"Androcles" Androcles@ MyPlace.org wrote in message .. .

[snip]

Answer: Because you are infinitely stupid.

Infinitely Stupid Revisited:
http://users.pandora.be/vdmoortel/di...lyStupid2.html

Dirk Vdm

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. 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 **** 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 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.

| 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 wris****ch", infinitely stupid dumb****.


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

"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.


| 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,
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.






|
| 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.
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 he
http://webexhibits.org/calendars/year-text-Galileo.html
and progress to he
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 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?
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 wris****ch", infinitely stupid dumb****.
|
|
| 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.

"Androcles" Androcles@ MyPlace.org wrote in message news
[snip almost 400 lines of dog ****]


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

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

"JanPB" wrote in message
oups.com...
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.

Bull****, they've only got one clock each. Learn to count.


| We seek tau(x,y,z,t) which is a function

Then what the **** are you using tau(x',y',z',t') for, dumb****?

Androcles.

Androcles wrote:
"JanPB" wrote in message
oups.com...
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.

Bull****, 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.

| We seek tau(x,y,z,t) which is a function

Then what the **** are you using tau(x',y',z',t') for, dumb****?

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

"JanPB" wrote in message
oups.com...
| Androcles wrote:
| "JanPB" wrote in message
| oups.com...
| 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.
|
| Bull****, 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.
Likewise you are a phuckwit and a ****wit.



|
| | We seek tau(x,y,z,t) which is a function
|
| Then what the **** are you using tau(x',y',z',t') for, dumb****?
|
| 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?
**** off, you are drunk, indecent, dishonest and a troll.
Androcles

Androcles wrote:
"JanPB" wrote in message
oups.com...
| Androcles wrote:
|
| Bull****, 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 **** are you using tau(x',y',z',t') for, dumb****?
|
| 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

"JanPB" wrote in message
oups.com...
| Androcles wrote:
| "JanPB" wrote in message
| oups.com...
| | Androcles wrote:
| |
| | Bull****, 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 **** are you using tau(x',y',z',t') for, dumb****?
| |
| | 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.

****ing idiot.
Stupid ****.
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 ****: 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 ****ing idiot: 4 = 80/(5+3)

****ing 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, ****wit.


|
| 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.