Jin He, who might have some basic training in photometrics, is of more use as meat than a physicist. But
Jin He doesn't address Autodynamics at all. What Jin He mostly demonstrates is how to abuse arix.org -- the pre-print server. In Jin He's most recent paper he assumes someone has actually performed an experiment in reaction to his unpublished paper.
Jin He is completely mistaken that a vertical orientation of the Brillet-Hall experiment will test Local Lorentz Invarience.
C.M. Will's paper remains the gold-standard.
http://arxiv.org/abs/gr-qc/0510072http://relativity.livingreviews.org/Articl...06-3/fig_2.htmlQUOTE
One of these is the Brillet-Hall experiment [48], which used a Fabry-Perot laser interferometer. In a recent series of experiments, the frequencies of electromagnetic cavity oscillators in various orientations were compared with each other or with atomic clocks as a function of the orientation of the laboratory [297, 169, 191, 14, 249]. These placed bounds on (c^−2 − 1) at the level of better than a part in 10^9.
Has truth2k been kicked off the forum?
Dr. Ricardo Carezani, author of Autodynamics, embraces RelativityIn the flawed page
Galilean Simplification, on the Autodynamics web site, you can clearly see the author embrace the principle of Galilean relativity. So let's create a test-theory of Galilean relativity and see what we learn.
Let's start with "All inertial frames are “equivalent”, which is where the “Relativity” in Special Relativity comes from. It makes no sense to say that something is moving; it only makes sense to say that one thing is moving with respect to another. Outside of geocentric theories, it is hard to imagine a “reasonable” universe where the relativity postulate does not hold. Galileo of course was not a geocentrist, and the aberration of starlight as show in the 1600's seemed to kill geocentricism.
Galileo says for co-moving frames, x = x' + v t' and t = t', which is pretty simple. Let's make a test theory of Galilean relativity and use experiment to test it.
Lets start with x = A(v) x' + b(v) t' AND t = c(v) t' + d(v) x'
Physically, these two frames must travel with respect to each other with speed v. And if S sees S' move at speed v, then S' should see S move at speed -v with every thing being the same up to a minus sign. Any time dilation should be symmetrical, because otherwise one of S or S' would be a preferred frame.
Since S sees S' move at speed v, then x'=0 must result in x/t = b(v)/c(v) = v, so b(v) = v c(v)
Since S' sees S move at speed -v, then x=0 implies x'/t' = - v c(v)/A(v), therefore A(v) = c(v)
Right now, we have
x = A(v) x' + v A(v) t' AND t = A(v) t' + d(v) x'
If there is a time dilation effect, it needs to be symmetric. For x' = 0, t = A(v) t', but for x = 0, we have
x' = - v t' which means t = A(v) t' -v d(v) t' or t' = t / (A(v) - v d(v) )
Setting these two time-dilations as equal to each other, we have
A(v) = 1 / (A(v) - v d(v)) or d(v) = (1/v) (A(v) - 1/A(v))
Therefore our test theory of Galilean relativity is
x = A(v)(x' + v t') AND t = A(v)(t' + (1/v) (1 - A(v)^-2) x')
which starts with the most basic generalization of Galilean relativity with just internal consistency.
Now, A(v) is some number, but it's a little inconvenient for our purposes, so let's define B(v)^2 = v^2/(1 - A(v)^-2)
In which case A(v) = 1/sqrt(1 - v^2/B(v)^2)
And our relations are
x = (1/sqrt(1 - v^2/B(v)^2))(x' + v t') AND t = (1/sqrt(1 - v^2/B(v)^2))(t' + (v/B(v)^2) x')
or
x = A(v)(x' + v t') AND t = A(v)(t' + (v/B(v)^2) x')
But since "All inertial frames are “equivalent”, we can compare our results at two different values of v, v1 and v2.
That leads to two different values of B1 = B(v1) and B2 = B(v2) and A1 = 1/sqrt(1 - v1^2/B1^2) and A2 = 1/sqrt(1 - v2^2/B2^2)
The result must be in the same form as our relativity transformation.
Applying our relations two times, first for v1 and then for v2, we have
x' = A1(x'' + v1 t'') AND t' = A1(t'' + (v1/B1^2) x'')
x = A2(x' + v2 t') = A1A2(x'' + v1 t'' + v2 t'' + v2 (v1/B1^2) x'') = A1A2((1 + v1v2/B1^2)x'' + (v1+v2) t'' )
t = A2 (t' + (v2/B2^2) x') = A1A2 (t'' + (v1/B1^2) x'' + (v2/B2^2) x'' + (v2/B2^2) v1 t'') = A1A2 ((1 + v1v2/B2^2)t'' + ((v1/B1^2) + (v2/B2^2)) x'' )
But since the coefficients of x'' and t'' will be different unless B1^2 = B2^2, then B(v)^2 must be a constant, independent of v. Let's write B(v)^2 = C^2.
So what is the value of C^2 ?
If C^2 is infinite, then A(v) = 1 and the equations of Galilean Relativity result.
If C^2 is finite and positive, then we have Lorentz transformations where sqrt(C^2) is the limiting speed of objects.
If C^2 is zero, then any non-zero v makes A(v) imaginary and infinite. Nothing could ever move.
If C^2 is negative, then we get a weird result where the possibility of going backwards in time results. This is apparently not physical.
So is C^2 finite or infinite?
sqrt(C^2) >= c because we have measured things moving at c. Indeed, every experiment is consistent with sqrt(C^2) = c, the speed of light.
Therefore, if Carezani embraces experiment and logic, then Carezani embraces Relativity.
Thanks to Pentcho Valev for raising this question and thanks to David Morin for answering it.
David Morin.
THERE ONCE WAS A CLASSICAL THEORY…: Introductory Classical Mechanics, with Problems and Solutions, 2004
http://www.courses.fas.harvard.edu/~phys16/Textbook/ch10.pdfhttp://www.courses.fas.harvard.edu/~phys16..._appendices.pdf