Re: What is Proper Time?

In sci.physics.relativity, HW@....(Dr. Henri Wilson)
<HW@>
wrote
on Fri, 07 Mar 2008 22:08:00 GMT
<8ue3t3dlj5le0r4rqhhvgsg6ii7mqfscj2@xxxxxxx>:
On Thu, 6 Mar 2008 20:59:16 -0800, The Ghost In The Machine
<ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote:

In sci.physics.relativity, HW@....(Dr. Henri Wilson)
<HW@>

We can tell that t12 = t'12 and t23 = t'23 but that tells us absolutely
nothing about the relationship of t12 and t23, or t'12 and t'23.

If one assumes absolute time, the fact that TWLS experiments show the
same speed indicates that there's no "headwind". The path there
is S-M-S, and the S-M path goes at speed c-v (velocity v-c) with the
M-S path c+v if one has a headwind with speed v. The total transit time
is therefore

t_h = d/(c-v) + d/(c+v)

(or (-d)/(v-c) + d/(v+c) )

= (d(c+v) + d(c-v))/(c^2-v^2) = 2dc/(c^2-v^2) = 2(d/c)/(1-v^2/c^2)

That's what aether theory states.
There is no aether, therefore it's wrong. Why do you even bother with the
maths?

You will, of course, have alternate maths for BaTH, I trust?

the math is obvious.
In BaTh the whole MM apparatus may be considered 'at rest'.

tAB=L/c
tBA=L/c

A little more complicated than that, as the MMX measures
(or attempts to measure) crabwinds versus headwinds;
however, both BaTH and nBat can (and do) assume that the
MMX apparatus is at rest, since the aether "surface" (if
there is one) is essentially frictionless.

Hence my terminology of "hockey pucks", and any "aether
wind" is ignorable since the puck's velocity is not
affected thereby.

For its part SR isn't quite as simple to characterize, as
the light photon is moving at c no matter who's observing
it; time and space twist in odd fashions (and are provable
from this hypothesis more or less alone, as Einstein showed
long ago) in order to assure this constancy. It turns
out, mathematically anyway, that the "aether wind" doesn't
affect the photon either, in SR.


However, TWLS shows nothing regarding nBaT (and presumably
BaTh); the frictionless hockey puck encounters only the
reflecting mirror and is always moving at speed c.

Light normally moves at c wrt it source. ...and all components of the
interferometer.

Only if the source is not moving with respect thereto, in BaTH.

Of course the bloody light source of an interferometer is not movingr.

I would assume that someone's done a variant where MMX is pointing
towards, say, Venus. I'll admit I can't say without looking though,
and proper setup would be difficult unless the MMX is performed
outside of the influence of Earth's gravity. Low earth orbit might
be good enough -- though of course Gravity Probe B was looking for far
subtler effects anyway.

If one uses a spinning device to feed the MMX, of size
about 10cm in diameter and running at 12,000 RPM, the edge
speed is about 63 m/s, or 2.1 * 10^-7 c. Bouncing light
off this device (assuming the edge is rotating in the
opposite direction of the beam, in an attempt to slow it
down) should result in something like the following:

lambda/lambda_0[*] nu/nu_0 c/c_0

nBAT: 1 1 - 2.1 * 10^-7 1 - 2.1 * 10^-7
SR: 1 + 2.1 * 10^-7 1 - 2.1 * 10^-7 1
Androcles: 1 - 2.1 * 10^-7 1 1 - 2.1 * 10^-7
BaTH: ? ? ?

In BaTH the question marks are because the values are
distance- or time-dependent (I have no idea which). Over short
distances I presume BaTH approximates nBAT; over long distances
it approximates SR.

The question I have is how short and how long "short" and "long" are.
One presumes a priori that these have a wavelength dependency,
since light is an oscillatory motion.



For christ's sake will you give up this aether nonsense. You are supposed to
be a ballistician, remember

But SR is an aether theory; Androcles has "proven" that. ;-)

Well he should know, he's a great believer in aether theories.

No no...he doesn't believe in SR. If anything, he vehemently disputes
it. He's more in line with BaTH.

Have you seen his silly animation of tAB and tBA.

No, I haven't. Of course since he assumes tAB != tBA, but refuses to
say how they *are* related, I can't draw many useful conclusions.


It is straight out of the aether theory book.



Henri Wilson. ASTC,BSc,DSc(T)
www.users.bigpond.com/hewn/index.htm

Einstein's Relativity is easy to understand if one has the IQ of a parrot and a gullibility index >0.95.

[*] to appease Dirk, I must explain that this is *not* the same as
rod length, as the endpoints of a wave are at different times,
unlike the rigid rod, which always shrinks.

--
#191, ewill3@xxxxxxxxxxxxx
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Relevant Pages

  • Re: What is Proper Time?
    ... There is no aether, therefore it's wrong. ... In BaTh the whole MM apparatus may be considered 'at rest'. ... reflecting mirror and is always moving at speed c. ... if the observers are moving at different speeds. ...
    (sci.physics.relativity)
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