Four years ago I derived the Andersen Transforms:

As it is now:
"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

" -4.0 = 0.866 * 4.619 " -Paul B. Andersen


As it was then:

===============================================
In article , Androcles
writes

Androcles: I have used x' = x - (-v)t when I changed the direction of the
mirror, and simply wrote it as x'=x+vt.

Charles Francis, BA, Ph.D., Cambridge: No you didn't. If you had you would
have ended with only one formula and no contraction.

Androcles: ...you have quite surprised me. Now you can apologize for your
remark "you do not know how to use a minus sign", for it should be clear to
all who is the one that doesn't understand how to use a minus sign.
Where did you get your degree, anyway?

Charles Francis, BA, Ph.D., Cambridge: Cambridge.

The fact that you think you can preach when you do not even know the
definition of a magnitude is not at all funny. Plonk, btw.
Regards
Charles Francis



In article , Marty Fouts
writes

Charles Francis writes:

In article , Androcles
writes

[snip]

Androcles: One cannot willy-nilly change "+' to '-' on a whim, but one
change the direction of motion of the mirror on a whim, and describe that
mathematically.

Charles Francis, BA, Ph.D., Cambridge: What I am convinced of is that you do
not know how to use a minus sign.

Marty Fouts : Actually, Androcles used the minus sign just fine in its
derivation. It even managed to get the result it should have. Where it
screwed up was in not recognizing that by changing the sign of v at the
start, the changed result was what it should have gotten. There is no
contradiction in Androcles "fumble", only a classic undergraduate example of
getting confused about sign conventions.

Charles Francis, BA, Ph.D., Cambridge: Undergraduate? I remember people
getting confused about this sort of thing prior to 'O' level (exams we take
at 16). No one who was making mistakes of this sort would have gone on to do
A level maths or physics (exams we take at 18), and would certainly not have
become an undergraduate.

(Amazingly, I'm a graduate... but not from Cambridge, thank goodness. I
really do wonder what the Lucasian chair holders, past and present, would
think of Francis)

Charles Francis, BA, Ph.D., Cambridge: I have no great problem with people
making mistakes, and certainly no problem with people coming on the NG
(newsgroup) to learn. But these days there are huge numbers of out and out
idiots who only seem to come on the NG only to create noise and display
their stupidity. Kill filing is not good enough. To restore the ng as a
discussion group for physics we need a regularly maintained list of posters
who should not be responded to.

Regards
Charles Francis

"Charles Francis" wrote in message
...

In article , Androcles
writes

"Charles Francis" wrote in message
...


Charles Francis, BA, Ph.D., Cambridge: Androcles is not even capable of
using a minus sign correctly.

Androcles: This is a good one, Henry, but the funniest I've heard was
better.
It goes like this:
"The two velocities are in opposite direction, and we can without loss of
generality assume that they have the same magnitude,.. "Sverker Johansson"
...

Charles Francis, BA, Ph.D., Cambridge: Actually it is not funny. You
understand less maths than a 12 year old school boy, and spend a huge amount
of time advertising your incredible incompetence on the NG.


Androcles: I have used
x' = x - (-v)t
when I changed the direction of the mirror, and simply wrote it as x'=x+vt.
Okay, that wasn't obvious to you, but now it should be. There is no change
in the formula, but maybe I'll change the page to make it clearer. (and
maybe I'll leave it right here)
v changes direction, so I'm compelled to change its sign.
If you prefer,
Let u = -v.
then x' = x-ut,
1/2[tau(0,0,0,t) + tau(0,0,0,t+x'/(c-u)+x'/(c+u))] = tau(x', 0,0,t+x'/(c-u))
and substituting back,
x' = x+vt,
1/2[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 since
x'/(c+v)+x'/(c-v) = x'/(c-v)+x'/(c+v),
tau = (x+vt/c^2)*beta as I have shown.
Given that time does not march backwards (or we have no agreement
whatsoever), there are two results. Why there two results I have shown also,
although the pedantic proof may be lacking for that. It still remains that
two results are obtained.


Dirk van de moortel: (the yapping little dog that wants to run with the
hounds after the fox.)

Again showing what real imbecility is all about.
Over and over.
This is absolutely INCREDIBLE (sorry for the capitals).



Dirk van de moortel: I don't find this funny anymore. It is getting very sad
indeed. I *do* like cynics. I don't like an androcles.

(Which is only the proof of the statement, "Nobody likes a cynic." See
http://www.i-cynic.com)

(If Charles Francis, BA, Ph.D., Cambridge doesn't find a so-called physicist
claiming speed is the same as velocity as funny, then Charles Francis, BA,
Ph.D., Cambridge must have had a sadly warped childhood. The Holy Church of
Relativity seems to have produced any number of acolytes prepared to deny
even the most basic physics in defense of their religion. Readers are
invited to read the reason behind the remarks Charles Francis, BA, Ph.D.,
Cambridge has made, and the reason behind the support given to SJ and from
the imbecile that can only yap like a dog in a pack, Dirk Vdm.
==============================================
And back to now again:

Relativists obviously do not know how to use a minus sign.
I wonder where Charles Francis, B.A., Ph.D. is working today?
As janitor at Agder University College, perhaps?

Student Mobility Advisor
Telecommunication:
Paul B. Andersen
Assistant Professor
Telephone +47 37 25 30 00, telefax +47 37 25 30 01
e-mail:

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Windows NT SP 5 indeed! LOL!

Androcles

Before working on the minus sign, which is far too much advanced maths
for you, you'd better try to manage the xor operator :

http://users.pandora.be/vdmoortel/di...Gibberish.html

"YBM" wrote in message ...

Before working on the minus sign, which is far too much advanced maths
for you, you'd better try to manage the xor operator :

http://users.pandora.be/vdmoortel/di...Gibberish.html

Ah oui, et alors après cela, il peut
"easily make use of (a b) and legitimately conclude either a = 0 or b = 0":
http://users.pandora.be/vdmoortel/di...es/YesBut.html

Dirk Vdm

"YBM" snipped what he had no answer for in
...
Androcles

Androcles says...

"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

" -4.0 = 0.866 * 4.619 " -Paul B. Andersen

First of all, when you write something inside quotes,
it must be *literally* what was said. Paul Andersen
did not say that -4.0 = 0.866 * 4.619, did he?

Second, why don't we go through the correct calculation
to determine the elapsed time for the travelling twin.
Rather than writing xi and tau, let me just use primes:

(x,t) = coordinate system for stay-at-home twin
(x',t') = coordinate system for travelling twin on outward journey
(x'',t'') = coordinate system for travelling twin on return journey

Let t1 = time travelling twin departs earth (according to stay-at-home twin)
Let t2 = time travelling twin turns around (according to stay-at-home twin)
Let t3 = time travelling twin returns to earth (according to stay-at-home twin)

Then the elapsed time according to the travelling twin is given by:

T_total = T_outward + T_return
= (t2' - t1') + (t3'' - t2'')

The numbers for our particular example a

t1 = x1 = 0
t2 = 4.619, x2 = 4
t3 = 9.238, x3 = 0
gamma = 2
v = .866

The transformation equations are

t' = gamma (t - vx/c^2)
t'' = gamma (t + vx/c^2)

So
t1' = t1'' = 0
t2' = 2 (4.619 - .866 * 4) = 2.31
T_out = 2.31 - 0 = 2.31

t2'' = 2 (4.619 + .866 * 4) = 16.166
t3'' = 2 (9.238 + .866 * 0) = 18.476
T_return = 18.476 - 16.166 = 2.31

T_total = T_out + T_return = 4.62

That's the *correct* calculation, Androcles.

--
Daryl McCullough
Ithaca, NY

"Androcles" wrote in message . ..
Four years ago I derived the Andersen Transforms:

No, you did not.

Four years ago, you derived the Adroclean transform in the posting:
http://groups.google.com/groups?q=g:....pa.h ome.com
From where I quote:
|
| Watch the signs, carefully.
|
| 1/2[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 mirror moves the opposite way.)
| Hence, if x' be chosen infinitesimally small,
| 1/2(1/(c+v) + 1/(c-v) ) @tau/@t = @tau/@x' + 1/(c+v) @tau/@t'
| or
| @tau/@x' - v/(c^2 -v ^2)@tau/@t = 0
|
| and (ta daaaaa)
| beta = 1/sqrt(1+v^2/c^2)
|
| You understand Einstein's derivation? What a joke! You probably dont
even
| know that sqrt(1) has two answers, 1 and -1. The correct answer to
the
| question I gave, and you guessed at, is
| tau = (t + vx/c^2)/sqrt(1 (PLUS) v^2/c^2)
| xi = (x + vt/c^2)/sqrt(1 + v^2/c^2)
| eta = y
| zeta = z

Ta daaaa - that's how to use a minus sign.

Well done, Androcles.

Paul

"Paul B. Andersen" wrote in message
om...
| "Androcles" wrote in message
. ..
| Four years ago I derived the Andersen Transforms:
|
| No, you did not.
|
| Four years ago, you derived the Adroclean transform in the posting:
|
http://groups.google.com/groups?q=g:....pa.h ome.com
| From where I quote:
| |
| | Watch the signs, carefully.
| |
| | 1/2[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 mirror moves the opposite way.)
| | Hence, if x' be chosen infinitesimally small,
| | 1/2(1/(c+v) + 1/(c-v) ) @tau/@t = @tau/@x' + 1/(c+v) @tau/@t'
| | or
| | @tau/@x' - v/(c^2 -v ^2)@tau/@t = 0
| |
| | and (ta daaaaa)
| | beta = 1/sqrt(1+v^2/c^2)
| |
| | You understand Einstein's derivation? What a joke! You probably dont
| even
| | know that sqrt(1) has two answers, 1 and -1. The correct answer to
| the
| | question I gave, and you guessed at, is
| | tau = (t + vx/c^2)/sqrt(1 (PLUS) v^2/c^2)
| | xi = (x + vt/c^2)/sqrt(1 + v^2/c^2)
| | eta = y
| | zeta = z
|
| Ta daaaa - that's how to use a minus sign.
|
| Well done, Androcles.
|
| Paul
I did, didn't I? That was how I got you to produce the Andersen Transforms.
Well done me.
Actually, I was referring above to my own derivation:
http://www.androc1es.pwp.blueyonder....lesfrancis.htm
Androcles.

"Androcles" wrote in message ...

"Paul B. Andersen" wrote in message
om...
| "Androcles" wrote in message
| . ..
| Four years ago I derived the Andersen Transforms:
|
| No, you did not.
|
| Four years ago, you derived the Adroclean transform in the posting:
|
http://groups.google.com/groups?q=g:....pa.h ome.com
| From where I quote:
| |
| | Watch the signs, carefully.
| |
| | 1/2[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 mirror moves the opposite way.)
| | Hence, if x' be chosen infinitesimally small,
| | 1/2(1/(c+v) + 1/(c-v) ) @tau/@t = @tau/@x' + 1/(c+v) @tau/@t'
| | or
| | @tau/@x' - v/(c^2 -v ^2)@tau/@t = 0
| |
| | and (ta daaaaa)
| | beta = 1/sqrt(1+v^2/c^2)
| |
| | You understand Einstein's derivation? What a joke!
| | You probably dont even
| | know that sqrt(1) has two answers, 1 and -1.
| | The correct answer to the
| | question I gave, and you guessed at, is
| | tau = (t + vx/c^2)/sqrt(1 (PLUS) v^2/c^2)
| | xi = (x + vt/c^2)/sqrt(1 + v^2/c^2)
| | eta = y
| | zeta = z
|
| Ta daaaa - that's how to use a minus sign.
|
| Well done, Androcles.
|
| Paul

I did, didn't I? That was how I got you to produce the Andersen Transforms.

Quite.
The posting you wrote July 7, 2001 made me produce
the following more than a year before.
Time is known to run backwards in Androcles' world.
That's how to use a minus sign.

Paul B. Andersen wrote Mars 3, 2000:
| The _form_ of the Lorentz transform
| depend on how the frames of references are chosen.
| There are a number of different possibilities:
|
| Choice #1:
| =============
| All axes parallel, origo of K' moving in positive direction
| along the x-axis of K
|
| ----|---------------- x' - v
| ----|---------------- x
|
| With this choice , the LT is:
| forward, K-K' inverse, K'-K
| x' = g*(x - v*t) x = g*(x' + v*t')
| t' = g*(t - x*v/c^2) t = g*(t' + x'*v/c^2)
|
| Choice #2:
| =============
| All axes parallel, origo of K moving in positive direction
| along the x'-axis of K'
|
| ----|---------------- x - v
| ----|---------------- x'
|
| With this choice, the LT is:
| forward, K'-K inverse, K-K'
| x = g*(x' - v*t') x' = g*(x + v*t)
| t = g*(t' - x'*v/c^2) t' = g*(t + x*v/c^2)
|
| Choice #3:
| [ etc.]
| ....
|
| However, Einstein used "choice #1" in his paper.
| When you interchange the frames of references,
| you get "choice #2" above.
| Finding it self contradictory that that the transform equations
| change as shown above is - not very smart.

Well done me.

Indeed.
The credit is all yours, obviously.

Paul

(Daryl McCullough) wrote in message ...
Androcles says...

"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

" -4.0 = 0.866 * 4.619 " -Paul B. Andersen

First of all, when you write something inside quotes,
it must be *literally* what was said. Paul Andersen
did not say that -4.0 = 0.866 * 4.619, did he?

Of course he didn't. :-)

I said that if the moving twin starts at the coordinate
x = 0 at the time t = 0, and is moving in the negative
x-direction at the speed v = 0.86c, he will reach
the coordinate x = -4LY at the time t = 4.619 years.

But Androcles insists that when the twin is moving in
the negative x-direction, time runs backwards because
t = x/v = -4.619 years. So he will arrive at the destination
years before he started.

My failure to understand that has made Androcles
roll farting on the floor. He thinks that by doing so,
he is making the "SRians" look foolish.

Paul

"Paul B. Andersen" wrote in message om...
(Daryl McCullough) wrote in message ...
Androcles says...

"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

" -4.0 = 0.866 * 4.619 " -Paul B. Andersen

First of all, when you write something inside quotes,
it must be *literally* what was said. Paul Andersen
did not say that -4.0 = 0.866 * 4.619, did he?

Of course he didn't. :-)

I said that if the moving twin starts at the coordinate
x = 0 at the time t = 0, and is moving in the negative
x-direction at the speed v = 0.86c, he will reach
the coordinate x = -4LY at the time t = 4.619 years.

But Androcles insists that when the twin is moving in
the negative x-direction, time runs backwards because
t = x/v = -4.619 years. So he will arrive at the destination
years before he started.

Nicely documented by The Master Himself
http://users.pandora.be/vdmoortel/di...s/NegTime.html
http://users.pandora.be/vdmoortel/di.../NegTime2.html

Dirk Vdm



My failure to understand that has made Androcles
roll farting on the floor. He thinks that by doing so,
he is making the "SRians" look foolish.

Paul