- #1
Ken S. Tucker
[SOLVED] Re: Kaluza-Klein help needed
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.
John Baez wrote:
> 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
See ref, Eq.(17.61). John's statement looks accurate summed
over 4D.
> 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.
> 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
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.
> 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
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?
> 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!
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.
> 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.
> 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?
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
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.
John Baez wrote:
> 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
See ref, Eq.(17.61). John's statement looks accurate summed
over 4D.
> 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.
> 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
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.
> 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
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?
> 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!
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.
> 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.
> 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?
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