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<p>Have neither of you looked at my drawing of an orbiting charged
particle around another heavier charged particle and</p>
<p>That all particles are actually finite and all finite particles
even macro particle will have their charge and mass centers
displaced and therefore</p>
<p>consume energy and is a completely un appreciated effect in all
accelerated particles by eM means in a gravito-electric field.<br>
</p>
<p>I do not know were the drawing went its in Grahame's 8/26/2017
4;13PM REPLY<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">wolf@NascentInc.com</a></pre>
<div class="moz-cite-prefix">On 8/28/2017 8:21 AM, Albrecht Giese
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:6da07d65-e10d-d813-a457-f54b8abe03ca@a-giese.de">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<meta content="text/html; charset=utf-8"
http-equiv="Content-Type">
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<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<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 type="cite"
cite="mid:0C943D193FFC4E568685FEB833FA0E59@vincent">
<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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