Re: Einstein's moving system coordinates aren't. ROFFMFAO!


"Hexenmeister" <vanquish@xxxxxxxxxxxxxx> wrote in message news:Dsr3g.12689$tc.5404@xxxxxxxxxxxxxxxxxxxxxxxxxxxx

"Paul B. Andersen" <paul.b.andersen@xxxxxxxxxxxxxxxx> wrote in message
news:e2lapo$5qo$1@xxxxxxxxxxxxxxxxxxx
| eleaticus wrote:
| >
| > The real question:
| >
| > is your post because you didn't take some (prescribed) drug or because
you
| > did take some (illegal) drug?
| >
| > Try again when you are (a) back on your meds (b) off the illegal (c)
have
| > removed your head from up your ass.
| >
| > eleaticus
| > ee-lee-aT-i-cus
|
| Since you respond in such a friendly manner, I will
| take the time to go through it again.
|
| Eleaticus wrote:
|| Let there be at your stationary system origin a hundred mile-long highway
|| extending along your x-axis to, of course, x=100.
||
|| As the moving system moves by at v~.866c (so g=gamma= 2), when its origin
is
|| at your x=0 you set your clock to zero and calculate what the moving
|| observer will see: x' = g(x-vt). t=0 and g=2 so: x'=200.
|
| And this is quite correct.
| I, being in the stationary system, can calculate
| that when the moving observer "see" the event
| "the end of the road x = 100, at the time t = 0",
| then the coordinates of this event in the moving frame are:
| x' = g(x - vt) = 200
| t' = g(t - xv/cc) = -173.2/c
|
| That is, as you so correctly stated, the moving observer
| will "see" the end of the road at his x' = 200.
| He will also see that his own clock is t' = -173.2/c.

ahahaha..HAHAHAHA...hahaha...


| You didn't mention this, but I am sure that a clever guy
| like you were aware of this.

ahaha...HAHAHA... hahaha...
Can Eleaticus have a diploma?


|
| Since the moving system moves in the positive x direction
| with the speed v, it is pretty obvious that seen from
| the moving system, the stationary system moves in the negative
| x' direction with the same speed v.

So the clock runs backwards...
ahaha...HAHAHA... hahaha...

|
| So since the end of the road (x = 100) moves in the negative
| x' direction with the speed v, it is easy to calculate where
| the end of the road will be at t' = 0:
| x' = 200 - (173.2/c)(0.866c) = 50
|
| That is, when the moving observer's clock shows t' = 0,
| he will "see" the end of the road at x' = 50.
|
| Isn't this rather obvious, Eleaticus?

Sure... clocks run backwards all the time..
ahaha...HAHAHA... hahaha...

| Or is it something you don't agree to so far?

No, of course not. Everyone knows clocks run backwards.
ahaha...HAHAHA... hahaha...

|
|| But consider the 'actual' moving system measurement. He too has set his
|| clock to zero and he too sees gamma=g as being 2. Our x=100 is his x' =
100
|| = g(x'-vt'), so with t' =0 he gets, x'=50.
|
| So the question is, when the moving observer sets his clock to zero,
| t' = 0, where does he then "see" the end of the road (x = 100)?
| We need only one of the transform equations to answer that question:
| x = g(x'+ v t') where x = 100, t' = 0 and g = 2.
| So the answer is, as you so correctly stated, x = 50
|
| When the moving observer's clock show t' = 0, he "sees" the end
| of the road (x = 100) at x' = 50.
|
| So we agree on everything.
| Or don't we?
|
| Is there a problem somewhere?

Maybe...
"He will also see that his own clock is t' = -173.2/c." -- might be
slightly questionable.
ROFLMAO!
Androcles.

Precious little series:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NegTime4.html
together with
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NegTime3.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NegTime2.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NegTime.html

Dirk Vdm


.

Relevant Pages

  • OT: Einssteins BooBoos - A^2
    ... In his seminal 1905 paper - title "On the Electrodynamics of Moving ... of the BEER to argue that a moving clock slows (dilates) ... stationary system observer "sees" that the moving clock "says' at some ... Every single-value examination of the time transformation equation demands ...
    (rec.gambling.poker)
  • Re: Einsteins moving system coordinates arent. ROFFMFAO!
    ... | eleaticus wrote: ... || Let there be at your stationary system origin a hundred mile-long highway ... | that when the moving observer "see" the event ... | He will also see that his own clock is t' = -173.2/c. ...
    (sci.physics.relativity)
  • Re: Confusion about time.
    ... "ON THE ELECTRODYNAMICS OF MOVING BODIES" Section 4 ... time t when at rest relatively to the stationary system, ... What is the rate of this clock, ... moving at v relative to the observer. ...
    (sci.physics.relativity)
  • Re: Einsteins moving system coordinates arent. ROFFMFAO!
    ... eleaticus wrote: ... I, being in the stationary system, can calculate ... that when the moving observer "see" the event ... He will also see that his own clock is t' = -173.2/c. ...
    (sci.physics.relativity)
  • Re: Some Contradictory Claims in SR:
    ... >> physical length of a moving ruler is the same as that of the ... we observe that a clock records different elapsed time after a journey. ... >> different length in different frames. ... >> do physics will result different physical laws for different frames. ...
    (sci.physics)