kilopascal wrote:
>
> 2002-06-13
>
> This reminds me of the common belief that if out of a pair of twins, one who
> travels in space and another who remains on earth, the returning traveller
> will find his twin who remained behind older looking. This is known as the
> twin paradox. But, as we know, nature abhors paradoxes, so the way the
> SCI-FI writers have presented it is wrong.
>
> As Joe pointed out, the observer (Twin remaining on Earth) (A) will observe
> his twin in space (B) ageing less as his clock is running slower as he moves
> away at the speed of light. The common belief is that when B returns to
> Earth, he is noticeably younger. The truth is, that as the ship turns
> around and moves back towards the earth at the speed of light, the opposite
> occurs. B's clock now appears to move faster than on earth and B's age
> catches up to that of A on the Earth. When B steps off the ship, he is the
> same age as his brother. The only difference will be the normal ageing that
> took place because the trip would have consumed some time. Thus time is
> conserved and nature remains in balance.
No offense, John, but you've got it quite wrong. The return trip does
not "undo" the effects of the outward bound leg. There is no such thing
as "conservation of time"!
In the Spring of 2001, I gave my PHYS 102 students this problem on
their third test (of five):
The first crew of the International Space Station, commanded by Bill
Shepherd, has just returned to Earth after spending 136.000 days in
space---as measured in NASA's reference frame. Assume that Capt Shepherd
wore a perfectly running watch and left another perfectly running watch
at home during his trip. The watches had maintained exactly the same
time prior to the trip. Also assume that the space station orbited at an
altitude of 354 km above Earth's surface. When Captain Shepherd returned
home, how much difference in time was there between the two watches and
which one was behind the other?
Of course, they had to first calculate the orbital speed at that
altitude (7698.85 m/s). Then they had to calculate the dilation of time
by the theory of special relativity and that showed that Shepherd's
watch in space lost 3.875 ms. Thus he was that much younger than if he
had not made the 136 day trip. (Notice that he did not regress in age,
nor did the liftoff or landing have any significant effect on this.) For
extra credit, the students could have calculated the effects of the
reduced gravitational field in space and, by the correspondance
principle, could have shown that there would have been a mitigating
"blue shift" which would mitigate this time loss by 0.431 ms, thus
causing him to lose only 3.444 ms. Nobody tried the latter calculation
but a very large portion of those who selected this problem to solve
(they got to choose two problems out of three) managed the first part.
Not bad for non-physics majors and especially for the non-science majors
in the class, in my opinion!
I've never published this but I'll hang it out quietly on my web sited
for a few days so those who wish to can see the solution and the
approximations used. Go to
http://www.cofc.edu/~frysingj/RelProbISS/test3prob3.html
Sorry, folks, but I just don't have the time at the moment to answer a
bunch of questions about this solution. Please visit a text book. OTOH,
if you catch a mistake, I would appreciate hearing about it.
Now, what I cannot imagine is someone going through that math in ifp
units! That would truly tax my abilities and I don't think my students
could have handled it in anything but SI.
Jim
--
Metric Methods(SM) "Don't be late to metricate!"
James R. Frysinger, CAMS http://www.metricmethods.com/
10 Captiva Row e-mail: [EMAIL PROTECTED]
Charleston, SC 29407 phone/FAX: 843.225.6789
