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<p><font size="-1" face="Helvetica, Arial, sans-serif">Hi Grahame,</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">you are right
that you did not attribute dilation to acceleration. Sorry if I
was not reading carefully. But on the other hand you understand
that the twins A and B will attribute their acceleration to
gravity and so explain why they observe dilation in the system
of the other one as it is. But also this cannot explain any
dilation as observed by the twins, as the influence of gravity
to be assumed here, is too small to explain what has to be
explained by the twins observing each other. <br>
</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">Your example
of the twins moving on half-circles towards a common point is a
quite interesting one. But it is not a simple example as both, A
and B, are permanently accelerated. This is not covered by Standard-SRT
to say it this way. <br>
</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">To find an
explanation by only using SRT I would like to describe first
another, also symmetric, experiment.<br>
</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">This is a
twin experiment which is symmetric in contrast to the usual one
(not yours). (I have discussed this earlier with Wolf). Here both
twins will move from a position at rest into opposite directions
with a constant speed v(1). Then, after a fixed time, t(1), they
will both turn quickly and return to each other. How does dilation
</font><font size="-1" face="Helvetica, Arial, sans-serif">behave
</font><font size="-1" face="Helvetica, Arial, sans-serif">here? -
Both twins will see themselves at rest and see the other one
moving off with the speed v(1+1) (i.e. adding the speeds
relativistically). Now let's look at the view of A. A assumes in
the beginning to be at rest. And for A the other one, B, moves off
with speed v(1+1). After the time t(1) A will now start to move
towards B. A will measure his acceleration and accelerate until
he reaches v(1+1) with respect to his previous state. B has
turned in the same moment, and in the view of A he will have a
speed of 0 with respect to the previous state of A. Now the
summary as seen from A: A moved in the first period with speed 0
as this was his frame at rest. For the second period he has
moved with v(1+1). B has in the view of A moved in the first
period with a speed v(1+1) off, in the second period with speed
0 with respect to the original frame of A. (B will see this whole
process mirrored.) So now, with respect to dilation, they have
both undergone a symmetric history as seen from the original
position of each one. And when they meet again, they will not be
surprised to have the same clock indications.</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">Now a little
modification. Assume that in this experiment both twins do not
move at constant speed but at a variable speed, but in a way
that at same times both move at the same speed, only in different
directions. Also this experiment would have the same result.
This can be understood if we decompose the variable speed into
infinitesimal pieces of constant speed, determine the dilation
for these parts, and integrate the parts for the whole process.
<br>
</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">The only way
in my view to solve your (circular) example in the scope of SRT
is to decompose the motion process into infinitesimal linear
pieces and to treat these pieces by SRT. Now, in the view of A,
it moves for an infinitesimal distance straight. And at the same
time also B moves at the same time in infinitesimal distance.
Then, after treating and then integrating all infinitesimal
pieces we get the results of both paths until the meeting point.</font><font
size="-1" face="Helvetica, Arial, sans-serif"> But first, we
have to treat as an example a typical infinitesimal piece for
both twins. For this, we can decompose the slant direction of
each infinitesimal piece into a component so that the one
component (for both twins) points straight away from each other,
parallel but in opposite directions. And the other component now
is perpendicular to this one and points roughly into the
direction of the final point. And now, let's treat both pairs of
components separately. <br>
</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">The
components towards the destination (but parallel to each other)
are similar for A and B. So from the view of A of B the other
one has the same speed vector and there is no reason for dilation
in the view of both. <br>
</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">And now the
other component. That is a situation discussed in my example
above. In that other dimension both twins move away from each
other and then return to each other in a symmetric way. And as
visible in my example above the will understand to have similar
clock indications at the end.</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">So, this is
an explanation for your experiment of circular motion which is
here treated only using SRT (no need for gravity). It it shows
that nothing happens which appears to the twins as a paradox. <br>
</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">A further
little remark to your mail text below: It is true that in the
view of Einstein an observer cannot distinguish between the effect
of gravity and of acceleration </font><font size="-1"
face="Helvetica, Arial, sans-serif"><font size="2" face="Arial"
color="#000080">("Einstein's 'man in a box' thought
experiment")</font>. But that is falsified meanwhile. Because
if the observer has a charged object with him, this will radiate
at acceleration but not at rest in a gravitational field. <br>
</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">I hope that
my explanations were understandable as I did not present a drawing.
- Or should I present one?</font></p>
<p><font size="-1" face="Helvetica, Arial, sans-serif">Best regards<br>
Albrecht<br>
</font></p>
<br>
<br>
<div class="moz-cite-prefix">Am 29.08.2017 um 12:41 schrieb Dr
Grahame Blackwell:<br>
</div>
<blockquote type="cite"
cite="mid:85328F3496EE444CAE0459B5C85FD35F@vincent">
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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">
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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">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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