Bile awski the Janitor.
Androcles
From: "JanPB" <fil...@gmail.com>
Newsgroups: sci.physics.relativity
Sent: Monday, September 12, 2005 12:04 AM
Subject: Re: why lorentz transformation?
: Androcles wrote:
: >
: > Hey phuckwit! Deal with the math or shut the fuck up.
:
: Math? Since you know almost nothing about it, I find it odd that you
: should lecture people (or more like, INSULT) on that very subject.
:
: --
: Jan Bielawski
How come you ran away, Bielawski? Three posts you have not answered,
and all you can do is insult. Phuckwit! Deal with the math or shut the
fuck up.
"JanPB" <fil...@gmail.com> wrote in message
news:1125371177.7...@o13g2000cwo.googlegroups.com...
| Henry Haapalainen wrote:
| >
| > "Francisco", I understand what you are trying to say, and you are
absolutely
| > right. But for many of these people relativity is a religion, and
you cannot
| > change their mind with any scientific argument.
|
| Where people get this stoopid idea I'll never know. Get real, please.
| Relativity is simply a theory derived from certain postulates.
Bollocks. It's based on a stoopid definition of time.
End of
| story, no religion.
Of course it's a religion, dumbfuck. You BELIEVE Einstein's
definition of time on FAITH, it's not science, it's religion.
|It's OK to object to those postulates on
| philosophical grounds, as I do for example. It is also true that this
| theory has an excellent experimental and predictive quantitative track
| record.
Irrelevant bullshit, the function tau is not linear, the objection is on
mathematical grounds.
|
| In other words: it's fine to object against relativity (or any other
| theory) *honestly*.
So done. You are dishonest in denying your FAITH is a religion.
| What is dishonest is calling relativity "religion"
| just because someone defends the way it's derived from the postulates.
|
Bullshit, it is derived from
[quote]
we establish by definition that the "time" required by light to travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
and your BLIND FAITH in that is a religion.
Androcles.
| --
| Jan Bielawski
|
"JanPB" <fil...@gmail.com> wrote in message
news:1125302090.6...@g44g2000cwa.googlegroups.com...
| RP wrote:
| >
| > You'll lose the argument,
|
| Oh, I know very well that Androcles won't understand a word I'm
saying.
Bullshit, I know you will attempt circularity.
| One always "loses" arguments on Usenet in that funny sense, I've done
| this before with many others as I'm always curious about limits of
| human capacity for withstanding plain contradictions and the means
| people use to protect themselves from certain inconvenient truths
| regarding themselves.
LOL! You are talking about yourself, phuckwit.
| In this case the guy would rather assume that 100 years and
generations
| of physicists "missed" an idiotic high-school-level "error" than admit
| to himself that he has no talent for that sort of thing and doesn't
| understand squat.
You are so ridiculous its incredible. I'm always bored by the limits of
human capacity for withstanding plain contradictions and the means
people use to protect themselves from certain inconvenient truths
regarding themselves.
| He would invent ridiculous Hollywoodesque plots
| involving Einstein and his supposed "intent" to "deceive" scientists -
| anything, *anything*, except the plain truth that Androcles has no
case
| and is wasting his life away on chimeras.
I'm always bored by the limits of
human capacity for withstanding plain contradictions and the means
people use to protect themselves from certain inconvenient truths
regarding themselves.
|
| > BTW, your first mistake that Andro
| > will tear to shreds is the fact that you can't derive something from
| > assumptions without arriving at yet further assumptions. That's just
| > simple logic. He'll be right on that point. Better try another
approach.
See, RP knows.
You on the other hand are too stupid to know that, which is why
you can't win. I'm always bored by the limits of
human capacity for withstanding plain contradictions and the means
people use to protect themselves from certain inconvenient truths
regarding themselves.
| The assumptions are all standard physics
What is "standard" physics? Vector addition of velocities?
| + Einstein's postulates.
He's only got one. Having trouble counting to one, are you?
The PoR is standard physics and Galileo's. Einstein's ONE
postulate is "light is always propagated in empty space with
a definite velocity c which is independent of the state of motion
of the emitting body. " No numerical value permitted in the proof
though. You can use a numerical value for explanation if need be.
I'm generously permitting you to use his definition of time,
although I do not accept it.
So
| unless Androcles insists on something exotic like the universe being
| non-isotropic or observers depending on their entire motion histories
| and the like, then he has no case
Quit bragging and start proving.
| since it's all just a mathematical
| derivation at that point. It would be interesting to find out what
sort
| of random assumption he brings up this time.
It'll be boring listening to you bragging and bringing up such total
irrelevancies as your argument, such as "would rather *assume*
that 100 years and generations of physicists "missed" an idiotic
high-school-level "error" than admit to himself that he has no talent
for that sort of thing and doesn't understand squat.
I dont assume, phuckwit, but you do, and that's why you have no
talent for this sort of thing and don't understand squat.
| I've seen one of them
| already - the idea that SR "ignores" the time it takes for light to
| reflect off a mirror!
| --
| Jan Bielawski
Are you claiming the mirror doesn't move, or that a photon
has no momentum? Which?
That's why you have no talent for this sort of thing and don't
understand squat.
Androcles.
"JanPB" <fil...@gmail.com> wrote in message
news:1125299145....@g14g2000cwa.googlegroups.com...
| Androcles wrote:
| > "JanPB" <fil...@gmail.com> wrote in message
| > news:1125257544.9...@g47g2000cwa.googlegroups.com...
| >
| > | Specifically,
| > | you let x' tend to zero on one side of the equation but not on the
| > | other. When the result is 0=1 you complain loudly.
| >
| > The fuck I do, but since you claimed to be able to derive
| > the cuckoo transformations without calculus, go ahead and do
| > it.
|
| Sure, no problem but I need to get some sleep first.
|
| > You are to use Einstein's light postulate, Galileo's relativity
| > exemplified by Einstein, "the reciprocal electrodynamic action
| > of a magnet and a conductor. The observable phenomenon here
| > depends only on the relative motion of the conductor and the
magnet",
| > Einstein's definition of time "we establish by definition that the
| > "time"
| > required by light to travel from A to B equals the "time" it
requires
| > to travel from B to A", and NO assumptions.
| > You will not succeed, I shall rip your "proof" to shreds.
|
| You'll not be able to even touch it, let alone rip to shreads. I'll
| derive the Lorentz transform from the assumption of linearity of the
| transform, from the "tau" equation, and the usual postulates.
Circularity is not permitted.
You are to use Einstein's light postulate, Galileo's relativity,
Einstein's definition of time.
Androcles.
| --
| Jan Bielawski
|
Daryl McCullough
Androcles, you are an idiot.
--
Daryl McCullough
Ithaca, NY
JanPB
>
> How come you ran away, Bielawski? Three posts you have not answered,
> and all you can do is insult. Phuckwit! Deal with the math or shut the
> fuck up.
I was busy preparing video editing work I had promised someone. This
bboard has always been a low-priority entertainment for me, sorry to
disappoint.
Another reason I didn't post the Lorentz transform derivation according
to Einstein is that you adopted a defensive attitude which is logically
untenable and leading nowhere. It goes something like this:
1. Let's say I post a detailed examination of Einstein's derivation of
the Lorentz transform from the tau-equation.
2. Your response is going to be: "I'm not interested in YOUR
derivation, I'm interested in EINSTEIN'S derivation!" In other words,
you are going to dismiss any argument explaining Einstein's work merely
on the basis of using different *English words*. For some reason the
difference between a THEORY and a mere DESCRIPTION of such escapes you.
Or perhaps it's an intentional debating tactics on your part.
What I find interesting, BTW, is that Einstein's derivation of the
Lorentz transform has several gaps in it (as well as one instance of
certain notational sloppiness) which were NEVER addressed by any of the
resident armchair relativity critics here. My guess is they never read
Einstein's paper with any sort of understanding and promptly get stuck
on some silly irrelevancy they trumpet on this bboard to high heaven.
If I post a detailed explanation of Einstein's derivation I shall
highlight those gaps (they are not important gaps, of course, as
Einstein was writing his paper for professionals, but they are worth
pointing out to students).
--
Jan Bielawski
Androcles
| Androcles wrote:
| >
| > How come you ran away, Bielawski? Three posts you have not answered,
| > and all you can do is insult. Phuckwit! Deal with the math or shut
the
| > fuck up.
|
| I was busy preparing video editing work I had promised someone. This
| bboard has always been a low-priority entertainment for me, sorry to
| disappoint.
|
| Another reason I didn't post the Lorentz transform derivation
according
| to Einstein is that you adopted a defensive attitude which is
logically
| untenable and leading nowhere. It goes something like this:
|
| 1. Let's say I post a detailed examination of Einstein's derivation of
| the Lorentz transform from the tau-equation.
|
| 2. Your response is going to be: "I'm not interested in YOUR
| derivation, I'm interested in EINSTEIN'S derivation!"
That is correct. I am not interested in your personal derivation, or
any other contributor to this newsgroup. Einstein's dervivation is
flawed
and hence yours must be as well.
In other words,
| you are going to dismiss any argument explaining Einstein's work
merely
| on the basis of using different *English words*.
Bullshit, if you actually knew any mathematics instead of mere symbol
manipulation you would know what a definition was and you'd know
what a prooof was. You don't though.
For some reason the
| difference between a THEORY and a mere DESCRIPTION of such escapes
you.
Bullshit. You don't know mathematics.
| Or perhaps it's an intentional debating tactics on your part.
Bullshit.
| What I find interesting, BTW, is that Einstein's derivation of the
| Lorentz transform has several gaps in it (as well as one instance of
| certain notational sloppiness) which were NEVER addressed by any of
the
| resident armchair relativity critics here.
Bullshit. Time is not a vector.
| My guess is they never read
| Einstein's paper with any sort of understanding and promptly get stuck
| on some silly irrelevancy they trumpet on this bboard to high heaven.
YOU have never read Einstein's paper with any understanding, and
central to that derivation is
稼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 half coming from the defintion (in words)
[quote]
we establish by definition that the "time" required by light to travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
| If I post a detailed explanation of Einstein's derivation I shall
| highlight those gaps (they are not important gaps, of course, as
| Einstein was writing his paper for professionals, but they are worth
| pointing out to students).
Go ahead, start proving.
Androcles
JanPB
So let me start by stating clearly my understanding of what it is I'm
supposed to do.
-------------------------------------------
I'll prove by *merely going over Einstein's own argument in more
detail* that the "tau-equation" together with the usual assumptions
(Einstein's two postulates, the clock synchronisation definition, etc.)
plus the assumption of transform linearity IMPLIES that the transform
is of the Lorentz type:
tau = phi(v)beta(v)(t - xv/c^2)
xi = phi(v)beta(v)(x - vt)
eta = phi(v)y
zeta = phi(v)z
...where phi(v) is a velocity-dependent constant still to be
determined, and:
beta(v) = 1/sqrt(1 - v^2/c^2)
-------------------------------------------
Am I correct in assuming that this is your objection?
--
Jan Bielawski
Androcles
No, you are not correct.
My objection is to the half in
稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
which comes from
[quote]
we establish by definition that the "time" required by a turtle to
travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
Einstein can prove nothing can go faster than a turtle.
Oops!... Did I say 'a turtle'? Sorry...'light'.
Androcles can prove your brain cannot go faster than a turtle's.
JanPB
>
> which comes from
>
> [quote]
> we establish by definition that the "time" required by a turtle to
> travel
> from A to B equals the "time" it requires to travel from B to A.
> [end quote]
Well, it does come from it. So what is it you claim is incorrect then?
(You did claim before that the Lorentz transform could not be derived
from the tau-equation, that's why I assumed that's what your objection
was.)
--
Jan Bielawski
Androcles
>
> which comes from
>
> [quote]
> we establish by definition that the "time" required by a turtle to
> travel
> from A to B equals the "time" it requires to travel from B to A.
> [end quote]
Well, it does come from it. So what is it you claim is incorrect then?
(You did claim before that the Lorentz transform could not be derived
from the tau-equation, that's why I assumed that's what your objection
was.)
Above is the tau equation I refer to.
Consider:
Light (or a turtle, makes no difference) is emitted from the
caboose of a train, reflects at the engine and returns to the caboose.
Einstein's equation says it will travel the length of the train and back
again, and will reflect at the engine at tau(x',0,0,t+x'/(c-v))
But now we do it this way.
Light is emitted from the engine, reflects at the caboose and
returns to the engine.
The time of reflection at the caboose is now tau(0,0,0,t+x'/(c PLUS v))
Diagramatically,
(fixed font needed).
|
|
| C'
| /
| B /
| ____________Mirror
| /\ /
| / \ /
C / \ /
|\ / \ /
| \ / \A'
| \ / | /
| \ / /
| \ / /
| \ / | /
| \ / /
| \/ |
| /\ /
| / \ / |
| / \ /
| / \ / |
| / ____\/__________Mirror
| /
| / D |
|/
/ ____________|____|________________
A D B A' C'
[quote]
we establish by definition that the "time" required by light to travel
from A to B equals the "time" it requires to travel from B to [A'].
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
We establish by definition that the "time" required by light to travel
from C to D equals the "time" it requires to travel from D to C'.
Distance between mirrors is x'
Einstein's equation:
稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
What it means in the diagram:
稼tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
稼tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
So the time at B, the engine, equals the time at D, the caboose,
but it doesn't. Ergo Einstein was a phuckwit.
Androcles
JanPB
>
> What it means in the diagram:
>
> ½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
Correct.
> So the time at B, the engine, equals the time at D, the caboose,
According to the clock sync of the moving system.
> but it doesn't.
According to the clock sync of the stationary system. And your point
is?
> Ergo Einstein was a phuckwit.
Stop repeating childish crap. If you have trouble understanding
something - ask. Your belittling Einstein is just so dumb. And your
belief in a high-school-level mistake being undetected already by any
random 1905 physics grad student (let alone Einstein, let alone any old
Annalen der Physik reviewer) is just too dumb for words.
--
Jan Bielawski
Androcles
>
> What it means in the diagram:
>
> 稼tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
Correct.
> So the time at B, the engine, equals the time at D, the caboose,
According to the clock sync of the moving system.
There is no "clock sync" of the moving system, fool!
Time in the moving system (tau) can *supposedly* be different to time
in the stationary system (t), but it can't be different to itself!
You've got tau, you've got t. That's it.
> but it doesn't.
According to the clock sync of the stationary system.
There is no "clock sync" of the stationary system, you clown!
Time in the stationary system (t) can *supposedly* be different to time
in the moving system (tau), but it can't be different to itself!
You've got tau, you've got t. That's it.
> And your point is?
And my point is you are an utterly useless incompetent phuckwit, a
clown, an idiot,
a moron, a fucking joke, just like your idiot tin god Einstein!
> Ergo Einstein was a phuckwit.
Stop repeating childish crap. If you have trouble understanding
something - ask.
You pompous egotistical fucking cretin! "clock sync" tau with tau!
Androcles.
Bilge
>Androcles, you are an idiot.
That is an insult to idiots.
JanPB
> > tau(x',0,0,t+x'/(c-v))
> >
> > What it means in the diagram:
> >
> > ½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
>
> Correct.
>
> > So the time at B, the engine, equals the time at D, the caboose,
>
> According to the clock sync of the moving system.
>
> There is no "clock sync" of the moving system, fool!
> Time in the moving system (tau) can *supposedly* be different to time
> in the stationary system (t), but it can't be different to itself!
You are hopelessly confused. Your equations:
> > ½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
> > ½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
simply follow from the clock synchronisation in the moving system.
There is nothing contradictory about them although you wrote them a bit
sketchily. Written out explicitly they are:
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))
1/2(tau(x',0,0,t) + tau(x',0,0,t+x'/(c-v)+x'/(c+v))) =
tau(0,0,0,t+x'/(c+v))
Nothing wrong with them.
> You've got tau, you've got t. That's it.
>
> > but it doesn't.
>
> According to the clock sync of the stationary system.
>
> There is no "clock sync" of the stationary system, you clown!
> Time in the stationary system (t) can *supposedly* be different to time
> in the moving system (tau), but it can't be different to itself!
Who says it's different than itself? You begin with two systems:
stationary (K) using t for time and moving (k) using tau for time. Both
agree to synchronise their clocks according to the Einstein convention.
We simply seek the formula relating tau (and xi, eta, zeta) to x,y,z,t.
--
Jan Bielawski
Androcles
> > tau(x',0,0,t+x'/(c-v))
> >
> > What it means in the diagram:
> >
> > 稼tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
>
> Correct.
>
> > So the time at B, the engine, equals the time at D, the caboose,
>
> According to the clock sync of the moving system.
>
> There is no "clock sync" of the moving system, fool!
> Time in the moving system (tau) can *supposedly* be different to time
> in the stationary system (t), but it can't be different to itself!
| You are hopelessly confused.
You are infinitely stupid.
| Your equations:
> > 稼tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
> > 稼tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
| simply follow from the clock synchronisation in the moving system.
| There is nothing contradictory about them although you wrote them a
bit
| sketchily. Written out explicitly they are:
| 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))
| 1/2(tau(x',0,0,t) + tau(x',0,0,t+x'/(c-v)+x'/(c+v))) =
| tau(0,0,0,t+x'/(c+v))
| Nothing wrong with them.
Put some numbers in, then.
c = 5 cars a second (32 car train, each car is 60 metres).
v = 3 = 0.6c
x' = 32 cars, the train length won't change by tomorrow, x' is
independent of time.
t = midnight = 0
1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)
I gave you the diagram.
Still nothing wrong, or are you still infinitely stupid?
> You've got tau, you've got t. That's it.
>
> > but it doesn't.
>
> According to the clock sync of the stationary system.
>
> There is no "clock sync" of the stationary system, you clown!
> Time in the stationary system (t) can *supposedly* be different to
> time
> in the moving system (tau), but it can't be different to itself!
|Who says it's different than itself?
This does:
tau(32,0,0,16) the time at the engine is 16 microseconds past midnight.
tau(0,0,0,4) the time at the caboose is 4 microseconds past midnight.
Run the train in reverse at midnight tomorrow night.
tau(0,0,0,16) the time at the caboose is 16 microseconds past midnight.
tau(32,0,0,4) the time at the engine is 4 microseconds past midnight.
You've said my equations are correct.
It follows from these results that you are infintiely stupid.
Want to put the cuckoo transformation in?
tau = (t-vx/c^2)/sqrt(1-v^2/c^2)
tau(32,0,0,16) = ?
tau(0,0,0,4) = ?
Go ahead, stupid, make it work and prove tau is a linear function,
because I don't know how. Show me.
| You begin with two systems:
| stationary (K) using t for time and moving (k) using tau for time.
Both
| agree to synchronise their clocks according to the Einstein
convention.
| We simply seek the formula relating tau (and xi, eta, zeta) to
x,y,z,t.
--
Jan Bielawski
You forget the third system, x',y,z,t with x', an end point in the
moving system independent of time which is not xi and not 0.
It is not zero because that would make
1/2(tau(0,0,0,t) + tau(0,0,0,t+0/(c-v)+0/(c+v))) =
1/2(tau(0,0,0,t) + tau(0,0,0,t)) = tau(0,0,0,t) which is trivially true,
half of nothing is nothing.
It is not xi because
x'/xi = sqrt(1-v^2/c^2) not equal 1 (unless v = 0, again a triviality)
Yet if we accelerate the separation between xi and x' increases.
This is the cuckoo length "contraction", and
for x' = 0, xi = 0 at any velocity,
for v = c, xi = x'/sqrt(1-1) = undefined.
For v = 0.866c, xi = 2x' and your stupidity is doubled.
Being infinitely stupid this will not trouble you, because
to you it's interesting - what are the limits of withstanding blatant
contradictions (infinite, OBVIOUSLY).
Androcles.
JanPB
> | Your equations:
>
> > > ½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
> > > ½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
>
> | simply follow from the clock synchronisation in the moving system.
> | There is nothing contradictory about them although you wrote them a
> bit
> | sketchily. Written out explicitly they are:
>
> | 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))
>
> | 1/2(tau(x',0,0,t) + tau(x',0,0,t+x'/(c-v)+x'/(c+v))) =
> | tau(0,0,0,t+x'/(c+v))
>
> | Nothing wrong with them.
>
> Put some numbers in, then.
Typical :-)
> c = 5 cars a second (32 car train, each car is 60 metres).
> v = 3 = 0.6c
> x' = 32 cars, the train length won't change by tomorrow, x' is
> independent of time.
> t = midnight = 0
>
> 1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
> 1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)
Correct.
> I gave you the diagram.
> Still nothing wrong, or are you still infinitely stupid?
Still nothing wrong. It would be more efficient if you pointed out
precisely where you see a contradiction.
> > You've got tau, you've got t. That's it.
> >
> > > but it doesn't.
> >
> > According to the clock sync of the stationary system.
> >
> > There is no "clock sync" of the stationary system, you clown!
> > Time in the stationary system (t) can *supposedly* be different to
> > time
> > in the moving system (tau), but it can't be different to itself!
>
> |Who says it's different than itself?
>
> This does:
> tau(32,0,0,16) the time at the engine is 16 microseconds past midnight.
According to the stationary clock at that spot.
> tau(0,0,0,4) the time at the caboose is 4 microseconds past midnight.
According to the stationary clock at that spot.
> Run the train in reverse at midnight tomorrow night.
> tau(0,0,0,16) the time at the caboose is 16 microseconds past midnight.
According to the stationary clock at that spot.
> tau(32,0,0,4) the time at the engine is 4 microseconds past midnight.
According to the stationary clock at that spot.
Still no contradiction.
> You've said my equations are correct.
> It follows from these results that you are infintiely stupid.
They are correct.
> tau = (t-vx/c^2)/sqrt(1-v^2/c^2)
> tau(32,0,0,16) = ?
> tau(0,0,0,4) = ?
>
> Go ahead, stupid, make it work and prove tau is a linear function,
> because I don't know how. Show me.
The tau in the first line is a different function than the tau in the
two lines that follow. Remember that the first tau is a function of
(x,y,z,t), the two other tau are functions of (x',y,z,t), where
x'=x-vt.
So let's write the first tau in terms of (x',y,z,t) which is what the
numbers were given in (I'm using gamma for 1/sqrt(1-v^2/c^2)):
tau = gamma(t-vx/c^2) = gamma(t - v(x'+vt)/c^2)
Now we can plug in your numbers. In your case gamma = 1.25, v=3, c=5.
The two other tau are:
tau(32,0,0,16) = 1.25 * (16 - 3(32+3*16)/25) = 8
tau(0,0,0,4) = 1.25 * (4 - 3(0+3*4)/25) = 3.2
(assuming I didn't make a numerical mistake somewhere).
This function tau can be rewritten as:
tau = Ax' + Bt
where:
A = -v/(c*sqrt(c^2-v^2))
B = sqrt(c^2-v^2)/c
...are constant (because v = const.), i.e. tau is a linear function of
(x',y,z,t).
> | You begin with two systems:
> | stationary (K) using t for time and moving (k) using tau for time.
> Both
> | agree to synchronise their clocks according to the Einstein
> convention.
> | We simply seek the formula relating tau (and xi, eta, zeta) to
> x,y,z,t.
>
> You forget the third system, x',y,z,t with x', an end point in the
> moving system independent of time which is not xi and not 0.
No, I remembered it but wanted to start with the clear setup of what it
is we are after: a transformation between K and k.
--
Jan Bielawski
Androcles
Androcles wrote:
> "JanPB" <fil...@gmail.com> wrote in message
> news:1127198662.0...@o13g2000cwo.googlegroups.com...
> | Your equations:
>
> > > 稼tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
> > > 稼tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
>
> | simply follow from the clock synchronisation in the moving system.
> | There is nothing contradictory about them although you wrote them a
> bit
> | sketchily. Written out explicitly they are:
>
> | 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))
>
> | 1/2(tau(x',0,0,t) + tau(x',0,0,t+x'/(c-v)+x'/(c+v))) =
> | tau(0,0,0,t+x'/(c+v))
>
> | Nothing wrong with them.
>
> Put some numbers in, then.
| Typical :-)
Androcles: Relativity is fucked up.
Bile awski: Can't be, someone would have seen it before.
Androcles: It's the 1/2.
Bile awski: Go read a book and learn.
Androcles: Here's a simple diagram showing why.
Bile awski: Show the math
Androcles: Here's the equations.
Bile awski: Nothing wrong with them.
Androcles: Put some numbers in, then.
Bile awski: Typical + stupid ape grin.
Androcles: You are a cunt, your brain is fucked. Fuck off.
> c = 5 cars a second (32 car train, each car is 60 metres).
> v = 3 = 0.6c
> x' = 32 cars, the train length won't change by tomorrow, x' is
> independent of time.
> t = midnight = 0
>
> 1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
> 1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)
| Correct.
> I gave you the diagram.
> Still nothing wrong, or are you still infinitely stupid?
| Still nothing wrong. It would be more efficient if you pointed out|
| precisely where you see a contradiction.
One step at a time, I'm dealing with a fucked up infinitely stupid ape.
> > You've got tau, you've got t. That's it.
> >
> > > but it doesn't.
> >
> > According to the clock sync of the stationary system.
> >
> > There is no "clock sync" of the stationary system, you clown!
> > Time in the stationary system (t) can *supposedly* be different to
> > time
> > in the moving system (tau), but it can't be different to itself!
>
> |Who says it's different than itself?
>
> This does:
> tau(32,0,0,16) the time at the engine is 16 microseconds past
> midnight.
| According to the stationary clock at that spot.
See what I mean? Infinitely stupid.
I have no idea what stationary clock you are talking about
or what spot you are talking about.
But, one step at a time. Let's deal with the stationary spot first.
That's 80, the engine moved from spot 32 at speed 3 for 16 time.
32+ 3*16 = 80.
Does 80 appear in either equation?
No.
Did anyone put a stationary clock at 80?
No.
Is there a "stationary" clock on the engine?
No; there is a clock that moves with the engine.
What are the limits of Bile awski's stupidity? Infinite.
Fuck off, cunt.
Androcles.
Dirk Van de moortel
"Androcles" <Androcles@ MyPlace.org> wrote in message news:yHbYe.4795$MD4....@fe1.news.blueyonder.co.uk...
That is probably what all apes think when they are dealing
with humans.
>
>
> > > You've got tau, you've got t. That's it.
> > >
> > > > but it doesn't.
> > >
> > > According to the clock sync of the stationary system.
> > >
> > > There is no "clock sync" of the stationary system, you clown!
> > > Time in the stationary system (t) can *supposedly* be different to
> > > time
> > > in the moving system (tau), but it can't be different to itself!
> >
> > |Who says it's different than itself?
> >
> > This does:
> > tau(32,0,0,16) the time at the engine is 16 microseconds past
> > midnight.
>
> | According to the stationary clock at that spot.
>
> See what I mean? Infinitely stupid.
> I have no idea what stationary clock you are talking about
> or what spot you are talking about.
the stationary clock that says t = 16 when the engine passes.
Gee, how difficult.
> But, one step at a time. Let's deal with the stationary spot first.
> That's 80, the engine moved from spot 32 at speed 3 for 16 time.
> 32+ 3*16 = 80.
>
> Does 80 appear in either equation?
> No.
> Did anyone put a stationary clock at 80?
> No.
> Is there a "stationary" clock on the engine?
> No; there is a clock that moves with the engine.
And there is a stationary clock at each point.
These clocks show t-time of events.
>
> What are the limits of Bile awski's stupidity? Infinite.
> Fuck off, cunt.
> Androcles.
The main problem is that you have no idea whatsoever
about the meaning of the variables that appear in the
equations.
Title: I have no idea what you are talking about
so YOU are infinitely stupid:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/InfinitelyStupid.html
Jan, you *do* know with what kind of sub-ape you
are dealing here, I hope?
Dirk Vdm
JanPB
> "Androcles" <Androcles@ MyPlace.org> wrote in message news:yHbYe.4795$MD4....@fe1.news.blueyonder.co.uk...
> >
> > Fuck off, cunt.
> > Androcles.
>
> The main problem is that you have no idea whatsoever
> about the meaning of the variables that appear in the
> equations.
I wonder why he does this. SR I mean.
--
Jan Bielawski
JanPB
> Androcles wrote:
[...]
> > c = 5 cars a second (32 car train, each car is 60 metres).
> > v = 3 = 0.6c
> > x' = 32 cars, the train length won't change by tomorrow, x' is
> > independent of time.
> > t = midnight = 0
> >
> > 1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
> > 1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)
>
> | Correct.
> >
> > |Who says it's different than itself?
> >
> > This does:
> > tau(32,0,0,16) the time at the engine is 16 microseconds past
> > midnight.
>
> | According to the stationary clock at that spot.
>
> See what I mean? Infinitely stupid.
> I have no idea what stationary clock you are talking about
> or what spot you are talking about.
Perhaps that's your problem. What you don't understand is not
automatically stupid, you know. I meant the stationary clock the engine
is passing when that clock shows 16 microseconds. At this instant the
clock *on* the engine shows tau(32,0,0,16) microseconds.
> But, one step at a time. Let's deal with the stationary spot first.
> That's 80, the engine moved from spot 32 at speed 3 for 16 time.
You now switched coordinates - now you are talking about x, not x'. The
coordinate x' is always 32 at the engine. Changing from (x',y,z,t) to
(x,y,z,t) changes the function tau.
> 32+ 3*16 = 80.
>
> Does 80 appear in either equation?
> No.
Of course not, the equation is for tau(x'y'z't), not for tau(x,y,z,t).
> Did anyone put a stationary clock at 80?
> No.
Yes. Clock distribution is the basis of Einstein's paper.
> Is there a "stationary" clock on the engine?
> No;
Not *on* the engine. At every instant the engine passes a stationary
clock.
> there is a clock that moves with the engine.
Yes.
--
Jan Bielawski
Androcles
Wonderment, or curiousity, is a natural consequence of all.
Never finding the answer is an inevitable consequence of the infinitely
stupid.
Fuck off, cunt.
Androcles
Androcles
| Androcles wrote:
| > "JanPB" <fil...@gmail.com> wrote in message
| > news:1127297583....@g47g2000cwa.googlegroups.com...
| > Androcles wrote:
| [...]
| > > c = 5 cars a second (32 car train, each car is 60 metres).
| > > v = 3 = 0.6c
| > > x' = 32 cars, the train length won't change by tomorrow, x' is
| > > independent of time.
| > > t = midnight = 0
| > >
| > > 1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
| > > 1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)
| >
| > | Correct.
| [...]
| > >
| > > |Who says it's different than itself?
| > >
| > > This does:
| > > tau(32,0,0,16) the time at the engine is 16 microseconds past
| > > midnight.
| >
| > | According to the stationary clock at that spot.
| >
| > See what I mean? Infinitely stupid.
| > I have no idea what stationary clock you are talking about
| > or what spot you are talking about.
|
| Perhaps that's your problem.
Nope, your problem.
What you don't understand is not
| automatically stupid, you know.
You are automatically stupid.
I meant the stationary clock the engine
| is passing when that clock shows 16 microseconds.
Yes, you dumb fuck.
It is at (80,0,0,16) which is not part of the equation.
At this instant the
| clock *on* the engine shows tau(32,0,0,16) microseconds.
So who gives a shit about a stationary clock or what spot
you are talking about, since it isn't part of the equation?
(rhetorical question, you are too stupid to answer sensibly.)
|
| > But, one step at a time. Let's deal with the stationary spot first.
| > That's 80, the engine moved from spot 32 at speed 3 for 16 time.
|
| You now switched coordinates - now you are talking about x, not x'.
You did that by prattling on about a "stationary clock at that spot".
That "spot" is x, not x'.
| The
| coordinate x' is always 32 at the engine.
Yes, dumbfuck.
Changing from (x',y,z,t) to
| (x,y,z,t) changes the function tau.
Why are you trying to do it then, dumbfuck?
| > 32+ 3*16 = 80.
| >
| > Does 80 appear in either equation?
| > No.
|
| Of course not, the equation is for tau(x'y'z't), not for tau(x,y,z,t).
Correct.
| > Did anyone put a stationary clock at 80?
| > No.
|
| Yes. Clock distribution is the basis of Einstein's paper.
Infinitely stupid dumbfuck, there are no clocks in the equation, let
alone one at 80.
| > Is there a "stationary" clock on the engine?
| > No;
|
| Not *on* the engine.
Right. There is a "moving" clock on the engine that is at rest with the
engine.
We call it the "driver's wristwatch", infinitely stupid dumbfuck.
At every instant the engine passes a stationary
| clock.
Irrelevant, all the stationary clocks are not part of the equation.
| > there is a clock that moves with the engine.
|
| Yes.
Ok, so what are you prattling on about a "stationary clock at that spot"
for?
Answer: Because you are infinitely stupid.
Androcles.
Dirk Van de moortel
"Androcles" <Androcles@ MyPlace.org> wrote in message news:WqRYe.278$Am6...@fe3.news.blueyonder.co.uk...
[snip]
> Answer: Because you are infinitely stupid.
Infinitely Stupid Revisited:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/InfinitelyStupid2.html
Dirk Vdm
JanPB
[...]
> I meant the stationary clock the engine
> | is passing when that clock shows 16 microseconds.
>
> Yes, you dumb fuck.
> It is at (80,0,0,16) which is not part of the equation.
Aha, I think I see what your problem is. You keep insisting on using an
extra (*third*) set of moving clocks. This (redundant) system of clocks
moves together with the moving clocks but is not Einstein-synchronised.
Instead, at each moment such clock displays the same time as the
stationary clock that just happens to momentarily coincide with it:
x'=x-vt
y=y
z=z
t=t
If you prefer I can use this system although it's unecessary and
Einstein doesn't use it. If we both agree to use this extra system then
some of our differences are merely terminological.
Be it as it may, you still haven't produced the promised contradiction
in the tau equation and the derivation of the Lorentz transform from
it.
> At this instant the
> | clock *on* the engine shows tau(32,0,0,16) microseconds.
>
> So who gives a shit about a stationary clock or what spot
> you are talking about, since it isn't part of the equation?
But it *is* the part of the equation - except this fact is obscured in
your view by using the third system of clocks. But the connection
between the stationary and moving clocks is the heart of the matter:
tau(32,0,0,16) is the time read by the moving clock on the engine at
the instant it passes the *stationary* clock which displays 16 at that
very moment. You express the same idea by introducing instead the third
system of moving clocks which are synced up to agree with the
stationary clocks.
Let's call this third system of clocks "x'-clocks". Then I could say
the same thing like so: "tau(32,0,0,16) is the time read by the moving
clock on the engine at the instant the moving x'-clock at the same spot
displays t=16".
Still no error in sight.
> (rhetorical question, you are too stupid to answer sensibly.)
All your questions are very easy. The most complex portion by far of
preparing my postings is the proofreading.
> | > But, one step at a time. Let's deal with the stationary spot first.
> | > That's 80, the engine moved from spot 32 at speed 3 for 16 time.
> |
> | You now switched coordinates - now you are talking about x, not x'.
>
> You did that by prattling on about a "stationary clock at that spot".
> That "spot" is x, not x'.
That's because I use only two systems of clocks: the stationary t and
the moving tau. The x'-clocks you use on top of those are not incorrect
but they are unecessary and they complicate the trminology. But I'm
willing to use this system, no problem.
> | > Does 80 appear in either equation?
> | > No.
> |
> | Of course not, the equation is for tau(x'y'z't), not for tau(x,y,z,t).
>
> Correct.
Well, so we agree.
> | > Did anyone put a stationary clock at 80?
> | > No.
> |
> | Yes. Clock distribution is the basis of Einstein's paper.
>
> Infinitely stupid dumbfuck, there are no clocks in the equation, let
> alone one at 80.
Clocks are everywhere. That's how the time coordinate is assigned to
events.
> | > Is there a "stationary" clock on the engine?
> | > No;
> |
> | Not *on* the engine.
>
> Right. There is a "moving" clock on the engine that is at rest with the
> engine.
> We call it the "driver's wristwatch", infinitely stupid dumbfuck.
>
>
> At every instant the engine passes a stationary
> | clock.
>
> Irrelevant, all the stationary clocks are not part of the equation.
On the contrary, it's 100% relevant because the tau equation is an
expression of a constraint *between* the stationary clocks (you
introduce the third redundant moving x'-clock system for the same
thing) and the standard moving tau-clocks. (By "standard" I mean
Einstein-synchronised.)
> | > there is a clock that moves with the engine.
> |
> | Yes.
>
> Ok, so what are you prattling on about a "stationary clock at that spot"
> for?
> Answer: Because you are infinitely stupid.
Back to square one then. We agreed on the terminology and I can use the
x'-clocks (although they are an unecessary burden) but you still
haven't produced the promised error in the tau equation and in the
derivation of the Lorentz transform from it.
--
Jan Bielawski
Androcles
| Androcles wrote:
| > "JanPB" <fil...@gmail.com> wrote in message
| > news:1127461664.3...@g14g2000cwa.googlegroups.com...
| [...]
| > I meant the stationary clock the engine
| > | is passing when that clock shows 16 microseconds.
| >
| > Yes, you dumb fuck.
| > It is at (80,0,0,16) which is not part of the equation.
|
| Aha, I think I see what your problem is.
I don't have a problem, I know what's wrong with SR.
You are not smart enough too see that.
| You keep insisting on using an
| extra (*third*) set of moving clocks.
Aha, I think I see what your problem is. You refuse to
use the real moving clocks and insist on using some
fake tau clocks.
| This (redundant) system of clocks
| moves together with the moving clocks but is not
Einstein-synchronised.
The redundant fake tau clocks are not Newton synchronized.
| Instead, at each moment such clock displays the same time as the
| stationary clock that just happens to momentarily coincide with it:
Yes, isn't it a wonderful concept?
This makes them INDEPENDENT of distance divided by time,
whereas the fake redundant tau clocks were velocity dependent.
This independence concept is really a good idea, don't you think?
Then we don't have to confuse apples with oranges or distance with
time anymore.
|
| x'=x-vt
| y=y
| z=z
| t=t
There you go, you've successfully transformed a stationary
system of coordinates K to a moving system of coordinates k.
Well done!
Now, why do you want to transform a moving system of coordinates
k to a fake redundant moving system of coordinates kappa
by multiplying by some fake redundant gamma?
Was it to deliberately confuse yourself?
|
| If you prefer I can use this system although it's unecessary and
| Einstein doesn't use it.
I'd prefer you used it. IN FACT, Einstein depends on it but tries
to hide it, hoping to sweep it under the carpet. It's the magician's
trick
of misdirection, the legs of the other girl on the box he saws in half.
Did you really think Einstein can saw a woman in half? He's
much better than David Copperfield, isn't he? Lots of smoke, mirrors
out in the open where you can see them... brilliant! Pity it's not
physics.
Then again, we don't really want the girl in the box injured, do we?
We just want to leave the theatre, having paid our MONEY, amazed
at Einstein's magic, right?
| If we both agree to use this extra system then
| some of our differences are merely terminological.
I agree to use it. Shall we give it a name? Part of the psychology
of hiding things is not to name them. Once you give it a name people
will remember it. Einstein doesn't want you to remember it. That's why
x' doesn't appear in the cuckoo transformations, and the other girl
in the box who's feet you see that is sawn in half never appears on
the stage. It would ruin the illusion if she did.
Let's call it k', shall we? That's an easy name to remember.
Now we have a transformation from stationary K to moving k',
x' = x-vt
y' = y
z' = z
t' = t
and a magical transformation from the moving k' to the moving kappa,
xi = x'/sqrt(1-v^2/c^2)
eta = y'
zeta = z'
tau = t' * sqrt(1 - v^2/c^2)
Well.... almost. Better to divide instead of multiply, it's more
confusing.
Let's see now...
tau = t' * sqrt(1 - v^2/c^2)
= t' * (1-v^2/c^2) / sqrt(1 - v^2/c^2)
= (t' - t'v^2/c^2) / sqrt(1 - v^2/c^2)
Ooops... two terms with t' in them.
Well.... v = x/t, so we can get rid of one
= (t' - t'* (x/t) * v /c^2) / sqrt(1 - v^2/c^2)
And t' = t, we want to hide that...
= (t - xv /c^2) / sqrt(1 - v^2/c^2)
Hide the x', so we write x-vt instead.
Voila!
Behold the cuckoo transformations, we've sawn the girl in half.
xi = (x-vt) /sqrt(1-v^2/c^2)
tau = (t-vx/c^2) /sqrt(1-v^2/c^2)
Nobody will notice how that was done!
I do admire the guy, he was the greatest huckster in history.
But.... 100 years is enough, joke's over.
So yes, I'd prefer we used it, if only to be decent and honest.
| Be it as it may, you still haven't produced the promised contradiction
| in the tau equation and the derivation of the Lorentz transform from
| it.
1/2 mass(treetop, apple+cherry) = mass(ground, apple)
is an example of the contradiction.
The mass of an apple is independent of height.
Then there is this one:
V = (c+v)/(1+v/c) = c (composition of velocities, section 5 )
Substituting c+v for it's value,
稼tau(0,0,0,t)+tau(0,0,0,t+x'/c+x'/c)] = tau(x',0,0,t+x'/c)
稼tau(0,0,0,t)+tau(0,0,0,t+2x'/c)] = tau(x',0,0,t+x'/c)
Now derive the cuckoo transformations from that.
|
| > At this instant the
| > | clock *on* the engine shows tau(32,0,0,16) microseconds.
| >
| > So who gives a shit about a stationary clock or what spot
| > you are talking about, since it isn't part of the equation?
|
| But it *is* the part of the equation
No it isn't.
Yes it is.
No it isn't.
Yes it is.
No it isn't.
Yes it is.
No it isn't.
PHUCKWIT!
The domain of the function tau is k', not K.
The tau transformation is from the moving frame to the moving frame.
There is no velocity involved.
- except this fact is obscured in
| your view by using the third system of clocks.
PHUCKWIT! Einstein obscured the real clocks.
| But the connection
| between the stationary and moving clocks is the heart of the matter:
Yep. They read the same time and are independent of motion.
"If we place x'=x-vt, it is clear that a point at rest in the system k
must have a system of values x', y, z, independent of time."
Careful avoidance of x',y',z',t'. Careful not to NAME the SECOND frame.
Name k as the third. The misdirection of the master magician and
huckster.
What you do NOT say is as important as what you do say, ask any
politician.
| tau(32,0,0,16) is the time read by the moving clock on the engine at
| the instant it passes the *stationary* clock which displays 16 at that
| very moment.
Yep. The ray reaches the engine when the engine is at 80, stationary
frame. The engine is ALWAYS at 32, moving frame.
Coordinate 32 (x'), moving frame k', is independent of time.
The time for a turtle to travel from the caboose to the engine (16
hours) equals the time for the turtle to travel from the engine to the
caboose (16 hours), IFF the turtle is on the train. True or false?
If the turtle is on the track he'll never make it to the engine, the
caboose will leave the turtle.
Just because light is faster than a turtle doesn't change the algebra,
and velocity doesn't change time.
| You express the same idea by introducing instead the third
| system of moving clocks which are synced up to agree with the
| stationary clocks.
I didn't introduce the unnamed k' frame, Einstein did.
"If we place x'=x-vt, it is clear that a point at rest in the system k
must have a system of values x', y, z, independent of time. "-- Albert
Huckster Einstein.
Don't blame me for it, I'm the messenger.
You've tried to shoot the messenger, the messenger shoots back and calls
you names.
|
| Let's call this third system of clocks "x'-clocks".
Ok. I don't care what name you give them, so long as you recognise
their existence.
| Then I could say
| the same thing like so: "tau(32,0,0,16) is the time read by the moving
| clock on the engine at the instant the moving x'-clock at the same
spot
| displays t=16".
|
| Still no error in sight.
Very good. There is no difference in time for the one way
speed of light or the one way speed of a turtle.
16 seconds from caboose to engine.
16 seconds from station to point 80 on the track.
Still no error is sight.
Now send the turtle back again. What time does it reach the caboose
and what time does it reach the station? Are these the same events?
|
| > (rhetorical question, you are too stupid to answer sensibly.)
|
| All your questions are very easy. The most complex portion by far of
| preparing my postings is the proofreading.
Good.
You are starting to think for the first time in your life.
Keep it up, I'm pro-education.
|
| > | > But, one step at a time. Let's deal with the stationary spot
first.
| > | > That's 80, the engine moved from spot 32 at speed 3 for 16 time.
| > |
| > | You now switched coordinates - now you are talking about x, not
x'.
| >
| > You did that by prattling on about a "stationary clock at that
spot".
| > That "spot" is x, not x'.
|
| That's because I use only two systems of clocks: the stationary t and
| the moving tau.
Yes, I know. That's why the correct clocks to use are the x'-clocks.
Then I could say the same thing like so: "tau(32,0,0,16) is the time
read by the moving clock on the engine at the instant the moving
x'-clock at the same spot
displays t=16".
Still no error in sight.
| The x'-clocks you use on top of those are not incorrect
Good!
| but they are unecessary and they complicate the trminology.
Not good, they are very necessary. Not using them makes it difficult
to understand the hoax. Not using a second girl in the box to
wiggle her toes makes it difficult to understand how to cut one girl in
half.
You are not allowed to see inside the box, you have to deduce how the
trick
is done. Or you could leave the theatre believing she was cut in two...
lots
of phuckwits fall for that.
| But I'm
| willing to use this system, no problem.
Good. We call it Newtonian Mechanics. Would you like me to teach
you this new-fangled idea, or shall I teach you how to cut women in half
instead?
Start here:
http://webexhibits.org/calendars/year-text-Galileo.html
and progress to here:
http://members.tripod.com/~gravitee/
|
Androcles: | > | > Does 80 appear in either equation?
| > | > No.
| > |
| > | Of course not, the equation is for tau(x'y'z't), not for
tau(x,y,z,t).
| >
Jan: | > Correct.
|
| Well, so we agree.
Yep. The domain of tau is the moving frame k' and the image (codomain
to us British) is the moving frame kappa. We agree at long last.
|
| > | > Did anyone put a stationary clock at 80?
| > | > No.
| > |
| > | Yes. Clock distribution is the basis of Einstein's paper.
| >
| > Infinitely stupid dumbfuck, there are no clocks in the equation, let
| > alone one at 80.
|
| Clocks are everywhere. That's how the time coordinate is assigned to
| events.
Ah, but what event, the turtle arriving at the caboose, or the turtle
arriving at the station?
16 hours to get back to the caboose (0',0',0',16'), but it will take
much
longer to walk home (0,0,0,400) if you take it for a ride.
That's why I don't like the 1/2.
|
| > | > Is there a "stationary" clock on the engine?
| > | > No;
| > |
| > | Not *on* the engine.
| >
| > Right. There is a "moving" clock on the engine that is at rest with
the
| > engine.
| > We call it the "driver's wristwatch", infinitely stupid dumbfuck.
| >
| >
| > At every instant the engine passes a stationary
| > | clock.
| >
| > Irrelevant, all the stationary clocks are not part of the equation.
|
| On the contrary, it's 100% relevant because the tau equation is an
| expression of a constraint *between* the stationary clocks (you
| introduce the third redundant moving x'-clock system for the same
| thing) and the standard moving tau-clocks. (By "standard" I mean
| Einstein-synchronised.)
You are repeating yourself. I did NOT introduce
"If we place x'=x-vt, it is clear that a point at rest in the system k
must have a system of values x', y, z, independent of time", that was
Albert Huckster Einstein.
All I'm doing is pointing it out and you've agreed to use it.
The time for a turtle to go from caboose to engine is the time from
engine to caboose, but the time to go from the station to the engine is
different to the time
to go from the engine to the station, the turtle hitched a ride.
|
| > | > there is a clock that moves with the engine.
| > |
| > | Yes.
| >
| > Ok, so what are you prattling on about a "stationary clock at that
spot"
| > for?
| > Answer: Because you are infinitely stupid.
|
| Back to square one then. We agreed on the terminology and I can use
the
| x'-clocks (although they are an unecessary burden) but you still
| haven't produced the promised error in the tau equation and in the
| derivation of the Lorentz transform from it.
Yes I have, it's the half, but back to square one then.
Androcles.
Dirk Van de moortel
"Androcles" <Androcles@ MyPlace.org> wrote in message news:plaZe.2755$Am6....@fe3.news.blueyonder.co.uk...
[snip almost 400 lines of dog shit]
>
> Yes I have, it's the half, but back to square one then.
> Androcles.
Perhaps *now* Jan understands why it is better to be in your
alleged killfile.
Dirk Vdm
JanPB
> "JanPB" <fil...@gmail.com> wrote in message
> news:1127543551....@g14g2000cwa.googlegroups.com...
> | Androcles wrote:
> | > "JanPB" <fil...@gmail.com> wrote in message
> | > news:1127461664.3...@g14g2000cwa.googlegroups.com...
> | [...]
> | > I meant the stationary clock the engine
> | > | is passing when that clock shows 16 microseconds.
> | >
> | > Yes, you dumb fuck.
> | > It is at (80,0,0,16) which is not part of the equation.
> |
> | Aha, I think I see what your problem is.
>
> I don't have a problem, I know what's wrong with SR.
> You are not smart enough too see that.
>
>
> | You keep insisting on using an
> | extra (*third*) set of moving clocks.
>
> Aha, I think I see what your problem is. You refuse to
> use the real moving clocks and insist on using some
> fake tau clocks.
The setup is that there are two observers (called "stationary" and
"moving", or K and k) and each uses clocks synchronised by the Einstein
convention. We seek tau(x,y,z,t) which is a function giving the reading
of the moving (k) clock at the instant it is at K's coordinate (x,y,z)
AND K's clock at that instant at that same spot reads t. None of this
is "fake".
> | This (redundant) system of clocks
> | moves together with the moving clocks but is not
> Einstein-synchronised.
>
> The redundant fake tau clocks are not Newton synchronized.
Why "redundant" all of a sudden? Why "fake"? We do discuss the tau
equation and how the Lorentz follows from it, don't we? Has the subject
changed again? Of course they are not Newton-synchronised. I thought we
were discussing SR, not Newtonian mechanics.
> | Instead, at each moment such clock displays the same time as the
> | stationary clock that just happens to momentarily coincide with it:
>
> Yes, isn't it a wonderful concept?
> This makes them INDEPENDENT of distance divided by time,
> whereas the fake redundant tau clocks were velocity dependent.
How are they "fake"? If you have problem with even setting up the
Einstein synchronisation on the moving clocks then say so. I thought
your beef was with the derivation of the Lorentz from the tau equation.
IOW, we take the tau equation as given.
> This independence concept is really a good idea, don't you think?
> Then we don't have to confuse apples with oranges or distance with
> time anymore.
>
> |
> | x'=x-vt
> | y=y
> | z=z
> | t=t
>
> There you go, you've successfully transformed a stationary
> system of coordinates K to a moving system of coordinates k.
A moving system, yes. Not the k one - that one uses clocks synchronised
differently and it is this synchronisation we are interested in.
> Well done!
> Now, why do you want to transform a moving system of coordinates
> k to a fake redundant moving system of coordinates kappa
> by multiplying by some fake redundant gamma?
Because of the way the whole SR premise is set up: you have two systems
in inertial motion wrt one another, both using identical procedure to
set up their clocks and you want to express one set of coordinates in
terms of the other.
It turns out that the coordinate relationship is the Lorentz one.
> | If you prefer I can use this system although it's unecessary and
> | Einstein doesn't use it.
>
> I'd prefer you used it. IN FACT, Einstein depends on it but tries
> to hide it, hoping to sweep it under the carpet. It's the magician's
> trick
> of misdirection, the legs of the other girl on the box he saws in half.
Nonsense. He doesn't "try to hide" anything and the whole x' business
isn't really necessary, it only makes the notation cleaner.
> | If we both agree to use this extra system then
> | some of our differences are merely terminological.
>
> I agree to use it. Shall we give it a name? Part of the psychology
> of hiding things is not to name them. Once you give it a name people
> will remember it. Einstein doesn't want you to remember it.
Whatever gave you that idea? He clearly defines x' and then uses it
right and left for two pages straight almost.
> That's why
> x' doesn't appear in the cuckoo transformations,
Well, of course it doesn't - the goal is to find the relation between
(x,y,z,t) and (xi,eta,zeta,tau). Einstein introduced a third system
(x',y,z,t) only for convenience: to make the tau equation and the
subsequent derivation shorter.
> Let's call it k', shall we? That's an easy name to remember.
Good.
> Now we have a transformation from stationary K to moving k',
> x' = x-vt
> y' = y
> z' = z
> t' = t
Yes.
> and a magical transformation from the moving k' to the moving kappa,
>
> xi = x'/sqrt(1-v^2/c^2)
> eta = y'
> zeta = z'
> tau = t' * sqrt(1 - v^2/c^2)
The last equation is incorrect. We haven't done the derivation yet but
the correct expression for tau comes from the tau equation:
tau = (t' - (x'+vt')v/c^2) / sqrt(1 - v^2/c^2)
The rest of your derivation of the "cuckoo" is therefore incorrect.
> Voila!
> Behold the cuckoo transformations, we've sawn the girl in half.
>
> xi = (x-vt) /sqrt(1-v^2/c^2)
> tau = (t-vx/c^2) /sqrt(1-v^2/c^2)
>
> Nobody will notice how that was done!
>
> I do admire the guy, he was the greatest huckster in history.
> But.... 100 years is enough, joke's over.
Hahaha! You really think you are dealing with something difficult here,
do you?
> | Be it as it may, you still haven't produced the promised contradiction
> | in the tau equation and the derivation of the Lorentz transform from
> | it.
>
> 1/2 mass(treetop, apple+cherry) = mass(ground, apple)
> is an example of the contradiction.
> The mass of an apple is independent of height.
How about skipping metaphors and finally telling us all what's wrong
with saying that:
b - a = c - b
implies
2*b = a+c
or
1/2*(a+c) = b
> Then there is this one:
> V = (c+v)/(1+v/c) = c (composition of velocities, section 5 )
>
> Substituting c+v for it's value,
> ½[tau(0,0,0,t)+tau(0,0,0,t+x'/c+x'/c)] = tau(x',0,0,t+x'/c)
>
> ½[tau(0,0,0,t)+tau(0,0,0,t+2x'/c)] = tau(x',0,0,t+x'/c)
>
> Now derive the cuckoo transformations from that.
These equations are wrong. The relevant elapsed times as measured by k'
are x'/(c-v) and x'/(c+v), not x'/c.
> | > At this instant the
> | > | clock *on* the engine shows tau(32,0,0,16) microseconds.
> | >
> | > So who gives a shit about a stationary clock or what spot
> | > you are talking about, since it isn't part of the equation?
> |
> | But it *is* the part of the equation
>
> No it isn't.
Semantics. You insist on using k'. Fine.
> Yes it is.
> No it isn't.
> Yes it is.
> No it isn't.
> Yes it is.
> No it isn't.
> PHUCKWIT!
> The domain of the function tau is k', not K.
The auxiliary tau - yes. Not the tau we *really* are after. You confuse
yourself with using the same letter tau to denote two different
functions:
tau(x,y,z,t) -- the one we are really after,
and:
tau(x',y,z,t) -- an auxiliary.
> The tau transformation is from the moving frame to the moving frame.
> There is no velocity involved.
There is velocity involved in *the difference between clock
synchronisation* between k' and k. The two systems are at rest with
respect to one another but they use clocks synchronised differently and
that difference has v built into it. Remember that "frame" is not just
the spatial coordinates, it also involves the clocks (the time
coordinate).
> | But the connection
> | between the stationary and moving clocks is the heart of the matter:
>
> Yep. They read the same time and are independent of motion.
>
> "If we place x'=x-vt, it is clear that a point at rest in the system k
> must have a system of values x', y, z, independent of time."
>
> Careful avoidance of x',y',z',t'. Careful not to NAME the SECOND frame.
Simply because it would be overly pedantic and a waste of time. The
readers of this paper were scientists who could do manipulations like
these in their sleep.
> Name k as the third. The misdirection of the master magician and
> huckster.
Again, you talk as if you were discussing an issue of great subtlety.
> | tau(32,0,0,16) is the time read by the moving clock on the engine at
> | the instant it passes the *stationary* clock which displays 16 at that
> | very moment.
>
> Yep. The ray reaches the engine when the engine is at 80, stationary
> frame. The engine is ALWAYS at 32, moving frame.
> Coordinate 32 (x'), moving frame k', is independent of time.
Fine.
> The time for a turtle to travel from the caboose to the engine (16
> hours) equals the time for the turtle to travel from the engine to the
> caboose (16 hours), IFF the turtle is on the train. True or false?
True (I'm assuming "turtle" means "light") according to the moving k
(tau) clocks.
> If the turtle is on the track he'll never make it to the engine, the
> caboose will leave the turtle.
> Just because light is faster than a turtle doesn't change the algebra,
> and velocity doesn't change time.
>
> | You express the same idea by introducing instead the third
> | system of moving clocks which are synced up to agree with the
> | stationary clocks.
>
> I didn't introduce the unnamed k' frame, Einstein did.
> "If we place x'=x-vt, it is clear that a point at rest in the system k
> must have a system of values x', y, z, independent of time. "-- Albert
> Huckster Einstein.
> Don't blame me for it, I'm the messenger.
> You've tried to shoot the messenger, the messenger shoots back and calls
> you names.
That's because it frustrates you that you don't understand this.
> | Let's call this third system of clocks "x'-clocks".
>
> Ok. I don't care what name you give them, so long as you recognise
> their existence.
>
>
> | Then I could say
> | the same thing like so: "tau(32,0,0,16) is the time read by the moving
> | clock on the engine at the instant the moving x'-clock at the same
> spot
> | displays t=16".
> |
> | Still no error in sight.
>
> Very good. There is no difference in time for the one way
> speed of light or the one way speed of a turtle.
> 16 seconds from caboose to engine.
According to the k' clocks.
> 16 seconds from station to point 80 on the track.
According to the K clocks.
> Still no error is sight.
> Now send the turtle back again. What time does it reach the caboose
> and what time does it reach the station? Are these the same events?
It reaches the caboose at (x',y,z,t)=(0,0,0,20) and at time
tau(0,0,0,20) according to the moving k (tau) clocks. If by station you
mean the location (x,y,z)=(0,0,0) then the light reaches it later, 12
seconds after it passed the caboose according to K.
> | The x'-clocks you use on top of those are not incorrect
>
> Good!
>
> | but they are unecessary and they complicate the trminology.
>
> Not good, they are very necessary. Not using them makes it difficult
> to understand the hoax.
There is no hoax. This stuff it way too easy and transparent to contain
any hoax of that type.
> Androcles: | > | > Does 80 appear in either equation?
> | > | > No.
> | > |
> | > | Of course not, the equation is for tau(x'y'z't), not for
> tau(x,y,z,t).
> | >
> Jan: | > Correct.
> |
> | Well, so we agree.
>
> Yep. The domain of tau is the moving frame k' and the image (codomain
> to us British) is the moving frame kappa. We agree at long last.
That was never a disagreement, I thought. It simply never ocurred to me
that you wanted to waste time on such excessive hair splitting. But
this is OK, it's not incorrect, just a bit sophomoric.
> | > | > Did anyone put a stationary clock at 80?
> | > | > No.
> | > |
> | > | Yes. Clock distribution is the basis of Einstein's paper.
> | >
> | > Infinitely stupid dumbfuck, there are no clocks in the equation, let
> | > alone one at 80.
> |
> | Clocks are everywhere. That's how the time coordinate is assigned to
> | events.
>
> Ah, but what event, the turtle arriving at the caboose, or the turtle
> arriving at the station?
The caboose. And what's up with the turtle? Where did the light go? Do
turtles know how to bounce off mirrors?
> 16 hours to get back to the caboose (0',0',0',16'), but it will take
> much
> longer to walk home (0,0,0,400) if you take it for a ride.
> That's why I don't like the 1/2.
But the tau equation records the time of arrival at the caboose, not at
the station. Note that caboose=(x',y,z)=(0,0,0) while
station=(x,y,z)=(0,0,0).
> The time for a turtle to go from caboose to engine is the time from
> engine to caboose,
...according to k (tau) clocks.
> but the time to go from the station to the engine is
> different to the time
> to go from the engine to the station, the turtle hitched a ride.
If by "turtle" you mean "light" then according to the station (K) or k'
clocks these times are equal.
> | Back to square one then. We agreed on the terminology and I can use
> the
> | x'-clocks (although they are an unecessary burden) but you still
> | haven't produced the promised error in the tau equation and in the
> | derivation of the Lorentz transform from it.
>
> Yes I have, it's the half, but back to square one then.
The half follows from the following:
1. system K (stationary) synchronises its clocks according to Einstein,
2. system k (moving) synchronises its clocks according to Einstein,
3. system k' (moving) synchronises its clocks so they agree with K
clocks.
4. The Einstein synchronisation in K and in k implies that light in
both satisfies the condition that it takes the same time to go from A
to B as from B to A. In particular, in the moving system k:
time(reflection) - time(emission) = time(absorption) -
time(reflection)
or
2*time(reflection) = time(emission) + time(absorption)
or
time(reflection) = 1/2*(time(emission) + time(absorption))
Hence 1/2.
--
Jan Bielawski
Androcles
Androcles wrote:
> "JanPB" <fil...@gmail.com> wrote in message
> news:1127543551....@g14g2000cwa.googlegroups.com...
> | Androcles wrote:
> | > "JanPB" <fil...@gmail.com> wrote in message
> | > news:1127461664.3...@g14g2000cwa.googlegroups.com...
> | [...]
> | > I meant the stationary clock the engine
> | > | is passing when that clock shows 16 microseconds.
> | >
> | > Yes, you dumb fuck.
> | > It is at (80,0,0,16) which is not part of the equation.
> |
> | Aha, I think I see what your problem is.
>
> I don't have a problem, I know what's wrong with SR.
> You are not smart enough too see that.
>
>
> | You keep insisting on using an
> | extra (*third*) set of moving clocks.
>
> Aha, I think I see what your problem is. You refuse to
> use the real moving clocks and insist on using some
> fake tau clocks.
| The setup is that there are two observers (called "stationary" and
| "moving", or K and k) and each uses clocks synchronised by the
Einstein
| convention.
Bullshit, they've only got one clock each. Learn to count.
| We seek tau(x,y,z,t) which is a function
Then what the fuck are you using tau(x',y',z',t') for, dumbshit?
Androcles.
JanPB
Huh??? Unless you misspoke, this instantly invalidates everything you
say because it violates the entire basic premise and experimental
setup of Einstein's paper.
> | We seek tau(x,y,z,t) which is a function
>
> Then what the fuck are you using tau(x',y',z',t') for, dumbshit?
Because this function is easier to work with (less printer's ink used)
than tau(x,y,z,t). And once tau(x',y,z,t) is derived, it's very easy to
switch back to what you really want, namely tau(x,y,z,t). It's net
savings. For example, the tau equation expressed directly in terms of
(x,y,z,t) would have read:
1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
tau(x+vx/(c-v)),0,0,t+x/(c-v))
...a somewhat messier (although equivalent) form compared to
Einstein's.
This equation would yield a slightly different constraint for
infinitesimal x than "the x' version":
@tau/@x * c^2 + @tau/@t * v = 0
which would eventually lead to tau(x,y,z,t)=beta(v)phi(v)(t-xv/c^2) -
exactly as before, except x' would have never been introduced or
mentioned.
--
Jan Bielawski
Androcles
I didn't misspeak.
There is a "moving" tau-clock and a "stationary" t-clock, any other
clock
is a copy clock of one of those.
Likewise you are a phuckwit and a fuckwit.
|
| > | We seek tau(x,y,z,t) which is a function
| >
| > Then what the fuck are you using tau(x',y',z',t') for, dumbshit?
|
| Because this function is easier to work with
Ah... I see.
We seek mass(treetop, apple) which is a function, and
1/2[mass(treetop, apple)+mass(treetop,orange)] = mass(ground,
windfall(apple))
Stupid idiot.
(less printer's ink used)
| than tau(x,y,z,t). And once tau(x',y,z,t) is derived, it's very easy
to
| switch back to what you really want, namely tau(x,y,z,t). It's net
| savings. For example, the tau equation expressed directly in terms of
| (x,y,z,t) would have read:
|
| 1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
| tau(x+vx/(c-v)),0,0,t+x/(c-v))
x' = vx now, does it?
No wonder you can't understand relativity, you can't even substitute
x' = x-vt correctly. Whoever tried to teach you basic algebra was
having a very hard and unsuccessful time.
Hint:
x+vx = x(1+v)
|
| ...a somewhat messier (although equivalent) form compared to
| Einstein's.
x+vx is equivalent to x-vt, is it?
Fuck off, you are drunk, indecent, dishonest and a troll.
Androcles
JanPB
> | Androcles wrote:
> |
> | > Bullshit, they've only got one clock each. Learn to count.
> |
> | Huh??? Unless you misspoke, this instantly invalidates everything you
> | say because it violates the entire basic premise and experimental
> | setup of Einstein's paper.
>
> I didn't misspeak.
> There is a "moving" tau-clock and a "stationary" t-clock, any other
> clock is a copy clock of one of those.
But you have to define what "a copy clock" is. Again, that's the heart
of the theory. The whole point of relativity was to examine carefully
assumptions of that sort and make them explicit so they don't get swept
under the rug.
> | > | We seek tau(x,y,z,t) which is a function
> | >
> | > Then what the fuck are you using tau(x',y',z',t') for, dumbshit?
> |
> | Because this function is easier to work with
>
> Ah... I see.
> We seek mass(treetop, apple) which is a function, and
> 1/2[mass(treetop, apple)+mass(treetop,orange)] = mass(ground,
> windfall(apple))
> Stupid idiot.
Not sure what you mean. Even done any algebra? Ever introduced
temporary expressions to save some computation effort here and there?
>
> | And once tau(x',y,z,t) is derived, it's very easy to
> | switch back to what you really want, namely tau(x,y,z,t). It's net
> | savings. For example, the tau equation expressed directly in terms of
> | (x,y,z,t) would have read:
> |
> | 1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
> | tau(x+vx/(c-v)),0,0,t+x/(c-v))
>
> x' = vx now, does it?
No, there is no x' here at all, only (x,y,z,t) is used throughout. So
the first slot is for x. On the left hand side we have the emission and
the absorption events. These take place at the caboose which moves at v
with respect to the stationary system K.
We assume that at time t the caboose is at the origin of K and that the
mirror is at x (that it, x units of K system to the right of the
origin).
Then the time it takes for light to reach the mirror according to the
system K is the usual expression x/(c-v). The time for light to travel
from the mirror back to the caboose is x/(c+v). So the total light
travel time is:
x/(c-v) + x/(c+v)
During this time the caboose moved away from the origin of K by the
distance:
velocity * light travel time = v * (x/(c-v) + x/(c+v)) =
= vx/(c-v) + vx/(c+v)
...hence the appearance of this quantity on the LHS inside the
absorption event at the caboose.
Similar consideration applies to the RHS which records the reflection:
light travel time = x/(c-v)
caboose x coordinate at that instant = velocity * light travel time =
= v * x/(c-v) = vx/(c-v)
thus the engine x coordinate = x + caboose x coordinate =
= x + vx/(c-v)
> No wonder you can't understand relativity, you can't even substitute
> x' = x-vt correctly.
I am not using any x' here. Just writing the Einstein sync relationship
(aka. the tau equation) in terms of (x,y,z,t).
> Whoever tried to teach you basic algebra was
> having a very hard and unsuccessful time.
> Hint:
> x+vx = x(1+v)
I have no "x+vx" anywhere. Perhaps I should have been more explicit
with my parentheses. When I wrote on the RHS:
x+vx/(c-v)
I meant:
x+[vx/(c-v)]
I also noticed I have a typo there - one unpaired parenthesis. The
equation should read then:
1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
tau(x+[vx/(c-v)],0,0,t+x/(c-v))
> x+vx is equivalent to x-vt, is it?
So there is no "x+vx" (the units would be wrong for a start).
--
Jan Bielawski
Androcles
| Androcles wrote:
| > "JanPB" <fil...@gmail.com> wrote in message
| > news:1127968741.0...@o13g2000cwo.googlegroups.com...
| > | Androcles wrote:
| > |
| > | > Bullshit, they've only got one clock each. Learn to count.
| > |
| > | Huh??? Unless you misspoke, this instantly invalidates everything
you
| > | say because it violates the entire basic premise and experimental
| > | setup of Einstein's paper.
| >
| > I didn't misspeak.
| > There is a "moving" tau-clock and a "stationary" t-clock, any other
| > clock is a copy clock of one of those.
|
| But you have to define what "a copy clock" is.
Footnote 3:
We shall not here discuss the inexactitude which lurks in the concept of
simultaneity of two events at approximately the same place, which can
only be removed by an abstraction.
Again, that's the heart
| of the theory.
No it's not. Again, you are a phuckwit.
| The whole point of relativity was to examine carefully
| assumptions of that sort and make them explicit so they don't get
swept
| under the rug.
Then it failed and so have you.
|
| > | > | We seek tau(x,y,z,t) which is a function
| > | >
| > | > Then what the fuck are you using tau(x',y',z',t') for, dumbshit?
| > |
| > | Because this function is easier to work with
| >
| > Ah... I see.
| > We seek mass(treetop, apple) which is a function, and
| > 1/2[mass(treetop, apple)+mass(treetop,orange)] = mass(ground,
| > windfall(apple))
| > Stupid idiot.
|
| Not sure what you mean.
Of course you are not sure what I mean. You are a phuckwit that doesn't
know what a function with two variables is, let alone four.
| Even done any algebra?
Not your kind, x' = x-vt, therefore x =vx.
ROFLMAO!
| Ever introduced
| temporary expressions to save some computation effort here and there?
I do that properly.
What's the mass of an apple at the top of a tree and why is it different
when it has fallen?
| >
| > | And once tau(x',y,z,t) is derived, it's very easy to
| > | switch back to what you really want, namely tau(x,y,z,t). It's net
| > | savings. For example, the tau equation expressed directly in terms
of
| > | (x,y,z,t) would have read:
| > |
| > | 1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
| > | tau(x+vx/(c-v)),0,0,t+x/(c-v))
| >
| > x' = vx now, does it?
|
| No, there is no x' here at all, only (x,y,z,t) is used throughout.
Fucking idiot.
Stupid cunt.
Moron.
Imbecile.
Deranged lunatic.
JanPB.
(temporary expressions to save some computation effort here and there,
and my printer's ink)
| So
| the first slot is for x.
It's got x' in it, deranged lunatic.
It is boring to find out how far Bielawski can be pushed into
irrelevancies and nonsense. - what are the limits of withstanding
blatant contradictions (infinite, apparently).
| On the left hand side we have the emission and
| the absorption events. These take place at the caboose which moves at
v
| with respect to the stationary system K.
So does the engine.
|
| We assume that at time t the caboose is at the origin of K and that
the
| mirror is at x (that it, x units of K system to the right of the
| origin).
You can assume all you want, deranged lunatic. Einstein doesn't mention
a mirror, only a reflection at x' (WHICH IS NOT x) that I'll call the
"engine".
The argument to the function tau is
1/2[tau(caboose, emission) + tau(caboose, reception)] = tau(engine,
reflection)
In K coordinates,
1/2[tau(0,0,0,0) + tau(60,0,0,20)] = tau(80, 16)
in k coordinates
1/2[tau(0,0,0,0) + tau(0,0,0,20)] = tau(32, 16)
|
| Then the time it takes for light to reach the mirror according to the
| system K is the usual expression x/(c-v).
Bile lawski the stupid cunt: 16 = 80/(5-3)
Imbecile, the time it takes for light to reach the mirror according to
the
system K is the usual expression t = x/c.
Androcles' granddaughter, age 11: 16 = 80/5
| The time for light to travel
| from the mirror back to the caboose is x/(c+v).
Bile lawski the fucking idiot: 4 = 80/(5+3)
Fucking idiot, the time for light to travel from the mirror back to the
caboose is
t = (x-vt)/c.
Androcles' granddaughter, age 11: 4 = (80-3*20)/ 5
I see now why Polacks have a rep in the USA for being stupid.
They are easily suckered.
So the total light
| travel time is:
|
| x/(c-v) + x/(c+v)
Bile lawski the drooling fart: 50 = 80/2 + 80/8
Androcles' granddaughter: 20 = 32/(5-3) + 32/(5+3)
You seem to think the caboose is still at the station, fuckwit.
|
| During this time the caboose moved away from the origin of K by the
| distance:
|
| velocity * light travel time = v * (x/(c-v) + x/(c+v)) =
| = vx/(c-v) + vx/(c+v)
The correct answer is vt.
Notice how much printer's ink that saves.
[snip crap]
Androcles.
Dirk Van de moortel
"Androcles" <Androcles@ MyPlace.org> wrote in message news:TnW_e.10673$VI6....@fe1.news.blueyonder.co.uk...
[snipping everything but Androdrool]
>
> Footnote 3:
> We shall not here discuss the inexactitude which lurks in the concept of
> simultaneity of two events at approximately the same place, which can
> only be removed by an abstraction.
---------------
> No it's not. Again, you are a phuckwit.
---------------
> Then it failed and so have you.
---------------
> Of course you are not sure what I mean. You are a phuckwit that doesn't
> know what a function with two variables is, let alone four.
---------------
> Not your kind, x' = x-vt, therefore x =vx.
> ROFLMAO!
---------------
> I do that properly.
> What's the mass of an apple at the top of a tree and why is it different
> when it has fallen?
---------------
> Fucking idiot.
> Stupid cunt.
> Moron.
> Imbecile.
> Deranged lunatic.
> JanPB.
> (temporary expressions to save some computation effort here and there,
> and my printer's ink)
---------------
> It's got x' in it, deranged lunatic.
>
> It is boring to find out how far Bielawski can be pushed into
> irrelevancies and nonsense. - what are the limits of withstanding
> blatant contradictions (infinite, apparently).
---------------
> So does the engine.
---------------
> You can assume all you want, deranged lunatic. Einstein doesn't mention
> a mirror, only a reflection at x' (WHICH IS NOT x) that I'll call the
> "engine".
> The argument to the function tau is
> 1/2[tau(caboose, emission) + tau(caboose, reception)] = tau(engine,
> reflection)
> In K coordinates,
> 1/2[tau(0,0,0,0) + tau(60,0,0,20)] = tau(80, 16)
> in k coordinates
> 1/2[tau(0,0,0,0) + tau(0,0,0,20)] = tau(32, 16)
---------------
> Bile lawski the stupid cunt: 16 = 80/(5-3)
>
> Imbecile, the time it takes for light to reach the mirror according to
> the
> system K is the usual expression t = x/c.
>
> Androcles' granddaughter, age 11: 16 = 80/5
---------------
> Bile lawski the fucking idiot: 4 = 80/(5+3)
>
> Fucking idiot, the time for light to travel from the mirror back to the
> caboose is
> t = (x-vt)/c.
>
> Androcles' granddaughter, age 11: 4 = (80-3*20)/ 5
>
> I see now why Polacks have a rep in the USA for being stupid.
> They are easily suckered.
---------------
> Bile lawski the drooling fart: 50 = 80/2 + 80/8
>
> Androcles' granddaughter: 20 = 32/(5-3) + 32/(5+3)
>
> You seem to think the caboose is still at the station, fuckwit.
---------------
> The correct answer is vt.
> Notice how much printer's ink that saves.
>
> [snip crap]
> Androcles.
Jan, do you realize what he is after?
Do you know how you can accomplish with him
what he is trying to accomplish with you?
Dirk Vdm
JanPB
>
No, actually. I find it somewhat entertaining and mildly interesting.
> Do you know how you can accomplish with him
> what he is trying to accomplish with you?
I'm all ears :-) (It's not like I'm in pain or anything)
--
Jan Bielawski
JanPB
>
> | So
> | the first slot is for x.
>
> It's got x' in it, deranged lunatic.
I was talking about my equation. My point was that the whole x'
business was introduced by Einstein only for convenience. Then to
support this claim I said that in fact one could have done the entire
derivation without ever referring to x'. Which is what I did - I wrote
the tau equatiion in terms of (x,y,z,t), NOT (x',y,z,t).
That's why the first slot is always x in the expression *I* wrote.
> [...]
> You can assume all you want, deranged lunatic. Einstein doesn't mention
> a mirror, only a reflection at x' (WHICH IS NOT x)
Einstein uses (x',y,z,t) to refer to events, I used straight (x,y,z,t)
for the same purpose. One event referred by two different coordinate
systems.
> that I'll call the
> "engine".
> The argument to the function tau is
> 1/2[tau(caboose, emission) + tau(caboose, reception)] = tau(engine,
> reflection)
> In K coordinates,
> 1/2[tau(0,0,0,0) + tau(60,0,0,20)] = tau(80, 16)
> in k coordinates
> 1/2[tau(0,0,0,0) + tau(0,0,0,20)] = tau(32, 16)
You confuse yourself by using numbers instead of expressions.
[invalid arguments follow]
--
Jan Bielawski
Dirk Van de moortel
"JanPB" <fil...@gmail.com> wrote in message news:1128019300.7...@g43g2000cwa.googlegroups.com...
> Dirk Van de moortel wrote:
> >
> > Jan, do you realize what he is after?
>
> No, actually. I find it somewhat entertaining and mildly interesting.
Oh, well, in that case, do carry on ;-)
Dirk Vdm
Androcles
I'll tell you. Honesty. You two have no idea what that is.
From: "Paul B. Andersen" <paul.b.ander...@deletethishia.no>
Date: Wed, 14 Sep 2005 14:00:34 +0200
Local: Wed, Sep 14 2005 1:00 pm
Subject: Re: Spectrum!
"But the two stars of Algol have different mass, radius and
density, and the B8 is well outside of the Roche limit
of the K2, while the K2 is just at the Roche limit of the B8.
That is, the K2 fills its Roche lobe completely, and mass
is transferred to the B8. So the K2 IS torn apart and there
is an accretion disk around the B8 akin to the rings of Saturn.
(This accretion disk is not stable, though. It is a transient
disk; the mass transferred from the K2 bounces off the surface
of the B8 and eventually falls back to the surface.) "
Being a B8, the surface the accretion disk bounces off looks like this:
http://sohowww.nascom.nasa.gov/hotshots/2003_11_04/c2w.gif
Any idea what this means?
Didn't think so.
Androcles.
Androcles
| Androcles wrote:
| > "JanPB" <fil...@gmail.com> wrote in message
| > news:1128011476.9...@g44g2000cwa.googlegroups.com...
| >
| > | So
| > | the first slot is for x.
| >
| > It's got x' in it, deranged lunatic.
|
| I was talking about my equation.
Oh, do you have one?
Not interested in YOUR equations, sorry.
I'm discussing Einstein's equation.
| My point was
That you can't read and don't understand schoolboy algebra?
Yes, we have that figured out.
| that the whole x'
| business was introduced by Einstein only for convenience.
Not very convenient to your equation, is it?
This is simply insane. You know ABSOLUTELY zero about the subject and
yet you write... THIS?? Why even do you waste your time on this?
What are the limits of withstanding blatant contradictions (infinite,
apparently).
news:1127968741.0...@o13g2000cwo.googlegroups.com
We seek tau(x,y,z,t) which is a function
> Then what the fuck are you using tau(x',y',z',t') for, dumbshit?
Because this function is easier to work with (less printer's ink used)
An error in Relativity "would be like Stephen Hawking dividing by zero
or
something equally trivial." -- Bielawski.
It's WAY too simple-minded.-- Bielawski.
"would have been caught immediately by the AdP reviewer." -- Bielawski.
It is boring to find out how far Bielawski can be pushed into
irrelevancies and nonsense.
Fuck off.
Androcles.
JanPB
> "JanPB" <fil...@gmail.com> wrote in message
> news:1128019906.7...@g43g2000cwa.googlegroups.com...
> | Androcles wrote:
> | > "JanPB" <fil...@gmail.com> wrote in message
> | > news:1128011476.9...@g44g2000cwa.googlegroups.com...
> | >
> | > | So
> | > | the first slot is for x.
> | >
> | > It's got x' in it, deranged lunatic.
> |
> | I was talking about my equation.
>
> Oh, do you have one?
> Not interested in YOUR equations, sorry.
> I'm discussing Einstein's equation.
You really must brush up on your reading comprehension skills. All I
was saying was that x' was not an essential part of Einstein's argument
but only a convenience. In fact - so I kept pontificating - one could
repeat Einstein's argument almost verbatim without *ever* referring to
the (x',y,z,t) system. And I provided a hint how to do it by writing
Einstein's tau equation in terms of (x,y,z,t) only:
1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
tau(x+[vx/(c-v)],0,0,t+x/(c-v))
Now we can go back to Einstein's way of doing things using x' if you
want.
<Back to work>
--
Jan Bielawski
Androcles
| Androcles wrote:
| > "JanPB" <fil...@gmail.com> wrote in message
| > news:1128019906.7...@g43g2000cwa.googlegroups.com...
| > | Androcles wrote:
| > | > "JanPB" <fil...@gmail.com> wrote in message
| > | > news:1128011476.9...@g44g2000cwa.googlegroups.com...
| > | >
| > | > | So
| > | > | the first slot is for x.
| > | >
| > | > It's got x' in it, deranged lunatic.
| > |
| > | I was talking about my equation.
| >
| > Oh, do you have one?
| > Not interested in YOUR equations, sorry.
| > I'm discussing Einstein's equation.
|
| You really must brush up on your reading comprehension skills. All I
| was saying was that x' was not an essential part of Einstein's
argument
| but only a convenience.
You really must brush up on your reading comprehension skills.
I'm < that means Androcles.
NOT < a negation
interested in
YOUR < Janitor Bielawski
equations.
Go back to your mop and bucket, you don't belong here.
In fact - so I kept pontificating - one could
| repeat Einstein's argument almost verbatim without *ever* referring to
| the (x',y,z,t) system. And I provided a hint how to do it by writing
| Einstein's tau equation in terms of (x,y,z,t) only:
|
| 1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
| tau(x+[vx/(c-v)],0,0,t+x/(c-v))
Ok... let's look at the RHS.
tau(x+[vx/(c-v)],0,0,t+x/(c-v))
Focus in on the time.
x/(c-v)
Put it in numbers, like I showed you.
80/(5-3)
= 40.
Does 16 = 40?
No.
Now fuck off.
Androcles.
JanPB
>
> | You really must brush up on your reading comprehension skills. All I
> | was saying was that x' was not an essential part of Einstein's
> argument
> | but only a convenience.
>
> You really must brush up on your reading comprehension skills.
> I'm < that means Androcles.
> NOT < a negation
> interested in
> YOUR < Janitor Bielawski
> equations.
Non sequitur. If you claim X than you must be ready to hear arguments
against X. If those arguments against X happen to be equations - so be
it.
In this case the counterargument consisted of demonstrating that x' was
*in principle* unnecessary. The very extistence of such equation
(doesn't matter who wrote it) disproves your claim that x' is essential
- that's the point.
As I said in my previous post we can stop debating this tangential
issue and go back to using x' and simply follow Einstein's derivation
of the Lorentz transform which is your main objection on this thread.
--
Jan Bielawski
Androcles
| Androcles wrote:
| > "JanPB" <fil...@gmail.com> wrote in message
| > news:1128027695....@g43g2000cwa.googlegroups.com...
| >
| > | You really must brush up on your reading comprehension skills. All
I
| > | was saying was that x' was not an essential part of Einstein's
| > argument
| > | but only a convenience.
| >
| > You really must brush up on your reading comprehension skills.
| > I'm < that means Androcles.
| > NOT < a negation
| > interested in
| > YOUR < Janitor Bielawski
| > equations.
|
| Non sequitur.
That's right. Your equations are non sequitur.
Now fuck off.
| If you claim X than you must be ready to hear arguments
| against X.
Einstein claimed X, not me. You can claim Y until the cows come home,
I'm not interested.
Now fuck off.
| If those arguments against X happen to be equations - so be
| it.
|
| In this case the counterargument consisted of demonstrating that x'
was
| *in principle* unnecessary.
Non sequitur. I'm not interested in your Y claims.
Now fuck off.
The very extistence of such equation
| (doesn't matter who wrote it) disproves your claim that x' is
essential
| - that's the point.
Non sequitur. Einstein claimed
X: x' is a point at rest in system k independent of time, not equal xi.
Now fuck off.
|
| As I said in my previous post we can stop debating this tangential
| issue and go back to using x' and simply follow Einstein's derivation
| of the Lorentz transform which is your main objection on this thread.
You can't derive the cuckoo transform until you resolve the X that
Einstein claimed.
Now fuck off.
Androcles.
Daryl McCullough
[stuff deleted]
Androcles is a strange case. He has convinced himself
that Einstein made a mistake in his derivation of the
Lorentz transformations, and he will not even *listen*
to a counterargument. His reasoning is that any
counterargument, if it goes beyond what Einstein said
doesn't count as *Einstein's* argument, it counts as
Jan's argument or Daryl's argument, or whoever. And
Androcles has no interest in the arguments of anyone
but Einstein.
Maybe there is a certain sense in which Androcles
is right. Einstein's argument is a failure in the
sense that it fails to persuade Androcles. And for
Androcles, that's the only standard for success that
counts.
--
Daryl McCullough
Ithaca, NY
JanPB
back to you faster.
Androcles wrote:
> "JanPB" <fil...@gmail.com> wrote in message
> news:1128027695....@g43g2000cwa.googlegroups.com...
> In fact - so I kept pontificating - one could
> | repeat Einstein's argument almost verbatim without *ever* referring to
> | the (x',y,z,t) system. And I provided a hint how to do it by writing
> | Einstein's tau equation in terms of (x,y,z,t) only:
> |
> | 1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
> | tau(x+[vx/(c-v)],0,0,t+x/(c-v))
>
> Ok... let's look at the RHS.
> tau(x+[vx/(c-v)],0,0,t+x/(c-v))
> Focus in on the time.
> x/(c-v)
> Put it in numbers, like I showed you.
> 80/(5-3)
> = 40.
> Does 16 = 40?
> No.
You should have x=32, not 80 in the numerator:
x/(c-v) = 32/(5-3) = 16
Remember that (x,y,z,t) in the tau equation express a relationship
between the light emission, reflection, and absorption in terms of
*initial position of the mirror* in the stationary system K.
(It's exactly the same with Einstein's tau equation except the
x'-coordinates of the caboose and the engine do not change with time so
his version is shorter. For example, he can keep 0 for the first slot
in both the emission and the absorption at the caboose whereas I had to
switch from 0 (emission) to vx/(c-v)+vx/(c+v) (absorption) there.)
Using your numbers t=0, x=32 (and y=z=0).
Perhaps it would have been better to denote these by (x0, y0, z0, t0)
but using (x,y,z,t) this way is the standard abuse of notation in order
to avoid excessive pedantry. I also wanted to mimic Einstein's own
abuse of notation here where he has (x',y,z,t) instead of the more
pedantic (x0', y0, z0, t0).
In the (x,y,z,t) coordinates the three events are:
emission at the caboose: (0,0,0,0)
reflection at the engine: (32+3*16,0,0,0+16) = (80,0,0,16)
absorption at the caboose: (3*16+3*4,0,0,16+4) = (60,0,0,20)
...so this version of the tau equation is:
1/2*(tau(0,0,0,0) + tau(60,0,0,20)) = tau(80,0,0,16)
--
Jan Bielawski
Androcles
Looks to me like
(16/20)*(tau(0,0,0,0) + tau(60,0,0,20)) = tau(80,0,0,16)
is correct and the 1/2 is bullshit.
Maybe you didn't learn arithmetic, Einstein didn't either.
Incidentally, why are you computing the time in the stationary frame?
That seems like excessive pedantry to me.
Androcles
JanPB
[...]
> | In the (x,y,z,t) coordinates the three events are:
> |
> | emission at the caboose: (0,0,0,0)
> | reflection at the engine: (32+3*16,0,0,0+16) = (80,0,0,16)
> | absorption at the caboose: (3*16+3*4,0,0,16+4) = (60,0,0,20)
> |
> | ...so this version of the tau equation is:
> |
> | 1/2*(tau(0,0,0,0) + tau(60,0,0,20)) = tau(80,0,0,16)
>
>
> Looks to me like
> (16/20)*(tau(0,0,0,0) + tau(60,0,0,20)) = tau(80,0,0,16)
> is correct and the 1/2 is bullshit.
This tau equation just says that according to the moving clocks it took
as much time for the light to go from the caboose to the mirror as it
took to go back:
tau(80,0,0,16) - tau(0,0,0,0) = tau(60,0,0,20) - tau(80,0,0,16)
When you collect the terms you get the 1/2.
> Maybe you didn't learn arithmetic, Einstein didn't either.
> Incidentally, why are you computing the time in the stationary frame?
> That seems like excessive pedantry to me.
Well, isn't the point of the whole derivation to find out the
relationship between the stationary coordinates and the moving
coordinates? The tau equation is just *a* relationship between the two,
expressed in terms of the yet unkown function tau(x,y,z,t).
This equation is forced on us by the assumption of clock
synchronisation and it is not obvious at first that this constraint is
useful and strong enough to allow us to derive that relationship (the
transform) but it turns out that it is (assuming the transform we seek
is linear, i.e. of the form:
tau(x,y,z,t) = Ax + By + Cz + Dt
...where A, B, C, D are constants, plus the usual assumptions regarding
the symmetry of space).
If you want to continue with Einstein's way of deriving things perhaps
it would be better to switch back to his slightly different function
tau(x',y,z,t) and his version of the tau equation so we don't talk
about two slightly different points of view at the same time.
My only point for writing exclusively in terms of (x,y,z,t) and
(xi,eta,zeta,tau) was to mention that the third system (x',y,z,t) is
not *in principle* important and the derivation is 95% identical
without it. Of course (x',y,z,t) is what Einstein uses and one cannot
just remove it without making that 5% change.
--
Jan Bielawski
Androcles
| Androcles wrote:
| > "JanPB" <fil...@gmail.com> wrote in message
| > news:1128389986.7...@z14g2000cwz.googlegroups.com...
| [...]
| > | In the (x,y,z,t) coordinates the three events are:
| > |
| > | emission at the caboose: (0,0,0,0)
| > | reflection at the engine: (32+3*16,0,0,0+16) = (80,0,0,16)
| > | absorption at the caboose: (3*16+3*4,0,0,16+4) = (60,0,0,20)
| > |
| > | ...so this version of the tau equation is:
| > |
| > | 1/2*(tau(0,0,0,0) + tau(60,0,0,20)) = tau(80,0,0,16)
| >
| >
| > Looks to me like
| > (16/20)*(tau(0,0,0,0) + tau(60,0,0,20)) = tau(80,0,0,16)
| > is correct and the 1/2 is bullshit.
|
| This tau equation just says that according to the moving clocks it
took
| as much time for the light to go from the caboose to the mirror as it
| took to go back:
|
| tau(80,0,0,16) - tau(0,0,0,0) = tau(60,0,0,20) - tau(80,0,0,16)
|
| When you collect the terms you get the 1/2.
Prove it.
|
| > Maybe you didn't learn arithmetic, Einstein didn't either.
| > Incidentally, why are you computing the time in the stationary
frame?
| > That seems like excessive pedantry to me.
|
| Well, isn't the point of the whole derivation to find out the
| relationship between the stationary coordinates and the moving
| coordinates?
Of course, and you have only stationary coordinates.
You are missing the moving coordinates entirely.
So what if it takes a turtle 16 seconds to go from 0 to 80 at speed 5
and 4 seconds from 80 to 60 at velocity -5?
Still, if you think 20/2 = 16 you are as crazy as Hammond <shrug>
| The tau equation is just *a* relationship between the two,
| expressed in terms of the yet unkown function tau(x,y,z,t).
Your tau equation is just *a* relationship between the ONE,
expressed by an idiot.
| This equation is forced on us by the assumption of clock
| synchronisation and it is not obvious at first that this constraint is
| useful and strong enough to allow us to derive that relationship (the
| transform) but it turns out that it is (assuming the transform we seek
| is linear, i.e. of the form:
|
| tau(x,y,z,t) = Ax + By + Cz + Dt
|
| ...where A, B, C, D are constants, plus the usual assumptions
regarding
| the symmetry of space).
Crap. You have no idea what you are babbling about.
|
| If you want to continue with Einstein's way of deriving things perhaps
| it would be better to switch back to his slightly different function
| tau(x',y,z,t) and his version of the tau equation so we don't talk
| about two slightly different points of view at the same time.
I don't know why you wanted to switch anyway.
Obviously you haven't a clue.
|
| My only point for writing exclusively in terms of (x,y,z,t) and
| (xi,eta,zeta,tau) was to mention that the third system (x',y,z,t) is
| not *in principle* important and the derivation is 95% identical
| without it.
Bullshit. You are babbling.
| Of course (x',y,z,t) is what Einstein uses and one cannot
| just remove it without making that 5% change.
Fucking drivel, you worthless babbling moron.
Androcles.
JanPB
> "JanPB" <fil...@gmail.com> wrote in message
> news:1128420620....@o13g2000cwo.googlegroups.com...
> | Androcles wrote:
> | > "JanPB" <fil...@gmail.com> wrote in message
> | > news:1128389986.7...@z14g2000cwz.googlegroups.com...
> | [...]
> | > |
> | > | 1/2*(tau(0,0,0,0) + tau(60,0,0,20)) = tau(80,0,0,16)
> | >
> | >
> | > Looks to me like
> | > (16/20)*(tau(0,0,0,0) + tau(60,0,0,20)) = tau(80,0,0,16)
> | > is correct and the 1/2 is bullshit.
> |
> | This tau equation just says that according to the moving clocks it
> took
> | as much time for the light to go from the caboose to the mirror as it
> | took to go back:
> |
> | tau(80,0,0,16) - tau(0,0,0,0) = tau(60,0,0,20) - tau(80,0,0,16)
> |
> | When you collect the terms you get the 1/2.
>
> Prove it.
I don't quite see why you ask such an obvious question be here it goes
anyway:
Starting with:
tau(80,0,0,16) - tau(0,0,0,0) = tau(60,0,0,20) - tau(80,0,0,16)
Move tau(80,0,0,16) to the left and tau(0,0,0,0) to the right:
2 * tau(80,0,0,16) = tau(0,0,0,0) + tau(60,0,0,20)
Divide by 2:
tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
This is it.
> | > Maybe you didn't learn arithmetic, Einstein didn't either.
> | > Incidentally, why are you computing the time in the stationary
> frame?
> | > That seems like excessive pedantry to me.
> |
> | Well, isn't the point of the whole derivation to find out the
> | relationship between the stationary coordinates and the moving
> | coordinates?
>
>
> Of course, and you have only stationary coordinates.
> You are missing the moving coordinates entirely.
Not sure why you say that - the tau equation refers to selected
coordinate quantities from *both* systems: (x,y,z,t) and
(xi,eta,zeta,tau).
> So what if it takes a turtle 16 seconds to go from 0 to 80 at speed 5
> and 4 seconds from 80 to 60 at velocity -5?
> Still, if you think 20/2 = 16 you are as crazy as Hammond <shrug>
The 16 and 4 seconds are times as measured by the stationary clocks
(t). The corresponding readings by the clocks sitting on the train
(tau) are different than those t-readings and they are equal in both
directions.
> | This equation is forced on us by the assumption of clock
> | synchronisation and it is not obvious at first that this constraint is
> | useful and strong enough to allow us to derive that relationship (the
> | transform) but it turns out that it is (assuming the transform we seek
> | is linear, i.e. of the form:
> |
> | tau(x,y,z,t) = Ax + By + Cz + Dt
> |
> | ...where A, B, C, D are constants, plus the usual assumptions
> regarding
> | the symmetry of space).
>
> Crap. You have no idea what you are babbling about.
Could you back it up. I simply stated the starting point of the
derivation argument.
> | My only point for writing exclusively in terms of (x,y,z,t) and
> | (xi,eta,zeta,tau) was to mention that the third system (x',y,z,t) is
> | not *in principle* important and the derivation is 95% identical
> | without it.
>
> Bullshit. You are babbling.
No, I can write both derivations side by side and the differences are
small.
--
Jan Bielawski
Androcles
So half of 80 is 60 and half of 20 is 16
Well done.
|
| > | > Maybe you didn't learn arithmetic, Einstein didn't either.
| > | > Incidentally, why are you computing the time in the stationary
| > frame?
| > | > That seems like excessive pedantry to me.
| > |
| > | Well, isn't the point of the whole derivation to find out the
| > | relationship between the stationary coordinates and the moving
| > | coordinates?
| >
| >
| > Of course, and you have only stationary coordinates.
| > You are missing the moving coordinates entirely.
|
| Not sure why you say that
Of course you're not. You're a phuckwit.
- the tau equation refers to selected
| coordinate quantities from *both* systems: (x,y,z,t) and
| (xi,eta,zeta,tau).
x= 80
y=0
z=0
t = 16 = 80/5 = x/c
x = 60
y = 0
z = 0
t = 4 = (80-60)/5 = (x2 - x1)/c
Where the fuck is v?
| > So what if it takes a turtle 16 seconds to go from 0 to 80 at speed
5
| > and 4 seconds from 80 to 60 at velocity -5?
| > Still, if you think 20/2 = 16 you are as crazy as Hammond <shrug>
|
| The 16 and 4 seconds are times as measured by the stationary clocks
| (t).
Yes ducky, and the moving train has fuck-all to do with that.
The corresponding readings by the clocks sitting on the train
| (tau) are different than those t-readings and they are equal in both
| directions.
There is no moving train in your equation. It doesn't exist.
No sign of it anywhere.
Light leaves 0, goes to 80, takes 16 seconds.
Light leaves 80, goes to 60, takes 4 seconds.
No train involved.
| > Crap. You have no idea what you are babbling about.
|
| Could you back it up. I simply stated the starting point of the
| derivation argument.
No you didn't. You gave track coordinates and track time.
The train is in the siding, not moving on the track.
Light leaves 0, goes to 80, takes 16 seconds, velocity of light 5.
Light leaves 80, goes to 60, takes 5 seconds, velocity of light -5.
16/20 * [ f(0,0,0,0) + f(60,0,0, 80/5 + (60-80)/-5)] = f(80,0,0,16)
No train, f is linear, no half either.
f(t) = t
| > | My only point for writing exclusively in terms of (x,y,z,t) and
| > | (xi,eta,zeta,tau) was to mention that the third system (x',y,z,t)
is
| > | not *in principle* important and the derivation is 95% identical
| > | without it.
| >
| > Bullshit. You are babbling.
|
| No, I can write both derivations side by side and the differences are
| small.
No you can't. You think 16 is half of 20 and 60 is half of 80.
Androcles
Daryl McCullough
Jan, this is incredibly painful to watch. Androcles isn't
paying any attention to what you are saying. He's only
calling you names and ignoring what you say.
Dirk Van de moortel
"Androcles" <Androcles@ MyPlace.org> wrote in message news:YDu0f.71458$RW.4...@fe2.news.blueyonder.co.uk...
>
> "JanPB" <fil...@gmail.com> wrote in message
> news:1128427205.0...@z14g2000cwz.googlegroups.com...
[snip]
> | tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
> |
> | This is it.
>
> So half of 80 is 60 and half of 20 is 16
> Well done.
No, you should say:
"So half of 60 is 80 and half of 20 is 16"
You can't even get faulty algebra right.
[snip]
> | No, I can write both derivations side by side and the differences are
> | small.
>
> No you can't. You think 16 is half of 20 and 60 is half of 80.
No, you think he thinks16 is half of 20 and 80 is half of 60.
Dirk Vdm
Paul B. Andersen
> | Starting with:
> |
> | tau(80,0,0,16) - tau(0,0,0,0) = tau(60,0,0,20) - tau(80,0,0,16)
> |
> | Move tau(80,0,0,16) to the left and tau(0,0,0,0) to the right:
> |
> | 2 * tau(80,0,0,16) = tau(0,0,0,0) + tau(60,0,0,20)
> |
> | Divide by 2:
> |
> | tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
> |
> | This is it.
>
> So half of 80 is 60 and half of 20 is 16
> Well done.
I will say this for you, Androcles.
You never cease to amaze.
Keep it up, the sky is the limit.
I am sure you can do even better.
Paul
Androcles
"Paul B. Andersen" <paul.b....@hia.no> wrote in message
news:1128434842.6...@g47g2000cwa.googlegroups.com...
"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
we find the Paul B. Andersen transforms are
t' = (t+xv/c^2)/sqrt(1-v^2/c^2)
x' = (x+vt) / sqrt(1-v^2/c^2)
which are the OTHER cuckoo transforms and which also approximate the
Galilean transform x' = x+vt.
They are of course blatantly obvious, quite, exactly, prove me wrong.
I will say this for you, tusselad.
You never cease to amaze.
Keep it up, the sky is the limit.
I am sure you can do even better.
Oh wait, you did:
Algol is a B8 and K2, the K2 being a frisbee that bounces off the
surface
off something. Roberts has observed the K2 near a black hole.
Androcles
David McAnally
>"Androcles" <Androcles@ MyPlace.org> wrote in message news:WqRYe.278$Am6...@fe3.news.blueyonder.co.uk...
>[snip]
>> Answer: Because you are infinitely stupid.
>Infinitely Stupid Revisited:
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/InfinitelyStupid2.html
Here are some of the facts that I have recently picked up about Androcles.
Androcles does not like changes in notation. Specifically, Androcles does
not like the fact that we use x' to denote the variable that Einstein
denoted by \xi, and that we have no symbol for the variable that Einstein
denoted by x'. In fact, Androcles is so averse to the fact that we have
no symbol for the variable that Einstein denoted by x', that he will
pretend that x' is being simultaneously used to denote both variables
(our usage and Einstein's usage), and then use this to attempt to show a
contradiction in relativity (of course, this is very dishonest, but such
dishonesty on Androcles' part should not surprise) - and he is willing to
do this, even after being explicitly informed that we have no symbol for
the variable denoted by Einstein as x', and therefore that any such
pretence holds no water.
Androcles' persistent claim that mathematicians "can't count to three" is
because he is upset at the reduction of what were effectively Einstein's
three sets of coordinates ((x,y,z,t), (x',y,z,t), (\xi,\eta,\zeta,\tau),
in the notation that he used in his 1905 paper) to two sets of coordinates
((x,y,z,t), (x',y',z',t'), in the notation that we use now). He also
finds his slur about mathematicians not being able to count to three
useful, since he thinks he can keep repeating it like a mantra in spite of
anything that is said to him, and as a consequence, he thinks that it will
allow him to evade having to answer the hard questions. In other words,
he finds it easier to insult others than to answer questions, and so he
will continue to insult, rather than answer, no matter what anybody says
to him.
Androcles also has a strange attitude to the notation x'. He has, as much
as, said that x' is a notation for a function of x, and that the symbol 1'
denotes the value taken by x' when x = 1. This is in spite of being told,
more than once, that x' is the indivisible symbol for a variable, for
example, in a paragraph such as below:
:: x, y, z, and t are variables which take values in the real numbers
:: (such as 0, 1, 2, 3, 8.5, etc). x' is the symbol for ONE variable
:: (which is denoted by the entire string, x') which takes values in the
:: real numbers (such as 0, 1, 2, 3, 8.5, etc). 0', 1', 2', 3', 8.5', are
:: not known symbols for any numbers - they are just ordinary numbers with
:: primes attached to them for affectation. The equality x' = 1 means
:: something concrete (just as the equalities x = 1 and y = 1 mean
:: something concrete). The formula x' = 1' doesn't mean a thing since
:: nobody has ever told us what 1' represents. These considerations are
:: independent of anything that you could ever say about coordinate
:: systems.
Another thing that I have noticed is that Androcles expresses the
two-argument function f(x,t) = x-vt as a function of one argument alone
(i.e. he will write f(x) = x-vt without the least concern about the
nontrivial dependence of the right-hand side on t). He suggested that I
did not know what a frame or a function was. I gave the following
definition of function in response, along with the observation that his
mathematical education is probably too deficient for him to have seen such
a definition, and urging him to look it up before spewing venom in
response (advice which I have no doubt he ignored).
:: A function is a set of ordered pairs in which no two distinct ordered
:: pairs have the same first component (i.e. a function f is a set of
:: ordered pairs such that for all x, y, z, if (x,y) and (x,z) are both
:: elements of f, then y = z; for any x in the domain of f (i.e. any x
:: which occurs in the first component of one of the elements of f), f(x)
:: is uniquely defined by the requirement that (x,f(x)) is an element of
:: f).
I have also noted that he has a strong impulse to automatically assume
that all functions are functions from R to R (i.e. he has a strong impulse
to assume that, for any function f, the domain of f is R and the range of
f is a subset of R). That not all functions are functions from R to R would
no doubt shock him.
As a consequence, he does not recognize the fact that a Lorentz
Transformation is a bijective function from R^4 to R^4 (as a function from
R^4 to R^4, a Lorentz Transformation is a four-component object, each
component of which is a real-valued function of four arguments). Further,
there is a function mapping (-c,c) to the space of Lorentz
Transformations, with the dependent variable v being treated as a
parameter of the Transformation.
I also specified that an inertial frame (in special relativity) is a
coordinate system in which inertial motion is represented by a linear
relationship between the coordinates.
It is of interest to go through the relevant part of Einstein's
derivation, just to see what is happening. This is not to show Androcles
what is happening, or for his benefit, since Androcles will just ignore
the reasoning here. The following is for independent intellectual
interest for those who actually use their brains for thinking.
After Einstein sets up his frames K (with coordinates (x,y,z,t)) and k
(with coordinates (\xi,\eta,\zeta,\tau)), he notes that a point which is
stationary in k has constant values for the variables x', y, z, where
x' = x-vt. Since x', y, z, t, are sufficient to determine a specific
event in spacetime, then \tau is expressible as a function of four
arguments, \tau(x',y,z,t), with dependence on an additional parameter v
(v is treated as a parameter, since v determines the relation between
frames K and k, i.e. as v changes, so does the frame k).
We can now take two clocks which are stationary relative to frame k, clock
1, for which x' = 0, y = 0, z = 0, for all time, and clock 2, for which
x' = X, y = 0, z = 0, for all time. If the clocks are showing time \tau,
then they are synchronized by the rule that Einstein gave in section 1.
Light leaves clock 1 at time t, relative to K, when clock 1 shows time
\tau_0, is reflected at clock 2, when clock 2 shows time \tau_1, and
returns to clock 1 when clock 1 shows time \tau_2. By the fact that the
clocks are stationary in frame k, and are synchronized in frame k, then
(\tau_0 + \tau_2)/2 = tau_1. Note that \tau_0 = \tau(0,0,0,t).
If X > 0, then the light is reflected at clock 2 at time t + X/(c-v),
relative to K.
This can be seen as follows: the light leaves clock 1 at x = vt, y = 0,
z = 0 at time t, and arrives at clock 2 at time t + T_1, for some T_1.
Since the light arrives at clock 2 at time t + T_1, then the coordinates
of clock 2 in K are x = vt + v T_1 + X, y = 0, z = 0, at that time.
Since light travels at speed c, then the coordinates are x = vt + c T_1,
y = 0, z = 0, at that time. It follows that v T_1 + X = c T_1, and
therefore that T_1 = X/(c-v).
It follows that \tau_1 = \tau(X,0,0,t + X/(c-v)).
By a similar argument, the light arrives back at clock 1 at time
t + X/(c-v) + X/(c+v), relative to K. It follows that
\tau_2 = \tau(0,0,0,t + X/(c-v) + X/(c+v)).
Since \tau_1 = (1/2) (\tau_0 + \tau_2), since the clocks are synchronized
in frame k, then
\tau(X,0,0,t + X/(c-v))
= (1/2) (\tau(0,0,0,t) + \tau(0,0,0,t + X/(c-v) + X/(c+v)).
Take the derivative of both sides with respect to X, and then set X = 0:
d\tau/dx' + (1/(c-v)) d\tau/dt = (1/2) (1/(c-v) + 1/(c+v)) d\tau/dt,
and so
d\tau/dx' = - (1/2) (1/(c-v) - 1/(c+v)) d\tau/dt = - v/(c^2-v^2) d\tau/dt.
On the other hand, if X < 0, then the light is reflected at clock 2 at
time t - X/(c+v), relative to K.
It follows that \tau_1 = \tau(X,0,0,t - X/(c+v)).
By a similar argument, the light arrives back at clock 1 at time
t - X/(c+v) - X/(c-v), relative to K. It follows that
\tau_2 = \tau(0,0,0,t - X/(c+v) - X/(c-v)).
Since \tau_1 = (1/2) (\tau_0 + \tau_2), since the clocks are synchronized
in frame k, then
\tau(X,0,0,t - X/(c+v))
= (1/2) (\tau(0,0,0,t) + \tau(0,0,0,t - X/(c+v) - X/(c-v)).
Take the derivative of both sides with respect to X, and then set X = 0:
d\tau/dx' - (1/(c+v)) d\tau/dt = - (1/2) (1/(c-v) + 1/(c+v)) d\tau/dt,
and so
d\tau/dx' = - (1/2) (1/(c-v) - 1/(c+v)) d\tau/dt = - v/(c^2-v^2) d\tau/dt.
In both cases, we reach the conclusion that
d\tau/dx' + (v/(c^2-v^2)) d\tau/dt = 0.
We now take the new case of two clocks which are stationary in frame k,
clock 1, for which x' = 0, y = 0, z = 0, for all time, and clock 2, for
which x' = 0, y = Y, z = 0, for all time. If the clocks are showing time
\tau, then they are synchronized by the rule that Einstein gave in section
1.
Light leaves clock 1 at time t, relative to K, when clock 1 shows time
\tau_0, is reflected at clock 2, when clock 2 shows time \tau_1, and
returns to clock 1 when clock 1 shows time \tau_2. By the fact that the
clocks are synchronized in frame k, then (\tau_0 + \tau_2)/2 = tau_1.
Note that \tau_0 = \tau(0,0,0,t).
If Y > 0, then the light is reflected at clock 2 at time
t + Y/sqrt(c^2-v^2), relative to K.
This can be seen as follows: the light leaves clock 1 at x = vt, y = 0,
z = 0 at time t, and arrives at clock 2 at time t + T_1, for some T_1.
Since the light arrives at clock 2 at time t + T_1, then the coordinates
of clock 2 in K are x = vt + v T_1, y = Y, z = 0, at that time.
Since light travels at speed c, then light travels a distance of c T_1 in
that time, and so c^2 T_1^2 = v^2 T_1^2 + Y^2, and so
T_1^2 = Y^2/(c^2-v^2).
It follows that \tau_1 = \tau(0,Y,0,t + Y/sqrt(c^2-v^2)).
By a similar argument, the light arrives back at clock 1 at time
t + 2 Y/sqrt(c^2-v^2, relative to K. It follows that
\tau_2 = \tau(0,0,0,t + 2 Y/sqrt(c^2-v^2)).
Since \tau_1 = (1/2) (\tau_0 + \tau_2), since the clocks are synchronized
in frame k, then
\tau(0,Y,0,t + Y/sqrt(c^2-v^2))
= (1/2) (\tau(0,0,0,t) + \tau(0,0,0,t + 2 Y/sqrt(c^2-v^2)).
Take the derivative of both sides with respect to Y, and then set Y = 0:
d\tau/dy + (1/sqrt(c^2-v^2)) d\tau/dt = (1/sqrt(c^2-v^2)) d\tau/dt,
and so d\tau/dy = 0.
Similarly, d\tau/dz = 0, and so
\tau(x',y,z,t) = f(t - (v/(c^2-v^2)) x'),
for some function f.
\tau is a linear function, by which is meant that \tau is a linear
function of its arguments - since the only requirement is a linear
relation between the coordinates of K and the coordinates of k, then there
is no requirement that \tau be a linear function of the parameter v, as
different values of v determine different values of the frame k. It
follows that
\tau = a (t - (v/(c^2-v^2)) x'),
for some constant a, whose value may depend on the parameter v, where for
the origin of k, \tau = 0 when t = 0.
If light propagates in the positive x-direction, so that x = ct, then
\xi = c \tau (assuming that the origin of K coincides with the origin of k
when \tau = 0). It follows that for such a light ray,
\xi = a c (t - (v/(c^2-v^2) x').
Since light travels at speed c in frame K, then
t = x'/(c-v),
and so
\xi = a (c^2/(c^2-v^2)) x'.
Since x', y,z, are constant for any object which is stationary in k, then
this is true for all values of \xi.
If light propagates in the positive \eta-direction from the origin in
frame k at time \tau = 0, then
\eta = c \tau = a c (t - (v/(c^2-v^2)) x').
Since light travels at speed c in frame K, then
x' = 0, y = sqrt(c^2-v^2) t,
so that x' = 0, t = y/sqrt(c^2-v^2), and so \eta = a c y/sqrt(c^2-v^2).
Since x', y, z, are constant for any object which is stationary in k, then
this is true for all values of \eta.
Similarly, \zeta = a c z/sqrt(c^2-v^2).
If we set \phi(v) = a/sqrt(1-v^2/c^2), so that a = \phi(v) sqrt(1-v^2/c^2),
then
\tau = a (c^2 t - v x)/(c^2-v^2)
= \phi(v) (t - vx/c^2)/sqrt(1-v^2/c^2),
\xi = \phi(v) (x-vt)/sqrt(1-v^2/c^2),
\eta = \phi(v) y,
\zeta = \phi(v) z.
Since K is travelling at velocity -v in the \eta-direction, relative to k,
then
t = \phi(-v) (\tau + v \xi/c^2)/sqrt(1-v^2/c^2)
= \phi(-v) \phi(v) t,
x = \phi(-v) (\xi + v \tau)/sqrt(1-v^2/c^2) = \phi(-v) \phi(v) x,
y = \phi(-v) \eta = \phi(-v) \phi(v) y,
z = \phi(-v) \zeta = \phi(-v) \phi(v) z.
It follows that \phi(-v) \phi(v) = 1.
The interpretation of \phi(v) is now seen by taking a stationary rod of
length L in the frame k, between \xi = 0, \eta = 0, \zeta = 0, and
\xi = 0, \eta = L, \zeta = 0 (so that the rod is moving relative to K in a
direction perpendicular to itself). It follows that in frame K at time t,
the rod is between x = vt, y = 0, z = 0, and x = vt, y = L/\phi(v), z = 0,
and so the length of the rod relative to K is L/\phi(v). That is, if a
rod of rest length L is moving, relative to K, at speed v in a direction
perpendicular to itself, then its length in K is L/\phi(v). This result
is independent of what direction perpendicular to the rod that the rod is
travelling, so that L/\phi(v) = L/\phi(-v), so 1/\phi(v) = 1/\phi(-v),
and so \phi(v) = \phi(-v).
Since \phi(v) = \phi(-v) and \phi(v) \phi(-v) = 1 and \phi(v) is positive,
the \phi(v) = 1. It follows that
\tau = (t - v c/c^2)/sqrt(1-v^2/c^2),
\xi = (x-vt)/sqrt(1-v^2/c^2),
\eta = y,
\zeta = z.
-----
Androcles
"David McAnally" <D.McAnally@i'm_a_gnu.uq.net.au> wrote in message
news:dhu7ut$8be$1...@bunyip2.cc.uq.edu.au...
| "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
writes:
|
| >"Androcles" <Androcles@ MyPlace.org> wrote in message
news:WqRYe.278$Am6...@fe3.news.blueyonder.co.uk...
|
| >[snip]
|
| >> Answer: Because you are infinitely stupid.
|
| >Infinitely Stupid Revisited:
| >
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/InfinitelyStupid2.html
|
| Here are some of the facts that I have recently picked up about
Androcles.
|
| Androcles does not like changes in notation. Specifically, Androcles
does
| not like the fact that we use x' to denote the variable that Einstein
| denoted by \xi, and that we have no symbol for the variable that
Einstein
| denoted by x'.
x' = x-vt, from
"If we place x'=x-vt, it is clear that a point at rest in the system k
must have a system of values x', y, z, independent of time."
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
\xi = x'/sqrt(1-v^2/c^2)
Specifically, Androcles does not like the fact that YOU use x' to denote
the variable that Einstein denoted by \xi, since they are not the same,
you infinitely stooopid cunt.
Androcles
Dirk Van de moortel
"Androcles" <Androcles@ MyPlace.org> wrote in message news:gSx0f.53353$VI6....@fe1.news.blueyonder.co.uk...
>
> "David McAnally" <D.McAnally@i'm_a_gnu.uq.net.au> wrote in message
> news:dhu7ut$8be$1...@bunyip2.cc.uq.edu.au...
> | "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
> writes:
> |
> | >"Androcles" <Androcles@ MyPlace.org> wrote in message
> news:WqRYe.278$Am6...@fe3.news.blueyonder.co.uk...
> |
> | >[snip]
> |
> | >> Answer: Because you are infinitely stupid.
> |
> | >Infinitely Stupid Revisited:
> | >
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/InfinitelyStupid2.html
> |
> | Here are some of the facts that I have recently picked up about
> | Androcles.
> |
> | Androcles does not like changes in notation. Specifically, Androcles
> | does
> | not like the fact that we use x' to denote the variable that Einstein
> | denoted by \xi, and that we have no symbol for the variable that
> | Einstein denoted by x'.
Actually, Androcles does not understand the concept of
coordinates, so he has no chance of ever understanding
any of equations involving coordinates.
And alas, to make it even worse, he doesn't even grasp
the concept of "variables in equations" as he proves once
more right here:
>
> x' = x-vt, from
>
> "If we place x'=x-vt, it is clear that a point at rest in the system k
> must have a system of values x', y, z, independent of time."
> Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
>
> \xi = x'/sqrt(1-v^2/c^2)
>
> Specifically, Androcles does not like the fact that YOU use x' to denote
> the variable that Einstein denoted by \xi, since they are not the same,
> you infinitely stooopid cunt.
> Androcles
Sometimes he pretends to be stupid on purpose, but even
that he can't properly manage. I honestly think it is a case
of clear cut brain rot.
Dirk Vdm
JanPB
> "JanPB" <fil...@gmail.com> wrote in message
> news:1128427205.0...@z14g2000cwz.googlegroups.com...
> | Androcles wrote:
> | > "JanPB" <fil...@gmail.com> wrote in message
> | > news:1128420620....@o13g2000cwo.googlegroups.com...
> | > | tau(80,0,0,16) - tau(0,0,0,0) = tau(60,0,0,20) - tau(80,0,0,16)
> | > |
> | > | When you collect the terms you get the 1/2.
> | >
> | > Prove it.
> |
> | I don't quite see why you ask such an obvious question be here it goes
> | anyway:
> |
> | Starting with:
> |
> | tau(80,0,0,16) - tau(0,0,0,0) = tau(60,0,0,20) - tau(80,0,0,16)
> |
> | Move tau(80,0,0,16) to the left and tau(0,0,0,0) to the right:
> |
> | 2 * tau(80,0,0,16) = tau(0,0,0,0) + tau(60,0,0,20)
> |
> | Divide by 2:
> |
> | tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
> |
> | This is it.
>
> So half of 80 is 60 and half of 20 is 16
> Well done.
Do you know the difference between:
(80,0,0,16)
and:
tau(80,0,0,16) ?
Define both.
> | > So what if it takes a turtle 16 seconds to go from 0 to 80 at speed
> 5
> | > and 4 seconds from 80 to 60 at velocity -5?
> | > Still, if you think 20/2 = 16 you are as crazy as Hammond <shrug>
> |
> | The 16 and 4 seconds are times as measured by the stationary clocks
> | (t).
>
> Yes ducky, and the moving train has fuck-all to do with that.
>
> The corresponding readings by the clocks sitting on the train
> | (tau) are different than those t-readings and they are equal in both
> | directions.
>
> There is no moving train in your equation. It doesn't exist.
> No sign of it anywhere.
Complete nonsense. The tau equation expresses a constraint between the
stationary coordinates and one moving coordinate (namely, tau). Without
either of those systems there is no tau equation.
> Light leaves 0, goes to 80, takes 16 seconds.
> Light leaves 80, goes to 60, takes 4 seconds.
> No train involved.
The train clocks are involved. What do you think "tau" means in that
equation? I suppose you could get rid of the train and just assume the
clocks are levitating or something - that's fine with me. But the
mental picture of the train is a convenient shortcut.
> | > Crap. You have no idea what you are babbling about.
> |
> | Could you back it up. I simply stated the starting point of the
> | derivation argument.
>
> No you didn't. You gave track coordinates and track time.
> The train is in the siding, not moving on the track.
> Light leaves 0, goes to 80, takes 16 seconds, velocity of light 5.
> Light leaves 80, goes to 60, takes 5 seconds, velocity of light -5.
>
> 16/20 * [ f(0,0,0,0) + f(60,0,0, 80/5 + (60-80)/-5)] = f(80,0,0,16)
> No train, f is linear, no half either.
Nope. The point is that these times (16 seconds and 4 seconds) are in
terms of the stationary clocks which is *not* what tau(...) are. They
are equal to certain *moving* clock readings. At the beginning of the
derivation we don't know these values - all we know at first is that
they are constrained by the tau equation. It turns out this constraint
is strong enough to produce the formula for these moving clock
readings.
> f(t) = t
Are you saying that:
tau(80,0,0,16) = 16
tau(0,0,0,0) = 0
tau(60,0,0,20)) = 20 ?
> | > | My only point for writing exclusively in terms of (x,y,z,t) and
> | > | (xi,eta,zeta,tau) was to mention that the third system (x',y,z,t)
> is
> | > | not *in principle* important and the derivation is 95% identical
> | > | without it.
> | >
> | > Bullshit. You are babbling.
> |
> | No, I can write both derivations side by side and the differences are
> | small.
>
> No you can't. You think 16 is half of 20 and 60 is half of 80.
No, what I think is:
tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
...which is a *totally* different thing than "16 is half of 20 and 60
is half of 80".
--
Jan Bielawski
JanPB
> JanPB says...
>
> Jan, this is incredibly painful to watch.
It's interesting to me to see how far he can go with this. Right now we
are reduced to elementary arithmetic (b-a=c-b implying 2b=a+c) and
elements of functions. With a bit more work we'll get down to the
nitty-gritty of the empty set soon.
> Androcles isn't
> paying any attention to what you are saying. He's only
> calling you names and ignoring what you say.
But I'm having all the fun :-)
--
Jan Bielawski
JanPB
> >
> > "JanPB" <fil...@gmail.com> wrote in message
> > news:1128427205.0...@z14g2000cwz.googlegroups.com...
> > | No, I can write both derivations side by side and the differences are
> > | small.
> >
> > No you can't. You think 16 is half of 20 and 60 is half of 80.
>
> No, you think he thinks 16 is half of 20 and 80 is half of 60.
I think so too. (I mean I also think that he thinks that I think that
16 is half of 20 and 80 is half of 60.)
--
Jan Bielawski
Dirk Van de moortel
"JanPB" <fil...@gmail.com> wrote in message news:1128496738.4...@g49g2000cwa.googlegroups.com...
:-)
So you see, he can't even properly do *faulty* algebra.
This is really brilliant
Dirk Vdm
Dirk Van de moortel
"JanPB" <fil...@gmail.com> wrote in message news:1128496606.6...@f14g2000cwb.googlegroups.com...
Yes, but obviously he thinks that you merely *think* that
you are having all the fun, while he is *really* having it.
I think that's very interesting and worth watching :-)
Dirk Vdm
Androcles
Yes.
|
| Define both.
(80,0,0,16) -A vector (as long as the 16 doesn't represent time, which
is not a vector)
tau(80,0,0,16) - A single-valued function in four variables.
Examples of a vector:
(0,0,0,0) + (60,0,0,20) = (60,0,0,20)
2*(40,0,0,8) = (80,0,0,16)
Examples of a phuckwit: 2*(60,0,0,20) = (80,0,0,16)
2* (80,0,0,16) = (60,0,0,20)
Example of a single-valued function in 4 variables:
$20 = price(apple+orange+cherry,$16*retail/wholesale)
Do you know the difference between a phuckwit and a mathematician?
Define both. (Not possible for a phuckwit)
| > | > So what if it takes a turtle 16 seconds to go from 0 to 80 at
speed
| > 5
| > | > and 4 seconds from 80 to 60 at velocity -5?
| > | > Still, if you think 20/2 = 16 you are as crazy as Hammond
<shrug>
| > |
| > | The 16 and 4 seconds are times as measured by the stationary
clocks
| > | (t).
| >
| > Yes ducky, and the moving train has fuck-all to do with that.
| >
| > The corresponding readings by the clocks sitting on the train
| > | (tau) are different than those t-readings and they are equal in
both
| > | directions.
| >
| > There is no moving train in your equation. It doesn't exist.
| > No sign of it anywhere.
|
| Complete nonsense. The tau equation expresses a constraint between the
| stationary coordinates and one moving coordinate (namely, tau).
Without
| either of those systems there is no tau equation.
Stationary coordinates:
Emission (0,0,0,0)
Reflection (80,0,0,16)
Reception (60,0,0,20)
Moving coordinates:
Emission (0,0,0,0)
Reflection (32,0,0,16)
Reception (0,0,0,20)
You don't have a 32 anywhere.
So you don't have a tau equation <shrug>.
You ARE simply insane. You know practically zero about the subject and
yet you write... THIS?? Why even do you waste your time on this?
|
| > Light leaves 0, goes to 80, takes 16 seconds.
| > Light leaves 80, goes to 60, takes 4 seconds.
| > No train involved.
|
| The train clocks are involved. What do you think "tau" means in that
| equation?
Time of the stationary system.
| I suppose you could get rid of the train and just assume the
| clocks are levitating or something - that's fine with me. But the
| mental picture of the train is a convenient shortcut.
Locking you in a padded cell would be convenient shortcut.
|
| > | > Crap. You have no idea what you are babbling about.
| > |
| > | Could you back it up. I simply stated the starting point of the
| > | derivation argument.
| >
| > No you didn't. You gave track coordinates and track time.
| > The train is in the siding, not moving on the track.
| > Light leaves 0, goes to 80, takes 16 seconds, velocity of light 5.
| > Light leaves 80, goes to 60, takes 5 seconds, velocity of light -5.
| >
| > 16/20 * [ f(0,0,0,0) + f(60,0,0, 80/5 + (60-80)/-5)] = f(80,0,0,16)
| > No train, f is linear, no half either.
|
| Nope. The point is that these times (16 seconds and 4 seconds) are in
| terms of the stationary clocks which is *not* what tau(...) are.
You don't have a point, you don't even have a neuron.
The point is that these prices ($16 and $4) are in terms of the
stationary wholesalers which is *not* what retail price(apple, orange)
are.
I dont know how you survive in the real world.
| They
| are equal to certain *moving* clock readings.
You don't have ANYTHING moving, not a 32 in sight.
At the beginning of the
| derivation we don't know these values
Yes we do.
Stationary coordinates:
Emission (0,0,0,0)
Reflection (80,0,0,16)
Reception (60,0,0,20)
Moving coordinates:
Emission (0,0,0,0)
Reflection (32,0,0,16)
Reception (0,0,0,20)
You've left out
Reflection (32,0,0,16)
You are simply insane. You know practically zero about the subject and
yet you write... THIS?? Why even do you waste your time on this?
It's WAY too simple-minded.-- Bielawski.
- all we know at first is that
| they are constrained by the tau equation. It turns out this constraint
| is strong enough to produce the formula for these moving clock
| readings.
|
| > f(t) = t
|
| Are you saying that:
|
| tau(80,0,0,16) = 16
| tau(0,0,0,0) = 0
| tau(60,0,0,20)) = 20 ?
Yes.
|
| > | > | My only point for writing exclusively in terms of (x,y,z,t)
and
| > | > | (xi,eta,zeta,tau) was to mention that the third system
(x',y,z,t)
| > is
| > | > | not *in principle* important and the derivation is 95%
identical
| > | > | without it.
| > | >
| > | > Bullshit. You are babbling.
| > |
| > | No, I can write both derivations side by side and the differences
are
| > | small.
| >
| > No you can't. You think 16 is half of 20 and 60 is half of 80.
|
| No, what I think is:
|
| tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
|
| ...which is a *totally* different thing than "16 is half of 20 and 60
| is half of 80".
But you don't think. You are not capable of thinking.
You are simply insane.
Even Einstein would have said (and did say, effectively)
tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
v(x,y,z,t) = x/t
c(x,y,z,t) = x/t
v(32,0,0,16) = 32/16 = 2.
c(80,0,0,16) = 80/16 = 5.
Your equation is the raving of a lunatic.
Seek psychiatric help (and fuck off).
Androcles
Dirk Van de moortel
"Androcles" <Androcles@ MyPlace.org> wrote in message news:imP0f.73401$RW.2...@fe2.news.blueyonder.co.uk...
>
> "JanPB" <fil...@gmail.com> wrote in message
> news:1128496083.7...@g14g2000cwa.googlegroups.com...
[snip]
> | Do you know the difference between:
> |
> | (80,0,0,16)
> |
> | and:
> |
> | tau(80,0,0,16) ?
>
> Yes.
> |
> | Define both.
>
> (80,0,0,16) -A vector (as long as the 16 doesn't represent time, which
> is not a vector)
This and the entire remainder saved for eternity under the
title: "Seek psychiatric help (and fuck off)":
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SeekHelp.html
Dirk Vdm
Androcles
|
Tell McCullough I'm taking careful note of what you say.
As of right now we are at:
Einstein:
稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
,
which is (with McCullough's choice of values,
c = 5,
v = 3,
x' = 32,
in FACT)
稼tau(0,0,0,t)+tau(0,0,0, 32/2+32/8)] = tau(32,0,0,32/2) , or
稼tau(0,0,0,0)+tau(0,0,0,20)] = tau(32,0,0,16).
Bile-wacky:
稼tau(0,0,0,t)+tau(60,0,0,20)] = tau(80,0,0,16),
and I'm having all the fun (copy, paste, Google record).
It can't be painful for McCullough to watch his precious numbers
reduced to the nitty-gritty empty-headed set
{McCullough, moortel, Bielawski}, because:
"No creature smarts so little as a fool." -- Alexander Pope
Androcles
David McAnally
So Androcles thinks that because Einstein defined the variable x' = x-vt
in his paper, then we should be stuck with that notation for all time.
How ridiculous. And the laughable thing about this is that Einstein
himself, in the very same paper, after he no longer needed x' to denote
the variable x' = x-vt, subsequently used x' to denote a completely
different variable. So, in effect, Einstein himself led the way when it
came to changing the meaning of x'.
>> \xi = x'/sqrt(1-v^2/c^2)
>>
>> Specifically, Androcles does not like the fact that YOU use x' to denote
>> the variable that Einstein denoted by \xi, since they are not the same,
What aren't the same? The symbol \xi and the symbol x'? Is the objection
really that trivial?
>> you infinitely stooopid cunt.
>> Androcles
>Sometimes he pretends to be stupid on purpose, but even
>that he can't properly manage. I honestly think it is a case
>of clear cut brain rot.
True. And he obviously thinks that vulgarisms and insults are an art
form, because he keeps expressing himself in a vulgar fashion and tossing
off insults in an effort to hide his own inadequacies and ignorance of the
matter in hand.
>Dirk Vdm
-----
Dirk Van de moortel
"David McAnally" <D.McAnally@i'm_a_gnu.uq.net.au> wrote in message news:di0q1v$1p0q$1...@bunyip2.cc.uq.edu.au...
Please note that I did not write this sentence.
He was using the third person.
>
> What aren't the same? The symbol \xi and the symbol x'? Is the objection
> really that trivial?
>
> >> you infinitely stooopid cunt.
> >> Androcles
>
> >Sometimes he pretends to be stupid on purpose, but even
> >that he can't properly manage. I honestly think it is a case
> >of clear cut brain rot.
>
> True. And he obviously thinks that vulgarisms and insults are an art
> form,
Actually, whether vulgarisms and insults are an art form
highly depends on the personality of the "artist" ;-)
> because he keeps expressing himself in a vulgar fashion and tossing
> off insults in an effort to hide his own inadequacies and ignorance of the
> matter in hand.
Yes.
Actually, sometimes he deliberately browns his pants
to conceal the fact that he can't hold his water ;-)
Dirk Vdm
Androcles
Nah, only until \xi is derived. Then you can pretend it is anything you
like.
| How ridiculous. And the laughable thing about this is that Einstein
| himself, in the very same paper, after he no longer needed x' to
denote
| the variable x' = x-vt, subsequently used x' to denote a completely
| different variable.
And the proof of this ridiculous assertion from the brain rotted fuckwit
McAnally is?
| So, in effect, Einstein himself led the way when it
| came to changing the meaning of x'.
And the proof of this ridiculous assertion from the brain rotted fuckwit
McAnally is?
|
| >> \xi = x'/sqrt(1-v^2/c^2)
| >>
| >> Specifically, Androcles does not like the fact that YOU use x' to
denote
| >> the variable that Einstein denoted by \xi, since they are not the
same,
|
| What aren't the same?
x' and \xi are not the same, \xi = x'/sqrt(1-v^2/c^2) (unless
sqrt(1-v^2/c^2) = 1)
The symbol \xi and the symbol x'?
Yes. Well done, the fuckwit McAnally might catch on in a year or two.
| Is the objection
| really that trivial?
YES! It really is that trivial.
Unless sqrt(1-v^2/c^2) = 1, as it must be since there is no relative
motion between x' and \xi, because \xi = x' / sqrt(1-v^2/c^2)
How ridiculous of McAnally to think there is some velocity between
\xi and x'.
|
| >> you infinitely stooopid cunt.
| >> Androcles
|
| >Sometimes he pretends to be stupid on purpose, but even
| >that he can't properly manage. I honestly think it is a case
| >of clear cut brain rot.
|
| True. And he obviously thinks that vulgarisms and insults are an art
| form, because he keeps expressing himself in a vulgar fashion and
tossing
| off insults in an effort to hide his own inadequacies and ignorance of
the
| matter in hand.
I learnt that trick from Dinky Vdm and Uncle Al a long time ago.
They are both stupid trolls.
> > "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote
> > in message news:2EVEb.87288$6K2.3...@phobos.telenet-ops.be...
> > some garbage to propagate his smear campaign.
> >
> > So you really don't want to discuss relativity, then, Dinky?
>
> Discuss relativity?
> With a load of crap like you?
>
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/LoadCrap.html
> HAHAHAHAHAHAHAHAHAHAHAHAHA!
> You are about as funny as Wilson Rabbidge!
>
> Dirk Vdm
A fuckwit troll like McAnally is too stooopid to know who the trolls
really are.
Androcles.
David McAnally
Sorry. I did not mean to give the wrong impression. I was aware that
Androcles wrote the comment to which you are referring. My question was
addressed to Androcles.
>> What aren't the same? The symbol \xi and the symbol x'? Is the objection
>> really that trivial?
>>
>> >> you infinitely stooopid cunt.
>> >> Androcles
>>
>> >Sometimes he pretends to be stupid on purpose, but even
>> >that he can't properly manage. I honestly think it is a case
>> >of clear cut brain rot.
>>
>> True. And he obviously thinks that vulgarisms and insults are an art
>> form,
>Actually, whether vulgarisms and insults are an art form
>highly depends on the personality of the "artist" ;-)
>> because he keeps expressing himself in a vulgar fashion and tossing
>> off insults in an effort to hide his own inadequacies and ignorance of the
>> matter in hand.
>Yes.
>Actually, sometimes he deliberately browns his pants
>to conceal the fact that he can't hold his water ;-)
:-)
>Dirk Vdm
-----
JanPB
> As of right now we are at:
> Einstein:
> ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
> ,
>
> which is (with McCullough's choice of values,
> c = 5,
> v = 3,
> x' = 32,
> in FACT)
>
> ½[tau(0,0,0,t)+tau(0,0,0, 32/2+32/8)] = tau(32,0,0,32/2) , or
>
> ½[tau(0,0,0,0)+tau(0,0,0,20)] = tau(32,0,0,16).
>
> Bile-wacky:
>
> ½[tau(0,0,0,t)+tau(60,0,0,20)] = tau(80,0,0,16),
Both equations are correct - they simply involve different functions -
both of which you denoted by the same symbol "tau".
Normally people don't use plugged-in numbers but general expressions of
the given variables so this type of purely notational confusion is
avoided.
(Incidentally, I had t=0 along with x=32, etc.)
--
Jan Bielawski
Androcles
Androcles wrote:
> As of right now we are at:
> Einstein:
> 絒tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] =
> tau(x',0,0,t+x'/(c-v))
> ,
>
> which is (with McCullough's choice of values,
> c = 5,
> v = 3,
> x' = 32,
> in FACT)
>
> 絒tau(0,0,0,t)+tau(0,0,0, 32/2+32/8)] = tau(32,0,0,32/2) , or
>
> 絒tau(0,0,0,0)+tau(0,0,0,20)] = tau(32,0,0,16).
>
> Bile-wacky:
>
> 絒tau(0,0,0,t)+tau(60,0,0,20)] = tau(80,0,0,16),
| Both equations are correct - they simply involve different functions -
| both of which you denoted by the same symbol "tau".
ROFLMAO!
You do struggle to be convincing, Bile-wacky!
You are as crazy as Hammond, he's "proven" his god exists.
| Normally people don't use plugged-in numbers but general expressions
of
| the given variables so this type of purely notational confusion is
| avoided.
| (Incidentally, I had t=0 along with x=32, etc.)
x = 32, eh? ROFLMAO!!!
Let you into a little secret, Bile-wacky. I often leave little clues
like
"絒tau(0,0,0,t)+... " instead of "絒tau(0,0,0,0)+" to see anyone will
notice.
Go ahead, Bile-wacky! "Derive" the wacky transforms from
the Bile-wacky tau equations, let's have more laughs.
The fuckwits mad moortel, Phuckwit Duck, Maniacal McCullough,
McAnally the Insane and all the other incompetent silly bastards
think x' is \xi, you think it is x. Go on, give us another class act.
Prove it to them, keep working, we'll get down to the nitty-gritty of
the empty set soon. Have more fun, I know you are enjoying yourself
almost as much as I am :-)
LOL!
Androcles.
JanPB
> "JanPB" <fil...@gmail.com> wrote in message
> news:1128542389.7...@o13g2000cwo.googlegroups.com...
> Androcles wrote:
>
> > As of right now we are at:
> > Einstein:
> > tau(x',0,0,t+x'/(c-v))
> > ,
> >
> > which is (with McCullough's choice of values,
> > c = 5,
> > v = 3,
> > x' = 32,
> > in FACT)
> >
> >
> > ½[tau(0,0,0,0)+tau(0,0,0,20)] = tau(32,0,0,16).
> >
> > Bile-wacky:
> >
>
> | Both equations are correct - they simply involve different functions -
> | both of which you denoted by the same symbol "tau".
>
> ROFLMAO!
> You do struggle to be convincing, Bile-wacky!
> You are as crazy as Hammond, he's "proven" his god exists.
Both equations are correct - they simply involve different functions.
You used the same symbol "tau" for two different things:
THING NUMBER ONE: tau(x',y,z,t)
THING NUMBER TWO: tau(x,y,z,t)
The first equation:
½[tau(0,0,0,0)+tau(0,0,0,20)] = tau(32,0,0,16)
...refers to tau(x',y,z,t).
The second equation:
½[tau(0,0,0,0)+tau(60,0,0,20)] = tau(80,0,0,16)
...refers to tau(x,y,z,t).
> | Normally people don't use plugged-in numbers but general expressions
> of
> | the given variables so this type of purely notational confusion is
> | avoided.
>
> | (Incidentally, I had t=0 along with x=32, etc.)
>
> x = 32, eh? ROFLMAO!!!
My mistake, I meant to say just "I had t=0".
> Let you into a little secret, Bile-wacky. I often leave little clues
> like
> "½[tau(0,0,0,t)+... " instead of "½[tau(0,0,0,0)+" to see anyone will
> notice.
And the point of this curious little exercise is...?
> Go ahead, Bile-wacky! "Derive" the wacky transforms from
> the Bile-wacky tau equations, let's have more laughs.
Sure, very easy to type it in. But dinner first.
> The fuckwits mad moortel, Phuckwit Duck, Maniacal McCullough,
> McAnally the Insane and all the other incompetent silly bastards
> think x' is \xi,
No, in fact they have just posted specifically stating the opposite.
Also, they'd be organically unable to say anything like you suggest
because they understand the material.
> you think it is x.
Whatever gave you that idea? In fact I spent last several posts
(including this one) doing exactly the opposite: explaining the
difference between the two uses of tau (namely, tau(x',y,z,t) and
tau(x,y,z,t)), explaining the difference between the systems (x',y,z,t)
and (x,y,z,t), between the clocks they use, and so on. How all this can
be misconstrued to mean its total opposite ("you think [x'] is x")
remains a mystery. Any takers?
--
Jan Bielawski
JanPB
> "JanPB" <fil...@gmail.com> wrote in message
> news:1128496083.7...@g14g2000cwa.googlegroups.com...
> | Androcles wrote:
> | > "JanPB" <fil...@gmail.com> wrote in message
> | > news:1128427205.0...@z14g2000cwz.googlegroups.com...
> | > |
> | > | I don't quite see why you ask such an obvious question be here it
> goes
> | > | anyway:
> | > |
> | > | Starting with:
> | > |
> | > | tau(80,0,0,16) - tau(0,0,0,0) = tau(60,0,0,20) - tau(80,0,0,16)
> | > |
> | > | Move tau(80,0,0,16) to the left and tau(0,0,0,0) to the right:
> | > |
> | > | 2 * tau(80,0,0,16) = tau(0,0,0,0) + tau(60,0,0,20)
> | > |
> | > | Divide by 2:
> | > |
> | > | tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
> | > |
> | > | This is it.
> | >
> | > So half of 80 is 60 and half of 20 is 16
> | > Well done.
> |
> | Do you know the difference between:
> |
> | (80,0,0,16)
> |
> | and:
> |
> | tau(80,0,0,16) ?
>
> Yes.
> |
> | Define both.
>
> (80,0,0,16) -A vector (as long as the 16 doesn't represent time, which
> is not a vector)
> tau(80,0,0,16) - A single-valued function in four variables.
That's just a generic definition. I meant physically, in this physical
setup with the light emitter, mirror, clocks, etc.
> Examples of a vector:
> (0,0,0,0) + (60,0,0,20) = (60,0,0,20)
> 2*(40,0,0,8) = (80,0,0,16)
Correct.
> Examples of a phuckwit: 2*(60,0,0,20) = (80,0,0,16)
> 2* (80,0,0,16) = (60,0,0,20)
Incorrect. Needless to say, I never wrote such a thing. Do you see the
difference between:
2* (80,0,0,16) = (60,0,0,20)
and:
2 * tau(80,0,0,16) = tau(0,0,0,0) + tau(60,0,0,20)
(Hint: the difference is huge, like night and day.)
> | > The corresponding readings by the clocks sitting on the train
> | > | (tau) are different than those t-readings and they are equal in
> both
> | > | directions.
> | >
> | > There is no moving train in your equation. It doesn't exist.
> | > No sign of it anywhere.
> |
> | Complete nonsense. The tau equation expresses a constraint between the
> | stationary coordinates and one moving coordinate (namely, tau).
> Without
> | either of those systems there is no tau equation.
>
> Stationary coordinates:
> Emission (0,0,0,0)
> Reflection (80,0,0,16)
> Reception (60,0,0,20)
>
> Moving coordinates:
> Emission (0,0,0,0)
> Reflection (32,0,0,16)
> Reception (0,0,0,20)
Correct, except "moving" here means the (x',y,z,t) one, not the one we
are really after, namely (xi,eta,zeta,tau). In *that* moving system we
have (using the auxiliary moving coordinates as Einstein did it):
Emission: (xi(0,0,0,0),eta(0,0,0,0),zeta(0,0,0,0),tau(0,0,0,0))
Reflection:
(xi(32,0,0,16),eta(32,0,0,16),zeta(32,0,0,16),tau(32,0,0,16))
Reception: (xi(0,0,0,20),eta(0,0,0,20),zeta(0,0,0,20),tau(0,0,0,20))
> | > Light leaves 0, goes to 80, takes 16 seconds.
> | > Light leaves 80, goes to 60, takes 4 seconds.
> | > No train involved.
> |
> | The train clocks are involved. What do you think "tau" means in that
> | equation?
>
> Time of the stationary system.
I think you should go and reread Einstein's paper. If you don't know
what tau is I see little point in continuing.
> | > | > Crap. You have no idea what you are babbling about.
> | > |
> | > | Could you back it up. I simply stated the starting point of the
> | > | derivation argument.
> | >
> | > No you didn't. You gave track coordinates and track time.
> | > The train is in the siding, not moving on the track.
> | > Light leaves 0, goes to 80, takes 16 seconds, velocity of light 5.
> | > Light leaves 80, goes to 60, takes 5 seconds, velocity of light -5.
> | >
> | > 16/20 * [ f(0,0,0,0) + f(60,0,0, 80/5 + (60-80)/-5)] = f(80,0,0,16)
> | > No train, f is linear, no half either.
> |
> | Nope. The point is that these times (16 seconds and 4 seconds) are in
> | terms of the stationary clocks which is *not* what tau(...) are.
>
> You don't have a point, you don't even have a neuron.
> The point is that these prices ($16 and $4) are in terms of the
> stationary wholesalers which is *not* what retail price(apple, orange)
> are.
> I dont know how you survive in the real world.
tau is not t, nothing can change this fact.
> | They
> | are equal to certain *moving* clock readings.
>
> You don't have ANYTHING moving, not a 32 in sight.
Einstein has 3 systems, two of them are moving:
1. (x,y,z,t)
2. (x',y,z,t) - an auxiliary
3. (xi,eta,zeta,tau)
1 is stationary, 2 and 3 are moving identically but using differently
synchronised clocks. The goal of this part of Einstein's paper is to
find the transformation formula between 1 and 3.
> At the beginning of the
> | derivation we don't know these values
>
> Yes we do.
> Stationary coordinates:
> Emission (0,0,0,0)
> Reflection (80,0,0,16)
> Reception (60,0,0,20)
>
> Moving coordinates:
> Emission (0,0,0,0)
> Reflection (32,0,0,16)
> Reception (0,0,0,20)
These are not the values I was referring to. Here it is once again with
crayons:
At the beginning of the derivation we don't know the following values:
tau(32,0,0,16),
tau(0,0,0,20).
We do know that tau(0,0,0,0)=0 because that's how we set up the mirror
experiment. The only thing we know initially about tau(32,0,0,16) and
tau(0,0,0,20) is that they are some numbers which fit into the
following equation:
1/2*(tau(0,0,0,0)+tau(0,0,0,20)) = tau(32,0,0,16)
Of course to be able to derive the transform one needs the general
statement of this equation, not just some random plugged in numbers:
1/2*(tau(0,0,0,0)+tau(0,0,0,x'/(c-v)+x'/(c+v))) =
tau(x',0,0,x'/(c-v))
> - all we know at first is that
> | they are constrained by the tau equation. It turns out this constraint
> | is strong enough to produce the formula for these moving clock
> | readings.
> |
> | > f(t) = t
> |
> | Are you saying that:
> |
> | tau(80,0,0,16) = 16
> | tau(0,0,0,0) = 0
> | tau(60,0,0,20)) = 20 ?
>
> Yes.
WOW. I don't think there is a point in continuing our discussion then.
Until you understand the *description* of Einstein's experimental setup
you have no moral right to complain about anything in that paper. It's
only fair.
> | > | No, I can write both derivations side by side and the differences
> are
> | > | small.
> | >
> | > No you can't. You think 16 is half of 20 and 60 is half of 80.
> |
> | No, what I think is:
> |
> | tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
> |
> | ...which is a *totally* different thing than "16 is half of 20 and 60
> | is half of 80".
>
> But you don't think. You are not capable of thinking.
> You are simply insane.
> Even Einstein would have said (and did say, effectively)
> tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
No, he said:
tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(0,0,0,20))
Anyway, as I said earlier - the two equations express a property of
*two different functions*, so naturally they are different. You gave
two different functions the same *name* and now are complaining about
some "contradiction".
> v(x,y,z,t) = x/t
Meaningless.
> c(x,y,z,t) = x/t
Meaningless.
--
Jan Bielawski
Androcles
Oh...
Physically.. hmmm...
Well, PHYSICALLY, x is measured along the ground, say from
LA (=0) to Berlin (=80), and x' (=32) is the length of the plane
from rudder( =0) to pilot (x') when it's on the ground at the
departure gate.
After take-off and when at cruising altitude
(I'm skipping acceleration down the runway, that's GR) ,
the flight attendant comes round with the drinks trolley,
but you are already "high" on cocaine or whatever is it you
fucked your brain up with, and if it's an old plane with a
projector she puts the movie on.
Now \xi is where the pilot sits, you are in seat x', for this reason:
the speed of light has to be 5 in all frames of reference!
So the plane stretches from 32(seat x') to 40, (seat \xi)
so that we on the ground see it length contracted, or it would look
funny. Now it takes t = 8 for the light to get from the projector
to the movie screen, where the pilot is, but we keep him up front
to mind the plane and not watch the movie (he's seen it before).
You have to be behind the screen where the light reflects back
to your eyes in seat x', and it also reaches the people in coach
at the back of the plane. So the round trip time of the light
on the plane is 16, as Einstein wants.
Proof:
\xi = x' /sqrt(1-v^2/c^2)
= 32/0.8
= 40
8 * 5 = 40, which is 1/2 of 80, the distance from LA to Berlin.
\tau = t * sqrt(1-v^2/c^2)
= 20 * 0.8
=16
You may have heard of metal fatigue in older planes,
because they stretch by SR whenever the movie is shown
and so new planes no longer have movie projectors, they
use TV screens instead.
When the front of the plane reaches Berlin, it stops,
but the back of the plane is still at 40, half-way to Berlin,
so it keeps going until the plane is 32 once again, which moves
your seat x' forward to the door and you can get off.
That's what happens physically. Showing the movie
stretches the plane to 40, but we on the ground see it length
contracted to 32.
|
| > Examples of a vector:
| > (0,0,0,0) + (60,0,0,20) = (60,0,0,20)
| > 2*(40,0,0,8) = (80,0,0,16)
|
| Correct.
Yes, I know.
| > Examples of a phuckwit: 2*(60,0,0,20) = (80,0,0,16)
| > 2* (80,0,0,16) = (60,0,0,20)
|
| Incorrect. Needless to say, I never wrote such a thing.
So you do see yourself as a phuckwit, even though you didn't write that.
I must be getting good at psychiatry after all.
| Do you see the
| difference between:
|
| 2* (80,0,0,16) = (60,0,0,20)
|
| and:
|
| 2 * tau(80,0,0,16) = tau(0,0,0,0) + tau(60,0,0,20)
|
| (Hint: the difference is huge, like night and day.)
Hmm.... let's see if I can work it out myself... don't tell me, ok?
2*(80,0,0,16) = (160,0,0,32) which is not equal to (60,0,0,2)
so that is FALSE.
tau(60,0,0,20) is... err... according to the wacky transforms ...
tau(x,0,0,t) = (t - vx/c^2) /sqrt(1 - v^2/c^2)
= (20 - 3*60/25)/sqrt(1- 9/25)
= (20-180/25)/sqrt(16/25)
= 12.8/0.8
= 16.
tau(0,0,0,0) = (0 - 3*0/25)/0.8
= 0
So tau(0,0,0,0)+tau(60,0,0,20) = 16.
2* tau(80,0,0,16) = 2* (16 - 3*80/25)/ 0.8
= 2* (16 - 9.6)/0.8
= 32
32 = 16.
So that is FALSE.
I see no difference between FALSE and FALSE.
Please tell me what the huge difference like night and day is.
Yes of course, we on the ground want to see the movie too,
smoke your wacky baccy and snort your cocaine.
Well, I don't. I've seen what does to your brain, but your
fellow relativists do.
|
| > | > Light leaves 0, goes to 80, takes 16 seconds.
| > | > Light leaves 80, goes to 60, takes 4 seconds.
| > | > No train involved.
| > |
| > | The train clocks are involved. What do you think "tau" means in
that
| > | equation?
| >
| > Time of the stationary system.
|
| I think you should go and reread Einstein's paper. If you don't know
| what tau is I see little point in continuing.
Oh, are you going to fuck off at last, you that thinks x' is x?
Everybody knows x' is \xi, just ask McAnally or Dinky Van de phuckwit.
In YOUR eqution, tau is the time of the stationary system, in Einstein's
equation, tau is the time on the stretched 747. Boeing evn built
stretch 747s to try to overcome the SR stretching, but physics is
physics, the plane still stretches if you use a projector for the movie.
There is no way anyone can overcome the LAWS of physics.
http://www.bizjournals.com/wichita/stories/1999/12/06/daily13.html
"Boeing Co. is not finding any immediate interest from its customers for
a stretch version of its 747 jet"
See, it's automatically stretched by SR.
Trains with 32 boxcars get 40 boxcars when they travel at speed 3.
|
| > | > | > Crap. You have no idea what you are babbling about.
| > | > |
| > | > | Could you back it up. I simply stated the starting point of
the
| > | > | derivation argument.
| > | >
| > | > No you didn't. You gave track coordinates and track time.
| > | > The train is in the siding, not moving on the track.
| > | > Light leaves 0, goes to 80, takes 16 seconds, velocity of light
5.
| > | > Light leaves 80, goes to 60, takes 5 seconds, velocity of
light -5.
| > | >
| > | > 16/20 * [ f(0,0,0,0) + f(60,0,0, 80/5 + (60-80)/-5)] =
f(80,0,0,16)
| > | > No train, f is linear, no half either.
| > |
| > | Nope. The point is that these times (16 seconds and 4 seconds) are
in
| > | terms of the stationary clocks which is *not* what tau(...) are.
| >
| > You don't have a point, you don't even have a neuron.
| > The point is that these prices ($16 and $4) are in terms of the
| > stationary wholesalers which is *not* what retail price(apple,
orange)
| > are.
| > I dont know how you survive in the real world.
|
| tau is not t, nothing can change this fact.
It's a fact that planes stretch when you show the movie, too.
It's also a fact that you are as crazy as George Hammond.
|
| > | They
| > | are equal to certain *moving* clock readings.
| >
| > You don't have ANYTHING moving, not a 32 in sight.
|
| Einstein has 3 systems, two of them are moving:
|
| 1. (x,y,z,t)
| 2. (x',y,z,t) - an auxiliary
| 3. (xi,eta,zeta,tau)
|
| 1 is stationary, 2 and 3 are moving identically but using differently
| synchronised clocks. The goal of this part of Einstein's paper is to
| find the transformation formula between 1 and 3.
Yes, I know. I told you that. British Airways is doubling its fleet,
courtesy of SR.
Yes, now put your crayons away and your straight-jacket back on, go back
to your padded cell and come back when you've finished smoking that
weed.
1/2*(tau(0,0,0,0)+tau(0,0,0,20)) = tau(32,0,0,16)
tau(0,0,0,0) = 0
1/2*(0+tau(0,0,0,20)) = tau(32,0,0,16)
1/2*tau(0,0,0,20) = tau(32,0,0,16)
I've simplified it for you,
1/2*(tau(0,0,0,x'/(c-v)+x'/(c+v))) = tau(x',0,0,x'/(c-v))
The time in Berlin is what.. err ... 10 hours ahead of LA?
| > - all we know at first is that
| > | they are constrained by the tau equation. It turns out this
constraint
| > | is strong enough to produce the formula for these moving clock
| > | readings.
| > |
| > | > f(t) = t
| > |
| > | Are you saying that:
| > |
| > | tau(80,0,0,16) = 16
| > | tau(0,0,0,0) = 0
| > | tau(60,0,0,20)) = 20 ?
| >
| > Yes.
|
| WOW. I don't think there is a point in continuing our discussion then.
GOOD! I told you ages ago to fuck off, you are a crazy.
| Until you understand the *description* of Einstein's experimental
setup
| you have no moral right to complain about anything in that paper. It's
| only fair.
Nature isn't fair. Look how she gave you vacuum for brains.
|
| > | > | No, I can write both derivations side by side and the
differences
| > are
| > | > | small.
| > | >
| > | > No you can't. You think 16 is half of 20 and 60 is half of 80.
| > |
| > | No, what I think is:
| > |
| > | tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
| > |
| > | ...which is a *totally* different thing than "16 is half of 20 and
60
| > | is half of 80".
| >
| > But you don't think. You are not capable of thinking.
| > You are simply insane.
| > Even Einstein would have said (and did say, effectively)
| > tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
|
| No, he said:
|
| tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(0,0,0,20))
Well done, you've caught me in an error. Ok, the huckster Einstein
who stretches Boeing 747s in flight said
tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(0,0,0,20))
which is not what the phuckwit Bile-wacky said, which is
tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
|
| Anyway, as I said earlier - the two equations express a property of
| *two different functions*, so naturally they are different. You gave
| two different functions the same *name* and now are complaining about
| some "contradiction".
Me? Complain? A contradiction? Surely not. Everyone knows planes stretch
in flight.
|
| > v(x,y,z,t) = x/t
|
| Meaningless.
|
| > c(x,y,z,t) = x/t
|
| Meaningless.
|
| --
| Jan Bielawski
Yes, of course... Now PLEASE fuck off.
Androcles.
Androcles
Androcles wrote:
> "JanPB" <fil...@gmail.com> wrote in message
> news:1128542389.7...@o13g2000cwo.googlegroups.com...
> Androcles wrote:
>
> > As of right now we are at:
> > Einstein:
> > tau(x',0,0,t+x'/(c-v))
> > ,
> >
> > which is (with McCullough's choice of values,
> > c = 5,
> > v = 3,
> > x' = 32,
> > in FACT)
> >
> >
> > 絒tau(0,0,0,0)+tau(0,0,0,20)] = tau(32,0,0,16).
> >
> > Bile-wacky:
> >
>
> | Both equations are correct - they simply involve different
> functions -
> | both of which you denoted by the same symbol "tau".
>
> ROFLMAO!
> You do struggle to be convincing, Bile-wacky!
> You are as crazy as Hammond, he's "proven" his god exists.
| Both equations are correct - they simply involve different functions.
| You used the same symbol "tau" for two different things:
| THING NUMBER ONE: tau(x',y,z,t)
| THING NUMBER TWO: tau(x,y,z,t)
Why not change the names instead, if they a different functions?
THING NUMBER ONE: cuckoo(x',y,z,t)
THING NUMBER TWO: wacky(x,y,z,t)
Now it's clear.
The first equation:
絒cuckoo(0,0,0,0)+cuckoo(0,0,0,20)] = cuckoo(32,0,0,16)
...refers to cuckoo(x',y,z,t).
The second equation:
絒wacky(0,0,0,0)+wacky(60,0,0,20)] = wacky(80,0,0,16)
...refers to wacky(x,y,z,t).
> | Normally people don't use plugged-in numbers but general expressions
> of
> | the given variables so this type of purely notational confusion is
> | avoided.
>
> | (Incidentally, I had t=0 along with x=32, etc.)
>
> x = 32, eh? ROFLMAO!!!
My mistake, I meant to say just "I had t=0".
Ok... x = 80, right?
> Let you into a little secret, Bile-wacky. I often leave little clues
> like
> "絒tau(0,0,0,t)+... " instead of "絒tau(0,0,0,0)+" to see anyone will
> notice.
And the point of this curious little exercise is...?
Way over your head.
> Go ahead, Bile-wacky! "Derive" the wacky transforms from
> the Bile-wacky tau equations, let's have more laughs.
| Sure, very easy to type it in. But dinner first.
> The fuckwits mad moortel, Phuckwit Duck, Maniacal McCullough,
> McAnally the Insane and all the other incompetent silly bastards
> think x' is \xi,
| No, in fact they have just posted specifically stating the opposite.
================================
"David McAnally" <D.McAnally@i'm_a_gnu.uq.net.au> wrote in message
news:di0q1v$1p0q$1...@bunyip2.cc.uq.edu.au...
"Specifically, Androcles does not like the fact that we use x' to denote
the variable that Einstein denoted by \xi".
================================
So IN FACT, the fuckwits mad moortel, Phuckwit Duck, Maniacal
McCullough,
McAnally the Insane and all the other incompetent silly bastards think
x' is \xi.
That proves you to be a lying cunt, IN FACT.
Anything wrong with being indecent and HONEST?
| Also, they'd be organically unable to say anything like you suggest
| because they understand the material.
That proves you to be a lying cunt, IN FACT.
Anything wrong with being indecent and HONEST?
> you think it is x.
| Whatever gave you that idea?
===================================
"JanPB" <fil...@gmail.com> wrote in message
news:1128027695....@g43g2000cwa.googlegroups.com...
"And I provided a hint how to do it by writing
Einstein's tau equation in terms of (x,y,z,t) only:
1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
tau(x+[vx/(c-v)],0,0,t+x/(c-v))
=====================================
gave me that KNOWLEDGE that you think x' is x, as demonstrated
in the last term -- x/(c-v).
So, WankAnally and his chronies think x' = \xi and Bilewacky thinks
x' = x, proven.
Changed your mind, have you?
Try not to dope the new up, it'll rot like the old one.
In fact I spent last several posts
(including this one) doing exactly the opposite: explaining the
difference between the two uses of tau (namely, tau(x',y,z,t) and
tau(x,y,z,t)), explaining the difference between the systems (x',y,z,t)
and (x,y,z,t), between the clocks they use, and so on. How all this can
be misconstrued to mean its total opposite ("you think [x'] is x")
remains a mystery. Any takers?
I doubt it, the sane ones are all laughing at you, the insane ones
are mumbling between themselves.
Wanna see it again?
"JanPB" <fil...@gmail.com> wrote in message
news:1128027695....@g43g2000cwa.googlegroups.com...
"And I provided a hint how to do it by writing
Einstein's tau equation in terms of (x,y,z,t) only:
1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
tau(x+[vx/(c-v)],0,0,t+x/(c-v))
So go ahead and do it after dinner, wackywit, very easy to type it in.
Derive the wacky transforms and get the same answer as the cuckoo
transforms.
ROFLMAO!
Androcles.
Paul B. Andersen
> | Do you see the
> | difference between:
> |
> | 2* (80,0,0,16) = (60,0,0,20)
> |
> | and:
> |
> | 2 * tau(80,0,0,16) = tau(0,0,0,0) + tau(60,0,0,20)
> |
> | (Hint: the difference is huge, like night and day.)
>
> Hmm.... let's see if I can work it out myself... don't tell me, ok?
>
>
> 2*(80,0,0,16) = (160,0,0,32) which is not equal to (60,0,0,2)
> so that is FALSE.
>
>
> tau(60,0,0,20) is... err... according to the wacky transforms ...
>
> tau(x,0,0,t) = (t - vx/c^2) /sqrt(1 - v^2/c^2)
>
> = (20 - 3*60/25)/sqrt(1- 9/25)
> = (20-180/25)/sqrt(16/25)
> = 12.8/0.8
> = 16.
>
> tau(0,0,0,0) = (0 - 3*0/25)/0.8
> = 0
>
> So tau(0,0,0,0)+tau(60,0,0,20) = 16.
>
> 2* tau(80,0,0,16) = 2* (16 - 3*80/25)/ 0.8
> = 2* (16 - 9.6)/0.8
> = 32
So 2*8 = 32 ?
Advanced math like this is difficult, isn't it? :-)
Paul
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hia.no> wrote in message news:1128632544....@g44g2000cwa.googlegroups.com...
After yesterday's entry
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SeekHelp.html ,
I didn't feel like wading through this one again.
So thanks for having spotted this :-)
Appropriate title:
"I see no difference between FALSE and FALSE":
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/FALSE.html
Dirk Vdm
Paul B. Andersen
Quite right.
Glad to see you agree.
But how could you not, it IS blatantly obvious, isn't it?
>
> we find the Paul B. Andersen transforms are
>
> t' = (t+xv/c^2)/sqrt(1-v^2/c^2)
> x' = (x+vt) / sqrt(1-v^2/c^2)
>
> which are the OTHER cuckoo transforms and which also approximate the
> Galilean transform x' = x+vt.
Quite correct. You are learning.
If the primed system move in the negative x-direction,
the Lorentz transform is:
t' = (t+xv/c^2)/sqrt(1-v^2/c^2)
x' = (x+vt)/sqrt(1-v^2/c^2)
and the Galilean transform is:
t' = t
x' = x+vt
> They are of course blatantly obvious, quite, exactly, prove me wrong.
Quite.
Glad to see that you finally find the obvious obvious.
>
> I will say this for you, tusselad.
> You never cease to amaze.
> Keep it up, the sky is the limit.
> I am sure you can do even better.
> Oh wait, you did:
>
> Algol is a B8 and K2, the K2 being a frisbee that bounces off the
> surfac e off something.
Really?
Is that because the K2 has broken apart like droplets of mercury?
You know, like you claim the Moon would do if it were fluid?
You knowledge never cease to amaze.
Paul
Androcles
|
| Advanced math like this is difficult, isn't it? :-)
Yes indeed, very difficult.
32 is the time for light to go from 0 to 80 and back again at speed 5,
(80+80)/5 = 32.
Let us remember that Bile wacky is using
2 * tau(x = 80,0,0,t = 16)
and not
Eistein's tau(x' = 32,0,0, t= 16) = 16
So 2 * 8 = 32.
Get it now, or just slow to catch on?
Androcles.
Daryl McCullough
>So tau(0,0,0,0)+tau(60,0,0,20) = 16.
>
>2* tau(80,0,0,16) = 2* (16 - 3*80/25)/ 0.8
> = 2* (16 - 9.6)/0.8
> = 32
Uh, no.
2*(16 - 9.6)/0.8
= 2*(6.4)/0.8
= 12.8/0.8
= 16
>32 = 16.
>So that is FALSE.
If you'd done the arithmetic correctly,
you would get 16 = 16, which is TRUE.
--
Daryl McCullough
Ithaca, NY
Androcles
Which is why there is no twin paradox of course.
If Stella comes back faster, it will take longer.
The faster she goes, the later she'll arrive.
When she's up to speed c, the distance to travel is infinite.
Glad to see you agree.
But how could you not, it IS blatantly obvious, isn't it?
Androcles.
Dirk Van de moortel
"Daryl McCullough" <stevend...@yahoo.com> wrote in message news:di44i...@drn.newsguy.com...
He can't even review his calculations on *command* :-)
It must be some kind of red fog in his eyes.
Dirk Vdm
schoe...@gmail.com
Dirk Van de moortel wrote:
This coming from a guy who believes proper units change in noninertial
frames. What a dope you are.
> Dirk Vdm
Dirk Van de moortel
<schoe...@gmail.com> wrote in message news:1128672098.0...@z14g2000cwz.googlegroups.com...
In the eyes of an imbecile who declares that
ds^2 = d(ict)^2 + dx^2 + dy^2 + dz^2
is positive definite, everyone must look like a dope.
No argument on that one.
Keep digging pig ;-)
Dirk Vdm
schoe...@gmail.com
Noble lord Dick[head], the tensor elements of that metric are defined
by the matrix:
[1, 0, 0, 0]
[0, 1, 0, 0]
[0, 0, 1, 0]
[0, 0, 0, 1]
http://mathworld.wolfram.com/MinkowskiSpace.html
"Minkowski space is a four-dimensional space possessing a Minkowski
metric
dtau^2==-(dx^0)^2+(dx^1)^2+(dx^2)^2+(dx^3)^2.
Alternatively (but less desirably), it can be considered to have a
Euclidean metric but with imaginary time coordinate x^0==ict ..."
> Keep digging pig ;-)
Nice one, Dick. Say have you realized yet why proper units of
noninertial frames don't change? What a dope you are.
>
> Dirk Vdm
Dirk Van de moortel
<schoe...@gmail.com> wrote in message news:1128674236.5...@f14g2000cwb.googlegroups.com...
=====================================================
"Positive definite":
Message-ID: <1128483437.7...@o13g2000cwo.googlegroups.com>
Schoenfeld:
| > >> >The context of my statement was clearly to
| > >> >show that you cannot isometrically embed pseudo-riemannian manifold
| > >> >into strictly positive-definte euclidean space.
| > >>
Answer:
| > >> Minkowsky space isn't positive-definite.
| > >
Schoenfeld:
| > >hint: ds^2 = d(ict)^2 + dx^2 + dy^2 + dz^2 is positive-definite,
| > >Euclidean metric.
| >
Answer:
| > No it isn't. If you choose t large enough, then ds^2 becomes
| > negative.
|
|
Schoenfeld:
| Don't take my word on it.
|
| http://mathworld.wolfram.com/MinkowskiSpace.html
| "Alternatively (but less desirably), it can be considered to have a
| Euclidean metric.."
=====================================================
"Confusion":
Message-ID: <a98beaaa.04041...@posting.google.com>
Other:
| > You still have a sign error. The radiation reaction force Fr is
| > proportional to -jerk.
|
Schoenfeld:
| There is no error, Tom.
Other:
|
| >
| > > For example, if gravitational force was proportional to
| > > the r^2, then Frad would be in the direction of the acceleration,
| > > since it is inversly proportional, it is in the opposite direction.
| >
| > For a falling object dr/dt < 0, so it is falling into a region of
| > INCREASING force in the -r direction, which means increasing
| > acceleration DOWNWARD, so the jerk is pointing DOWNWARD. Fr must of
| > course point upward.
|
Schoenfeld:
| Tom, you are wrong.
|
| If a(r) = 1/r^2, then the jerk is in the OPPOSITE DIRECTION. Think
| about it.
|
| Or, do the math.
|
| d a(r)/dr = - 1/r^3
|
| The snap is in the same direction as the acceleration, and the crackle
| the same as the jerk. But what about the pop?
=====================================================
"DisCont":
Message-ID: <a98beaaa.04011...@posting.google.com>
| > Special relativity and quantum theory get along just fine.
|
Schoenfeld:
| Please define the axioms of a space which is simultaneously discrete
| and continuous.
=====================================================
"Thanks":
Message-ID: <1125172474.5...@z14g2000cwz.googlegroups.com>
| Robert Low wrote:
| > Schoenfeld wrote:
| > > Suppose A was the rank-2 tensor:
| > >
| > > A = [1, 2, 3]
| > > [4, 5, 6]
| > > [7, 8, 9]
| > >
| > > The "embedded diagonal rank-1 tensor" here is [1, 5, 9]. What is the
| > > proper name for this entity?
| >
| > 'Not a tensor', as another poster already told you. A rank 1 tensor
| > has to be either a vector or a covector, and this is neither.
|
Schoenfeld:
| It is something relevant, but obviously there is no point continuing
| discussion. Thanks for you help, please avoid any more in future
=====================================================
schoe...@gmail.com
> > [Dirk snips his recalcitrant ignorance and replaces with 3 year old
> > quote mining expedition]
>
And in characteristic style, noble lord Dick[head] surreptitiously
snips the post that highlights his refractory incompetence and
recalcitrant ignorance, doesn't answer the school-boy question posed to
him, then finishes off with a quote-mining expedition of 3 year old
posts to see if he can misquote his way out of the inevitable derision
destined for his foul face.
Well noble lord dick[head], why don't you tell us what the metric
signature of ds^2 = d(ict)^2 + dx^2 + dy^2 + dz^2 is?
hint: You can start by looking at the matrix which defines the tensor
elements of that metric:
[1, 0, 0, 0]
[0, 1, 0, 0]
[0, 0, 1, 0]
[0, 0, 0, 1]
hint2: they don't say it has (or 'can be considered to have') a
Euclidean metric unless this metric has positive-definite signature
(since noble lord dick[head], all Euclidean metrics are
positive-definite).
hint3:
http://mathworld.wolfram.com/MinkowskiSpace.html
"Minkowski space is a four-dimensional space possessing a Minkowski
metric dtau^2==-(dx^0)^2+(dx^1)^2+(dx^2)^2+(dx^3)^2.
Alternatively (but less desirably), it can be considered to have a
Euclidean metric but with imaginary time coordinate x^0==ict..."
Will Noble lord Dick[head] now answer simple question posed to him?
====
Do proper units change in noninertial frames? [Yes/No]
====
Hey by the way, I heard you went to the butcher and asked for 10 cents
worth of dog meat and the butcher asked if you wanted it wrapped up or
if you would eat it on the spot. Figures..
Dirk Van de moortel
<schoe...@gmail.com> wrote in message news:1128676735.6...@g49g2000cwa.googlegroups.com...
makes it easy for himself:
http://groups.google.com/group/sci.physics.relativity/msg/0dbec364adaa9ab3
Dirk Vdm
Androcles
| > "Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:di44i...@drn.newsguy.com...
| > > Androcles says...
| > >
| > > >So tau(0,0,0,0)+tau(60,0,0,20) = 16.
| > > >
| > > >2* tau(80,0,0,16) = 2* (16 - 3*80/25)/ 0.8
| > > > = 2* (16 - 9.6)/0.8
| > > > = 32
| > >
| > > Uh, no.
Uh, yes.
Time for light to go from 0 to 80 at speed 5 is 16.
[quote]
we establish by definition that the "time" required by light to travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
Time back again, 16.
16+16 = 32.
The idiots moortel and Bilewacky think
tau(80,0,0,16) = 8 and tau(80-60,0,0,t+4) = 8
means tau is a linear function.
Androcles.
Androcles
<schoe...@gmail.com> wrote in message
news:1128674236.5...@f14g2000cwb.googlegroups.com...
You are wasting your time with the moron.
From the diagram below you can see at a glance that
Einstein's definition of time is nonsense and the function tau
is not linear:
|
|
| C'
| /
| B /
| ____________Mirror
| /\ /
| / \ /
C / \ /
|\ / \ /
| \ / \A'
| \ / | /
| \ / /
| \ / /
| \ / | /
| \ / /
| \/ |
| /\ /
| / \ / |
| / \ /
| / \ / |
| / ____\/__________Mirror
| /
| / D |
|/
/ ____________|____|________________
A D B A' C'
[quote]
we establish by definition that the "time" required by light to travel
from A to B equals the "time" it requires to travel from B to [A'].
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
We establish by definition that the "time" required by light to travel
from C to D equals the "time" it requires to travel from D to C'.
Distance between mirrors is x'
Einstein's equation:
稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
What it means in the diagram:
稼tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
稼tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
So the time at B, the engine, equals the time at D, the caboose,
but it doesn't. Ergo Einstein was a phuckwit.
Androcles.
JanPB
> "JanPB" <fil...@gmail.com> wrote in message
> news:1128563892....@g43g2000cwa.googlegroups.com...
> | Androcles wrote:
> | > "JanPB" <fil...@gmail.com> wrote in message
> | > news:1128496083.7...@g14g2000cwa.googlegroups.com...
> | > |
> | > | Do you know the difference between:
> | > |
> | > | (80,0,0,16)
> | > |
> | > | and:
> | > |
> | > | tau(80,0,0,16) ?
> | >
> | > Yes.
> | > |
> | > | Define both.
> | >
> | > (80,0,0,16) -A vector (as long as the 16 doesn't represent time,
> which
> | > is not a vector)
> | > tau(80,0,0,16) - A single-valued function in four variables.
> |
> | That's just a generic definition. I meant physically, in this physical
> | setup with the light emitter, mirror, clocks, etc.
>
> Oh...
> Physically.. hmmm...
>
> Well, PHYSICALLY, x is measured along the ground, say from
> LA (=0) to Berlin (=80), and x' (=32) is the length of the plane
> from rudder( =0) to pilot (x') when it's on the ground at the
> departure gate.
...measured in the stationary units (that's a very important caveat).
BTW, you keep switching the names of the objects in the expewriment. So
far we had:
1. the light source and the mirror (that's in Einstein's paper),
2. the caboose and the engine,
3. the rudder and the pilot.
It would be nice if we could just stick to some fixed terminology.
The rest of your answer is hopelessly confused. For example, you write:
> [...]
> Now \xi is where the pilot sits, you are in seat x', for this reason:
This is completely wrong, you are mixing units now. What you wrote
above is exactly like saying that the temperature difference between LA
and Berlin is 70 because it is 90 degrees Fahrenheit in LA and 20
degrees Celsius in Berlin.
All I was asking was: what is the physical meaning of:
(80,0,0,16)
vs.
tau(80,0,0,16)
Hint number two: one is a specific event coordinate (location and
instant in time), the other is a time reading. What are they
specifically?
> | > Examples of a vector:
> | > (0,0,0,0) + (60,0,0,20) = (60,0,0,20)
> | > 2*(40,0,0,8) = (80,0,0,16)
> |
> | Correct.
>
> Yes, I know.
>
>
> | > Examples of a phuckwit: 2*(60,0,0,20) = (80,0,0,16)
> | > 2* (80,0,0,16) = (60,0,0,20)
> |
> | Incorrect. Needless to say, I never wrote such a thing.
>
>
> So you do see yourself as a phuckwit, even though you didn't write that.
I never said that 2*(60,0,0,20) = (80,0,0,16) or that 2*(80,0,0,16) =
(60,0,0,20).
> | Do you see the
> | difference between:
> |
> | 2* (80,0,0,16) = (60,0,0,20)
> |
> | and:
> |
> | 2 * tau(80,0,0,16) = tau(0,0,0,0) + tau(60,0,0,20)
> |
> | (Hint: the difference is huge, like night and day.)
>
> Hmm.... let's see if I can work it out myself... don't tell me, ok?
>
>
> 2*(80,0,0,16) = (160,0,0,32) which is not equal to (60,0,0,2)
> so that is FALSE.
Yes, it is false. That's why the tau equation doesn't claim this.
> tau(60,0,0,20) is... err... according to the wacky transforms ...
At this point we don't know yet what the exact value of this
tau(60,0,0,20) is (we haven't derived the transform yet). My question
was what the equation 2*tau(80,0,0,16)=tau(0,0,0,0)+tau(60,0,0,20)
meant.
Of course if we pretend we have derived the transform already then we
can do what you did below (although I rather suspect you are plugging
in the numbers without understanding what's going on). Also, using the
Lorentz transform formula implies you are using tau(x,y,z,t), not
tau(x',y,z,t) - which is fine as long as you know which is which. It
would be *much* simpler and better not to plug in the numbers, it
obscures everything:
> tau(x,0,0,t) = (t - vx/c^2) /sqrt(1 - v^2/c^2)
>
> = (20 - 3*60/25)/sqrt(1- 9/25)
> = (20-180/25)/sqrt(16/25)
> = 12.8/0.8
> = 16.
Correct.
> tau(0,0,0,0) = (0 - 3*0/25)/0.8
> = 0
Correct.
> So tau(0,0,0,0)+tau(60,0,0,20) = 16.
>
> 2* tau(80,0,0,16) = 2* (16 - 3*80/25)/ 0.8
> = 2* (16 - 9.6)/0.8
> = 32
Incorrect. 2* (16 - 9.6)/0.8 = 16, not 32.
> 32 = 16.
> So that is FALSE.
It's 16 = 16, so that is TRUE.
> | > | > Light leaves 0, goes to 80, takes 16 seconds.
> | > | > Light leaves 80, goes to 60, takes 4 seconds.
> | > | > No train involved.
> | > |
> | > | The train clocks are involved. What do you think "tau" means in
> that
> | > | equation?
> | >
> | > Time of the stationary system.
> |
> | I think you should go and reread Einstein's paper. If you don't know
> | what tau is I see little point in continuing.
>
> Oh, are you going to fuck off at last, you that thinks x' is x?
No, I said that already: x' is not x. Where did you get that idea from?
tau is *not* the time of the stationary system. It is the time of the
moving system. It differs from the time of the *auxiliary* moving
system (x',y,z,t) because the auxiliary system uses the same clock
readings as the *stationary* system (which means its clocks are *not*
Einstein-synchronised in the moving system), while the moving system
(xi,eta,zeta,tau) uses its own set of *Einstein-synchronised* clocks.
> In YOUR eqution, tau is the time of the stationary system, in Einstein's
> equation, tau is the time on the stretched 747.
No, the values of tau are readings of *moving Einstein-synchronised
clocks* in *both* setups: tau(x,y,z,t) and tau(x',y,z,t). These values
are *different* from readings of the *auxiliary* moving system's
clocks.
> | > - all we know at first is that
> | > | they are constrained by the tau equation. It turns out this
> constraint
> | > | is strong enough to produce the formula for these moving clock
> | > | readings.
> | > |
> | > | > f(t) = t
> | > |
> | > | Are you saying that:
> | > |
> | > | tau(80,0,0,16) = 16
> | > | tau(0,0,0,0) = 0
> | > | tau(60,0,0,20)) = 20 ?
> | >
> | > Yes.
> |
> | WOW. I don't think there is a point in continuing our discussion then.
>
> GOOD! I told you ages ago to fuck off, you are a crazy.
>
>
> | Until you understand the *description* of Einstein's experimental
> setup
> | you have no moral right to complain about anything in that paper. It's
> | only fair.
>
> Nature isn't fair.
That doesn't mean you can claim someone is an idiot just because you
don't understand something.
Look, I don't mind discussing the logic of an argument but I don't want
to go into explaining in detail the very basics of the experimental
setup Einstein uses. I assumed you knew at least that much. Find
someone who can explain it to you in person - an electronic bboard is
just too slow for that.
> | > Even Einstein would have said (and did say, effectively)
> | > tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
> |
> | No, he said:
> |
> | tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(0,0,0,20))
>
> Well done, you've caught me in an error. Ok, the huckster Einstein
>
> who stretches Boeing 747s in flight said
>
> tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(0,0,0,20))
>
> which is not what the phuckwit Bile-wacky said, which is
>
> tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
As I said before, let's just use *Einstein's* tau all the time so we
don't get confused by your inserting the numbers.
But I don't know now what the point of this discussion is given the
fact you don't know how it differs from tau. Until this is cleared up
all the debates regarding the tau equation and the subsequent
derivation are pointless.
--
Jan Bielawski
Androcles
Not at all. You asked for a PHYSICAL definition, not a cuckoo
definition.
| BTW, you keep switching the names of the objects in the expewriment.
So
| far we had:
|
| 1. the light source and the mirror (that's in Einstein's paper),
| 2. the caboose and the engine,
| 3. the rudder and the pilot.
|
| It would be nice if we could just stick to some fixed terminology.
|
| The rest of your answer is hopelessly confused. For example, you
write:
|
| > [...]
| > Now \xi is where the pilot sits, you are in seat x', for this
reason:
|
| This is completely wrong, you are mixing units now.
Don't be silly, \xi is moving and so is x', they are not the same,
\xi = x'/sqrt(1-v^2/c^2) according to the cuckoo transforms.
What they are by the wacky transforms I don't know, you've
not derived them after dinner yet.
| What you wrote
| above is exactly like saying that the temperature difference between
LA
| and Berlin is 70 because it is 90 degrees Fahrenheit in LA and 20
| degrees Celsius in Berlin.
Nonsense, temperature isn't part of the cuckoo transforms. Totally
irrelevant.
| All I was asking was: what is the physical meaning of:
|
| (80,0,0,16)
|
| vs.
|
| tau(80,0,0,16)
I've told you.
If you can't read that's your problem <shrug>
|
| Hint number two: one is a specific event coordinate (location and
| instant in time), the other is a time reading. What are they
| specifically?
Answer to hint number 2 that seems to be a question:
Specifically, (80,0,0,16) is an event coordinate with a time,
the other is the time of an event with a time. tau(80,0,0,16) = 16.
c = 5 (given), 80/5 = 16.
Wasn't hard, was it?
|
| > | > Examples of a vector:
| > | > (0,0,0,0) + (60,0,0,20) = (60,0,0,20)
| > | > 2*(40,0,0,8) = (80,0,0,16)
| > |
| > | Correct.
| >
| > Yes, I know.
| >
| >
| > | > Examples of a phuckwit: 2*(60,0,0,20) = (80,0,0,16)
| > | > 2* (80,0,0,16) =
(60,0,0,20)
| > |
| > | Incorrect. Needless to say, I never wrote such a thing.
| >
| >
| > So you do see yourself as a phuckwit, even though you didn't write
that.
|
| I never said that 2*(60,0,0,20) = (80,0,0,16) or that 2*(80,0,0,16) =
| (60,0,0,20).
Ok, no need to repeat yourself, I only mentioned that you see yourself
as the phuckwit I was referring to, since I said those were the examples
of a phuckwit and you responded as if I meant you.
| > | Do you see the
| > | difference between:
| > |
| > | 2* (80,0,0,16) = (60,0,0,20)
| > |
| > | and:
| > |
| > | 2 * tau(80,0,0,16) = tau(0,0,0,0) + tau(60,0,0,20)
| > |
| > | (Hint: the difference is huge, like night and day.)
| >
| > Hmm.... let's see if I can work it out myself... don't tell me, ok?
| >
| >
| > 2*(80,0,0,16) = (160,0,0,32) which is not equal to (60,0,0,2)
| > so that is FALSE.
|
| Yes, it is false. That's why the tau equation doesn't claim this.
|
| > tau(60,0,0,20) is... err... according to the wacky transforms ...
|
| At this point we don't know yet what the exact value of this
| tau(60,0,0,20) is (we haven't derived the transform yet).
Yeah, you were going to do that after dinner... maybe you haven't
eaten for 2 days. I don't know who the 'we' is though, I'm not
going to derive the wacky transforms, and I've had two dinners,
two breakfasts and two lunches since then.
I think you'll find it has two values, 12 and 20. 12 when the light got
to 60 the first time and 20 when it returned after reflection at 80.
Make sure wacky() is a linear function, won't you?
My question
| was what the equation 2*tau(80,0,0,16)=tau(0,0,0,0)+tau(60,0,0,20)
| meant.
It means 2 * 16 = 0+20, of course. I think it's silly, but go ahead and
derive the wacky transforms, this is quite funny.
| Of course if we pretend we have derived the transform already then we
| can do what you did below (although I rather suspect you are plugging
| in the numbers without understanding what's going on).
Why not pretend you had dinner and go ahead and derive the wacky
transforms instead of all this chatter?
It might take your changed mind off your hunger.
Also, using the
| Lorentz transform formula implies you are using tau(x,y,z,t), not
| tau(x',y,z,t) - which is fine as long as you know which is which.
I'm not using the cuckoo transforms at all, they can't be derived.
I may be able to use the wacky transforms if you finish dinner
and get on with deriving them.
| It
| would be *much* simpler and better not to plug in the numbers, it
| obscures everything:
That's ok, take the half (½) out, it's a number and obscures everything.
I've told you before I object to the half. Use (t1-t0)/(t2-t0) instead.
| > tau(x,0,0,t) = (t - vx/c^2) /sqrt(1 - v^2/c^2)
| >
| > = (20 - 3*60/25)/sqrt(1- 9/25)
| > = (20-180/25)/sqrt(16/25)
| > = 12.8/0.8
| > = 16.
|
| Correct.
|
| > tau(0,0,0,0) = (0 - 3*0/25)/0.8
| > = 0
|
| Correct.
|
| > So tau(0,0,0,0)+tau(60,0,0,20) = 16.
| >
| > 2* tau(80,0,0,16) = 2* (16 - 3*80/25)/ 0.8
| > = 2* (16 - 9.6)/0.8
| > = 32
|
| Incorrect. 2* (16 - 9.6)/0.8 = 16, not 32.
Nah, the time for light to go from 0 to 80 is 16, because
(80-0)/16, which is x/t = c and c is 5.
Then we have Einstein's definition,
" we establish by definition that the "time" required by a turtle to
travel
from A to B equals the "time" it requires to travel from B to A."
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
2* 16 = 32. <shrug>
Of course you haven't derived the wacky transformation yet,
so maybe you think 60 = 0 before dinner and something else
after dinner.
|
| > 32 = 16.
| > So that is FALSE.
|
| It's 16 = 16, so that is TRUE.
Nope... it's 16 = 32, FALSE.
(80-0)/5 = 16, but that's only one way. (0-80)/-5 = 16 going back.
I don't know where you get 60 from in your wacky function, nothing
startling is happening there. Had dinner yet?
Why not ask Dinky, he might know and be able to help you derive
the wacky transforms.
|
| > | > | > Light leaves 0, goes to 80, takes 16 seconds.
| > | > | > Light leaves 80, goes to 60, takes 4 seconds.
| > | > | > No train involved.
| > | > |
| > | > | The train clocks are involved. What do you think "tau" means
in
| > that
| > | > | equation?
| > | >
| > | > Time of the stationary system.
| > |
| > | I think you should go and reread Einstein's paper. If you don't
know
| > | what tau is I see little point in continuing.
| >
| > Oh, are you going to fuck off at last, you that thinks x' is x?
|
| No, I said that already: x' is not x. Where did you get that idea
from?
From you, of course. "JanPB" <fil...@gmail.com> wrote in message
news:1128027695....@g43g2000cwa.googlegroups.com...
"And I provided a hint how to do it by writing
Einstein's tau equation in terms of (x,y,z,t) only:
1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
tau(x+[vx/(c-v)],0,0,t+x/(c-v))
gave me that KNOWLEDGE that you think x' is x, as demonstrated
in the last term -- x/(c-v).
Nobody knows what v is in your equation....
| tau is *not* the time of the stationary system. It is the time of the
| moving system.
What moving system? You haven't got one.
| It differs from the time of the *auxiliary* moving
| system (x',y,z,t) because the auxiliary system uses the same clock
| readings as the *stationary* system
Well yeah... if there was an *auxiliary* moving system (x',y,z,t),
but your equation
1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
tau(x+[vx/(c-v)],0,0,t+x/(c-v))
doesn't have x' in it, right?
(which means its clocks are *not*
| Einstein-synchronised in the moving system),
What moving system?
while the moving system
| (xi,eta,zeta,tau) uses its own set of *Einstein-synchronised* clocks.
What moving system?
|
| > In YOUR eqution, tau is the time of the stationary system, in
Einstein's
| > equation, tau is the time on the stretched 747.
|
| No, the values of tau are readings of *moving Einstein-synchronised
| clocks* in *both* setups: tau(x,y,z,t) and tau(x',y,z,t).
Yeah, that's ok, the time for light to go from the caboose to the engine
is 16,
and the time for light to go from the station to 80 is 16.
So there is no problem with the time for light to go back to 80 being
16.
The problem is it passes the caboose on the way back at x= 60 , t = 20
because the caboose moved.
So 1/2 of 16+4 = 16, right? As I told you, my objection is the 1/2
Leave the numbers in, it clarifies everything.
That way we don't have any ghost moving system (xi,eta,zeta,tau)
and the *auxiliary* moving system (x',y,z,t) is the ONLY moving system.
| These values
| are *different* from readings of the *auxiliary* moving system's
| clocks.
Silly, isn't it?
| > | > - all we know at first is that
| > | > | they are constrained by the tau equation. It turns out this
| > constraint
| > | > | is strong enough to produce the formula for these moving clock
| > | > | readings.
| > | > |
| > | > | > f(t) = t
| > | > |
| > | > | Are you saying that:
| > | > |
| > | > | tau(80,0,0,16) = 16
| > | > | tau(0,0,0,0) = 0
| > | > | tau(60,0,0,20)) = 20 ?
| > | >
| > | > Yes.
| > |
| > | WOW. I don't think there is a point in continuing our discussion
then.
| >
| > GOOD! I told you ages ago to fuck off, you are a crazy.
| >
| >
| > | Until you understand the *description* of Einstein's experimental
| > setup
| > | you have no moral right to complain about anything in that paper.
It's
| > | only fair.
| >
| > Nature isn't fair.
|
| That doesn't mean you can claim someone is an idiot just because you
| don't understand something.
Yes I can, you've invented two moving systems, both called k, idiot.
|
| Look, I don't mind discussing the logic of an argument but I don't
want
| to go into explaining in detail the very basics of the experimental
| setup Einstein uses.
Go ahead and derive the wacky transforms from
1/2*(tau(0,0,0,t) + tau(vx/(c-v)+vx/(c+v),0,0,t+x/(c-v)+x/(c+v)) =
tau(x+[vx/(c-v)],0,0,t+x/(c-v))
then.
Look, I don't mind discussing the logic of an argument but I'm
willing to go into explaining in detail the very basics of the
experimental setup Einstein uses to a phuckwit, as I have patiently
done.
If you don't like numbers, take the half out.
| I assumed you knew at least that much.
Much better than you ever will.
| Find
| someone who can explain it to you in person - an electronic bboard is
| just too slow for that.
So fuck off. I've told you that as well. I object to arseholes like
you spreading Einstein's fucking lies, so leave the electronic bboard
alone and be decent and honest, you don't understand the light
passes the caboose at tau(60,0,0,20) and carries on to tau(0,0,0,32), so
that
1/2 [tau(0,0,0,0) + tau(0,0,0,32)] = tau(80,0,0,16) and tau = t.
|
| > | > Even Einstein would have said (and did say, effectively)
| > | > tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
| > |
| > | No, he said:
| > |
| > | tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(0,0,0,20))
| >
| > Well done, you've caught me in an error. Ok, the huckster Einstein
| >
| > who stretches Boeing 747s in flight said
| >
| > tau(32,0,0,16) = 1/2*(tau(0,0,0,0) + tau(0,0,0,20))
| >
| > which is not what the phuckwit Bile-wacky said, which is
| >
| > tau(80,0,0,16) = 1/2*(tau(0,0,0,0) + tau(60,0,0,20))
|
| As I said before, let's just use *Einstein's* tau all the time so we
| don't get confused by your inserting the numbers.
I'll compromise.
Take the 1/2 out, use (t1-t0)/(t2-t0) or leave my numbers in.
All or nothing.
Then I'll stop calling you names.
Is that fair enough?
Is that decent and honest?
| But I don't know now what the point of this discussion is given the
| fact you don't know how it differs from tau. Until this is cleared up
| all the debates regarding the tau equation and the subsequent
| derivation are pointless.
No more insults if we take ALL the numbers out. Is that decent and
honest?
I'm being as fair as I can be, but I'm not giving in.
Suffer the slings and arrows of outrageous insult, or reason with
me.
Androcles.
PD
Daryl McCullough wrote:
> Androcles says...
>
> >So tau(0,0,0,0)+tau(60,0,0,20) = 16.
> >
> >2* tau(80,0,0,16) = 2* (16 - 3*80/25)/ 0.8
> > = 2* (16 - 9.6)/0.8
> > = 32
>
> Uh, no.
> 2*(16 - 9.6)/0.8
> = 2*(6.4)/0.8
> = 12.8/0.8
> = 16
>
> >32 = 16.
> >So that is FALSE.
Oh, good Lord, Androcles really said that? And he refused to admit the
arithmetic error? How Androcles of him!