Why does SRT and LET have identical Math???

jem wrote:
kenseto wrote:
"Sam Wormley" wrote in message
news:y4aoh.341693$1i1.296817@attbi_s72...

kenseto wrote:


There is no experimental support that B will predict that A's clock is
running slow.


Seto is wrong again!

Clocks A and B are separated by some non zero distance. There is a
non zero constant velocity between clocks A and B, such that |v| 0
and dv/dt = 0 .

The observer in the frame of the clock (A) measures the clock (B)

running

slow (time dilation as modeled by SR). Empirically so!

The observer in the frame of the clock (B) measures the clock (A)

running

slow (time dilation as modeled by SR). Empirically so!


Hey idiot assertion is not emperical.
GPS clock is B and ground clock is A:
From A's point of view: The SR effect on B as predicted by A is 7 us/day
running slow.
From B's point of view: The SR effect on A as predicted by B is 7 us/day
running fast.



SR's /reciprocal/ time dilation effect applies only in the case of
*Inertial* clocks. How many times have you been told this? 50? 100?
How many more times will you have to be told?

Since no frame on earth is inertial then SR must be wrong on earth. How
many time you have been told that??

kenseto wrote:

jem wrote:

kenseto wrote:

"Sam Wormley" wrote in message
news:y4aoh.341693$1i1.296817@attbi_s72...


kenseto wrote:



There is no experimental support that B will predict that A's clock is
running slow.


Seto is wrong again!

Clocks A and B are separated by some non zero distance. There is a
non zero constant velocity between clocks A and B, such that |v| 0
and dv/dt = 0 .

The observer in the frame of the clock (A) measures the clock (B)

running


slow (time dilation as modeled by SR). Empirically so!

The observer in the frame of the clock (B) measures the clock (A)

running


slow (time dilation as modeled by SR). Empirically so!


Hey idiot assertion is not emperical.
GPS clock is B and ground clock is A:
From A's point of view: The SR effect on B as predicted by A is 7 us/day
running slow.
From B's point of view: The SR effect on A as predicted by B is 7 us/day
running fast.



SR's /reciprocal/ time dilation effect applies only in the case of
*Inertial* clocks. How many times have you been told this? 50? 100?
How many more times will you have to be told?


Since no frame on earth is inertial then SR must be wrong on earth. How
many time you have been told that??

The theory itself (i.e. what it says) has nothing /whatsoever/ to do
with how well the theory duplicates the measurements it's intended to
predict. In SR, when a clock orbits an Inertial clock (an approximation
to the GPS situation), time dilation is not reciprocal.

So there's the first additional time you needed to be told.

"jem" wrote in message
...
kenseto wrote:

jem wrote:

kenseto wrote:

"Sam Wormley" wrote in message
news:y4aoh.341693$1i1.296817@attbi_s72...


kenseto wrote:



There is no experimental support that B will predict that A's clock
is
running slow.


Seto is wrong again!

Clocks A and B are separated by some non zero distance. There is a
non zero constant velocity between clocks A and B, such that |v| 0
and dv/dt = 0 .

The observer in the frame of the clock (A) measures the clock (B)

running


slow (time dilation as modeled by SR). Empirically so!

The observer in the frame of the clock (B) measures the clock (A)

running


slow (time dilation as modeled by SR). Empirically so!


Hey idiot assertion is not emperical.
GPS clock is B and ground clock is A:
From A's point of view: The SR effect on B as predicted by A is 7
us/day
running slow.
From B's point of view: The SR effect on A as predicted by B is 7
us/day
running fast.



SR's /reciprocal/ time dilation effect applies only in the case of
*Inertial* clocks. How many times have you been told this? 50? 100?
How many more times will you have to be told?


Since no frame on earth is inertial then SR must be wrong on earth. How
many time you have been told that??

The theory itself (i.e. what it says) has nothing /whatsoever/ to do
with how well the theory duplicates the measurements it's intended to
predict. In SR, when a clock orbits an Inertial clock (an approximation
to the GPS situation), time dilation is not reciprocal.

But you said that the GPS situation is due to that the orbiting clock is not
inertial. Since you now deny that so what is the cause that the SR time
dilation effect on the GPS is not reciprocal???

Ken Seto

kenseto wrote:

"jem" wrote in message
...

kenseto wrote:


jem wrote:


kenseto wrote:


"Sam Wormley" wrote in message
news:y4aoh.341693$1i1.296817@attbi_s72...



kenseto wrote:




There is no experimental support that B will predict that A's clock

is

running slow.


Seto is wrong again!

Clocks A and B are separated by some non zero distance. There is a
non zero constant velocity between clocks A and B, such that |v| 0
and dv/dt = 0 .

The observer in the frame of the clock (A) measures the clock (B)

running



slow (time dilation as modeled by SR). Empirically so!

The observer in the frame of the clock (B) measures the clock (A)

running



slow (time dilation as modeled by SR). Empirically so!


Hey idiot assertion is not emperical.
GPS clock is B and ground clock is A:

From A's point of view: The SR effect on B as predicted by A is 7

us/day

running slow.

From B's point of view: The SR effect on A as predicted by B is 7

us/day

running fast.



SR's /reciprocal/ time dilation effect applies only in the case of
*Inertial* clocks. How many times have you been told this? 50? 100?
How many more times will you have to be told?


Since no frame on earth is inertial then SR must be wrong on earth. How
many time you have been told that??

The theory itself (i.e. what it says) has nothing /whatsoever/ to do
with how well the theory duplicates the measurements it's intended to
predict. In SR, when a clock orbits an Inertial clock (an approximation
to the GPS situation), time dilation is not reciprocal.


But you said that the GPS situation is due to that the orbiting clock is not
inertial. Since you now deny that so what is the cause that the SR time
dilation effect on the GPS is not reciprocal???


In my example, the /orbiting/ clock is not Inertial - the /orbited/
clock is.

"jem" wrote in message
...
kenseto wrote:

"jem" wrote in message
...

kenseto wrote:


jem wrote:


kenseto wrote:


"Sam Wormley" wrote in message
news:y4aoh.341693$1i1.296817@attbi_s72...



kenseto wrote:




There is no experimental support that B will predict that A's clock

is

running slow.


Seto is wrong again!

Clocks A and B are separated by some non zero distance. There is a
non zero constant velocity between clocks A and B, such that |v|
0
and dv/dt = 0 .

The observer in the frame of the clock (A) measures the clock (B)

running



slow (time dilation as modeled by SR). Empirically so!

The observer in the frame of the clock (B) measures the clock (A)

running



slow (time dilation as modeled by SR). Empirically so!


Hey idiot assertion is not emperical.
GPS clock is B and ground clock is A:

From A's point of view: The SR effect on B as predicted by A is 7

us/day

running slow.

From B's point of view: The SR effect on A as predicted by B is 7

us/day

running fast.



SR's /reciprocal/ time dilation effect applies only in the case of
*Inertial* clocks. How many times have you been told this? 50? 100?
How many more times will you have to be told?


Since no frame on earth is inertial then SR must be wrong on earth. How
many time you have been told that??

The theory itself (i.e. what it says) has nothing /whatsoever/ to do
with how well the theory duplicates the measurements it's intended to
predict. In SR, when a clock orbits an Inertial clock (an approximation
to the GPS situation), time dilation is not reciprocal.


But you said that the GPS situation is due to that the orbiting clock is
not
inertial. Since you now deny that so what is the cause that the SR time
dilation effect on the GPS is not reciprocal???


In my example, the /orbiting/ clock is not Inertial - the /orbited/
clock is.

ROTFLOL.....so you can assign *inertial* and *not inertial* to a clock to
suit your theory?

In sci.physics.relativity, kenseto

wrote
on Tue, 9 Jan 2007 06:56:34 -0500
:

"The Ghost In The Machine" wrote in message
...
In sci.physics.relativity, kenseto

wrote
on Mon, 8 Jan 2007 13:04:41 -0500
:

"The Ghost In The Machine" wrote in
message
...
In sci.physics.relativity, kenseto

wrote
on Mon, 8 Jan 2007 11:19:30 -0500
:

"Sam Wormley" wrote in message
news:gatoh.342690$1i1.309253@attbi_s72...
kenseto wrote:


Hey idiot.....he selected the frame that sees all the clocks
moving
wrt
the
observer are running slow. That means that the observer is in a
preferred
frame.


No, Seto, no "preferred" frames required. Special relativity
predicts
that
clocks in relative motion wrt the observer will measure t' = t *
gamma.
This
is routinely observed.

ROTFLOL.....wormy doesn't even understand SR. In SR the observer will
measure t'=t/gamma.

You're both wrong. :-)

SR predicts t' = sqrt(1-v/c)/sqrt(1+v/c), since both sides are using
lightspeed Dopper measurements; there are no other methods of
measurement.

Sigh....you are incredibly stupid. Doppler measurement is not the rate
of an
observed clock.


It is as far as the observer's concerned. How else is he supposed to
know what the rate is, jump frames and whip out a slide rule? :-P

Hey idiot you don't measure the rate of a moving clock. You use the SR time
dilation to predict its rate.


That's not a measurement; that's a prediction.

--
#191,
Does anyone else remember the 1802?

--
Posted via a free Usenet account from http://www.teranews.com

"The Ghost In The Machine" wrote in message
...
In sci.physics.relativity, kenseto

wrote
on Tue, 9 Jan 2007 06:56:34 -0500
:

"The Ghost In The Machine" wrote in
message
...
In sci.physics.relativity, kenseto

wrote
on Mon, 8 Jan 2007 13:04:41 -0500
:

"The Ghost In The Machine" wrote in
message
...
In sci.physics.relativity, kenseto

wrote
on Mon, 8 Jan 2007 11:19:30 -0500
:

"Sam Wormley" wrote in message
news:gatoh.342690$1i1.309253@attbi_s72...
kenseto wrote:


Hey idiot.....he selected the frame that sees all the clocks
moving
wrt
the
observer are running slow. That means that the observer is in a
preferred
frame.


No, Seto, no "preferred" frames required. Special relativity
predicts
that
clocks in relative motion wrt the observer will measure t' = t
*
gamma.
This
is routinely observed.

ROTFLOL.....wormy doesn't even understand SR. In SR the observer
will
measure t'=t/gamma.

You're both wrong. :-)

SR predicts t' = sqrt(1-v/c)/sqrt(1+v/c), since both sides are using
lightspeed Dopper measurements; there are no other methods of
measurement.

Sigh....you are incredibly stupid. Doppler measurement is not the
rate
of an
observed clock.


It is as far as the observer's concerned. How else is he supposed to
know what the rate is, jump frames and whip out a slide rule? :-P

Hey idiot you don't measure the rate of a moving clock. You use the SR
time
dilation to predict its rate.


That's not a measurement; that's a prediction.

Right.....there is no way to measure the actual rate of a moving clock. You
can confirm your prediction after the clocks are reunited. Measuring the
doppler shift is not the actual rate of the clock. Why? Because the rate of
a moving clock doesn't change when it is in a state of uniform motion.

In sci.physics.relativity, kenseto

wrote
on Thu, 11 Jan 2007 08:55:51 -0500
:

"The Ghost In The Machine" wrote in message
...
In sci.physics.relativity, kenseto

wrote
on Tue, 9 Jan 2007 06:56:34 -0500
:

"The Ghost In The Machine" wrote in
message
...
In sci.physics.relativity, kenseto

wrote
on Mon, 8 Jan 2007 13:04:41 -0500
:

"The Ghost In The Machine" wrote in
message
...
In sci.physics.relativity, kenseto

wrote
on Mon, 8 Jan 2007 11:19:30 -0500
:

"Sam Wormley" wrote in message
news:gatoh.342690$1i1.309253@attbi_s72...
kenseto wrote:


Hey idiot.....he selected the frame that sees all the clocks
moving
wrt
the
observer are running slow. That means that the observer is in a
preferred
frame.


No, Seto, no "preferred" frames required. Special relativity
predicts
that
clocks in relative motion wrt the observer will measure t' = t
*
gamma.
This
is routinely observed.

ROTFLOL.....wormy doesn't even understand SR. In SR the observer
will
measure t'=t/gamma.

You're both wrong. :-)

SR predicts t' = sqrt(1-v/c)/sqrt(1+v/c), since both sides are using
lightspeed Dopper measurements; there are no other methods of
measurement.

Sigh....you are incredibly stupid. Doppler measurement is not the
rate
of an
observed clock.


It is as far as the observer's concerned. How else is he supposed to
know what the rate is, jump frames and whip out a slide rule? :-P

Hey idiot you don't measure the rate of a moving clock. You use the SR
time
dilation to predict its rate.


That's not a measurement; that's a prediction.

Right.....there is no way to measure the actual rate of a moving clock. You
can confirm your prediction after the clocks are reunited. Measuring the
doppler shift is not the actual rate of the clock. Why? Because the rate of
a moving clock doesn't change when it is in a state of uniform motion.


As opposed to what, a state of non-motion? Be very precise here, and
include the acceleration:

[1] the clock as measured by someone by its elbow, as the clock
accelerates from 0 m/s (relative to the "stationary" clock) to the final
velocity).

[2] the "stationary" clock measurements of the moving clocks' light
signals, a combination of Doppler effect and SR gamma correction.

[3] the calculated value of the moving clocks' time ticks, correcting
for the Doppler effect.

My computations suggest that [1] indicates "no change",
[2] will indicate a clock speedup or slowdown because of
the Doppler (though for most problems of this type it's
a slowdown), and [3] will always indicate a slowdown in SR,
if done properly.

--
#191,
fortune: not found

--
Posted via a free Usenet account from http://www.teranews.com

"The Ghost In The Machine" wrote in message
news
In sci.physics.relativity, kenseto

wrote
on Thu, 11 Jan 2007 08:55:51 -0500
:

"The Ghost In The Machine" wrote in
message
...
In sci.physics.relativity, kenseto

wrote
on Tue, 9 Jan 2007 06:56:34 -0500
:

"The Ghost In The Machine" wrote in
message
...
In sci.physics.relativity, kenseto

wrote
on Mon, 8 Jan 2007 13:04:41 -0500
:

"The Ghost In The Machine" wrote
in
message
...
In sci.physics.relativity, kenseto

wrote
on Mon, 8 Jan 2007 11:19:30 -0500
:

"Sam Wormley" wrote in message
news:gatoh.342690$1i1.309253@attbi_s72...
kenseto wrote:


Hey idiot.....he selected the frame that sees all the clocks
moving
wrt
the
observer are running slow. That means that the observer is
in a
preferred
frame.


No, Seto, no "preferred" frames required. Special
relativity
predicts
that
clocks in relative motion wrt the observer will measure t'
= t
*
gamma.
This
is routinely observed.

ROTFLOL.....wormy doesn't even understand SR. In SR the
observer
will
measure t'=t/gamma.

You're both wrong. :-)

SR predicts t' = sqrt(1-v/c)/sqrt(1+v/c), since both sides are
using
lightspeed Dopper measurements; there are no other methods of
measurement.

Sigh....you are incredibly stupid. Doppler measurement is not the
rate
of an
observed clock.


It is as far as the observer's concerned. How else is he supposed
to
know what the rate is, jump frames and whip out a slide rule? :-P

Hey idiot you don't measure the rate of a moving clock. You use the
SR
time
dilation to predict its rate.


That's not a measurement; that's a prediction.

Right.....there is no way to measure the actual rate of a moving clock.
You
can confirm your prediction after the clocks are reunited. Measuring the
doppler shift is not the actual rate of the clock. Why? Because the rate
of
a moving clock doesn't change when it is in a state of uniform motion.


As opposed to what, a state of non-motion? Be very precise here, and
include the acceleration:

[1] the clock as measured by someone by its elbow, as the clock
accelerates from 0 m/s (relative to the "stationary" clock) to the final
velocity).

The rate of the accelerated clcok would run slower than the stay at home
clock.

[2] the "stationary" clock measurements of the moving clocks' light
signals, a combination of Doppler effect and SR gamma correction.

Doppler effect is not the rate of the observed clock.

[3] the calculated value of the moving clocks' time ticks, correcting
for the Doppler effect.

The rate of an observed clcok does not change if it is in uniform motion (no
acceleration).

Ken Seto

My computations suggest that [1] indicates "no change",
[2] will indicate a clock speedup or slowdown because of
the Doppler (though for most problems of this type it's
a slowdown), and [3] will always indicate a slowdown in SR,
if done properly.

--
#191,
fortune: not found

--
Posted via a free Usenet account from http://www.teranews.com