Post by beda pietanzaPost by Odd BodkinPost by Ken SetoNo. The clock never ticks "faster" (or "slower") -- all clocks ALWAYS
tick at their usual rate, independent of their motion or location. This
is MUCH more subtle than clocks simply ticking faster or slower.
This seems to imply that a clock second is a universal interval of time....
No it doesn’t imply that. As is often the case, you manage to misunderstand
the simplest statements made on this newsgroup. This is why it is a futile
gesture to try to learn relativity by asking questions on the newsgroup.
Tom’s statement is that all clocks ALWAYS tick at their usual rate,
independent of their motion or location, IN THEIR OWN INERTIAL REFERENCE
FRAME. This does not mean that any clock ticks at the same rate in any
other inertial reference frame. Nor does it mean that if the clock is
measured to tick at a different rate in a different reference frame, then
the clock is now ticking at a different rate in its own frame.
beda
odd, are you sure that what you wrote is meaningful??
clock A is tickling as its usual rate
clock B, differently moving, is, also, tickling at its usual rate
Yes.
Post by beda pietanzato make it simple we cancel the frame where they are at rest,
They are different rest frames. The clocks are moving relative to each
other. How do you “cancel” different frames?
Post by beda pietanzaof course since nobody interfere with them, A and B keep their usual rate, correct??
if word have any meaning this means that the two clock are running at same rate.
No, it doesn’t. Perhaps it would help to understand better what “ticking at
its usual rate” means in terms of metrology. It does not mean that the rate
of A is compared to B and is found to be ticking at the same rate. There is
no comparison of clocks in different frames made in determining “ticking at
its usual rate.”
What it means is that, comparing the clock readings to local and at-rest
physical processes, those processes continue to take the same amount of
time as measured on the local clock. Concretely, you might have a quartz
piezoelectric oscillator and a spring-mass oscillator next to clock A; and
you might have a fast-decaying radioactive sample and a magnetic pendulum
next to clock B. Clock A continues to tick at its usual rate if the number
of oscillations of the piezoelectric oscillator and the spring-mass
oscillator still have the same rate per tick as always. Clock B continues
to tick at its usual rate if the half-life of the sample and the
oscillations of the magnetic pendulum still have the value per tick as
always. There is no comparison between A and B. There are comparisons
between the clocks and local physics processes.
So you CANNOT say that A and B have the same rate. What you CAN say with
some confidence is that, though the physical processes that run the
piezoelectric crystal, the spring-mass system, the radioactive sample, and
the magnetic pendulum are all completely different than each other, it is
still true that the laws of the physics that run all those processes are
the same in the rest frame of A and the rest frame of B. That’s what the
principle of relativity says: the laws of physics are the same in different
inertial frames. Interestingly, this does NOT force A and B to have the
same rate, because here you’ve not compared them with each other at all.
Post by beda pietanzathat is impossible because seeing them both by a third observer they are tickling at
different rate.
something is wrong with your assertion
I give you a hint: in the SR procedure, the clock at rest in its frame,
is °°assumed** tickling
at its usual rate (instead it is tickling at a "lower hidden absolute rate")
just like the ruler, it is assumed to remain unchanged but it has an
"hidden absolute contraction"
just like the two ways speed of light along the ruler, it is assumed to
be invariant locally, but it is really
longer and longer as the ruler' speed is higher.
the poor local observer that you posit into a windowless room in order to
fool him and to fool yourself,
is ridiculous, the ruler, the clock, the two way SOL are different
according to the speed of them.
and the lenght of the ruler, the rate of the clock, and the two ways of
the speed of light are different, and any observer can see that,
regardless if they are aware of it, or regardless of any make beleive of
your crooked logic.
Well, let’s see, if I can parse what you attempted to say, maybe you’re
asking what if it’s the case that whatever it is that is slowing clock B
(relative to A) is ALSO slowing the physical process behind the radioactive
decay and is ALSO slowing the physical process behind the magnetic
pendulum? And so there’s some hypothesis that, though the specific physical
processes used to benchmark clock B were selected at random, it JUST SO
HAPPENS that these particular processes are modified to slow at EXACTLY the
same factor as clock B. And since they were randomly chosen, others could
have been chosen, and so you’d come to the mysterious conclusion that NO
MATTER WHAT process is used to benchmark clock B, that process is
mysteriously slowed by exactly the same factor to follow the clock.
And so you’d rather believe that motion has a deep effect on ALL physical
processes to produce EXACTLY the same kind of slow-down that clock B is
experiencing, and that’s why clock B still APPEARS to tick at the same
rate, even though it is in fact slowed down absolutely. Is that what you’re
asking me to entertain?
Post by beda pietanzaat end, yes the clock A is measured going slower by a relatevely moving
SR frame of yours, but this doesn't mean that clock A runs at its usual
rate, it is just that you are pretending not to know that it is running
at the its absolute "hidden" rate determined by its absolute "hidden" speed.
but you keep pretending not to know...to fool yourself
cheers
beda
Post by Odd BodkinNone of this is going to make any sense to you, until you change your
stance and start reading beginning physics books checked out from the
library for free. I know this is something that you can’t stand the thought
of. And so maybe the thing to come to terms with is that you are never
going to understand relativity, because you can’t do the prep work.
Post by Ken SetoIOW, the passage of a clock second in A’s frame is corresponded to the
passage of a clock second in B’s frame. But such implication disagrees
Delta(tB)=Gamma*Delta(tA)
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables