Re: Simultan events within the bubble of SR can take place pretty much anywhere in space

In sci.physics.relativity, jt64@xxxxxxxx
<jt64@xxxxxxxx>
wrote
on 24 Mar 2007 15:40:50 -0700
<1174776050.574690.145650@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
On 24 Mar, 22:50, The Ghost In The Machine
<e...@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
In sci.physics.relativity, j...@xxxxxxxx
<j...@xxxxxxxx>
wrote
on 24 Mar 2007 13:47:45 -0700
<1174769265.226660.295...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:





On 24 Mar, 21:24, "Eric Gisse" <jowr...@xxxxxxxxx> wrote:
On Mar 24, 6:39 am, j...@xxxxxxxx wrote:

On 24 Mar, 10:17, "Dirk Van de moortel" <dirkvandemoor...@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
"Eric Gisse" <jowr...@xxxxxxxxx> wrote in messagenews:1174725457.082237.137200@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Mar 23, 2:07 pm, j...@xxxxxxxx wrote:
How come that simultan events for an object within SR can for a moving
object within the theory take place at different spaces.

...because simultaneity is relative.

This is discussed in every SR textbook. Try reading one.

but preferably not before the age of 13 :-)

Dirk Vdm

So this is where the dumbfucks gather up trying to scare an opponent
away by imtimidation but actually not put forward any argument lol.

Physics textbooks _are_ scary. So much math!

It is so unreasonable of us to expect you to understand a little
mathematics so you could answer your own questions!

Well i am used you brainless lowIQ morons, only 'TOTAL INCAPACITY OF
ANY STRAIGHT LINE OF THOUGHT* could lead anyone to beleive that two
*SIMULTANEOUS EVENTS IN SAME FRAME* actually can take place at two
different places in space. I don't care how fucking much you want to
dilate the time in the frame, *IDIOTS THE UNSYMMETRIC RATIO WILL BE
THE SAME*. Because you most stupid fuckers do no realise that you deal
with an vector dependent transformation.

8 years of whining about relativity, and you still haven't learned
anything. Impressive.

YOUR FUCKING SHIP HAVE TWO CLOCKS, ONE IN THE BACK AND ONE IN THE
FRONT AND THEY WILL SHOW SAME TIME UNTIL THE FUCKING SHIP HALTS

*****AND THAT IS BECAUSE......................
THE FUCKING SHIP CAN NOT STOP AT TWO PLACES IN THE SAME MOMENT*******

So now you low IQ morons go fucking masturbate dream of Captain Kirk
and wormholes. You fucking loo diggers.

*giggles*

My physics degree is a year or so from completion, what about you? Are
you still stuck in primary school?

JT- Dölj citerad text -

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No i am stuck with relativists at sci.physics.relativity but thank god
they will never let a theoretical phycisist have any important job at
NASA.

The fucking stupid theoretical physicians is stuck at institution
teaching small sci-fi loving morons, become fullblown idiots like
Erich Gisse.

Interesting notion. So, since Special Relativity is
apparently being perpetuated by physics professors in
the context of a corrupted system, it should not be taught?

I might agree with you that far, except that the Universe
is laughing at all of us, and producing, at least as of
right now, results generally consistent with Special and
General Relativity (there is an exception but I can't
say I understand it; I'll refer you to Uncle Al regarding
chirality -- no, it's nowhere near Newtonian/Galilean).

As for clocks on a ship -- the discrepancy is on the
order of 3.3 nanoseconds per meter (since that's about
how long it takes a light ray to travel 1 meter) at the
very very most. You'll have to restate your thought-problem
to this "low-IQ moron" (me) in order for me to analyze it.

Yeah you low IQ moron have to start with a simpler excercise, if a
ship pass any object in space at 0.9999 c and at that *EXACT* moment a
light is sent out from the ship how long has the lightfront expanded
in front of the ship *AFTER ONE SECOND PASSED IN THE REST SPACE OF
THE OBJECT*

0.0001 light second, of course. Did you expect a different answer?


What is the spatial separation between the object and the ship and the
object and the lightfront after one second in the *REST SPACE OF THE
OBJECT*.

Again, 0.0001 light second.


You see i have to take it slowly with low IQ morons who do not realise
that a lighfront within the ships back will expand symmetrically
forward with a lightfront in the ships front if sent symmetrically.

No, it won't. The light wave in *back* of the ship will
be 1.9999 light-second behind the spacecraft, as measured
by the object -- assuming the object can in fact measure such;
see below.


It further follows that a lightfront expanding from the ships inner
fronthull in a backward direction will expand symmetrically backward
to the backhull with a front sent from the backend of ship and
backwards lightpaths sent from fronthull and back of the ship must
have same length of if puls same spatial separation.

Are you able to follow the reasoning and answer the question so far.

You'll have to clarify your reasoning, mostly because
there are several problems.

[1] One cannot observe lightwaves unless they hit the observer.
Mirrors can be used to direct them if necessary.

[2] A ship passing an object will have to be some distance
away from it unless one wants it to actually hit the
object. Therefore, the notion of a light wave starting
from the ship exactly as it passes an object needs to be
clarified a small amount, though for problems of this type
it's probably not worth the bother.

[3] 0.9999c is impossible anyway, but never mind that.

So, a modified experiment might be along the following.

A spacecraft is moving at 0.9999 c past three marker
buoys. The outer marker buoys have big mirrors (or
secondary relay cameras) which can direct light waves
from the ship's path into the middle buoy. The ship
is 300 m in length, with the ends of the ship equipped
with flashing lights. A third light amidships flashes
such that an amidships observer will see all three
lights flash simultaneously.

The mirrors can pick up the signals from all three
lights; the middle buoy can see the reflections.
The middle buoy can also observe all lights directly,
using two cameras each covering about 90 degrees
field-of-view. (For purposes of this problem the
actual size and placement of these cameras are ignored.
A real camera 3m in length might introduce a 10 ns error.)

The other two buoys are 30 km distant, motionless with
respect to the central buoy, and all three buoys are
collinear to a line parallel to and a small distance
from the ship's path.

For additional clarity all lights are sodium affairs,
with a readily identifiable double wavelength D line.
For simplicity we'll merely assume 589 nm.

At the exact time the middle light flashes, it is
passing by the middle buoy; the middle buoy can see
this directly. What will the buoy see regarding the
other two lights, and what will the buoy see regarding
all three lights as reflected by the other buoys'
mirrors?

SR predicts the following.

(a) For clarity, ship time will be designated subscript-S,
and buoy time subscript-O. Time zero is the exact
point at which the midpoint of the spacecraft passes
the middle buoy. The displacement between buoy and
ship at the moment of passage is for now ignored.

(b) The events are (0m,0s)_S, (150m,-5*10^-4s)_S, and (-150m,-5*10^-4)_S,
as specified. For simplicity, m will be replaced by u, with 1u
being 1 lightsecond. The events are therefore (0,0)_S,
(5*10^-4,-5*10^-4)_S, and (-5*10^-4,-5*10^-4)-S. These transform
into (0,0)_O, (g*9.9995*10^-4,-g*9.9995*10^-8)_O, and
(-g*5*10^-8,-g*5*10^-8)_O, where g=1/sqrt(1-v^2/c^2)
is approximately 70.712. The middle buoy will therefore
see two flashes, at 0 and at 7.0709 * 10^-2, as the
light wave from forward has to travel back.

(c) The front marker buoy is 30 km distant, which in our new
units means it's 10^-4u distant. Therefore, the
leading flash is far ahead of the buoy; the middle
and rear lights will reflect back showing up as a
single flash but with two disparate wavelengths, at
2 * 10^-4 past zero.

(d) The back marker buoy will send one signal at
7.0909 * 10^-2, as the front light's waves have
to travel that much father back before being
reflected forward. The middle light will show up
at 2 * 10^-4 again. The rear light has a bit
of a head start and its reflection will be observed
by the middle buoy at about 1.929 * 10^-4 -- *before*
the reflection from the middle light, as the middle
light is farther away.

(e) The marker buoy will see three wavelengths. 589 nm
is only seen for the middle light, and only by the
direct camera, not by the relays. The forward light
is seen far into the infrared by the direct camera --
83.3 microns. The tail light is somewhere in UV or
even soft X-ray territory at 4.16 nm, as observed
by the direct camera. The determination of the
wavelengths from the mirrors is left as an exercise,
but here's a hint: one can get either 83.3 microns or
4.16 nm, and nothing else.

(Note that 83.3 microns = 589 nm * sqrt(1+.9999)/sqrt(1-.9999).)

Now, is this clear to everybody? I certainly hope so;
specified in this fashion it's reasonably clear to me.
Since this is a thought experiment I'm not sure how to
easily verify these predictions, but various experiments
corroborate both SR and GR.



JT
JT

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