Re: Please help with relativity math

From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 09/22/04 

Date: Wed, 22 Sep 2004 04:44:42 GMT

"Eli Botkin" <elibotkin@optonline.net> wrote in message
news:1095816613.Od8QGXcw1pPHBDAQwR4O9g@teranews...
|
| "John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
| news:vaL$TNIiAJUBFwKO@kennaugh2435hex.freeserve.co.uk...
| > In message <41372918.173861755@news.houston.sbcglobal.net>,
| > seppala51@sbcglobal.net writes
| > >Would someone please be kind enough to post the formula for the
| > >following problem? A space craft is on the x-axis. At time t0=0 as
| > >measured by the craft's clock, the craft fires a laser in the
| > >x-direction, and simultaneously starts accelerating along the x-axis.
| > >The accelerometer on the craft reads g. The acceleration stops at
| > >time t1= N seconds as measured by an ideal clock on board the craft.
| > >How far away is the pulse of light from this craft (as measured by
| > >observers on the craft) when the acceleration stops?
| >
| > "It is important to emphasise that special relativity purports to
| > describe certain physical phenomena only relative to (or from the point
| > of view of) inertial reference systems, and the speed of a clock
| > relative to one of these systems determines its timekeeping behaviour
| > (G. Builder, 1958)."
| >
| > My understanding is that present dogma says that you cannot apply SR
| > unless you define an inertial frame to which all measurements can be
| > referred. Then it is only measurements relative to that inertial frame
| > which are relevant. If at to=0 there was an observer stationary w.r.t
| > the craft (before it accelerated) you can define how far away the pulse
| > is from that observer. It is not until the acceleration stops and the
| > moving observer becomes an inertial observer that you can treat it as a
| > mathematical problem covered by relativity. The history involving
| > acceleration is junked. You have two inertial FoR and you know how far
| > the pulse of light is in one and can work out where it is in the other.
| >
| > But then I could be wrong. Bilge says I have no understanding of
| > relativity.
| >
| > Nailing relativity is like trying to nail fog.
| > --
| > John Kennaugh
| > to email convert the number from hex to decimal
|
| Hi John:
| Before I give you the "formula" I'll write a few words to maybe straighten
| out some of your SR misconceptions.
| Yes, the Lorentz transformation equations apply only to transformations
| between inertial coordinate frames. Therefore the frame-fixed observers
are
| inertial. But these observers can observe (measure and record) the
motions
| of ALL objects, inertial and accelerated, that are moving with respect to
| their inertial frame. At each event of an accelerating object there is an
| instantaneous comoving inertial frame.
|
| In the problem you pose, when the acceleration, g, ends at time t (by the
| spacecraft's clock), that event's comoving frame allows you to make the
| transformation from the initial frame (actually you need to do some
algebra

Oh... 'Do' some algebra. Wow!
Is that rigorous algebra, or just some sloppy old algebra?

For quotations following, reference:
 http://www.fourmilab.ch/etexts/einstein/specrel/www/
 ("On the Electrodynamics of Moving Bodies" by Albert Einstein)

1) "light is always propagated in empty space with a definite velocity c
which is independent of the state of motion of the emitting body",
a totally unproven assumption without any evidence to support it.

2) "In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c to be a universal constant- the velocity of light in empty
space.",
an admitted assumption that is quite worthless when there is any
relative motion between A and B, yet essential to the derivation of the
remainder of Einstein's nonsense.

3) The equation
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)) ,
the ½ of which is derived from 2) above and is tantamount to saying
(1/3 + 2/3)/2 = 1/3.

4) The missing 0' from that equation, since x' = x-vt, hence 0' = 0-vt,
and the equation should be
½[tau(-vt,0,0,t)+tau(-vt,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
at the very least.

5) The further assumption "IF we place x' = x-vt ... " without considering
IF we place x' = x+vt, from which we derive (using Einstein's method)
  tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
  xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen

6) The statements
 "But the ray moves relatively to the initial point of k,
when measured in the stationary system, with the velocity c-v..."
and
"It follows, further, that the velocity of light c cannot be altered by
composition with a velocity less than that of light. For this case we obtain
V = (c+w)/(1+w/c) = c."
which are contradictory, the first being Galilean, the second being
contrary to the vector addition of velocities, an axiom of a vector space.

7) The lack of a check to verify the theory is self-consistent by feeding
the new PoR given in 6) into the equation given in 3) and finding a total
failure.
Check:
(t1-t)/(t2-t)*[tau(-vt,0,0,t)+tau(-vt,0,0,t+x'/V+x'/V)] = tau(x',0,0,t+x'/V)

Androcles.

| to find an intersection between two lines first, then apply the
contraction
| function, and not just plug values into the standard Lorentz equations).
|
| At that time, t, the spacecraft speed is V = c*tanh(gt/c). In the
initial
| frame the acceleration ends at
| T = (c/g)*sinh(gt/c) and X = (c^2/g)*[cosh(gt/c)-1]. At that instant
the
| distance (as measured by the comoving frame's observer, and therefore, the
| spacecraft's observer) from the spacecraft to the light-pulse is delta-x =
| (Vc/g)*sqrt[(1+V/c)/(1-V/c)] (provided I haven't made some algebraic
error).
|
| Keep at it
| Eli Botkin
|
|

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