Discussion:
Kaluza-Klein help needed
(too old to reply)
John Baez
2005-04-30 08:30:31 UTC
Permalink
For a course on a classical mechanics I decided to have my
students work out the geodesics on a 5-dimensional manifold
M x U(1) with the metric h given by

h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1

where i,j = 1,2,3,4, g is a metric on M and A is a 1-form
on M describing the electromagnetic vector potential.

I was hoping to get the equation for the motion of
a charged particle in a electromagnetic field, namely

m (D^2q/Dt^2)^i = e F^i_j (dq/dt)^j

where:

dq/dt is the derivative of the path q(t),

D^2q/Dt^2 is its covariant 2nd derivative,

F_{ij} = d_i A_j - d_j A_i is the electromagnetic field,

m is the particle's mass, and

e is its charge.

And indeed, I get this assuming that the particle moves
around the circle U(1) at velocity e/m:

(dq/dt)^5 = e/m

HOWEVER, I don't find that the velocity of the particle around
the circle is constant!

(dq/dt)^5 seems to have nonzero derivative, which is annoying:
the particle's effective charge-to-mass ratio changes with time!

Am I making a mistake or what?

It's all the more annoying because the spacetime M x U(1) has
rotational symmetry in the U(1) coordinate, so by Noether's theorem,
the momentum in the 5 direction is conserved.

However, the velocity in the 5 direction appears not to be
conserved, basically because we obtain the velocity from the
momentum by raising an index:

(dq/dt)^5 = h^{5a} (dq/dt)_5

and h^{5a} is time-dependent.

I wish I were making some mistake here - am I?
René Meyer
2005-05-01 08:15:15 UTC
Permalink
For the time-dependent e/m:

You started with taking e/m constant, right? So taking the time
derivative of (dq/dt)^5=e/m you get d/dt (dq/dt)^5 = 5 (dq/dt)^4
d^2q/dt^2 = 0.

René.
Lubos Motl
2005-05-01 17:14:28 UTC
Permalink
Dear John,

let me first say how the actual electromagnetism arises from the
five-dimensional KK Universe.

The charge of the particle "Q" is a quantized entity that is conserved,
and therefore it is identified with the total momentum along the extra
circle, not the total velocity. The momentum is conserved; the velocity is
not.

Second, the four-dimensional mass satisfies

m4^2 = m5^2 + p5^2

and it has an extra contribution from the momentum along the fifth
direction, i.e. from the charge. Note that for a fixed circumference of
x^5, "p5" is conserved, and therefore a constant value of m5 is the same
thing as the constant value of m4.

Third, your chosen metric
Post by John Baez
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
is morally OK, but it's not the most natural "right" choice at the
non-linear level. Even if you decide to stabilize (and omit) the
KK-dilaton (g_{55}, which would appear as a pre-factor of the second
term in the equation below), it is better to consider the metric of the
form

ds^2 = dx_i^2 + (dx^5 + A_i.dx^i)^2 (**)

Note that *this* has the right gauge invariance inherited from the
diffeomorphisms. You can consider the periodicity of x^5 to be constant,
say 2.pi, and the gauge transformation is

x^5 -> x^5 + lambda (x^i) (##)

Note that if you transform your coordinates as in (##) and you
simultaneously transform

A_i -> A_i - d_i lambda (x^j),

then the metric (**) is gonna be invariant. Your metric does not have this
property if you're accurate: the x^5-rotating diffeomorphism does not act
as an electromagnetic U(1) transformation because it influences not only
h_{i5} but also your h_{ij}.

Note that "my" more accurate metric (**) diffes from yours - for example,
h_{ij} is no longer independent of A_{i} because it contains an extra
correction of the form A_{i}.A_{j}.

I am confident that if you redo the calculation with the "improved" metric
and with the assumption that the momentum_5 is the (conserved) charge, you
will obtain exactly the correct electromagnetic force.

All the best
Lubos
______________________________________________________________________________
E-mail: ***@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
eFax: +1-801/454-1858 work: +1-617/384-9488 home: +1-617/868-4487 (call)
Webs: http://schwinger.harvard.edu/~motl/ http://motls.blogspot.com/
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Lubos Motl
2005-05-01 17:34:49 UTC
Permalink
I forgot to mention a trivial comment. The momentum "p" - in
electromagnetism - is what corresponds to "-i.hbar.derivative" with
respect to "x". But its action on the wavefunction is not gauge-invariant.
Of course, the U(1) gauge-invariant quantity is (p-eA), if you allow me to
use the non-relativistic formula, and this is the quantity that is equal,
in the nonrelativistic theory, to "mv" where "v" is velocity. In a
(special) relativistic theory, (p-eA) is not exactly velocity, because of
the gamma factors.

Once again about the identification of the charge and the 5-momentum. In
quantum theory, it is easy - the wavefunction depends on x^5 as
exp(i.Q.x^5) if the circumference of x^5 is taken to be 2.pi. Also, "Q"
defined in this way is conserved in the quantum theory. It's clear that in
the classical theory, this will correspond to a conserved feature of the
classical trajectory. The classical trajectory is a geodesic in 5
dimensions, obtained by a parallel transport of a small 5-vector p^M whose
(constant) invariant "p^M.p^N.g_{MN}" equals "m_5^2". And the conserved
momentum will come from the second term in the metric in (**) in the
previous message - this is the U(1) gauge-invariant quantity. But of
course dx^5/dx^0 has no reason to be conserved.
______________________________________________________________________________
E-mail: ***@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
eFax: +1-801/454-1858 work: +1-617/384-9488 home: +1-617/868-4487 (call)
Webs: http://schwinger.harvard.edu/~motl/ http://motls.blogspot.com/
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Arkadiusz Jadczyk
2005-05-01 21:47:32 UTC
Permalink
On Sat, 30 Apr 2005 08:30:31 +0000 (UTC),
Post by John Baez
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
You should take:

h_{ij} = g_{ij}+A_iA_i
h_{i5} = A_i
h_{55} = 1

because with your

h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1

h is not necessarily invertible and even when it is, d/dx^5 is not a
Killing vector.

ark

--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--
Arkadiusz Jadczyk
2005-05-01 21:48:13 UTC
Permalink
On Sat, 30 Apr 2005 08:30:31 +0000 (UTC),
Post by John Baez
For a course on a classical mechanics I decided to have my
students work out the geodesics on a 5-dimensional manifold
M x U(1) with the metric h given by
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
where i,j = 1,2,3,4, g is a metric on M and A is a 1-form
on M describing the electromagnetic vector potential.
As I wrote in my previous post, it should be

h_{ij} = g_{ij}+A_iA_j
Post by John Baez
I was hoping to get the equation for the motion of
a charged particle in a electromagnetic field, namely
m (D^2q/Dt^2)^i = e F^i_j (dq/dt)^j
dq/dt is the derivative of the path q(t),
D^2q/Dt^2 is its covariant 2nd derivative,
F_{ij} = d_i A_j - d_j A_i is the electromagnetic field,
Here one must be careful. We want to write geodesic equation in 5D.
It reads

D^2q/ds^2 = 0

PROVIDED "s" is the length parameter in 5D (or an affine variation of
it).

But "length" in 5D is not the same as "length in 4D - where we want to
have just standard Lorentz force and nothing more using 4D length.

The simplest way to get the desired result quickly is by noticing that the
problem is "gauge invariant" and that we can always choose the gauge (that
is x^5 coordinate) in such a way that the A_i vanish identically on the
trajectory (there are no obstruction for making A_i to vanish along
1-dimensional curve).

ark





--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--
g***@lisi.com
2005-05-02 05:24:07 UTC
Permalink
Post by John Baez
For a course on a classical mechanics I decided to have my
students work out the geodesics on a 5-dimensional manifold
M x U(1) with the metric h given by
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
where i,j = 1,2,3,4, g is a metric on M and A is a 1-form
on M describing the electromagnetic vector potential.
Hey, nice assignment. One doesn't usually encounter KK theory so early
on...

But, umm, there's your mistake right there on the first line... it
needs to be:

h_{ij} = g_{ij} + A_i A_j
h_{i5} = - A_i
h_{55} = 1

I don't know why people do KK theory with a metric. It's MUCH nicer to
do with a vielbein. The first four 1-form vielbein vectors are then
just the four for spacetime,
e^a = dx^i e_i^a
and the fifth one is
e^5 = -A + ek^5
= -dx^i A_i + dx^5

This vielbein gives the correct metric above. And for bigger KK spaces
than U(1) it's
e^p = -A^B xi_B^p + ek^p
= -dx^i A_i^B xi_B^p + dx^m ek_m^p
in which xi_B is a 1-form corresponding to a Killing vector of the KK
space. I wrote up some notes on this stuff a few years ago:
http://sifter.org/~aglisi/Physics/KKUrs.pdf

To get the equation of motion, as a geodesic, it's easiest to extremize
the path length (or actually the square of the path length) -- a good
mechanics problem. Things should work out right and relieve your
befuddlement when you put in that correct metric. :)

-Garrett
Arkadiusz Jadczyk
2005-05-02 05:27:27 UTC
Permalink
On Sat, 30 Apr 2005 08:30:31 +0000 (UTC),
Post by John Baez
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
You should take:

h_{ij} = g_{ij}+A_iA_i
h_{i5} = A_i
h_{55} = 1


becuase with your

h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1

h is not necessarily invertible and even when it is, d/dx^5 is not a
Killing vector.

ark



--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--
Arkadiusz Jadczyk
2005-05-02 13:16:07 UTC
Permalink
I wrote nonsense about the Killing property. Sorry.

d/dx^5 is a Killing vector anyway. That is not the problem. The problem
is that we want to have g_{ij} to be the metric induced by projecting
vectors orthogonal to d/dx^5.

Anyway the trick with taking A_i=0 on the trajectory should then work.

ark


On Mon, 2 May 2005 05:27:27 +0000 (UTC), Arkadiusz Jadczyk
Post by Arkadiusz Jadczyk
Post by John Baez
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
h_{ij} = g_{ij}+A_iA_i
h_{i5} = A_i
h_{55} = 1
becuase with your
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
h is not necessarily invertible and even when it is, d/dx^5 is not a
Killing vector.
--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--
Arkadiusz Jadczyk
2005-05-04 06:57:28 UTC
Permalink
I wrote nonsense about the Killing property. Sorry.

d/dx^5 is a Killing vector anyway. That is not the problem. The problem
is that we want to have g_{ij} to be the metric induced by projecting
vectors orthogonal to d/dx^5.

Anyway the trick with taking A_i=0 on the trajectory should then work.

ark


On Mon, 2 May 2005 05:27:27 +0000 (UTC), Arkadiusz Jadczyk
Post by Arkadiusz Jadczyk
Post by John Baez
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
h_{ij} = g_{ij}+A_iA_i
h_{i5} = A_i
h_{55} = 1
becuase with your
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
h is not necessarily invertible and even when it is, d/dx^5 is not a
Killing vector.
--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--
Ken S. Tucker
2005-05-04 06:55:23 UTC
Permalink
To John et al...
I agree with everyone else who posted.
For a few comments, I'll ref to P.G. Bergmann's
"Intro to...Relativity", he details KK's 5D stuff.
Post by John Baez
For a course on a classical mechanics I decided to have my
students work out the geodesics on a 5-dimensional manifold
M x U(1) with the metric h given by
:-)
Post by John Baez
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
See ref, Eq.(17.61). John's statement looks accurate summed
over 4D.
Post by John Baez
where i,j = 1,2,3,4, g is a metric on M and A is a 1-form
on M describing the electromagnetic vector potential.
I was hoping to get the equation for the motion of
a charged particle in a electromagnetic field, namely
m (D^2q/Dt^2)^i = e F^i_j (dq/dt)^j
Yes, see Eq.(12.57).
The last term on the LHS, can also be zero and
yield information, that corresponds to John's
RHS above. If john's LHS above vanishes then it
is a constraint on motion.
Post by John Baez
dq/dt is the derivative of the path q(t),
D^2q/Dt^2 is its covariant 2nd derivative,
F_{ij} = d_i A_j - d_j A_i is the electromagnetic field,
m is the particle's mass
Stop, same ref, see (6.14..19). A particles mass has
3 possible components, covariant energy-momentum "p_0",
invariant "p", and contravariant "p^0".
(Mixing g-fields with EM-fields, IMO does requires
care to define the meaning of mass).
Presuming John mean's a conserved "p" as defined
by (6.19), let's good ahead.
Post by John Baez
e is its charge.
And indeed, I get this assuming that the particle moves
(dq/dt)^5 = e/m
Above John pointed out h_{55}=1, that's a constant!
How can variables exist in the direction of the 5th
dimension if it's metric is constant?
There cannot be a basis for acceleration in the
direction of 5th dimension, i.e.

(dq/dt)^5 = 0 , (Tucker suggests)

IMO, that's a constraint in 5D. Look at it physically,
if you have charges going off into the 5th dimension
won't that rather make the other 4 lonely?
Post by John Baez
HOWEVER, I don't find that the velocity of the particle around
the circle is constant!
the particle's effective charge-to-mass ratio changes with time!
I think if John employs the quantum theory, (meaning energy changes
discontinuously, applied to the "m", then the classical
type predictions, made on the basis of Maxwell, of particles,
like electrons spiralling into the nucleus will not happen.

It looks like dm =0 but is not a constant.
Post by John Baez
Am I making a mistake or what?
Well, by experiment we have found, "power" is
not continuously radiated, hence a spirally orbiting
electron descending into a nucleus obeys the vector
dot product,

q*E.V = 0

(q=charge, E=E-field, V-Velocity),

IOW's it can't spiral. Otherwise a spiral type orbit
would have

q*E.V =/=0,

as the charge sinks into the nucleus, continuously
radianting power, that is classically predicted but
physically not apparent and gave birth to Quantum T.

In the case of the Lorentz force John introduce above,
the component,

f_0 = q*F_i0 U^i == q*E.V =0.

I think that means there is no force in the direction of
time.
Post by John Baez
It's all the more annoying because the spacetime M x U(1) has
rotational symmetry in the U(1) coordinate, so by Noether's theorem,
the momentum in the 5 direction is conserved.
However, the velocity in the 5 direction appears not to be
conserved, basically because we obtain the velocity from the
(dq/dt)^5 = h^{5a} (dq/dt)_5
and h^{5a} is time-dependent.
I wish I were making some mistake here - am I?
Don't know, but IMHO, one should give due consideration
to the quantization of energy exchange. AE's GR gives
a good foundation for that following his GR1916 Eq.(65a),
where he suggests the Lorentz force vanishes. I provided
one sample "f_0=0" above, but many disagree.
If you ref to Dover's PoR GR1916, pg 156, and read past
Eq.(66) you'll see, "if kappa_sigma vanishes", (that's
Lorentz force in Eq(65)), and find that's how EM-forces
are included into the T_uv side of G_uv=T_uv, and I think
ok with quantum theory.

Regards
Ken S. Tucker
m***@yahoo.com
2005-05-20 08:54:38 UTC
Permalink
Post by John Baez
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
HOWEVER, I don't find that the velocity of the particle around
the circle is constant!
You got stumped by this problem?

That's easy. You've got the wrong metric. It's

h_{ij} = g_{ij} + K A_i A_j
h_{i5} = h_{5i} = K A_i
h_{55} = K

The Lagrangian may be written (using the summation convention) as:
L = 1/2 (g_{ij} u^i u^j) + K/2 (A_i u^i + u^5)^2
The conjugate momenta are:
p_i = @L/@u^i = g_{ij} u^j + K A_i (A_j u^j + u^5)
p_5 = @L/@z = A_i u^i + u^5 = e,
thus
p_i = g_{ij} u^j + e A_i
as expected.

The canonical forces are:
F_i = @L/@x^i
= 1/2 g_{jk,i} u^j u^k + e A_{j,i} u^j + K,i e^2/2K
F_5 = @L/@u^5 = 0.
The latter assumes the metric is gauge invariant, so that @g/@x^5 = 0
and @A/@x^5 = 0 and @K/@x^5. Note that in the most general case, on a
principal bundle a gauge-invariant metric will still allow for gauge
metric K to be dependent on the base space. This provides a kind of
geometric running of the coupling constant.

It would be interesting to work out the 5-vacuum solution with
spherical symmetry, stationarity and gauge-symmetry, just to see what
goes on with K. In fact, it will NOT be constant in general.
Reissner-Nordestroem will NOT be recovered without removing one of the
3 conditions above. The point source has a "running U(1) coupling" K,
built in, of an entirely geometric nature.

Anyway...

Thus the equations of motion are
dp_i/ds = F_i
dp_5/ds = F_5 = 0
which yields
de/ds = 0 --> e preserved.
and
g_{ij,k} u^j u^k + g_{ij} d(u^j)/ds + A_{i,j} u^j e
= 1/2 g_{jk,i} u^j u^k + e A_{j,i} u^j + K,i e^2/2K.
Using the expressions for the Christoffel symbols
{ij,k} = 1/2 (g_{ik,j} + g_{jk,i} - g_{ij,k})
and the field strength
F_{ij} = A_{j,i} - A_{i,j}
this results in:
g_{ij} d(u^j)/ds + {i,jk} u^j u^k = e F_{ij} u^j + K,i e^2/2K.
The quantity g_{ij} u^i u^j isn't conserved if K varies, but rather
there is an exchange between the charge Hamiltonian
H_e = e^2/2K
and the mass hamiltonian
H_m = 1/2 g_{ij} u^i u^j
a kind of mass-charge conversion.

If you restrict K to be constant (which Reissner-Nordestroem's 5-D
extension has) then you get the usual Lorentz force law.

As an exercise (both you and your students) generalize this to
Yang-Mills fields A_i -> A^a_i with a Lie index, allowing the gauge
part of the metric k_{ab} to remain variable in the base space
coordinates. Compare the two sets of solutions:
(A) Kaluza-Klein:
G_{ij} = 0 for i,j = base space indices
+ k_{ab} = constant
vs.
(B) K-K with running coupling:
k is adjoint-invariant (semi-simple groups: this means
is reduces to a coupling constant for each non-Abelian
simple factor + quadratic form with the Abelian factors)
+ G_{ij} = 0 for ALL indices i,j

Trying to develop the general case for spherical symmetry (stationarity
and gauge invariance).

What does the geodesic look like for the U(1) type-(B) solution where K
may "run"? The more general case for non-Abelian groups?

Another exercise: compare the extra "dark energy" terms in the U(1)
type (B) vacuum solutions to those which are present in an actual dark
energy model, e.g. Parker's VCDM scalar field model. See any
resemblances? (Hint, hint!)
m***@yahoo.com
2005-06-01 16:50:17 UTC
Permalink
Post by m***@yahoo.com
Post by John Baez
h_{ij} = g_{ij}
h_{i5} = A_i
h_{55} = 1
HOWEVER, I don't find that the velocity of the particle around
the circle is constant!
You got stumped by this problem?
That's easy. You've got the wrong metric. It's
h_{ij} = g_{ij} + K A_i A_j
h_{i5} = h_{5i} = K A_i
h_{55} = K
L = 1/2 (g_{ij} u^i u^j) + K/2 (A_i u^i + u^5)^2
[etc]

This can be worked out in a transparent fashion (at least when not
reduced to ASCII form!) in the more general case, within the principal
bundle formalism, using the *left-quotient* operator. The derivative of
the quotient completely captures in a much more general and clearer
context the notion of a connection, making everything easier to work
with.

So, some background and notation first. Let P be acted on by a group
G. This gives you a product operation with signature
P.G -> P
and properties
pe = p (e = group identity);
p(gh) = (pg)h,
with cosets
pG = { pg: g in G }

One can then define the set
p\q = { g in G: pg = q }.
The stablizer of point p is just p\p. The group acts transitively on P
if all the p\q are non-empty. It acts freely if p\q has at most one
member -- in which case one may consider the quotient as a partial
operation
p\q: defined for all q in pG
with the properties:
p p\q = q
p\p = e
(p\q)^{-1} = q\p
p\q q\r = p\q
(pg)\(qh) = g^{-1} p\q h.

When these spaces are also manifolds, one can consider the actions of
these operators under derivatives. Recalling that the key property of
tangent vectors on a manifold M is that for a curve r(t)
r'(t) is in T_{r(t)}(M),
we'll use the notation denoting by m' (with subscripts) a general
element of T_m(M).

With the coset notation, a local SECTION on the space P is defined as a
map
s: P/G = M -> P
such that
s(m)G = m.
A principal bundle is thus a manifold P acted on by a symmetry group G
in such a way that its cosets M = P/G can be arranged into a manifold,
which can be coordinatized by differentiable local sections whose
domains cover M.

In the following, for a function f: M -> N, I'll use the notation Df:
TM -> TN to denote its deriative. So, for f: R^m -> R^n, Df is just
the mxn Jacobian matrix of f. Its product with a vector v in TM is
denoted Df[v] in TN.

The product P.G -> P, naturally induces one with signatures:
p.T_g(G), T_p(P).g -> T_{pq}(G)
with the property
(pg)' = p'g + pg'.

Differentiating the relation,
s(m)gG = s(m)G = m
one finds:
ds_m[m']G = ds_m[m']gG + s(m)g'G = ds_m[m']G = m'.
Thus
ds_m[m']G = m'; s(m)g' = 0.

The corresponding action for quotients
p\T_q(P), T_p(P)\q -> T_{p\q}(G)
is not given at the outset, but must be postulated; and is assumed to
have the property
(p\q)' = p'\q + p\q'.
The connection one-form is just
omega_p[p'] = p\p'.
The horizontal lift of a vector m', m = pG, to a point p is
L_p[m'] = s(m) ds_m[m']\p + ds_m[m'] s(m)\p
or more briefly
L = s ds\p + ds s\p
The section-relative connection is
B = s\ds
or more precisely:
B_m[m'] = s(m)\ds_m[m'].

Its action under a global transformation s -> sg is
B -> (sg)\d(sg) = g^{-1} s\ds g = g^{-1} B g
and its action under a local transformation
s(m) -> s(m) x(m)
is
B -> (sx)\d(sx) = x^{-1} s\(ds x + s x')
= x^{-1} s\ds x + x^{-1} x'
= x^{-1} B x + x^{-1} x'.

In a similar way, one may define natural actions of the group on the
cotangent space through the properties:
g.T*_p(P).h -> T*_{gph}(P)
(gwh)[p'] = w[g^{-1} p' h^{-1}]
as well as a quotient
p\T*_q(P) -> T*_{p\q}(P)
p\w[q'] = w[p q'].
This time, it's the product that requires the additional structure of
the connection, with
p.T*_g(G) -> T*_{pg}(P)
p.a[q'] = a[p\q'].

(One can also define operations for the tangent and cotangent spaces:
T_p(P)\T*_q(P) -> R
p'\w = w[p']
).

So with the background out of the way...

Given a metric k on G that is invariant under left and right products
k_{hgf}(h g'_1 f, h g'_2 f) = k_g(g'_1, g'_2)
and a metric g on the base space M, define the bundle metric on P by
h = g + k(omega, omega)
or more precisely:
h_p(p'_1, p'_2) = g_{pG}(p'_1G, p'_2G) + k_{pG}(p\p'_1, p\p'_2).
In the general case, the scaling of k may be positionally dependent on
the points in the base space M.

Given a section s, one may decompose P and TP as follows:
p = s(m) g; where m = pG; g = s(pG)\p
with
p' = ds[m'] g + s g'.
One finds that
p'G = ds[m']gG + s g'G = m' + 0 = m',
and
p\p' = g^{-1} s\(ds g + s g') = g^{-1} B_m[m'] g + g^{-1} g'
or
p\p' = g^{-1} u g
introducing the horizontal velocity
u = B_m[m'] + v
and gauge velocity
v = g' g^{-1}.

In terms of the decomposition, the metric can be written as:
h_p(p'_1, p'_2) = g_m(m'_1, m'_2) + k_m(u'_1, u'_2).

In component form, this becomes:
h_{AB}p'_1^A p'_2^B = g_{mn} m'_1^m m'_2^n + k_{ab} u'_1^a u'_2^b.
using the summation convention on the indices.

The components of the total metric are thus:
h_{mn} = g_{mn} + k_{ab} B^a_m B^b_n
h_{mb} = k_{ab} B^a_m
h_{an} = k_{ab} B^b_n
h_{ab} = k_{ab}.

So, from this you can readily find the geodesic equations of motion.
These arise from the Lagrangian:
L(p,p') = 1/2 h_p(p',p').

Variation of L gives you the desired results.

Working within the decomposition, p = s(m) g as above, define the
variation of g by
Delta(g) = D g.
then
Delta(g^{-1}) = -g^{-1} D.
and the variation of v will be:
Delta(v) = Delta(g' g^{-1})
= (Delta g)' g^{-1} - g' g^{-1} D
= (Dg)' g^{-1} - v D
= D' g g^{-1} + D v - v D
= D' + [D,v].

The variations with respect to D give you the charge and precession
equation (Wong's equation) for the charge. The variation with respect
to m' gives you the ordinary momentum and the Lorentz force law coupled
with the charge.

First, employing the derivative notation alluded to before:
Delta(u) = Delta(B_m[m'] + v)
= DB_m[Delta(m)][m'] + B_m[Delta(m')] + D' + [D,v].
Thus
Delta(L) = Delta(1/2 g_m(m',m') + 1/2 k_m(u,u))
= 1/2 Dg_m[Delta(m)](m',m') + 1/2 Dk_m[Delta(m)](u,u)
+ g_m(m',Delta(m')) + k_m(u,Delta(u)).

Picking out the variation with respect to D', one finds the
corresponding momentum
p_D = k_m(u, ())
which is just u with its (Lie) index lowered.

Picking out the variation with respect to D, one finds the force term
F_D:
F_D = k_m(u, [(),v]) = p_D[[(),v]].

Since p_D is a cotangent vector in T*G, it's natural to extend the Lie
bracket to the cotangent space. So, let's consider first how the Lie
bracket is defined. For a Lie vector e' in the tangent space T_e(G) =
L, one has
g e' g^{-1} in T_{g e g^{-1})(G) in L.
Therefore, L is closed under the adjoint operator
ad_g(e') = g e' g^{-1},
and derivatives of this operator make sense. The Lie bracket is then
just
[g',e'] = (g e' g^{-1})'.
Explicitly:
[g'(s), e'(t)] = d^2(g(s) e(t) g(s)^{-1})/ds dt.

So, over the cotangent space T_e*(G) = L*, one has the coadjoint action
defined in the same way
co_g(w) = g w g^{-1}
and Lie bracket
[g', w] = (g w g^{-1})'.
Applying this to a Lie vector e', one finds
[e_1',w][e_2'] = (e_1 w e_1^{-1})'[e_2']
= (e_1 w e_1^{-1}[e_2'])'
= (w[e_1^{-1} e_2' e_1])'
= w[(e_1^{-1} e_2' e_1)'].
Since
e_1^{-1}'(t) = -e_1^{-1}(t) e_1'(t) e_1^{-1}(t)
which evaluated at the given t where e_1(t) = e is just
e_1^{-1}'(t) = -e_1'(t),
then the Lie bracket for e_1^{-1} is the negative of that for e_1.
Thus
[e_1',w][e_2'] = -w[[e_1',e_2']]
= w[[e_2',e_1']].
Or, in more prosaic notation:
[a,w][b] = w[[b,a]], for lie vectors a, b in L.


Since the metric k is assumed to be invariant under action of G to the
left and right, one has:
d/ds (k_{m}(g(s) h_1' g(s)^{-1}, g(s) h_2' g(s)^{-1}))
= d/ds (k_h(h_1', h_2') = 0.
Applying this to the individual vectors using the Leibnitz property,
one gets
d/ds (k(g h_1' g^{-1}, g h_2' g^{-1}))
= k(d(g h_1' g^{-1})/ds, g h_2' g^{-1})
+ k(g h_1' g^{-1}, d(g h_2' g^{-1})/ds)
= k([g', h_1'], h_2') + k(h_1', [g', h_2']).
Thus for Lie vectors, a, b, c in L:
k([a,b],c) + k(b,[a,c]) = 0.
In particular, since the metric is symmetric, then for b = c,
k([a,b],b) = 0.
Therefore, the force can be rewritten as:
F_D = k_m(u, [(),v])
= k_m(u, [(),u - B])
= k_m(u, [(),u]) - k_m(u, [(),B])
= -k_m(u, [(),B])
= -p_D([(),B])
= -[B,p_D] = [p_D, B].

The Wong equation for the precession of the charge is thus
d(p_D)/dt = [p_D, B[dm/dt]].

The variation with respect to m' yields the momentum
p = g_m(m', ()) + k_m(u, B_m())
= g_m(m', ()) + p_D(B_m())
= g_m(m', ()) + p_D.B_m.
In component form
p_m = g_{mn} m'^n + p_D_a B^a_m,
which is the usual expression for the canonical momentum in the
presence of a Lorentz force given by the potential Lie vector B coupled
to the charge Lie vector p_D.

The corresponding force is the variation with respect to m:
F = 1/2 Dg_m[](m',m') + 1/2 Dk_m[](u,u) + k_m(u,DB_m()[m']).
= 1/2 Dg_m[](m',m') + 1/2 Dk_m[](u,u) + p_D(DB_m()[m']).

The corresponding equation of motion is dp/ds = F. Differentiating p,
we find
dp/ds = Dg_m[m'](m',()) + g_m(m'',()) + p_D'[B_m()] + p_D[dB_m[m'][]]
= Dg_m[m'](m',()) + g_m(m'',()) + p_D[[B[m'],B[]]] +
p_D[dB_m[m'][]].
Thus
Dg_m[m'](m',()) + g_m(m'',()) + p_D[[B[m'],B[]]] + p_D[dB_m[m'][]]
= 1/2 Dg_m[](m',m') + 1/2 Dk_m[](u,u) + p_D(DB_m[][m']).

Defining the Yang-Mills force
G(v, w) = DB_m[v][w] - DB_m[w][v] + [B_m[v], B_m[w]]
and Christoffel coefficients
Gamma(u,v,w) = 1/2 (Dg[u](v,w) + Dg[v](u,w) - Dg[w](u,v))
the equations of motion become
g_m(m'', ()) + Gamma(m', m', ()) = G((), m') + 1/2 Dk_m[](u,u).

Applying this to the vector m', and noting that
(g(m', m'))' = 2 g(m'', m') + 2 Gamma(m', m', m')
we get
(g(m', m'))' = Dk_m[m'](u, u).

If the gauge metric has an inverse, k^{-1}, one may define the charge
magnitude
|p_D|^2 = k^{-1}(p_D, p_D).
= p_D[u]
= k(u, u)
and
Dk_m[m'](u, u) = -D{k^{-1}_m)(p_D, p_D).
Adjoint invariance applies to this metric as well
k^{-1}([a,b], c) + k^{-1}(b, [a,c]) = 0.
Therefore, under the derivative, its action becomes
(|p_D|^2)' = D(k^{-1})[m'](p_D, p_D) + 2 k^{-1}(p_D, [p_D, B])
= D(k^{-1})[m'](p_D, p_D)
= -Dk_m[m'](u, u).

Thus, one arrives at the constant of motion:
g(m', m') + |p_D|^2.
If the gauge metric is constant, then each part is a constant of motion
separately.
m***@yahoo.com
2005-06-07 07:29:10 UTC
Permalink
Post by m***@yahoo.com
Post by m***@yahoo.com
L = 1/2 (g_{ij} u^i u^j) + K/2 (A_i u^i + u^5)^2
[etc]
This can be worked out in a transparent fashion (at least when not
reduced to ASCII form!) in the more general case, within the principal
bundle formalism, using the *left-quotient* operator.
Currently to be found at
http://federation.g3z.com/Physics/Index.htm

under the title:
The Equation of Motion for Yang-Mills Particles

A reworking of the somewhat well-known Duviryak reference, which
generalizes it to general relativistic particles and also employs the
left-quotient is under the title:

Wong's Equations: General Relativistic Yang-Mills Particles

Duviryak's treatment, however, does not readily admit an extension to
the case where the gauge metric becomes a function of base space.

Also of interest is the article listed under the title:
Yang-Mills Fields and Mass-Charge Relations

which comprises some concluding commentary regarding the recent
articles posted by Yablon on the prospect of a mass-charge formula, as
well as touching on some matters also brought up in the other 2
writeups.

All 3 are (currently) in Word format (apologies if everything isn't
rendered correctly on your client; I'm still on a learning curve for
TeX/MathML-based formatters).
Ken S. Tucker
2005-06-07 22:47:34 UTC
Permalink
Hi Mark and all...
Post by m***@yahoo.com
Post by m***@yahoo.com
Post by m***@yahoo.com
L = 1/2 (g_{ij} u^i u^j) + K/2 (A_i u^i + u^5)^2
[etc]
This can be worked out in a transparent fashion (at least when not
reduced to ASCII form!) in the more general case, within the principal
bundle formalism, using the *left-quotient* operator.
Currently to be found at
http://federation.g3z.com/Physics/Index.htm
The Equation of Motion for Yang-Mills Particles
A reworking of the somewhat well-known Duviryak reference, which
generalizes it to general relativistic particles and also employs the
Wong's Equations: General Relativistic Yang-Mills Particles
Duviryak's treatment, however, does not readily admit an extension to
the case where the gauge metric becomes a function of base space.
Yang-Mills Fields and Mass-Charge Relations
which comprises some concluding commentary regarding the recent
articles posted by Yablon on the prospect of a mass-charge formula, as
well as touching on some matters also brought up in the other 2
writeups.
All 3 are (currently) in Word format (apologies if everything isn't
rendered correctly on your client; I'm still on a learning curve for
TeX/MathML-based formatters).
According to the Lorentz Transformation, lengths
perpendicular to motion are unaffected, IOW's a
velocity in direction dx will leave the length Y
invariant.

Given all possible motions in 4D space and time,
we can set a finite length perpendicular to dx,dy
dz,dt that will remain invariant, but will appear,
physically , as a finite constant. IMO that finite
constant, is none other than the electrical charge
"e". It appears as perpendicular to 4D spacetime,
and can then be considered as a 5th *finite* dimension.

The failure that occurred (IMO) in Kaluza's 5D is
the "e" was regarded as absolute, when it's really
relative. By relative, I mean, a fundamental charge
exists only in relation to another charge.

For example from basic classical EM theory the energy
of a pair of charges is,

energy = q1*q2/r.

Removing either charge, gives

voltage= q/r

which is relative to some ground, aka gauge invariant.

It's energy we're really measuring in physics, that's
what's metered, aka Kilowatt hours.

Returning to 5D, the correction to Kaluza requires
the principle of the relativity of charge. It takes
some thinking to understand "c" is an invariant yet
velocity is relative. Likewise, charge is relative
but "e" is an invariant.

In view of the principle of the relativity of charge,
it stands to reason that two distinct charges are
required in 5D, q1 and q2.

The original purpose of 5D was to incorporate quantum
effects into the 4D spacetime field. The separation of
q1 and q2 does that by providing a finite length.
Regards
Ken S. Tucker

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