Re: Bile awski the Janitor.



Androcles wrote:
> "JanPB" <filmart@xxxxxxxxx> wrote in message
> news:1127198662.061600.115280@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

> | Your equations:
>
> > > ½[tau(A,t)+tau(A',t+x'/(c-v)+x'/(c+v))] = tau(B,t+x'/(c-v))
> > > ½[tau(C,t)+tau(C',t+x'/(c-v)+x'/(c+v))] = tau(D,t+x'/(c PLUS v))
>
> | simply follow from the clock synchronisation in the moving system.
> | There is nothing contradictory about them although you wrote them a
> bit
> | sketchily. Written out explicitly they are:
>
> | 1/2(tau(0,0,0,t) + tau(0,0,0,t+x'/(c-v)+x'/(c+v))) =
> | tau(x',0,0,t+x'/(c-v))
>
> | 1/2(tau(x',0,0,t) + tau(x',0,0,t+x'/(c-v)+x'/(c+v))) =
> | tau(0,0,0,t+x'/(c+v))
>
> | Nothing wrong with them.
>
> Put some numbers in, then.

Typical :-)

> c = 5 cars a second (32 car train, each car is 60 metres).
> v = 3 = 0.6c
> x' = 32 cars, the train length won't change by tomorrow, x' is
> independent of time.
> t = midnight = 0
>
> 1/2[tau(0,0,0,0) + tau(0,0,0,16+4)] = tau(32,0,0,16)
> 1/2[tau(32,0,0,0) + tau(32,0,0,20)] = tau(0,0,0,4)

Correct.

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

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

> > You've got tau, you've got t. That's it.
> >
> > > but it doesn't.
> >
> > According to the clock sync of the stationary system.
> >
> > There is no "clock sync" of the stationary system, you clown!
> > Time in the stationary system (t) can *supposedly* be different to
> > time
> > in the moving system (tau), but it can't be different to itself!
>
> |Who says it's different than itself?
>
> This does:
> tau(32,0,0,16) the time at the engine is 16 microseconds past midnight.

According to the stationary clock at that spot.

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

According to the stationary clock at that spot.

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

According to the stationary clock at that spot.

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

According to the stationary clock at that spot.

Still no contradiction.

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

They are correct.

> tau = (t-vx/c^2)/sqrt(1-v^2/c^2)
> tau(32,0,0,16) = ?
> tau(0,0,0,4) = ?
>
> Go ahead, stupid, make it work and prove tau is a linear function,
> because I don't know how. Show me.

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

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

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

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

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

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

This function tau can be rewritten as:

tau = Ax' + Bt

where:

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

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

> | You begin with two systems:
> | stationary (K) using t for time and moving (k) using tau for time.
> Both
> | agree to synchronise their clocks according to the Einstein
> convention.
> | We simply seek the formula relating tau (and xi, eta, zeta) to
> x,y,z,t.
>
> You forget the third system, x',y,z,t with x', an end point in the
> moving system independent of time which is not xi and not 0.

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

--
Jan Bielawski

.



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