right) ^{(5)} = \phi^{1/3}\left(\begin{array}{cc} {g^{\alpha \beta} & {-B^\alpha } \\ {-B^\beta} & {\phi^{-1} + B_\gamma B^\gamma } }\end{array}\right)$$

$$= \phi^{1/3}\left(\begin{array}{cc} {g^{\alpha \beta} & {-\kappa A^\alpha } \\ {-\kappa A^\beta} & {\phi^{-1} + \kappa^2 A _\gamma A^\gamma } }\end{array}\right)$$

Given: $g(4) = \det [g_{\alpha \beta}](4) = g$, then the proof that g(5) = g for \phi = 1, follows:

Assume $(g_{\alpha \beta })(4)$ is a diagonal matrix - say for spherical co-ordinates, use determinant $\det [B_a B_b](5) = 0$, throw in the $g_{\beta\beta}$ terms to this mix. Because each $(g_{a5})(4) \equiv 0$, only the last term in () contributes. In the following expression I have dropped the (4) for 4d (so here, to save you repeating one of my many miscalculations, $g_{55} = g_{55}(4) = 0 \neq g_{55}(5) = 1$).

$$g(5) = ( g_{00} g_{11} g_{22} g_{33} g_{55}) + ( B_0 B_0 g_{11} g_{22} g_{33} g_{55}) + (g_{00} B_1 B_1 g_{22} g_{33} g_{55})$$

$$+ (g_{00} g_{11} B_2 B_2 g_{33} g_{55}) + (g_{00} g_{11} g_{22} B_3 B_3 g_{55}) + (g_{00} g_{11} g_{22} g_{33} B_5 B_5)$$

$$= g( B_\gamma B^\gamma g_{55} + B_5^2) = g (0 + 1 ) = g$$

If you really want the correct expression where $\phi \neq 1$, then multiply every $B_a \ by \ \phi^{1/2}$ and every $g_{ab} \ by \ \phi^{-1/3}$ so the grand result is:

$g(5) = \phi^{2/2}\phi^{-5/3} g = \phi^{-2/3} g$.




Appendix 2.
This successfully cast L(5) into the form of the known Klein Gordon ‘KG’ Lagrangian L(4) which is as reassuring as showing that canonical momentum p(5) = (p(4), m).

But I made an oversight by glossing over the following:

In general, physics (of what actually happens) lies in the equations of motion that a Lagrangian (density) can generate rather than the physics (of symmetry) within the Lagrangian itself. So L(5) = L(4) is not such an obvious blessing unless the determinants g(4) = g(5) which is true for \phi =1. This is needed for the ‘classical’ equations of motion to be consistent between CM & QM. Off the top of my head, this suggests that (like g_00(4)), \phi differs very little in magnitude from 1 and packs in even more physics.

Back in the late 1960’s in 6th form, I was fortunate enough to be taught ‘A’ level applied maths (which was basically Newton’s laws of motion) by an enigmatic man. He started most lessons by slowly stroking his beard, pausing, then he chalked on the black board using 3d vectors - integrating force = mass x acceleration to produce the conservation of energy equation. This was an example of expressing the same physics but in two different ways. (When he described planet motion, he used complex numbers.) As a result of this misspent youth, I later became rather interested in the links between: equation of motion, Hamiltonian, Lagrangian and expressing ideas in different ways. I attended a postgrad GR course for a term, later struggled to read Thorne-Misner-Wheeler, found Felsager very helpful and Zee enlightening but abstract.


First, for \phi =1, where Action (for 4d, or 5d when you include x^5):
$$S = \int L d(vol) = \int L \sqrt{-g} d^4x dx^5$$

Then:
$$\sqrt{-g}\partial L / \partial \psi^* = \partial_a [ \sqrt{-g}\partial L / \partial (\partial_a \psi ^*) ]$$

We know this works in 4d to give ‘KG’. Let's try for 5d and assume g(5) = g(4) = g.
$L(5) = L(\psi^*_{,a}, \psi_{,a})$, so LHS = 0. RHS gives ‘KG’

See sections 2, 5 & 8 of http://en.wikipedia.org/wiki/Lagrangian

 

[country boy]

The moral of the story is that
1. classical kinematics & electro-magnetism can be linked using a vary simple g(5) and q/m.
2. simple connections between QM & CM can be established without resort to feeding in A_j ‘by hand’ into L. I must admit F_ij F^ij/4 still has to be fed in by hand if you want the full blown Lagrangian density. This is not an attempt to put QM/CM on the same footing - but just to show consistency.
3. Plank mass does not appear in these L. I leave that to the ‘Kaluza Klein miracle’ using L(5 R) of GR curvatures mentioned earlier.
4. So could someone tell me why 1 & 2 get away with a 2nd type of KK miracle allowing any old mass with no need for the x^5 having to be curled up.
5. Could a g(6) on similar lines be any help to those looking at Yang Mills etc? Would it not be great to be able to have a baryon, quark or lepton mass rather than a Plank mass as ‘lowest’ mass starting point?

You may have thought of this, but let me mention it anyway:

The reason Klein introduced the curled-up fifth dimension was to explain the quantization of the electric charge. To do this he had to make the dimension close to the Planck length. Using this length as a wavelength then gives the Planck mass. But if you don't care about quantizing anything, then you can choose the charge and mass to be anything you want. The Kaluza "miracle" of deriving electromagnetism from a fifth dimension in GR was pre-QM and didn't restrict the size of the extra dimension.

It may be interesting to note that the Planck mass can be made much smaller (even similar to an observed particle mass) if the gravitational constant is allowed to become extremely large under conditions of particle formation.

 

[MikeL#]

charge & mass

1. Country boy, thanks for your physics comments about allowing any charge, mass & x^5 (pre-QM). This encourages me to persist in clarifying what I have been referring to as the metric from Albrecht.

2. The comment that Klein’s idea was to quantize charge etc. (post-QM) gently warns me that I may be wrong about the error of using K (e.g. by Markwho4 in post #10). So I still need to cautiously proceed in checking the consequences of K.

3. I still need to grind through http://en.wikipedia.org/wiki/Einstein-Hilbert_action (or what I have been calling the GR Lagrangian) for 5d before I can build on $x^5 \approx 1/Plankmass$.

4. I still have more refs to follow up from post #10.

So I am beginning to see my way forward on 4 fronts for the next few weeks.

As for relationship between Plank mass, G & 2 extra dimensions, there is an interesting powerpoint presentation by Fulvia de Fazio www.ba.infn.it/~gruppo4/Talks/DeFazio_04052006.ppt[/URL] - see around page 9 to 12.

Thanks, best wishes, Mike.
$$m = m = m = m = m = m = m = m = m = m = m = m = m = m = m = m = m$$

P.S. To find out more about the \phi radion field (not dilation as I said in #10) dip into p 9-15 of Overduin & Wesson [URL]http://xxx.lanl.gov/PS_cache/gr-qc/pdf/9805/9805018.pdf[/URL]

What I called the metric from ‘Albrecht’ is consistent with the above. He has a little mix up on p 6 & 7 where he can't make up his mind whether \kappa = 8 \pi G or \kappa^2 = 16 \pi G (I prefer the latter). On p 7, I think his equation for R(5) should use \kappa^2 instead of \kappa, (and his equation for S(5) should refer to 2 \pi r rather than 2 \pi R).

http://th.physik.uni-frankfurt.de/~lxd/intern/KK/
Look at the ansatz of Kaluza in 1919 and Klein in 1926. You can see how errors crept into the John Biaz homework question.

 

[MikeL#]

Reason why \kappa so small when using Lagrangian with GR curvature.

This post is a progress report where I attempt to sketch an answer to my original question (somewhat rephrased). I confess it was not so simple (as it assumes a basic intro to EM, QM, Lagrangian, curvature & GR). Then I sketch Appendix-2b on Fariday & wave-function.

First, here goes with some nerdy corrections to previous posts before the above sketches. I intend to attach at rich text version of this thread to correct errors in situ (in case anyone wishes to quote these ideas).

(update for #6): From now on I intend to replace the (5) superscript with $\tilde g$ etc. for 5d. So I must not use any more:
$$\tilde A^\alpha$$ as $A^\alpha = g^{\alpha \beta} A_\beta$ are 4d concepts.

(update for #10):
I use:
Lectures on Physics, by Feynman, Leighton, Sands (F-I to III), Addison Wesley, 1965,
Gravitation by Misner, Thorne & Wheeler (MTW), Freeman, 1973,
Geometry, Particles & Fields by Bjorn Felsager (BF), Springer, 1997.
Kaluza-Klein Gravity, Overduin & Wesson 1998 http://xxx.lanl.gov/PS_cache/gr-qc/pdf/9805/9805018.pdf
A Unified Grand Tour of Theoretical Physics by Ian Lawrie (IDL), Institute of Physics Publishing, 2001,
The Landscape of Theoretical Physics - A Global view, by Matej Pavsic (MP), 2001, ISBN 0-7923-7006-6.
(The last 2 are in British Library, London, top floor Science, PQ65.)It turns out after all from KK-GR that $\kappa^2 = 4 \pi \epsilon_0 G = (4 \pi \epsilon_0)(G) = 1 \ if \ 4 \pi \epsilon_0 = 1 = G$ but at this stage of my understanding I really don’t like this convention and confusion abounds over the 4pi. (See http://en.wikipedia.org/wiki/Planck_units or http://en.wikipedia.org/wiki/Gravitational_coupling_constant .)
With \hbar = 1 = c (which I can cope with), this reduces all quantities to numbers (which I don’t find helpful at all) e.g. Plank mass = 1,

electron mass $= m_e = 9.10956 \times 10^{-31} kg / 2.177 \times 10^{-8} kg = 4.184 \times 10^{-23}$.

electron charge $= q_e = -137.036^{-0.5} = -0.0854245$.

Instead, I prefer to follow the convention of Feynman-I-32-3, and take

electron charge $= q_e \ where \ e^2 = q^2_e/(4 \pi \epsilon_0) = \hbar c \alpha \approx \hbar c /137$.

Because of the confusion over the ‘dimension’ of \kappa, I will use:

Permittivity of free space $= \epsilon_0 = q^2_e/(4 \pi \alpha \hbar c) = 8.854 \times 10^{-12} coulomb^2 / joule.m$.

Gravitation (force) constant $= G = \hbar c / m^2_{Planck} = 6.673 \times 10^{-11} joule.m / kg^2$ and not set them to 1.

‘Electric force constant’ $= \frame{E} = 1/(4 \pi \epsilon_0) = \alpha \hbar c / q^2_e = 8.988 \times 10^9 joule.m / coulomb^2$.

So $\kappa^2 = G / \frame{E} = q^2_e / ( \alpha m^2_{Planck})$.(My original question - somewhat rephrased!)
Physics preamble:
1. GR incorporates EM by bolting in the EM energy tensor into the total energy tensor T_ij that feeds (& is fed by) the 4d curvature tensor G_ij via the equation of motion: $G_{\alpha \beta} = 8 \pi G T_{\alpha \beta}$.
2. The standard KK extension to GR incorporates A_j within a 5d metric to make the curvature tensor G_ab include EM energy (with opposite sign) and save bolting it into T_ab. An extension of this idea also saves bolting in mass as well, so T_ab = 0, so in theory, the universe could be described entirely by 5d curvature.
3. 4d Lagrangians describe simple EM-SR-CM & Klein-Gordon-QM. Applying principles of gauge symmetry bolts in EM vector potentials A_j to the Lagrangians.
4. The application of the same form of 5d metric saves having to bolt on A_j via gauge symmetry.

Maths preamble:
The 5d KK metric in 2 $= \tilde g_{ab} = g_{ab} + B_{ab}$ where:
$$g_{5 \beta} = 0 ; \ B_{ab} = B_a B_b; \ B_5 = 1; \ B_\beta = \kappa A_\beta; \ \kappa^2 = 4 \pi \epsilon_0 G = q^2_e / ( \alpha m^2_{Planck})$$

The 5d metric in 4 is the same apart from $B_\beta = \lambda A_\beta; \ \lambda = q_e / m_e$ when describing the mechanics of an electron (or = Q/M for a classical particle of mass M & charge Q).

$$\lambda_e = \sqrt{\alpha}(m_{Planck}/ m_e)\times \kappa = 2.042 \times 10^{21} \times \kappa$$
I am starting to use \lambda for the ‘larger’ brother of \kappa.


Questions:
1. Why are the 5d metrics so similar in form?
2. Why is \lambda so much larger than \kappa? (Very sketchy answer to question 2)
Rather than use the equation of motion & Einstein tensor G, I will play with the Lagrangian and Ricci curvature R (as it is quicker). I am being rather careless with \sqrt{-g}, part of me wants to keep it in as g is rather relevant, another part of me wants to drop it as I my original question is only concerned with simple Lagrangian where g reduces to \eta of Cartesian co-ordinates. I struggled myself to reduce $\tilde R \longrightarrow (R + \kappa^2 F_{\alpha \beta} F^{\alpha \beta})$ but only at expense of ignoring a factor of 4 (and terms which doubtfully just might cancel if I considered the integration limits carefully). My only excuse is that this is 21 orders of magnitude better than my previous attempt.

$$\tilde L = \int (\tilde R \sqrt{- \tilde g } dy) d^4x /(16 \pi G \int dy) = \int (R + \kappa^2 F_{\alpha \beta} F^{\alpha \beta})\sqrt{- g } d^4x/(16 \pi G) = L$$

But we know from EM that L contains $\int ( F_{\alpha \beta} F^{\alpha \beta} \epsilon_0/4) \sqrt{- g } d^4x$.

So $\kappa^2 /(16 \pi G) = \epsilon_0/4 \longrightarrow \kappa^2 = 4 \pi \epsilon_0 G = q^2_e / ( \alpha m^2_{Planck})$.

So why is this $\kappa^2$ so small?

Easy: In GR we are considering the effect of mass-energy (tensor) on curvature. So instead of thinking about the force or potential on a charge in an electric field, think of the electrostatic energy contained in that field & divide it by c^2 to get a ‘mass equivalent’. GR is interested in this ‘gravitational mass equivalent’ effect which is 42 orders of magnitude weaker than the electrostatic effect. (I hope Douglas Adams would have approved of this ‘explanation’.)

Note: I think of the very large m_Planck just as a device to make G and \kappa very small. But instead, I think it has come to symbolise the paradigm that unification of GR & QM demands m_Planck to be some sort of enormously unfriendly quantum of matter. I am out of my depth here.

Conversely: For the Simple (non GR) Lagrangians, if we want to use a metric to describe the potential on a charge in an electric field (instead of recourse to gauge symmetry) we are considering the strong electrostatic effect and not the weak ‘gravitational mass equivalent’ effect. Hence the need here for the more powerful $B_\beta = \lambda A_\beta; \ \lambda = q_e / m_e$.Now I am going to push the boat out in an attempt to answer question 1:

When we apply this \lambda metric to GR Lagrangian, the FF term explodes into $F_{\alpha \beta} F^{\alpha \beta} \lambda^2 / 4$, so we have to somehow eliminate or reduce it.
One way to do this is to consider a 6d \lambda metric:

$$\tilde g_{ab} = g_{ab} + B_{ab}; \ where \ \bar\alpha, \bar\beta$$ run from 5 to 6:

$$g_{\bar\alpha \beta} = g^{\bar\alpha \beta} = 0; \ B_{\alpha \beta} = B_{6 \beta} = B^{5 \beta} = B_{66} = 0^* ; \ B_{56} = B^{56} = 1^*$$

$$B_{5 \beta} = \lambda A_\beta; \ B^{6 \beta} = - \lambda A^\beta; \ B^{66} = \lambda^2 A_\gamma A^\gamma - B_{55}$$

I think if we set $\ B_{55} = 0$ then we eliminate the enormous FF term.
I think if we set $\ B_{55} = \kappa / \lambda \approx 5 \times 10^{-22}$ then we recover the FF term with correct magnitude.

* Another more symmetric 6d metric could involve $\sin \chi \ and \ \cos \chi \ where \ \chi = \pi / 4 \ but \ B_{\bar\alpha \bar\beta}$ terms would be more involved.

If this works, it could explain why the 5d metric is only an approximation if you don’t worry too much about the problem 21 orders of magnitude. But then the similar idea in 6d may resolve this problem.Appendix-2b - Fariday & wave-function.
While I am in fantasy-land, I may as well sketch in another idea. For now, forget GR & \kappa and go back to simple 5d Lagrangian with simple $B_{ab} = B_a B_b; \ B_{\bar\beta} = B_5 = 1; \ B_\beta = \lambda A_\beta$. But now add in the FF term, which I omitted before when I was only interested in the matter equation of motion (and not the EM field equation of motion).

Then:
$$4 \tilde L = \epsilon_0 ( \tilde F^*_{ab} \tilde F^{ab} + 2 \partial_a A^*_{\bar\beta} \tilde \partial^{\bar\beta} A^a ) = \epsilon_0 (F_{\alpha \beta} F^{\alpha \beta} + 2 \partial_a A^*_{\bar\beta} \tilde \partial^a A^{\bar\beta} ) = 4 L$$
$$= \epsilon_0 (F_{\alpha \beta} F^{\alpha \beta} + 2 D^*_\alpha A^*_{\bar\beta} D^\alpha A^{\bar\beta} + 2 m^2 A^*_{\bar\beta} A^{\bar\beta} )$$

where:
$$\tilde F_{\alpha \beta} = F_{\alpha \beta}; \tilde F^{ab} = \tilde \partial^b A^a - \tilde \partial^a A^b$$ etc. I could expand later.

The main point is that in 5d where you have to mess about with complex conjugate,
$$\epsilon_0^{-1/2}(A^*_5, A^5) = ( \psi^*, \psi)$$

And for 6d where $B_{\bar\beta} = 2^{-1/2} \ and \ \psi_5 + \imath \psi_6 = \psi$ of the 5d:
$$\epsilon_0^{-1/2}(A_5, A_6) = \epsilon_0^{-1/2}(A^5, A^6) = ( \psi_5, \psi_6)$$

If you are fussy and don’t like $B_{\bar\beta} = 2^{-1/2} \ (or \ 3^{-1/2}$ for 7d) merely because it makes the metric determinant zero (and the geometry meaningless), then use the 6d metrics mentioned earlier.

Hidden in all these sums is the notion that the wave-function could be considered as simply being proportional to the 5th component of the vector potential. And using 6d could split this into 2 real fields.

$$m = m = m = m = m = m = m = m = m = m = m = m = m = m = m = m = m$$

 

[MikeL#]

Update. R(5) to R(4) + EM help needed

1. On more carefully working through original Kaluza ideas, I am having a problem with the standard reduction of the 5d GR Lagrangian expression into 4d GR + EM.
I am using $B_\beta = \kappa A_\beta$ and for now putting aside the non-standard idea (of \lambda = q/m) as this particular forum concerns documented standard physics.
Currently I am getting familiar with the subset of LaTex used here, while writing up and re-checking this problem. All the many EM terms cancel, apart from the very one I want to keep which has the wrong sign!
Please help me to spot the silly mistake. I suppose an expert could understand abstract methods (pull-backs, fibre bundles etc.) to solve this in a few lines, but please bear with me and use the old fashioned maths (that I understood from my undergraduate physics days). Even if the new way is better, I want to know that the old way also works.

2. To help with this, I have tried to make nomenclature and calculations more clear as follows:
Implicitly set \phi =1 of post #6 (as did Kaluza) by not mentioning it.
Continue to use B (a number) instead of A (a quantity) for 4d.
Make up to 6d using C (a number) to highlight its dimensional nature and similarities between B & C.
Continue to use matrix representations for g & \Gamma to calculate ‘elements’.
These ‘elements’ are themselves matrices: 4x4, 4x2 & 2x2.
Calculate \Gamma
Calculate R ‘element’ by ‘element’ (rather than attempt 6x6 matrix multiplication) to avoid errors.
Explicitly reduce to 5d by setting all C to 1 at end to reproduce Kaluza’s picture.

3. Explain factor-of-4 errors in posts #10 #13.

$$m = m = m = m = m = m = m = m = m = m = m = m = m = m = m = m = m$$


Nomenclature - g, h; B, C; A, F; Gamma, R.

The \bar is used as follows:
$$\bar B^2 \equiv B_\epsilon B^\epsilon \ \ \& \ \ \bar C^2 \equiv C_\epsilon C^\epsilon$$

I use brackets around ( X_jk ) to imply subscripted flavour of matrix X. Whereas (X)_jk would imply element X_jk of matrix (X).
The \vec is used to emphasise > 4d (as it is more clearly readable here than \bar or \tilde).
When \vec is not used, then we are usually dealing with 4d. An exception to this rule is in the following split where the terms are obviously transient and leading to fuller terms.

$$\begin{quote} \def\v{\vec} \def\a{\alpha} \def\b{\beta} \def\c{\gamma} \newcommand{\Bra[1]}{\left(\begin{array}} \newcommand{\Ket}{\end{array}\right)} \def\nl{\newline} \mbox{Split (vec g) into a gravitational (g) part + electromagnetic (h) part:} \nl = 0+ (\v g_{ab}) \nl = (\v g_{ab}) = (g_{ab} + h_{ab}) = \Bra{cc} g_{\a\b} & 0 \\ 0 & g_{\v\a \v\b}\Ket + \Bra{cc} {\bar C^2 B_\a B_\b} & {B_\a C_{\v\b}} \\ {C_{\v\a} B_\b} & {0}\Ket \nl (\v g^{ab}) = (g^{ab} + h^{ab}) = \Bra{cc} g^{\a\b} & 0 } \\ 0 & g^{\v\a \v\b} \Ket + \Bra{cc} 0 & -B^\a C^\v\b \\ {-C^{\v\a} B^\b} & {\bar B^2 C^{\v\a} C^{\v\b}} \Ket \nl \mbox{These definitions were chosen so that:} \nl \v g^{ab}\v g_{bc} = \delta^a_{\; c} \longrightarrow B^\a = g^{\a\b} B_\b \ \ \& \ \ C^{\v\a} = g^{\v\a \v\b} C_{\v\b} \;\;\; \mbox{and their inverses.} \nl \mbox{Please note changes in nomenclature to previous posts:} \nl \mbox{the 4x2 matrices: } (g_{\a \v\b}) \; \& \; (g_{\v\a \b}) \;\; \mbox{are null as before} \nl \mbox{but the 2x2 matrix is not null: } (g_{\v\a \v\b}) \end{quote}$$


$$\begin{quote} \def\v{\vec} \def\a{\alpha} \def\b{\beta} \def\nl{\newline} h_{\a\b} \; \mbox{is used instead of } B_{\a\b} \mbox{ to save confusion with:} \nl 2B_{(\a,\b)} \equiv B_{\a,\b} + B_{\b,\a} \nl 2B_{[\a,\b]} \equiv B_{\a,\b} - B_{\b,\a} \nl B^{[\mu}_{\;,\b]} \equiv g^{\mu\a} B_{[\a,\b]} \nl \nl \mbox{Define Fariday in terms of vector potentials:} \nl F_{\a\b} = \partial_\a A_\b - \partial_\b A_\a = A_{\b,\a } - A_{\a,\b} = 2 A_{[\b,\a]} \nl \nl \mbox{So: } \kappa F_{\a\b} = 2 B_{[\b,\a]} \nl \underline {\kappa^2 F_{\a\b} F^{\a\b}/4 = B_{[\a,\b]}B^{[\a,\b]} = B_{\a,\b}B^{[\a,\b]} } \nl \nl \mbox{as } B_{[\b,\a]}B^{[\b,\a]} = B_{[\a,\b]}B^{[\a,\b]} \nl = (B_{\a,\b} - B_{\b,\a})(B^{\a,\b} - B^{\b,\a})/4 \nl = (B_{\a,\b} B^{\a,\b} - B_{\a,\b} B^{\b,\a})/2 = B_{\a,\b} B^{[\a,\b]} \end{quote}$$


Calculate Gamma (without u)
My previous approach using Classical Mechanics Lagrangian with proper velocity ‘u’ led to the 10**21 ‘error’. A better approach would be to develop the method of Markwh04 below.

But for now, this calculation will pull Gamma out of a hat, extend from 4d to 5d or 6d, and assume the ‘cylindrical condition’ that each element of $\vec g_{ab} \ \& \ B_\alpha$ (since B is included in \vec g) is independent of $x^{\vec\gamma}$.

$$\begin{quote} \def\G{\Gamma} \def\v{\vec} \def\a{\alpha} \def\b{\beta} \def\c{\gamma} \def\nl{\newline} \mbox{Define: } \G^\mu_{\:\b \c} \equiv g^{\mu \a} \G_{\a\b\c} \nl \mbox{and: }\G_{\a(\b\c)} \equiv g_{\a(\b,\c)} - \frac{1}{2} g_{(\b\c),\a} \nl \nl \mbox{This expresses a symmetry. Rewriting more simply:} \nl \G_{\a\b\c} = (g_{\a\b,\c} + g_{\a\c, \b} - g_{\b\c, \a})/2 \nl \nl \mbox{To extend into 6d, simply rewrite as:} \nl \v\G_{abc} = (\v g_{ab,c} + \v g_{ac,b} - \v g_{bc,a})/2 \nl = \G_{abc} + (h_{ab,c} + h_{ac,b} - h_{bc,a})/2 \nl \nl \mbox{The ‘cylindrical condition’ ensures: } \nl g_{\v\a \v\b, c} = 0 = h_{ab, \v\c} \nl \mbox{And for any ONE of (a,b,c)=} (\v\a,\v\b,\v\c), \; \; \G_{abc} = 0 \end{quote}$$


$$\begin{quote} \def\v{\vec} \def\G{\Gamma} \def\a{\alpha} \def\b{\beta} \def\c{\gamma} \def\nl{\newline} \mbox{So: the 6d version - 4d version of Gamma (first symbol breaks down)} \nl = 0+ \vec\Gamma_{\a\b\c} - \G_{\a\b\c} \nl = \v\Gamma_{\a\b\c} - \Gamma_{\a\b\c} = \bar C^2[ (B_{\a,\c} B_\b + B_\a B_{\b,\c}) + (B_{\a,\b} B_\c + B_\a B_{\c,\b }) - (B_{\b,\a} B_\c + B_\b B_{\c,\a }) ]/2 \nl = \bar C^2(B_{[\a,\c]} B_\b + B_\a B_{(\b,\c)} + B_{[\a,\b]} B_\c) \nl = \bar C^2(\Delta_{\a\b\c} + B_\a B_{(\b,\c)}) \nl \nl \mbox{where: } \Delta_{\a\b\c} \equiv B_\b B_{[\a,\c]} + B_{[\a,\b]} B_\c \nl \mbox{and: } \Delta^\a _{\:\b\c} \equiv B_\b B^{[\a}_{\;,\c]} + B^{[\a}_{\;,\b]} B_\c \nl \nl \v\G_{\a \v\b \c} = (h_{\a \v\b, \c}+ 0 - h_{\v\b \c, \a} )/2 = C_{\v\b} B_{[\a,\c]} \nl \v\G_{\a\b \v\c} = (0 + h_{\a \v\c,\b} - h_{\b \v\c, \a} )/2 = C_{\v\c} B_{[\a,\b]} \nl \v\G_{\v\a \b \c} = (h_{\v\a \b,\c}+h_{\v\a\c, \b} - 0)/2 = C_{\v\a} B_{(\b,\c)} \nl \v\G_{\v\a \v\b \v\c} = (0 + 0 - 0)/2 = 0 = \v\G_{\v\a \v\b \c} \ \ etc. \nl \mbox{so for any TWO of (a,b,c)=} (\v\a,\v\b,\v\c), \; \; \v\G_{abc} = 0 \end{quote}$$


Nomenclature to summarise Gamma with lowered indices

If we think of 2nd rank square matrices, they can be multiplied easily (row by column), they happily occupy 2 dimensions on paper and symmetric ones are pleasing to the eye. On the down side, the matrices may not commute, but if only a trace is needed (& elements commute) then the trace is the same either way.
This is a luxury as the Einstein summation rule can be applied by carefully keeping track of the indices without recourse to the easy multiplication rule.

But 3rd rank ‘cube’ matrices can be messy. The indices have to be carefully studied, and perhaps it is simpler to reduce the problem down to using a number of square matrices - or even forgetting about matrix manipulations. Never the less I will give it a cautions try with reservations.

$$\begin{quote} \def\G{\Gamma} \def\v{\vec} \def\a{\alpha} \def\b{\beta} \def\c{\gamma} \def\SE{\searrow} \newcommand{\Bra[1]}{\left(\begin{array}} \newcommand{\Ket}{\end{array}\right)} \def\nl{\newline} \mbox{Introducing intuitive nomenclature to multiply 3rd rank matrices: } \nl (( )) \ \& \ (( ) \SE ) \nl ((\vec\Gamma_{a b c})) = \Bra{c} {(\v\G_{\a b c}}) \\ {(\v\G_{\v\a b c})} \Ket = \Bra{c} \Bra{cc} {\v\G_{\a\b\c}} & {\v\G_{\a\b \vec\c}} \\ {\v\G_{\a \v\b \c}} & {\v\G_{\a \v\b \v\c}} \Ket \\ \Bra{cc} {\v\G_{\v\a\b\c}} & {\v\G_{\v\a \b \v\c}} \\ {\v\G_{\v\a \v\beta \c}} & {\v\G_{\v\a \v\b \vec\c}} \Ket \Ket \nl = \Bra{c} \Bra{cc} {\G_{\a\b\c} + \bar C^2(\Delta_{\a\b\c} + B_\a B_{(\b,\c)}) & {B_{[\a,\b]}C_{\v\c}} \\ {C_{\v\b}B_{[\a,\c]}} & {0} \Ket \\ C_{\vec\alpha} \Bra{cc} {B_{(\b,\c)}} & {0} \\ {0} & {0} \Ket \Ket \nl \nl \mbox{Converting ((across, down), down) to ((across, across), down) gives: } \nl ((\v\G_{a b c}) \SE) = \Bra{cc} {(\v\G_{\a bc}) & \SE} \\ {(\v\G_{\v\a b c})} & \SE \Ket =\Bra{ccccc} {\v\G_{\a\b\c}} & {\v\G_{\a\b \v\c}} & \SE & {\v\G_{\a \v\b \c}} & {\v\G_{\a \v\b \v\c}} \\ {\v\G_{\v\a \b\c}} & {\v\G_{\v\a \b \v\g}} & \SE & {\v\G_{\v\a \v\b \c}} & {\v\G_{\v\a \v\b \v\c}} \Ket \nl =\Bra{ccccc} {\G_{\a\b\c}} + {\bar C^2(\Delta_{\a\b\c} + B_\a B_{(\b,\c)})} & {B_{[\a,\b]}C_{\v\c}} & \SE & {C_{\v\b}B_{[\a,\c]}} & {0} \\ C_{\v\a}[{B_{(\b,\c)}} & {0} & \SE & {0} & {0}] \Ket \end{quote}$$

This is useful for some multiplication especially when showing how to raise the first index of Gamma, which is our next step.


Raise the first index of Gamma
To check that no silly errors are made, this could be calculated element by element in a similar way to the above and only then summarising into a more pleasing symmetric matrix at the end.
But in fact, this is just a matter of separately calculating each column of the ‘SE arrow’ matrix version of Gamma. This will be shown separately element by element for the first column.

$$\begin{quote} \def\G{\Gamma} \def\v{\vec} \def\a{\alpha} \def\b{\beta} \def\c{\gamma} \def\ul{\underline} \newcommand{\Bra[1]}{\left(\begin{array}} \newcommand{\Ket}{\end{array}\right)} \def\nl{\newline} \def\SE{\searrow} \mbox{Raising first index:} \nl ((\v\G^m_{\: bc}) \SE) \nl =((\v\G^m_{\: bc}) \SE) =\Bra{cc} {(\v\G^\mu_{\: bc})} & \SE \\ {(\v\G^{\v\mu}_{\: bc}) & \SE } \Ket =\Bra{cc} {g^{\mu \a} & -B^\mu C^{\v\a}} \\ {-C^{\v\mu} B^\a & g^{\v\mu \v\a} + \bar B^2 C^{\v\mu} C^{\v\alpha} } \Ket \times \nl \; \; \times \Bra{ccccc} {\G_{\a\b\c} + \bar C^2(\Delta_{\a\b\c} + B_\a B_{(\b,\c)}) & {B_{[\a,\b]}C_{\v\c}} & \SE & {C_{\v\b}B_{[\a,\c]}} & {0} \\ C_{\v\a} [{B_{(\b,\c)}} & {0} & \SE & {0} & {0}] \Ket \nl \nl \mbox{Calculating first column underlining cancelling terms:} \nl (\v\G^m_{\:\b\c}) = \Bra{c} {\v\G^\mu_{\:\b\c}} \\ {\v\G^{\v\mu}_{\:\b\c} } \Ket \\ =\Bra{c} {\G^\mu_{\:\b\c} + \bar C^2(\Delta^\mu_{\:\b\c} + \ul{B^\mu B_{(\b,\c)}}) - \ul{\bar C^2 B^\mu B_{(\b,\c)}} } \\ {- C^{\v\mu} B^\a({\G_{\a \b\c} + \bar C^2(\Delta_{\a\b\c} + \ul{B_\a B_{(\b,\c)}}) ) + C^{\v\mu} (1 + \ul{\bar C^2 \bar B^2}) B_{(\b,\c)}} \Ket \nl \nl \mbox{Continuing with 8 columns:} \nl \Bra{cc} {(\v\G^\mu_{\: bc})} & \SE \\ {(\v\G^{\v\mu}_{\: bc}) & \SE } \Ket =\Bra{ccccc} { \G^\mu_{\:\b\c} + \bar C^2\Delta^\mu_{\:\b\c} } & { B^{[\mu}_{\;,\b]} C_{\v\c} } & \searrow & { C_{\v\b}B^{[\mu}_{\;,\c]} } & 0 \\ C^{\v\mu} [ { -B_\a \v\G^\a_{\:\b\c} + B_{(\b,\c)} } & { -B^\a B_{[\a,\b]}C_{\v\c} } & \SE & { -C_{\v\b}B^\a B_{[\a,\c]} } & 0] \Ket \end{quote}$$

$$\begin{quote} \def\G{\Gamma} \def\v{\vec} \def\a{\alpha} \def\b{\beta} \def\c{\gamma} \newcommand{\Bra[1]}{\left(\begin{array}} \newcommand{\Ket}{\end{array}\right)} \def\nl{\newline} \mbox{Summarising \& rewriting in a more familiar form:} \nl (\vec g_{ab}) \nl = (\v g_{ab}) = (g_{ab} + h_{ab}) = \Bra{cc} g_{\a\b} & 0 \\ 0 & g_{\v\a \v\b}\Ket + \Bra{cc} {\bar C^2 B_\a B_\b} & {B_\a C_{\v\b}} \\ {C_{\v\a} B_\b} & {0}\Ket \nl \nl (\vec g^{bc}) = (g^{bc} + h^{bc}) = \Bra{cc} g^{\b\c} & 0 } \\ 0 & g^{\v\b \v\c} \Ket + \Bra{cc} 0 & -B^\b C^\v\c \\ {-C^{\v\b} B^\c} & {\bar B^2 C^{\v\b} C^{\v\c}} \Ket \nl \nl \v g_{ab}\v g^{bc} = \delta_a^{\: c} \longrightarrow B^\a = g^{\a\b} B_\b \ \ \& \ \ C^{\v\a} = g^{\v\a \v\b} C_{\v\b} \;\;\; \mbox{and inverses.} \nl \Delta^\mu_{\:\b\c} = B_\b B^{[\mu}_{\;,\c]} + B^{[\mu}_{\;,\b]} B_\c \nl \Bra{c} {(\v\G^\mu_{\: bc})} \\ {(\v\G^{\v\mu}_{\: bc}) } \Ket =\Bra{c}\Bra{cc} { \G^\mu_{\:\b\c} + \bar C^2\Delta^\mu_{\:\b\c} } & { B^{[\mu}_{\;,\b]} C_{\v\c} } & { C_{\v\b}B^{[\mu}_{\;,\c]} } & 0 \Ket \\ C^{\v\mu} \Bra{cc} [ { -B_\a \v\G^\a_{\:\b\c} + B_{(\b,\c)} } & { -B^\a B_{[\a,\b]}C_{\v\c} } & { -C_{\v\b}B^\a B_{[\a,\c]} } & 0] \Ket \Ket \end{quote}$$



Appendix-0 Calculate R
The longest calculation, yet the basis for the whole topic.
In progress …

$$\begin{quote} \def\G{\Gamma} \def\v{\vec} \def\a{\alpha} \def\b{\beta} \def\c{\gamma} \def\nl{\newline} \def\l{\lambda} \def\m{\mu} \def\n{\nu} \def\s{\sigma} \def\d{\partial} \mbox{Take these definitions from MTW (p 419) or Lawrie (p 502):} \nl R = g^{\a\b} R_{\a\b} \ \ & \ \ R_{\a\b} = R^\l_{\:\a\l\b} \\ R^\l_{\:\a\m\b} = ( \d_\m\G^\l_{\:\a\b} - \d_\b\G^\l_{\:\a\m}) + ( \G^\l_{\:\s\m} \G^\s_{\:\a\b} - \G^\l_{\:\s\b} \G^\s_{\:\a\m} ) \nl \nl \mbox{Contract \& change (a,b,l,m,s) to (b,c,m,m,n):} \nl R^\m_{\:\b\m\c} = ( \d_\m\G^\m_{\:\b\c} - \d_\c\G^\m_{\:\b\m}) + (\G^\m_{\:\n\m} \G^\n_{\:\b\c} - \G^\m_{\:\n\c}\G^\n_{\:\b\m} ) \nl \nl \mbox{Split \& increase dimensions:} \nl \v R = \v g^{bc} \v R_{bc} = \v R1 + \v R2 + \v R3 \nl \v R1 = \v g^{bc} ( \d_m\v\G^m_{\:bc} - \d_c\v\G^m_{\:bm}) \nl \v R2 = \v g^{bc} (\v\G^m_{\:nm} \v\G^n_{\:bc})= \v\G^m_{\:nm}(\v g^{bc} \v\G^n_{\:bc}) \nl \v R3 = -\v g^{bc}(\G^m_{\:nc}\G^n_{\:bm}) \nl \nl \mbox{Expanding the first one:} \nl \v R1 = \v g^{bc} \d_\m\v\G^\m_{\:\b\c} - \v g^{b \c} \d_\c( \v\G^\m_{\:b\m} + \v\G^{\v\m}_{\:b\v\m}) \end{quote}$$

to be continued



Factor-of-four and other errors

I made factor-of-4 errors in posts #10 #13 by misinterpreting implied conventions.
$$\kappa^2 = 16 \pi \epsilon_0 G = 4 q^2_e / ( \alpha m^2_{Planck} )$$

I neither like making \kappa = 2 using implied convention C2 nor \kappa = 1 using C1, but there it is.

C2: $\kappa^2 = 4(4 \pi \epsilon_0)(G) = 4 \ if \ 4 \pi \epsilon_0 = 1 = G$

C1: $\kappa^2 = (16 \pi G) ( \epsilon_0) = 1 \ if \ 16 \pi \ G = 1 = \epsilon_0$

Markwh04
C1 is needed to make sense of PF examples in post #10 (where B_j=A_j, B_5=1, (i,j)=(\alpha, \beta), (g,h) as defined) and I have subsequently realized that conceptually K=\phi of post #6, rather than \kappa :
e.g. Markwh04 https://www.physicsforums.com/showthread.php?t=136063
h_{ij} = g_{ij} + K A_i A_j
h_{i5} = h_{5i} = K A_i
h_{55} = K

Postscript
7f. Another mystery, sometimes the first symbol does not get expanded properly. When it happens, I try to prefix it with 0+, repeat it, start on newline and hope for the best.