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
