All Einsteinians, both profane and initiated, should immediately take
their heads out of the sand and resolve the following problem. In
Chapter 23 in his "Relativity" Einstein claims that a clock at rest
undergoes time contraction, that is, runs fast by a factor of gamma
relative to a clock situated at the edge of a rotating disk. The
Juggler also claims that this follows from Lorentz transforms. However
Lorentz transforms predict no time contraction - rather, the only time
distortion they predict is symmetrical time dilation for two inertial
systems. So Einsteinians should prove that their god is not a Juggler
and derive, RIGOROUSLY, the time contraction factor (gamma) from
Lorentz transforms.

Pentcho Valev

"Pentcho Valev" wrote in message ups.com...
All Einsteinians, both profane and initiated, should immediately take
their heads out of the sand and resolve the following problem. In
Chapter 23 in his "Relativity" Einstein claims that a clock at rest
undergoes time contraction, that is, runs fast by a factor of gamma
relative to a clock situated at the edge of a rotating disk. The
Juggler also claims that this follows from Lorentz transforms. However
Lorentz transforms predict no time contraction - rather, the only time
distortion they predict is symmetrical time dilation for two inertial
systems. So Einsteinians should prove that their god is not a Juggler
and derive, RIGOROUSLY, the time contraction factor (gamma) from
Lorentz transforms.

Since this has been explained to you before (at least
10 times), you can stop reading here.

For the others:

The symmetry of the Lorentz transformation exists
between two inertially moving clocks. The clock at the
edge of the disk is not inertial.

If you want to know the total elapsed time of a non
inertially moving clock, you must integrate.
The total proper time T of a non-inertially moving clock
as calculated form the point of view of an inertial clock is
given by
T = Integral{ t1...t2; sqrt( 1 - [v(t)/c]^2 ) dt }
where v(t) is the speed of the non-inertial clock as a
function of the time t on the inertial clock.
You immediately see that
T = t2 - t1

You can *not* turn this around to something like
t = Integral{ T1...T2; sqrt( 1 - [v(T)/c]^2 ) dT }
since the T-clock is not moving inertially, so there is no
symmetry.

Dirk Vdm

Pentcho Valev wrote:
All Einsteinians, both profane and initiated, should immediately take
their heads out of the sand and resolve the following problem. In
Chapter 23 in his "Relativity" Einstein claims that a clock at rest
undergoes time contraction, that is, runs fast by a factor of gamma
relative to a clock situated at the edge of a rotating disk. The
Juggler also claims that this follows from Lorentz transforms. However
Lorentz transforms predict no time contraction - rather, the only time
distortion they predict is symmetrical time dilation for two inertial
systems. So Einsteinians should prove that their god is not a Juggler
and derive, RIGOROUSLY, the time contraction factor (gamma) from
Lorentz transforms.

Pentcho Valev

The charge that he is a "Juggler" may we well
deserved;
http://www.ncf.ca/~ek867/feynman.jugling.jpg
it seems to go with the territory,
but it is much clerarer in this case
that he is including the path as a part of the
clock mechanism as seen by a remote observer.

We now ask ourselves whether both clocks
go at the same rate from the standpoint
of the non-rotating Galileian reference-body K.
As judged from this body, the clock at the centre
of the disc has no velocity, whereas the clock at
the edge of the disc is in motion relative to K
in consequence of the rotation.
http://www.bartleby.com/173/23.html

The same practice in your previous post was
easily clarified by Harald by putting the
word "appear" in a statement where simple
logic dictates it had to be. It didn't change
the meaning, it only aided the reader in a
logical interpretation.

The theorist has previouly established the delay
caused by a non-zero length optical path so there
is no justifcation for the reader to fail to
apply such delay when reading the phrase:
"As judged from"

Sue...

BTW I admit to being profane but suggesting
I might be an "Einsteinian" is way over the
top. )

Dirk Van de moortel wrote:

"Pentcho Valev" wrote in message ups.com...
All Einsteinians, both profane and initiated, should immediately take
their heads out of the sand and resolve the following problem. In
Chapter 23 in his "Relativity" Einstein claims that a clock at rest
undergoes time contraction, that is, runs fast by a factor of gamma
relative to a clock situated at the edge of a rotating disk. The
Juggler also claims that this follows from Lorentz transforms. However
Lorentz transforms predict no time contraction - rather, the only time
distortion they predict is symmetrical time dilation for two inertial
systems. So Einsteinians should prove that their god is not a Juggler
and derive, RIGOROUSLY, the time contraction factor (gamma) from
Lorentz transforms.

Since this has been explained to you before (at least
10 times), you can stop reading here.

For the others:

The symmetry of the Lorentz transformation exists
between two inertially moving clocks. The clock at the
edge of the disk is not inertial.

If you want to know the total elapsed time of a non
inertially moving clock, you must integrate.
The total proper time T of a non-inertially moving clock
as calculated form the point of view of an inertial clock is
given by
T = Integral{ t1...t2; sqrt( 1 - [v(t)/c]^2 ) dt }
where v(t) is the speed of the non-inertial clock as a
function of the time t on the inertial clock.
You immediately see that
T = t2 - t1

You can *not* turn this around to something like
t = Integral{ T1...T2; sqrt( 1 - [v(T)/c]^2 ) dT }
since the T-clock is not moving inertially, so there is no
symmetry.


You should start from Lorentz transforms and obtain, rigorously and by
giving a detailed explanation of each step, the time contraction factor
gamma. Again (for a profane): LORENTZ TRANSFORMS - INFERENCE - GAMMA.
It is this inference that Einstein mentions in Chapter 23 in his
"Relativity" but forgets to give the steps. So try again. Your
arguments above are irrelevant.

Pentcho Valev

In sci.physics, Pentcho Valev

wrote
on 30 Jun 2005 05:08:41 -0700
. com:


Dirk Van de moortel wrote:

"Pentcho Valev" wrote in message ups.com...
All Einsteinians, both profane and initiated, should immediately take
their heads out of the sand and resolve the following problem. In
Chapter 23 in his "Relativity" Einstein claims that a clock at rest
undergoes time contraction, that is, runs fast by a factor of gamma
relative to a clock situated at the edge of a rotating disk. The
Juggler also claims that this follows from Lorentz transforms. However
Lorentz transforms predict no time contraction - rather, the only time
distortion they predict is symmetrical time dilation for two inertial
systems. So Einsteinians should prove that their god is not a Juggler
and derive, RIGOROUSLY, the time contraction factor (gamma) from
Lorentz transforms.

Since this has been explained to you before (at least
10 times), you can stop reading here.

For the others:

The symmetry of the Lorentz transformation exists
between two inertially moving clocks. The clock at the
edge of the disk is not inertial.

If you want to know the total elapsed time of a non
inertially moving clock, you must integrate.
The total proper time T of a non-inertially moving clock
as calculated form the point of view of an inertial clock is
given by
T = Integral{ t1...t2; sqrt( 1 - [v(t)/c]^2 ) dt }
where v(t) is the speed of the non-inertial clock as a
function of the time t on the inertial clock.
You immediately see that
T = t2 - t1

You can *not* turn this around to something like
t = Integral{ T1...T2; sqrt( 1 - [v(T)/c]^2 ) dT }
since the T-clock is not moving inertially, so there is no
symmetry.


You should start from Lorentz transforms and obtain, rigorously and by
giving a detailed explanation of each step, the time contraction factor
gamma. Again (for a profane): LORENTZ TRANSFORMS - INFERENCE - GAMMA.
It is this inference that Einstein mentions in Chapter 23 in his
"Relativity" but forgets to give the steps. So try again. Your
arguments above are irrelevant.

Pentcho Valev


The Lorentz is only useful for inertial reference frames. A
circular merry-go-round/Sagnac disc is not an inertial reference
frame.

Nevertheless, it is possible, given the hypotheses:

[1] The laws of physics are the same for all observers.
[2] Lightspeed is one of those laws.

to derive time dialation.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

--
#191,
It's still legal to go .sigless.

"Pentcho Valev" wrote in message ups.com...


Dirk Van de moortel wrote:

"Pentcho Valev" wrote in message ups.com...
All Einsteinians, both profane and initiated, should immediately take
their heads out of the sand and resolve the following problem. In
Chapter 23 in his "Relativity" Einstein claims that a clock at rest
undergoes time contraction, that is, runs fast by a factor of gamma
relative to a clock situated at the edge of a rotating disk. The
Juggler also claims that this follows from Lorentz transforms. However
Lorentz transforms predict no time contraction - rather, the only time
distortion they predict is symmetrical time dilation for two inertial
systems. So Einsteinians should prove that their god is not a Juggler
and derive, RIGOROUSLY, the time contraction factor (gamma) from
Lorentz transforms.

Since this has been explained to you before (at least
10 times), you can stop reading here.

For the others:

The symmetry of the Lorentz transformation exists
between two inertially moving clocks. The clock at the
edge of the disk is not inertial.

If you want to know the total elapsed time of a non
inertially moving clock, you must integrate.
The total proper time T of a non-inertially moving clock
as calculated form the point of view of an inertial clock is
given by
T = Integral{ t1...t2; sqrt( 1 - [v(t)/c]^2 ) dt }
where v(t) is the speed of the non-inertial clock as a
function of the time t on the inertial clock.
You immediately see that
T = t2 - t1

You can *not* turn this around to something like
t = Integral{ T1...T2; sqrt( 1 - [v(T)/c]^2 ) dT }
since the T-clock is not moving inertially, so there is no
symmetry.


You should start from Lorentz transforms and obtain, rigorously and by
giving a detailed explanation of each step, the time contraction factor
gamma. Again (for a profane): LORENTZ TRANSFORMS - INFERENCE - GAMMA.
It is this inference that Einstein mentions in Chapter 23 in his
"Relativity" but forgets to give the steps. So try again. Your
arguments above are irrelevant.

Pentcho Valev

Weren't you in write-only mode?
I explicitly said "For the others".
So that seems to have worked.

Matrix form of Lorentz transformation with movement in arbitrary
direction between momentarily co-moving inertial frame with
coordinates (T,X,Y,Z) of non-inertial clock and coordinates (t,x,y,z)
of inertial clock:
( dT ) ( g -g*v*nx -g*v*ny -g*v*nz ) ( dt )
( dX ) ( -g*v*nx (g-1)*nx^2+1 (g-1)*ny*nx (g-1)*nz*nx ) ( dx )
( dY ) = ( -g*v*ny (g-1)*nx*ny (g-1)*ny^2+1 (g-1)*nz*ny ) ( dy )
( dZ ) ( -g*v*nz (g-1)*nx*nz (g-1)*ny*nz (g-1)*nz^2+1 ) ( dz )
whe
we better look at this in a fixed font,
we work in units where c = 1,
(nx,ny,nz) = unit vector in direction of velocity (constantly changing) ,
v = amplitude of velocity (constant in this case)
= sqrt( [v nx]^2 + [v ny]^2 + [v nz]^2 )
= sqrt( [dx/dt]^2 + [dy/dt]^2 + [dz/dt]^2 ) ,
g = 1/sqrt(1-v^2) = gamma.

Since we are only intersted in events on the rotating clock, we
always have dX=dY=dZ=0 and we can concentrate on the time
equation only
(and now writing multiplication with spaces in stead of with "*"):
dT = g dt - g v nx dx - g v ny dy - g v nz dz
= g ( 1 - v nx dx/dt - v ny dy/dt - v nz dz/dt ) dt
= g ( 1 - v nx v nx - v ny v ny - v nz v nz ) dt
= g ( 1 - (v nx)^2 - - (v ny)^2 - (v nz)^2 ) dt
= g ( 1 - v^2 ( nx^2 + ny^2 + nz^2 ) ) dt
= g ( 1 - v^2 ) dt
= sqrt( 1 - v^2 ) dt
= 1/g dt

So
T = Integral{ t1...t2; sqrt( 1 - [v(t)]^2 ) dt }
= Integral{ t1...t2; sqrt( 1 - [v]^2 ) dt }
= sqrt( 1 - [v]^2 ) Integral{ t1...t2; dt }
= sqrt( 1 - [v]^2 ) ( t2-t1 )
= 1/g ( t2-t1 )
and turning it around
t2-t1 = g T
so
t2-t1 T
For two events on the rotating clock, the clock at rest shows
more elapsed time t2-t1 than the rotating clock (T), i.o.w.
the rotating clock "runs slow" as compared to the other.

Exercise:
Find the the expression starting from the invariant interval:
(dT)^2 = (dt)^2 - (dx)^2 - (dy)^2 - (dz)^2

Dirk Vdm

Pentcho Valev wrote:
All Einsteinians, both profane and initiated, should immediately take
their heads out of the sand and resolve the following problem. In
Chapter 23 in his "Relativity" Einstein claims that a clock at rest
undergoes time contraction, that is, runs fast by a factor of gamma
relative to a clock situated at the edge of a rotating disk.


Time [flow] is affected by relative velocity and gravitation--
precisely modeled by SR and GTR. There has never been to date
a prediction of SR or GTR that was contradicted by an observation.

No progress, although I should admit your diligence. Discuss the
problem with the initiated and try again.

Pentcho Valev


Dirk Van de moortel wrote:

"Pentcho Valev" wrote in message ups.com...


Dirk Van de moortel wrote:

"Pentcho Valev" wrote in message ups.com...
All Einsteinians, both profane and initiated, should immediately take
their heads out of the sand and resolve the following problem. In
Chapter 23 in his "Relativity" Einstein claims that a clock at rest
undergoes time contraction, that is, runs fast by a factor of gamma
relative to a clock situated at the edge of a rotating disk. The
Juggler also claims that this follows from Lorentz transforms. However
Lorentz transforms predict no time contraction - rather, the only time
distortion they predict is symmetrical time dilation for two inertial
systems. So Einsteinians should prove that their god is not a Juggler
and derive, RIGOROUSLY, the time contraction factor (gamma) from
Lorentz transforms.

Since this has been explained to you before (at least
10 times), you can stop reading here.

For the others:

The symmetry of the Lorentz transformation exists
between two inertially moving clocks. The clock at the
edge of the disk is not inertial.

If you want to know the total elapsed time of a non
inertially moving clock, you must integrate.
The total proper time T of a non-inertially moving clock
as calculated form the point of view of an inertial clock is
given by
T = Integral{ t1...t2; sqrt( 1 - [v(t)/c]^2 ) dt }
where v(t) is the speed of the non-inertial clock as a
function of the time t on the inertial clock.
You immediately see that
T = t2 - t1

You can *not* turn this around to something like
t = Integral{ T1...T2; sqrt( 1 - [v(T)/c]^2 ) dT }
since the T-clock is not moving inertially, so there is no
symmetry.


You should start from Lorentz transforms and obtain, rigorously and by
giving a detailed explanation of each step, the time contraction factor
gamma. Again (for a profane): LORENTZ TRANSFORMS - INFERENCE - GAMMA.
It is this inference that Einstein mentions in Chapter 23 in his
"Relativity" but forgets to give the steps. So try again. Your
arguments above are irrelevant.

Pentcho Valev

Weren't you in write-only mode?
I explicitly said "For the others".
So that seems to have worked.

Matrix form of Lorentz transformation with movement in arbitrary
direction between momentarily co-moving inertial frame with
coordinates (T,X,Y,Z) of non-inertial clock and coordinates (t,x,y,z)
of inertial clock:
( dT ) ( g -g*v*nx -g*v*ny -g*v*nz ) ( dt )
( dX ) ( -g*v*nx (g-1)*nx^2+1 (g-1)*ny*nx (g-1)*nz*nx ) ( dx )
( dY ) = ( -g*v*ny (g-1)*nx*ny (g-1)*ny^2+1 (g-1)*nz*ny ) ( dy )
( dZ ) ( -g*v*nz (g-1)*nx*nz (g-1)*ny*nz (g-1)*nz^2+1 ) ( dz )
whe
we better look at this in a fixed font,
we work in units where c = 1,
(nx,ny,nz) = unit vector in direction of velocity (constantly changing) ,
v = amplitude of velocity (constant in this case)
= sqrt( [v nx]^2 + [v ny]^2 + [v nz]^2 )
= sqrt( [dx/dt]^2 + [dy/dt]^2 + [dz/dt]^2 ) ,
g = 1/sqrt(1-v^2) = gamma.

Since we are only intersted in events on the rotating clock, we
always have dX=dY=dZ=0 and we can concentrate on the time
equation only
(and now writing multiplication with spaces in stead of with "*"):
dT = g dt - g v nx dx - g v ny dy - g v nz dz
= g ( 1 - v nx dx/dt - v ny dy/dt - v nz dz/dt ) dt
= g ( 1 - v nx v nx - v ny v ny - v nz v nz ) dt
= g ( 1 - (v nx)^2 - - (v ny)^2 - (v nz)^2 ) dt
= g ( 1 - v^2 ( nx^2 + ny^2 + nz^2 ) ) dt
= g ( 1 - v^2 ) dt
= sqrt( 1 - v^2 ) dt
= 1/g dt

So
T = Integral{ t1...t2; sqrt( 1 - [v(t)]^2 ) dt }
= Integral{ t1...t2; sqrt( 1 - [v]^2 ) dt }
= sqrt( 1 - [v]^2 ) Integral{ t1...t2; dt }
= sqrt( 1 - [v]^2 ) ( t2-t1 )
= 1/g ( t2-t1 )
and turning it around
t2-t1 = g T
so
t2-t1 T
For two events on the rotating clock, the clock at rest shows
more elapsed time t2-t1 than the rotating clock (T), i.o.w.
the rotating clock "runs slow" as compared to the other.

Exercise:
Find the the expression starting from the invariant interval:
(dT)^2 = (dt)^2 - (dx)^2 - (dy)^2 - (dz)^2

Dirk Vdm

Pentcho Valev wrote:

All Einsteinians, both profane and initiated, should immediately take
their heads out of the sand and resolve the following problem. In
Chapter 23 in his "Relativity" Einstein claims that a clock at rest
undergoes time contraction, that is, runs fast by a factor of gamma
relative to a clock situated at the edge of a rotating disk.


Time [flow] is affected by relative velocity and gravitation--
precisely modeled by SR and GTR. There has never been to date
a prediction of SR or GTR that was contradicted by an observation.

"Pentcho Valev" wrote in message ups.com...
No progress, although I should admit your diligence. Discuss the
problem with the initiated and try again.

So what's the matter?
Can't make the exercise?

Well, I told you to stop reading, didn't I?
It's not my fault that you are so frighteningly stupid, is it?

What a dope.

Dirk Vdm