<div dir="ltr"><div><div><div>Hi Carl,<br><br></div> I think your point is already well known - that all observers "stationary" at sea level experience the same rate of time dilation - I wrote for example in my 2010 article on time dilation: <br><br><b><span style="line-height:115%;font-family:"arial","sans-serif";font-size:14pt"><span lang="EN-US"><span class="gmail-help1"><span lang="EN-US"><font face="Times New Roman"><font face="Times New Roman">"Applying
a model of time dilation to the earth is considerably more complicated.
Einstein in his </font></font></span><font face="Times New Roman"><font face="Times New Roman"><a href="http://www.conspiracyoflight.com/pdf/Einstein-On_the_Electrodynamics_of_Moving_Bodies.pdf"><span lang="EN-US">1905 article</span></a><span lang="EN-US"> argued that an observer on the
equator would experience more time dilation due to their rotation at </span>v<sup>2</sup>/2c<sup>2</sup><span lang="EN-US">,
as compared to an observer at the pole. This has been now shown not to
be true; - the rate of clocks at sea level (or more precisely, on </span><a href="http://en.wikipedia.org/wiki/Geoid"><span lang="EN-US">earth's geoid</span></a><span lang="EN-US">
surface) all count at the same rate all over the globe. The reason for
this is that the two competing time dilation effects, gravitational
time dilation and velocity time dilation, cancel out. The conventional
argument goes that earth is an oblate sphere due to the equator bulging
out from the centrifugal force. Since the pole is closer to the center
than the equator, the pole is at a larger negative gravitational
potential -GM/r, and thus experiences a larger time dilation gh/c</span><sup>2</sup><span lang="EN-US">
than the equator. However, since the equator observer is rotating at
the tangential velocity v, they experience a counter-acting velocity
time dilation </span>v<sup>2</sup>/2c<sup>2</sup><span lang="EN-US">,
which is sometimes referred to as the centripetal term. The height
effect at the pole is almost twice as large as the velocity effect at
the equator; the argument is then that the distribution of mass in the
earth due to its oblateness creates additional gravitational forces
that serve to balance this out. This last term is the gravitational
quadrupole term. When all three terms are added together, the time
dilation on the geoid is the same rate everywhere on the globe. " - this explanation was derived directly from Ashby's papers as I recall. <br><br></span></font></font></span></span></span></b></div><b><span style="line-height:115%;font-family:"arial","sans-serif";font-size:14pt"><span lang="EN-US"><span class="gmail-help1"><font face="Times New Roman"><font face="Times New Roman"><span lang="EN-US">I think the real problem is that you dispute the existence of the Sagnac effect - this is something different than the above, namely the Sagnac effect arises when a beam of light, or a clock, rotates with respect to the "fixed stars", and if that rate of rotation differs from your "stationary" observer on the earth, then there is an observable effect (clocks change their rate, light beams change their arrival time). The Sagnac effect is not relativistic as it dominates at low velocities. <br><br></span></font></font></span></span></span></b></div><b><span style="line-height:115%;font-family:"arial","sans-serif";font-size:14pt"><span lang="EN-US"><span class="gmail-help1"><font face="Times New Roman"><font face="Times New Roman"><span lang="EN-US">Doug<br></span></font></font></span></span></span></b></div><div class="gmail_extra"><br><div class="gmail_quote">On Wed, Nov 2, 2016 at 11:12 AM, <span dir="ltr"><<a href="mailto:cj@mb-soft.com" target="_blank">cj@mb-soft.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><u></u>
<div bgcolor="#ffffff">
<div><font size="2" face="Arial">Your group certainly has many creative ideas
regarding Einstein and Relativity. Unfortunately, nearly all of the
assumptions are not correct.</font></div>
<div><font size="2" face="Arial"></font> </div>
<div><font size="2" face="Arial">The Earth provides wonderful evidence for us,
moostly because we have really accurate data rwegarding its (Equatorial)
diameter and its precise sidereal rotation period.</font></div>
<div><font size="2" face="Arial"></font> </div>
<div><font size="2" face="Arial">Specifically regarding your
discussions:</font></div>
<div><font size="2" face="Arial"></font> </div>
<div><font size="2" face="Arial">Someone standing at the North Pole does not have
"velocity" nor "acceleration" and so the Lorentz-Fitzgerald (beta) factor is
exactly ONE. (velocity is ZERO).</font></div>
<div><font size="2" face="Arial"></font> </div>
<div><font size="2" face="Arial">Fine. The more interesting situation is for a
man standing at the Equator. He DOES have a precisely known (v) there
and so it is easy to calculate the Time Dilation Factor to be <strong><font size="5" face="Times New Roman">0.999 999 999 998 796
560</font></strong></font></div>
<div><strong></strong> </div>
<div><font size="2" face="Arial">And so everyone ASSUMES that he would EXPERIENCE
Time Dilation there. There are TWO major errors in those
assumptions. (a) HE would not experience any Relativity issues, and only
the OBSERVER at the North Pole (motionless) would SEE any such effects.
Nearly everyone is wrong about this error.</font></div>
<div><font size="2" face="Arial"></font> </div>
<div><font size="2" face="Arial">(b) the other error in assumptions is due to
everyone IGNORING General Relativity, which happens to have its own time-rate
effect. It is called the Equivalency Principle. (Surprisingly) this
effect happens to be the OPPOSITE of Special Relativity's Time Dilation.
Thaht calculation (onn Earth) is also easy and precise to do, and it is
<strong><font size="5" face="Times New Roman">1.000 000 000 001 203
440</font></strong></font></div>
<div><strong><font size="5"></font></strong> </div>
<div><font size="2" face="Arial">Oddly enough, since BOTH Relativity time-rate
effects must constantly apply to our guy att the Equator, we must MULTIPLY these
two effects to find the ACTUAL Relativity time-rate effect for that guy at the
Equator. That simple product is EXACTLY ONE.</font></div>
<div><font size="2" face="Arial"></font> </div>
<div><font size="2" face="Arial">In other words, the ACTUAL effect due to BOTH forms
of Relativity exactly cancel each other out. There is NO DIFFERENCE in
time rate between the guy at the North Pole nd anyone at the Equator, or
anywhere else on Earth.</font></div>
<div><font size="2" face="Arial"></font> </div>
<div><font size="2" face="Arial">Only when someone is in a Non-Inertial Reat Frame,
that is, accelerating, will an OBSERVER see any time-rate effects.</font></div>
<div><font size="2" face="Arial"></font> </div>
<div><font size="2" face="Arial">A complete explanation is at <a href="http://mb-soft.com/public4/dilation.html" target="_blank">http://mb-soft.com/public4/<wbr>dilation.html</a></font></div>
<div><font size="2" face="Arial"></font> </div>
<div><font size="2" face="Arial">Carl Johnson</font></div>
<div><font size="2" face="Arial">(A Theoretical Physicist, educated at the
University of Chicago)</font></div>
<div><font size="2" face="Arial"></font> </div></div>
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