You were doing pretyty well until you came upwith this bit of
nonsense. You reckon someone is going to make a gravitational
wave generator, do you?

No LISA the gravitational wave detector. It has been sheduled by NASA
but gooness knows when it wil go up. LISA like all scientific
experiments requires an ultra stable platform. Manned space flight can
never provide this.

"Hatunen" wrote in message ...
On Sat, 20 Aug 2005 10:48:35 GMT, "Dirk Van de moortel"
wrote:


"Bill Hobba" wrote in message ...

"Puppet_Sock" wrote in message
oups.com...
Pentcho Valev wrote:
In his book "Science in a Free Society" P. Feyerabend states: "Today
science prevails not because of its comparative merits but because the
show has been rigged in its favour".
[snip]

You know, I must not know the right people. Because I've been
a nuclear physicist for 15 years, and a science student for
10 years before that, and I've never been invited to one of
*those* parties where this rigging goes on.

Although not a scientist I too studied it at university and graduate school
and I was never invited nor knew of anyone who was. I was friends with some
scientists who worked at the CSIRO - they were never invited either. I have
been invited to parties by IT companies who obviously wanted to influence my
opinion, to all sorts of parties for all sorts of reasons, but never to one
rigging science.


It sounds like a hell of a lot of fun. If anybody *does* know
of such an event, I'd sure like an invite.
Socks

Well said.

Agreed :-)

Did I not get an invite becaue I didn't join Sigma Pi Sigma when
I was elected?

Nah, you're confusing with
http://images.google.com/images?q=phi+zappa+crappa

Dirk Vdm

Androcles wrote:
"JanPB" wrote in message
ups.com...
Androcles wrote:
On the court docket, Science v Einstein.

Oh dear. Not again :-)

(Quote attributions are slightly off again.)

Prosecution opens:
Ladies and Gentlemen of the jury.

Reference: http://www.fourmilab.ch/etexts/einstein/specrel/www/

The first transformation we are given is the Galilean,

x' = x-vt
y = y
z = z
t = t

Keep in mind this transformation, we'll come back to it.

You have to agree with that, Einstein states:

"If we place x'=x-vt, it is clear that a point at rest in the system k
must have a system of values x', y, z, independent of time."

We have completed the transform from the stationary system, K, to the
moving system
which I'm going to name k' because Einstein doesn't give it a name.
You'll see why shortly.

For all x in K, x' in k', (x',y,z,t) = g(x,y,z,t).

You cannot possible disagree with that, you can only object to my
choice
of name.

OK so far.

No, no no! How can you withstand such a blatant contradiction?

It is clear, so ist klar, in agreement with experience, and because
Einstein says so, a point at rest in system k' is independent of time.

Meaning: a point at rest in k' has its x' coordinate independent of t.
OK.

No, no no! How can you withstand such a blatant agreement?

We have now completed the transformation from K to k', the function g,
and can place K on the back burner.

You cannot possibly disagree with that. (I know of one dumb relativist
that does...he insists the system of coordinates k' doesn't exist.
Such
is the mentality I deal with here.)

OK, OK, fine. Just skip the attacks, they take too much space.


Since you don't understand this stuff, why don't you refrain from
posting on this thread? What's your point? Anything wrong with being
decent and honest? -Jan Bielawski


One might object to it on philosophical grounds but not on mathematical
ones (its mathematics is very easy, BTW, this is not where the
difficulty
with this paper lies, and this is not where you'll ever find anything
wrong). -- Jan Bielawski


I still can prove that Einstein's 1905 paper
has no mistakes in it. -- Jan Bielawski

Don't like having your nose rubbed in your own ****?

I have no idea what you're talking about. I have never objected to
anything you wrote above, it's correct, always has been.

It's what we do to little peeing puppies that have milk teeth and can't
bark yet.
Oh dear, not again :-)

Now we come to Einstein's transformation.
Not Lorentz's, not Galileo's, but Einstein's.

For all x in k', xi in kappa, (xi, eta,zeta,tau) = cuckoo(x',y,z,t)

Never seen a "cuckoo" function before but OK otherwise.

I can't withstand such a blatant agreement.

:-)

You can now begin disagreeing out of pure phuckwittery.

No, out of your error. Details follow.

As I said, pure phuckwittery. Details follow.

We have a transformation from the moving system k' to the
moving system kappa.

Einstein would have you believe that

tau = cuckoo_tau(g(x,y,z,t))
xi = cuckoo_xi(g(x,y,z,t))

is called the "Lorentz transformation".

OK, fine.

I call it the cuckoo transformation, there is no relative motion
between
k' and kappa, the time in k' has been found to be x'/(c-v) and
x'/(c+v),
the only purpose to the function cuckoo_tau is to satisfy Einstein's
fraudulent whim,

Whatever.
I can't withstand such a blatant agreement. You must be wrong.

"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."

As Counsel for the Physicists, I rest my case.

As Counsel for the Mathematicians, we have yet to prove that
cuckoo_tau
is not a linear function.

But I did that already, so I'll repeat it with additional comment for
the incompetent.

Here it is algebraically:
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))
(given)

Doubling both sides:
tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v)) = 2 *
tau(x',0,0,t+x'/(c-v))

Taking out the t for 3:00pm on a Friday afternoon:

tau(0,0,0,0)+tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))

Synchronize clocks at t = 0, tau(0,0,0,0) = 0, we remove tau(0,0,0,0)+

tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))

OK so far.

Taking coordinate x' as infinitessimally small, as Einstein says,
you not quite realizing x' is both a coordinate and a distance,
he does that to differentiate, so we leave the distance alone,
dx/dt = x/t anyway with a constant velocity.

It's OK although I told you twice already how very simply to rewrite
this equation so that x' is ONLY a ccordinate, never a distance
(according to your terminology).


I'm glad you've told me three times now.
Since you don't understand this stuff, why don't you refrain from
posting on this thread? What's your point? Anything wrong with being
decent and honest? -Jan Bielawski

Well, never mind all that. Tell me where exactly in the following
equation is x' treated as distance:

tau(0,0,0,0)+tau(0,0,0,(x'-0)/(c-v)+(x'-0)/(c+v)) = 2 *
tau(x',0,0,(x'-0)/(c-v))

See? They are all coordinates now.


But wait!
WHY is coordinate x' zero, other than the reason I'd given?
Very simple. There is no relative motion between k' and kappa,

Right.

I can't withstand such a blatant contradiction. You must be right.


the coordinate x' is independent of time.

In other words, "objects at rest in kappa have x' coordinate
independent of t". OK.

We do not have
xi = x'-ut or x' = x'+ut or any other function xi = ****up(x')
for Lorentz's sake, there is no u, v, w or velocity between system k'
and system kappa.

Here your first mistake creeps in.

k' and kappa use different sets of clocks

Really! My shorter car has a different clock to my car.
Gotcha, now I can see where I was insane. Thank you for curing
me of that insanity.

Well, you wrote it yourself, to quote:

x' = x-vt
y = y
z = z
t = t

....Look at the last equation. It says that k' and K (the stationary
system) use the same clocks. In other words, k' and kappa use
*different* clocks. Namely, kappa uses a set of clocks at rest in
kappa, synchronised in kappa. And k' uses a set of clocks *not* at rest
in k' (they are at rest in K), *not* synchronised in k' (they are
synchronised in K).

to define their times (k' uses t-clocks, kappa uses tau-clocks).

Yes, of course. I see it now.

This means kappa and k' - although at rest wrt one another - use
different length unit along the X axis.

Yes, I see it now. The kappa clock uses Greek marathons and the
k' clock uses Roman miles.

Something like that :-)

This means you cannot in principle rule out a relationship of the type:

xi = some_function(x', v)

where v is the relative velocity of K and k'.

Yes, of course. My desk is shorter than my desk when a bus goes by
outside because I have two different clocks on it. I see where my
mistake crept in.

No, you don't have two different clocks on it, you (the system k')
merely use someone else's (the system K) clocks for your measurements.
This offset is automatically factored into the "tau" equation provided
you don't ruin it by arbitrary resetting one of the variable values.

It's true though that some_function does not depend on t, I grant you
this much :-)

Oh thank you so much! How very kind of you.
Be careful, though, if you give me a inch I might take a day.

In units where c = 1, heh...

The time at zero is the same time at x', same at xi;

You need to make this more precise.


Ah, I seeeeee......
Let's try a footnote to make it more precise and duck the issue:
"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."
Is that precise enough?

No. It's just hand waving.

As I just said the clocks of kappa
and k' will read different things at, say, the place where the mirror
is situated or at the common origin of k' and kappa.

Yes, you did, didn't you?
And I ducked it with a footnote. My mistake crept in.
Phuckwittery is so much fun, isn't it?
You do realize your buddy moortel doesn't come up to the lofty
heights of an IQ of 10, in double figures, like yours?

I never had my IQ tested so I wouldn't know.

Perhaps he's moving faster than you and had it Lorentz-contracted.
He still thinks I'm the troll, I don't want to disappoint him.
Sorry, I was adding a little curve to spacetime there.


no translation between frames, this is the moving frame only, the
stationary frame K is simmering on the back burner.

No movement between frames but there is still a non-identity
transformation between them.

Like this non-identity matrix?
[ 1 0 ]
[ 0 1 ]

No, you have to compute this matrix by completing the derivation. It's
not a Lorentz-form transformation because one of the systems (k') is
using clocks which are not synchronised in it. The system k' is only an
auxiliary. Anyway, the transformation from k' to kappa is:

[ gamma 0 0 0 ]
[ 0 1 0 0 ]
[ 0 0 1 0 ]
[ gamma*v/c^2 0 0 1/gamma ]

(I'm using the coordinate order (x',y,z,t))

This transform is not needed for the derivation Einstein is after but
since you asked here it is. Point is, despite k' and kappa being at
rest wrt one another, it's not an identity matrix and you can see that
both the X coordinate and the time coordinate is different in both
systems, most importantly, the time in kappa changes as you change x' -
that's why you cannot reset x' to 0 in that one slot of the "tau"
equation.

I'll be sure to keep two clocks on my desk. I did notice that 30
centimetres and 12 inches on my ruler were not precisely the same
length, but I never realized it was because of special relativity.


Hence:

tau(0,0,0, x'/(c-v)+x'/(c+v)) = 2 * tau(0,0,0,x'/(c-v))

That's wrong, you can't reset x' to 0 in just one instance and not the
others.


Ah, I see. Ok, the time at zero is different to the time at x',

It is, because k' insists on using K's clocks which are not
synchronised according to him.

different
again at xi, even though x' is infinitessimally small and we shall not
discuss the inexactitude that the lurkers might perceive, and now I need
three clocks on my desk, one caesium, one balance-wheel and a cuckoo
clock with a pendulum because it's WRONG. Gotcha.

And the ring dial sun clock (there is one on my shelf).

--
Jan Bielawski

On Sat, 20 Aug 2005 10:48:35 GMT, "Dirk Van de moortel"
wrote:


"Bill Hobba" wrote in message ...

"Puppet_Sock" wrote in message
oups.com...
Pentcho Valev wrote:
In his book "Science in a Free Society" P. Feyerabend states: "Today
science prevails not because of its comparative merits but because the
show has been rigged in its favour".
[snip]

You know, I must not know the right people. Because I've been
a nuclear physicist for 15 years, and a science student for
10 years before that, and I've never been invited to one of
*those* parties where this rigging goes on.

Although not a scientist I too studied it at university and graduate school
and I was never invited nor knew of anyone who was. I was friends with some
scientists who worked at the CSIRO - they were never invited either. I have
been invited to parties by IT companies who obviously wanted to influence my
opinion, to all sorts of parties for all sorts of reasons, but never to one
rigging science.


It sounds like a hell of a lot of fun. If anybody *does* know
of such an event, I'd sure like an invite.
Socks

Well said.

Agreed :-)

Did I not get an invite becaue I didn't join Sigma Pi Sigma when
I was elected?

************* DAVE HATUNEN ) *************
* Tucson Arizona, out where the cacti grow *
* My typos & mispellings are intentional copyright traps *

"JanPB" wrote in message
oups.com...
Androcles wrote:
"JanPB" wrote in message
ups.com...
Androcles wrote:
On the court docket, Science v Einstein.

Oh dear. Not again :-)

(Quote attributions are slightly off again.)



It must be some HTML or other creeping in, I don't get it
with other posts.




Prosecution opens:
Ladies and Gentlemen of the jury.

Reference: http://www.fourmilab.ch/etexts/einstein/specrel/www/

The first transformation we are given is the Galilean,

x' = x-vt
y = y
z = z
t = t

Keep in mind this transformation, we'll come back to it.

You have to agree with that, Einstein states:

"If we place x'=x-vt, it is clear that a point at rest in the system
k
must have a system of values x', y, z, independent of time."

We have completed the transform from the stationary system, K, to
the
moving system
which I'm going to name k' because Einstein doesn't give it a name.
You'll see why shortly.

For all x in K, x' in k', (x',y,z,t) = g(x,y,z,t).

You cannot possible disagree with that, you can only object to my
choice
of name.

OK so far.

No, no no! How can you withstand such a blatant contradiction?

It is clear, so ist klar, in agreement with experience, and because
Einstein says so, a point at rest in system k' is independent of
time.

Meaning: a point at rest in k' has its x' coordinate independent of t.
OK.

No, no no! How can you withstand such a blatant agreement?

We have now completed the transformation from K to k', the function
g,
and can place K on the back burner.

You cannot possibly disagree with that. (I know of one dumb
relativist
that does...he insists the system of coordinates k' doesn't exist.
Such
is the mentality I deal with here.)

OK, OK, fine. Just skip the attacks, they take too much space.


Since you don't understand this stuff, why don't you refrain from
posting on this thread? What's your point? Anything wrong with being
decent and honest? -Jan Bielawski


One might object to it on philosophical grounds but not on
mathematical
ones (its mathematics is very easy, BTW, this is not where the
difficulty
with this paper lies, and this is not where you'll ever find anything
wrong). -- Jan Bielawski


I still can prove that Einstein's 1905 paper
has no mistakes in it. -- Jan Bielawski

Don't like having your nose rubbed in your own ****?

I have no idea what you're talking about. I have never objected to
anything you wrote above, it's correct, always has been.

It's what we do to little peeing puppies that have milk teeth and
can't
bark yet.
Oh dear, not again :-)

Now we come to Einstein's transformation.
Not Lorentz's, not Galileo's, but Einstein's.

For all x in k', xi in kappa, (xi, eta,zeta,tau) = cuckoo(x',y,z,t)

Never seen a "cuckoo" function before but OK otherwise.

I can't withstand such a blatant agreement.

:-)

You can now begin disagreeing out of pure phuckwittery.

No, out of your error. Details follow.

As I said, pure phuckwittery. Details follow.

We have a transformation from the moving system k' to the
moving system kappa.

Einstein would have you believe that

tau = cuckoo_tau(g(x,y,z,t))
xi = cuckoo_xi(g(x,y,z,t))

is called the "Lorentz transformation".

OK, fine.

I call it the cuckoo transformation, there is no relative motion
between
k' and kappa, the time in k' has been found to be x'/(c-v) and
x'/(c+v),
the only purpose to the function cuckoo_tau is to satisfy Einstein's
fraudulent whim,

Whatever.
I can't withstand such a blatant agreement. You must be wrong.

"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."

As Counsel for the Physicists, I rest my case.

As Counsel for the Mathematicians, we have yet to prove that
cuckoo_tau
is not a linear function.

But I did that already, so I'll repeat it with additional comment
for
the incompetent.

Here it is algebraically:
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))
(given)

Doubling both sides:
tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v)) = 2 *
tau(x',0,0,t+x'/(c-v))

Taking out the t for 3:00pm on a Friday afternoon:

tau(0,0,0,0)+tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))

Synchronize clocks at t = 0, tau(0,0,0,0) = 0, we remove
tau(0,0,0,0)+

tau(0,0,0,x'/(c-v)+x'/(c+v)) = 2 * tau(x',0,0,x'/(c-v))

OK so far.

Taking coordinate x' as infinitessimally small, as Einstein says,
you not quite realizing x' is both a coordinate and a distance,
he does that to differentiate, so we leave the distance alone,
dx/dt = x/t anyway with a constant velocity.

It's OK although I told you twice already how very simply to rewrite
this equation so that x' is ONLY a ccordinate, never a distance
(according to your terminology).


I'm glad you've told me three times now.
Since you don't understand this stuff, why don't you refrain from
posting on this thread? What's your point? Anything wrong with being
decent and honest? -Jan Bielawski

Well, never mind all that. Tell me where exactly in the following
equation is x' treated as distance:

tau(0,0,0,0)+tau(0,0,0,(x'-0)/(c-v)+(x'-0)/(c+v)) = 2 *
tau(x',0,0,(x'-0)/(c-v))

See? They are all coordinates now.




Exactly this.

"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."

Einstein is saying there is no function f such that x' = f(t).
That's what independent of time means.
Hence there can be no inverse function t = f^-1 (x').
tau(0,0,0,t) does not exist.
I'll allow tau(t). Now it must be proven.





But wait!
WHY is coordinate x' zero, other than the reason I'd given?
Very simple. There is no relative motion between k' and kappa,

Right.

I can't withstand such a blatant contradiction. You must be right.


the coordinate x' is independent of time.

In other words, "objects at rest in kappa have x' coordinate
independent of t". OK.

We do not have
xi = x'-ut or x' = x'+ut or any other function xi = ****up(x')
for Lorentz's sake, there is no u, v, w or velocity between system
k'
and system kappa.

Here your first mistake creeps in.

k' and kappa use different sets of clocks

Really! My shorter car has a different clock to my car.
Gotcha, now I can see where I was insane. Thank you for curing
me of that insanity.

Well, you wrote it yourself, to quote:

x' = x-vt
y = y
z = z
t = t

....Look at the last equation. It says that k' and K (the stationary
system) use the same clocks.



No it doesn't. It says that the time measured by the moving clock
is identical to the time measured by the stationary clock,
"in agreement with experience", "it is clear", "it is at once apparent".
K is on the back burner, and so is it's clock.
..
It also says rulers do not change length when you move them.

But, if you want to be picky-picky,

xi = x-vt
eta = y
zeta = z
tau = t

I thought that might be confusing.



In other words, k' and kappa use
*different* clocks. Namely, kappa uses a set of clocks at rest in
kappa, synchronised in kappa. And k' uses a set of clocks *not* at rest
in k'




Oh yes they are at rest in k', and x' is independent of them, being
independent of time.




(they are at rest in K), *not* synchronised in k' (they are
synchronised in K).

Yes they are synchronized in k', and remain synchronized,
"in agreement with experience", "it is clear", "it is at once apparent".



to define their times (k' uses t-clocks, kappa uses tau-clocks).

Yes, of course. I see it now.

This means kappa and k' - although at rest wrt one another - use
different length unit along the X axis.

Yes, I see it now. The kappa clock uses Greek marathons and the
k' clock uses Roman miles.

Something like that :-)

This means you cannot in principle rule out a relationship of the
type:

xi = some_function(x', v)

where v is the relative velocity of K and k'.

Yes, of course. My desk is shorter than my desk when a bus goes by
outside because I have two different clocks on it. I see where my
mistake crept in.

No, you don't have two different clocks on it, you (the system k')
merely use someone else's (the system K) clocks for your measurements.



No I don't, I synchronized my wris****ch with a new battery
installation,
my computer is linked via the internet to the system K.
They remain synchronized - or would if my wris****ch were perfect.
Anyway, 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.




This offset is automatically factored into the "tau" equation provided
you don't ruin it by arbitrary resetting one of the variable values.

It's true though that some_function does not depend on t, I grant you
this much :-)

Oh thank you so much! How very kind of you.
Be careful, though, if you give me a inch I might take a day.

In units where c = 1, heh...


I use units where c = 5, v = 3.
I use a train that is 32 units long on a track that is 80 units,
so that 80 = 5 * 16 and 80 = 3*16 +32.
I reflect the light at 80 K (32 k') and it meets the end of the
train at 60 K (0 k') so that (80-60)/5 = 4 seconds and
(32-0)/(3+5) = 4 also, and (32-0)/(5-3) = 16.
Then I add 8 more cars to the train, apply the cuckoo definition
and make the light-time 8 each way.
Now the longer train, 40, divided by 8 gives me 5, the speed of
light on the train, same as the track.
The trouble is I can't make it fit with "The observable phenomenon
here depends only on the relative motion of the conductor and the
magnet, whereas the customary view draws a sharp distinction
between the two cases in which either the one or the other of these
bodies is in motion. "
I don't take the customary view, you see. The track is moving, the
speed of light is 5 on the train, so now I have to stretch the track
beyond 80 to 80*gamma = 100, then recycle.

Of course I send my train right around the xmas tree, bouncing the
light back and forth between the caboose and engine until it's grown
big enough to sell to British Rail and carry passengers. The steel
rails I sell for scrap. I've been thinking of buying gold rails...

I got the idea from Sagnac, where the customary view is that
the turntable rotates. Not taking the customary view, I make the
observer run around the turntable. Thn the speed of light is
c+v, c-v for him and c on the "stationary" turntable. the trouble
is, his watch isn't slowing down like it should, he's seeing
a beat frequency, it is clear, in agreement with experience.






The time at zero is the same time at x', same at xi;

You need to make this more precise.


Ah, I seeeeee......
Let's try a footnote to make it more precise and duck the issue:
"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."
Is that precise enough?

No. It's just hand waving.




Yes, I rather thought SR was just hand-waving bull****.



As I just said the clocks of kappa
and k' will read different things at, say, the place where the mirror
is situated or at the common origin of k' and kappa.

Yes, you did, didn't you?
And I ducked it with a footnote. My mistake crept in.
Phuckwittery is so much fun, isn't it?
You do realize your buddy moortel doesn't come up to the lofty
heights of an IQ of 10, in double figures, like yours?

I never had my IQ tested so I wouldn't know.


Don't bother, its mostly about what you can remember anyway.


Perhaps he's moving faster than you and had it Lorentz-contracted.
He still thinks I'm the troll, I don't want to disappoint him.
Sorry, I was adding a little curve to spacetime there.


no translation between frames, this is the moving frame only, the
stationary frame K is simmering on the back burner.

No movement between frames but there is still a non-identity
transformation between them.

Like this non-identity matrix?
[ 1 0 ]
[ 0 1 ]

No, you have to compute this matrix by completing the derivation. It's
not a Lorentz-form transformation because one of the systems (k') is
using clocks which are not synchronised in it. The system k' is only an
auxiliary. Anyway, the transformation from k' to kappa is:

[ gamma 0 0 0 ]
[ 0 1 0 0 ]
[ 0 0 1 0 ]
[ gamma*v/c^2 0 0 1/gamma ]

(I'm using the coordinate order (x',y,z,t))


How pretty! And just what is v between k' and kappa?


This transform is not needed for the derivation Einstein is after but
since you asked here it is. Point is, despite k' and kappa being at
rest wrt one another, it's not an identity matrix and you can see that
both the X coordinate and the time coordinate is different in both
systems, most importantly, the time in kappa changes as you change x' -
that's why you cannot reset x' to 0 in that one slot of the "tau"
equation.

I'll be sure to keep two clocks on my desk. I did notice that 30
centimetres and 12 inches on my ruler were not precisely the same
length, but I never realized it was because of special relativity.


Hence:

tau(0,0,0, x'/(c-v)+x'/(c+v)) = 2 * tau(0,0,0,x'/(c-v))

That's wrong, you can't reset x' to 0 in just one instance and not the
others.


Ah, I see. Ok, the time at zero is different to the time at x',

It is, because k' insists on using K's clocks which are not
synchronised according to him.


I'll buy k' a watch for Christmas, I want to grow more trains.
Ya gotta invest to make money.
You wouldn't have any gold rails, would you?


different
again at xi, even though x' is infinitessimally small and we shall not
discuss the inexactitude that the lurkers might perceive, and now I
need
three clocks on my desk, one caesium, one balance-wheel and a cuckoo
clock with a pendulum because it's WRONG. Gotcha.

And the ring dial sun clock (there is one on my shelf).

Mine's bigger than yours!

http://www.amherst.edu/~ermace/sth/birdseye.jpeg

Androcles


--
Jan Bielawski

Androcles wrote:
"JanPB" wrote in message
oups.com...
Androcles wrote:
"JanPB" wrote in message
ups.com...
Androcles wrote:
On the court docket, Science v Einstein.

Oh dear. Not again :-)

(Quote attributions are slightly off again.)


It must be some HTML or other creeping in, I don't get it
with other posts.

I use Google to post, don't know what causes it.

[snip]
"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."

Einstein is saying there is no function f such that x' = f(t).

No, x' is a function of it: x'=x-vt after all. Einstein is saying point
at rest in k has a constant x' coordinate.

That's what independent of time means.
Hence there can be no inverse function t = f^-1 (x').
tau(0,0,0,t) does not exist.
I'll allow tau(t). Now it must be proven.

You need to explain this - as written it's nonsensical:

1. there is the inverse: t = (x-x')/v
2. tau(0,0,0,t) exists - it's the reading of the k (tau) clock sitting
at the origin at the instant the K-clock momentarily coincident with it
reads t seconds.
3. tau is a function of four variables, hence no tau(t).

[snip]
k' and kappa use different sets of clocks

Really! My shorter car has a different clock to my car.
Gotcha, now I can see where I was insane. Thank you for curing
me of that insanity.

Well, you wrote it yourself, to quote:

x' = x-vt
y = y
z = z
t = t

...Look at the last equation. It says that k' and K (the stationary
system) use the same clocks.



No it doesn't. It says that the time measured by the moving clock
is identical to the time measured by the stationary clock,

Same thing. Another way of saying that k' uses K-clocks for its time
measurements is to say that k' does carry his own set of clocks but
each of them always reads the same as the momentarily coincident
K-clock. Either way - k' and kappa use differently synchronised clocks.

"in agreement with experience", "it is clear", "it is at once apparent".
K is on the back burner, and so is it's clock.
.
It also says rulers do not change length when you move them.

But, if you want to be picky-picky,

xi = x-vt
eta = y
zeta = z
tau = t

I thought that might be confusing.

Tau is not t. Differently synchronised clocks. You had it correctly
befo t=t.


In other words, k' and kappa use
*different* clocks. Namely, kappa uses a set of clocks at rest in
kappa, synchronised in kappa. And k' uses a set of clocks *not* at rest
in k'




Oh yes they are at rest in k', and x' is independent of them, being
independent of time.

x'=x-vt, dependent of time. What's independent of time is the x'
coordinate of an object at rest in k'.



(they are at rest in K), *not* synchronised in k' (they are
synchronised in K).

Yes they are synchronized in k', and remain synchronized,
"in agreement with experience", "it is clear", "it is at once apparent".

No, t=t, remember? The dependence of tau on t is yet unknown so we
cannot just assume they are equal. (In fact they come out unequal at
the end:

tau = (t-vx/c^w)*gamma

....but that comes later.

[snip]

--
Jan Bielawski

"Pentcho Valev" schrieb im Newsbeitrag
oups.com...
In his book "Science in a Free Society" P. Feyerabend states: "Today
science prevails not because of its comparative merits but because the
show has been rigged in its favour". Also, "the apostles of science
were the more determined conquerors" who "materially suppressed the
bearers of alternative cultures". Sounds realistic and yet the horror
story of George Orwell is much closer to the realities of the
situation:



Yes it is not nice whats going on. From a system one for all we run straight
ahead into all againt one.
Is that the american copy ? How rich must the richest and mightiest become
to have enough ?


"In the end the Party would announce that two and two made five, and
you would have to believe it. It was inevitable that they should make
that claim sooner or later: the logic of their position demanded it.
Not merely the validity of experience, but the very existence of
external reality, was tacitly denied by their philosophy. The heresy of
heresies was common sense. And what was terrifying was not that they
would kill you for thinking otherwise, but that they might be right.
For, after all, how do we know that two and two make four? Or that the
force of gravity works? Or that the past is unchangeable? If both the
past and the external world exist only in the mind, and if the mind
itself is controllable what then?"


Wasnt that always ? Or do you think most men have gone to war because they
had fun ?
Of shure some of them had fun. But most of them just had no other choice.

Joe

Pentcho Valev

"JanPB" wrote in message
oups.com...
| Androcles wrote:
| "JanPB" wrote in message
| oups.com...
| Androcles wrote:
| "JanPB" wrote in message
| ups.com...
| Androcles wrote:
| On the court docket, Science v Einstein.
|
| Oh dear. Not again :-)
|
| (Quote attributions are slightly off again.)
|
|
| It must be some HTML or other creeping in, I don't get it
| with other posts.
|
| I use Google to post, don't know what causes it.

I'm using Outlook Express. This time you are correctly indented.
Beats me as well.


|
| [snip]
| "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."
|
| Einstein is saying there is no function f such that x' = f(t).
|
| No, x' is a function of it: x'=x-vt after all. Einstein is saying
point
| at rest in k has a constant x' coordinate.

Jeez.... x' is the width of my desk. I will not be different tomorrow,
it was the same yesterday, it won't change if I carry it outside.
It is independent of time.
x = x'+vt. x is NOT independent of time, it is the changing position
of the train as it moves. The length of the train doesn't change.

There IS a function f such that x = f(t), namely x = vt +(a constant
x').
There is NO function f such that x' = f(t) because x' is a constant!
x' is a length, not a coordinate.
You can MAKE it a coordinate IF-AND-ONLY-IF you make the
other end of the length zero.
I still have one inch between 0 and 1, 1 and 2, 2 and 3.
I can change coordinate x, but I have to change coordinate
x-1 as well.
I cannot change x', it's a constant.

If Einstein had not deliberately confused you this would be
so much easier.

Let's try this.
"If we place L =x-vt, it is clear that a point at rest in the system k
must have a system of values L, y, z, independent 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."

Now I'll use A,B and L. L = B-A.

½[tau(A,t)+tau(A, t+L/(c-v)+L/(c+v))] = tau(B, t+L/(c-v))

Hence, if L be taken infinitesimally small... ???
Hence, if A be taken infinitesimally small... ???
Hence, if B be taken infinitesimally small... ???

Jeez... How can you be so thick?


|
| That's what independent of time means.
| Hence there can be no inverse function t = f^-1 (x').
| tau(0,0,0,t) does not exist.
| I'll allow tau(t). Now it must be proven.
|
| You need to explain this - as written it's nonsensical:

I can feed you but I can't **** for you, you have to digest
it for yourself and do that.


| 1. there is the inverse: t = (x-x')/v

t = (x-L)/v
L is a constant. We'll use a point particle, an electron.
L = 0
t = x/v. Whoopee....?
That's the Galilean g() transformation.
We are DONE with v and K.

Next up, the cuckoo transformation, k' to kappa.
The time for light to travel the length L equals the time
for light to travel the length -L.

a) t = (L + (-L) )/ c

b) c = (L + (-L) )/ t

c) L+ (-L) = ct.

(Androcles spoon-feeding)

Get a grip!


| 2. tau(0,0,0,t) exists - it's the reading of the k (tau) clock sitting
| at the origin at the instant the K-clock momentarily coincident with
it
| reads t seconds.

There is no k-clock, it's a kappa-clock and a k' -clock. (sigh...)


| 3. tau is a function of four variables, hence no tau(t).
|
| [snip]
| k' and kappa use different sets of clocks
|
| Really! My shorter car has a different clock to my car.
| Gotcha, now I can see where I was insane. Thank you for curing
| me of that insanity.
|
| Well, you wrote it yourself, to quote:
|
| x' = x-vt
| y = y
| z = z
| t = t
|
| ...Look at the last equation. It says that k' and K (the stationary
| system) use the same clocks.
|
|
|
| No it doesn't. It says that the time measured by the moving clock
| is identical to the time measured by the stationary clock,
|
| Same thing. Another way of saying that k' uses K-clocks for its time
| measurements is to say that k' does carry his own set of clocks but
| each of them always reads the same as the momentarily coincident
| K-clock. Either way - k' and kappa use differently synchronised
clocks.
|
| "in agreement with experience", "it is clear", "it is at once
apparent".
| K is on the back burner, and so is it's clock.
| .
| It also says rulers do not change length when you move them.
|
| But, if you want to be picky-picky,
|
| xi = x-vt
| eta = y
| zeta = z
| tau = t
|
| I thought that might be confusing.
|
| Tau is not t. Differently synchronised clocks. You had it correctly
| befo t=t.

Jeez... feeding candy to a baby is easier than this.

Just how the **** do you synchronize a clock?
Say my watch to your watch?
tau = t = 15:10 pm, done at xi - x = 0, no distance between them.
We can do that in London or LA or Hong Kong.

Then we let them run.

You go you way, I go mine, we meet again somewhere and you think
because you went to the Moon and back while I went to the pub they'll
be showing different times.

Get a grip on reality and pay for my beer, you can pay for your own
trip to the moon as well, and the caesium clocks. Can I keep mine as a
memento of your trip?


|
| In other words, k' and kappa use
| *different* clocks. Namely, kappa uses a set of clocks at rest in
| kappa, synchronised in kappa. And k' uses a set of clocks *not* at
rest
| in k'
|
|
|
|
| Oh yes they are at rest in k', and x' is independent of them, being
| independent of time.
|
| x'=x-vt, dependent of time.

Dingbat, you are disagreeing with Einstein now. L = x-vt.
L is a constant, x' is a constant. It used to be k was a constant, c
was a constant, so Einstein used those up and pretended c was
from celaritas, Latin for speed, and he wants the speed of light
constant. Not in his first German paper, though, there it is V.
A minor slip he soon corrected, he can't have you using c or k
as a constant, you might get wise to his chicanery.

| What's independent of time is the x'
| coordinate of an object at rest in k'.

Yes. One end of a rod. The other end is 0' = 0-vt.
The length of the rod is L = x'-0'.
There is no v between kappa and k', you used it up
with x' = x-vt.
It's gone, there is no more v-candy left. Don't cwy,
I've got some special treats for you when you grow up.
Real physics, would you like that?

|
|
| (they are at rest in K), *not* synchronised in k' (they are
| synchronised in K).
|
| Yes they are synchronized in k', and remain synchronized,
| "in agreement with experience", "it is clear", "it is at once
apparent".
|
| No, t=t, remember?

Of course, and tau = t as well. You owe me a beer. At this rate
you'll owe me a dinner.


The dependence of tau on t is yet unknown so we
| cannot just assume they are equal.

We are not assuming they are equal, we are proving that
they cannot be unequal.

(In fact they come out unequal at
| the end:
|
| tau = (t-vx/c^w)*gamma
|
| ...but that comes later.
|
| [snip]

IN FACT, they do not.

v = 0, you ate it with x' = x-vt, there is no motion between kappa and
k'.
c = 0, c = (x' - x')/t by Einstein's definition, repeated as
****wittery in

"In agreement with experience we further assume the quantity

2AB/ (t'A - tA) = c",

which should be

(AB + BA) / (t'A - tA) = c = 0.

That is in agreement with experience.

tau = (t- 0*x/0^2)* sqrt(1 - 0^2/0^2)

Einstein conned you all the way through, you owe me a beer, sucker.

Androcles.

Einstein in fact had feet of clay in a number of respects. "Gott
wuefelt nichts" was a famous quotation of his relating to Quantum
theory. In fact you get uyncertainty with chaos without even using the
Q word. "Aber das Scmetterling" - a butterfly in Japan can cause a
hurricane in the Gulf. The future, even classically, is a 4 dimensional
box of uncertainty. You see you have to measure more and more precisely
to make predictions. With the Q word there is, of course, an implicit
limit.

I have had some further thoughts.Perhaps for Science to advance we need
feet of clay. Science can be cruel. In many ways it is a pity Leverrier
ended his career the way he did. He corresponded with Adams as a young
man and in 1846 Neptune was disconered. The search for Vulcan is very
much a footnote to General Relativity. I personally much prefer to
think of Neptune rather than the friutless hunt for Vulcan.

Yet the hunt was not totally fruitless, it was a vital ingredient of
GR.

In societies that are NOT free we get Science dictated by the ruler's
whim. The inheritance of acquired characteristcs (Lysenko) being a case
in point. Also unfree socities tend to channel their resources into big
macho prestige projects. Like a man on Mars. I think that LISA and the
Shuttle shows us the way NOT to allocate resources.

The astrnaut corps is basically faschistic. They want $40e9 (probably
more like $100e9 to go to Mars. Why should the rest of society be
forced to provide that level of resources.