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From: mpalenik on 4 Mar 2010 13:17
On Mar 4, 1:12 pm, Ste <ste_ro...(a)hotmail.com> wrote:
> On 4 Mar, 17:20, mpalenik <markpale...(a)gmail.com> wrote:
>
>
>
> > On Mar 4, 12:02 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > On 4 Mar, 16:29, 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.
>
> > > > I don't know what happened to this thread, or why google registers
> > > > several different copies of this thread, each with only a few messages
> > > > in it, but I'm going to describe the thought experiment I posted
> > > > before in greater detail.
>
> > > > In order for the ^ to receive the pulse of light from both emitters at
> > > > the same time and half way between the two emitters, the light must be
> > > > emitted BEFORE the ^ is half way between the two emitters.
>
> > > Indeed.
>
> > > > From the stationary frame, the two emitters emit light at the same
> > > > time and the ^ moves vertically until it is half way between the two
> > > > of them, when it receives both pulses.
>
> > > Ok.
>
> > > > But from the moving frame, the ^ sees the emitters emit light when it
> > > > is not half way between them, and then it sees the emitters move until
> > > > he is exactly half way between them.  However, the speed of light is
> > > > the same in all frames, and unaffected by the motion of the emitters.
>
> > > > So the the emitter emit while one is closer to him than the other.
> > > > The fact that they are moving does not affect the way light
> > > > propagates.  If the light signal from the closer one reaches him at
> > > > the same time as the light signal from the farther one, that means the
> > > > farther one must have emitted first.  The fact that he is half way
> > > > between the emitters at the end doesn't matter, because in his frame,
> > > > it is the emitters that are moving, which does not affect the speed of
> > > > light.
>
> > > This seems implausible. You cannot possibly (that is, physically) have
> > > the situation you describe, in the way you describe it. The receiver,
> > > when it receives the pulse of light, cannot possibly be in more than
> > > one position relative to the two sources.
>
> > It isn't.
>
> > > It either receives when the
> > > sources are equidistant, or it receives when one source is closer.
>
> > The sources are equidistant.
>
> > > It
> > > cannot receive when it is equistant *and* when one source is closer,
> > > for that would be an obvious contradiction.
>
> > What you don't understand is the fact that the sources were not
> > equidistant when they emitted the pulse.  
>
> Indeed, but it's neither here nor there where the sources where at the
> time of emission.

Not in the rest frame. But it is in the moving frame.

>
> The question is where the sources are at the time of reception,

Not in the ^ frame. The fact that the sources have moved does not
affect the way the light propagates. They could run over the moons of
jupiter and back, and it wouldn't affect the way the light propagates.

Lets say that you have two sets of emitters, A and B. One set is
moving and the other set is not. If they both emit

They both emit like this
A B


A B

But the B set is moving like this

A
B

A
B

The light from the A set and B set propagates exactly the same way.
It doesn't matter that the B set has moved.
From: Ste on 4 Mar 2010 13:18
On 4 Mar, 18:04, mpalenik <markpale...(a)gmail.com> wrote:
> On Mar 4, 1:00 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> If it's a constant velocity in one frame, it's a constant velocity in
> every frame, even though that constant velocity is zero in one
> particular frame.  Constant velocity implies that no force is being
> applied, since F = dP/dt.

But as I say, acceleration is absolute, not relative, whereas a
measure of velocity could be interpreted to be relative.



> Traveling on inertia kind of invokes the mideivil idea that something
> is required to keep the object moving, as if the object contains some
> kind of inertia that if taken away would cause it to stop.

I was treating inertia simply as "the tendency of matter to maintain
its absolute velocity through space".
From: Ste on 4 Mar 2010 13:19
On 4 Mar, 18:03, JT <jonas.thornv...(a)hotmail.com> wrote:
>
> 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.

Lol. One has to laugh.
From: PD on 4 Mar 2010 13:20
On Mar 4, 12:00 pm, Ste <ste_ro...(a)hotmail.com> wrote:
> 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?

Movement relative to an observer.

>
> > 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: PD on 4 Mar 2010 13:21
On Mar 4, 12:18 pm, Ste <ste_ro...(a)hotmail.com> wrote:
> On 4 Mar, 18:04, mpalenik <markpale...(a)gmail.com> wrote:
>
> > On Mar 4, 1:00 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > If it's a constant velocity in one frame, it's a constant velocity in
> > every frame, even though that constant velocity is zero in one
> > particular frame.  Constant velocity implies that no force is being
> > applied, since F = dP/dt.
>
> But as I say, acceleration is absolute, not relative, whereas a
> measure of velocity could be interpreted to be relative.
>
> > Traveling on inertia kind of invokes the mideivil idea that something
> > is required to keep the object moving, as if the object contains some
> > kind of inertia that if taken away would cause it to stop.
>
> I was treating inertia simply as "the tendency of matter to maintain
> its absolute velocity through space".

Where did you get that definition? It is terrible. There is no
absolute velocity through space.
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