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From: PD on 4 Mar 2010 12:46
On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote:
> On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote:
>
>
>
> > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your
> > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR.
>
> > > > > > > > > > You must just go through the entire thread and not pay any attention
> > > > > > > > > > to what anybody says.  Ever.
>
> > > > > > > > > > 1) What you've stated above is not an effect of SR.  It is an effect
> > > > > > > > > > of propagation delay, which was used to calculate c from the motion of
> > > > > > > > > > the moons of jupiter hundreds of years ago.
>
> > > > > > > > > Ok.
>
> > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run
> > > > > > > > > > faster.  But SR says nothing about whether you are moving toward or
> > > > > > > > > > away from an object.
>
> > > > > > > > > <suspicious eyebrow raised> Ok.
>
> > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT
> > > > > > > > > > from the amount that SR predicts the clock *actually* slows down
>
> > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR
> > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown?
> > > > > > > > > And for example, if we racked up the value of 'c' to near infinity,
> > > > > > > > > would SR still predict an "actual" slowdown, even though the
> > > > > > > > > propagation delays would approach zero?
>
> > > > > > > > With what you have described, I checked just to be sure, even though I
> > > > > > > > was already pretty sure what the answer would be, the time you read
> > > > > > > > moving away the clock would be:
>
> > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x
>
> > > > > > > > and when you move toward the clock
>
> > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x
>
> > > > > > > > so moving away from the clock:
> > > > > > > > dt2/dt = 1-v/c
> > > > > > > > and toward
> > > > > > > > dt2/dt = 1-v/c
>
> > > > > > > > Special relativity predicts that the moving clock will always slow
> > > > > > > > down as
> > > > > > > > dt2/dt = sqrt(1-v^2/c^2)
>
> > > > > > > > What you *measure* is a combination of the actual slow down predicted
> > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation
> > > > > > > > delays (which depend on the direction of motion).
>
> > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a
> > > > > > > certain distance, synchronise them when they are both stationary, and
> > > > > > > then accelerate them both towards each other (and just before they
> > > > > > > collide, we bring them stationary again). Are you seriously saying
> > > > > > > that both clocks report that the other clock has slowed down, even
> > > > > > > though they have both undergone symmetrical processes? Because there
> > > > > > > is obviously a contradiction there.
>
> > > > > > Yes, that is correct.  Both will report a slow down.  And in fact,
> > > > > > which ever one breaks the inertial frame to match speed with the other
> > > > > > is the one that will be "wrong".  This is still within the realm of
> > > > > > SR, not GR.
>
> > > > > What if they both "break the inertial frame"?
>
> > > > Then whichever frame they both accelerate into will be the one that
> > > > has measured the "correct" time dilation.
>
> > > So in other words, the clocks will register the same time, but will
> > > have slowed in some "absolute sense"?
>
> > Yes--assuming they both accelerated by the same amount (that is to
> > say, assuming they both broke the inertial frame in a symmetric way).
> > Otherwise, they will register different times.
>
> Agreed.
>
> So let's explore an extension of this scenario. Let's say you have two
> clocks, and you accelerate both of them up to a common speed, and
> after they have travelled a certain distance, you turn them around and
> return them to the starting point. The only difference is that one
> clock goes a certain distance, and the other clock goes twice that
> distance, but they *both* have the same acceleration profile - the
> only difference is that one clock spends more time travelling on
> inertia.
>
> Obviously, one clock will return to the starting point earlier than
> the other. But when both have returned, are their times still in
> agreement with each other, or have they changed?

Agreement. Both of them will agree, but will be showing a time earlier
than a third clock that was left behind at the starting point.
From: mpalenik on 4 Mar 2010 12:49
On Mar 4, 12:46 pm, PD <thedraperfam...(a)gmail.com> wrote:
> On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
>
>
> > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your
> > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR.
>
> > > > > > > > > > > You must just go through the entire thread and not pay any attention
> > > > > > > > > > > to what anybody says.  Ever.
>
> > > > > > > > > > > 1) What you've stated above is not an effect of SR.  It is an effect
> > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of
> > > > > > > > > > > the moons of jupiter hundreds of years ago.
>
> > > > > > > > > > Ok.
>
> > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run
> > > > > > > > > > > faster.  But SR says nothing about whether you are moving toward or
> > > > > > > > > > > away from an object.
>
> > > > > > > > > > <suspicious eyebrow raised> Ok.
>
> > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT
> > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down
>
> > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR
> > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown?
> > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity,
> > > > > > > > > > would SR still predict an "actual" slowdown, even though the
> > > > > > > > > > propagation delays would approach zero?
>
> > > > > > > > > With what you have described, I checked just to be sure, even though I
> > > > > > > > > was already pretty sure what the answer would be, the time you read
> > > > > > > > > moving away the clock would be:
>
> > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x
>
> > > > > > > > > and when you move toward the clock
>
> > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x
>
> > > > > > > > > so moving away from the clock:
> > > > > > > > > dt2/dt = 1-v/c
> > > > > > > > > and toward
> > > > > > > > > dt2/dt = 1-v/c
>
> > > > > > > > > Special relativity predicts that the moving clock will always slow
> > > > > > > > > down as
> > > > > > > > > dt2/dt = sqrt(1-v^2/c^2)
>
> > > > > > > > > What you *measure* is a combination of the actual slow down predicted
> > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation
> > > > > > > > > delays (which depend on the direction of motion).
>
> > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a
> > > > > > > > certain distance, synchronise them when they are both stationary, and
> > > > > > > > then accelerate them both towards each other (and just before they
> > > > > > > > collide, we bring them stationary again). Are you seriously saying
> > > > > > > > that both clocks report that the other clock has slowed down, even
> > > > > > > > though they have both undergone symmetrical processes? Because there
> > > > > > > > is obviously a contradiction there.
>
> > > > > > > Yes, that is correct.  Both will report a slow down.  And in fact,
> > > > > > > which ever one breaks the inertial frame to match speed with the other
> > > > > > > is the one that will be "wrong".  This is still within the realm of
> > > > > > > SR, not GR.
>
> > > > > > What if they both "break the inertial frame"?
>
> > > > > Then whichever frame they both accelerate into will be the one that
> > > > > has measured the "correct" time dilation.
>
> > > > So in other words, the clocks will register the same time, but will
> > > > have slowed in some "absolute sense"?
>
> > > Yes--assuming they both accelerated by the same amount (that is to
> > > say, assuming they both broke the inertial frame in a symmetric way).
> > > Otherwise, they will register different times.
>
> > Agreed.
>
> > So let's explore an extension of this scenario. Let's say you have two
> > clocks, and you accelerate both of them up to a common speed, and
> > after they have travelled a certain distance, you turn them around and
> > return them to the starting point. The only difference is that one
> > clock goes a certain distance, and the other clock goes twice that
> > distance, but they *both* have the same acceleration profile - the
> > only difference is that one clock spends more time travelling on
> > inertia.
>
> > Obviously, one clock will return to the starting point earlier than
> > the other. But when both have returned, are their times still in
> > agreement with each other, or have they changed?
>
> Agreement. Both of them will agree, but will be showing a time earlier
> than a third clock that was left behind at the starting point.

Wait, maybe I'm confused by Ste's setup. Didn't he say that one
travels twice as far as the other? But then he also says that you
turn them both around and return them to the start after traveling a
certain distance. Have they moved different distances in his scenario
or not?
From: mpalenik on 4 Mar 2010 12:55
On Mar 4, 12:46 pm, PD <thedraperfam...(a)gmail.com> wrote:
> On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
>
>
> > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your
> > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR.
>
> > > > > > > > > > > You must just go through the entire thread and not pay any attention
> > > > > > > > > > > to what anybody says.  Ever.
>
> > > > > > > > > > > 1) What you've stated above is not an effect of SR.  It is an effect
> > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of
> > > > > > > > > > > the moons of jupiter hundreds of years ago.
>
> > > > > > > > > > Ok.
>
> > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run
> > > > > > > > > > > faster.  But SR says nothing about whether you are moving toward or
> > > > > > > > > > > away from an object.
>
> > > > > > > > > > <suspicious eyebrow raised> Ok.
>
> > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT
> > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down
>
> > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR
> > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown?
> > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity,
> > > > > > > > > > would SR still predict an "actual" slowdown, even though the
> > > > > > > > > > propagation delays would approach zero?
>
> > > > > > > > > With what you have described, I checked just to be sure, even though I
> > > > > > > > > was already pretty sure what the answer would be, the time you read
> > > > > > > > > moving away the clock would be:
>
> > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x
>
> > > > > > > > > and when you move toward the clock
>
> > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x
>
> > > > > > > > > so moving away from the clock:
> > > > > > > > > dt2/dt = 1-v/c
> > > > > > > > > and toward
> > > > > > > > > dt2/dt = 1-v/c
>
> > > > > > > > > Special relativity predicts that the moving clock will always slow
> > > > > > > > > down as
> > > > > > > > > dt2/dt = sqrt(1-v^2/c^2)
>
> > > > > > > > > What you *measure* is a combination of the actual slow down predicted
> > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation
> > > > > > > > > delays (which depend on the direction of motion).
>
> > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a
> > > > > > > > certain distance, synchronise them when they are both stationary, and
> > > > > > > > then accelerate them both towards each other (and just before they
> > > > > > > > collide, we bring them stationary again). Are you seriously saying
> > > > > > > > that both clocks report that the other clock has slowed down, even
> > > > > > > > though they have both undergone symmetrical processes? Because there
> > > > > > > > is obviously a contradiction there.
>
> > > > > > > Yes, that is correct.  Both will report a slow down.  And in fact,
> > > > > > > which ever one breaks the inertial frame to match speed with the other
> > > > > > > is the one that will be "wrong".  This is still within the realm of
> > > > > > > SR, not GR.
>
> > > > > > What if they both "break the inertial frame"?
>
> > > > > Then whichever frame they both accelerate into will be the one that
> > > > > has measured the "correct" time dilation.
>
> > > > So in other words, the clocks will register the same time, but will
> > > > have slowed in some "absolute sense"?
>
> > > Yes--assuming they both accelerated by the same amount (that is to
> > > say, assuming they both broke the inertial frame in a symmetric way).
> > > Otherwise, they will register different times.
>
> > Agreed.
>
> > So let's explore an extension of this scenario. Let's say you have two
> > clocks, and you accelerate both of them up to a common speed, and
> > after they have travelled a certain distance, you turn them around and
> > return them to the starting point. The only difference is that one
> > clock goes a certain distance, and the other clock goes twice that
> > distance, but they *both* have the same acceleration profile - the
> > only difference is that one clock spends more time travelling on
> > inertia.
>
> > Obviously, one clock will return to the starting point earlier than
> > the other. But when both have returned, are their times still in
> > agreement with each other, or have they changed?
>
> Agreement. Both of them will agree, but will be showing a time earlier
> than a third clock that was left behind at the starting point.

If they have the same acceleration profile but one travels twice as
far as the other because it is allowed to move at a constant speed
longer (as I understand his question--although he seems to contradict
this at the beginning), they will NOT be in agreement.

If he accelerates them up to a common speed and turns them both around
at the same time and stops them at the same time, so they both travel
the SAME distance, which is what he says at the beginning (but then
contradicts this later on), then the WILL be in agreement.
From: Ste on 4 Mar 2010 13:00
On 4 Mar, 17:23, mpalenik <markpale...(a)gmail.com> wrote:
> On Mar 4, 12:17 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
>
>
>
>
> > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your
> > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR.
>
> > > > > > > > > > > You must just go through the entire thread and not pay any attention
> > > > > > > > > > > to what anybody says.  Ever.
>
> > > > > > > > > > > 1) What you've stated above is not an effect of SR.  It is an effect
> > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of
> > > > > > > > > > > the moons of jupiter hundreds of years ago.
>
> > > > > > > > > > Ok.
>
> > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run
> > > > > > > > > > > faster.  But SR says nothing about whether you are moving toward or
> > > > > > > > > > > away from an object.
>
> > > > > > > > > > <suspicious eyebrow raised> Ok.
>
> > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT
> > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down
>
> > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR
> > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown?
> > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity,
> > > > > > > > > > would SR still predict an "actual" slowdown, even though the
> > > > > > > > > > propagation delays would approach zero?
>
> > > > > > > > > With what you have described, I checked just to be sure, even though I
> > > > > > > > > was already pretty sure what the answer would be, the time you read
> > > > > > > > > moving away the clock would be:
>
> > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x
>
> > > > > > > > > and when you move toward the clock
>
> > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x
>
> > > > > > > > > so moving away from the clock:
> > > > > > > > > dt2/dt = 1-v/c
> > > > > > > > > and toward
> > > > > > > > > dt2/dt = 1-v/c
>
> > > > > > > > > Special relativity predicts that the moving clock will always slow
> > > > > > > > > down as
> > > > > > > > > dt2/dt = sqrt(1-v^2/c^2)
>
> > > > > > > > > What you *measure* is a combination of the actual slow down predicted
> > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation
> > > > > > > > > delays (which depend on the direction of motion).
>
> > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a
> > > > > > > > certain distance, synchronise them when they are both stationary, and
> > > > > > > > then accelerate them both towards each other (and just before they
> > > > > > > > collide, we bring them stationary again). Are you seriously saying
> > > > > > > > that both clocks report that the other clock has slowed down, even
> > > > > > > > though they have both undergone symmetrical processes? Because there
> > > > > > > > is obviously a contradiction there.
>
> > > > > > > Yes, that is correct.  Both will report a slow down.  And in fact,
> > > > > > > which ever one breaks the inertial frame to match speed with the other
> > > > > > > is the one that will be "wrong".  This is still within the realm of
> > > > > > > SR, not GR.
>
> > > > > > What if they both "break the inertial frame"?
>
> > > > > Then whichever frame they both accelerate into will be the one that
> > > > > has measured the "correct" time dilation.
>
> > > > So in other words, the clocks will register the same time, but will
> > > > have slowed in some "absolute sense"?
>
> > > Yes--assuming they both accelerated by the same amount (that is to
> > > say, assuming they both broke the inertial frame in a symmetric way).
> > > Otherwise, they will register different times.
>
> > Agreed.
>
> > So let's explore an extension of this scenario. Let's say you have two
> > clocks, and you accelerate both of them up to a common speed, and
> > after they have travelled a certain distance, you turn them around and
> > return them to the starting point. The only difference is that one
> > clock goes a certain distance, and the other clock goes twice that
> > distance, but they *both* have the same acceleration profile - the
> > only difference is that one clock spends more time travelling on
> > inertia.
>
> > Obviously, one clock will return to the starting point earlier than
> > the other. But when both have returned, are their times still in
> > agreement with each other, or have they changed?
>
> They are not in agreement because one was traveling faster longer.

It does not travel faster, only longer. I assume that is what you
meant.

So what you're saying is that mere movement causes a time dilation
effect?



> Acceleration only appears indirectly in SR.  Velocity is what appears
> in all the SR equations, it just happens that velocity is a function
> of acceleration, which is why acceleration is important.
>
> Also, I would avoid using the term "traveling on intertia".  It just
> physically isn't a very meaningful thing to say.  You could simply say
> "moving" are "traveling at a constant velocity," or something like
> that.

Indeed, it was just convenient to emphasise that no force was being
applied, whereas "constant velocity" seems to beg the question of
"relative to what"..
From: JT on 4 Mar 2010 13:03
On 4 mar, 18:49, mpalenik <markpale...(a)gmail.com> wrote:
> On Mar 4, 12:46 pm, PD <thedraperfam...(a)gmail.com> wrote:
>
>
>
>
>
> > On Mar 4, 11:17 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > On 4 Mar, 16:49, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > On Mar 4, 11:45 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > On 4 Mar, 16:32, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > On Mar 4, 11:28 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > On 4 Mar, 16:20, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > On Mar 4, 10:31 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > On 4 Mar, 13:40, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > > > On Mar 4, 3:12 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > > > On 3 Mar, 20:01, mpalenik <markpale...(a)gmail.com> wrote:
>
> > > > > > > > > > > > On Mar 3, 12:52 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > > > > > > > > > No. In SR, clocks *appear* to run slower as you are increasing your
> > > > > > > > > > > > > distance from the clock. The effect is entirely apparent in SR.
>
> > > > > > > > > > > > You must just go through the entire thread and not pay any attention
> > > > > > > > > > > > to what anybody says.  Ever.
>
> > > > > > > > > > > > 1) What you've stated above is not an effect of SR.  It is an effect
> > > > > > > > > > > > of propagation delay, which was used to calculate c from the motion of
> > > > > > > > > > > > the moons of jupiter hundreds of years ago.
>
> > > > > > > > > > > Ok.
>
> > > > > > > > > > > > 2) If you were to move TOWARD the clock, it would appear to run
> > > > > > > > > > > > faster.  But SR says nothing about whether you are moving toward or
> > > > > > > > > > > > away from an object.
>
> > > > > > > > > > > <suspicious eyebrow raised> Ok.
>
> > > > > > > > > > > > 3) The amount that the clock would appear to slow down is DIFFERENT
> > > > > > > > > > > > from the amount that SR predicts the clock *actually* slows down
>
> > > > > > > > > > > Really? I'm growing increasingly suspicious. In what way does SR
> > > > > > > > > > > predict the "actual" slowdown, as opposed to the "apparent" slowdown?
> > > > > > > > > > > And for example, if we racked up the value of 'c' to near infinity,
> > > > > > > > > > > would SR still predict an "actual" slowdown, even though the
> > > > > > > > > > > propagation delays would approach zero?
>
> > > > > > > > > > With what you have described, I checked just to be sure, even though I
> > > > > > > > > > was already pretty sure what the answer would be, the time you read
> > > > > > > > > > moving away the clock would be:
>
> > > > > > > > > > t2 = t - (x+vt)/c = t(1-v/c) - x
>
> > > > > > > > > > and when you move toward the clock
>
> > > > > > > > > > t2 = t + (x+vt)/c = t(1+v/c) + x
>
> > > > > > > > > > so moving away from the clock:
> > > > > > > > > > dt2/dt = 1-v/c
> > > > > > > > > > and toward
> > > > > > > > > > dt2/dt = 1-v/c
>
> > > > > > > > > > Special relativity predicts that the moving clock will always slow
> > > > > > > > > > down as
> > > > > > > > > > dt2/dt = sqrt(1-v^2/c^2)
>
> > > > > > > > > > What you *measure* is a combination of the actual slow down predicted
> > > > > > > > > > by SR (sqrt(1-v^2/c^2) and whatever changes occur due to propagation
> > > > > > > > > > delays (which depend on the direction of motion).
>
> > > > > > > > > Ok. So let us suppose that we take two clocks. Separate them by a
> > > > > > > > > certain distance, synchronise them when they are both stationary, and
> > > > > > > > > then accelerate them both towards each other (and just before they
> > > > > > > > > collide, we bring them stationary again). Are you seriously saying
> > > > > > > > > that both clocks report that the other clock has slowed down, even
> > > > > > > > > though they have both undergone symmetrical processes? Because there
> > > > > > > > > is obviously a contradiction there.
>
> > > > > > > > Yes, that is correct.  Both will report a slow down.  And in fact,
> > > > > > > > which ever one breaks the inertial frame to match speed with the other
> > > > > > > > is the one that will be "wrong".  This is still within the realm of
> > > > > > > > SR, not GR.
>
> > > > > > > What if they both "break the inertial frame"?
>
> > > > > > Then whichever frame they both accelerate into will be the one that
> > > > > > has measured the "correct" time dilation.
>
> > > > > So in other words, the clocks will register the same time, but will
> > > > > have slowed in some "absolute sense"?
>
> > > > Yes--assuming they both accelerated by the same amount (that is to
> > > > say, assuming they both broke the inertial frame in a symmetric way).
> > > > Otherwise, they will register different times.
>
> > > Agreed.
>
> > > So let's explore an extension of this scenario. Let's say you have two
> > > clocks, and you accelerate both of them up to a common speed, and
> > > after they have travelled a certain distance, you turn them around and
> > > return them to the starting point. The only difference is that one
> > > clock goes a certain distance, and the other clock goes twice that
> > > distance, but they *both* have the same acceleration profile - the
> > > only difference is that one clock spends more time travelling on
> > > inertia.
>
> > > Obviously, one clock will return to the starting point earlier than
> > > the other. But when both have returned, are their times still in
> > > agreement with each other, or have they changed?
>
> > Agreement. Both of them will agree, but will be showing a time earlier
> > than a third clock that was left behind at the starting point.
>
> Wait, maybe I'm confused by Ste's setup.  Didn't he say that one
> travels twice as far as the other?  But then he also says that you
> turn them both around and return them to the start after traveling a
> certain distance.  Have they moved different distances in his scenario
> or not?- Dölj citerad text -
>
> - Visa citerad text -

lol you framejumping grasshoppers have just have no idea what is
***REALLY*** going on have you. Don't forget u can always use the
fudgefactor.

JT
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