Re: Roberts' persusaive rhetoric.
Dirk Van de moortel
"Androcles" <du...@dummy.net> wrote in message news:9fAbd.61468$ay5....@fe1.news.blueyonder.co.uk...
> Roberts writes:
Whatever Tom writes, is way beyond your mental and intellectual
capabilities:
http://users.pandora.be/vdmoortel/dirk/Physics/ImmortalFumbles.html
Dirk Vdm
Titan Point
> Roberts writes:
>
> Relativity - it is IMPOSSIBLE to construct an alternative description
> without violating one of the postulates or disregarding a very large
> body of experimental evidence. If you truly believe that Special
> Relativity simply must be false (for whatever reason), go back and
> review the four Postulates, and find a hole in them."
>
> This is why most (if not all) physicists today are failed mathematicians-
> it is IMPOSSIBLE for special relativity to be anything other than false.
> There is nothing wrong with the group postulates, but SR violates the
> second.
> " 2. Any transformation has an inverse, which is also a transformation."
> So what is the inverse transformation to
> x' = (x-vt)/sqrt(1-v^2/c^2) ?
> A little algebra and
> x' * sqrt(1-v^2/c^2) = x-vt
> x = x' * sqrt(1-v^2/c^2) + vt.
...which ISN'T the inverse transformation.
>
> That is not an equation you'll ever see in SR.
That's because it was written by a non-mathematical imbecile.
> However, the PoR must hold.
>
>
> Consider v = 0.866c, and this lasts for a period of time of 1 year;
...in whose frame of reference?
> c, of
> course, is one light-year per year.
> By definition, in the frame of S', we must move a distance from (0',0')
> to (0.866', 1.0')
0 is not a variable. It is zero in all frames of reference. There is
no such thing as 0'
> if we have a velocity v = 0.866.
In whose frame of reference? In the frame of S' the rocket has NOT
MOVED AT ALL.
> We know ask what this distance is in S, and to do that we use
> x = x' * sqrt(1-v^2/c^2) +vt.
> = 0.866 * 0.5 + 0.866*.. err... t? .... [1]
> ah, we have a problem. We do not yet have t.
We do have 0' so that makes a difference *rolls eyes*
> Well,
> t' = (t-vx/c^2) / sqrt(1-v^2/c^2), so
>
> t' * sqrt(1-v^2/c^2) = t-vx/c^2,
> t = t' * sqrt(1-v^2/c^2) + vx/c^2.
> = 0.866 * 0.5 + 0.866 * ?....[2], again we have a problem.
>
> It was x we were trying to find.
> It looks as if we'll need to use simultaneous equations.
>
> x = 0.433 + 0.866 * t
> t = 0.433 + 0.866 * x
>
> We subtract one equation from the other...
> x-t = 0.866t - 0.866x, and then add t to each side.
> x = 0.866t- 0.866x+t, then add 0.866x to each side,
> x + 0.866x = 0.866t+t, or
> 1.866 x = 1.866 t. Dividing both sides by 1.866,
> x = t.
Hey doofus, from
x+ 0.866x = 0.866t + t
if you divide both sides by 0.866 you get
(x + 0.866x)/0.866 = (0.866t+t)/0.866
rationalising
x(1+0.866)/0.866 = t(1+0.866)/0.866
which is not the same as
1.866x =1.866t
Any competent schoolchild can do algebra but you can't because you're
a doofus.
> Now we can substitute into [2]
> t = 0.866 * 0.5 + 0.866 * t
> t = 0.433 + 0.866 * t
> t - 0.866 * t = 0.433
> t(1-0.866) = 0.433
> t = 0.433 /(1-0.866)
> = 0.433/0.134
> = 3.23
> And of course since x = t, x = 3.23 also.
>
> So... our intrepid traveller travels a distance of 0.866 light years
> in one year, by his own clock, at speed v = 0.866c.
In other words he moved at the speed of light since
0.866 ly/0.866 yr = 1 ly/y = c
Is that what you're claiming?
>
> We, however, determine that he travelled 3.23 light years and
> it took him 3.23 years to do so.
...and only because you can't do basic algebra AND DON'T UNDERSTAND
SPECIAL RELAIVITY AT ALL.
>
> Therefore he was moving at the speed of light.
Yes, of course he was, you moron.
>
> Hmm.. well, as Roberts (and Einstein) says,
> V = (c+v) / (1+v/c), this is a transformation, so let us solve for v to find
> the inverse according to:
> " 2. Any transformation has an inverse, which is also a transformation."
>
> V * (1+v/c) = c+v
>
> v = V * (1+v/c) - c
> = V+Vv/c - c
> v - Vv/c = V - c
> v(1-V/c) = V-c
> v = (V-c) / (1-V/c)
> = (0.866 - 1) / (1- 0.866/c)
> = -0.134 / 0.134
> = -1.0c.
> How very odd.
Not at all. Go back to school and learn some algebra.
>
>
> The relativist will now declare that I must have made a mistake somewhere.
> Let him find it.
Found them. The problem is definitely between chair and keyboard.
> I would submit that the first relativist has made a mistake somewhere.
> I have found it.
Yes, replying to a non-mathematical doofus who makes basic mistakes in
algebra is a mistake.
>
> For quotations following, reference:
> http://www.fourmilab.ch/etexts/einstein/specrel/www/
> ("On the Electrodynamics of Moving Bodies" by Albert Einstein)
[snipped more algebraic dyslexia]
AllYou!
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:D_Bbd.278700$3T7.14...@phobos.telenet-ops.be...
And the need for you to attack instead of ignore continues............
--
"Therefore the two clocks S1 and S2 do not both give the *time*
correctly." -- Albert Einstein
Tom Roberts
> " 2. Any transformation has an inverse, which is also a transformation."
> So what is the inverse transformation to
> x' = (x-vt)/sqrt(1-v^2/c^2) ?
> A little algebra and
> x' * sqrt(1-v^2/c^2) = x-vt
> x = x' * sqrt(1-v^2/c^2) + vt.
>
It's not wrong, it's just not normally displayed because it is merely an
incidental step on the path to the inverse transformation, but is not
the inverse transformation itself (nowhere close). The inverse transform
must relate (x,t) on the LHS to (x',t') on the RHS, and you still have a
"t" on the RHS. You need to continue with the algebra, eliminating all
references to (x,t) on the RHS, and express both x and t on the LHS as
expressions using (x',t') on the RHS. Note that this inherently requires
two equations, and you must invert them together.
Perhaps you can demonstrate your ability to actually do algebra
by displaying your steps.... And perhaps not....
If you perform this, you'll find that the inverse transform is identical
to the original transform except that v is replaced by -v. That, of
course, is so because the relative motion of the two frames is reciprocal.
Tom Roberts tjro...@lucent.com
Dirk Van de moortel
"Tom Roberts" <tjro...@lucent.com> wrote in message news:i7Xbd.9810$Rf1....@newssvr19.news.prodigy.com...
I already have rubbed his fat nose in it:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html
Rubbing their nose in it, you can teach cats and dogs not to
piss on the carpet. With Androcles it doesn't help.
Dirk Vdm
Androcles
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:i7Xbd.9810$Rf1....@newssvr19.news.prodigy.com...
Snipped what you had no answer for, Roberts?
Typical relativist tactics, that. Here it is again.
Roberts writes:
"This is why most (if not all) physicists today believe in Special
Relativity - it is IMPOSSIBLE to construct an alternative description
without violating one of the postulates or disregarding a very large
body of experimental evidence. If you truly believe that Special
Relativity simply must be false (for whatever reason), go back and
review the four Postulates, and find a hole in them."
This is why most (if not all) physicists today are failed mathematicians-
it is IMPOSSIBLE for special relativity to be anything other than false.
There is nothing wrong with the group postulates, but SR violates the
second.
" 2. Any transformation has an inverse, which is also a transformation."
So what is the inverse transformation to
x' = (x-vt)/sqrt(1-v^2/c^2) ?
A little algebra and
x' * sqrt(1-v^2/c^2) = x-vt
x = x' * sqrt(1-v^2/c^2) + vt.
That is not an equation you'll ever see in SR.
However, the PoR must hold.
Consider v = 0.866c, and this lasts for a period of time of 1 year; c, of
course, is one light-year per year.
By definition, in the frame of S', we must move a distance from (0',0')
to (0.866', 1.0') if we have a velocity v = 0.866.
We know ask what this distance is in S, and to do that we use
x = x' * sqrt(1-v^2/c^2) +vt.
= 0.866 * 0.5 + 0.866*.. err... t? .... [1]
ah, we have a problem. We do not yet have t.
Well,
t' = (t-vx/c^2) / sqrt(1-v^2/c^2), so
t' * sqrt(1-v^2/c^2) = t-vx/c^2,
t = t' * sqrt(1-v^2/c^2) + vx/c^2.
= 0.866 * 0.5 + 0.866 * ?....[2], again we have a problem.
It was x we were trying to find.
It looks as if we'll need to use simultaneous equations.
x = 0.433 + 0.866 * t
t = 0.433 + 0.866 * x
We subtract one equation from the other...
x-t = 0.866t - 0.866x, and then add t to each side.
x = 0.866t- 0.866x+t, then add 0.866x to each side,
x + 0.866x = 0.866t+t, or
1.866 x = 1.866 t. Dividing both sides by 1.866,
x = t.
Now we can substitute into [2]
t = 0.866 * 0.5 + 0.866 * t
t = 0.433 + 0.866 * t
t - 0.866 * t = 0.433
t(1-0.866) = 0.433
t = 0.433 /(1-0.866)
= 0.433/0.134
= 3.23
And of course since x = t, x = 3.23 also.
So... our intrepid traveller travels a distance of 0.866 light years
in one year, by his own clock, at speed v = 0.866c.
We, however, determine that he travelled 3.23 light years and
it took him 3.23 years to do so.
Therefore he was moving at the speed of light.
Hmm.. well, as Roberts (and Einstein) says,
V = (c+v) / (1+v/c), this is a transformation, so let us solve for v to find
the inverse according to:
" 2. Any transformation has an inverse, which is also a transformation."
V * (1+v/c) = c+v
v = V * (1+v/c) - c
= V+Vv/c - c
v - Vv/c = V - c
v(1-V/c) = V-c
v = (V-c) / (1-V/c)
= (0.866 - 1) / (1- 0.866/c)
= -0.134 / 0.134
= -1.0c.
How very odd.
The relativist will now declare that I must have made a mistake somewhere.
Let him find it.
I would submit that the first relativist has made a mistake somewhere.
I have found it.
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
―[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 ― 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
―[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)
where V = (c+v)/(1+v/c) as required by the redefined PoR.
Androcles.
Dirk Van de moortel
"Androcles" <du...@dummy.net> wrote in message news:iAccd.92428$BI5....@fe2.news.blueyonder.co.uk...
>
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:i7Xbd.9810$Rf1....@newssvr19.news.prodigy.com...
>
> Snipped what you had no answer for, Roberts?
Can't solve a system of two equations with two unknows, Parker?
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html
Dirk Vdm
sal
> Roberts writes:
>
> "This is why most (if not all) physicists today believe in Special
> Relativity - it is IMPOSSIBLE to construct an alternative description
> without violating one of the postulates or disregarding a very large body
> of experimental evidence. If you truly believe that Special Relativity
> simply must be false (for whatever reason), go back and review the four
> Postulates, and find a hole in them."
>
> This is why most (if not all) physicists today are failed mathematicians-
> it is IMPOSSIBLE for special relativity to be anything other than false.
> There is nothing wrong with the group postulates, but SR violates the
> second.
> " 2. Any transformation has an inverse, which is also a transformation."
> So what is the inverse transformation to x' = (x-vt)/sqrt(1-v^2/c^2) ?
> A little algebra and
> x' * sqrt(1-v^2/c^2) = x-vt
> x = x' * sqrt(1-v^2/c^2) + vt.
>
> That is not an equation you'll ever see in SR. However, the PoR must hold.
>
>
> Consider v = 0.866c, and this lasts for a period of time of 1 year; c, of
> course, is one light-year per year.
> By definition, in the frame of S', we must move a distance from (0',0') to
> (0.866', 1.0') if we have a velocity v = 0.866.
SO: It is clearly stated:
x' = 0.866
t' = 1.0
> We know ask what this
> distance is in S, and to do that we use x = x' * sqrt(1-v^2/c^2) +vt.
> = 0.866 * 0.5 + 0.866*.. err... t? .... [1]
> ah, we have a problem. We do not yet have t. Well,
> t' = (t-vx/c^2) / sqrt(1-v^2/c^2), so
>
> t' * sqrt(1-v^2/c^2) = t-vx/c^2,
> t = t' * sqrt(1-v^2/c^2) + vx/c^2.
> = 0.866 * 0.5 + 0.866 * ?....[2], again we have a problem.
You just plugged in 0.866 for t'. But up above, you clearly stated that
t' = 1.0. Your statement here should read
t = 1.0 * 0.5 + 0.866 * ?....[2], again we have a problem.
This correction will need to be carried through the rest of the
calculations to obtain a correct result.
--
I can be contacted through http://www.physicsinsights.org
mike3
> Roberts writes:
>
> "This is why most (if not all) physicists today believe in Special
> Relativity - it is IMPOSSIBLE to construct an alternative description
> without violating one of the postulates or disregarding a very large
> body of experimental evidence. If you truly believe that Special
> Relativity simply must be false (for whatever reason), go back and
> review the four Postulates, and find a hole in them."
>
> This is why most (if not all) physicists today are failed mathematicians-
> it is IMPOSSIBLE for special relativity to be anything other than false.
> There is nothing wrong with the group postulates, but SR violates the
> second.
> " 2. Any transformation has an inverse, which is also a transformation."
> So what is the inverse transformation to
> x' = (x-vt)/sqrt(1-v^2/c^2) ?
> A little algebra and
> x' * sqrt(1-v^2/c^2) = x-vt
> x = x' * sqrt(1-v^2/c^2) + vt.
>
> That is not an equation you'll ever see in SR.
> However, the PoR must hold.
>
BWAHAHAHAHA! You have DISPROVEN your own argument! You showed the
transformation has an inverse, therefore it does NOT violate the law
you brought up. Because it has one, it is VALID under that law. It
doesn't matter whether it appears in the theory or not. It is still
VALID!
>
> Consider v = 0.866c, and this lasts for a period of time of 1 year; c, of
> course, is one light-year per year.
> By definition, in the frame of S', we must move a distance from (0',0')
> to (0.866', 1.0') if we have a velocity v = 0.866.
> We know ask what this distance is in S, and to do that we use
> x = x' * sqrt(1-v^2/c^2) +vt.
> = 0.866 * 0.5 + 0.866*.. err... t? .... [1]
> ah, we have a problem. We do not yet have t.
> Well,
> t' = (t-vx/c^2) / sqrt(1-v^2/c^2), so
>
> t' * sqrt(1-v^2/c^2) = t-vx/c^2,
> t = t' * sqrt(1-v^2/c^2) + vx/c^2.
> = 0.866 * 0.5 + 0.866 * ?....[2], again we have a problem.
>
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:pan.2004.10.17....@nospam.org...
> On Thu, 14 Oct 2004 19:15:17 +0000, Androcles wrote:
>
> > Roberts writes:
> >
> > "This is why most (if not all) physicists today believe in Special
> > Relativity - it is IMPOSSIBLE to construct an alternative description
> > without violating one of the postulates or disregarding a very large body
> > of experimental evidence. If you truly believe that Special Relativity
> > simply must be false (for whatever reason), go back and review the four
> > Postulates, and find a hole in them."
> >
> > This is why most (if not all) physicists today are failed mathematicians-
!!!
Thanks sal, for pointing this out:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Persuasive.html
Title: "Roberts' persuasive rhetoric. We have a problem."
I specially like his opening phrase
"This is why most (if not all) physicists today are failed mathematicians"
Absolutely brilliant :-))
Much obliged - keep 'em coming. This saves *me* the trouble to wade
through his excrement.
These ones I *will* accept through private email, I promise ;-)
Dirk Vdm
Dirk Van de moortel
"mike3" <mike...@yahoo.com> wrote in message news:1d54b7e4.04101...@posting.google.com...
I just found out that the problem is *much* worse:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Persuasive.html
He can't even substitute a variable.
And the equation you see isn't part of the inverted transformation
at all. See also
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html
Dirk Vdm
Androcles
x = x' * sqrt(1-v^2/c^2) + vt.
>> We know ask what this
>> distance is in S, and to do that we use x = x' * sqrt(1-v^2/c^2) +vt.
>> = 0.866 * 0.5 + 0.866*.. err... t? .... [1]
>> ah, we have a problem. We do not yet have t. Well,
>> t' = (t-vx/c^2) / sqrt(1-v^2/c^2), so
>>
>> t' * sqrt(1-v^2/c^2) = t-vx/c^2,
>> t = t' * sqrt(1-v^2/c^2) + vx/c^2.
>> = 0.866 * 0.5 + 0.866 * ?....[2], again we have a problem.
>
> You just plugged in 0.866 for t'. But up above, you clearly stated that
> t' = 1.0. Your statement here should read
>
> t = 1.0 * 0.5 + 0.866 * ?....[2], again we have a problem.
>
> This correction will need to be carried through the rest of the
> calculations to obtain a correct result.
>
>
Oops. I did. You are right.
So, to avoid making that kind of blunder again, I'll fall back on
the inverse LT's that Andersen computed for me,
"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
and Roberts says exist by the group axioms.
The moving clock, receding, runs slow.
One would expect it to appear to from doppler red shift, but this shift
is magnified by gamma.
The moving clock, approaching, runs fast.
One would expect it to appear to from doppler blue shift, but this shift
is magnified by gamma.
How can we be certain of this?
From Einstein's Doppler equation, of course, and the Andersen Transform
above.
As should be obvious by now, I do make blunders, so perhaps you'd be kind
enough to prove for me that 1/tau = nu, 1/t = f, and the frequency transform
Einstein gives,
nu' = nu. [1-cos(phi) .v/c] / sqrt(1-v^2/c^2)
produces the correct shift by making phi = 0 and phi = pi , thereby proving
Andersen was correct?
I'd do it myself, but I've somehow forgotten that cos(0) = 1 and cos(pi)
= -1.
Androcles.
jem
> Thanks sal, for pointing this out:
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Persuasive.html
> Title: "Roberts' persuasive rhetoric. We have a problem."
> I specially like his opening phrase
> "This is why most (if not all) physicists today are failed mathematicians"
> Absolutely brilliant :-))
>
> Much obliged - keep 'em coming. This saves *me* the trouble to wade
> through his excrement.
> These ones I *will* accept through private email, I promise ;-)
>
> Dirk Vdm
What sort of person gets pleasure from belittling others, and takes
pride in doing it?
Dirk Van de moortel
"jem" <x...@xxx.xxx> wrote in message news:_Ttcd.104909$a85.81994@fed1read04...
It depend on who these "others" are, what exactly they do,
and why they do it ;-)
Cheers,
Dirk Vdm
Dirk Van de moortel
"Androcles" <du...@dummy.net> wrote in message news:MOtcd.78955$ay5....@fe1.news.blueyonder.co.uk...
Well, stupid hog, they aren't ;-)
But one of these days you might shit a pearl. So keep trying.
Dirk Vdm
sal
No, don't switch to the "Andersen transform". The "blunder" was a very
small mistake, really, and the so-called "Andersen transform" is wrong, as
you know perfectly well.
Please, continue your original analysis. Carry it through, with correct
values plugged in, and see what answer you get. Show us the result!
Judging by other responses in this thread, almost nobody reading your
post realized that what you were doing was a perfectly acceptable way of
inverting the transform. Show them that they're wrong, and you're
right -- finish the calculation and show the result.
Of course, if you do that, you may find that it works out without any
"paradox" or contradiction, which would show that the Lorentz transforms
are invertible, just like Tom Roberts said. But you wouldn't hesitate to
post the result just because of that, would you?
>
> "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
>
> and Roberts says exist by the group axioms.
>
> The moving clock, receding, runs slow. One would expect it to appear to
> from doppler red shift, but this shift is magnified by gamma.
>
> The moving clock, approaching, runs fast. One would expect it to appear
> to from doppler blue shift, but this shift is magnified by gamma.
>
> How can we be certain of this?
> From Einstein's Doppler equation, of course, and the Andersen Transform
> above.
>
> As should be obvious by now, I do make blunders, so perhaps you'd be
> kind enough to prove for me that 1/tau = nu, 1/t = f, and the frequency
> transform Einstein gives,
> nu' = nu. [1-cos(phi) .v/c] / sqrt(1-v^2/c^2) produces the correct shift
> by making phi = 0 and phi = pi , thereby proving Andersen was correct?
> I'd do it myself, but I've somehow forgotten that cos(0) = 1 and cos(pi)
> = -1.
>
> Androcles.
--
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:1098102962.Ii0CDLqlzOP6+K8B4QZvTg@teranews...
> On Sun, 17 Oct 2004 12:44:28 +0000, Androcles wrote:
[snip]
> > So, to avoid making that kind of blunder again, I'll fall back on the
> > inverse LT's that Andersen computed for me,
>
> No, don't switch to the "Andersen transform". The "blunder" was a very
> small mistake, really, and the so-called "Andersen transform" is wrong, as
> you know perfectly well.
>
> Please, continue your original analysis. Carry it through, with correct
> values plugged in, and see what answer you get. Show us the result!
>
> Judging by other responses in this thread, almost nobody reading your
> post realized that what you were doing was a perfectly acceptable way of
> inverting the transform.
Sal, surely you must know that no one (except you who seems
to be looking for that pearl,) *reads* his posts.
Dirk Vdm
AllYou!
And if you had your arrogant way, not only would you not read *his* posts,
but no one would. Why else would you always try to discourage others from
having dialog with those whom you regard as beneath you?
The fact is that you get very pissed when you don't know how to address many
issues raised by others because it shows you up, or you get very pissed when
others won't agree with you because again, in your view, it questions your
unquestionable certainty.
You're a joke and add nothing to this NG but vitriol.
--
"I don't know what happens in an atomic clock, but I know that it's not
motion." ---- D Moortel
Dirk Van de moortel
"AllYou!" <ida...@conversent.net> wrote in message news:7ZydnU0uZLO...@conversent.net...
Fabricating quotes again, dishonest coward?
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Cowardice.html
Keep it up ;-)
Dirk Vdm
AllYou!
Thanks for the invite......I will. You can't produce any evidence that I'm
any of the things you've asserted about me. I've never said anything
bigoted or dishonest, and if I am a coward, I'm certainly not afraid to
tangle with you!
But speaking of cowardice, you're so afraid of what I'll do to you that you
dare not even offer a definition of a clock! LOL!
Go ahead, prove your bravery.......try it. LOL!
Virtually none of your posts these days are related to physics. Instead,
virtually all of them are aimed at tearing into others. Care to explain
why?
You make this way too easy.
--
"We don't know what happens in an atomic clock, but I know that it's not
motion." ---- D Moortel
Dirk Van de moortel
"AllYou!" <ida...@conversent.net> wrote in message news:ItydnYIcyos...@conversent.net...
>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:3VRcd.282223$9g1.14...@phobos.telenet-ops.be...
> >
> > "AllYou!" <ida...@conversent.net> wrote in message
> news:7ZydnU0uZLO...@conversent.net...
[snip]
> > > "I don't know what happens in an atomic clock, but I know that it's not
> > > motion." ---- D Moortel
> >
> > Fabricating quotes again, dishonest coward?
> > http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Cowardice.html
> > Keep it up ;-)
>
> Thanks for the invite......I will. You can't produce any evidence that I'm
> any of the things you've asserted about me. I've never said anything
> bigoted or dishonest, and if I am a coward,
If you weren't, then you would use your real name to sign your
posts in stead of hiding behind "Allyou!".
Dirk Vdm
AllYou!
There are plenty of other reasons to maintain anonymity here and none of
them have to do with cowardice, and certainly not of you. But you'll hang
your hat on anything you can to deflect from the reality that your
accusations are lies.
And if you weren't a coward, you'd not snip my posts. Here's what you left
out.
But speaking of cowardice, you're so afraid of what I'll do to you that you
dare not even offer a definition of a clock! LOL! Go ahead, prove your
bravery.......try it. LOL!
Virtually none of your posts these days are related to physics. Instead,
virtually all of them are aimed at tearing into others. Care to explain
why?
You make this way too easy.
--
Time in S' as compared to S:
[T= T'*g]
Time on a linear clock in S' as compared to it's twin in S as both observed
from S:
[T = g*(T'+vL'/c^2)]
Androcles
I blundered. I admit it. You found my blunder.
I don't think Andersen blundered with his inverse LT, though.
If Roberts is right and the LTs are invertible, Andersen has done it.
What Andersen doesn't like is that the LT says the faster you go,
the sooner you'll arrive, and the AT says the faster you return, the
later you'll arrive. That makes a perverse kind of sense to me, the kind
of thing I'd expect from an inverse. Moving clocks run slow going away
and run fast when approaching.
Proof:
t = 1/f, so f = 1/t.
(See? Inverses.)
nu = f .sqrt([1+v/c]/[1-v/c]) approaching, *
nu = f .sqrt([1-v/c]/[1+v/c]) receding,
tau = 1/nu.
Therefore tau increases one way and decreases the other.
What's so wrong about that?
This is about mathematics, not about intuition or likes and dislikes.
*[I'm using nu and f rather than primes to avoid confusion of frames
and to be consistent with t and tau rather than t and t'. As proof I have
them the right way around, I quote Einstein:
"In the system K, very far from the origin of co-ordinates, let there be a
source of electrodynamic waves....We wish to know the constitution of
these waves, when they are examined by an observer at rest in the
moving system k." ]
Surely you recall the contra-revolving satellites?
phi is constantly changing in the equation
nu = f. (1-v.cos(phi)/c) / sqrt(1-v^2/c^2)
so we'd need to write d(phi)/dt, right?
Since tau = 1/nu, f = 1/t, and since we are discussing inversion,
tau = t . sqrt(1-v^2/c^2)/(1- v.cos(phi)/c)
unless I've blundered again. Probably I did.
d(sin(omega.t))/dt = omega.cos(t), doesn't it?
Unless I've forgotten the chain rule again.
Doesn't matter, you'll find my blunder (or moortel will and gloat).
I rather like this equation in the way it relates to the direction.
If you recall, I showed that x = vt by
tau = (t-vx / c^2)/ sqrt(1-v^2/c^2)
= (t- tv^2 / c^2)/ sqrt(1-v^2/c^2)
= t(1-v^2/c^2)/sqrt(1-v^2/c^2)
= t . sqrt(1-v^2/c^2)
but of course that assumed cos(phi) = 0.
When phi = pi, we have cos(pi) = -1,
tau = t . sqrt(1-v^2/c^2)/(1+ v/c)
To reiterate, Roberts' challenge is review the four postulates [of a group]
and to find a hole in them. I've reviewed them. No hole found.
"2. Any transformation has an inverse, which is also a transformation."
So now you prove the Andersen Transform is wrong as you claim.
I say it is as correct as the LT.
Androcles.
Dirk Van de moortel
"Androcles" <du...@dummy.net> wrote in message news:u8Tcd.105001$BI5....@fe2.news.blueyonder.co.uk...
>
> "sal" <pragm...@nospam.org> wrote in message
> news:1098102962.Ii0CDLqlzOP6+K8B4QZvTg@teranews...
> > On Sun, 17 Oct 2004 12:44:28 +0000, Androcles wrote:
[snip]
> > Of course, if you do that, you may find that it works out without any
> > "paradox" or contradiction, which would show that the Lorentz transforms
> > are invertible, just like Tom Roberts said. But you wouldn't hesitate to
> > post the result just because of that, would you?
>
> I blundered. I admit it. You found my blunder.
Bravo.
Next comes the pearl.
> I don't think Andersen blundered with his inverse LT, though.
> If Roberts is right and the LTs are invertible,
"If Roberts is right and the LTs are invertible"
How dense did you say you were?
Dirk Vdm
sal
> And if you had your arrogant way, not only would you not read *his* posts,
> but no one would.
I don't think it works that way. People are more inclined to read things
others have responded to already.
Take my posts, for instance. It used to be nobody looked at them. But
these days, if I start a thread, I know for sure, in advance, that there
will be at least THREE replies to it.
1) Androcles will respond, pointing out that I'm wrong.
2) Dirk will respond to Androcles, pointing out that he's icky.
3) AllYou! will respond to Dirk, pointing out that he's just posting
insults.
It's not exactly a heavy physics discussion, but hey, it gives me
something to do...
AllYou!
There's nothing wrong with predictability. OK, I'll admit that the idiot
has gotten under my skin and that I'm taking advantage of every opportunity
he gives me to show what an low-life he is. But that usually doesn't last
too long with me because despicable characters like him tend to bore me
after a while.
sal
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
> wrote in message news:1lScd.282260$SI7.14...@phobos.telenet-ops.be...
>>
[Dirk:]
>> > > Fabricating quotes again, dishonest coward?
[AllYou!]
>> > You can't produce any evidence that I'm any of the things you've
>> > asserted about me. I've never said anything bigoted or dishonest,
>> > and if I am a coward,
[Dirk:]
>> If you weren't, then you would use your real name to sign your posts in
>> stead of hiding behind "Allyou!".
[AllYou!]
> There are plenty of other reasons to maintain anonymity here and none of
> them have to do with cowardice, and certainly not of you.
This accusation -- "You use a pseudonym, so you're a coward!" -- gets
thrown around from time to time.
It may be true that most of the sensible posts come from people with real
names. However, it's also true that some of the most cogent,
well-informed posts in this group have come from anonymized, well-nigh
untraceable pseudonyms. And at the same time, some of the most irrational,
sub-human gibbering has come from creatures who took no pains at all to
hide their identities.
So, I don't see that you can conclude much from use of a pseudonym.
Stalking is rare but it does happen and some people worry about it more
than others.
[AllYou!]
> Virtually none of your posts these days are related to physics.
Sound like you've been around for a while. Your wording suggests you've
seen more physics from Dirk in the past than what he's been putting into
his recent posts.
I have that impression too, but I haven't counted posts or anything to try
to back it up. Maybe he's just too busy to put a lot of time into Usenet
these days.
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:1098123500.HGG7LDUa/V394B9d4MO+XA@teranews...
> On Mon, 18 Oct 2004 10:50:56 -0400, AllYou! wrote:
>
> > And if you had your arrogant way, not only would you not read *his* posts,
> > but no one would.
>
> I don't think it works that way. People are more inclined to read things
> others have responded to already.
>
> Take my posts, for instance. It used to be nobody looked at them. But
> these days, if I start a thread, I know for sure, in advance, that there
> will be at least THREE replies to it.
>
> 1) Androcles will respond, pointing out that I'm wrong.
>
> 2) Dirk will respond to Androcles, pointing out that he's icky.
Had to look it up...
hm, I like that word, specially in this context. Very appropriate
indeed.
>
> 3) AllYou! will respond to Dirk, pointing out that he's just posting
> insults.
>
> It's not exactly a heavy physics discussion, but hey, it gives me
> something to do...
Surely you must have better things to do :-)
Dirk Vdm
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:1098124286.nz/1kvNmi2lScgetquCR2A@teranews...
> On Mon, 18 Oct 2004 12:55:21 -0400, AllYou! wrote:
>
>
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
> > wrote in message news:1lScd.282260$SI7.14...@phobos.telenet-ops.be...
> >>
> [Dirk:]
> >> > > Fabricating quotes again, dishonest coward?
>
> [AllYou!]
> >> > You can't produce any evidence that I'm any of the things you've
> >> > asserted about me. I've never said anything bigoted or dishonest,
> >> > and if I am a coward,
>
> [Dirk:]
> >> If you weren't, then you would use your real name to sign your posts in
> >> stead of hiding behind "Allyou!".
>
> [AllYou!]
> > There are plenty of other reasons to maintain anonymity here and none of
> > them have to do with cowardice, and certainly not of you.
>
> This accusation -- "You use a pseudonym, so you're a coward!" -- gets
> thrown around from time to time.
>
> It may be true that most of the sensible posts come from people with real
> names. However, it's also true that some of the most cogent,
> well-informed posts in this group have come from anonymized, well-nigh
> untraceable pseudonyms. And at the same time, some of the most irrational,
> sub-human gibbering has come from creatures who took no pains at all to
> hide their identities.
>
> So, I don't see that you can conclude much from use of a pseudonym.
It was concluded from a combination of the pseudonym and a few
years of experience with this "person". How many years did you say
you had with "Allyou!"?
Hey, this sounds familiar. We've been here before, haven't we?
Dirk Vdm
AllYou!
And if you look back through what happened a couple of years ago, you'll see
that as soon as I dared question your posts and you saw that we were headed
to a point where you'd be in over your head, like trying to reconcile how
time can be what a clock says it is with the puzzle of a linear clack, you
began the most aggressive of personal attacks.
But then when I resurfaced this time, you barely skipped a beat and tried to
discredit me again with similar attacks, hoping to keep what you saw as the
secret of your inadequacy from being exposed.
And then there were the butt-munch, asswhipe buddies of yours' like Hobba
who have no mind of their own and fell right into line, but then there were
a certain number of others who took me at face value.
So now your secret is out and your incompetence has been exposed. You're so
damned afraid of an uneducated no-nothing like me that you refuse to even
venture a guess as to the definition of a clock.
Amazing.
sal
[ vitriol ]
Hey, would you please stop setting the followups to alt.morons?
You've succeeded in teleporting 2 of my replies and about a dozen of your
own into that group, but it hasn't slowed Dirk down one whit. I think
he's got his news reader set to ignore the followups field.
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:1098127032.M6LWZuf6orZeUsTykN6cAQ@teranews...
> On Mon, 18 Oct 2004 14:53:51 -0400, AllYou! wrote:
>
> [ vitriol ]
>
> Hey, would you please stop setting the followups to alt.morons?
>
> You've succeeded in teleporting 2 of my replies and about a dozen of your
> own into that group, but it hasn't slowed Dirk down one whit. I think
> he's got his news reader set to ignore the followups field.
I can do magic with my news reader ;-)
Dirk Vdm
sal
What can I say? I'm a fan of H. P. Lovecraft.
AllYou!
> On Mon, 18 Oct 2004 14:53:51 -0400, AllYou! wrote:
>
> [ vitriol ]
>
> Hey, would you please stop setting the followups to alt.morons?
>
> You've succeeded in teleporting 2 of my replies and about a dozen of your
> own into that group, but it hasn't slowed Dirk down one whit. I think
> he's got his news reader set to ignore the followups field.
Actually, I learned at the feet of the master. I was setting him up and you
got caught in the trap. OK, I'll stop, but only because you asked so
nicely.
Dirk Van de moortel
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:1098126568.QEKgFVuGkDowVhxzCfe3IQ@teranews...
> On Mon, 18 Oct 2004 18:35:52 +0000, Dirk Van de moortel wrote:
[snip]
> > Hey, this sounds familiar. We've been here before, haven't we?
>
> What can I say? I'm a fan of H. P. Lovecraft.
Had a google-look ... never heard of him before... Interesting.
Do you have a "One Size Fits All" entry point for Lovecraft?
Dirk Vdm
Paul B. Andersen
"sal" <pragm...@nospam.org> skrev i melding news:1098102962.Ii0CDLqlzOP6+K8B4QZvTg@teranews...
> On Sun, 17 Oct 2004 12:44:28 +0000, Androcles wrote:
>
> >
> "paradox" or contradiction, which would show that the Lorentz transforms
> are invertible, just like Tom Roberts said. But you wouldn't hesitate to
> post the result just because of that, would you?
If you really think that you can teach Androcles anything
about transformations and their inverse, you should have a look at the following:
http://groups.google.com/groups?q=g:thl3103600516d&dq=&hl=no&lr=&selm=38CA6965.F185F814%40hia.no
and the thread it is in.
That's more than four years ago, and the discussion is still going on. :-)
Paul
Androcles
> On Mon, 18 Oct 2004 10:50:56 -0400, AllYou! wrote:
>
>> And if you had your arrogant way, not only would you not read *his*
>> posts,
>> but no one would.
>
> I don't think it works that way. People are more inclined to read things
> others have responded to already.
>
> Take my posts, for instance. It used to be nobody looked at them. But
> these days, if I start a thread, I know for sure, in advance, that there
> will be at least THREE replies to it.
>
> 1) Androcles will respond, pointing out that I'm wrong.
LMAO! I'm surprised you challenged Andersen's Transforms,
I have to admit. I'd have thought the fumble mumbler would have
been all over you on that one, Andersen is his hero. I don't know
if he has, I don't read his posts. Still, divide and conquer.
>
> 2) Dirk will respond to Androcles, pointing out that he's icky.
Well, of course. He's been stalking me for years and still thinks
-1 = sqrt( (-1)^2) because sqrt() is always positive.
> 3) AllYou! will respond to Dirk, pointing out that he's just posting
> insults.
I don't really understand the purpose of that. Obviously Dinky is
ineducable with a following of equally ineducable gorillas, so why even try
to educate the puppy? He seems to get his kicks from hurling insults,
thinks he's doing the world a favour, will never reach adulthood, so
leave the spermless one to his own de-vices or vices.
> It's not exactly a heavy physics discussion, but hey, it gives me
> something to do...
I burned your pellet in the middle, BTW, so that the two halves
separated.
Androcles
Androcles
It was cribbed it from Bilge. You don't imagine he has any originality, do
you?
Androcles.
sal
Uh, not really. Main thing is to avoid the "collaborations" with August
Derleth -- they were done posthumously and it's not clear HP contributed
more than a couple words scribbled on a napkin to any of them.
"The Lurking Fear" is a good place to start if you're looking for
sub-human gibberrers.
"The Color out of Space" is one of his creepiest, though too much science
in your background may make it hard to swallow. It's also lost a lot of
its shock value since Love Canal happened.
"At the Mountains of Madness" is one of my personal favorites, as is
"The Dream Quest of Unknown Kadeth". Those are both (short) novels; most
of what he wrote was short stories.
I'm afraid he's just a little too recent to be available on Gutenberg,
unfortunately. Only a few of his stories seem to be available online.
But see also:
http://www.locksley.com/cthulhu/necron.htm
Dirk Van de moortel
"Androcles" <du...@dummy.net> wrote in message news:30Wcd.105597$BI5....@fe2.news.blueyonder.co.uk...
>
> "sal" <pragm...@nospam.org> wrote in message
> news:1098123500.HGG7LDUa/V394B9d4MO+XA@teranews...
> > On Mon, 18 Oct 2004 10:50:56 -0400, AllYou! wrote:
> >
> >> And if you had your arrogant way, not only would you not read *his*
> >> posts,
> >> but no one would.
> >
> > I don't think it works that way. People are more inclined to read things
> > others have responded to already.
> >
> > Take my posts, for instance. It used to be nobody looked at them. But
> > these days, if I start a thread, I know for sure, in advance, that there
> > will be at least THREE replies to it.
> >
> > 1) Androcles will respond, pointing out that I'm wrong.
>
> LMAO! I'm surprised you challenged Andersen's Transforms,
> I have to admit. I'd have thought the fumble mumbler would have
> been all over you on that one, Andersen is his hero. I don't know
> if he has, I don't read his posts. Still, divide and conquer.
>
> >
> > 2) Dirk will respond to Androcles, pointing out that he's icky.
>
> Well, of course. He's been stalking me for years and still thinks
> -1 = sqrt( (-1)^2) because sqrt() is always positive.
Gee.... what an incredible stupid piece of crap you are :-)
So you still think that
sqrt(x^2) = x for real x > 0
sqrt(x^2) = -x for reall x < 0
implies this?
You *still* don't understand it?
HAHAHAHAHAHAHAHA!
Dirk Vdm
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:pan.2004.10.18....@nospam.org...
I'll have a look in the library next time...
Thanks.
Dirk Vdm
Dirk Van de moortel
"Androcles" <du...@dummy.net> wrote in message news:30Wcd.105597$BI5....@fe2.news.blueyonder.co.uk...
>
> news:1098123500.HGG7LDUa/V394B9d4MO+XA@teranews...
[snip]
> > 2) Dirk will respond to Androcles, pointing out that he's icky.
>
> Well, of course. He's been stalking me for years and still thinks
> -1 = sqrt( (-1)^2) because sqrt() is always positive.
This simply *screams* for a new entry:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/STILL.html
Title: "I STILL DON'T GET IT!!!"
Do they actually let you come near your grandchildren? Gee.
Dirk Vdm
sal
>
> "sal" <pragm...@nospam.org> wrote in message
> news:1098123500.HGG7LDUa/V394B9d4MO+XA@teranews...
>> On Mon, 18 Oct 2004 10:50:56 -0400, AllYou! wrote:
>>
>>> And if you had your arrogant way, not only would you not read *his*
>>> posts,
>>> but no one would.
>>
>> I don't think it works that way. People are more inclined to read
>> things others have responded to already.
>>
>> Take my posts, for instance. It used to be nobody looked at them. But
>> these days, if I start a thread, I know for sure, in advance, that there
>> will be at least THREE replies to it.
>>
>> 1) Androcles will respond, pointing out that I'm wrong.
>
> LMAO! I'm surprised you challenged Andersen's Transforms, I have to admit.
I've looked them over in the past (as you no doubt recall). They are, as
I recall, the Lorentz transform with a sign flipped (i.e., wrong-way).
They provide the wrong answer, very reliably, every time. I seem to
recall you were delighted because Paul Andersen supposedly posted the
transform, with incorrect sign, at some point, and you've been quoting it
ever since.
As far as I'm concerned, it's old news -- Paul Andersen either made a
mistake or didn't, you either quoted him correctly or didn't, and either
way I don't much care. Counting his typos isn't one of my hobbies. As
I've said in earlier posts, months ago, I don't do so-called "Andersen
transforms" -- at the point where you say "I'm using the Andersen
transform" I stop reading.
Eric Gisse
How was this man an engineer at any point in his life?
Androcles
I notice you didn't bother to provide the multiplicative or additive
inverses
to the LTs that Roberts says have to exist.
Hint:
1-1 = 0 Group (set R, operator +), identity
1/1 = 1 (Group (set R, operator *), identity.
I thought you like puzzles.
Maybe only those you create and know the answers too, huh?
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:cl19l7$ba3$1...@dolly.uninett.no...
>
> "sal" <pragm...@nospam.org> skrev i melding news:1098102962.Ii0CDLqlzOP6+K8B4QZvTg@teranews...
> > On Sun, 17 Oct 2004 12:44:28 +0000, Androcles wrote:
[snip probably something utterly stupid]
> > Of course, if you do that, you may find that it works out without any
> > "paradox" or contradiction, which would show that the Lorentz transforms
> > are invertible, just like Tom Roberts said. But you wouldn't hesitate to
> > post the result just because of that, would you?
>
> If you really think that you can teach Androcles anything
> about transformations and their inverse, you should have a look at the following:
> http://groups.google.com/groups?q=g:thl3103600516d&dq=&hl=no&lr=&selm=38CA6965.F185F814%40hia.no
> and the thread it is in.
>
> That's more than four years ago, and the discussion is still going on. :-)
They can't be more imbecile than Andocles.
According to Androcles the pair of equations
sqrt( x^2 ) = x for all real x > 0
sqrt( x^2 ) = -x for all real x < 0
implies
-1 = sqrt( (-1)^2 )
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/STILL.html
etc...
He can't even handle one (1!) equation.
Ask him to say something about a quadratic equation.
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Quadratic.html
Ask him to solve a set of equations:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html
I think it's the blood in his wild eyes that prevents him from seeing.
Someone should do the right thing and finally put him to sleep.
Dirk Vdm
sal
You're kidding, right? Any SR text has these -- even Martin Gardner must
have shown them. Are you really asserting they don't exist?
If frame S' is moving in the +X direction relative to frame S, at velocity
v, and if g = 1/sqrt(1-v^2), and (0,0) -> (0',0'), and c=1, then
t' = g * (t - vx)
x' = g * (-vt + x)
Inverse transform is
t = g * (t' + vx')
x = g * (vt' + x')
Compose them and see what you get.
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:1098188826.rvmapfHxAXI0iDrV6yh5Pg@teranews...
> On Tue, 19 Oct 2004 05:31:17 +0000, Androcles wrote:
[snip]
> > I notice you didn't bother to provide the multiplicative or additive
> > inverses to the LTs that Roberts says have to exist.
>
> You're kidding, right? Any SR text has these -- even Martin Gardner must
> have shown them. Are you really asserting they don't exist?
>
> If frame S' is moving in the +X direction relative to frame S, at velocity
> v, and if g = 1/sqrt(1-v^2), and (0,0) -> (0',0'), and c=1, then
>
> t' = g * (t - vx)
> x' = g * (-vt + x)
>
> Inverse transform is
>
> t = g * (t' + vx')
> x = g * (vt' + x')
>
> Compose them and see what you get.
Hundreds of times this has been show to him (and to others).
He always goes berserk when he has to handle the square roots.
He can't decide whether they are positive or negative ;-)
Dirk Vdm
Paul B. Andersen
"sal" <pragm...@nospam.org> skrev i melding news:1098154892.OPjjUzqGjauqy3e6VJ+ZIQ@teranews...
> On Mon, 18 Oct 2004 20:50:07 +0000, Androcles wrote:
> > LMAO! I'm surprised you challenged Andersen's Transforms, I have to admit.
For once I agree with Androcles.
I am surprised as well! :-)
> I've looked them over in the past (as you no doubt recall). They are, as
> I recall, the Lorentz transform with a sign flipped (i.e., wrong-way).
> They provide the wrong answer, very reliably, every time. I seem to
> recall you were delighted because Paul Andersen supposedly posted the
> transform, with incorrect sign, at some point, and you've been quoting it
> ever since.
I wrote it with the right signs, and it provides
the right answer, very reliable, every time.
Please read it again carefully, you will see that it is correct.
> "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
The quotation is correct, and my words are correct, even if
they may be a bit ambiguous, taken out of context as they are.
But I think it _should_ be obvious that "reversing the direction"
refers to reversing the direction of relative motion of the frames
compared to the "conventional direction", in which case
the signs will be opposite to the "conventional form" of the LT.
But here is the context it is taken from:
Einstein let the k-system (greek letters) move in the positive
direction along the x-axis of the K system (latin letters)
with xi- and x-axes aligned.
|-> v
--------|-------------> xi k moving
-----|----------------> x K stationary
In this case the Lorentz transform is:
xi = (x - vt)/sqrt(1-v^2/c^2)
tau = (t - xv/c^2)/sqrt(1-v^2/c^2)
or:
x = (xi + v*tau)/sqrt(1-v^2/c^2)
t = (tau + xi*v/c^2)/sqrt(1-v^2/c^2)
But there are alternative ways to define the frames of reference.
We can reverse the directions of motion of k
v <-|
-----|----------------> xi k moving
--------|-------------> x K stationary
which is the same as interchanging the frames,
(as compared to Einstein's convention)
|-> v
--------|-------------> x K moving
-----|----------------> xi k stationary
which 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)
It should be an obvious triviality that the signs in the LT
depend on how the frames of reference are defined.
(You can find two more combinations by letting the axes
point in the opposite directions as well.)
The story behind this is that Androcles - several years
ago - reversed the direction of motion of the k-system,
and made a big number of the fact that the signs in the LT
changed. He finds it contradictory, and insists that both
forms of the LT must be valid _simultaneously_.
I have indeed told him "a LOT of times" that it is not
contradictory but an obvious triviality - to no avail.
That guy is simply unable to learn anything!
That's why he keep quoting my correct words - thinking
that he is making a fool of me when doing so.
You can read all about it in:
http://groups.google.com/groups?q=g:thl3103600516d&dq=&hl=no&lr=&selm=38CA6965.F185F814%40hia.no
and the thread it is in.
I am confident that you now know who he is making a fool of. :-)
Paul
sal
>
> "sal" <pragm...@nospam.org> skrev i melding
> news:1098154892.OPjjUzqGjauqy3e6VJ+ZIQ@teranews...
>> On Mon, 18 Oct 2004 20:50:07 +0000, Androcles wrote:
>> > LMAO! I'm surprised you challenged Andersen's Transforms, I have to
>> > admit.
>
> For once I agree with Androcles.
> I am surprised as well! :-)
Um -- I had the impression that when A. referred to the "Andersen
transform" he was talking about a version of the Lorentz transform with
incorrect signs. Certainly its application at his hands invariably
produces incorrect results.
>> I've looked them over in the past (as you no doubt recall). They are,
>> as I recall, the Lorentz transform with a sign flipped (i.e.,
>> wrong-way). They provide the wrong answer, very reliably, every time. I
>> seem to recall you were delighted because Paul Andersen supposedly
>> posted the transform, with incorrect sign, at some point, and you've
>> been quoting it ever since.
>
> I wrote it with the right signs, and it provides the right answer, very
> reliable, every time.
>
> Please read it again carefully, you will see that it is correct.
OK.
>> "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
Yes, as stated that's certainly correct, assuming you started with a
"right-moving" Greek frame before you interchanged them.
The issue that comes up, I think, is that you reversed the direction while
keeping v unchanged. If Androcles tries to apply it while
simultaneously changing v ---> -v then it doesn't work.
Alternatively, if he tries to apply it to a set of frames where the Greek
frame is moving in the +x direction at v, it also doesn't work.
The key here is that the Greek frame is moving in the -X direction at
velocity v.
> It should be an obvious triviality that the signs in the LT depend on
> how the frames of reference are defined. (You can find two more
> combinations by letting the axes
> point in the opposite directions as well.)
>
> The story behind this is that Androcles - several years ago - reversed
> the direction of motion of the k-system, and made a big number of the
> fact that the signs in the LT changed. He finds it contradictory, and
> insists that both forms of the LT must be valid _simultaneously_.
>
> I have indeed told him "a LOT of times" that it is not contradictory but
> an obvious triviality - to no avail. That guy is simply unable to learn
> anything!
>
> That's why he keep quoting my correct words - thinking that he is making
> a fool of me when doing so.
>
> You can read all about it in:
> http://groups.google.com/groups?q=g:thl3103600516d&dq=&hl=no&lr=&selm=38CA6965.F185F814%40hia.no
> and the thread it is in.
>
> I am confident that you now know who he is making a fool of. :-)
Indeed.
In general when Androcles says he is "using the Andersen transform" he
follows it with a mis-application of the Lorentz transform. After walking
through it once or twice and pointing out that he'd gotten a sign
flipped, I simply stopped looking at it.
>
> Paul
Dirk Van de moortel
"Eric Gisse" <fs...@uaf.edu> wrote in message news:fd0fc2fa.04101...@posting.google.com...
Impossible.
Either he lied about it, or he has eaten too much scrap meat.
Dirk Vdm
Androcles
I concluded that the (ahem..., he whose name must not be mentioned, but he
told me LOTS of times) transforms were the inverse of the Lorentz
Transforms.
So no, I'm not asserting they don't exist.
As for Gardner, he mumbled something about
"If two people stand on opposite sides of a huge concave lens, each sees
the other as smaller, but that is not the same as saying that each is
smaller."
I concluded from that the two people were really were the same size and
we were talking about an illusion, and as experiment I put a not so huge
concave len in front of a mirror to see if I looked smaller than I really
am.
Actually, I do this quite often to shave, I wear concave eyeglasses, and
I really do look smaller than I really am. So I found a convex lens and
that makes me look bigger than I am. So the see myself at the right size,
I figured I'd need one of each and that convex lens is the inverse transform
of the concave lens.
Other than that, no, I didn't see anything that corresponded to the inverse
(ahem) transforms... dang, I almost said it.
>
> If frame S' is moving in the +X direction relative to frame S, at velocity
> v, and if g = 1/sqrt(1-v^2), and (0,0) -> (0',0'), and c=1, then
>
> t' = g * (t - vx)
> x' = g * (-vt + x)
>
> Inverse transform is
>
> t = g * (t' + vx')
> x = g * (vt' + x')
>
> Compose them and see what you get.
Hmm... here's the problem. I'm standing midway between two mirrors, A and B.
A-------------- ME -------------B
-x----------------0--------------+x
I shine two lights simultaneously at A and at B (really only one light bulb
in a tube)
as I move away from 0 with velocity v, so that x' = +x-vt and -x' = -x-vt.
The light bounces off the mirrors back to (well, I'm not sure), Einstein
says to
ME in his words, but his equation says to 0.
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))
Now, I don't change direction, the light does, so I don't see where he gets
his t1 = x'/(c-v) and t2 = x'/(c+v) from.
I think I can solve it, though, if we consider x' as distance and not as
a coordinate. Then it would be t1 = -x'/(v-c) and t2 = x'/(v+c).
If you prefer, t1-t0 = -x'/(v-c) and t2-t0 = x'/(v+c).
We have two positive time intervals because |c| > v.
Thus -x'/(v-c) = x'/(c-v) and we are all happy.
The next hard part is the light reaching the other mirror.
"But the ray moves relatively to the initial point of k, when measured in
the stationary system, with the velocity c-v, so that x'/(c-v) = t."
For the other mirror, we have
"But the ray moves relatively to the initial point of k, when measured in
the stationary system, with the velocity c+v, so that x'/(c+v) = t."
and that makes the RHS of the equation different.
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))
Now I get, for my watch, (I'm the one doing the moving)
tau = g * (t + vx)
xi = g * (vt + x) for one mirror and
tau = g * (t - vx)
xi = g * (x') for the other, x' = x-vt being previously defined, of
course.
He who shall not be named said:
"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)"
and that is what I have for the other mirror.
Recall that I am now displaced from zero at tau2, so the light, travelling
at
invariant c in the stationary frame, reaches me from one mirror before the
light from the other mirror.
Hence tau(x',0,0,t+x'/(c-v)) = tau(x',0,0,t+x'/(c+v))
Since the light returns to 0 and not 0' = 0-vt, v = 0.
To answer your question "Are you kidding?": No, I'm not kidding at all.
Einstein omitted a prime.
Androcles
sal
You made no comment. Did that mean you agree that these are inverses?
>
> Hmm... here's the problem. I'm standing midway between two mirrors, A and
> B. A-------------- ME -------------B
> -x----------------0--------------+x
> I shine two lights simultaneously at A and at B (really only one light
> bulb in a tube)
> as I move away from 0 with velocity v, so that x' = +x-vt and -x' = -x-vt.
There are some problems with this statement. You've mixed
coordinate values with coordinate names. For any particular value of the
x coordinate, the value of the x' coordinate is x-vt. That was Einstein's
definition of the first coordinate in the k frame. But you're also using
both x and x' to refer to specific locations on the x and x' axes.
That is confusing and can lead to errors.
> The light bounces off the mirrors back to (well, I'm not sure), Einstein
> says to
> ME in his words, but his equation says to 0.
No, that is not correct. He says the flashes return to ME in his
equations, too. Third paragraph, page 44, Dover, he says,
"We first define tau as a function of x', y, z, and t"
Tau is a function of mixed coordinates. The first argument is not the x
coordinate in the K frame; it's the x' coordinate in the k frame.
> 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))
Coordinates (0,0,0,t) are the starting coordinates for ME.
Coordinates (0,0,0,t+x'/(c-v)+x'/(c+v)) are the ending coordinates for ME.
Coordinates (x',0,0,t+x'/(c-v)) are the coordinates of the mirror at the
moment of impact, in the k frame (with K's time value).
> Now, I don't change direction, the light does, so I don't see where he
> gets his t1 = x'/(c-v)
Time for the light flash to catch up with the moving mirror, located at x'
in the k frame, as measured in the K frame. Tau is a function of time in
the K frame, and space coordinates in the k frame, as I said.
> and t2 = x'/(c+v) from.
Time to get from x' back to ME, in the k frame, as measured by clocks in
the K frame. ME is (bad grammar) moving toward the right, as the flash
comes back to the left, so it takes less time to close the gap.
> I think I can solve it,
> though, if we consider x' as distance and not as a coordinate.
But it's not a distance -- it's a coordinate. I'll keep reading and
see where this comes in, and see if your assumption that it's a distance
causes a problem.
> Then it
> would be t1 = -x'/(v-c) and t2 = x'/(v+c). If you prefer, t1-t0 =
> -x'/(v-c) and t2-t0 = x'/(v+c).
A little odd but not totally wrong. This is still time in the K
frame.
> We have two positive time intervals
> because |c| > v.
Since c ~ 300000 k/sec > 0, we also have |c| = c. Whatever, that's still
correct.
> Thus -x'/(v-c) = x'/(c-v) and we are all happy.
You've successfully multiplied twice by -1/-1. Not sure I see the point,
but carry on.
> The next hard part is the light reaching the other mirror.
> "But the ray moves relatively to the initial point of k, when measured
> in
> the stationary system, with the velocity c-v, so that x'/(c-v) = t." For
> the other mirror, we have
> "But the ray moves relatively to the initial point of k, when measured
> in
> the stationary system, with the velocity c+v, so that x'/(c+v) = t." and
> that makes the RHS of the equation different.
Yes, it does.
> 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))
Your RHS is not correct. Your second mirror is located at -x', not at x',
and the values in this formula are coordinates, not distances. So, your
equation for the second mirror must read
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 this is where the problem creeps in from your assumption that x' was
a distance, not a coordinate.
Refigure the rest of this with the correct coordinate value, which is -x',
and see what you get. From here down it just looks to me like the flipped
sign on the x' coordinate is the source of the problems.
> Now I get, for my watch, (I'm the one doing the moving)
> tau = g * (t + vx)
> xi = g * (vt + x) for one mirror and
> tau = g * (t - vx)
> xi = g * (x') for the other, x' = x-vt being previously defined, of
> course.
>
> He who shall not be named said:
> "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)"
> and that is what I have for the other mirror.
>
> Recall that I am now displaced from zero at tau2, so the light,
> travelling at
> invariant c in the stationary frame, reaches me from one mirror before
> the light from the other mirror.
> Hence tau(x',0,0,t+x'/(c-v)) = tau(x',0,0,t+x'/(c+v)) Since the light
> returns to 0 and not 0' = 0-vt, v = 0. To answer your question "Are you
> kidding?": No, I'm not kidding at all. Einstein omitted a prime.
>
> Androcles
--
Dirk Van de moortel
"Androcles" <du...@dummy.net> wrote in message news:Hhcdd.84675$ay5....@fe1.news.blueyonder.co.uk...
>
> "sal" <pragm...@nospam.org> wrote in message
> news:1098188826.rvmapfHxAXI0iDrV6yh5Pg@teranews...
> > On Tue, 19 Oct 2004 05:31:17 +0000, Androcles wrote:
[snip]
> > If frame S' is moving in the +X direction relative to frame S, at velocity
> > v, and if g = 1/sqrt(1-v^2), and (0,0) -> (0',0'), and c=1, then
> >
> > t' = g * (t - vx)
> > x' = g * (-vt + x)
> >
> > Inverse transform is
> >
> > t = g * (t' + vx')
> > x = g * (vt' + x')
> >
> > Compose them and see what you get.
>
> Hmm... here's the problem.
You don't even understand what "compose them" means, do you?
Why don't you just ask: "sal, what are you talking about?"
> I'm standing midway between two mirrors, A and B.
> A-------------- ME -------------B
> -x----------------0--------------+x
> I shine two lights simultaneously at A and at B (really only one light bulb
> in a tube)
> as I move away from 0 with velocity v, so that x' = +x-vt and -x' = -x-vt.
Geeeeeeee ..... you don't know what these letters *represent* ;-)
> The light bounces off the mirrors back to (well, I'm not sure), Einstein
> says to
> ME in his words, but his equation says to 0.
> 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))
How would a complete idiot, who can't even handle a simple
square root, understand a fucking monster like this???
Stop it, you make us laugh *so* hard... It hurts!
Dirk Vdm
Dirk Van de moortel
"sal" <pragm...@nospam.org> wrote in message news:pan.2004.10.19...@nospam.org...
Hadn't seen your reply before I posted mine.
Are you aiming at Sainthood?
:-)
Dirk Vdm
Androcles
> On Tue, 19 Oct 2004 15:44:52 +0200, Paul B. Andersen wrote:
>
>>
>> "sal" <pragm...@nospam.org> skrev i melding
>> news:1098154892.OPjjUzqGjauqy3e6VJ+ZIQ@teranews...
>>> On Mon, 18 Oct 2004 20:50:07 +0000, Androcles wrote:
>>> > LMAO! I'm surprised you challenged Andersen's Transforms, I have to
>>> > admit.
>>
>> For once I agree with Androcles.
>> I am surprised as well! :-)
>
> Um -- I had the impression that when A. referred to the "Andersen
> transform" he was talking about a version of the Lorentz transform with
> incorrect signs. Certainly its application at his hands invariably
> produces incorrect results.
>
>
>>> I've looked them over in the past (as you no doubt recall). They are,
>>> as I recall, the Lorentz transform with a sign flipped (i.e.,
>>> wrong-way). They provide the wrong answer, very reliably, every time. I
>>> seem to recall you were delighted because Paul Andersen supposedly
>>> posted the transform, with incorrect sign, at some point, and you've
>>> been quoting it ever since.
>>
>> I wrote it with the right signs, and it provides the right answer, very
>> reliable, every time.
>>
>> Please read it again carefully, you will see that it is correct.
>
> OK.
Please read this carefully, you will see it is correct.
We have two frames and two directions we can move,
toward the mirror and away from the mirror.
That is four possible outcomes.
They are
1)
xi = (x-vt)*g
tau = (t -vx)*g
2)
xi = (x+vt)*g
tau = (t+vx)*g
3)
x = (x'+v.tau)*g
t = (v.xi+tau)*g
4)
x = (xi-v.tau)*g
t = (tau-v.xi)*g
Andersen has successfully pointed out that 1) and 3) are equivalent.
Nobody wants to look at 2) and 4), which are also equivalent.
You can transform between frames AND change direction, but
refusing to retain the existing frames while changing direction
doesn't solve the twin paradox, because that is exactly what happens.
The proof is in the blue shift from the returning clock.
Simply compute tau = 1/nu from the doppler equation.
Recall that nu = f.[1-v.cos(phi)] *g and you are going to rotate
phi only, you are not going to exchange the greek and roman frames.
You will find, if you do this correctly, that Stella will have sent
Terence exactly the right number of birthday greetings as
Terence sends Stella. Recall that they are twins.
They simply will not arrive on their joint birthday.
Receding clocks run slow. Approaching clocks run fast.
Which ignores the Greek frame moving in the -X direction at velocity v
I'm not trying to make a fool of Andersen, he can do that all by himself,
and so can you. Time dilation is directional (or would be if it existed).
Einstein has made fools of the lot of you.
Androcles.
Androcles
It depends on the operator. We are discussing Roberts use of inverse.
As you know, the set R and the operator ' + ' has an inverse ' - ' with the
identity 0.
We can transform x to x' with x' = x +1 and back again with x = x'-1, which
is analogous to what you are doing.
For the same set R and the operator '*', we can transform x to x' with x' =
2x and back again with x = x'/2.
I consider
x' = g * (t-vx) to have an inverse
x = (t-x'/g) / v.
>>
>> Hmm... here's the problem. I'm standing midway between two mirrors, A and
>> B.
>> A-------------- ME -------------B
>> -x----------------0--------------+x
>> I shine two lights simultaneously at A and at B (really only one light
>> bulb in a tube)
>> as I move away from 0 with velocity v, so that x' = +x-vt and -x'
>> = -x-vt.
>
> There are some problems with this statement. You've mixed
> coordinate values with coordinate names. For any particular value of the
> x coordinate, the value of the x' coordinate is x-vt.
It's quite easy, do it on the back of two envelopes.
Overlay one envelope on the other, pinprick through twice, and then mark
one envelope k and the other K. At the leftmost pinprick for K write
0, and mark x for the rightmost. On the k envelope, mark a corresponding
0' and x'. Now slide one envelope past the other with velocity v and time it
until the 0' hole aligns with the x hole, and bingo, 0' = 0 - vt, x' = x -
vt.
0-------------------x t = 0.
0'------------------x' tau = 0.
--------------------0-------------------x t = 1.
--------------------|<-------+vt----->| v = dx/dt
0'------------------x' tau = 1.
see, x' = x-vt.
Now repeat but move the opposite way, until x' aligns with 0.
x' = x+vt, 0' = 0+vt.
0------------------x
--------------------|<-------(-vt)---->|
--------------------0'------------------x'
see, x' = x+vt, 0' = 0+vt.
> That was Einstein's
> definition of the first coordinate in the k frame.
Sure, I've drawn a picture. Do it yourself on paper. (agreement 1)
> But you're also using
> both x and x' to refer to specific locations on the x and x' axes.
Yeah, that's right... x' and 0' are coordinates in k, origin 0',0',0', tau
and x and 0 are coordinates in K, origin 0,0,0,t. (agreement 2)
>
> That is confusing and can lead to errors.
Well, yes, of course. I totally agree. (agreement 3)
Einstein was deliberately confusing. He refused to label the k
origin as 0',0',0', tau. He called it 0,0,0,t to confuse you, and you
fell for it. :-)
That's why it leads to the error of the Lorentz Transform.
>
>> The light bounces off the mirrors back to (well, I'm not sure), Einstein
>> says to
>> ME in his words, but his equation says to 0.
>
> No, that is not correct. He says the flashes return to ME in his
> equations, too. Third paragraph, page 44, Dover, he says,
>
> "We first define tau as a function of x', y, z, and t"
That's right. That is what he says. (agreement 4)
> Tau is a function of mixed coordinates. The first argument is not the x
> coordinate in the K frame; it's the x' coordinate in the k frame.
But he's also using both 0 and 0 to refer to specific locations on the x and
x' axes.
That is confusing and can lead to errors. Not only that, but you've
blundered, the first argument is not the x' coordinate in the k frame, it is
the 0'
coordinate in the k-frame that he calls 0 because it coincides with 0 in the
K-frame
when t=t. Until you have learned to label ALL coordinates correctly you'll
run into
errors.
>
>> 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))
>
> Coordinates (0,0,0,t) are the starting coordinates for ME.
>
> Coordinates (0,0,0,t+x'/(c-v)+x'/(c+v)) are the ending coordinates for ME.
That's right. (agreement 5)
I (ME) start at (0,0,0, t )and finish at (0,0,0,t+ ...), time has passed and
I haven't moved. v= 0. If I HAD moved, I'd be at (vt,0,0, t+...) .
(Well, I am, actually, because v = 0.)
> Coordinates (x',0,0,t+x'/(c-v)) are the coordinates of the mirror at the
> moment of impact, in the k frame (with K's time value).
That's right, (agreement 6),only the mirror is moving relative to K.
I'm not, I end up at (0,0,0,t+...). Since I remain in the K-frame, tau = t.
>
>> Now, I don't change direction, the light does, so I don't see where he
>> gets his t1 = x'/(c-v)
>
> Time for the light flash to catch up with the moving mirror, located at x'
> in the k frame, as measured in the K frame.
Yep. (agreement 7)
The mirror is running away from me, too.
Do you think I should chase after it?
> Tau is a function of time in
> the K frame, and space coordinates in the k frame, as I said.
That's right. (agreement 8)
>> and t2 = x'/(c+v) from.
>
> Time to get from x' back to ME, in the k frame, as measured by clocks in
> the K frame.
Ahh. We have our first MAJOR disagreement.
Ain't no way that moving mirror is going to impart a reflected velocity of
c+v. That'd violate the First Amendment to the Same Laws of Electrodynamics
and Optics will be valid for all frames of reference for which the equations
of mechanics hold good, (c+v)/(1 + v/c) = c.
I'll stop there and we'll pick up when we are in agreement once more.
Androcles.
Dirk Van de moortel
"Androcles" <du...@dummy.net> wrote in message news:9_fdd.85781$ay5....@fe1.news.blueyonder.co.uk...
>
> "sal" <pragm...@nospam.org> wrote in message
> news:pan.2004.10.19...@nospam.org...
> > On Tue, 19 Oct 2004 17:37:43 +0000, Androcles wrote:
> >
> >>
> >> "sal" <pragm...@nospam.org> wrote in message
> >> news:1098188826.rvmapfHxAXI0iDrV6yh5Pg@teranews...
[snip]
> >>> If frame S' is moving in the +X direction relative to frame S, at
> >>> velocity v, and if g = 1/sqrt(1-v^2), and (0,0) -> (0',0'), and c=1,
> >>> then
> >>>
> >>> t' = g * (t - vx)
> >>> x' = g * (-vt + x)
> >>>
> >>> Inverse transform is
> >>>
> >>> t = g * (t' + vx')
> >>> x = g * (vt' + x')
> >>>
> >>> Compose them and see what you get.
> >
> > You made no comment. Did that mean you agree that these are inverses?
>
> It depends on the operator. We are discussing Roberts use of inverse.
> As you know, the set R and the operator ' + ' has an inverse ' - ' with the
> identity 0.
> We can transform x to x' with x' = x +1 and back again with x = x'-1, which
> is analogous to what you are doing.
> For the same set R and the operator '*', we can transform x to x' with x' =
> 2x and back again with x = x'/2.
> I consider
> x' = g * (t-vx) to have an inverse
> x = (t-x'/g) / v.
Title: "Solving a set of 2 equations with 2 unknowns - Revisited"
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve2.html
Remember this one?
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html
God job, pig.
Dirk Vdm
Paul B. Andersen
"Androcles" <du...@dummy.net> skrev i melding news:Hhcdd.84675$ay5....@fe1.news.blueyonder.co.uk...
>
> "sal" <pragm...@nospam.org> wrote in message
> news:1098188826.rvmapfHxAXI0iDrV6yh5Pg@teranews...
> > On Tue, 19 Oct 2004 05:31:17 +0000, Androcles wrote:
> >> I notice you didn't bother to provide the multiplicative or additive
> >> inverses
> >> to the LTs that Roberts says have to exist.
> >
> > You're kidding, right? Any SR text has these -- even Martin Gardner must
> > have shown them. Are you really asserting they don't exist?
>
> I concluded that the (ahem..., he whose name must not be mentioned, but he
> told me LOTS of times) transforms were the inverse of the Lorentz
[..]
> He who shall not be named said:
> "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)"
So Androcles doesn't even know what the inverse transform is. :-)
Paul
Paul B. Andersen
"Androcles" <du...@dummy.net> skrev i melding news:Xqddd.85340$ay5....@fe1.news.blueyonder.co.uk...
>
>
> I'm not trying to make a fool of Andersen, he can do that all by himself,
> and so can you.
You mean like this?
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/MakeFool.html
Paul
mike3
> "mike3" <mike...@yahoo.com> wrote in message news:1d54b7e4.04101...@posting.google.com...
> > "Androcles" <du...@dummy.net> wrote in message news:<9fAbd.61468$ay5....@fe1.news.blueyonder.co.uk>...
> > > Roberts writes:
> > >
> > > "This is why most (if not all) physicists today believe in Special
> > > Relativity - it is IMPOSSIBLE to construct an alternative description
> > > without violating one of the postulates or disregarding a very large
> > > body of experimental evidence. If you truly believe that Special
> > > Relativity simply must be false (for whatever reason), go back and
> > > review the four Postulates, and find a hole in them."
> > >
> > > This is why most (if not all) physicists today are failed mathematicians-
> > > it is IMPOSSIBLE for special relativity to be anything other than false.
> > > There is nothing wrong with the group postulates, but SR violates the
> > > second.
> > > " 2. Any transformation has an inverse, which is also a transformation."
> > > So what is the inverse transformation to
> > > x' = (x-vt)/sqrt(1-v^2/c^2) ?
> > > A little algebra and
> > > x' * sqrt(1-v^2/c^2) = x-vt
> > > x = x' * sqrt(1-v^2/c^2) + vt.
> > >
> > > That is not an equation you'll ever see in SR.
> > > However, the PoR must hold.
> > >
> >
> >
> > transformation has an inverse, therefore it does NOT violate the law
> > you brought up. Because it has one, it is VALID under that law. It
> > doesn't matter whether it appears in the theory or not. It is still
> > VALID!
>
> I just found out that the problem is *much* worse:
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Persuasive.html
> He can't even substitute a variable.
>
> And the equation you see isn't part of the inverted transformation
> at all. See also
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html
>
> Dirk Vdm
Ooooooh.... he failed to see that it was a SYSTEM of equations and
that this solutions were circular! Like saying that the solution to
(1/2x) + x = 3 is
x = 1/(6 - 2x)) !!!