Re: Stupid question for Dork Van de merde

In sci.physics.relativity, Sorcerer
<Headmaster@xxxxxxxxxxxxxxxxxx>
wrote
on Mon, 13 Nov 2006 20:48:17 GMT
<lS46h.125520$3x1.21895@xxxxxxxxxxxxxxxxxxxxxxxxx>:

Given:
The second is the duration of 9 192 631 770 periods of the radiation
corresponding to the transition between the two hyperfine levels of the
ground state of the cesium 133 atom. -- NIST, Dork Van de Psycho.

A year is 9,192,631,770 * 60 * 60 * 24 * 365.2425 periods of the
radiation corresponding to the transition between the two hyperfine levels
of the ground state of the cesium 133 atom. -- Wackypedia

"So if T = 5 years and v = 0.8c, then the stay at home twin will
have aged 10 years while his travelling twin sister will have aged
6 years." -- Dork Van de Psycho.
(5 years is 10 years or 6 years, I'm not sure which).

"Time is what a clock says" Dork Van de shithead.


Now for the stupid question.
What happened to the missing 1,160,365,758,078,260,000 periods of
the radiation corresponding to the transition between the two hyperfine
levels of the ground state of the cesium 133 atom's prodigal twin sister?



FWIW, here's an attempted answer to your question. I hope it is
enlightening.

At time 0 of course both clocks are emitting pulses. We accelerate
clock B to 0.8c instantaneously, which is of course impossible but never
mind.

Clock A now observes pulses coming from clock B once every 2.134 or so
seconds (sqrt(41)/3). Clock B observes pulses coming from clock A
once every 2.134 or so seconds.

I should note that Dirk van de Moortel has made a small
error here, which is unfortunate (IMO he should have used
0.6c or 0.96c). Therefore, I will generalize this result
by simply setting v = 0.8c and using algebra.

So...The Gospel According to SR:

Clock A will see pulses from clock B once every
sqrt(1+v/c)/sqrt(1-v/c) seconds. (This result can be
derived from the Lorentz -- sorry, I mean, the "cuckoo
transform" -- and the assumption that lightspeed between
A and B is always c.) Clock B will see pulses from clock
A once every sqrt(1+v/c)/sqrt(1-v/c) seconds as well,
by reciprocity.

Once B reaches its destination, it will turn around.
Clock A will then see pulses from B once every
sqrt(1-v/c)/sqrt(1+v/c) seconds; clock B will see the same
from A.

Since B takes the same amount of time (by its
lights) to fly both trip legs, it will see
(L'/v)*(sqrt(1+v/c)/sqrt(1-v/c) + sqrt(1-v/c)/sqrt(1+v/c))
from A for some L'. (It is naive to assume L = L'.)

These are A clock ticks, of course, and all must be accounted for;
therefore one must equate (letting g = 1/sqrt(1-v^2/c^2)):

2L/v = (L'/v)*(sqrt(1+v/c)/sqrt(1-v/c) + sqrt(1-v/c)/sqrt(1+v/c))
= (L'/v)*((1+v/c + 1 - v/c))/sqrt(1-v^2/c^2))
= 2L'g/v

or L' = L/g.

For its part A observes L'/v clock ticks from B during the outbound leg,
and L'/v ticks during the return leg. These are of course unequal
in length because B is observed to be ticking more slowly during
the outbound leg. The amount of time A observes is therefore

2L/v = (L'/v) *(sqrt(1+v/c)/sqrt(1-v/c) + sqrt(1-v/c)/sqrt(1+v/c))
= 2L'g/v again.

For a 5 year round trip at 0.8c (L = 2 light-years) B will "lose"
about (L/v)*(1 - 1/g) or 1 years' worth of ticks.

This is clearly a highly ridiculous result -- but then
the premises were rather silly. In particular, the best
rockets I can dream up might barely make 0.05c with the
boron-proton reaction, and that's assuming they can refuel
at the other end and can extract all of the energy from
the reaction to throw out the helium nuclei generated
by the reaction. If they cannot refuel, one might make
about 0.025c = 1/40 c, and the trip will take 160 years
objective, 159.95 years subjective. Any real trip will
require acceleration; at 100 N/kg accelerating to 1/40c
will take about 20 hours 50 minutes, or 0.00238 years;
the total time accelerating and decelerating would take
0.01 year. 100 N/kg is a brutal acceleration experienced
for only a short time by fighter jocks; a more reasonable
acceleration might be 20 N/kg (2 'g's) which wipes out
any relativistic correction right there, or 10 N/kg which
takes twice as much time to accelerate and decelerate
as the relativistic correction will shave off during the
free-flight period.

But after the trip is done the two clocks will still show
the discrepancy.

--
#191, ewill3@xxxxxxxxxxxxx
Error 16: Not enough space on file system to delete file(s)

--
Posted via a free Usenet account from http://www.teranews.com

.

Quantcast