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There can be no defence against the twin paradox

There can be no defence against the twin paradox - Physics Forum

There can be no defence against the twin paradox - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


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  #1  
Old 10-28-2003, 04:09 PM
Stephen Bint
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Default There can be no defence against the twin paradox



I would like to clarify the paradox by describing a simpler experiment, in
which clear readings are taken as two spaceships fly toward a meeting point
at 0.75c.

The experiment:

My twin and I, called A and B, are in spaceships, at rest with respect to
eachother and exactly one light-year apart. We know this by triangulation. I
synchronize my clock to the one on B's spaceship by setting mine exactly one
year ahead of his, as I see it through my telescope.

We intend to travel towards eachother very quickly and watch eachother's
clocks and speedometers and use triangulation to keep track of our apparent
speed, as we travel.

At an agreed time, we accelerate rapidly towards one another, until our
speedometers say 0.75c. We travel at 0.75c most of the way, then decellerate
until we are at rest with respect to eachother, at a meeting point.


Before I move,

1. His speed, using triangulation, appears to be 0.
2. His clock appears to be going at exactly the same rate as mine.
3. His speedometer says 0.


while I am accelerating,

1. His speed, using triangulation, rises to slightly less than 1c.
2. His clock appears to be getting slower and slower.
3. His speedometer rises to less than 0.75c.

while I am travelling at 0.75c, after allowing the image of his speedometer
saying 0.75c to reach me across space,

1. His speed, using triangulation, appears to be less than 1c.
2. His clock appears to be going slower than mine, at a constant rate.
3. His speedometer says 0.75c.


while I am deccelerating,

1. His speed, using triangulation, shrinks from near 1c, to 0.
2. His clock's rate improves from slower than mine, to the same as mine.
3. His speedometer's reports fall from 0.75c to 0.

When we get out and shake hands,

1. His speed, using triangulation, appears to be 0.
2. His clock appears to be going at exactly the same rate as mine.
3. His speedometer says 0.

Is that what would happen? If it is, we have an inescapable paradox. If I
witnessed those events, then my twin must be younger than me. His clock ran
at varying degrees of slowness, but never faster than mine, so his clock
must be behind mine.

But of course, because my twin behaved identically to me, he must have had
an identical experience and my clock must be behind his.

These contradictory and mutually exclusive predictions cannot both be true,
nor can only one, because the twins have behaved identically. Which leaves
"neither perdiction is true", as the only conceivable possibility.

If I have made a mistake in describing the experience of time dilation in
this example, please point it out.

Stephen






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  #2  
Old 10-28-2003, 04:17 PM
Sam Wormley
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Default There can be no defence against the twin paradox


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  #3  
Old 10-28-2003, 05:08 PM
Stephen Bint
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Default There can be no defence against the twin paradox


"Sam Wormley" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
[Only registered users see links. ]

Thank you for those. The first link is no use to me in this context, though
it does describe Einstein's theories very well.

The page at the second link attempts to solve the paradox using the
acceleration argument, which does not survive the problem as I have laid it
out in this thread. It only works when one twin remains stationary.

The third one is interesting. It never occurred to me that obsevations
relevant to relativity would be produced as a spin-off of GPS. I didn't read
them thoroughly, so I was left unsure as to whether the clocks are brought
forward in order to compensate for the sum of the effects of speed and low
gravity. Are they?

Will you hazard a guess at why my example leads to an inescapable paradox?

Stephen



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  #4  
Old 10-28-2003, 05:31 PM
Sam Wormley
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Default There can be no defence against the twin paradox

Stephen Bint wrote:

The NAVSTAR GPS clocks are set to 10.22999999543 MHz vs 10.23000000000 MHz
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  #5  
Old 10-28-2003, 05:33 PM
Uncle Al
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Default There can be no defence against the twin paradox

Stephen Bint wrote:
[snip]


You're already screwed. A relativistic universe has four distinct
distances: luminosity (inverse square), angular diameter, parallax,
and proper motion. No two of them need agree to maintain
consistency. Clocks can only be synchronized by being local.

There is no paradox. One twin travels relativistically, one twin
stays at home. When they reunite the traveling twin is seen to have
aged much less than its genetic double. The only "paradox" is the
asymmetry (not the effect). If they *both* travel, you work out the
relative hyperbolic rotation in 4-space vs. an outside observer to
both.

Acceleration breaks the symmetry of who ages faster. To accomplish
that, the acceleration can occur before the clocks (or the twins)
exist. Only reference frames matter.

Inertial frames with relative *velocities* pursue different paths
through spacetime in Special Relativity. 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). Acceleration is
irrelevant in SR to the running of the clocks (as opposed to
Equivalence Principle acceleration in GR). Acceleration is necessary
at some arbitrary time not associated with the experiment itself for
breaking the symmetry of clock observation. Acceleration defines
which reference frame takes what path through spacetime - even if it
occurs when the clocks are *off* (or not even constructed yet, or
destroyed) - so the situation is NOT symmetric. There is a difference
between the reference frame and any clocks in it.

1) Acceleration is an absolute measurement and it does not require a
clock to make the measurement (e.g, simultaneous displacement of three
independent orthogonally cantilevered masses). There is no doubt who
was accelerated even if a clock was not running/existing during
aceleration. Any past accelerated reference frame has a different
mixture of space and time from an unaccelerated frame.

2) 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. This is an important subtlety and the key to the
whole thing. You cannot synchronize clocks except by having them
local. That's what Relativity demands. If they are local at the
start, you can tell who was naughty thereafter without needing a clock
to do the acceleration measurement. Accelerometers are not clocks.

EXAMPLE: We have 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 the 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! We have an accelerometer and
they have an acelerometer. We know who accelerated to set up the
experiment even if there wasn't a clock present when it happened.

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.

Acceleration is arguably General Relativity, as we did setting up the
experiment. It is irrelevant to the clocks. No clock is running or
even exists during acceleration. Numbers written on slips of paper
are unaffected by Special or General Relativity. One could as easily
build the clocks from their component parts after setting up the
experiment. No clock exists during acceleration up or down. The
*reference frame* has accelerated in the past, and that changes its
mix of space and time relative to an unaccelerated frame. The clocks
are passive observers in a presently unaccelerated setting.

Finally.... compare elapsed times. Elapsed time #2=#3 (straight line
motion for both traveling clocks, no acceleration!), but 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 (or,
Triplets) without any running clock having been accelerated - or
having even existed during acceleration up or down.


--
Uncle Al
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(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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  #6  
Old 10-28-2003, 05:44 PM
kenseto
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Default There can be no defence against the twin paradox


"Stephen Bint" <[Only registered users see links. ]> wrote in message
news:3f9e944a$0$116$[Only registered users see links. ].. .
point
I
one

There is no way to synchronize the two clocks as you said.

Ken Seto




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  #7  
Old 10-28-2003, 05:58 PM
Stephen Bint
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Default There can be no defence against the twin paradox


"kenseto" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
in
to
triangulation.
Why not? I have cameras at opposite sides of my spaceship, so by measuring
the angles they adopt when trained on a point of the other ship, I can
triangulate the distance. Knowing the distance, I can account for the time
elapsed since the image of the clock I can see began its journey to my eye.

At what point do you think I would be doing the impossible?


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  #8  
Old 10-28-2003, 06:39 PM
sal
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Default There can be no defence against the twin paradox



Stephen Bint wrote:

You're not doing anything impossible. It's a perfectly good gedanken
experiment, which could be carried in principle if we only had the
equipment.

An "inertial frame" should not be confused with an "observer". An
"observer" tends to imply a single point location, or a single event; an
inertial frame fills all space and time and can encompass numerous
events. All objects in an inertial frame are at rest with respect to
each other and are not accelerating, and by implication all the clocks
in one frame can be synchronized, just as you proposed.

The fun begins when the twins accelerate, but that event can be made
arbitrarily brief and hence can be neglected in the analysis of it.
(So, we should be able to figure it all out without invoking GR.)

All that's necessary to resolve the apparent paradox is to grind through
the transformations and see what comes out. Of course, we know the
answer in advance: just place an observer in the middle, in the same
inertial frame as the one the twins started out in. The situation is
totally symmetric, as viewed by that observer, and hence the twins must
have aged the same amount at the final event, which all three observers
participate in.

I'm going to take a stab at running the math on it, mostly as an
exercise (since we really do already know the result). Note, of course,
that proving the result is consistent may not help with the feeling that
something just isn't right here! Such are the limits of intuition.

(To email me directly replace nospam with foobox.)

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  #9  
Old 10-28-2003, 06:43 PM
Stephen Bint
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Default There can be no defence against the twin paradox


"sal" <believer@nospam.com> wrote in message
news:3F9EB7CA.4010400@nospam.com...

Thanks man, that's above and beyond the call of duty. I hope you make it to
the other side


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  #10  
Old 10-28-2003, 06:54 PM
sal
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Default There can be no defence against the twin paradox



Stephen Bint wrote:

There is a key point which you may have overlooked.

The clocks are in sync until the twins accelerate. Assume the
acceleration is arbitrarily brief (unless you care to integrate over a
curved manifold), and that they both check the time again after they've
accelerated. At that point, when they are both at speed, when they look
at each other's clocks, they'll find that they are _no_ _longer_ in sync.

The problem, of course, is that you can't synchronize clocks between
frames moving at different velocities. Both twins changed reference
frames after syncing their clocks.

Plug and chug in the transformations to get the degree to which they're
out of sync.

Haven't worked the math through on it yet (I have a day job here, too)
but I'm poking at it. If I get a quantitative result I'll email it.

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