Re: Ton of Bricks Paradox/Contradiction?
- From: stevendaryl3016@xxxxxxxxx (Daryl McCullough)
- Date: 18 Sep 2007 08:56:26 -0700
Alen says...
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.
That depends on how you measure "relative motion", but
that's not what I said. I said that one of the observers
is accelerating in the direction of the other observer,
and one of the observers is accelerating in the direction
*away* from the other observer.
Imagine having a heavy metal ball suspended from the
corners of a cubical frame using springs. If you accelerate
to the right, then the metal ball will move to the left
(relative to the frame). If you accelerate to the left,
the metal ball will move to the right.
In this way, the direction of acceleration is measurable.
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.
How do the observers accelerate, if not by using
rockets?
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.
A frame doesn't have a "proper time". Proper time
is a function of a *path* through spacetime. Frames
can have a *coordinate* time, but it doesn't make
any sense to talk about proper time for a frame.
To talk about coordinate time, you have to bring up
the issue of how clocks are synchronized. As I have
explained before, clock synchronization is frame-dependent.
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.
There is no meaning to synchronization other than
from the perspective a particular synchronization procedure.
To say that two watches are synchronized means
that the event at which one watch shows time tau=0
is simultaneous with the event at which another
watch shows time tau=0. To say that two events
are simultaneous is to say that the time coordinate
assigned to one event has the same value as the
time coordinate assigned to the other event. So
talking about "synchronized watches" necessarily
involves talking about coordinate systems, which
necessarily involves talking about synchronization
procedures.
But there is an implicit postulate that, after the procedure
is complete, the clocks REMAIN synchronised.
Well, that's false. Accelerating watches that were synchronized
according to their initial frame will no longer be synchronized
according to their new frame.
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.
You can say it, but it's false.
Give me an *operational* or *physical* meaning to the
claim that two watches are synchronized. What does it
*mean* to be synchronized? I have no idea what you mean
by that. I know what *I* mean by "synchronized", and my
meaning is frame-dependent.
Two clocks A and B that are at rest in frame F are
said to be synchronized in frame F if a light signal
sent from A at time tau (according to A) arrives at
B at time tau + D/c (according to B), where D is given
by
D = 1/(2c) * time required for light to travel from A to B
and back
Alternatively, two clocks A and B that are at rest in frame F
are said to be synchronized in frame F if
a clock is transported slowly from clock A at time tau1
(according to clock A) to clock B at time tau2 (according
to clock B) will show elapsed time tau satisfying
tau is approximately tau2 - tau1
(where the approximation gets better the slower the
third clock is transported).
If you want to propose a frame-independent notion of
clocks being synchronized, then go ahead, but explain
what it means *physically*.
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.
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.
As I have explained, there is no such thing as "proper
time of the moving frame". There is only *coordinate*
time of the moving frame, and coordinate time requires
a synchronization procedure.
The proper time shown on one of the clocks is *not*
the coordinate time of the moving frame. According to
the moving frame, one observer (the one on the right)
decelerates earlier than the other observer, and
because of this, the proper time on the right observer's
watch is different from the proper time on the left
observer's watch.
The clock records are identical
As a function of proper time, but not as a function
of coordinate time.
and show that the observers started accelerating
simultaneously,
What does "simultaneously" mean? It has no meaning
*except* the meaning given by a synchronization
procedure.
--
Daryl McCullough
Ithaca, NY
.
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