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<p>Wolf,</p>
<p>just a short answer to this mail.</p>
<p>Einstein's example is a simplified situation. It is simplified in
so far that the moving clock comes back to the clock at rest. So
in the equation which I have given:</p>
<p>t' = (t-xv/c<sup>2</sup>)* (1-c<sup>2</sup>/v<sup>2</sup>)<sup>-1/2</sup></p>
<p>the distance x becomes 0 (zero). Then your equation can be used.
But if two moving clocks are compared in a state where they are at
different positions then the full equation (above) has to be used.</p>
<p>The short version is also applicable for the twin experiment
because at the end both twins meet again at the same position.</p>
<p>In which respect is there a paradox?</p>
<p>Best<br>
Albrecht<br>
</p>
<br>
<div class="moz-cite-prefix">Am 30.08.2017 um 07:54 schrieb Wolfgang
Baer:<br>
</div>
<blockquote type="cite"
cite="mid:7ebc358f-41ce-7496-1862-edc6d5bea447@nascentinc.com">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<p>I've now looked up the reference from "on the Electrodynamics
of moving Bodies" by Albert Einstein translated from Annalen der
Physik 17,1905 in The Principle of Relativity by H.A. Lorentz,
A Einstein, H. , wit notes by A Sommerfeld p 49 in Section #4</p>
<p>" If we assume that the result proved for a polygonal line is
also valid for a continuously curved line, we arrive at this
result: If one of two synchronous clocks at A is moved in a
closed curve with constant velocity until it returns to A, the
journey lasting t seconds, then by the the clock remained at
rest the traveled clock on its arrival at A will be 1/2 t v<sup>2</sup>/c<sup>2</sup>
seconds slow." <br>
</p>
<p>Am I wrong in interpreting these words as implying a twin
paradox.? I'm not claiming that there is a twin paradox.</p>
<p>Only that the straight forward interpretation of Einstein's
words suggest there is a paradox <br>
</p>
<p>best</p>
<p>Wolf <br>
</p>
<p><br>
</p>
<pre class="moz-signature" cols="72">Dr. Wolfgang Baer
Research Director
Nascent Systems Inc.
tel/fax 831-659-3120/0432
E-mail <a class="moz-txt-link-abbreviated" href="mailto:wolf@NascentInc.com" moz-do-not-send="true">wolf@NascentInc.com</a></pre>
<div class="moz-cite-prefix">On 8/29/2017 3:41 AM, Dr Grahame
Blackwell wrote:<br>
</div>
<blockquote type="cite"
cite="mid:85328F3496EE444CAE0459B5C85FD35F@vincent">
<meta content="text/html; charset=utf-8"
http-equiv="Content-Type">
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<div><font size="2" face="Arial" color="#000080">Hi Albrecht,</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">Regrettably,
you appear to have misread my text. If you read it again
more carefully, you will see that at NO point do I propose,
or even suggest, that acceleration gives rise to time
dilation. I am well aware that, as you say, "gravity and
acceleration are different regarding [time] dilation" - so
your attempts to persuade me of this are quite unnecessary.</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">The whole point
of my text was, as I said at the outset, to resolve the
'twins going opposite directions around a circle' paradox,
with reference to classical SR (and GR, as it happens - bear
with me on this). For SR to be self-consistent (which I
believe it is - that's not the same as it being correct!)
there has to be an explanation that fits the terms of
Relativity which explains how it can be that both A and B
would expect their clocks to coincide on re-meeting - as
they clearly would from the perspective of a third observer,
static with respect to the circle centre, and so they must
of course coincide from everyone's perspective. If it can
be shown that they'd expect their clocks to be different
then Relativity is dead - but it is most definitely not that
simple! [That's why it's survived for over a century; it's
not just that thousands of other physicists over that
century have been incapable of such analysis!]</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">Relativity
states that any scenario can validly be assessed from the
perspective of any individual, who may consider themself to
be static - and that their assessment of that scenario is
equally 'correct' to any OTHER assessment from any other
frame of reference. SR restricts such assessment to
inertial frames, GR extends it to non-inertial frames - but
this same principle holds true.</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">We can add to
this the fact that if such an observer experiences what we
might refer to as a 'G-force' acting on them then they will
know that they must be in a non-inertial frame. The term
'G-force' is convenient for our purposes as it is used to
apply both to forces due to gravitation and to accelerating
forces; it is implicit in GR (through the Equivalence
Principle) that the observer will not know which of these
two applies (Einstein's 'man in a box' thought experiment),
but that if (as he fully validly may, under Relativity) he
considers himself to be at rest then he must necessarily
attribute such forces to gravitational effects (without
having to ascertain where those effects arise from - that
could be tricky in our example scenario!)</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">Please not that
I am NOT saying that these principles actually apply in our
physical reality - I am simply stating the mantra of
Relativity, both SR and GR, since that's the mathematical
framework in which I'm seeking to show self-consistency.
Others in the group are proposing that Relativity is
disproved by this 'twins thought experiment', I'm observing
that it is not; the truth or falsehood of Relativity as a
model of true reality is not what I'm about here - in fact
I'm seeking to show that Relativity CANNOT be disproved by
such a simple setup, it needs rather more thought than that!</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">Albrecht, I
think you misunderstand my purpose here. It's not my
intention EITHER to prove OR to disprove Relativity; my sole
intention is to show that this 'twins scenario' does NOT
show an inconsistency in Relativity - it is NOT a paradox.
In this respect the question of whether Relativity does or
does not match true objective reality is totally irrelevant;
the only question is whether or not Relativity agrees with
itself.</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">The importance
if this exercise shouldn't be underestimated: if we are to
challenge the fundamental premises of Relativity, it has to
be on FAR stronger ground than a proposed 'paradox' that has
been refuted time and time again over the past 100 years -
we do ourselves, and science, a serious disservice if we
convince fans of Relativity that our view that it's wrong is
based on a simplistic misunderstanding of its basics!</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">So, again:
external observer sees A and B perform mirror images of each
others' manoeuvres - so of course clocks will match on
re-meeting. So A and B will also see clocks coinciding -
and fully expect that to be the case. How come, given that
Relativity allows each to see their position in the universe
as static?</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">Simple: since
the external observer sees A (for example) as experiencing
acceleration towards the centre of the circle, A him/herself
will inevitably experience a G-force acting outward from the
centre of that observer's circle. Considering him/herself
static in space, A will have no option but to regard that as
a gravitational effect from some unknown source (note that
physicists have no trouble envisaging gravitation acting
from unknown sources - we're told that such sources make up
the vast majority of the mass-energy in our universe!).
Since A knows that gravitation causes time dilation (NOTE
THAT I AM <strong>NOT</strong> PROPOSING, HERE OR
ANYWHERE, THAT ACCELERATION CAUSES SUCH DILATION), he/she
will inevitably expect their clock to have been slowed, as
well as knowing that B's motion will have also slowed B's
clock. So matching of clocks on re-meeting is to be totally
expected by A (and B) - no paradox.</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">This is all
about perceptions from different perspectives, and the
assertion in Relativity that all such perceptions are
equally valid/true.</font></div>
<div> </div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">With regard to
assessing time and distance of B, as assessed by A: whilst
not relevant to this analysis, the question has arisen - so
let's look at it from A's perspective. A sends out a
broadcast radio signal in the general direction of B; on
receiving that signal, B sends a time-stamped response
(broadcast in A's general direction); From the time between
sending and receipt, 'knowing' such signals to travel both
ways at c relative to him/herself (according to SR), A can
calculate the distance to B at the time B responded - which
will be halfway between send and received, from A's
perspective; A will also have a record of B's clock-time at
that point halfway between A's send and receive - and so an
indication of how B's time is progressing compared with A's
[This is all according to SR 'rules', I'm not proposing that
A's assessments will in fact be correct in absolute terms -
though of course SR considers them to be equally correct to
any other view].</font></div>
<div> </div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">Having glanced
briefly at Wolf's latest response, I'd just say that
mass-energy considerations can also be very misleading in a
Relativistic scenario, unless handled exceedingly carefully
with full regard for different perspectives. As a very
simple illustration: A single photon observed from one
reference frame may be red- or blue-shifted when observed
from a different frame, and so carry different energy.
Extension of this to massive energetic particles, and
applying mass-energy equivalence, makes it clear that we
can't simply assess the mass-energy characteristics of an
object or system from one frame then simply carry those
measures across to another frame. I don't know whether this
has a bearing on Wolf's comments, I didn't get to see much
of what you sent previously, Wolf, for some reason.</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">Best regards,</font></div>
<div><font size="2" face="Arial" color="#000080">Grahame</font></div>
<div> </div>
<div> </div>
<div> </div>
<div> </div>
<div>----- Original Message ----- </div>
<blockquote style="BORDER-LEFT: #000080 2px solid; PADDING-LEFT:
5px; PADDING-RIGHT: 0px; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px">
<div style="FONT: 10pt arial; BACKGROUND: #e4e4e4; font-color:
black"><b>From:</b> <a title="phys@a-giese.de"
href="mailto:phys@a-giese.de" moz-do-not-send="true">Albrecht
Giese</a> </div>
<div style="FONT: 10pt arial"><b>To:</b> <a
title="general@lists.natureoflightandparticles.org"
href="mailto:general@lists.natureoflightandparticles.org"
moz-do-not-send="true">general@lists..natureoflightandparticles.org</a>
</div>
<div style="FONT: 10pt arial"><b>Sent:</b> Monday, August 28,
2017 4:21 PM</div>
<div style="FONT: 10pt arial"><b>Subject:</b> Re: [General]
[NEW] SRT twin Paradox</div>
<div><br>
</div>
<p><font size="-1" face="Arial">Hi Grahame,</font></p>
<p><font size="-1" face="Arial">sorry, but I find a very
fundamental error in your arguments: You describe a pair
of twins which observe each other in a situation where
they are permanently accelerated. And then you argue with
dilation caused by gravity. But that does not fit the
physical reality.</font></p>
<p><font size="-1" face="Arial">Gravity and acceleration are
different regarding dilation. Gravity causes dilation, no
question. But acceleration does not cause dilation. How
can one know? 1) You find this in every textbook about
special relativity; 2) it was experimentally proven in the
Muon storage ring at CERN. The extension of the life time
of the muons was only dependent on the actual speed of the
particles, not on the very strong acceleration in the
ring. If that would have been an effect according to an
equivalent gravitational field, their lifetime would have
to be extended by an additional factor of roughly 1000
compared to the results observed.<br>
</font></p>
<font size="-1" face="Arial"><br>
</font>
<div class="moz-cite-prefix"><font size="-1" face="Arial">Am
27.08.2017 um 22:18 schrieb Dr Grahame Blackwell:</font><br>
</div>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<meta name="GENERATOR" content="MSHTML 8.00.6001.23588">
<style></style>
<div><font size="2" face="Arial" color="#000080">Hi
Albrecht,</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">I'm afraid
I have to disagree with you on a couple of points.</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">First, I
agree completely that gravitation doesn't come under
SR. However the concept of gravitation is essential to
explanation of the 'twins going in opposite directions
around a circle and meeting on the far side'
(non-)paradox. [It may be that in your view this
scenario cannot then be simply a playing-out of SR, it
must be a GR issue?]</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">Consider:
Twin A and twin B each view themselves as being static,
with the other twin tracing out a path that takes them
away and then brings them back into proximity from a
different direction, having formed a loop of some kind;
however, from the point of view of an observer static
with respect to the centre of a large circle, A and B
have started together at some point on the perimeter of
that circle and have each followed opposite halves of
that circle to meet again on its other side. I.e. from
the perspective of that observer the motions of A and B
are symmetric, so their clocks (synchronised at the
start) will still be synchronised when they meet again.
[We're assuming here that this all takes place in deep
space, far from any gravitational influences.]</font></div>
</blockquote>
<font color="#000080"><font size="2"><font face="Arial">None
of the twins can view himself as being static, because
they are accelerated all the time and they will notice
that. So the laws of SR are not applicable for this
process in a simple way.</font></font></font><br>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<div> </div>
<div><font size="2" face="Arial" color="#000080">From A's
point of view, A has remained static and B has performed
a large loop in space, finally coming back alongside A.
According to SR, therefore, A will observe a
slowing-down of B's clock and so will expect B's clock
to have lost time, in real terms as measured in A's
frame (if it were an inertial frame). <br>
</font></div>
</blockquote>
<font color="#000080"><font size="2"><font face="Arial">No, it
is not an inertial frame.</font></font></font><br>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<div><font size="2" face="Arial" color="#000080">[We can
deal with the issue of A reading B's clock whilst B is
on the move by B digitally emitting their clock-time at
intervals, to be received by A who will assess those
transmissions on the basis of their crossing space at
speed c across the distance that A measures B to be from
him at times of transmission - this could be done fairly
easily by A keeping a record of B's distance at
all times as measured on A's clock.]</font></div>
</blockquote>
<font color="#000080"><font size="2"><font face="Arial">Also
this is not possible. A can receive signals from B, but
he does not know the distance. According to SR this
distance is not clearly defined because the assessment
of any distance depends on the motion state of the
observer. Which speed will A assume for himself? He
cannot assume to be static as he notices to be
accelerated.</font></font></font><br>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<div> </div>
<div><font size="2" face="Arial" color="#000080">B will have
a corresponding mirror-image experience of A's motion,
and so will expect A to have lost time in real (B-frame)
terms. This appears to suggest that both A and B would
each expect the other's clock to have fallen behind
their own - a paradox.</font></div>
</blockquote>
<font color="#000080"><font size="2"><font face="Arial">Also
regarding time a similar problem like for distance is
applicable. When are signals in different frames
synchronised or when is time is running faster or
slower? For any observer in different frames the result
of this question may be different. </font></font></font><br>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<div> </div>
<div><font size="2" face="Arial" color="#000080">However,
our external observer will have seen A performing a
circular course - so A will inevitably have experienced
a 'G-force' of some kind (centripetal, from our
observer's persective). Since A considers him/herself
to be static, he/she MUST attribute this to some
gravitational influence - indeed, from the SR/GR
perspective there must indeed be a gravitational
influence in A's frame, from the perspective of that
frame; one just does not get G-force without either
acceleration or gravitation. (Here, of course,
Relativity begins to become unravelled, as A is far from
any massive body that could give rise to a gravitational
field - maybe they'll need to start inventing their own
local 'dark matter'). Note that the scenario being
considered - A and B traversing opposite sides of a
circle - involves NO gravitational fields - BUT A and B
would HAVE TO PRESUME the existence of such a field in
their reference frame if they are to reconcile a force
they're experiencing with their assumption that they are
static (a totally valid assumption, in Relativity
terms).</font></div>
</blockquote>
<font color="#000080"><font size="2"><font face="Arial">As
said above, even if both, A and B, attribute the force
of acceleration to gravity, they are in error; and it
does anyway not help the situation. For your
consideration they need a gravitational field for
dilation, but this does not exist, and acceleration does
not replace it.</font></font></font><br>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<div> </div>
<div><font size="2" face="Arial" color="#000080">Resolution
of this (apparent) paradox, as I said before, rests on A
(and likewise B) considering themselves to have been
subject to a gravitational field - and experiment shows
us that gravitational fields slow time - so their own
clock will have slowed as well as the others. So they
will both expect their clocks to be synchronised on
re-meeting.</font></div>
</blockquote>
<font color="#000080"><font size="2"><font face="Arial">That
is anyway true also in the absence of dilation.</font></font></font><br>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<div> </div>
<div><font size="2" face="Arial" color="#000080">As I say,
this is where Relativity begins to become unravelled: A
and B will either each have to acknowledge that they are
NOT in fact static, or they will have to invent a
convincing explanation for a gravitational effect in the
absence of any 'ponderous mass' (to use Einstein's
term). But given that, synchronisation of clocks is not
an issue - as long as we allow A and B to each presume
existence of a gravitational field in their frame
(which, as you say, takes it into the sphere of GR).</font></div>
</blockquote>
<font color="#000080"><font size="2"><font face="Arial">Not
applicable as mentioned above.</font></font></font>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<div> </div>
<div><font size="2" face="Arial" color="#000080">Second
point: in your case of the travelling-twin versus the
stay-at-home twin, the traveller would again experience
G-force, which they could if they wish regard as a
gravitational effect (since under Relativity they are
free to consider themselves as static). They would
therefore expect their clock (including biological
clock) to have slowed (Pound-Rebka again), and so know
that they have actually been travelling more than one
year in 'objective' terms - whatever that might mean in
this context.</font></div>
</blockquote>
<font color="#000080"><font size="2"><font face="Arial">The
twin travelling, B, cannot assume that he is static
because he has to notice his acceleration. And that is
different from gravity. And even if it could be
identified with gravity this would not solve the example
which I have given.</font></font></font><br>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<div> </div>
<div><font size="2" face="Arial" color="#000080">But of
course the reality is that slowing of time is NOT
symmetric, it's a consequence of motion with respect to
the unique objectively-static universal reference
frame. Only when serious scientists start asking WHY
Relativity does (or appears to do) what it does will we
make any progress on this issue.</font></div>
</blockquote>
<font color="#000080"><font size="2"><font face="Arial">Which
progress to you expect? There is no symmetry in the case
where twin B returns and so you cannot conclude anything
from symmetry.</font></font></font><br>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<div> </div>
<div><font size="2" face="Arial" color="#000080">I think
we're agreed on the key issues. Perhaps it's time to
stop discussing how a self-consistent mathematical
system (which doesn't happen to match true reality)
copes with paradoxes of its own making!</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">Best
regards,</font></div>
<div><font size="2" face="Arial" color="#000080">Grahame</font></div>
</blockquote>
<font color="#000080"><font size="2"><font face="Arial">As I
have mentioned in the other mail: It is in conflict with
Einstein's relativity to compare clocks residing in
different frames. The result of any comparison depends
on the motion state of the observer. That is what
Einstein says.<br>
<br>
But the other solution is to follow the Lorentzian
relativity. In that case the imagination becomes easy
(in contrast to Einstein).<br>
<br>
Greetings back<br>
Albrecht.<br>
</font></font></font>
<blockquote
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent"
type="cite">
<div> </div>
<div> </div>
<div> </div>
<div>----- Original Message ----- </div>
<blockquote style="BORDER-LEFT: #000080 2px solid;
PADDING-LEFT: 5px; PADDING-RIGHT: 0px; MARGIN-LEFT: 5px;
MARGIN-RIGHT: 0px">
<div style="FONT: 10pt arial; BACKGROUND: #e4e4e4;
font-color: black"><b>From:</b> <a
title="phys@a-giese.de" href="mailto:phys@a-giese.de"
moz-do-not-send="true">Albrecht Giese</a> </div>
<div style="FONT: 10pt arial"><b>To:</b> <a
title="general@lists.natureoflightandparticles.org"
href="mailto:general@lists.natureoflightandparticles.org"
moz-do-not-send="true">general@lists..natureoflightandparticles.org</a>
</div>
<div style="FONT: 10pt arial"><b>Sent:</b> Sunday, August
27, 2017 7:48 PM</div>
<div style="FONT: 10pt arial"><b>Subject:</b> Re:
[General] [NEW] SRT twin Paradox</div>
<div><br>
</div>
<p>Hi Grahame,</p>
<p>without going into details of this discussion I only
want to point to the following fact:</p>
<p>Whereas you are of course right that the twin situation
is not a paradox but logically clean, what we all as I
think have sufficiently discussed here, the following is
not correct in my view:</p>
<p>The twin situation has absolutely NOTHING TO DO with
gravity.</p>
<p>Two arguments for this:</p>
<p>o The so called twin paradox is purely Special
Relativity. Gravity on the other hand, is General
Relativity. This is the formal point.</p>
<p>o From practical numbers it is visible that gravity
cannot be an explanation. Take the usual example saying
that one twin stays at home and the other one travels -
as seen from the twin at home - for twenty years away
and then twenty years back. From the view of the twin at
home, at the other ones return 40 years have gone. For
the travelling twin only one year has gone (This case is
theoretically possible if the proper speed is taken,
about 0.9997c)). Then the travelling twin would have
saved 39 years of life time. Now look at the possible
influence of gravity: Assume it takes the travelling
twin a year to change his speed from almost c to almost
- c , then, even if the speed of proper time would
decrease to zero, he would have saved only one year.
But, in this example, he has saved 39 years. How could
this work? No one in physics assumes that proper time
can run inversely. So this is no possible explanation.</p>
<p>How is it explained? I do not want to repeat again and
again the correct (but a bit lengthy) explanation, but I
attempt to give a short version: In Einstein's
relativity the run of time in different frames can
logically not be continuously compared, it can only be
compared at interaction points where two clocks (or
whatever) are at the same position. And the
determination of the situation at such common position
has to be done by the Lorentz transformation. And this
determination works, as many times said here, without
logical conflicts.</p>
<p>If you solve this problem using the Lorentzian SRT,
then the result is the same but the argument is
different, more physics-related, and also better for the
imagination. If wanted, I can of course explain it.</p>
<p>Albrecht<br>
</p>
<p><br>
</p>
<br>
<div class="moz-cite-prefix">Am 27.08.2017 um 01:13
schrieb Dr Grahame Blackwell:<br>
</div>
<blockquote
cite="mid:CCE2F4D7ECF5430E943629241B634443@vincent"
type="cite">
<meta name="GENERATOR" content="MSHTML 8.00.6001.23588">
<div><font size="2" face="Arial" color="#000080">I'm
sorry Wolf, but it seems that you're still not
getting it.</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">This
situation can be explained fully logically WITHOUT
either twin making any assumptions about SR or GR -
simply from their own observations and from
well-proven experimental findings.</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">If we
label the twins A and B, then their situations are
effectively symmetric* - so we'll consider the
scenario from the viewpoint of twin A.</font></div>
<div><font size="2" face="Arial" color="#000080">A
considers him/herself static, and all motion to be
attributable to twin B. So - and this agrees with
experimental observation of clocks at high speed (in
planes and in GPS satellites) - twin A will observe
twin B's clock running slow, if A's own clock is not
upset by any effect. HOWEVER, since A is actually
travelling in circular motion, (s)he will experience
a centripetal force; assuming him/herself to be
static, this will necessarily be attributed to
gravitational effects - and it's well known from
experiment (Pound-Rebka and successors) that
gravitational fields cause time dilation - so A will
expect their own clock to be running more slowly
also due to that 'gravitational' effect (note that
this is not any assumption of SR or GR, simply
inference from proven experimental results) [and so
also A's observation of B's clock, measured against
A's own clock, will not fit the standard SR
time-dilation model, for reasons that A will fully
comprehend]. For A, the cumulative time-dilation
for B's perceived relative speed and for A's own
perceived 'gravitational' effect exactly balance -
so A will fully expect both clocks to coincide when
the twins meet again (as B will also).</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">No
paradox.</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">* It
needs to be said that further study of causation of
'relativistic time dilation' leads to the
understanding that this is an objective effect due
to travelling at speed relative to the unique
objectively-static universal reference frame. So if
the centre of the circle traced out by A and B is
itself in motion relative to that reference frame
then it cannot be assumed that A's and B's motions
will be symmetric; in that case their clocks may
well not be precisely synchronised on their meeting
again. This is an observation relating to physical
reality, which in no way contradicts the
self-consistency of SR (or GR) as a mathematical
system.</font></div>
<div> </div>
<div><font size="2" face="Arial" color="#000080">Best
regards,</font></div>
<div><font size="2" face="Arial" color="#000080">Grahame</font></div>
<div> </div>
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