Re: Question: In SR is a clock second an interval of universal time?????


"The Ghost In The Machine" <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote in message
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In sci.physics, PD
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wrote
on 8 Jan 2007 06:39:26 -0800
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On Jan 8, 8:02 am, "kenseto" <kens...@xxxxxxxxxx> wrote:
"The Ghost In The Machine" <e...@xxxxxxxxxxxxxxxxxxxxxxx> wrote in
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In a way, kenseto is correct; in a way, he is not. If an
observer accelerates with his local clock (assuming the
clock and the observer can handle the acceleration without
changing its signal), neither will notice nothing out
of the ordinary (apart from the acceleration artifacts).
In its way, the clock isn't changing; certainly the clock
isn't going to be observed to change *in its reference
frame*.
Of course, another observer who is not accelerating will
have a slightly different opinion.

You are confused.....when a clock is accelerated its clock rate is
changed
to a slower rate compared to the clock that is not accelerated. What
this
mean is that the accelerated clock's clock second will contain a larger
amount of absolute time than the stay at home clock's clock second.
That is
the reason why the traveling clock accumulated less clock seconds after
it
is returned.

Ken Seto

Really, Ken?

Let's start with two spaceships, A and B, in relative motion.
Because of their relative motion, the observer in A says that B's clock
is slower.

I was talking about A and B started in the same frame.

Now B, hearing this report in a message from A, says, "Wait a minute.
Maybe I should do something about this. I'm just standing still here --
I haven't been doing a thing. I think I'll turn my rocket on and speed
up."
So B turns on the rocket and of course accelerates and in fact does so
that B is now keeping pace with A. As a result of B accelerating, the
*relative* speed between A and B is now zero. The observer in A now
looks at B's clock and sees that it is now going at the same rate as
his own clock. This means that B's clock must have *sped up*.

But B's rate is slowed after the acceleration when it is compared to a clock
in the original frame of B before the acceleration.

So here we have a clear case where as a result of B's acceleration, B's
clock (as seen by A) *speeds up*, not slows down, in conflict with your
explanation, Ken.

B does not speeds up compared to a clock that is in the original frame of B
before B accelerated away. .

Define "now". If B is some distance away A will not be
able to observe (using electromagnetic radiation as opposed
to a theoretical infinite-speed tachyonic device) that B's
clock has sped up until some time on A's clock has elapsed.

You have a comprehension problem. Measuriong doppler shift is not measuring
the rate of a clock moving relative to you.

Ken Seto


.

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