MJackson.Wbst@PARC-MAXC.ARPA (11/02/83)
AAAAARRRrrrrggggghhhhhhh!!!
It's amazing how --> *DUMB* <-- a response one can come up with by
shooting from the hip. It's astounding how many people one can make
oneself a fool of in front of through the miracle of electronic mail.
I haven't thought about problems like this in ten years--not much
relativity involved in building copiers--but that's no excuse for
flinging bull around.
Of course, a little thought reveals that the resolution of the twin
paradox has NOTHING TO DO WITH "clocks run slower in an accelerated
frame." This effect, while real, is much too small to be significant.
(Example: a four-year round-trip at constant 1g boost will generate a
quite nice 17.5% difference in "locally" elapsed time, but four years on
earth [1g] versus 4 years floating out near Saturn [0g] is detectably
different only with sensitive chronometry.)
Now the general conclusion (the accelerated twin ages more slowly) is
correct. The apparent symmetry between the two twins is in fact broken
by the (locally measurable) acceleration one undergoes. The problem can
be easily analyzed in the "stationary" frame because it is an inertial
(unaccelerated) frame; however, most books merely demonstrate the
results at the end of the trip (which resolves the paradox), glossing
over the details or hiding them in infinitesimal periods of infinite
acceleration.
The problem is hard to analyze in an accelerated frame, and the reasons
reflect the fact that the "inertial observer's coordinate system," a
space-filling latticework of meter sticks with clocks at the
intersections which every book on relativity beats to death, is more
than the pedantic fantasy it may appear. That construct, and the
synchronization of clocks it entails, represents a complete, consistent
methodology for compensation for "signal time-of-flight effects," which
correction I so cavalierly dismissed in the first paragraph of my
preceding message.
Such a system is impossible for an accelerating observer, since if he
has one a moment later it will be moving relative to him. Meter sticks
will shrink; worse, clocks which are synchronized in the frame will no
longer look synchronized to the observer.
Among other problems, actually measuring the rate at which time passes
for someone flying past REQUIRES synchronization of clocks in your
frame. If his clock ticks once per second in his frame, it will tick
now here and then there in yours; measuring the time-interval between
the ticks in your frame requires one to transport a measure of "now"
from "here" to "there" (or, equivalently, of "then" from "there" to
"here"). Since for the accelerating observer the relationship between
"now-here" and "now-there" is constantly changing, the analysis is
difficult; what the observer measures is determined locally and
confounded by changing simultaneity-relationships, and the results may
be counterintuitive.
The appendix gives an example in the spirit of the original question.
Both time dilation and time compression appear and there are no
discontinuities in the passage of time, as advertised. This exposition
owes quite a lot to *Spacetime Physics* by Taylor and Wheeler, which I
recommend to anyone who is willing to undertake some study; it answers,
or equips the reader to answer, virtually every relativistic "paradox"
known. I hope this clears things up, and I apologize again to anyone
led astray by my previous blitherings.
Mark
-------------------
APPENDIX:
The round-trip twin paradox
Assume an inertial observer i and an accelerated observer a. a departs
at a constant 1g acceleration (10 m/s/s) which he holds for Y years (1
year = 3E7 sec). He then reverses thrust for Y years to stop, continues
for Y years to head back, and reverses thrust again for Y years to stop
at the origin. Two general questions:
1) What happens?
2) Given time-marker broadcasts, what does it look like?
It is easiest to work with the velocity parameter (apologies; there's no
theta in the ASCII charset):
H = arctanh (v/c)
(arctanh = inverse hyperbolic tangent, c = 3E8 m/s). This is "easiest"
because changes in velocity parameter are *additive*, which is *not*
true of velocities when relativistic effects are taken into account.
1) Analyze in the i frame. Since acceleration is constant in the a
frame, differential change in velocity parameter is
dH = (g/c)*dta
(ta = time in the a frame). This immediately yields expressions valid
during the first Y years:
H = g*ta/c
v = c*tanh (g*ta/c)
dti = cosh (H)*dta, (ti = time in i frame) which integrates to
ti = (c/g)*sinh (g*ta/c)
dxi = tanh (H)*dti = tanh (H)*cosh (H)*dta = sinh (H)*dta,
(xi = separation in i frame) which integrates to
xi = ((c**2)/g)*(cosh (g*ta/c) - 1)
At the end of Y years (1/4 round trip) this yields (for Y = 1 year)
H = 1
v = .76c
ti = 1.175 years
xi = .54 ly
Now by symmetry the next three stages are quite similar. Thus when a
returns (after 4 years in his own frame, hence 4 years older) his twin
will have aged 4.7 years. For his effort, a will have gotten 1.08 ly
from earth.
(A more radical effect: let Y=5. Now at the 1/4 point
H = 5
v = 0.9999092c
ti = 74.2 years
xi = 73.2 ly
So a can get 146.4 ly from earth and return in 20 "local" years @ a mere
1g, but when he gets back 296.8 years will have elapsed on earth. There
goes the old neighborhood.)
2) To i, "what it looks like" is (1); we worked the problem in his
frame. Of course as a flies away the pulses come further apart, and as
he flies back they come closer together, but (time-delay) correction for
this is straightforward and consistent since i has (the possibility of)
a complete coordinate system.
To a, "what it looks like" is that the number of pulses he has received
is the number of pulses which have overtaken him, hence on the first 1/4
trip
pr = ti - xi/c
= (c/g)*(1 - exp(-g*ta/c))
In order to convert this information ("how many pulses have I
received?") into what is desired ("what time has elapsed in the i
frame?") a must account for time delay; as long as he is accelerating he
can't have a complete coordinate system (as noted above) so he must
*infer* the appropriate delay from the speed of light and the present
APPARENT distance to i:
delay = xa/c = xi/(cosh (H)*c)
elapsed ti as inferred by a = pr + delay
[I should say that a could, arguably, use a different method for
calculating the "delay"; this would implicitly redefine the
instantaneous inertial frame in which the delay is figured. All such
choices leave unchanged the essential conclusion in the last paragraph
below.]
For the case analyzed in (1), at the end of the first 1/4 trip (ta = 1
year)
pr = .63 years worth
inferred elapsed ti = .98 years
So a will see a slight time dilation on i's part. However, at the end
of the second 1/4 trip (ta = 2 years)
pr = 1.27 years worth
inferred elapsed ti = pr + xa/c = pr + xi/c (no relative motion)
= ti = 2.35 years
and time compression is evident! Note that a received the same number
of pulses in the second 1/4 as in the first, and the same amount of time
elapsed in his frame. What changed was his inference about the rate at
which time was passing for i.
Of course, on the second half of the trip a will encounter the rest of
i's pulses and the "delay" will go to zero; hence at the end a and i
will agree that i "aged" 4.7 years while a was flying around for 4
years.notes@ucbcad.UUCP (11/06/83)
#R:hpfclj:14500001:ucbcad:22400003:000:251 ucbcad!kalash Oct 30 15:46:00 1983 It is my understanding that the twin paradox (almost) goes away when you take acceleration (and general relitivity) into account. While you get interesting effects with inertial frames, getting up to C tends to play havoic with calculations. Joe
gwyn%brl-vld@sri-unix.UUCP (11/10/83)
From: Doug Gwyn (VLD/VMB) <gwyn@brl-vld> No, the situation is not symmetrical. In fact the "paradox" of the twins is not the time dilation for a moving twin, it is precisely the argument you were giving, namely, that each observer could consider himself at rest and the other as the one undergoing time dilation, apparently leading to two differing predictions as to the relative ages of the observers when the journey is over. The correct answer is that the moving twin is younger than the stationary one (according to their bodies' internal clocks) at the end of the journey. If you plot out the hypothetical trip on a space-time diagram, including the clock pulses, you can see what is happening.
dnc@dartvax.UUCP (11/12/83)
oh iget it, the flying twin is receiveng many signals per his seconds, so at the end he has received n signals, but only lived a fraction of thm of his sconds, he is greedily grabbing more time per second.