MST rest mass definition.

On Feb 23, 2:44 am, "Ken S. Tucker" wrote:

However counter-intuitive it may be, "rest mass" is certainly not an
invariant by definition...

You don’t have to search for your quest elsewhere while the answer is
right in your backyard. Consider the Schwarzschild metric below.

** ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

The geodesic equation associated with the temporal variable can easily
be derived as follows. It helps if you understand what a Langrangian
is after staring at Wienberg’s textbook for over four decades.
shrug

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

Where

** E, m = constants
** B^2 c^2 = (dr/dt)^2 / (1 – 2 U)^2 + r^2 (dO/dt)^2 / (1 – 2 U)

When the curvature of space is weak, and the speed is low, the above
equation simplifies into the ever so familiar Newtonian result below.

** E = m B^2 c^2 / 2 - m U c^2

Or

** E = m v^2 / 2 – m G / r

Where

** v^2 ~ (dr/dt)^2 + r^2 (dO/dt)^2

Back to the geodesic equation mentioned above, the rest mass is
dependent on the curvature of spacetime as described below

** m (1 – 2 U) = Rest mass

Where

** m = invariant mass

So far so good, right? Well, GR is only a few more steps away from
total disaster after that. shrug

If the company catches me talking to a teddy bear
I'll get fired, do have a real name I can put on
my time card?
Ken

On Feb 23, 1:55 pm, Koobee Wublee wrote:
On Feb 23, 2:44 am, "Ken S. Tucker" wrote:

However counter-intuitive it may be, "rest mass" is certainly not an
invariant by definition...

You don’t have to search for your quest elsewhere while the answer is
right in your backyard. Consider the Schwarzschild metric below.

** ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

The geodesic equation associated with the temporal variable can easily
be derived as follows. It helps if you understand what a Langrangian
is after staring at Wienberg’s textbook for over four decades.
shrug

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

Where

** E, m = constants
** B^2 c^2 = (dr/dt)^2 / (1 – 2 U)^2 + r^2 (dO/dt)^2 / (1 – 2 U)

When the curvature of space is weak, and the speed is low, the above
equation simplifies into the ever so familiar Newtonian result below.

** E = m B^2 c^2 / 2 - m U c^2

Or

** E = m v^2 / 2 – m G / r

Where

** v^2 ~ (dr/dt)^2 + r^2 (dO/dt)^2

Back to the geodesic equation mentioned above, the rest mass is
dependent on the curvature of spacetime as described below

** m (1 – 2 U) = Rest mass

Where

** m = invariant mass

So far so good, right? Well, GR is only a few more steps away from
total disaster after that. shrug

It is suggested that Ken Tucker to be checked in for a psychiatric
evaluation. After staring at Weinberg’s textbook for over four
decades, he has totally failed to understand what a Lagangian is.
Now, due to his personal incompetence, he thinks he is now talking to
stuffed animals. The man is getting worse. shrug

On Feb 23, 6:23*pm, "Ken S. Tucker" wrote:
If the company catches me talking to a teddy bear
I'll get fired, do have a real name I can put on
my time card?
Ken

On Feb 23, 1:55 pm, Koobee Wublee wrote:



On Feb 23, 2:44 am, "Ken S. Tucker" wrote:

However counter-intuitive it may be, "rest mass" is certainly not an
invariant by definition...

You don’t have to search for your quest elsewhere while the answer is
right in your backyard. *Consider the Schwarzschild metric below.

** *ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** *U = G M / c^2 / r
** *dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

The geodesic equation associated with the temporal variable can easily
be derived as follows. *It helps if you understand what a Langrangian
is after staring at Wienberg’s textbook for over four decades.
shrug

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

Where

** *E, m = constants
** *B^2 c^2 = (dr/dt)^2 / (1 – 2 U)^2 + r^2 (dO/dt)^2 / (1 – 2 U)

When the curvature of space is weak, and the speed is low, the above
equation simplifies into the ever so familiar Newtonian result below.

** *E = m B^2 c^2 / 2 - m U c^2

Or

** *E = m v^2 / 2 – m G / r

Where

** *v^2 ~ (dr/dt)^2 + r^2 (dO/dt)^2

Back to the geodesic equation mentioned above, the rest mass is
dependent on the curvature of spacetime as described below

** *m (1 – 2 U) = Rest mass

Where

** *m = invariant mass

So far so good, right? *Well, GR is only a few more steps away from
total disaster after that. *shrug- Hide quoted text -

- Show quoted text -

rest mass refers to the internal (angular) momenta
of atoms, of which there is none for a newtonian corpuscle,
even though it was never a theory per se. ("hypothesis non fingo,"
in deed; let Newton be, the mintmaster)

Orville Wright at 34 (left)
Wilbur Wright at 38 (right)

The difficulties which obstruct the pathway to success in flying
machine construction are of three general classes: (1)
Those which relate to the construction of the sustaining wings; (2)
Those which relate to the generation and application of the power
required
to drive the machine through the air; and (3)
Those relating to the balance and steering of the machine after it is
actually in flight.

Although the breakthrough achieved by the brothers would be in the
solution to the third problem addressed by Wilbur—the problem of
stability and control in powered flight—they would also considerably
advance the knowledge in the other two areas, as well, in the march to
their great achievement.

On Feb 23, 9:02*pm, rasterspace wrote:
rest mass refers to the internal (angular) momenta
of atoms, of which there is none for a newtonian corpuscle,
even though it was never a theory per se. *("hypothesis non fingo,"
in deed; let Newton be, the mintmaster)

Orville Wright at 34 (left)
Wilbur Wright at 38 (right)

The difficulties which obstruct the pathway to success in flying
machine construction are of three general classes: (1)
Those which relate to the construction of the sustaining wings; (2)
Those which relate to the generation and application of the power
required
to drive the machine through the air; and (3)
Those relating to the balance and steering of the machine after it is
actually in flight.

Although the breakthrough achieved by the brothers would be in the
solution to the third problem addressed by Wilbur—the problem of
stability and control in powered flight—they would also considerably
advance the knowledge in the other two areas, as well, in the march to
their great achievement.

How can their be rest mass beyond theory if there is No Abolute Rest?

Albert Einstein

all one needs is relative rest, y'know;
don't measure it zipping by at sublight velocity (I mean,
light has a speed, not a directed movement, probably because
it is not a newtonian God-am corpuscle.

if you're referring to the Eocene to (I forgot) transition,
which supposedly eventually resulted in glaciation --
CO2 removed from teh 1000ppm level by carbonic acid weathering
of the Himalayas unto the Quaternary Period --
it was explained simply by Hansen that the present rise
is a lot faster. so,
insofar as one abdures the srange paradigm of "global" warming,
when insolation is essentially zero at the poles (other
than the atmospheric part of it, twilight etc.)

most notable study on this is Shackleton et al, but
as I recall, they used oxygen isotopes in sediment corings, and
the CO2 part may have been interpolated, perhaps by others.

if erosion can cause a)
lowering of land, and b)
displacement of water,
how are these tabulated in this datum?...
I won't mention the old tide-gauge guy
on http://21stcenturysciencetech.com, ol'what's-his-name.

Well KW it's tough writing mathski (math in ascii), you've
done well below.
(Had to figure B as in Beta).

On Feb 23, 1:55 pm, Koobee Wublee wrote:
On Feb 23, 2:44 am, "Ken S. Tucker" wrote:

However counter-intuitive it may be, "rest mass" is certainly not an
invariant by definition...

You don’t have to search for your quest elsewhere while the answer is
right in your backyard. Consider the Schwarzschild metric below.

** ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

The geodesic equation associated with the temporal variable can easily
be derived as follows. It helps if you understand what a Langrangian
is after staring at Wienberg’s textbook for over four decades.
shrug

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

Where

** E, m = constants
** B^2 c^2 = (dr/dt)^2 / (1 – 2 U)^2 + r^2 (dO/dt)^2 / (1 – 2 U)

When the curvature of space is weak, and the speed is low, the above
equation simplifies into the ever so familiar Newtonian result below.

** E = m B^2 c^2 / 2 - m U c^2

Or

** E = m v^2 / 2 – m G / r

Where

** v^2 ~ (dr/dt)^2 + r^2 (dO/dt)^2

Back to the geodesic equation mentioned above, the rest mass is
dependent on the curvature of spacetime as described below

** m (1 – 2 U) = Rest mass

Where

** m = invariant mass

So far so good, right?

Yes, perfect. If you want (using your formula) have a look at

m = Rest mass / (1 - 2 U)

A derivative, dm/dr yeilds the variation of Mercury's
orbit, that is to say, a rotation of the semi-major
axis (43"/century), also called the "relativitic force
supplement" by celestrial mechanics.

Well, GR is only a few more steps away from
total disaster after that. shrug

Well GR math is still evolving, AE and team did a great
job, it's using the math in applications that gets trippy.
I find one needs to be very careful using tensors analysis,
especially if CS conditions are imposed and tensor processes
are used. That's oxymoronic as the processes apply to all
CS's, so even the introduction of a metric form (1,1,1,-1)
to begin with blows the process and generates artifacts.
Regards
Ken S. Tucker

On Feb 24, 9:02*am, "Ken S. Tucker" wrote:
[...]

A derivative, dm/dr yeilds the variation of Mercury's
orbit, that is to say, a rotation of the semi-major
axis (43"/century), also called the "relativitic force
supplement" by celestrial mechanics.

No, it does not. Don't make **** up.


Well, GR is only a few more steps away from
total disaster after that. *shrug

Well GR math is still evolving, AE and team did a great
job, it's using the math in applications that gets trippy.
I find one needs to be very careful using tensors analysis,
especially if CS conditions are imposed and tensor processes
are used. That's oxymoronic as the processes apply to all
CS's, so even the introduction of a metric form (1,1,1,-1)
to begin with blows the process and generates artifacts.
Regards
Ken S. Tucker

You do not know what you are talking about. In the slightest.

On Feb 24, 9:02 am, "Ken S. Tucker" wrote:
Koobee Wublee wrote:

You don’t have to search for your quest elsewhere while the answer is
right in your backyard. Consider the Schwarzschild metric below.

** ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

The geodesic equation associated with the temporal variable can easily
be derived as follows. It helps if you understand what a Langrangian
is after staring at Wienberg’s textbook for over four decades.
shrug

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

Sorry, my mistake. The above should read

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

Where

** E, m = constants
** B^2 c^2 = (dr/dt)^2 / (1 – 2 U)^2 + r^2 (dO/dt)^2 / (1 – 2 U)

When the curvature of space is weak, and the speed is low, the above
equation simplifies into the ever so familiar Newtonian result below.

** E = m B^2 c^2 / 2 - m U c^2

Or

** E = m v^2 / 2 – m G / r

Where

** v^2 ~ (dr/dt)^2 + r^2 (dO/dt)^2

Back to the geodesic equation mentioned above, the rest mass is
dependent on the curvature of spacetime as described below

** m (1 – 2 U) = Rest mass

Correction:

** m sqrt(1 – 2 U) = Rest mass

Where

** m = invariant mass

So far so good, right?

Yes, perfect. If you want (using your formula) have a look at

m = Rest mass / (1 - 2 U)

Correction:

Rest mass = (Intrinsic mass) * sqrt(1 – 2 U)

A derivative, dm/dr yeilds the variation of Mercury's
orbit, that is to say, a rotation of the semi-major
axis (43"/century), also called the "relativitic force
supplement" by celestrial mechanics.

That is not correct. The tale of Mercury’s orbital anomaly cannot be
derived that way. shrug

Well, GR is only a few more steps away from
total disaster after that. shrug

Well GR math is still evolving, AE and team did a great
job, it's using the math in applications that gets trippy.

This is just not true. GR math was already done 100 years ago.
shrug

Forget about Einstein the nitwit, the plagiarist, and the liar. The
nitwit knew nothing and was very clueless. The derivation of the
field equations was not justified with any known physical insight. It
was the brainchild of Hilbert, Nordstrom, Levi-Civita, Ricci, and a
few others. shrug

I find one needs to be very careful using tensors analysis,
especially if CS conditions are imposed and tensor processes
are used.

There is nothing magical or mysterious about the tensor math. After
all, the basics ware already developed by Leibniz about 350 years
ago. shrug

That's oxymoronic as the processes apply to all
CS's, so even the introduction of a metric form (1,1,1,-1)
to begin with blows the process and generates artifacts.

What is that again? shrug

Just some thoughts on the subject I prepared for another occasion but
will post here for feedback and to add to convrsation and hopefully
understanding.

To ask, ‘what makes particles out of radiation?” and say “another
particle, the Higgs”, is only kicking the ball down the road, for one
would then have to ask well, “what makes the Higgs itself, a
particle?

The origin of mass is the origin of the universe itself sense all is
interrelated but one could localize it and say that the more local
origin of that aspect of the interrelationship that we call mass is
the speed of light barrier itself and that the origin of rest mass is
that same barrier all around or the speed of light squared which is
the speed of light in circular and or spherical rotation which is what
gives energy rest mass.
In equations E=hf/c^2 E or energy = h which is a constant = to c
around which f the variable frequency oscillates as if orbiting, /c^2
which is analogous to and = to r^2 on quantum level. So E=hf/c^2 = and
is analogous to F=mm/r^2
And so c=h are the constants or ground states around which all waves
and particles oscillate or orbit and c^2 = h/2pi is constant around
which rest mass oscillates. Both these sets of constants are necessary
for each others existence as no motion could exist without both
expansion and contraction or both matter measured as c^2 = h/2pi and
space measured as c=h.

Contrary to popular belief you are correct, “rest mass” is not
invariable, both because nothing in the universe is absolutely at
rest, and because rest mass, according to geometrical interpretation
of E=mc^2, is just kinetic energy/relative mass, of photonic energy,
or electromagnetic energy, in circular and or spherical motion.

But to extend that even further, just as “rest mass” emerges from
electromagnetic energy, which before had a constant velocity of “c” in
linear direction, to the speed of light in circular and or spherical
rotation, the speed of light in the linear direction likewise is not
constant, as it does transform to this circular and or spherical
rotation which is “rest mass”.

And so, if there is an invariant mass, it is “h” in equation E=hf/c^2
and h/2pi/2, which is also equal to “c”, and as such the invariance is
in its existence in both relative motion and relative stillness as a
stability in motion.

The “speed of light barrier”, is the rest frame of the universe,
creating the drag which gives rise to waves, when the constant speed
of light in straight line, is compressed against like a spring,
displacing energy into angular direction, creating waves.
Anything moving at constant speed of light in straight line, is
equivalent to being still, according to Newton’s first law of motion.
Anything that deviates from this is accelerated motion subject to
F=ma=mv^2 and therefore generates inertial/gravity mass, even as waves
of E=hf/c^2 and of course at E=hf=mc^2 which is circular and or
spherical rotation of electromagnetic energy giving it rest mass.

Conrad J Countess