Re: Androcles, all variables must be substracted by the same value to preserve equation!!!!
From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 08/13/04
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Date: Fri, 13 Aug 2004 13:51:07 GMT
"Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message
news:cfidl4$d7c$1@dolly.uninett.no...
|
| "Androcles" <androc1es@nospamblueyonder.co.uk> skrev i melding
news:if2Tc.4523$pR.49322083@news-text.cableinet.net...
| >
| > "Paul B. Andersen" <paul.b.andersen@hia.no> wrote in message
| > news:cfi0ri$bj8$1@dolly.uninett.no...
| > | "Androcles" <androc1es@nospamblueyonder.co.uk> skrev i melding
| > news:reRSc.4274$VC2.46438548@news-text.cableinet.net...
| > | > Einstein's 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))
| > | > Subtracting t and removing the superfluous y and z values,
| > | > ½[tau(0,0)+tau(0,x'/(c-v)+x'/(c+v))] = tau(x',x'/(c-v))
| > | > and noting that Einstein defined x' = x-vt, hence 0' = 0-vt,
| > | > ½[tau(-vt,0)+tau(-vt, (x-vt)/(c-v)+(x-vt)/(c+v))] =
| > | >
| > tau((x-vt),(x-vt)/(c-v))
| > | > At x = vt, this further reduces to
| > | > ½[tau(-vt,0)+tau(-vt, 0)]= tau(x-vt,0)
| > | > and by adding vt = x,
| > | > ½[tau(0,0)+tau(0, 0)]= tau(x,0)
| > | > Adding back t,
| > | > ½[tau(0,t)+tau(0,t)]= tau(x,t)
| > | > So the time at 0 is the same as the time at x for all x.
| > | > Hence time is not equivalent to space, time being the same
everywhere.
| > |
| > | Androcles. these recurring demonstrations of your
| > | mathematical skills aren't really necessary, you know.
| > | We all know them by now.
| > |
| > | Paul
| >
| > Well, you may know them by now, but once again you are wrong, since you
are
| > not 'all'. The Andersen Transforms are a clear enough indication of the
| > nonsense of SR, quote:
| > "That is, we can reverse the directions of the frames
| > which is the same as interchanging the frames,
| > which - as I have told you a LOT of times,
| > OBVIOUSLY will lead to the transform:
| > t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
| > x = (xi - v*tau)/sqrt(1-v^2/c^2)
| > or:
| > tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
| > xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen,
|
| Ta-daa:
|
| > in which we find that the faster one travels, the later one arrives.
|
| You did it again!
|
| BTW, who are "we"?
| Androcles and .. who else?
|
| Paul
Well, I suppose it is too difficult for you to calculate, nTaul. Put the
damp sand in your bucket, pat it down, turn the bucket upside down and place
the little flag in the pile. Then you'll have a nice sand castle to play
with.
Androcles
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