Length Energy Conversion.

It seems like standard procedure to express
the time metric given by,

g_00 = 1 - 2*G*m/(r*c^2) as

g_00 = 1 - 2m/r.

Implicit within this assumption is

length units = G/c^4 * energy units

which I'll abbreviate to

L = K * E, K = G/c^4

On this, perform an SR transform to find

L' = K' * E'

where L' = L/ gamma , E' = E*gamma and

K' = K/ gamma^2.

Evidently G is not invariant.

Defining length by components X^u and energy by
E_v permits the relation,

X^u = k^uv E_v

where K = |k^uv|. The tensor k^uv is the
*gravitational constant* tensor.

On each side, outer product with E_u,
to produce the invariant,

E_u X^u = k^uv E_v E_u

The LHS produces the invariant,
E_0 * X^0 = E*t = action = h (Plancks constant)
(similiarily E_i*X^i = m*v*x = action, i=1,2,3)

For now the LHS assumes the sum 2*h, so

2*h = k^uv E_v E_u

Whats good about this is how Plancks constant *h*,
relates to the Newtonian gravitational relative-constant
*k^uv* and energies *E_v E_u*, generally relativistic.
All ambiguity is apparently removed.

For definition, in General Relativity, I suggest
Newtons Gravitational constant G, should be
regarded as a relative-constant defined in GR
by K = |k^uv| where K distinguishes the tensor
weight from any asumption that G is a scalar.
So notation G refers to the idea that Newtons
gravitational constant is invariant - as usual, but
notation K acknowledges the constant as relative.

Collapsing 2*h = k^uv E_v E_u

= 2*h = k*M^2 where M is invariant mass,

and defines k = 2*h/M^2 (c=1).

where k is the gravitational scalar constant.

Here's the quagmire, we have a number attached
to invariant h. We need to attach a number to the
gravititional invariant k to define invariant mass M,
or vis-versa.

To define how to do this, it is best to set h=1 and
k=1 then the invariant M = N*(delta M), where
(delta M) are indivisible quanta's of energy, and

(delta M) =2 when N=1.

This is reasonable. Emit one photon (N=1)
requires equal delta energies from two charges,
expresed by (delta M) =2, so that each charge
surrenders an equal amount of potential energy
(and mass M) to form the photon.
The positive charge and the negative charge
equally move toward one another, and one
contributes the -electric field and the other the
+electric field, to birth the photon.

Regards
Ken S. Tucker

"Ken S. Tucker" wrote in message
om...
| Length Energy Conversion.
|
| It seems like standard procedure to express
| the time metric given by,
|
| g_00 = 1 - 2*G*m/(r*c^2) as
|
| g_00 = 1 - 2m/r.
|
| Implicit within this assumption is
|
| length units = G/c^4 * energy units
|
| which I'll abbreviate to
|
| L = K * E, K = G/c^4
|
| On this, perform an SR transform to find
|
| L' = K' * E'
|
| where L' = L/ gamma , E' = E*gamma and
|
| K' = K/ gamma^2.
|
| Evidently G is not invariant.
|
| Defining length by components X^u and energy by
| E_v permits the relation,
|
| X^u = k^uv E_v
|
| where K = |k^uv|. The tensor k^uv is the
| *gravitational constant* tensor.
|
| On each side, outer product with E_u,
| to produce the invariant,
|
| E_u X^u = k^uv E_v E_u
|
| The LHS produces the invariant,
| E_0 * X^0 = E*t = action = h (Plancks constant)
| (similiarily E_i*X^i = m*v*x = action, i=1,2,3)
|
| For now the LHS assumes the sum 2*h, so
|
| 2*h = k^uv E_v E_u
|
| Whats good about this is how Plancks constant *h*,
| relates to the Newtonian gravitational relative-constant
| *k^uv* and energies *E_v E_u*, generally relativistic.
| All ambiguity is apparently removed.
|
| For definition, in General Relativity, I suggest
| Newtons Gravitational constant G, should be
| regarded as a relative-constant defined in GR
| by K = |k^uv| where K distinguishes the tensor
| weight from any asumption that G is a scalar.
| So notation G refers to the idea that Newtons
| gravitational constant is invariant - as usual, but
| notation K acknowledges the constant as relative.
|
| Collapsing 2*h = k^uv E_v E_u
|
| = 2*h = k*M^2 where M is invariant mass,
|
| and defines k = 2*h/M^2 (c=1).
|
| where k is the gravitational scalar constant.
|
| Here's the quagmire, we have a number attached
| to invariant h. We need to attach a number to the
| gravititional invariant k to define invariant mass M,
| or vis-versa.
|
| To define how to do this, it is best to set h=1 and
| k=1 then the invariant M = N*(delta M), where
| (delta M) are indivisible quanta's of energy, and
|
| (delta M) =2 when N=1.
|
| This is reasonable. Emit one photon (N=1)
| requires equal delta energies from two charges,
| expresed by (delta M) =2, so that each charge
| surrenders an equal amount of potential energy
| (and mass M) to form the photon.
| The positive charge and the negative charge
| equally move toward one another, and one
| contributes the -electric field and the other the
| +electric field, to birth the photon.

Now can you do this for a graviton? Four charges birth a graviton?

FrediFizzx

"Ken S. Tucker" wrote in message
. com...
| Length Energy Conversion.
|
| It seems like standard procedure to express
| the time metric given by,
|
| g_00 = 1 - 2*G*m/(r*c^2) as
|
| g_00 = 1 - 2m/r.
|
| Implicit within this assumption is
|
| length units = G/c^4 * energy units
|
| which I'll abbreviate to
|
| L = K * E, K = G/c^4
|
| On this, perform an SR transform to find
|
| L' = K' * E'
|
| where L' = L/ gamma , E' = E*gamma and
|
| K' = K/ gamma^2.
|
| Evidently G is not invariant.
|
| Defining length by components X^u and energy by
| E_v permits the relation,
|
| X^u = k^uv E_v
|
| where K = |k^uv|. The tensor k^uv is the
| *gravitational constant* tensor.
|
| On each side, outer product with E_u,
| to produce the invariant,
|
| E_u X^u = k^uv E_v E_u
|
| The LHS produces the invariant,
| E_0 * X^0 = E*t = action = h (Plancks constant)
| (similiarily E_i*X^i = m*v*x = action, i=1,2,3)
|
| For now the LHS assumes the sum 2*h, so
|
| 2*h = k^uv E_v E_u
|
| Whats good about this is how Plancks constant *h*,
| relates to the Newtonian gravitational relative-constant
| *k^uv* and energies *E_v E_u*, generally relativistic.
| All ambiguity is apparently removed.
|
| For definition, in General Relativity, I suggest
| Newtons Gravitational constant G, should be
| regarded as a relative-constant defined in GR
| by K = |k^uv| where K distinguishes the tensor
| weight from any asumption that G is a scalar.
| So notation G refers to the idea that Newtons
| gravitational constant is invariant - as usual, but
| notation K acknowledges the constant as relative.
|
| Collapsing 2*h = k^uv E_v E_u
|
| = 2*h = k*M^2 where M is invariant mass,
|
| and defines k = 2*h/M^2 (c=1).
|
| where k is the gravitational scalar constant.
|
| Here's the quagmire, we have a number attached
| to invariant h. We need to attach a number to the
| gravititional invariant k to define invariant mass M,
| or vis-versa.
|
| To define how to do this, it is best to set h=1 and
| k=1 then the invariant M = N*(delta M), where
| (delta M) are indivisible quanta's of energy, and
|
| (delta M) =2 when N=1.
|
| This is reasonable. Emit one photon (N=1)
| requires equal delta energies from two charges,
| expresed by (delta M) =2, so that each charge
| surrenders an equal amount of potential energy
| (and mass M) to form the photon.
| The positive charge and the negative charge
| equally move toward one another, and one
| contributes the -electric field and the other the
| +electric field, to birth the photon.

Now can you do this for a graviton? Four charges birth a graviton?
FrediFizzx

Wow, thanks Fizzx, thats an ingenious question,
and the 4-charge hint is very helpful.
With 4 charges I can form 2 energy systems,
since an electrical potential energy system
requires 2 charges (and baring entanglement
that would happen using 3 charges) , 4 charges
permit 2 separate energy systems.
(Remember one system is E = q*Q/r).

So we can set up two systems of energy,
A and B, and also define,

A_u B_v = - B_u A_v

(you can demonstrate this equation with the
help of a friend. While facing your friend B
have them rotate their arm ClockWise, to you
'A' the arm rotates CCW, so switching A and B
switches direction).

In place of the above (2*h = k^uv E_v E_u)
we put,

2*h = k^uv A_u B_v, (A_u B_v = -B_u A_v)

therefore k^uv = - k^vu, in the case of a
4 charge system.

The asymmetry of the *Gravitional constant*
tensor does indeed allow k^uv to have very
similiar relationships to the field tensor F_uv.

For example, the relation

F_uv,w + F_vw,u + F_wu,v = 0

in a vacuum describes classical EM waves,
and a similiar equation should exist for k^uv.

In my orginal post I suggested,

2*h = k*M^2

where M represented a 2 charge system
(and later this produced a photon).

Fizzx, by introducing a 4 charge system enabled
this to be expressed,

2*h = k*M*M'

with M defined by 2 charges and M' defined by
the other two charges, and both are invariant.

Thanks all, and especially Fizzx. couldn't do a
graviton, but got an interesting insight.
Ken S. Tucker
PS: Fizzx, do you really live in Berlin?

"Ken S. Tucker" wrote in message

| PS: Fizzx, do you really live in Berlin?

No. I am just using the *free* News.CIS.DFN.DE usenet newsgroup
server because it is the best one for the sci.physics.x groups that I
have found so far (my three different ISP's all have terrible usenet
servers). Well, at least for free servers. I haven't tried ones you
have to pay for. I am reporting from Los Angeles --cough, cough. We
had a pretty miserable week here last week.

FrediFizzx

"FrediFizzx" wrote in message ...
"Ken S. Tucker" wrote in message

| PS: Fizzx, do you really live in Berlin?

No. I am just using the *free* News.CIS.DFN.DE usenet newsgroup
server because it is the best one for the sci.physics.x groups that I
have found so far (my three different ISP's all have terrible usenet
servers). Well, at least for free servers. I haven't tried ones you
have to pay for. I am reporting from Los Angeles --cough, cough. We
had a pretty miserable week here last week.

FrediFizzx

Sorry, to strain the topic, but this spring I needed
to build a fire-proof building in the Okanogan
Valley, (I asked some questions in alt.architure).
Seems an improved UL or local building code is in
order, just so I build a non flammable house!
KST

Ken S. Tucker wrote:
Length Energy Conversion.
It seems like standard procedure to express
the time metric given by,
g_00 = 1 - 2*G*m/(r*c^2) as
g_00 = 1 - 2m/r.

Implicit within this assumption is
length units = G/c^4 * energy units
which I'll abbreviate to
L = K * E, K = G/c^4

The usual way to make that reduction is simply to choose units in which
G=c=1. Then it's obvious those symbols can be omitted from all
equations. The usual choice is for length to be the unitsful quantity,
so time is measured in meters (i.e. the time it takes for light to
travel 1 meter), and mass is also measured in meters (e.g. the sun has a
mass around 1.5 kilometers, IIRC).


On this, perform an SR transform to find
L' = K' * E'
where L' = L/ gamma , E' = E*gamma and
K' = K/ gamma^2.

It does not make sense to "transform" units. If one did that, none of
the standard textbooks of physics or engineering would be valid in any
moving frame (and please note the earth is moving wrt just about
anything of significance...). Components of tensors transform, not the
units (of course one can change length or time units and treat it as a
change in coordinates; but you are not doing that).

It also makes no sense to apply a Lorentz transform to one Schwarzschild
metric component -- g_00 is merely one component of the metric tensor,
and you MUST transform the SET OF COMPONENTS of the tensor, not just one
of them. Note that if you did that, it would NOT result in coordinates
in which the mass is moving at the velocity of the transform, it's more
complicated than that....(hint: changes in the metric components change
the projections of your measuring instruments onto the coordinates [yes,
echoes of Heisenberg here...]). Note also that in GR such a global
Lorentz transform has no particular significance, and one could use a
Galilean velocity transform and obtain a result with equal usefulness
(i.e. virtually none).

[... more nonsense based on KST's lack of understanding of what
tensors really are...]


Tom Roberts

Tom Roberts wrote in message ...
Ken S. Tucker wrote:
Length Energy Conversion.
It seems like standard procedure to express
the time metric given by,
g_00 = 1 - 2*G*m/(r*c^2) as
g_00 = 1 - 2m/r.

Implicit within this assumption is
length units = G/c^4 * energy units
which I'll abbreviate to
L = K * E, K = G/c^4

The usual way to make that reduction is simply to choose units in which
G=c=1.

The "usual" way is far from proven, look at all
the effort to prove c = constant and h = constant.
Setting G = constant is an ARBITUARY notion
from Newtonian concepts. It has not been
experimentally nor theoretically proven.

Then it's obvious those symbols can be omitted from all
equations. The usual choice is for length to be the unitsful quantity,
so time is measured in meters (i.e. the time it takes for light to
travel 1 meter), and mass is also measured in meters (e.g. the sun has a
mass around 1.5 kilometers, IIRC).

In simple SR mass transforms like

m' = m*gamma and x' = x/gamma

(relative along x and x').

On this, perform an SR transform to find
L' = K' * E'
where L' = L/ gamma , E' = E*gamma and
K' = K/ gamma^2.

It does not make sense to "transform" units.

What??? you use centimenters, I use inches.

You measure X' cms and I measure X inches.

Then X' = (&X'/&X) * X where

&X'/&X = 2.54

And please take a moment to reflect on the
source of the units and components, ie

vector x = e'_u X'^u = e_u X^u or

vector x = e'^u X'_u = e^u X_u.

The e are basis vectors, and assuming u=1
it makes sense to say x is a unitless length,
and it follows from the example,

e'_1 = 1/cm, e'^1 = cm likewise for e_u,

so that the Kronecker delta is unitless. Recall
the K delta is

delta^u_v = e^u * e_v

now set u=v =1 and find the required

(cm) * (1/cm) = 1.

If one did that, none of
the standard textbooks of physics or engineering would be valid in any
moving frame (and please note the earth is moving wrt just about
anything of significance...).

Well, sure, when the motion is significant
relativistic details (like covariant and contravariant)
are needed.

Components of tensors transform, not the
units (of course one can change length or time units and treat it as a
change in coordinates; but you are not doing that).

Any tensor or tensor component of rank 0 has
units as explained above, I'm transforming w.r.t
those units.

It also makes no sense to apply a Lorentz transform to one Schwarzschild
metric component -- g_00 is merely one component of the metric tensor,
and you MUST transform the SET OF COMPONENTS of the tensor, not just one
of them.

Well that is certainly true. I went for SR (avoiding GR)
to show how the relation L =G/c^4 *E causes some
LT issues even when g_00~1, for basic units L and E.

Note that if you did that, it would NOT result in coordinates
in which the mass is moving at the velocity of the transform, it's more
complicated than that....(hint: changes in the metric components change
the projections of your measuring instruments onto the coordinates [yes,
echoes of Heisenberg here...]).

Ok, best if the G was examined with more rigour,
ie. from a GR point of view, but you pretty much
lost me on the Heisenberg bend, I get the general
idea but, but I need a BIGGER HINT.

Note also that in GR such a global
Lorentz transform has no particular significance, and one could use a
Galilean velocity transform and obtain a result with equal usefulness
(i.e. virtually none).

[... more nonsense based on KST's lack of understanding of what
tensors really are...]

Can you prove G is a invariant constant?

Above you stated G=c=1 so you have as
much confidence in G=1 as c=1. Now c has
been scutinized and tested big time, MMX
to begin with, and has been examined math-
matically.
Why art thou so confident G=1 ?

Tom Roberts
Regards Ken S. Tucker