Post A.U=0 (kst)

Agreed, A.U=0 always. In components, we have

0 = A.U = A^u U_u == A^0 U_0 + A^i U_i

where i sums over 1,2,3.

I'll choose a CS where all U_i=0, then using algebra
I find A^0 =0 since U_0 0 in a weak field.

In general, A^u = D(U^u)/ds =g^uv D(U_v)/ds,
but we want to solve,

A^i = g^i0 D(U_0)/ds

presuming U_i=0 is a constant rendering D(U_i)=0.

Retaining the specified CS,

A^0 = g^00 D(U_0)/ds =0

therefore D(U_0) /ds =0, where g^00 0, as
specified in the weak field.

It follows that A^i =0, if U_i=0.

Hence A^i =0 is always true if a CS satisfying U_i=0
is always true.

To recap, A^i =0 is classically referred to as the
geodesic equation, and is true in all CS's where U_i=0.

A question arises, can we always find U_i=0 is true
in all relative circumstances?

Let's associate, U_i = g_iu U^u and expand to,

U_i =0 = g_i0 U^0 + g_ij U^j (sum j over 1,2,3).

Using algebra produces the requirement,

g_i0 = - g_ij dx^j/dx^0 .

for a geodesic.

In the metric ds^2 = g_uv dx^u dx^v the geodesic requirement
*collapses* the expanded metric from,

ds^2 = g_00 dx^0 dx^0 + 2*g_i0 dx^i dx^0 + g_ij dx^i dx^j

(using algebra) to,

ds^2 = g_00 dx^0 dx^0 - g_ij dx^i dx^j .

In the SR limit that becomes ds^2 = dt^2 - dr^2
when g_00...g_33 =1, (note positive signature).

Here's the funny part, Dr. Tucker has found a metric
where A^u=0 always, and then is obligated to provide
an explanation why Lorentz Force =0 is in accord
with Quantum Theory!

Thanks, Ken S. Tucker

Ken S. Tucker:
Post A.U=0 (kst)

Agreed, A.U=0 always. In components, we have

0 = A.U = A^u U_u == A^0 U_0 + A^i U_i


Rather than try to figure out what you've done, I'll just
provide a simple derivation. Given a four velocity, U^a,
U^a U_a = 1 so that the derivative with respect to the
proper time, d/d\tau is:

(d/d\tau) (U.U) = (d/d\tau) U^a U_a = 2 U^a dU_a/d(\tau) = 0

therefore, if A_u = dU_a/d\tau, A.U = 0

(Bilge) wrote in message ...
Ken S. Tucker:
Post A.U=0 (kst)

Agreed, A.U=0 always. In components, we have

0 = A.U = A^u U_u == A^0 U_0 + A^i U_i


Rather than try to figure out what you've done, I'll just
provide a simple derivation. Given a four velocity, U^a,
U^a U_a = 1 so that the derivative with respect to the
proper time, d/d\tau is:

(d/d\tau) (U.U) = (d/d\tau) U^a U_a = 2 U^a dU_a/d(\tau) = 0

therefore, if A_u = dU_a/d\tau, A.U = 0

Thanks Bilge and all my friends in sp. relativity.
I regret the home planet intends to beam me up,
so I won't be posting for awhile.
Thank you and best regards...
Ken S. Tucker

On Mon, 17 May 2004 13:45:22 -0700, Ken S. Tucker wrote:

(Bilge) wrote in message ...
Ken S. Tucker:
Post A.U=0 (kst)

Agreed, A.U=0 always. In components, we have

0 = A.U = A^u U_u == A^0 U_0 + A^i U_i


Rather than try to figure out what you've done, I'll just
provide a simple derivation. Given a four velocity, U^a,
U^a U_a = 1 so that the derivative with respect to the
proper time, d/d\tau is:

(d/d\tau) (U.U) = (d/d\tau) U^a U_a = 2 U^a dU_a/d(\tau) = 0

therefore, if A_u = dU_a/d\tau, A.U = 0

Thanks Bilge and all my friends in sp. relativity.
I regret the home planet intends to beam me up,
so I won't be posting for awhile.
Thank you and best regards...
Ken S. Tucker

Take care, Ken.

Jeff

--
Add an underscore between 'd' and 's' and remove the first three
letters of the alphabet for email.

(Bilge) wrote in message ...
Ken S. Tucker:
Post A.U=0 (kst)

Agreed, A.U=0 always. In components, we have

0 = A.U = A^u U_u == A^0 U_0 + A^i U_i


Rather than try to figure out what you've done, I'll just
provide a simple derivation. Given a four velocity, U^a,
U^a U_a = 1 so that the derivative with respect to the
proper time, d/d\tau is:

(d/d\tau) (U.U) = (d/d\tau) U^a U_a = 2 U^a dU_a/d(\tau) = 0

therefore, if A_u = dU_a/d\tau, A.U = 0

Thanks ((I've temporarily moved from
Ontario to Vancouver BC))
What I've done is to show how the geodesic
condition A^i = 0, (i=1,2,3) requires U_i =0,
((assuming my reasoning to be correct)).
However U^i may still be a variable like
dU^i/ds 0 (d is ordinary diff) and U^i 0.

That simplifies the geodesic requirement,
A^i=0 , equivalent to the requirement U_i=0,
neat eh?

Now we need to find a CS where U_i=0 always,
and all motion in GR is rendered geodesical,
ie. A^i=0 generally.

Use a DYNAMIC NON-ORTHOGONAL SPACETIME METRIC,
defined by,

g_0i = - g_ij dx^j/dx^0 (j=1,2,3),

and a positive metric signature (1,1,1,1)
((Tucker's Metric))

and U_i =0 always.

Those are valid substitutions in SR as they
produce from ds^2 = g_uv dx^u dx^v,

ds^2 = dt^2 - dx^2 - dy^2 - dz^2.

Regards
Ken S. Tucker

Ken S. Tucker wrote:
What I've done is to show how the geodesic
condition A^i = 0, (i=1,2,3) requires U_i =0,
((assuming my reasoning to be correct)).

The geodesic condition is really A^u = 0. This difference becomes
crucial below.

Throughout, index i runs over {1,2,3} while u runs
over {0,1,2,3}.

NOTE: the condition A^u=0 applies in any coordinate system whatsoever,
but your condition U_i=0 applies ONLY in certain specific coordinate
systems, one such system is the instantaneously-comoving inertial frame
of the object in question.

Remarkably, you are displaying the same error as Steve Bell in another
thread -- confusion over the limitations of coordinate notation.

All this is MUCH better expressed in a coordinate-independent manner;
here capital letters without indices represent the corresponding 4-vectors:

U is the tangent 4-vector of a timelike object's path,
\tau is its proper time
Definition of A: A = DU/d\tau
All paths satisfy: U.A = 0
Geodesic condition: A = 0


However U^i may still be a variable like
dU^i/ds 0 (d is ordinary diff) and U^i 0.

Sure. And you could replace i with u there.

But if your condition U_i=0 holds, then necessarily U^i=0 (and dU^i/ds=0
also), as long as you use orthogonal coordinates (specifically, if
g_0i=0 for all i).

When you make explicitly coordinate-dependent statements (e.g. ANY
equation using just i and not u), you MUST state what coordinates you
are using.

In addition, you are also quite lax in specifying whether
your equations hold only at a single point along the
trajectory, or over some region of the trajectory. I
am assuming the latter. Note, please, that the equations
I gave above hold everywhere along the path. (yours, for
instance, have the problem that the instantaneously-
comoving inertial frame changes for different points of
the path; you have completely ignored this).


That simplifies the geodesic requirement,
A^i=0 , equivalent to the requirement U_i=0,
neat eh?

Not "neat", simply wrong.

Simple counterexample: consider ANY non-geodesic path (so A is nonzero):
in the instantaneously-comoving inertial frame, U^i=0, and U_i=0, but A
is nonzero (specifically, some A^i is nonzero, because A^0=0 in this
frame). So U_i=0 does not imply the path is a geodesic.

And I repeat: the geodesic requirement is A=0. Not what
you wrote. A^i=0 is necessary but not sufficient for the
path to be a geodesic.


Now we need to find a CS where U_i=0 always,
and all motion in GR is rendered geodesical,
ie. A^i=0 generally.

You can easily find such coordinates, just use U as the time coordinate
and the object itself as the origin of the spatial coordinates -- then
U_i=0 always, even for non-geodesic paths. But selecting coordinates has
no power whatsoever to turn a non-geodesic path into a geodesic -- the
geodesic condition (A=0) is utterly independent of coordinates.

Your A^i=0 is NOT the geodesic condition. In component
notation the geodesic condition is A^u=0. A^i=0 is
necessary but not sufficient for the path to be a geodesic.


You have become confused through too strong a reliance on
component-based notation, while not bothering to specify your coordinates.


Here's another problem for you to consider about your whole approach:
You have repeatedly used the index i running over {1,2,3} -- how about
null coordinates? For them there are two spacelike coordinates and two
null coordinates, so i in {1,2,3} does not make sense at all.


[... further nonsense]


Tom Roberts

Ken S. Tucker says...

Post A.U=0 (kst)

Agreed, A.U=0 always. In components, we have

0 = A.U = A^u U_u == A^0 U_0 + A^i U_i

where i sums over 1,2,3.

I'll choose a CS where all U_i=0, then using algebra
I find A^0 =0 since U_0 0 in a weak field.

In general, A^u = D(U^u)/ds =g^uv D(U_v)/ds,
but we want to solve,

A^i = g^i0 D(U_0)/ds

Hmm. Why are you writing it in terms of U_u instead
of U^u?

presuming U_i=0 is a constant rendering D(U_i)=0.

No, that's not true. U_i = 0 certainly implies that dU_i/ds = 0,
but it *doesn't* imply DU_i/ds = 0.

As I've written several times, the equation for DU/ds involves
the connection coefficients G^u_vw. If G^u_vw is nonzero, then
DU/ds can be nonzero, even when the components of U are all
constant.

Specifically:

DU^u/ds = dU^u/ds + G^u_vw U^v U^w

or if you want to use 1-forms instead of vectors,

DU_u/ds = dU_u/ds - G_u^vw U_v U_w

So in a coordinate system in which U_0 = 1 and U_i = 0 (i = 1,2,3)

DU_0/ds = - G_0^00
DU_i/ds = - G_i^00

Yes, I know it's strange that a derivative of a constant
can be nonzero, but if the G's are nonzero, then U really
*isn't* a constant. Here's one way to think about it:

U = U^u e_u

where e_u = the u'th basis vector. So there are two ways
that the vector U can vary: (1) Its components may vary,
or (2) the basis vectors e_u may vary. The G's keep track
of how the basis vectors e_u vary from point to point.

--
Daryl McCullough
Ithaca, NY

Jeff Krimmel wrote in message . ..
On Tue, 01 Jun 2004 12:51:09 -0700, Ken S. Tucker wrote:

Tom Roberts wrote in message m...

[...]

Remarkably, you are displaying the same error as Steve Bell in another
thread -- confusion over the limitations of coordinate notation.

Tom, it's clear Steve Bell has displayed an exceptional
grasp of tensors, likely beyond your understanding, but
like me and others, we're sometimes pushing the frontiers
of that understanding and typos can occur.

[sounds horn]
Ken, please come back.

Thank you for your kind comments.
In a few days I won't be posting for awhile,
because of work to do at an isolated location,
but *I'll be back*.
Regards
Ken S. Tucker













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