Re: twin patadox question
- From: "rotchm@xxxxxxxxx" <rotchm@xxxxxxxxx>
- Date: 30 Aug 2005 10:38:36 -0700
Ok, here is my shot.
Posit: Spaceship 1 (s1) at "rest", spaceship 2 (s2) coincides with s1
(and synch to zero) and travels away at speed 0.99c wrt s1. Spaceship 3
(s3) coincides with s2 (grabs s2 time) at a distance of 10 ly wrt s1
and approaching s1 at speed 0.99c wrt s1.
That is the only info we have. We want to know what will s1 and s3
clocks indicate when they coincide.
Units, lightyear and year -> c = 1
Lets analyze this from 1) s1 point of view, 2) from s2 point of view.
g = gamma = 1/sqrt(1-0.99^2)
1) s1 POINT OF VIEW:
s2 travels at speed 0.99c wrt s1 up to a distance of 10 ly. Thus, the
time elapsed on s1 during this is
t1 = 10 ly / 0.99c =...= 10/0.99 = 10.101010101010... years. In words,
the time elapsed on s1 clock is 10.101.. years for s2 to reach the 10
ly mark.
Now, according to s1, s2's clock thus indicates 10.1010101/g =
1.424922826... years when s2 reaches the 10 ly mark.
s3 grabs this value.
For the return trip done by s3, the calculation is identical: It takes
a time of 10.101010101 ly for s3 to reach s1 as measured by s1 and a
time of 1.4249... years on s3 clock as observed by s1.
Hence, the total trip took 2x10.101010101 years on s1's clock (s1 =
20.20202020202 y) and s1 obseves s3 clock to indicate 2x1.4249... =
2.84984565... years, as they coincide.
2) s2 POINT OF VIEW:
s2 being well versed in SR and knowing that he is at the 10 ly mark IN
s1 FRAME, deduces that s1 is at a distance of 10/g ly from s2. That
is, as observed from s2, (as s2 and s3 coincide), s1 is at a distance
of 10/g.
Since s1 is travelling at speed 0.99c wrt s2, the time elapsed on s2
clock during this is t2=(10/g)/0.99 = 1.4249228... years. This is the
time indicated on s2 clock as s2 and s3 coincides. The time elapsed on
s1 clock is thus 1.4249228/g = 0.20101010101 according to s2.
Now, s1 is at a distance of 10/g from s3 (and s2) AND s1 has speed 0.99
wrt s2, s3 has speed (0.99+0.99)/(1+0.99^2) = 0.999949498... wrt s2 and
towards s1. Therefore, wrt s2 clock, it takes a time of
(10/g)/(0.99994948-0.99) = 141.7833943 years for s1 and s3 to coincide.
Thus, according to s2, s1 clock has elapsed for a time of 141.7833943/g
= 20.0010091 years. Again, according to s2, s3 clock elapsed for a
time of 141.7833943/g(0.99994948) = 1.4249, were g(0.99994948) =
1/sqrt(1-0.99994948^2).
Thus, according to s2, s1 clock indicates 0.20101010101 + 20.0010091 =
20.202020202 years and s3 clock indicates 1.4249 + 1.4249 = 2.8498...
Same values obtained in 1)
Whoah!!
---
If you want to be sure, then always doubt }:-)
.
- References:
- twin patadox question
- From: slefkowit
- twin patadox question
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