Daryl McCullough:

Androcles, you are an idiot.

That is an insult to idiots.

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

But now we do it this way.
Light is emitted from the engine, reflects at the caboose and
returns to the engine.
The time of reflection at the caboose is now tau(0,0,0,t+x'/(c PLUS
v))
Diagramatically,
(fixed font needed).

|
|
| C'
| /
| B /
| ____________Mirror
| /\ /
| / \ /
C / \ /
|\ / \ /
| \ / \A'
| \ / | /
| \ / /
| \ / /
| \ / | /
| \ / /
| \/ |
| /\ /
| / \ / |
| / \ /
| / \ / |
| / ____\/__________Mirror
| /
| / D |
|/
/ ____________|____|________________
A D B A' C'


[quote]
we establish by definition that the "time" required by light to travel
from A to B equals the "time" it requires to travel from B to [A'].
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

We establish by definition that the "time" required by light to travel
from C to D equals the "time" it requires to travel from D to C'.

Distance between mirrors is x'

Einstein's equation:

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))

What it means in the diagram:

½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))

Correct.

So the time at B, the engine, equals the time at D, the caboose,

According to the clock sync of the moving system.

There is no "clock sync" of the moving system, fool!
Time in the moving system (tau) can *supposedly* be different to time
in the stationary system (t), but it can't be different to itself!

You are hopelessly confused. Your equations:

½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))

simply follow from the clock synchronisation in the moving system.
There is nothing contradictory about them although you wrote them a bit
sketchily. Written out explicitly they a

1/2(tau(0,0,0,t) + tau(0,0,0,t+x'/(c-v)+x'/(c+v))) =
tau(x',0,0,t+x'/(c-v))

1/2(tau(x',0,0,t) + tau(x',0,0,t+x'/(c-v)+x'/(c+v))) =
tau(0,0,0,t+x'/(c+v))

Nothing wrong with them.

You've got tau, you've got t. That's it.

but it doesn't.

According to the clock sync of the stationary system.

There is no "clock sync" of the stationary system, you clown!
Time in the stationary system (t) can *supposedly* be different to time
in the moving system (tau), but it can't be different to itself!

Who says it's different than itself? You begin with two systems:
stationary (K) using t for time and moving (k) using tau for time. Both
agree to synchronise their clocks according to the Einstein convention.
We simply seek the formula relating tau (and xi, eta, zeta) to x,y,z,t.

--
Jan Bielawski

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

But now we do it this way.
Light is emitted from the engine, reflects at the caboose and
returns to the engine.
The time of reflection at the caboose is now tau(0,0,0,t+x'/(c PLUS
v))
Diagramatically,
(fixed font needed).

|
|
| C'
| /
| B /
| ____________Mirror
| /\ /
| / \ /
C / \ /
|\ / \ /
| \ / \A'
| \ / | /
| \ / /
| \ / /
| \ / | /
| \ / /
| \/ |
| /\ /
| / \ / |
| / \ /
| / \ / |
| / ____\/__________Mirror
| /
| / D |
|/
/ ____________|____|________________
A D B A' C'


[quote]
we establish by definition that the "time" required by light to
travel
from A to B equals the "time" it requires to travel from B to [A'].
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

We establish by definition that the "time" required by light to
travel
from C to D equals the "time" it requires to travel from D to C'.

Distance between mirrors is x'

Einstein's equation:

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))

What it means in the diagram:

½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))

Correct.

So the time at B, the engine, equals the time at D, the caboose,

According to the clock sync of the moving system.

There is no "clock sync" of the moving system, fool!
Time in the moving system (tau) can *supposedly* be different to time
in the stationary system (t), but it can't be different to itself!

| You are hopelessly confused.


You are infinitely stupid.


| Your equations:

½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))

| simply follow from the clock synchronisation in the moving system.
| There is nothing contradictory about them although you wrote them a
bit
| sketchily. Written out explicitly they a

| 1/2(tau(0,0,0,t) + tau(0,0,0,t+x'/(c-v)+x'/(c+v))) =
| tau(x',0,0,t+x'/(c-v))

| 1/2(tau(x',0,0,t) + tau(x',0,0,t+x'/(c-v)+x'/(c+v))) =
| tau(0,0,0,t+x'/(c+v))

| Nothing wrong with them.

Put some numbers in, then.
c = 5 cars a second (32 car train, each car is 60 metres).
v = 3 = 0.6c
x' = 32 cars, the train length won't change by tomorrow, x' is
independent of time.
t = midnight = 0

1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)

I gave you the diagram.
Still nothing wrong, or are you still infinitely stupid?

You've got tau, you've got t. That's it.

but it doesn't.

According to the clock sync of the stationary system.

There is no "clock sync" of the stationary system, you clown!
Time in the stationary system (t) can *supposedly* be different to
time
in the moving system (tau), but it can't be different to itself!

|Who says it's different than itself?

This does:
tau(32,0,0,16) the time at the engine is 16 microseconds past midnight.
tau(0,0,0,4) the time at the caboose is 4 microseconds past midnight.

Run the train in reverse at midnight tomorrow night.
tau(0,0,0,16) the time at the caboose is 16 microseconds past midnight.
tau(32,0,0,4) the time at the engine is 4 microseconds past midnight.

You've said my equations are correct.
It follows from these results that you are infintiely stupid.

Want to put the cuckoo transformation in?

tau = (t-vx/c^2)/sqrt(1-v^2/c^2)
tau(32,0,0,16) = ?
tau(0,0,0,4) = ?

Go ahead, stupid, make it work and prove tau is a linear function,
because I don't know how. Show me.


| You begin with two systems:
| stationary (K) using t for time and moving (k) using tau for time.
Both
| agree to synchronise their clocks according to the Einstein
convention.
| We simply seek the formula relating tau (and xi, eta, zeta) to
x,y,z,t.


--
Jan Bielawski


You forget the third system, x',y,z,t with x', an end point in the
moving system independent of time which is not xi and not 0.

It is not zero because that would make
1/2(tau(0,0,0,t) + tau(0,0,0,t+0/(c-v)+0/(c+v))) =
1/2(tau(0,0,0,t) + tau(0,0,0,t)) = tau(0,0,0,t) which is trivially true,
half of nothing is nothing.

It is not xi because
x'/xi = sqrt(1-v^2/c^2) not equal 1 (unless v = 0, again a triviality)

Yet if we accelerate the separation between xi and x' increases.
This is the cuckoo length "contraction", and
for x' = 0, xi = 0 at any velocity,
for v = c, xi = x'/sqrt(1-1) = undefined.
For v = 0.866c, xi = 2x' and your stupidity is doubled.

Being infinitely stupid this will not trouble you, because
to you it's interesting - what are the limits of withstanding blatant
contradictions (infinite, OBVIOUSLY).

Androcles.

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

| Your equations:

½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))

| simply follow from the clock synchronisation in the moving system.
| There is nothing contradictory about them although you wrote them a
bit
| sketchily. Written out explicitly they a

| 1/2(tau(0,0,0,t) + tau(0,0,0,t+x'/(c-v)+x'/(c+v))) =
| tau(x',0,0,t+x'/(c-v))

| 1/2(tau(x',0,0,t) + tau(x',0,0,t+x'/(c-v)+x'/(c+v))) =
| tau(0,0,0,t+x'/(c+v))

| Nothing wrong with them.

Put some numbers in, then.

Typical :-)

c = 5 cars a second (32 car train, each car is 60 metres).
v = 3 = 0.6c
x' = 32 cars, the train length won't change by tomorrow, x' is
independent of time.
t = midnight = 0

1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)

Correct.

I gave you the diagram.
Still nothing wrong, or are you still infinitely stupid?

Still nothing wrong. It would be more efficient if you pointed out
precisely where you see a contradiction.

You've got tau, you've got t. That's it.

but it doesn't.

According to the clock sync of the stationary system.

There is no "clock sync" of the stationary system, you clown!
Time in the stationary system (t) can *supposedly* be different to
time
in the moving system (tau), but it can't be different to itself!

|Who says it's different than itself?

This does:
tau(32,0,0,16) the time at the engine is 16 microseconds past midnight.

According to the stationary clock at that spot.

tau(0,0,0,4) the time at the caboose is 4 microseconds past midnight.

According to the stationary clock at that spot.

Run the train in reverse at midnight tomorrow night.
tau(0,0,0,16) the time at the caboose is 16 microseconds past midnight.

According to the stationary clock at that spot.

tau(32,0,0,4) the time at the engine is 4 microseconds past midnight.

According to the stationary clock at that spot.

Still no contradiction.

You've said my equations are correct.
It follows from these results that you are infintiely stupid.

They are correct.

tau = (t-vx/c^2)/sqrt(1-v^2/c^2)
tau(32,0,0,16) = ?
tau(0,0,0,4) = ?

Go ahead, stupid, make it work and prove tau is a linear function,
because I don't know how. Show me.

The tau in the first line is a different function than the tau in the
two lines that follow. Remember that the first tau is a function of
(x,y,z,t), the two other tau are functions of (x',y,z,t), where
x'=x-vt.

So let's write the first tau in terms of (x',y,z,t) which is what the
numbers were given in (I'm using gamma for 1/sqrt(1-v^2/c^2)):

tau = gamma(t-vx/c^2) = gamma(t - v(x'+vt)/c^2)

Now we can plug in your numbers. In your case gamma = 1.25, v=3, c=5.
The two other tau a

tau(32,0,0,16) = 1.25 * (16 - 3(32+3*16)/25) = 8
tau(0,0,0,4) = 1.25 * (4 - 3(0+3*4)/25) = 3.2

(assuming I didn't make a numerical mistake somewhere).

This function tau can be rewritten as:

tau = Ax' + Bt

whe

A = -v/(c*sqrt(c^2-v^2))
B = sqrt(c^2-v^2)/c

....are constant (because v = const.), i.e. tau is a linear function of
(x',y,z,t).

| You begin with two systems:
| stationary (K) using t for time and moving (k) using tau for time.
Both
| agree to synchronise their clocks according to the Einstein
convention.
| We simply seek the formula relating tau (and xi, eta, zeta) to
x,y,z,t.

You forget the third system, x',y,z,t with x', an end point in the
moving system independent of time which is not xi and not 0.

No, I remembered it but wanted to start with the clear setup of what it
is we are after: a transformation between K and k.

--
Jan Bielawski

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

| Your equations:

½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))

| simply follow from the clock synchronisation in the moving system.
| There is nothing contradictory about them although you wrote them a
bit
| sketchily. Written out explicitly they a

| 1/2(tau(0,0,0,t) + tau(0,0,0,t+x'/(c-v)+x'/(c+v))) =
| tau(x',0,0,t+x'/(c-v))

| 1/2(tau(x',0,0,t) + tau(x',0,0,t+x'/(c-v)+x'/(c+v))) =
| tau(0,0,0,t+x'/(c+v))

| Nothing wrong with them.

Put some numbers in, then.

| Typical :-)

Androcles: Relativity is ****ed up.
Bile awski: Can't be, someone would have seen it before.
Androcles: It's the 1/2.
Bile awski: Go read a book and learn.
Androcles: Here's a simple diagram showing why.
Bile awski: Show the math
Androcles: Here's the equations.
Bile awski: Nothing wrong with them.
Androcles: Put some numbers in, then.
Bile awski: Typical + stupid ape grin.
Androcles: You are a ****, your brain is ****ed. **** off.




c = 5 cars a second (32 car train, each car is 60 metres).
v = 3 = 0.6c
x' = 32 cars, the train length won't change by tomorrow, x' is
independent of time.
t = midnight = 0

1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)

| Correct.

I gave you the diagram.
Still nothing wrong, or are you still infinitely stupid?

| Still nothing wrong. It would be more efficient if you pointed out|
| precisely where you see a contradiction.

One step at a time, I'm dealing with a ****ed up infinitely stupid ape.


You've got tau, you've got t. That's it.

but it doesn't.

According to the clock sync of the stationary system.

There is no "clock sync" of the stationary system, you clown!
Time in the stationary system (t) can *supposedly* be different to
time
in the moving system (tau), but it can't be different to itself!

|Who says it's different than itself?

This does:
tau(32,0,0,16) the time at the engine is 16 microseconds past
midnight.

| According to the stationary clock at that spot.

See what I mean? Infinitely stupid.
I have no idea what stationary clock you are talking about
or what spot you are talking about.
But, one step at a time. Let's deal with the stationary spot first.
That's 80, the engine moved from spot 32 at speed 3 for 16 time.
32+ 3*16 = 80.

Does 80 appear in either equation?
No.
Did anyone put a stationary clock at 80?
No.
Is there a "stationary" clock on the engine?
No; there is a clock that moves with the engine.

What are the limits of Bile awski's stupidity? Infinite.
**** off, ****.
Androcles.

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

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

| Your equations:

½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))

| simply follow from the clock synchronisation in the moving system.
| There is nothing contradictory about them although you wrote them a
bit
| sketchily. Written out explicitly they a

| 1/2(tau(0,0,0,t) + tau(0,0,0,t+x'/(c-v)+x'/(c+v))) =
| tau(x',0,0,t+x'/(c-v))

| 1/2(tau(x',0,0,t) + tau(x',0,0,t+x'/(c-v)+x'/(c+v))) =
| tau(0,0,0,t+x'/(c+v))

| Nothing wrong with them.

Put some numbers in, then.

| Typical :-)

Androcles: Relativity is ****ed up.
Bile awski: Can't be, someone would have seen it before.
Androcles: It's the 1/2.
Bile awski: Go read a book and learn.
Androcles: Here's a simple diagram showing why.
Bile awski: Show the math
Androcles: Here's the equations.
Bile awski: Nothing wrong with them.
Androcles: Put some numbers in, then.
Bile awski: Typical + stupid ape grin.
Androcles: You are a ****, your brain is ****ed. **** off.




c = 5 cars a second (32 car train, each car is 60 metres).
v = 3 = 0.6c
x' = 32 cars, the train length won't change by tomorrow, x' is
independent of time.
t = midnight = 0

1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)

| Correct.

I gave you the diagram.
Still nothing wrong, or are you still infinitely stupid?

| Still nothing wrong. It would be more efficient if you pointed out|
| precisely where you see a contradiction.

One step at a time, I'm dealing with a ****ed up infinitely stupid ape.

That is probably what all apes think when they are dealing
with humans.



You've got tau, you've got t. That's it.

but it doesn't.

According to the clock sync of the stationary system.

There is no "clock sync" of the stationary system, you clown!
Time in the stationary system (t) can *supposedly* be different to
time
in the moving system (tau), but it can't be different to itself!

|Who says it's different than itself?

This does:
tau(32,0,0,16) the time at the engine is 16 microseconds past
midnight.

| According to the stationary clock at that spot.

See what I mean? Infinitely stupid.
I have no idea what stationary clock you are talking about
or what spot you are talking about.

the stationary clock that says t = 16 when the engine passes.
Gee, how difficult.

But, one step at a time. Let's deal with the stationary spot first.
That's 80, the engine moved from spot 32 at speed 3 for 16 time.
32+ 3*16 = 80.

Does 80 appear in either equation?
No.
Did anyone put a stationary clock at 80?
No.
Is there a "stationary" clock on the engine?
No; there is a clock that moves with the engine.

And there is a stationary clock at each point.
These clocks show t-time of events.


What are the limits of Bile awski's stupidity? Infinite.
**** off, ****.
Androcles.

The main problem is that you have no idea whatsoever
about the meaning of the variables that appear in the
equations.

Title: I have no idea what you are talking about
so YOU are infinitely stupid:
http://users.pandora.be/vdmoortel/di...elyStupid.html

Jan, you *do* know with what kind of sub-ape you
are dealing here, I hope?

Dirk Vdm

Dirk Van de moortel wrote:
"Androcles" Androcles@ MyPlace.org wrote in message ...

What are the limits of Bile awski's stupidity? Infinite.
**** off, ****.
Androcles.

The main problem is that you have no idea whatsoever
about the meaning of the variables that appear in the
equations.

I wonder why he does this. SR I mean.

--
Jan Bielawski

Androcles wrote:
"JanPB" wrote in message
ups.com...
Androcles wrote:
[...]
c = 5 cars a second (32 car train, each car is 60 metres).
v = 3 = 0.6c
x' = 32 cars, the train length won't change by tomorrow, x' is
independent of time.
t = midnight = 0

1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)

| Correct.
[...]

|Who says it's different than itself?

This does:
tau(32,0,0,16) the time at the engine is 16 microseconds past
midnight.

| According to the stationary clock at that spot.

See what I mean? Infinitely stupid.
I have no idea what stationary clock you are talking about
or what spot you are talking about.

Perhaps that's your problem. What you don't understand is not
automatically stupid, you know. I meant the stationary clock the engine
is passing when that clock shows 16 microseconds. At this instant the
clock *on* the engine shows tau(32,0,0,16) microseconds.

But, one step at a time. Let's deal with the stationary spot first.
That's 80, the engine moved from spot 32 at speed 3 for 16 time.

You now switched coordinates - now you are talking about x, not x'. The
coordinate x' is always 32 at the engine. Changing from (x',y,z,t) to
(x,y,z,t) changes the function tau.

32+ 3*16 = 80.

Does 80 appear in either equation?
No.

Of course not, the equation is for tau(x'y'z't), not for tau(x,y,z,t).

Did anyone put a stationary clock at 80?
No.

Yes. Clock distribution is the basis of Einstein's paper.

Is there a "stationary" clock on the engine?
No;

Not *on* the engine. At every instant the engine passes a stationary
clock.

there is a clock that moves with the engine.

Yes.

--
Jan Bielawski

"JanPB" wrote in message
oups.com...
| Dirk Van de moortel wrote:
| "Androcles" Androcles@ MyPlace.org wrote in message
...
|
| What are the limits of Bile awski's stupidity? Infinite.
| **** off, ****.
| Androcles.
|
| The main problem is that you have no idea whatsoever
| about the meaning of the variables that appear in the
| equations.
|
| I wonder why he does this. SR I mean.

Wonderment, or curiousity, is a natural consequence of all.
Never finding the answer is an inevitable consequence of the infinitely
stupid.
**** off, ****.
Androcles

"JanPB" wrote in message
oups.com...
| Androcles wrote:
| "JanPB" wrote in message
| ups.com...
| Androcles wrote:
| [...]
| c = 5 cars a second (32 car train, each car is 60 metres).
| v = 3 = 0.6c
| x' = 32 cars, the train length won't change by tomorrow, x' is
| independent of time.
| t = midnight = 0
|
| 1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
| 1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)
|
| | Correct.
| [...]
|
| |Who says it's different than itself?
|
| This does:
| tau(32,0,0,16) the time at the engine is 16 microseconds past
| midnight.
|
| | According to the stationary clock at that spot.
|
| See what I mean? Infinitely stupid.
| I have no idea what stationary clock you are talking about
| or what spot you are talking about.
|
| Perhaps that's your problem.

Nope, your problem.


What you don't understand is not
| automatically stupid, you know.

You are automatically stupid.


I meant the stationary clock the engine
| is passing when that clock shows 16 microseconds.

Yes, you dumb ****.
It is at (80,0,0,16) which is not part of the equation.


At this instant the
| clock *on* the engine shows tau(32,0,0,16) microseconds.


So who gives a **** about a stationary clock or what spot
you are talking about, since it isn't part of the equation?
(rhetorical question, you are too stupid to answer sensibly.)





|
| But, one step at a time. Let's deal with the stationary spot first.
| That's 80, the engine moved from spot 32 at speed 3 for 16 time.
|
| You now switched coordinates - now you are talking about x, not x'.

You did that by prattling on about a "stationary clock at that spot".
That "spot" is x, not x'.



| The
| coordinate x' is always 32 at the engine.

Yes, dumb****.

Changing from (x',y,z,t) to
| (x,y,z,t) changes the function tau.

Why are you trying to do it then, dumb****?




| 32+ 3*16 = 80.
|
| Does 80 appear in either equation?
| No.
|
| Of course not, the equation is for tau(x'y'z't), not for tau(x,y,z,t).

Correct.


| Did anyone put a stationary clock at 80?
| No.
|
| Yes. Clock distribution is the basis of Einstein's paper.

Infinitely stupid dumb****, there are no clocks in the equation, let
alone one at 80.


| Is there a "stationary" clock on the engine?
| No;
|
| Not *on* the engine.

Right. There is a "moving" clock on the engine that is at rest with the
engine.
We call it the "driver's 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.


| there is a clock that moves with the engine.
|
| Yes.

Ok, so what are you prattling on about a "stationary clock at that spot"
for?
Answer: Because you are infinitely stupid.

Androcles.