Re: The Cost of Relativity
From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 07/25/04
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Date: Sun, 25 Jul 2004 12:18:01 GMT
"Tom Potter" <tdp@earthlink.net> wrote in message
news:2mh0voFmqubbU5@uni-berlin.de...
|
| "Androcles" <androc1es@nospamblueyonder.co.uk> wrote in message
| news:TdcMc.6353$W05.65068143@news-text.cableinet.net...
| >
| > "Tom Potter" <tdp@earthlink.net> wrote in message
| > news:2mcriaFl7n0tU2@uni-berlin.de...
| > |
| > | "Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com>
| wrote
| > | in message news:40ff82a0$1@usenet01.boi.hp.com...
| > | >
| > | > "Ballisticus" <B@..> wrote in message
| > | news:6vstf0tb22m8ml2ldlpl27jgf54r4iuehp@4ax.com...
| > | >
| > | > [snip]
| > | >
| > | > > As Androcles said,
| > | >
| > | > This is the last thing I saw Androfart say:
| > | >
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html
| > | > Comment?
| > |
| > | Yes,
| > |
| > | Why do you embarrass yourself
| > | and your family
| > | by maintaining a web site
| > | for the purpose of slandering folks?
| > |
| > | Is your ego so weak,
| > | that you have to do this to feel adequate?
| > |
| > | Why not develop some specific skills,
| > | and accomplishments, so that you
| > | won't have the need to ego trip on other folks?
| > |
| > | --
| > | Tom Potter http://home.earthlink.net/~tdp
| > |
| > I hadn't see that one:-)
| > Notice how cleverly he leaves the white space to show the asymmetry?
|
| Androcles have pity on the man,
| and especially his family.
Dinky de Torquemada doesn't deserve pity.
|
| Obviously he has had few ego building experiences,
| and he has the compulsion to feed his ego,
| not by positive accomplishments,
| but by attacking folks who have larger egos
| and greater accomplishments.
|
Actually, if I had any ego, I'd be brandishing it with my everyday name.
The stalking fumble mumbler has taken it upon himself to discover it.
Still, I have discovered Einstein's divide-by-zero.
It IS an accomplishment, since as far as I know nobody else has previously
discovered it.
It's right here, but I seek no kudos.
By 'v' we mean dx/dt. It's simple enough, a small change in distance divided
by a small change in time.
When we multiply dx/dt by t, we get .... yep, x.
so vt = x.
Now Einstein writes: "If we place x' = x-vt..."
So, x' = x-x and that must equal 0, agreed?
Of course this is simply the coordinate of the origin of the moving frame,
which is always at 0-vt in the stationary frame.
So far so good.
But then Einstein goes on...
"From the origin of system k let a ray be emitted at the time tau0 along the
X-axis to x',"
But as we've just seen, the origin of system k and x' are one and the same.
The ray has zero distance to travel.
(Or we could suppose that the origin of system k is at -vt, from 0' = 0-vt,
but that doesn't appear in in the equation.)
"and at the time tau1 be reflected thence to the origin of the co-ordinates,
arriving there at the time tau2; we then must have ½(t0+ t2) = t1,"
Well, yes. We have ½(0+0) = 0. So what?
x' is not some point remote from the origin of k where the reflection takes
place, it is AT the origin of k.
Einstein proceeds:
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))
and takes partial derivatives. To do this he says (looking as if he knows
what he's saying)
"Hence, if x' be chosen infinitesimally small, "
but it is already zero!
1/2 [1/(c-v) + 1/(c+v)]dtau/dt = dtau/dx' + 1/(c-v) dtau/dt,
Which with a little manipulation is
1/2 [1/(c-v) + 1/(c+v)]dtau/dt - 1/(c-v) dtau/dt = dtau/dx'
dtau/dt (1/2 [1/(c-v) + 1/(c+v)] - 1/(c-v)) = dtau/dx'
= dtau/d(x-vt)
= dtau/d0
= dtau/0
and we have a divide by zero.
Here's the full text, verbatim.
"If we place x'=x-vt, it is clear that a point at rest in the system k must
have a system of values x', y, z, independent of time. We first define tau
as a function of x', y, z, and t. To do this we have to express in equations
that tau is nothing else than the summary of the data of clocks at rest in
system k, which have been synchronized according to the rule given in § 1.
>From the origin of system k let a ray be emitted at the time tau0 along the
X-axis to x', and at the time tau1 be reflected thence to the origin of the
co-ordinates, arriving there at the time tau2; we then must have
½(tau0+tau2) =tau1, or, by inserting the arguments of the function tau and
applying the principle of the constancy of the velocity of light in the
stationary system:-
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)).
Hence, if x' be chosen infinitesimally small,
1/2 [1/(c-v) + 1/(c+v)]dtau/dt = dtau/dx' + 1/(c-v) dtau/dt,"
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
Anyone that knows anything at all about mathematics would know that division
by zero is undefined.
Androcles.
| --
| Tom Potter http://home.earthlink.net/~tdp
|
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