"Sylvia Else" wrote in message u...
| Sorcerer wrote:
| "Sylvia Else" wrote in message ...
| | Sorcerer wrote:
| | "Sylvia Else" wrote in message u...
| | | Sorcerer wrote:
| | |
| | |
| | | Oh, so the stay-at-home triplet has Werner syndrome, or do all
| | | relativist dorks suffer from it?
| | |
| | | The stay at home triplet ages more than the travelling siblings,
| |
| | Oh really? Why? Werner syndrome perhaps, or some crazy
| | mathematics you do not understand? Or are you a crazy troll too?
| |
| |
| | The reason the stay at home person ages more in this situation has been
| | adequately explained ad nauseam.
|
| Only by ****wits.
| http://www.androcles01.pwp.blueyonde...minoEffect.GIF
| http://www.androcles01.pwp.blueyonde...winParadox.htm
| http://www.androcles01.pwp.blueyonde...ket/Rocket.htm
| http://www.androcles01.pwp.blueyonde...mart/Smart.htm
| http://www.androcles01.pwp.blueyonde...lgol/Algol.htm
| http://www.androcles01.pwp.blueyonder.co.uk/GPS/GPS.htm
| http://www.androcles01.pwp.blueyonde.../Clockgain.PNG
| http://www.androcles01.pwp.blueyonder.co.uk/E^2/EnergySquare.htm
| http://www.androcles01.pwp.blueyonde...er/Doppler.htm
| http://www.androcles01.pwp.blueyonde...lina/Drive.htm
| http://www.androcles01.pwp.blueyonde...nac/Sagnac.htm
| http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.htm
| http://www.androcles01.pwp.blueyonde...endy/Wendy.htm
|
|
|
| Why are you so angry about this? Worried about your inheritance?
|
| Sylvia.


Why are you a ****head? Eager for your inheritance, dork?

Sylvia Else yazdi:
maxwell wrote:

Hello Ikoluk:
Liked your paper, thanks for the reference & discussion of Einstein
trying to resolve his own paradox in 1918 of his 1905 problem with his
1916 'solution'. Incidentally, since Einstein spent the rest of his
life trying AND FAILING to reconcile his Special Theory (based on
Maxwell's electromagnetism) & his theory of gravity, why do people
continue to conjoin these irreconcilable theories? The only thing they
have in common is the buzzword 'relativity'. Even the String theorists
& Gravitational 'foam' gang continue to fail in this attempt at 'Grand
Unification.
Anyway, back to the Twins. Herbert Dingle in 1962 pointed out that one
can readily set up a totally symmetric twin situation where each
departs in opposite directions for an equally long time at very high
equal speed & then reverse (using the rocket thrusters in reverse which
they used to accelerate) & return to the starting point in space. Each
twin, with a telescope, will see the other's clock going slow through
both legs of the journey (the turnaround is unknown but is equal & can
be made insignificant, like the initial acceleration time). The only
acceptable symmetric solution (as this is a symmetric problem) is that
both have aged equally - so at the point of meeting they will agree
that they were viewing an 'optical illusion' through their telescopes.
The only remaining question is: do the people on Earth, who never went
on the journey, also age at the same rate?

On the outgoing journey, each twin will *see* the other's clock running
slowly. This a combination of the relativistic slowing of the other
clock, together with the red shift effect.

On the return journey, each twin will *see* the other's clock running
fast. This is a combination of the relativistic slowing of the other
clock and the blue shift effect, which more than offsets the slowing.

The net effects of the rates *seen* is that when the twins again meet,
they are unsurprised to find that their clocks are again synchronised.
There no optical illusions involved.

During each phase of the journey, if twin A calculates the rate of twin
B's clock, based on A's observations, allowing for the red shift, or
blue shift, then A will conclude that B's clock is running slowly.

A can also calculate what time is shown on B's clock at any instant in
A's frame of reference. During the outgoing leg, A will find that B's
clock is running behind A's (and getting more so, since it's running
slowly). During the ingoing leg, A will find that B's clock is running
ahead of A's (but getting less so since it's running slowly). At the
point where A changes direction he will note that the calculation of
what B's clock shows at that instant jumps forward abruptly. This does
not imply that anything unusual happens to B's clock, but just reflects
that the change of direction affects the meaning of "that instant" at a
place remote from A.

The same applies when B considers A.

There is no paradox here.

Sylvia.

The clock rate doesn't depend on red shift/blue shift. Let twin B's
clock send a light pulse at the end of each clock. If it sends n light
pulses, the twin A will observe exactly n ligth pulses whatever the
value of distance is, whatever the value of speed is(provided it is
less than c) and whatever the value of the frequency shift factor is.

However, while I was thinking about this I have noticed something: time
dilation and clock rate dilation are not the same thing! Each tick in a
clock is an event and an event's observed time can be different from
time dilation. For example one can set a clock by using a light pulse
and two mirrors. The pulse is reflected between the mirrors and the
time interval between the reflection times of mirror 1 can be
considered as one tick of this clock. If the light speed is source
dependent then the duration of each tick is the same regardless of the
speed of the clock and the time delation. I will explain this in the
other thread "relativity vs velocity addition". However, in SR this
time depends on the speed. Let the speed of the clock is v. Then the
total time would be

t=x'/(c-v)+x'/(c+v)=x'.2.c/(c^2-v^2)

Since x'=x.sqrt(1-v^2/c^2) it becomes

t=2.x/sqrt(c^2-v^2)

which is direction independent. There is a version of the symmetric
twin paradox where the paradox can only be solved only if we assume the
clock rate is the same in all inertial frames. This time the twin's
clocks can be started/stopped by a signal send by the observer C. After
some time the twin leaves the earth, C broadcast a start signal to
them. After a time interval t1, he/she send another signal to stop the
clocks of both twin. This occurs during outbound part of the journey. C
repeats this process during the inbound half. Let this time the time
interval be t2. After the end of experiment, the twins compare their
total time ticks. Since the cases are symmetric again, their total
ticks must be the same. From the point of view of twin A, his/her clock
and twin B's clock starts and stops at the same time because their
distances to the observer C are the same at all times. Since t1 and t2
are two arbitrary values and their inbound and outbound speeds can be
different, the only explanation for the correct result is that twin B's
clock rate is the same with his/her clock rate. So we can derive the
following result:

"The clock rates are the same in all inertial frames whatever the type
of clocks are"

This is actually againts my claim that the dilation factor should be
direction dependent and makes my speed formula that I have derived in
the other threat useless.

Lokman Kolukisa

wrote:

Sylvia Else yazdi:

maxwell wrote:


Hello Ikoluk:
Liked your paper, thanks for the reference & discussion of Einstein
trying to resolve his own paradox in 1918 of his 1905 problem with his
1916 'solution'. Incidentally, since Einstein spent the rest of his
life trying AND FAILING to reconcile his Special Theory (based on
Maxwell's electromagnetism) & his theory of gravity, why do people
continue to conjoin these irreconcilable theories? The only thing they
have in common is the buzzword 'relativity'. Even the String theorists
& Gravitational 'foam' gang continue to fail in this attempt at 'Grand
Unification.
Anyway, back to the Twins. Herbert Dingle in 1962 pointed out that one
can readily set up a totally symmetric twin situation where each
departs in opposite directions for an equally long time at very high
equal speed & then reverse (using the rocket thrusters in reverse which
they used to accelerate) & return to the starting point in space. Each
twin, with a telescope, will see the other's clock going slow through
both legs of the journey (the turnaround is unknown but is equal & can
be made insignificant, like the initial acceleration time). The only
acceptable symmetric solution (as this is a symmetric problem) is that
both have aged equally - so at the point of meeting they will agree
that they were viewing an 'optical illusion' through their telescopes.
The only remaining question is: do the people on Earth, who never went
on the journey, also age at the same rate?

On the outgoing journey, each twin will *see* the other's clock running
slowly. This a combination of the relativistic slowing of the other
clock, together with the red shift effect.

On the return journey, each twin will *see* the other's clock running
fast. This is a combination of the relativistic slowing of the other
clock and the blue shift effect, which more than offsets the slowing.

The net effects of the rates *seen* is that when the twins again meet,
they are unsurprised to find that their clocks are again synchronised.
There no optical illusions involved.

During each phase of the journey, if twin A calculates the rate of twin
B's clock, based on A's observations, allowing for the red shift, or
blue shift, then A will conclude that B's clock is running slowly.

A can also calculate what time is shown on B's clock at any instant in
A's frame of reference. During the outgoing leg, A will find that B's
clock is running behind A's (and getting more so, since it's running
slowly). During the ingoing leg, A will find that B's clock is running
ahead of A's (but getting less so since it's running slowly). At the
point where A changes direction he will note that the calculation of
what B's clock shows at that instant jumps forward abruptly. This does
not imply that anything unusual happens to B's clock, but just reflects
that the change of direction affects the meaning of "that instant" at a
place remote from A.

The same applies when B considers A.

There is no paradox here.

Sylvia.


The clock rate doesn't depend on red shift/blue shift

What the observer observes most definitely does, and a correct theory
only has to describe the totality of the observations.

Sylvia.