I was looking at the units of G - M^3/(S^2*kg)

And I was curious as to whether the value of G ( the gravitational
constant) be the same to different observer in different frames of
reference.


--
Only in movies do guys pick the engagement ring.

Observations of Bernard - No 89

BernardZ wrote:
I was looking at the units of G - M^3/(S^2*kg)

And I was curious as to whether the value of G ( the gravitational
constant) be the same to different observer in different frames of
reference.


--
Only in movies do guys pick the engagement ring.

Observations of Bernard - No 89

xxein: Yes, but obviously each moving inertial observer observes
differently. That is the key to any possible physical understanding.
With math, you can do anything that makes sense with a math. But, as
you see, there is much more than a mathematical satisfaction to be
desired. There is, after all, a physic.

Happy hunting.

BernardZ wrote:
I was looking at the units of G - M^3/(S^2*kg)

Well, it has those units in MKS. Other systems of units are common. In
GR the most common units have G=1 -- it is unitless. The second most
common units set G=1/(8pi) and again it is unitless.

We also set c=1 (unitless), which means in GR the only unit
is length, and cm is most common. This means that time, energy,
mass, and momentum are all measured in cm.


And I was curious as to whether the value of G ( the gravitational
constant) be the same to different observer in different frames of
reference.

In GR, G is a constant, and as such is of course completely independent
of any coordinates, be they a reference frame or other construct. Its
units, however, are conventional; different observers can use different
conventions.


Tom Roberts

In sci.physics.relativity, BernardZ

wrote
on Sun, 27 Nov 2005 10:31:55 +1100
MPG.1df39b6d251760119897c0@news:
I was looking at the units of G - M^3/(S^2*kg)

An alternative, slightly more meaningful formulation might
be N-m^2/kg^2. Since 1 newton is 1 kg dragged at 1 m/s/s
this works.


And I was curious as to whether the value of G ( the gravitational
constant) be the same to different observer in different frames of
reference.


It should be the same for everybody. :-)

--
#191,
It's still legal to go .sigless.

"BernardZ" wrote in message
news:MPG.1df39b6d251760119897c0@news...
I was looking at the units of G - M^3/(S^2*kg)

And I was curious as to whether the value of G ( the gravitational
constant) be the same to different observer in different frames of
reference.

This is one of the best questions I have seen in the past few months. The
following is my take on this subject.

In GR, the curvature of space time is a function of G as well as mass,
length, and speed of light. Since length is dependent on an observer, it is
rather silly to have the actual curvature of spacetime dependent on an
observer. The curvature of spacetime should be universally the same to all
observers. Thus, G must vary to accomodate for different observers.

Other than GR, the theories that allow variable and isotropic speed of light
in vacuum are ever more evident that G is observer dependent.

Koobee Wublee wrote:
In GR, the curvature of space time is a function of G as well as mass,
length, and speed of light.

And all other contributions to the energy-momentum tensor. And the
boundary conditions.


Since length is dependent on an observer, it is
rather silly to have the actual curvature of spacetime dependent on an
observer.

Yes, that would be silly. That's why curvature is described by the
various curvature tensors. Tensors, of course, are completely
independent of any observer or coordinates.

BTW, length in general does not depend on observer, either. But the
length of a specified object can do so. Length in general is determined
by the metric tensor, and is thus independent of observer or
coordinates. But specific measurements of an object's length will
project the invariant length between the endpoints of the measurement
onto a specific observer's coordinates, and that makes such measurements
observer dependent.


The curvature of spacetime should be universally the same to all
observers.

Yes, meaning the curvature tensors.


Thus, G must vary to accomodate for different observers.

No -- such "accomodation" is not needed, as the curvature TENSORS are
naturally independent of observer. If G did vary then the Lagrangian
would not have the appropriate symmetry.


Other than GR, the theories that allow variable and isotropic speed of light
in vacuum are ever more evident that G is observer dependent.

Nonsense. Theories are not "evidence" (or "evident").

There is no experimental support whatsoever for the notion that G
varies, either for different places or over time.


Tom Roberts

"Tom Roberts" wrote in message
...
Koobee Wublee wrote:
In GR, the curvature of space time is a function of G as well as mass,
length, and speed of light.

And all other contributions to the energy-momentum tensor. And the
boundary conditions.

The metric is the key. It plays the same role as a pair of glasses correct
one's eyesight. This metric corrects a distortion (observed curvature in
spacetime).

Since length is dependent on an observer, it is rather silly to have the
actual curvature of spacetime dependent on an observer.

Yes, that would be silly. That's why curvature is described by the various
curvature tensors. Tensors, of course, are completely independent of any
observer or coordinates.

Since the observed energy described by the spacetime equation with
Schwarzschild metric (I am sure now you know what I am talking about) is
indeed conserved, one can then write an equation to show how the observed
energy is a function of the mass (as if observed at no speed and no
curvature in spacetime), the measure of curvature in spacetime
(gravitational poentials), and the observed speed.

E = m c^2 sqrt(1 - 2 U) / sqrt(1 - B^2)

Where

** m = rest mass observed at no speed nor curvature in spacetime
** U c^2 = G M / r
** B^2 c^2 = (dr/dt)^2 / (1 - 2 U) + (r dH/dt)^2 / (1 -2 U)^2 +...

Since

E = m' c^2

Where

** m' = observed moving mass trapped in curvature of spacetime

One can relate this observed mass to

m' = m sqrt(1 - 2 U) / sqrt(1 - B^2)

So, as you can see, observed mass is also a function of U which measures the
curvature in spacetime. This means U has to be observer independent. The
catcher-22 situation is that U is a function of mass which is observer
dependent. Therefore, you have not resolved your silliness by calling the
curvature of spacetime observer independent, unless you allow G to vary
according to the observer's environmental condition such as his own
curvature in spacetime.

BTW, length in general does not depend on observer, either. But the length
of a specified object can do so. Length in general is determined by the
metric tensor, and is thus independent of observer or coordinates. But
specific measurements of an object's length will project the invariant
length between the endpoints of the measurement onto a specific observer's
coordinates, and that makes such measurements observer dependent.

The element of the metric describing the radial measure to the center of a
massive object is (1 / sqrt(1 - 2 U)). How blatant can you be by making the
statements above? After all, did you not describe an personing falling into
a black hole that he would observe himself quickly doing so? In the
meantime, some one else very far away would observe this event taking
forever where this hapless guy would take a long way to fall into the black
hole.

No -- such "accomodation" is not needed, as the curvature TENSORS are
naturally independent of observer. If G did vary then the Lagrangian
would not have the appropriate symmetry.

One Lagrangian one observer. It is a fucntion of observed speed, and
obviously this speed has to be observed by someone and one particular
someone.

Other than GR, the theories that allow variable and isotropic speed of
light in vacuum are ever more evident that G is observer dependent.

Nonsense. Theories are not "evidence" (or "evident").

Now, you are picking on my usage of vocabularies. Mind you that this is not
an English class. What I mean 'evident' is 'very clearly understandable'.

There is no experimental support whatsoever for the notion that G varies,
either for different places or over time.

Yes, there is no measurement of G done from difference curvature in
spacetime.