Natural Science Forum / Physics / General Physics / July 2008

  Your problem STATES constant velocity... so be definition the Twins
  pass each other (or collide in a puff of plasma and subatomic particles).
  If you actually intend for them to change relative velocity.... well see:

  Physics FAQ: The Twin Paradox
    http://edu-observatory.org/physics-faq/Relativity/SR/TwinParadox/twin_paradox.html

It would be a) except time dilation isn't even apparent, so it is b).

Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?

"Easy: he did NOT say that." - cretin harald.vanlintelButNotThis@epfl.ch
According to moron van lintel, Einstein did not write the equation he wrote.

According to xxein:
It is an artefactual/superficially imposed yin-yang of sorts.

According to Lamenting Shubert:
Why do you want to know?

" In neither system (meaning frame of reference in modern-day terminology)
is the speed of light c-v or c+v.  In both systems the speed of light is c."
-- cretin Jimmy Black fmlast3@organization.edu.
According to the imbecile Jimmy Black, Einstein did not write the equation
he wrote.

The inequality Einstein wrote:
 http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif

'we establish by definition that the "time" required by
light to travel from A to B equals the "time" it requires
to travel from B to A' because I SAY SO and you have to
agree because I'm the great genius, STOOOPID, don't you
dare question it. -- Rabbi Albert Einstein

"Counterfactual assumptions yield nonsense.
If such a thing were actually observed, reliably and reproducibly, then
relativity would immediately need a major overhaul if not a complete
replacement." -- Humpty Roberts.

What does the basic symmetry of the situation suggest?

If you want to survey bright minds it helps if
you know where to find them.

<< The key to understanding special relativity is
Einstein's relativity principle, which states that:

   All inertial frames are totally equivalent
   for the performance of all physical experiments.

In other words, it is impossible to perform a physical
experiment which differentiates in any fundamental
sense between different inertial frames. By definition,
Newton's laws of motion take the same form in all
inertial frames. Einstein generalized this result in his
special theory of relativity by asserting that all laws
of physics take the same form in all inertial frames. >>
--Richard. Fitzpatrick
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html

Sue...

[snip rest of crap]

Reality is not a peer vote, stooopid.  f.cking imbecile.  Since you
are too f.cking stooopid to read the FAQ, Uncle Al will spoon feed it
up your anus.  READ IT, JACKASS:  The ratio by which the twins have
aged at the end when they are back together again is the same in all
reference frames:

ratio = sqrt(t^2 - x^2 - y^2 - z^2)/t (with units of c=1)

Inertial frames with relative *velocities* pursue different paths
through spacetime. No clock anomaly is apparent in any of them until
clocks are compared (by all being local when you do it, initial
calibration then experiment).  The situation is NOT symmetric.

Past acceleration is irrelevant to the running of present clocks but
not to the mixture of space and time in the reference frame that said
clocks measure. You cannot synchronize clocks except by having them
local. If they are local at the start, you can tell who was naughty
thereafter without measuring acceleration.

Given three identical clocks that are off (a state of not running, or
of not even having been fabricated) and zeroed. Each clock has/will
have a very short toggle jiggger switch sticking out. We load them (or
their parts, or ore and a smelter and a machine shop) in individual
spaceships and set up then experiment.

CLOCK 1: That's our clock. It sits stationary in our inertial
reference frame with a little jigger sticking out. Touch the jigger
and the "off" state becomes "on" or the "on" state becomes "off."
Clock 1 is "off." Or we can build it from parts just before we need
it, and in the "off" state, zeroed.

CLOCK 2: In a spaceship traveling at 0.999c relative to our inertial
frame of reference. Clock 2 is "off." It was built after all
acceleration ceased, and set to zero. It skims past Clock 1 (our
clock), the jiggers touch, both Clocks 1 and 2 are now "on" and
locally synchronized by touching. Elapsed time accumulates in each
one. The situation is NOT symmetric!

CLOCK 3: In a spaceship traveling at 0.999c relative to our inertial
frame of reference, but 180 degrees counter in direction to Clock 2.
Clock 3 is zeroed and "off." It was built after all acceleration
ceased, and set to zero.

Some arbitrary time after Clocks 1 and 2 synchronize and turn "on" by
touching, Clocks 2 and 3 brush past each other, touching jiggers.
Clock 2 is now "off," Clock 3 is now "on." Write down the elapsed time
in now "off" Clock 2, then smash the clock with a sledgehammer. Or
melt it down, or toss it over the side. The spaceship with Clock 3 is
returning back over the path taken by the spaceship with Clock 2.

CLOCK 1: That's our clock. It sits stationary in our inertial
reference frame with a little jigger sticking out. Clock 3 rushes
past, jiggers touch. Clocks 3 and 1 are now off. All clocks are off.
No clock has accelerated while "on" or even while existing. Write down
elapsed times, smash each clock with a sledgehammer. Or melt them
down, or toss them.

BOTTOM LINE: Get all three slips of paper together... Accelerate as
you need. Or send all the results to all three folks by radio and
never decelerate. All clocks have been smashed, melted, tossed. Their
elapsed times were written down. The numbers on the papers won't
change when you accelerate or broadcast the data.

Finally.... compare elapsed times. Elapsed time #2+#3 does not equal
#1, the local stationary reference frame summation. The sum of #2+#3
elasped time is only about 4.5% that than of #1's accumulated elapsed
time. You have the Twin Paradox (Triplets) without any running clock
having been accelerated - or having even existed during acceleration
up or down.

Idiot.  Imagine being your undershorts given your crass inabilty to
find your butthole while squatting over a mirror with a flashlight.

f and g) The twins are both the same revolutions of
Earth wrt the Sun old.
So they are the same, yrs, days, minutes and seconds
and nanoseconds etc.. old
no matter what the clock on board the spaceships said.

What is each of their motions through space? If they accelerated they
felt weight and therefore could detect their motion.

Mitch Raemsch; Twice Nobel Laureate 2008

In sci.physics, GigaS
<GigaS.2d00838@physicsbanter.com>
wrote
on Fri, 25 Jul 2008 20:59:21 +0100
<GigaS.2d00838@physicsbanter.com>:

No data available, since you've not specified exactly when
they leave said planets.  Herewith a suggestion.

Let planets A, B, and C be lined up in a row, with distances
as diagrammed.

A----d---------B----------d'--------------C

At some point, B sends out a flash of light.  When A sees this light, it
launches twin 1.  When B sees this light, it launches twin 2.  How long
does it take both twins to reach a common point in space, assuming both
twins have identical rockets?

We assume B as the "absolute origin" for convenience.  B fires its
pulse at (0,0),  A receives it at time d (we choose units so that
c = 1 length unit per time unit).  C receives it at time d'.
The twins meet at a point D.

(0,0)_B -- laser fires
(-d,d)_B -- A receives pulse, launches 1
(-d+vtA,d+tA)_B -- Twin 1 is moving
(d',d')_B -- B receives pulse, launches 2
(d'-vtC,d'+tC)_B -- Twin 2 is moving

If we want to work out where and when they meet, then we
have a pair of equations:

d+tA = d'+tC
-d/v+tA = d'/v-tC

Adding these two equations yields

d-d/v+2*tA = d'+d'/v
tA = (d'+d'/v-d+d/v)/2
-d+vtA = -d+(d'v+d'-dv+d)/2
      = (d'v+d'-dv-d)/2

Subtracting them yields

d+d/v = d'-d'/v+2*tC
tC = (d+d/v-d'+d'/v)/2
d'-vtC = d' + (-dv-d+d'v-d')/2
      = (-dv-d+d'v+d')/2

As should be expected, these are equal, and define a point D.
Also, d+tA = d'+tC = (d'+d'/v+d+d/v)/2.

So much for working out B's view of the problem.  We now turn
to twin 1, and since he launches from A, we need to know
what A sees as well.  To that end, we have a simple translation:

(0,0)_A -- Twin 1 launches
(vt1, t1)_A -- Twin 1 is moving.

Of course Twin 1 will see things differently.  *Now* we invoke
the Lorentz, and get

(0, g*t1*(1-v^2))_1

where g is the usual 1/sqrt(1-v^2).

We still have some adjustments to make, as Twin 1 launches a bit
late relative to B, but that's pretty trivial.  Our final result
for Twin 1 (by simply equating t1 = tA, which is the time B thinks
twin 1 is actually traveling) is

t'1 = d + g*tA*(1-v^2) = d + b*tA = d + b*(d'+d'/v-d+d/v)/2
   = d*(1-b/2+b/(2v))+d'*(b+b/(2v))

where b = sqrt(1-v)/sqrt(1+v).

Twin 2 has his own ideas:

(-vt2,t2)_C
(0, g*t2*(1-v^2))_2
t'2 = d' + g*tC*(1-v^2) = d' + b*tC = d'+b*(-d'+d'/v+d+d/v)/2
   = d*(b+b/(2v))+d'*(1-b/2+b/(2v))

If B is exactly in the middle, then so is D, and d=d' and t'1 = t'2.
That's the obvious solution...but if d != d', we need to figure
out whether t'1 < t'2 or t'1 > t'2.

t'1 : t'2 ::
d*(1-b/2+b/(2v))+d'*(b+b/(2v)) : d*(b+b/(2v))+d'*(1-b/2+b/(2v)) ::
d*(1-b/2)+d'*(b) : d*(b)+d'*(1-b/2) ::
d*(1-b) : d'*(1-b) ::
d : d'

(Since b < 1, 1-b > 0, and we're perfectly safe.)

In other words, the twin that travels the longer distance
does not age as much.

Of course, all of this doesn't mean much unless we know
how the twins got to A and C in the first place, as it's
far from clear that they're the same age (as far as B
is concerned) *before* they've started their journeys.

As phrased, f).  You'll need to clarify your question.

When twin 1 views twin 2 on her planet, he sees a 'younger' twin - because
he is observing light that was emitted earlier.
The same applies to twin 2 viewing twin 1.

The best answer from above is (b), though I'm not too keen on that one.