Re: Acceleration of charge

sal wrote:
This is getting too long -- time for the chain saw.

ouch!!

On Sun, 26 Feb 2006 12:19:51 -0800, Ken S. Tucker wrote:

For "4D Spin" see,
http://groups.google.com/group/sci.math/browse_frm/thread/52f41cc86074262/c7dac8f7ab142b5c?lnk=st&q=%224d+curl%22&rnum=1&hl=en#c7dac8f7ab142b5c

John Baez has some good comments in that thread too, I refer to
those where "4D spin" is concerned, as they are parallel to my
understanding if you wish to discuss it further, I'd be glad to, and
that provides at least a commonality to start from a mathematical
basis.

I'll take a look at it; haven't yet, tho.

Not trying to blow you off, I think there's some smart
ideas in that thread, then we might discuss 4D curl.

sal wrote:
On Fri, 24 Feb 2006 12:05:27 -0800, Ken S. Tucker wrote:

Sal, studied your post carefully...

The body of references on this subject in dearth, and the math I
present is a simplification of what I've been able to glean.
Fredifizzx kindly helped me out at his site on this article,

http://www.vacuum-physics.com/KST/GR_Charge_Couple3.pdf

I took a look at it. The point seems to be that the energy of a
system of two charged particles which results solely from their
interaction is a form of energy and hence results in a
"gravitational field" (non-flat metric) in the area. But that's
obvious, and I don't really see how that leads to any need for
anything beyond a determination of the stress/energy tensor for
charged particles.

ok, try it yourself, it's called "The field of an electrically
charged mass point" (see Bergmann's Relativity, pg 204). It
contradicts Purcell (below), as it makes a single lone charge a
massive object.

Don't have Bergmann, but a single lone charge is of course a massive
object.

below...

Furthermore, you seem to have left out the energy of the E fields,

No such thing, the energy is stored in an interacting system, see
Prucells, "E&M" pg.8.

As it happens I do have Purcell. But it's an intro text, and on page
8 he's not addressing the energy stored in the fields themselves,
which is what I was referring to. To find the energy in the field you
need to integrate |E|^2 over space, IIRC, and that's going to
gravitate. This is quite apart from the system energy needed to
assemble the pair of charges.

Good, Purcell, pg 8 "only way...the interaction"
that eliminates singular charges.

The idea a single charge could masturbate itself to create
a self energy like an electron was dispelled many years
ago.

which, if you want to be consistent, must also contribute to
T. Normally that's rolled into the mass of the particle but you
explicitly set that to zero. So, to get a valid answer, it seems
like you really need to include a term for the local field
intensity in T.

It wasn't needed.

Well, maybe not, but if you actually had such a situation it would
surely be present.

Anyhow enough on that for now.

ok

[sni.i.i.p]


You're combining F and g into a single tensor, right...?

Yes in fact compelled too. Because "F" is very much a necessity
in determining EM energy and energy affects "g",

One does not follow from the other. EM energy enters T and that
affects g, but that doesn't have anything to do with the notion
that you need to merge F with g.

It certainly does when solving the EFE' using a "charge couple",
when quantum interactions are also solved. The +/- "charge couple"
functions as dipole to emit or receive radiation.

Not following. "g" affects everything, "F" is invisible to anything
without charge.

Right, that's why we do use a_uv = a*F_uv in the
nonsymetrical metric. Even more remarkably and
simplistic, is that a_uv demands a relation, and a
length, so it is equatable to curvature and in that
way, the curvature is built into the metric, but that
is also consistent with the continuum theory.

If you merge them, how do you separate the effects
again? How does the tensor know to affect charged particles
differently from non-charged ones? (I'm _really_ not following
this...)

Ok, as you know E=mc^2, and Energy is defined and
quantified in many ways, but in any of those ways, it's
equivalent to "m" which contributes to gravitation,
irrespective of the nature of the mass-energy.
So you may convert a description of Energy into any
form whatsoever and expect an equally gravitating
(and an inertial) effect.
So it follows we have a right and obligation to describe
Energy in terms of a "charge couple", as a generic form
of energy and solve for a gravitation effect, as demo'd in

http://www.vacuum-physics.com/KST/GR_Charge_Couple3.pdf

((Caution, that article predicts a null result for LIGO,
when using that description of energy, as the radiatives
will appear as EM-radiation) :-(.

I think I'll start an new thread for the rest.

[ snip ]


[sal points out an error:]

Fabc + F bca + Fcab = Jd

First of all, you left out some rather important punctuation here,

Sal, I specified Maxwell's 2nd set, so you knew what I was talking
about!

Call me a dunce but that equation really stopped me for a bit. It
_looked_ right but it clearly wasn't -- I don't see ME's in that exact
form all that often (they can be written about six different ways, of
course).
It was a while before I realized you'd left out the "*" and so had the
wrong tensor on the left.

Ok sorry, Let's look at, "F abc + F bca + F cab = Jd"
from the pov of induction based on electrical transformers.

((the covariance/contravariance doesn't matter in this
relation in my understanding, it's pretty flat spacetime)).

Haven't looked at the article but I'm familiar with the
higher-dimension cross product. As far as I know it's rarely used,
probably in part because it's confusing

The mathematicians seem to have good definitions.

Oh, absolutely, I just haven't seen it used much, as I said. (I've
read that it is indeed not used much in general, but that's a
different statement and I can't attest to its truth of my own
experience.)
And it is certainly a first cousin of the wedge product.

The "wedge" is based on the 2D

{x dy - y dx} .

I go out a foot from the origin to x=1, and y=0.
and x dy shows up as a point moving upwards.
Then goto y=1 and x=0 and dx moves the point
sideways, either dx =+/- depending on CW or
CCW rotation. That's a simple rotation.

Let's improve that by using

{x dy/ds - y dx/ds}

and then put in invariant electrical relation,
(a*b/s^2), and using the 4-velocity U^v to get,

(ab/s^2) * {x^u U^v - x^v U^u}

= a*F^uv = b*F^uv .

showing the symmetry of "a" to "b"

Ken

[ snip ]


Jay Yablon, has highlighted those relations as he applies flux in
GR to sub-atomic volumes, where confinement, and the analog to
Gausses Law in 4D is defined, using a more sophisticated
geometric application of the conservation of energy, where of
course the Divergence at a point is zero. In that application
(see gr-qc 0511050) he uses,

dV = sqrt(g) dx dy ... = invarariant.

In general you want dV to be a tensor to keep it invariant,
otherwise your volume integrals aren't well defined and that's
never good.

I think "dV" is a scalar, that's a standard derivation.

dV isn't the volume element? Well, with sqrt(g) on the right you've
got the equivalent of the determinant of the Jacobian present, which
is what you need to scale it to get the right answer.

The k-form isn't explicitly written out, but it's where that
determinant is coming from, in some sense.


Gauss' Law is, of course, just a special case of Stokes' theorem.
If w is a k-form, V is a volume, dw is the exterior derivative of
w, and dV is the boundary of V, then Stokes' theorem reads,

integral_dV (w) = integral_V (dw)

Gauss's law AKA the "divergence theorem" is just obtained from
Stokes' theorem by replacing the vector field you want to integrate
over a surface with its dual, which is a 2-form. The 2-form field
is of course a natural for integration over a surface. Its
exterior derivative, which you integrate over the volume, is a
3-form, and in 3 dimensions, a 3-form and a zero-form (which is its
3D dual) are isomorphic. In other words, there's only _one_ 3-form
in 3 dimensions, up to a scale factor, and we might as well just
treat it as a scalar, and that's what we call the divergence.

Yes, but you may be treating time as a scalar, as Newton does.

When you're working with Gauss's law, you're generally operating in 3
dimensions at a single point in time, and time is indeed behaving as a
scalar -- or rather it's not in the picture at all.

So the vector field becomes the 2-form "w" in Stokes' theorem as
stated above, and its divergence becomes the 3-form "dw".

It's well defined, but I think that involves "relative tensors" in
GR, (like densities), I don't see where a "3 form" comes from??

Here, I explain:

The "divergence theorem", in Euclidean 3-space, states that the
integral of a vector field over a closed surface equals the integral
of the divergence of the field over the volume enclosed by the
surface. In Gauss's law the "divergence" in question is the charge
density and the vector field is the E field, of course.

The vector field is integrated over the surface by dotting it into the
normal vector to the surface at each point.

The divergence theorem is a corollary of Stokes' theorem on manifolds.
Define a 2-form field which is the dual to the original vector field.
Take the integral of that 2-form field over the surface: just apply
the 2-form to the surface element at each point (a 2-form measures
surface area, of course). That integral is equal to the integral of
the original vector field over the surface.

Now, Stokes' theorem says the exterior derivative of that 2-form
field, integrated over the enclosed volume, is equal to the original
surface integral, too. The exterior derivative of that 2-form field
is a 3-form field.

However, we're in 3-space, recall. In 3-space any 3-form is just the
same as the volume element, times a scalar field. The scalar field is
the "dual" to the 3-form, and in this case it's the divergence of the
original vector field.

Clear (as mud)?

In general, if you can reduce the problem to an integral of a k-form
field over a hypersurface, then you can apply Stokes' theorem.



And you can apply this in as many dimensions as you like as long as
you're reasonably careful with the entities you're integrating.

Perhaps, simple examples, net based, words like "reasonably careful"
are a bit fluffly, though I'll compromise minor details to arrive at
the point in this forum.

Well the problem here is that my understanding of Stokes' theorem and
exterior derivatives is largely picture-based and intuitive (no I will
not say how long it took me to arrive at an intuitive understanding of
this somewhat opaque-seeming topic). That doesn't make it easy to explain
in words, and it doesn't help with the fact that I haven't applied them a
whole lot of real problems as a result of which I tend to mess up the
signs big-time whenever I try to write out the actual expressions.



That has special improtance because of his very gutsy decision to
apply "g" into sub-atomic problems and then that affects the flux
density of the bosons.

"g" makes a difference in subatomic problems??

Well I think so. To demo that lets make clear that g=|g_uv| ,
(determinant of the metric) is a relative tensor.

We (Jay may not agree) use the covariant derivative

g;u =0

and the partial derivative

g,u =/=0.

The g;u = 0 is obviously a scalar

It's not a scalar, it's a gradient, which is a rank 1 tensor which
happens, in this case, to be equal to zero.

and describes a "state" that can
only be varied incrementally such as by adding another scalar like
Plancks "h", as an example,

g;u + h = g';u

You can't add a scalar to a rank 1 tensor and get anything especially
useful out of it.


where the g';u is an *incremetally* changed metric due to a photon
(in big speak a boson) interaction, emitted or absorbed, that
changes the state.

OTOH, the g,u is NOT invariant and can be denoted

g,u = h/time = frequency.

The diff between g;u and g,u is in the Christoffel, and the diff
g;u- g,u (the Christoffel) may be decribed as a photon,

Say what?

g;u - g,u is a photon?

No comprendo.

(or
boson). To do that I needed to put Maxwell's equations inside of the
Christoffel, and works good, and so is consistent.

Best Ken
S. Tucker


--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org

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