Does gravitation really cause time dilation? Or does it cause the speed of
light to slow down instead?

Consider the world famous Minkowski Space-Time Diagram,

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

c = speed of light, locally
c' = speed of light far away from a mass
dt = time observed under gravitation
dt' = time observed far way
dr = local space under gravitation

Since my assumption is that dt = dt' because dr/dt is close to zero. There
should not be any time dilation.

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

Differentiate both sides,

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

2 c dc = - 2 (- G M / r^2) dr = 2 G M / r^2, where

G = gravitational constant
M = mass of gravitation
r = distance from center of M

Solve this very simple differential equation,

c^2 = K - 2 G M / r, where K = integration constant

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

c(r) = c' sqrt(1 - 2 G M / r)

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 proposed
by the General Theory of Relativity in which it fails to explain why matter
even falls into this curved space in the first place.

What about the gravitational red shift? It should be not different from
what we have already known.

h f - G M m / r = m c'^2, where

h = Planck's constant
m = mass of the test photon

f = (m c'^2 / h) (1 + G M / (r c"^2))

Since h f' = m c'^2, where

f' = frequency at r = oo

f = f'(1 + G M / (r c^2))

Yes, we still get our gravitational red shift.

How about the refraction of stars near the sun? Since the speed of light is
lower, it would cause additional 'bending' of light on top of the classical
gravitational effect on the photon. The experiment has a very large error
band anyway.

How about the Mercury's orbit? I don't know. Shall we say it is under
construction.

Dear Australopithecus Afarensis:

"Australopithecus Afarensis" wrote in message
news:sUh0b.7960$cj1.5338@fed1read06...
....
Does gravitation really cause time dilation?

Being located in spaces of different curvature show different rates of time
passage. Like on the surface of the Earth vs. in GPS orbit.

Or does it cause the speed of
light to slow down instead?

c is c for any local frame. c =/= c between frames in curved space

David A. Smith

The speed of light, as measured with upper elevation units of measurement,
slows down as elevation is reduced. This, in turn causes the units of
measurement for time to increase (clocks run slower). There is no such thing as
curved space. That idea is a bit of scientific fakery used to overcome the fact
that a childish mathematical error in the generation of GR invention made it
impossible to solve the mathematics.

Actually, this can all be generalized.

* * *

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.

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.

"Australopithecus Afarensis" wrote in message news:sUh0b.7960$cj1.5338@fed1read06...
Does gravitation really cause time dilation? Or does it cause the speed of
light to slow down instead?

[snip]

What about the gravitational red shift? It should be not different from
what we have already known.

h f - G M m / r = m c'^2, where

h = Planck's constant
m = mass of the test photon

f = (m c'^2 / h) (1 + G M / (r c"^2))

Since h f' = m c'^2, where

f' = frequency at r = oo

f = f'(1 + G M / (r c^2))

Yes, we still get our gravitational red shift.

[snip]


If you were to fire cannon balls into space from a gravity well, the
balls would slow down, but they would also get closer together as they
left the gravitational field. So their frequency would still be the
same in terms of cannon balls per second. So how would light be any
different? How would slowing light down change its frequency?

On the other hand, gravitational red shifting can be very nicely
explained if the light is originally emitted at a lower frequency
because of a time dilated emitter.

Double-A

This is the escape velocity problem. A particle with some energy has to
lose its potential energy to escape. This is also an orbital mechanics
problem.

E = (m - m0) c^2 - m U, where

E = total energy
m = relativistic mass
m0 = rest mass
c = speed of light
U = gravitational potential

If (m U) ((m - m0) c^2), then the particle is trapped. It would have an
orbit around the central mass that gives rise to the gravitational
potential.

If (m U) = ((m - m0) c^2), then the particle can escape out the
gravitational system.

A photon is no different. So,

E = m c^2 - m U since m0 = 0 for a photon

According to Planck,

Energy = h f = m c^2, where

h = Planck's constant
f = frequency of the photon

So a photon with energy of (m c^2 = h f) has to shed potential energy (m U)
to escape. Say

f ' = frequency of the photon after escape
m' = mass of the photon after escape, so

h f ' = m c^2 - m U = h f - m U, with

m = h f / c^2, solve for f / f '

f / f ' = 1 - U / c^2,

Since f / f ' 1, there would be a red shift. The gravitational red shift
is an energy conservation phenomenon not a time dilation one. The result is
similar to Doppler Effect.

I made a mistake in the previous post. So, I will correct it right here.

Let's say

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

c' = speed of light after escape

The mass of the photon would pick faster into the gravitational potential
than under c'^2 = c^2. That's all. So,

f / f ' = (1 - 3 U / c'^2) / (1 - 2 U / c'^2)

If gravitatioanl red shift is caused by time dilation due to gravity, is
energy also conserved?

----- Original Message -----
From: "Double-A"
Newsgroups: sci.physics.relativity
Sent: Monday, September 01, 2003 10:55 AM
Subject: None Occurring Time Dilation in Gravity
" Does gravitation really cause time dilation? Or does it cause the speed
of
light to slow down instead?

[snip]

What about the gravitational red shift? It should be not different from
what we have already known.

h f - G M m / r = m c'^2, where

h = Planck's constant
m = mass of the test photon

f = (m c'^2 / h) (1 + G M / (r c"^2))

Since h f' = m c'^2, where

f' = frequency at r = oo

f = f'(1 + G M / (r c^2))

Yes, we still get our gravitational red shift.

[snip]

If you were to fire cannon balls into space from a gravity well, the
balls would slow down, but they would also get closer together as they
left the gravitational field. So their frequency would still be the
same in terms of cannon balls per second. So how would light be any
different? How would slowing light down change its frequency?

On the other hand, gravitational red shifting can be very nicely
explained if the light is originally emitted at a lower frequency
because of a time dilated emitter.

Double-A

"Australopithecus Afarensis"
wrote in message 6MO4b.36321$cj1.6257@fed1read06

If gravitational red shift is caused by time dilation due to
gravity, is energy also conserved?
Of course (energy is always conserved, otherwise the
nature would have to know how to create energy,
which at least has been never observed).

For photons conservation of energy is trivial since they
don't have any rest mass and so they always have
the same energy hf (however because of time dilation
at the emitter it looks like different energy to different
observers, as any kinetic energy).

It is a little bit more tricky for those objects that have
rest mass. Then the change in speed of light keeps
the energy constant too. BTW, except of the change of
speed of light there is also time dilation (that causes
"gravitational acceleration" according to
dT/dt=1+gx/c^2, where T is proper time, t is oordinate
time) and space curvature too (hoe it does it is easy
to derive so I skip it). Without time dilation and space
curvature (or at least one) the speed of light cloudn't
change (since c=x/t), and without both effects in right
proportion (namely 50/50) the energy couldn't be
conserved. So while presence of masses slows down
the rate of time, it also increases amount of space
around masses since otherwise (slowing of time alone)
would create/destroy energy from/to nothing, which
nature can't do: if only time changed, the change of c
as proportional to time would make energy proportional
to the square of the change of time (mc^2), so small
change of space is needed to keep energy changing
proportionally to the time dilation. I let you have fun
of deriving the proper equations.

-- Jim

You are correct about the photon's role in conservation of energy is rather
trivial. However, all solid and good theory must incorporate both the
photons and non-photons equally with great effectiveness. So, from the
corrected orbital mechanics equation,

E = (m - m0) c^2 - m0 U,

m0 = 0 because a photon does not have a rest mass.

E = h f ' = m c^2 = h f c^2 / c'^2, because

It is necessary to modify the Planck's quantum energy equation from just
plain old E = h f to E = h f c^2 / c'^2.

In this case, the gravitational red shift with the constraint,

c'^2 = c^2 + 2 U, is given by

f = f ' / (1 - 2 U / c"^2), which is

In agreement with General Relativity without the introduction of time
dilation.

Do I have fun deriving this? Yes, I do since I have no other help other
than myself. Oh, I can thank my dog, Kublai. I came up with all this stuff
while walking with Kublai because it is the only time I can concentrate on
something totally irrelevant to my work or my everyday life.

----- Original Message -----
From: "Jim Jastrzebski"
Newsgroups: sci.physics.relativity
Sent: Monday, September 01, 2003 08:11 PM
Subject: None Occurring Time Dilation in Gravity

"Australopithecus Afarensis"
wrote in message 6MO4b.36321$cj1.6257@fed1read06

If gravitational red shift is caused by time dilation due to
gravity, is energy also conserved?
Of course (energy is always conserved, otherwise the
nature would have to know how to create energy,
which at least has been never observed).

For photons conservation of energy is trivial since they
don't have any rest mass and so they always have
the same energy hf (however because of time dilation
at the emitter it looks like different energy to different
observers, as any kinetic energy).

It is a little bit more tricky for those objects that have
rest mass. Then the change in speed of light keeps
the energy constant too. BTW, except of the change of
speed of light there is also time dilation (that causes
"gravitational acceleration" according to
dT/dt=1+gx/c^2, where T is proper time, t is oordinate
time) and space curvature too (hoe it does it is easy
to derive so I skip it). Without time dilation and space
curvature (or at least one) the speed of light cloudn't
change (since c=x/t), and without both effects in right
proportion (namely 50/50) the energy couldn't be
conserved. So while presence of masses slows down
the rate of time, it also increases amount of space
around masses since otherwise (slowing of time alone)
would create/destroy energy from/to nothing, which
nature can't do: if only time changed, the change of c
as proportional to time would make energy proportional
to the square of the change of time (mc^2), so small
change of space is needed to keep energy changing
proportionally to the time dilation. I let you have fun
of deriving the proper equations.

-- Jim