I have to walk away from

m c'^2 = m c^2 + 2 m U

Energy equating equation has to be independent of the Aether stress
equation.

However,

c'^2 = c^2 + 2 U, the Aether Stress Equation, or

c = c' sqrt(1 - 2 U / c'^2)

Is a still a keeper until proven wrong.

* * *

"Australopithecus Afarensis" wrote in message
news:UJY0b.9805$cj1.5657@fed1read06...
"Start with two identical clocks on the surface of the earth in a valley,
and set them to the same value; slowly transport one to a neighboring
mountain. Leave them there for a long time, and then slowly transport the
second clock to the mountain (in an identical manner to the first). The two
clocks no longer display the same time (the earlier one displays a later
time than the later one)."

Is some one forgetting about E = m c^2? c^2 would certainly affect the
Cesium clock!

You have raised a very interesting point. The time non-dilation due to
atomic clocks can be explained with the equation:

m c'^2 = m c^2 + 2 m U

For a complete derivation, see the 1st attached message below.

However, there is also a real time dilation due to speed.

The equation of time dilation is:

dt'/dt = 1 / sqrt(1 - (2 pi H / (D c))^2), where

dt' = time observed at our level
dt = time observed at altitude H above us
D = time in a sideral day
c = speed of light
pi = 3.1415926

Time dilation due to raising and finally lowering this clock is down in the
noise level, of course. If you want me to show you how to derive this, I am
happy to do so.

Why is the Principle of Relativity coming into play? See the 2nd attached
message of Twin's Paradox.

----- Original Message -----
From: "Australopithecus Afarensis"
Newsgroups: sci.physics.relativity
Sent: Tuesday, August 19, 2003 09:13 PM
Subject: None Occurring Time Dilation in Gravity


Consider the world famous Minkowski Space-Time Diagram,

(c dt)^2 = (c' dt')^2 - (dr')^2 = (c' dt')^2 - (dr'/dt')^2 (dt')^2, where

c = local. speed of light under gravity
c' = speed of light very far away
t = local time observed under gravity
t' = time observed very far way
r = local length observed under gravity
r' = length observed very far away

c^2 = c'^2 (dt'/dt)^2 - (dr'/dt')^2 (dt'/dt)^2

Since all time dilation of gravity all came down on the speed issues, I am
not convinced that gravity does cause time nor length dilations. Time and
length dilations are manifestation of Relativity (based on speed alone). If
an object is not observed to move at any speed, it is not under time
dilation even with gravity applied to it. Therefore,

dt = dt', dr = dr', or

c^2 = c'^2 - (dr/dt)^2

Differentiate both sides,

2 c (dc/dr) (dr/dt) = - 2 (dr/dt) (d^2r/dt^2) = - 2 (dr/dt) (dU/dr), where

U = gravitational potential, where
U = G M / r if r 2 G M / c'^2, where
G = gravitational constant
M = mass caused this gravitation in which
r = 0 is the center of M, therefore

2 c dc = - 2 (dU/dr) dr = - 2 dU

Solve this very, very simple differential equation,

c^2 = K - 2 U, where

K = integration constant

At r = oo, U = 0, c = c', therefore K = c'^2

c^2 = c'^2 - 2 U, or

c(r) = c' sqrt(1 - 2 U(r) / c'^2), or

m c'^2 = m c^2 + 2 m U, where

m = mass

What does it all mean? Matter that has mass causes distortion in this
Aether which 100 years the whole scientific community following the cult of
Einstein turned their backs on. The distortion is in the increase in the
permittivity of free space around it. That caused the speed of light to
decrease (c = 1 / sqrt(permittivity * permeability)). GRAVITY IS CAUSED BY
THE STRESS IN THE AETHER (GRADIENT in the PERMITTIVITY) and not by the
curved space mumbo jumbo proposed by the General Theory of Relativity in
which it even fails to explain why matter falls into this curved space in
the first place.

We can then come up with a U that fits observation, such as

** Star refraction under solar eclipse
** Precesion of perihelion of Mercury

Because the manefistation of this stress in the Aether is perceived as
gravity.

----- Original Message -----
From: "Australopithecus Afarensis"
Newsgroups: sci.physics.relativity
Sent: Monday, August 04, 2003 05:41 PM
Subject: The Traveling Twin: SR's ******* child.


By studying the Twins Paradox in a more detailed fashion, one can prove
whether if Einstein's or Poincare/Lorentz's interpretation to the theory of
Relativity is the correct one.

Either of these two interpretations tells us that each twin would observe
the other one's time slowing down. As long as they don't ever meet again,
there is no problem. What if they do meet with no relative speed between
them? Whose time is the correct one? The problem can be solved based on
either twin's point of view. However, if we focus on the traveling twin's
point of view, the problem becomes tremendously easier.

Start with F(t) = dp(t)/dt = m d/dt[v(t) (1 - v(t)^2 / c^2)^-1/2], where

F(t) = thrust of the starship
p(t) = momentum of the starship
m = rest mass of the starship
v(t) = speed of the starship
t = time observed by the traveling twin
c = speed of light

Let's also say m can deliver thrust F(t) without loosing any or negligible
mass. Then,

F(t) = m (1 - v(t)^2 / c^2)^-3/2 dv(t)/dt

Multiply both sides by v(t),

P = F(t) v(t) = m v(t) (1 - v(t)^2 / c^2)^-3/2 dv(t)/dt, where

P = power = constant = measure of twins' technology

Bring dt to the other side,

P dt = m v(t) (1 - v(t)^2 / c^)^-3/2 dv(t)

Solve this differential equation,

Kn + Pn t / (m c^2) = (1 - v(t)^2 / c^2)^-1/2, where

Pn = power delivered in the n'th phase = P, -P, or 0
Kn = integration constant for the n'th phase

There are 6 phases in the traveling twin's adventu t1, t2, t3, t4, t5,
t6 correspond to power delivered by the starship P1, P2, P3, P4, P5, P6,
respectively.

t1 = powering away with P1 = P from speed 0 to V
t2 = cruising away with P2 = 0 at speed V
t3 = powering away with P3 = -P from speed V to 0
t4 = powering home with P4 = -P from speed 0 to -V
t5 = cruising home with P5 = 0 at speed -V
t6 = powering home with P6 = P from speed -V to 0, then

V = maximum speed of the traveling twin in the whole trip

Similarly the twin at home also has these 6 phases of these events t1', t2',
t3', t4', t5', t6'.

According to Lorentz Transform,

tn' = integrate from (tn-1 to tn-1 + tn) of [(1 - v(t)^2 / c^2)^-1/2 dt],
where

tn = t1, t2, t3, t4, t5, t6 with t0 = 0
tn' = t1', t2', t3', t4', t5', t6'

Recall Kn + Pn t / (m c^2) = (1 - v(t)^2 / c^2)^-1/2, so

tn' = integrate from (tn-1 to tn-1 + tn) of {[Kn + Pn t / (m c^2)] dt}, or

tn' = tn [Kn + Pn (tn + 2 tn-1) / (2 m c^2)]

During phase t1 or 0 = t = t1, P1 = P,

K1 = 1 because v(0) = 0, so

t1' = t1 [1 + P t1 / (2 m c^2)]

Since v(t1) = V,

(1 - V^2 / c^2)^-1/2 = 1 + P t1 / (m c^2)

During interval t2 or t1 = t = t1 + t2, P2 = 0,

K2 = (1 - V^2 / c^2)^-1/2 = 1 + P t1 / (m c^2), so

t2' = t2 [1 + P t1 / (m c^2)]

During interval t3 or t1 + t2 = t = t1 + t2 + t3, P3 = -P,

(1 - V^2 / c^2)^-1/2 = K3 - P (t1 + t2) / (m c^2) at t = t1 + t2, or

1 + P t1 / (m c^2) = K3 - P (t1 + t2) / (m c^2), so

K3 = 1 + P (2 t1 + t2) / (m c^2)

Since v(t1 + t2 + t3) = 0,

1 = 1 + P (2 t1 + t2) / (m c^2) - P (t1 + t2 + t3) / (m c^2), or

t1 = t3, so

Since t1 = t3, then

t1 = t3 = t4 = t6 and t1' = t3' = t4' = t6' (warping intervals) as well as

t2 = t5 and t2' = t5' (coasting intervals)

t1' + t2' + t3' + t4' +t5' + t6' = 4 t1' + 2 t2'

4 t1' + 2 t2' = 4 t1 [1 + P t1 / (2 m c^2)] + 2 t2 [1 + P t1 / (m c^2)]

Define

Tg = 4 t1 = total time of powering experienced by the traveling twin
Tc = 2 t2 = total time of cruising experienced by the traveling twin
To = 4 t1' + 2 t2' = total time observed by the twin at home, then

To = Tg [1 + P Tg / (8 m c^2)] + Tc [1 + P Tg / (4 m c^)], or

When the traveling twin re-unites with his twin at home, the traveling twin
would be younger by:

To - (Tg + Tc) = P Tg (Tg + 2 Tc) / (8 m c^2)

Since the clock was synchronized before the traveling twin goes on for that
trip, there is no need to pay special attention to synchronize their clocks.
As you can see, the cruising time (when the starship is not accelerating)
does also play into the time dilation effect when the twins re-unite again
at the end. It make a difference to know who is actually doing the
traveling. Acceleration gives it away, or actually applying energy to one
of them gives it away. Notice the effect of time dilation due to the
cruising time is amplified by how long the traveling twin is under constant
power of thrust. Therefore, the absolute reference must exist some where.
That would mean Einstein's interpretation is dead wrong, and
Poincare/Lorentz's interpretation to the theory of Relativity is the correct
one after all.

* * *

Since traditional studies use constant g = F / m as the propulsion of choice
instead of constant power which is more accurate to describe the motion of
any vehicles including starships. In this case,

F = dp(t)/dt = m d/dt[v(t) (1 - v(t)^2 / c^2)^-1/2]

g = F / m = d/dt[v(t) (1 - v(t)^2 / c^2)^-1/2], or

g t = v(t) (1 - v(t)^2 / c^2)^-1/2, or

v(t) = g t / [1 + (g t / c)^2], or

(1 - v(t)^2 / c^2)^-1/2 = [1 + (g t / c)]^-1/2

Apply Lorentz Transform,

dt' = (1 - v(t)^2 / c^2)^-1/2 dt = [1 + (g t / c)^2]^1/2 dt

t' = (c / 2g) {arcsinh(g t / c) + (g t / c) [1 + (g t / c)^2]^1/2}

The traveling twin will be younger by

(2c / g) arcsinh( g Tg / 4c) + (Tc + Tg / 2) [1 + (g Tg / 4c)^2]^-1/2} -
Tg - Tc

If I plug in 4.5 billion years (4.5E+09) of earth history, I get a time
dilation of 1.0E+19 years as seen by any empty space of 0 g. It is
obviously not correct to say g = F / m in general.