Re: The Twin Paradox versus General Relativity.
- From: "Nicolaas Vroom" <nicolaas.vroom@xxxxxxxxxx>
- Date: Tue, 03 Oct 2006 14:06:15 GMT
"Nicolaas Vroom" <nicolaas.vroom@xxxxxxxxxx> schreef in bericht
news:W4hRg.92439$aj2.1184804@xxxxxxxxxxxxxxxxxxxxxxxx
"Paul B. Andersen" <paul.b.andersen@xxxxxxxxxxxxxxxx> schreef in bericht
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Nicolaas Vroom wrote:
Question:
Is it possible to explain the Twin Experiment from the moving point of
view
OK.
I will show a scenario where we see it from the travelling
twin's point of view.
But first:
An observer accelerating at a will measure the rate of
an instantly stationary clock at a distance d in the direction
of the acceleration to be: (1 + a*d/c^2)
See comment 4
SNIP
Paul
The central question to answer is:
In order to understand the Twin Paradox completely
do you need GR ? Yes or No.
Apparently this is a tricky question.
The most thoroughly discussion I found in the book:
"Einsteins theory of Relativity" by Max Born.
Specific the pages 354 - 356.
Which gives the mathematical details.
(There is a short discussion about this problem
in the book "Introducing Einstein's Relativity"
by Ray d'Inverno at page 38)
I also did a google search with:
Twin Paradox General Relativity
and I found the following links:
1. Twin paradox
From Wikipedia, the free encyclopedia
http://en.wikipedia.org/wiki/Twin_paradox
Specific:
Resolution of the Paradox in General Relativity
2. The Twin Paradox: The "General Relativity" Explanation:
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html
Neither of those two links gives any detail about the mathematics
involved.
Reading the above documents I have two comments:
1. The central argument of the GR explanation is based on
the acceleration around the return point.
However suppose you perform this experiment with 3 clocks:
Clock 1 at rest
Clock 2 with a speed 0.5 c to the right
Clock 3 with a speed 0.5 c to the left.
1a. The exepriment starts when Clock1 and Clock 2 meet.
At that moment you reset both clocks.
1b. The return point is defined When Clock 2 meets Clock 3.
The number of Counts of Clock 3 is set equal to the number
of counts of Clock 2
1c. The experiment is finished When Clock 3 meets Clock 1
1d. At that moment if Clock 1 has 400 counts you will see that
Clock 3 has 346.4 counts.
( 4 counts versus 2*sq(3) counts)
In this scenario there is no acceleration involved.
Neither at the begin, nor the end nor halfway at the return point.
2. The basic difference between the example by me described
in the sketch : goto : http://users.telenet.be/nicvroom/dirk3.htm
and the example by Paul (excluding accelaration is a factor 100.
In the sketch there are 6 important events:
A0,B0 start of the experiment.
A1 Observer A issues return signal
A2
B2 Return point
A3 Observer A receives return signal back
A4,B3 end of experiment.
In my sketch the return point is defined when B
receives signal 1 from Observer A.
Assume that this is signal 100 and that Observer
A will continue sending equaly spaced signals.
If you do that Observer A will in total issue 400 signals
instead of 4 when the two meet at A4,B3
That means B will receive over a period over 173 units
100 signals (unitl he reaches the return point P)
From there on B will receive 300 equally spaced
signals over a period of 173 units until the two meets.
For Observer A a simular scenarion exists.
First he will receive 173 equally spaced signals over
a period of 300 units (until point A3) and there
after 173 equally spaced signals over a period
of 100 units.
Accordingly to the above text by Paul during the acceleration
around point P A's clock should increase by 100 (101) units.
My main objection around this idea that B will not agree with
this. He will not observe/receive , during his trip backwards,
a sharp increase in signals.
(This should happen between the signals 150 and 250)
3. In the book by Born at page 356 the following is written:
"Thus the clock paradox is due to a false application of SR,
namely, to a case in which the mothods of GR theory
should be applied"
implying that GR is obliged.
This is a much stronger statement than you will read in the
above mentioned url's nor in the book by d'Inverno.
4. This is equation 101a in the book by Born at page 353.
5. For me the most difficult thing to comprehend is that
part of the explanation depents on gravitational fields
introduced as a result of celestial bodies.
I can understand that.
But than what happens if no stars or masses are involved ?
Apparently this is also a tricky question.
IMO you can not perform the twin experiment when there
is no mass (masses) involved.
In the simplest configuration there should at least be one mass
i.e. one star.
And if that is the case there is a gravitational field involved
That means there is no SR without GR.
But if that is the case this should als be true for both observers.
Nicolaas Vroom
http://users.pandora.be/nicvroom/
.
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