Androcles wrote:
"JanPB" wrote in message
ups.com...
| Androcles wrote:
| "JanPB" wrote in message
| oups.com...
| [...]
| I meant the stationary clock the engine
| | is passing when that clock shows 16 microseconds.
|
| Yes, you dumb ****.
| 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 **** 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 dumb****, 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