Re: Ton of Bricks Paradox/Contradiction?

On Sep 18, 1:21 am, stevendaryl3...@xxxxxxxxx (Daryl McCullough)
wrote:
Alen says:

The observers accelerate in the SAME direction.

Yes, but that direction is *towards* one of the observers,
and *away* from the other observer. So their situations
are not equivalent.

There is never any relative motion between the observers,
either in the stationary frame, or in their own frame.

The two observers are simultaneously accelerated
from the positions of the rockets,

That sentence doesn't mean anything unless you specify
*how* "simultaneously" is defined.

and carry their OWN clocks with them into the moving
frame, while the rockets themselved remain stationary.

I have no idea what that means. Are the rockets
accelerating, or not?

No, the rockets remain stationary, while the
two observers accelerate into the moving frame
from the locations of the rockets. The rockets
remain where they are.

The observers' clocks and the rocket clocks are all
initially synchronised within the stationary frame.
According to the observers' own, synchronised clocks,
they start accelerating at the same time in the
stationary frame. Since the observers' accelerations
relative to their respective stationary-frame
starting-points are identical, and they take identical
times, on their OWN clocks, to complete the identical
accelerations, they thus arrive in the moving frame
at identical times on their OWN clocks.

Agreed. The time shown on one observer's watch when
he stops accelerating will be the same as the time
shown on the other observer's watch when
that observer stops accelerating.

Yes, and they will both now be at velocity v, the
velocity of the moving frame, so the time on their
watches is now the moving frame proper time.

Since, according to their OWN clocks, they started off at
identical times, that will always be true for them.

What does that mean? *What* will always be true for them?
That their watches will always be synchronized? As I have
explained to you, "synchronized" is only meaningful if you
specify a synchronization procedure. The procedure for
synchronization for observers at rest in one frame is
*different* from the procedure for another frame, and
they give *different* answers.

Yes, but that views synchronisation only from the
perspective of the synchronisation procedure. But
there is an implicit postulate that, after the procedure
is complete, the clocks REMAIN synchronised. This
is like the principle that an object accelerated to
velocity v will remain at that velocity until something
alters it. In the same way it can be assumed that
clocks REMAIN synchronised until something
desynchronises them. I say that it cannot be supposed
that they can become desynchronised unless
something affects one clock in a way that is different
to the other. Identical accelerations affect them equally
in all respects, and therefore they can be assumed
to remain synchronised.

Whatever effects the accelerations produce in these
clocks, the effects are identical, since the accelerations
are identical, in terms of the laws of physics.

There are two different meanings to saying "the accelerations
are identical". One meaning is as a function of *proper* time,
and second is as a function of *coordinate* time.

If we plot a(tau) for each observer, where a is the "felt"
acceleration, and tau is the time shown on that observer's
watch, we will get identical graphs.

Yes

If we plot a(t) for each observer, where a is the "felt"
acceleration, and t is the coordinate time in some frame,
then depending on the frame, you will get *different* plots.
That's because, as I have explained many times, the two
observers start accelerating at the same *proper* time tau,
but they don't necessarily start at the same *coordinate*
time t. To insure that they start at the same coordinate
time, you *must* specify what synchronization procedure
is used, and synchronization procedures are specific to
a specific frame.

I know that that is your argument.

Since the two observers complete the identical accelerations,
in the SAME direction, in identical times, according to
their OWN clocks, their OWN clocks will automatically
remain synchronised when they arrive in the moving
frame, and have remained synchronised at all times
since the start of the accelerations, these being identical
in all respects, with nothing that could cause the
clocks to become desynchronised.

I've already explained to you that that is NOT the case.
The two observers undergo identical accelerations as a
function of *proper* time, but *not* as a function of
*coordinate* time, where the coordinates are those appropriate
to the "moving" frame.

--
Daryl McCullough
Ithaca, NY

The problem with that is that, when the observers
reach the moving frame, their proper time IS
now the proper time of the moving frame.

According to the observers' clocks, which now
represent the proper time in the moving frame
itself, they started simultaneously at identical
time intervals previous to the time when they
arrived in the moving frame. This contradicts
the 'coordinate time' perspective.

If the observers' times become the moving frame
proper time, it might be expected that there would
be a difference in their clock records, and their
velocity records, which would show that they did
not take equal times to reach the moving frame, thus
confirming the coordinate time assertion that they
did not start simultaneously, and also confirming that
the distance between them changed. But that is not
the case. The clock records are identical, and show
that the observers started accelerating simultaneously,
always had identical velocities, and achieved velocity
v simultaneously in their own frame, which becomes
the moving frame at velocity v, and that their
distance apart in their own frame, now the moving
frame, never changed.

This contradicts the 'coordinate' perspective, and
creates an inconsistency that requires solution.

It is no use simply insisting on the original
synchronisation process. How do you know, for
example, that the time dilations on the transported
clocks are permanent? How do you know that
they do not disappear when the transported
clocks decelerate and come to a stop in the
stationary frame? The permanence of the time
dilations, viewed from the moving frame, cannot
simply be declared, and everything else simply
dismissed.

Alen




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Relevant Pages

  • Re: Ton of Bricks Paradox/Contradiction?
    ... Yes, but that direction is *towards* one of the observers, ... either in the stationary frame, ... According to the observers' own, synchronised clocks, ... explained before, clock synchronization is frame-dependent. ...
    (sci.physics.relativity)
  • Re: Ton of Bricks Paradox/Contradiction?
    ... Yes, but that direction is *towards* one of the observers, ... frame, while the rockets themselved remain stationary. ... Since the observers' accelerations ... specify a synchronization procedure. ...
    (sci.physics.relativity)
  • Re: Einsteins Train Gedanken Re-visited
    ... in one inertial frame of reference, ... he uses that postulate to define synchronization with light rays as ... rod now moving at v along with two moving observers at points A and B.. ... Thus the clocks are not synchronized w.r.t. the moving observers. ...
    (sci.physics.relativity)
  • Re: Einsteins Train Gedanken Re-visited
    ... in one inertial frame of reference, ... he uses that postulate to define synchronization with light rays as ... rod now moving at v along with two moving observers at points A and B. ... Thus the clocks are not synchronized w.r.t. the moving observers. ...
    (sci.physics.relativity)
  • Re: A little challenge for relativists.
    ... >> three regions which cannot be connected by a lorentz transform. ... Spacelike intervals can always be made simultaneous, so all observers ... Lorentz had to assume this in order to define a frame ...
    (sci.physics.relativity)
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