Post by Julio Di Egidio"In the context of special relativity, we present a twins experiment that is
symmetric between the twins, so that a paradox appears inescapable, in the
form of a violation of the principle of causality. [...] Bottom line: if
special relativity is correct, it must be incomplete."
<http://seprogrammo.blogspot.co.uk/2015/07/symmetric-twins-paradox.html>
Your conclusion is wrong, basically because YOU DID NOT USE SR. You just made up
an answer.
Post by Julio Di EgidioIn the context of special relativity, we present a twins experiment that is
symmetric between the twins, so that a paradox appears inescapable, in the
form of a violation of the principle of causality.
Remember that "paradox" can mean "seemingly incorrect but discovered to be
correct upon detailed analysis" -- THAT is the sense in which the second word of
"twin paradox" is used. It is a "paradox" IN THAT SENSE, and not any sort of
inconsistency or contradiction or error.
There is no "violation of the principle of causality", because YOU DID NOT USE SR.
Post by Julio Di EgidioHere is the experiment (we would argue that effects of acceleration can be
Hmmm. the twins cannot separate and rejoin without acceleration. But then,
acceleration itself does not affect clocks directly.
In SR, the accumulated time on a clock depends upon its path through spacetime,
and the accelerations here serve to separate the paths of the clocks (twins).
That separation cannot be "made arbitrarily small".
As long as the two rockets are identical, and programmed to provide identical
thrust profiles (though in different directions), then any effects of
acceleration will cancel out. I assume this below.
Post by Julio Di Egidio== The twins, call them L and R, are each given a clock at birth and get the
clock fixed to their body: the two clocks have been previously synchronised.
Assume, for simplicity, that the common origin of space-time for the clocks
is set to the moment and place of the twins' birth, as well as a common
choice of coordinates is made, such that the twins, at that very moment,
share the same frame of reference (up to arbitrary precision). Just after
birth, the twins (with their clocks) are each embarked on a rocket. Assume
collinear motion for simplicity. The two rockets start in opposite direction
relative to the origin, carrying L and R respectively. The plan is for the
rockets to fly away from each other at some fixed (appropriately high)
constant speed for some fixed (appropriately long) proper time, then simply
invert course and fly back at opposite speed to the point of origin (in
space), and stop there. (Effects of acceleration can be minimised by making
the total proper time the rockets are subject to acceleration appropriately
small relative to the total proper time of the journey.) The twins at that
point rejoin, again sharing a common frame of reference (up to arbitrary
precision). We ask what the two clocks measure for each twin in this common
frame. ==
OK. With one addition this is just the usual symmetrical twins paradox. The
addition is that the "coordinates" you chose are an inertial frame, and the
locations of the twins' separation and rejoining are the same place in that
inertial frame. (Being the same place happens naturally since their rockets are
identical and they follow the same trajectories relative to that inertial frame
except for direction of travel.)
The answer is well known: when they rejoin, the two twins' clocks indicate the
same total elapsed proper time, for the simple reason that their trajectories
through spacetime are symmetric (relative to the initial inertial frame).
Post by Julio Di EgidioTo compute results, we apply special relativity. (We advise readers go
through this little but essential exercise by themselves.)
You ought to take your own advice!
Post by Julio Di EgidioWe find that R's clock looks late (i.e. back in time) to L, but, *at the
same time* (twins and clocks are in a common frame of reference), L's clock
looks late to R!
This is just plain wrong. The two clocks indicate the same elapsed proper time
after they rejoin.
Post by Julio Di EgidioThis conclusion is an inescapable consequence of the theory,
No, it isn't. It is just plain wrong. You made a gross mistake, and since you
did not show your math, we cannot show you where you made it.
Post by Julio Di Egidioand now the problem becomes how to make sense of these results. [... further
incorrect claims]
I suspect you have fallen victim of some books' claiming "moving clocks run
slow". That is a SHORTCUT, and you MUST understand the caveats that lead up to
it. In particular, this is only LOOSELY valid, and ONLY with respect to an
inertial frame. Note that neither twin is at rest in an inertial frame, and that
approach simply cannot be used.
One easily calculate the elapsed proper time of each twin,
using the initial inertial frame. We have:
T_e = integral sqrt(1 - v(t)^2/c^2) dt
Where T_e is the elapsed proper time of the clock in question,
t is the time coordinate of the inertial frame, and v(t) is
the clock's speed relative to that frame. T_e is OBVIOUSLY
the same for both twins, because v(t) is the same for them
(since only v^2 appears, the direction does not matter).
Note that for identical rockets, any acceleration will have
identical effects on the two twins' T_e.
Post by Julio Di Egidioif we had a third observer stationary with the origin, she and only she
would see L and R's clocks in synch with each other. In fact, adding
observers changes nothing: still the clocks would be late relative to her own
clock, and still the twins would see *her* clock late relative to their own.
You are VERY CONFUSED. After they rejoin, the two twins' clocks are co-located.
NO MATTER WHO LOOKS AT THEM their relationship MUST be the same. And it is: when
computed CORRECTLY, all observers calculate equal T_e for the two twins. This
includes each twin, and observer at rest in the inertial frame I introduced
above, and any other observer whatsoever.
It is true that adding observers changes nothing. In particular, it did not
change the fact that YOU MADE A GROSS MISTAKE.
Post by Julio Di Egidioit is now clear that special relativity is completely misunderstood/mystified
Yes. It is completely misunderstood BY YOU.
Post by Julio Di EgidioIs is also clear (but usually not acknowledged) that time dilation is
totally symmetrical between two frames of reference moving relative to each
other
This is "acknowledged" by any good book on relativity, and by anyone who
understands SR,; but it applies ONLY FOR INERTIAL FRAMES.
Post by Julio Di Egidioif you and I are moving relative to each other, *at the same time* your
clocks will look slower to me and my clocks will look slower to you.
Yes, FOR THE SPECIFIC METHOD OF MEASUREMENT IMPLICITLY ASSUMED. This method
requires each observer to use an assistant to measure the rate of the other
clock, because it is MOVING RELATIVE TO THE OBSERVER'S FRAME.
In order to avoid effects of propagation delays, we normally
insist that observations are made by someone co-located
with the event being observed. To observe a clock reading
does not require the observer to be co-moving with the clock,
just co-located. This means no single observer can measure
the rate of a clock moving relative to her, she needs an
assistant at rest in her inertial frame, with a clock
synchronized to hers, and pre-positioned along the path
the moving clock will follow in her frame. The assistant
records the value of the moving clock when it passes her,
as well as the value of her own clock. The observer herself
makes the same observations, and from the combination of
both the rate of the moving clock can be computed.
So while the two observers make the same sort of measurements, the two
measurements ARE DIFFERENT (they use different assistants and different
synchronizations of the clocks involved). There is no contradiction when these
two measurements both find the moving clock is ticking slower than their own clocks.
Post by Julio Di EgidioThat is indeed already enough to establish the paradox
Yes, IN THE SENSE ABOVE: to a quick look, SUCH AS YOURS, it seems contradictory;
but to a THOROUGH analysis that is CORRECT, there is no contradiction or
problem. The two measurements are just different.
Post by Julio Di EgidioOn the other hand, note that most presentations of special relativity
problems are set up in terms of a "rest" frame vs. a "moving" frame: that is
not incorrect but it is only half of the coin, because, at the same time and
in a perfectly symmetrical way, the "rest" frame is moving relative to the
"moving" frame. As said, the relations involved are symmetrical,
The problem here is that this only holds for INERTIAL frames. And neither twin
is at rest in an inertial frame. That's why I introduced the inertial frame in
which their separation and rejoin are at the same location. Using THAT frame,
which _IS_ inertial, the analysis is simple and straightforward.
It is possible to find analyses of this situation on the web. But
BEWARE OF FAKES -- there is lots of misinformation on the web,
SUCH AS YOUR SITE. You are MUCH better off to get a good textbook
on the subject, because it is clear that you are not knowledgeable
enough about SR to distinguish sense from nonsense. I recommend:
Taylor and Wheeler, _Spacetime_Physics_.
Do you really think that if your claims were true that nobody else would have
noticed it over the past 110 years? Assuming that thousands of physicists are
wrong and that you alone are right is a VERY BAD ASSUMPTION.
Tom Roberts
