The relativity of simultaneity
bill
Simultaneity.
In ‘Relativity the Special and General Theory’ Einstein points out
that from the stationary observer’s point of view (M) flash B (toward
which the train is moving) will reach the passenger before flash A.
Will flash B arrive at the passenger’s location at the same point
relatively to the platform as seen by the stationary observer?
I.e. if there is a second stationary observer (S) located at the point
where, according to M, flash B arrives at the passenger’s location
will the passenger see flash B when he is at the same location as S?
Sue...
The thought experiment doesn't use a model of light propagation
which we would use today.
http://en.wikipedia.org/wiki/Emitter_theory
Einstein’s theory of wavefronts versus
Einstein’s relativity of simultaneity
http://arxiv.org/abs/physics/0606233
So it has several ambiguous interpretations.
Your time will be better spent understanding
how formal space-time coordinate systems are
used. They don't need permission from the
philosophers of 1905 to use an imaginary
time axis. ;-)
Minkowski’s Four-Dimensional Space
http://www.bartleby.com/173/17.html
Space-time
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Then you have some tools for real problems.
Relativistic particle dynamics
http://farside.ph.utexas.edu/teaching/em/lectures/node126.html
Sue...
papa...@gmail.com
Yes, observer S being in the same frame as observer M will see light
signal coming from B arriving to the train observer M'.
Miguel Rios
bill
> Yes, observer S being in the same frame as observer M will see light
> signal coming from B arriving to the train observer M'.
>
> Miguel Rios
Thanks Miguel - but will the train passenger see the light signal
coming from B when he is at the same location as S?
bill
> On Jul 15, 7:58 pm, bill <cosmo...@optusnet.com.au> wrote:
>
> > Regarding Einstein’s train gedanken depiction of the Relativity of
> > Simultaneity.
>
> > In ‘Relativity the Special and General Theory’ Einstein points out
> > that from the stationary observer’s point of view (M) flash B (toward
> > which the train is moving) will reach the passenger before flash A.
>
> > Will flash B arrive at the passenger’s location at the same point
> > relatively to the platform as seen by the stationary observer?
>
> > I.e. if there is a second stationary observer (S) located at the point
> > where, according to M, flash B arrives at the passenger’s location
> > will the passenger see flash B when he is at the same location as S?
>
> The thought experiment doesn't use a model of light propagation
> which we would use today.
Thanks Sue, but my question was specifically in relation to that
particular thought experiment.
Androcles
"bill" <cosm...@optusnet.com.au> wrote in message
news:0d32297c-48e7-4961...@z66g2000hsc.googlegroups.com...
Regarding Einstein’s train gedanken depiction of the Relativity of
Simultaneity.
In ‘Relativity the Special and General Theory’ Einstein points out
that from the stationary observer’s point of view (M) flash B (toward
which the train is moving) will reach the passenger before flash A.
=============================================
Well, the idiot was wrong.
Why did Einstein say
the speed of light from A to B is c-v,
the speed of light from B to A is c+v,
the "time" each way is the same?
----------------------------------------------------------------------------------
Will flash B arrive at the passenger’s location at the same point
relatively to the platform as seen by the stationary observer?
I.e. if there is a second stationary observer (S) located at the point
where, according to M, flash B arrives at the passenger’s location
will the passenger see flash B when he is at the same location as S?
=============================================
Yes.
Now that we've established the idiot was a self-contradictory clown,
why keep prattling on about the moron?
harry
"bill" <cosm...@optusnet.com.au> wrote in message
news:0d32297c-48e7-4961...@z66g2000hsc.googlegroups.com...
: Simultaneity.
: In ‘Relativity the Special and General Theory’ Einstein points out
: that from the stationary observer’s point of view (M) flash B (toward
: which the train is moving) will reach the passenger before flash A.
Ok, in the case of two simultaneous flashes (at each side of the train)
according to two observers M and M' near or in the middle of the train at
the time of the flashes (note that that will be so according to *any*
stationary observer; it's just that the two flashes will arrive at M at the
same time).
A-----------------B
-> train
: Will flash B arrive at the passenger’s location at the same point
: relatively to the platform as seen by the stationary observer?
Sure - that is called an event. Events cannot differ between views, as that
would lead to contradictions: everyone must agree on what locally happens
(the simulateneity problem is only about distant events).
: I.e. if there is a second stationary observer (S) located at the point
: where, according to M, flash B arrives at the passenger’s location
: will the passenger see flash B when he is at the same location as S?
Yes, that's even a way to verify that statement. Note that that "when"
refers to measurements in the stationary system (you could write it t1, to
distinguish it from the passenger's time t1'). And of course, all this will
only work if the passenger doesn't use the stationary system time keeping
but made a recalibration for the moving system.
Regards,
Harald
rval...@gmail.com
bill ha escrito:
Yes.
RVHG (Rafael Valls Hidalgo-Gato)
rval...@gmail.com
Androcles ha escrito:
> "bill" <cosm...@optusnet.com.au> wrote in message
> news:0d32297c-48e7-4961...@z66g2000hsc.googlegroups.com...
> Regarding Einstein�s train gedanken depiction of the Relativity of
> Simultaneity.
>
> In �Relativity the Special and General Theory� Einstein points out
> that from the stationary observer�s point of view (M) flash B (toward
> which the train is moving) will reach the passenger before flash A.
> =============================================
> Well, the idiot was wrong.
>
Einstein was right.
> Why did Einstein say
> the sped of light from A to B is c-v,
False. Einstein said {But the ray moves relatively to M', when
measured in the stationary system (the embankment), with the velocity
c-v,...}. Relative velocity between two moving entities (the ray and
M') is NOT the speed of light c from A to B. YOU are the unique one
saying that "the sped of light from A to B is c-v,".
> the speed of light from B to A is c+v,
False. Einstein said {But the ray moves relatively to M', when
measured in the stationary system (the embankment), with the velocity c
+v,...}. Relative velocity between two moving entities (the ray and
M') is NOT the speed of light c from B to A. YOU are the unique one
saying that "the speed of light from B to A is c+v,".
> the "time" each way is the same?
>
Yes, is the same. Distance from A to B equals distance from B to A
(geometry is Euclidean), and the speed c is the same in both
directions (by postulate AND experience).
> ----------------------------------------------------------------------------------
>
>
> Will flash B arrive at the passenger�s location at the same point
> relatively to the platform as seen by the stationary observer?
>
> I.e. if there is a second stationary observer (S) located at the point
> where, according to M, flash B arrives at the passenger�s location
> will the passenger see flash B when he is at the same location as S?
>
> =============================================
> Yes.
>
> Now that we've established the idiot was a self-contradictory clown,
> why keep prattling on about the moron?
RVHG (Rafael Valls Hidalgo-Gato)
Martin Hogbin
Fistly, beware of the crackpots, there are many on this
group. Just ignore them.
Secondly, yes. What you are doing is setting up a
'stationary' reference frame in which events can be
assigned time and space ccordinates. This is the
conceptial mechanism by which M knows when the flash
reaches the passenger.
Flash B reaching the passenger is one event in all
frames. All that is different is the time and space
coordinates that differnt frames assign to that event.
Martin Hogbin
PD
> Regarding Einstein’s train gedanken depiction of the Relativity of
> Simultaneity.
>
> In ‘Relativity the Special and General Theory’ Einstein points out
> that from the stationary observer’s point of view (M) flash B (toward
> which the train is moving) will reach the passenger before flash A.
>
> Will flash B arrive at the passenger’s location at the same point
> relatively to the platform as seen by the stationary observer?
No, of course not. The train moves along the track, carrying the
passenger with it, while the light signal propagates.
Here, does this help?
http://www.youtube.com/watch?v=wteiuxyqtoM
Martin Hogbin
> On Jul 15, 6:58 pm, bill <cosmo...@optusnet.com.au> wrote:
>> Regarding Einstein’s train gedanken depiction of the Relativity of
>> Simultaneity.
>>
>> In ‘Relativity the Special and General Theory’ Einstein points out
>> that from the stationary observer’s point of view (M) flash B (toward
>> which the train is moving) will reach the passenger before flash A.
>>
>> Will flash B arrive at the passenger’s location at the same point
>> relatively to the platform as seen by the stationary observer?
>
> No, of course not. The train moves along the track, carrying the
> passenger with it, while the light signal propagates.
> Here, does this help?
> http://www.youtube.com/watch?v=wteiuxyqtoM
I think one of us may have misunderstood the question.
Martin Hogbin
bill
>
> I think one of us may have misunderstood the question.
>
> Martin Hogbin
On the basis that your response agrees with my line of thinking
it is obvious that you were not the one that misunderstood
the question :)
PD
Apparently, and I have no doubt who. Hopefully the link to the
animation will answer the question for the OP anyway.
Sue...
I have a video that shows the Wicked Witch of the West
melting so I have never doubted that witches can do that.
If you have a video that shows light moving as a massive
particle, then it must be true.
http://en.wikipedia.org/wiki/Emitter_theory
http://nobelprize.org/physics/articles/ekspong/index.html
http://www.quackwatch.org/01QuackeryRelatedTopics/pseudo.html
Sue...
PD
> On Jul 17, 12:23 am, PD <TheDraperFam...@gmail.com> wrote:
>
>
>
> > On Jul 16, 1:52 pm, Martin Hogbin <goatNOSP...@hogbin.org> wrote:
>
> > > PD wrote:
> > > > On Jul 15, 6:58 pm, bill <cosmo...@optusnet.com.au> wrote:
> > > >> Regarding Einstein’s train gedanken depiction of the Relativity of
> > > >> Simultaneity.
>
> > > >> In ‘Relativity the Special and General Theory’ Einstein points out
> > > >> that from the stationary observer’s point of view (M) flash B (toward
> > > >> which the train is moving) will reach the passenger before flash A.
>
> > > >> Will flash B arrive at the passenger’s location at the same point
> > > >> relatively to the platform as seen by the stationary observer?
>
> > > > No, of course not. The train moves along the track, carrying the
> > > > passenger with it, while the light signal propagates.
> > > > Here, does this help?
> > > >http://www.youtube.com/watch?v=wteiuxyqtoM
>
> > > I think one of us may have misunderstood the question.
>
> > Apparently, and I have no doubt who. Hopefully the link to the
> > animation will answer the question for the OP anyway.
>
> I have a video that shows the Wicked Witch of the West
> melting so I have never doubted that witches can do that.
>
> If you have a video that shows light moving as a massive
> particle, then it must be true.
But it doesn't move as a massive particle. Massive particles don't
move at c.
Sue...
Are you claiming light is a particle? There is money in
it it for you if you can prove it.
<<Now, does not the prize to Einstein imply
that the Academy recognised the particle
nature of light? The Nobel Committee says
that Einstein had found that the energy exchange
between matter and ether occurs by atoms emitting
or absorbing a quantum of energy,hv .
As a consequence of the new concept of light quanta
(in modern terminology photons) Einstein proposed the
law that an electron emitted from a substance by
monochromatic light with the frequency has to have
a maximum energy of E=hv-p, where p is the energy needed to
remove the electron from the substance. Robert Andrews
Millikan carried out a series of measurements over a
period of 10 years, finally confirming the validity of this
law in 1916 with great accuracy. Millikan had, however,
found the idea of light quanta to be unfamiliar and strange.
The Nobel Committee avoids committing itself to the
particle concept. Light-quanta or with modern terminology,
photons, were explicitly mentioned in the reports on
which the prize decision rested only in connection with
emission and absorption processes. The Committee says
that the most important application of Einstein's photoelectric
law and also its most convincing confirmation has come from
the use Bohr made of it in his theory of atoms, which explains
a vast amount of spectroscopic data. >>
http://nobelprize.org/physics/articles/ekspong/index.html
http://en.wikipedia.org/wiki/Emitter_theory
Sue...
http://www.quackwatch.org/01QuackeryRelatedTopics/pseudo.html
>
>
>
> >http://en.wikipedia.org/wiki/Emitter_theoryhttp://nobelprize.org/phys...
>
> > Sue...
bill
> > Will flash B arrive at the passenger’s location at the same point
> > relatively to the platform as seen by the stationary observer?
>
> No, of course not. The train moves along the track, carrying the
> passenger with it, while the light signal propagates.
> Here, does this help?http://www.youtube.com/watch?v=wteiuxyqtoM
>
Thanks for that reference but as far as I can see it says
nothing in relation to my question as it only relates to
the stationary observer and the passenger.
Bill
kenseto
> Regarding Einstein’s train gedanken depiction of the Relativity of
> Simultaneity.
>
> In ‘Relativity the Special and General Theory’ Einstein points out
> that from the stationary observer’s point of view (M) flash B (toward
> which the train is moving) will reach the passenger before flash A.
This assertion of Einstein is wrong. Why? Because it is based on
closing velocities between the light fronts from A and B and the train
observer M'. Closing velocities as perceived by the track observer (M)
has no effect on the propagation of light fronts in the train. Closing
velocities is not observed by the train observer M'. The speed of
light from different directions in the M' frame is isotropic. What
this mean is that the SR concept of relativity of simultaneity (RoS)
is based on a false premise. This also mean that any SR paradox (such
as the pole and the barn paradox) that uses RoS to resolve will remain
a paradox.
If the track observer wants to determine if the flashes from A and B
will arrival will arrive at M' simultaneously he uses the LT as
follows:
A and B are at equal distance of 0.5L from M'.
The contracted length of 0.5L is equal to 0.5L/gamma.
Therefore the arrival time (according to the train clock) of the
flashes from A and B is equal to 0.5L/c*gamma.
This means that according to the LT the flashes will arrive at the
train observer simultaneously. This also means that the SR concept of
RoS is bogus. Note that the LT does not use the closing velcities as
perceived by the track observer to determine the simultaneity of
events in the train frame.
Ken Seto
PD
Ken has been corrected on each and every one of the mistakes in this
post numerous times. Unfortunately, he can no longer retain what he
learns on Tuesday to change what he says on Wednesday. As a result, he
will continue to loop over and over again. Be warned that there is
little value in trying to explain anything to him.
PD
Sue...
Then go back to the old posts and correct them.
He has rigourous support for the content
of THIS post.
http://en.wikipedia.org/wiki/Emitter_theory
Einstein’s theory of wavefronts versus
Einstein’s relativity of simultaneity
http://arxiv.org/abs/physics/0606233
All you are offering for to his statements
is insults and camp-fire tales.
Sue...
>
> PD
kenseto
Hey idiot there is no mistake in what I said. Using closing velocities
to determine the arrival time of the light fronts from A and B is
wrong. Besides it violates the isotropy of the speed of light in the
train frame. You are stupid.
Ken Seto
>Unfortunately, he can no longer retain what he
> learns on Tuesday to change what he says on Wednesday. As a result, he
> will continue to loop over and over again. Be warned that there is
> little value in trying to explain anything to him.
>
> PD- Hide quoted text -
>
> - Show quoted text -
papa...@gmail.com
The only stupid brainless here is you.
You have been shown not one but numerous times that the closing speed
is measured by the track observer and not by the train observer. Why
you insist on placing those closing speeds in the train frame is only
your fault. You also are unable to provide any single formula and
diagram proving your nonsense.
OK, once more I will put here the complete derivation, showing the
relativity of simultaneity. I'll use some numbers, so you assert it is
wrong say exactly where you think it is wrong. I'm using the same
notation as Einstein in http://www.bartleby.com/173/9.html
v= 0.6c = 180000 km/sec
c= 300000 km/sec
sqrt(1-v^2/c^2) = 0.8
Let K be the frame of the track observer M. Also let K’ be the frame
of the train observer M’. The Lorentz transformation equations between
these two frames are:
x'=(x-vt) / (sqrt(1- v^2/c^2)) ; t’ = (t-vx/c^2) / (sqrt(1- v^2/c^2))
When, on frame K, (x,t) = (0,0) it follows that, on frame K’, (x’,t’)
= (0,0) , meaning observers M and M’ coincide. At that instant of time
(t = t' =0), there are two strikes at locations A, at x_A = -100000km,
and location B, at x_B= + 100000km
The back strike light signal moves along the track frame K as x = ct.
Substituting x in Lorentz equations, we obtain:
x'=(c-v)t/(sqrt(1-v^2/c^2)) ; t'=(1-v/c)t/(sqrt(1-v^2/c^2)) , hence
x'/t'=c.
The front strike light signal moves on the track frame K as x = -ct.
Substituting x in Lorentz equations, we obtain:
x'= - (c+v)t / (sqrt(1-v^2/c^2)) ; t'= (1+ v/c)t / (sqrt(1-v^2/
c^2)) , hence x'/t' = - c.
Then it follows light speed is isotropic in the train frame K’ and has
the same value c as on the track frame K.
Then the back strike signal moves on the train frame K’ as x’ = ct’
= (c-v)t / (sqrt(1-v^2/c^2))
Then the front strike signal moves on the train frame K’ as x’ = - ct’
= - (c-v)t / (sqrt(1-v^2/c^2))
Equation on frame K for the back strike light is x_A(t) = ct – 100000
Equation on frame K for the front strike light is x_B(t) = 100000 – ct
So at the track location we have x_A(t) = x_B(t) = 0 , which implies t
= 100000/c = 1/3 sec. So this is the time when the track observer sees
both strike light signals as simultaneous.
Now, according to the track observer M calculations, the front strike
light signal will reach the train observer location (which on the
train frame K’ is x’=0) at point C, while the back strike light signal
will reach the train observer location at point D (both as measured on
frame K), which are given by the following relations:
Point C: train observer M' speed relation on frame K: x_M' = vt =
180000t ; front strike light signal equation x_B(t) = 100000 - ct.
They intersect at t_C = 100000 / (c + v) = 0.208 sec, at location x_C
= vt_C = 37500 km. Here c+v is the closing speed between observer M'
and the light signal measured on frame K.
On frame K’ of the train observer, point C space-time coordinates are
given by the Lorentz equations, resulting in:
x’_C = (x_C-vt) / (sqrt(1- v^2/c^2)) = (37500 – 180000 * 0.208)/0.8 =
0 (as it should be)
t’_C = (t_C-vx_C/c^2) / (sqrt(1- v^2/c^2)) = (0.208 – (180000*37500)/
300000^2)/0.8 = 0.167 sec. This is the time, on frame K' when the
front strike light signal reaches the train observer M'.
Point D: train observer M' speed relation on frame K: x_M' = vt =
180000t ; back strike light signal equation x_B(t) = ct - 100000. They
intersect at t_D = 100000 / (c - v) = 0.833 sec, at location x_D =
vt_D = 150000 km.
On frame K’ of the train observer, point D coordinates are given by
Lorentz equations, resulting in:
x’_D = (x_D-vt) / (sqrt(1- v^2/c^2)) = (150000 – 180000 * 0.803)/0.8 =
0 (as it should)
t’_D = (t_D-vx_D/c^2) / (sqrt(1- v^2/c^2)) = (0.803 – (180000*150000)/
300000^2)/0.8 = 0.667 sec. This is the time, on frame K' when the back
strike light signal reaches the train observer M'.
Conclusions:
a) Track observer on frame K sees two simultaneous strikes at t = 1/3
sec, x = 0.
b) Train observer on frame K’ sees two non-simultaneous strikes, the
first at t’ = 0.167 sec, x’ = 0 and the second at t’ = 0.667 sec at x’
= 0.
This proves the relativity of simultaneity.
Miguel Rios
rval...@gmail.com
kenseto ha escrito:
Maybe this can help. Closing velocity between two moving entities is
not a relativistic invariant (same in all frames), it is frame
dependent. c-v depends on v. For the stationary frame (observer M) it
is c-v, for the moving frame (observer M’ in the train frame) it is c
in both senses (because v is zero in this case), without violating any
speed of light isotropy in the train frame. Don’t confuse closing
velocity (c-v) between the light ray and the moving M’ with light
velocity (c) (the two entities measured in the same frame, the
stationary or the moving one). See my response to Androcles in this
same thread.
> Ken Seto
>
RVHG (Rafael Valls Hidalgo-Gati)
kenseto
There is no confusion on my part. Closing speed is indeed observer
dependent. But the track observer cannot use his observed closing
speeds to determine if the events in the train frame is simultaneous.
The train observer must make this determination by himself. In the
train frame the speed of light is isotropic and therefore he does not
observe any closing speed between him and the light fronts. Therefore
if the strikes occur simultaneously at equal distance then he will
observe them to be simultaneous.
Ken Seto
> Ken Seto
>
> RVHG (Rafael Valls Hidalgo-Gati)- Hide quoted text -
PD
He doesn't. This has been pointed out to Seto dozens of times, but he
can't remember what he had for breakfast two days ago, let alone what
errors were corrected for him two days ago.
kenseto
No you and PD are the brainless runts of the SRians.
>
> You have been shown not one but numerous times that the closing speed
> is measured by the track observer and not by the train observer.
Hey idiot....I was the one who point this out to you. Closing speeds
between the light fronts and the train observer is observed only by
the track observer. However the track observer cannot use observed
closing speeds to determine the simultaneity of the strikes in the
train frame. The train observer must make this determination by
himself. If the track observer wants to determine if the train
observer will see the strikes to be simutlaneous he use the LT as
follows:
1. The speed of light in the train is isotropic.
2. The distance light needs to travel form both direction to reach the
track observer is 0.5L.
3. The track observer determined that the strikes were simultaneous to
begin with.
4. The distance the light fronts need to travel to reach the train
observer in the train frame is contracted to 0.5L/gamma.
5. Therefore the arrival time for both light fronts from A and B to
reach the train observer simultaneously is 0.5/c*gamma.
6. This calculation will agree with observation that the clock in the
train frame is running slow.
The rest of your ranting based on closing speeds naive and laughable.
Ken Seto
>Why
> you insist on placing those closing speeds in the train frame is only
> your fault. You also are unable to provide any single formula and
> diagram proving your nonsense.
>
> OK, once more I will put here the complete derivation, showing the
> relativity of simultaneity. I'll use some numbers, so you assert it is
> wrong say exactly where you think it is wrong. I'm using the same
> notation as Einstein inhttp://www.bartleby.com/173/9.html
> Miguel Rios- Hide quoted text -
kenseto
You are a lying sack of shit. Einstein did use closing speeds to
derive the concept of RoS. He said: The train observer is rushing
toward the light from the front (c+v) and receding away from the light
from the rear (c-v). That's why the train observer will see the light
from the front before he sees the light from the rear.
Ken Seto
>but he
> can't remember what he had for breakfast two days ago, let alone what
> errors were corrected for him two days ago.
>
>
>
> > The train observer must make this determination by himself. In the
> > train frame the speed of light is isotropic and therefore he does not
> > observe any closing speed between him and the light fronts. Therefore
> > if the strikes occur simultaneously at equal distance then he will
> > observe them to be simultaneous.
>
> > Ken Seto
>
> > > Ken Seto
>
> > > RVHG (Rafael Valls Hidalgo-Gati)- Hide quoted text -
>
> > > - Show quoted text -- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
PD
Watched this yet? http://www.youtube.com/watch?v=wteiuxyqtoM
In case you don't remember, I've suggested you watch this at least
twice before. Do you remember ever watching it, Ken? I realize you've
slept since then.
papa...@gmail.com
Your 6 points are total nonsense as a simple diagram, which you are
unable to draw, will show.
The following diagram shows the effect of the closing speed as
measured in the track frame.
< >
< * >
>< *
> / <
A>____________M’/______________<B
M C D ---> x
Where > is the A light signal moving away at c
< is the B light signal moving away at c
/ is the spacetime movement of the train observer following x’ = v*t’
C is the point over the x axis on the track frame where the B light
signal reaches observer M’, according to the track observer M.
D is the point over the x axis on the track frame where the A light
signal reaches observer M’, according to the track observer M.
* are the intersection points in the spacetime.
Miguel Rios
rval...@gmail.com
Read again Einstein’s text and be careful determining which are the
events considered simultaneous or not respect the two different
frames. The events are {two strokes of lightning A and B which are
simultaneous with reference to the railway embankment}(I put
Einstein’s word between { }). Observer M in the embankment will see
the two light rays arriving to him simultaneously by definition of the
experiment. But now we have a third event (the simultaneous arriving
of the two light ray to M) that must be present in all frames. This
excludes completely the possibility that the two light rays meet at
the centre point M’ in the moving train frame at the same time, as you
are suggesting. The same rays can’t be arriving at two different
points M and M’ at the same time in each of these points (no matter in
what frame are you observing the events), what implies that some ray
arrives to M’ (the B one) before the arriving of the other (the A
one). Which is your error? To suppose that lightning A and B,
simultaneous for the embankment frame by definition, are simultaneous
events also in the moving train frame [if the strikes occur
simultaneously at equal distance] (your words between [ ]), which is
precisely what you are trying to prove. Closing velocities (c-v) and (c
+v), different ones for each frame that is moving with a non zero
velocity v in the stationary frame, can be used without any problem to
show that the simultaneous arriving of the two light rays at the
centre point of the frame can occurs only in the stationary frame
where v equals zero.
> Ken Seto
>
>
>
> > Ken Seto
> >
> > RVHG (Rafael Valls Hidalgo-Gati)- Hide quoted text -
> >
> > - Show quoted text -
RVHG (Rafael Valls Hidalgo-Gato)
Bryan Olson
> "papar...@gmail.com" [Miguel Rios] wrote:
>> You have been shown not one but numerous times that the closing speed
>> is measured by the track observer and not by the train observer.
>
> Hey idiot....I was the one who point this out to you. Closing speeds
> between the light fronts and the train observer is observed only by
> the track observer. However the track observer cannot use observed
> closing speeds to determine the simultaneity of the strikes in the
The track-fixed observer can determine that in the rest frame
of the track, the two signals reach the train rider at different
times. Ken Seto stated that the train rider sees the signals at
the same time, which would be a contradictory result.
To see why, let's say the train rider is actually a device
consisting of two photo-detectors, an AND gate, a bomb. One
photo-detector is aimed forward; the other backward. Each
generate a voltage when the light from a lightning strike
reaches it, coming from the direction in which it is aimed. The
photo-detectors feed into the AND gate, which triggers the bomb.
Thus the bomb detonates if and only if the light from the two
lighting strikes reach the device simultaneously.
As we've seen, in the rest frame of the track, the light from
the front strike reaches the device before light from the rear
strike, so the bomb does not go off. According to Seto's stated
theory, in the rest frame of the train the lights reach the
device simultaneously, and thus the bomb would detonate.
Seto's idea leaves us with frames of reference differing in not
just the coordinates of various events, but in whether we have
a train versus mangled wreckage of what was formerly a train.
Of course after Seto's failure to comprehend any of the
previous explanations, we do not expects him to grasp this
version either.
--
--Bryan
Spaceman
<snipped>
> As we've seen, in the rest frame of the track, the light from
> the front strike reaches the device before light from the rear
> strike, so the bomb does not go off.
I've got a better one for you,
A group of men design a bomb that will only create an atomic
explosion when a few simultaenous events all occur at the same time.
A plane drops this bomb and you are watching it fall.
Will the simultaneous events occur?
Yes, no matter what you "the observer" sees.
Hint: This is how an atomic bombs actually work.
It is based upon a simultaneous explosion occuring
simultanously.
Because they needed simulteniety, Einstein was kicked
out of the room.
No simultaneity, no atomic boom.
Einstein would have argued it would never work according
to a moving observer and his RoS.
So Sheesh!
Simutenity.. Big ass boom, no matter what the stupid
ass observer says about "the observations"
This is also why Einstein was not in the final build team
of the damn bomb.
Get a clue.
What is seen is "observationally" is simply distance and motion
illusion and to find out about simultaneous stuff you need to
use absolute measurements and absolute timing to see if
it actually occured simultaneously.
What occurs can and does occur simulteaneously
if designed to, and does no matter about any stupid
ass observer frame.
Now,
I have dropped the damn atomic bomb on this thread
Simulteneity DOES IN FACT OCCUR!
BBBBBBOOOOOOOOMMMMMMMM
You are all pancakes and dust and burned,
no matter if you saw the simultenous
actions to occur simultenous or not.
The END.
ELOL
--
James M Driscoll Jr
Spaceman
Bryan Olson
> Bryan Olson wrote:
> <snipped>
>> As we've seen, in the rest frame of the track, the light from
>> the front strike reaches the device before light from the rear
>> strike, so the bomb does not go off.
>
> I've got a better one for you,
> A group of men design a bomb that will only create an atomic
> explosion when a few simultaenous events all occur at the same time.
> A plane drops this bomb and you are watching it fall.
> Will the simultaneous events occur?
> Yes, no matter what you "the observer" sees.
>
> Hint: This is how an atomic bombs actually work.
> It is based upon a simultaneous explosion occuring
> simultanously.
> Because they needed simulteniety, Einstein was kicked
> out of the room.
> No simultaneity, no atomic boom.
> Einstein would have argued it would never work according
> to a moving observer and his RoS.
In actual fact, Einstein argued in a letter to President
Roosevelt that the atomic bomb would work. Roosevelt
authorized the Manhattan Project largely as a result of
that letter.
> So Sheesh!
> Simutenity.. Big ass boom, no matter what the stupid
> ass observer says about "the observations"
> This is also why Einstein was not in the final build team
> of the damn bomb.
No; I looked it up. Einstein was not on the project because
the Army Intelligence Office denied him the needed security
clearance. Manhattan Project scientists were not allowed to
consult with Einstein or apprise him of their progress.
> Get a clue.
> What is seen is "observationally" is simply distance and motion
> illusion and to find out about simultaneous stuff you need to
> use absolute measurements and absolute timing to see if
> it actually occured simultaneously.
> What occurs can and does occur simulteaneously
> if designed to, and does no matter about any stupid
> ass observer frame.
>
> Now,
> I have dropped the damn atomic bomb on this thread
> Simulteneity DOES IN FACT OCCUR!
> BBBBBBOOOOOOOOMMMMMMMM
> You are all pancakes and dust and burned,
> no matter if you saw the simultenous
> actions to occur simultenous or not.
> The END.
> ELOL
If you can take a break from congratulating yourself, you
might check what real Manhattan Project scientists taught
about relativity. They actually build the awesome weapons,
which requires a lot of theory that works in reality. You
convinced yourself that you understand such weapons, which
requires a theory that works in your own head. Vastly
different standards.
--
--Bryan
kenseto
> In case you don't remember, I've suggested you watch this at least
> twice before. Do you remember ever watching it, Ken? I realize you've
> slept since then.
Hey idiot...the video is the same as Einstein's gedanken. It uses
closing velocities to argue that the flashes were not simultaneous in
the train. This violates the isotropy of the speed of light in the
train and violates the fact that the track observer determined that
the flashes were simultaneous to begin with.
Ken Seto
kenseto
Hey fucking idiot....how many times do I have to tell you that closing
speeds as perceived by the track observer have no effect on the
isotropy of the speed of light in the train. Also the train observer
must determine if the strikes are simultaneous by himself...not by
what the track observer sees as closing speeds between he and the
light fronts. The only factors that determine the simultaneity of the
strikes in the train frame are as follows:
1. Were the strikes occurred at equal distance from the train
observer? The answer is yes.
2. Were the strikes occurred simultaneously to begin with? The answer
is yes. This is determined by the track observer.
Ken Seto
>
> < >
> < * >
> >< *
> > / <
> A>____________M’/______________<B
> M C D ---> x
>
> Where > is the A light signal moving away at c
> < is the B light signal moving away at c
> / is the spacetime movement of the train observer following x’ = v*t’
> C is the point over the x axis on the track frame where the B light
> signal reaches observer M’, according to the track observer M.
> D is the point over the x axis on the track frame where the A light
> signal reaches observer M’, according to the track observer M.
> * are the intersection points in the spacetime.
>
kenseto
I suggested no such thing. Here's what I suggested.
From the track frame's point of view:
1. The strikes occur at equal distance of 0.5L m from him.
2. The speed of light in the track frame is isotropic.
3. Einstein stipulated that the track observer sees the flashes to be
simultaneous.
4. From items 1,2 and 3 the track observer determined that the flashes
occur simultaneously to begin with.
5. The track observer says: The speed of light is isotropic in the
train frame.
6. The track observer said: The flashes occur at equal distance of
0.5L/gamma from the strikes.
7. The track observer said: The flashes occur simultaneously to begin
with.
8. Therefore the track observer must also said: The train observer
must also sees the strikes to be simultaneous at a time of 0.5L/
gamma*c second according to the train clock. This conclusion will
agree with the fact that the train clock is running slow.
Ken Seto
>The same rays can’t be arriving at two different
> points M and M’ at the same time in each of these points (no matter in
> what frame are you observing the events), what implies that some ray
> arrives to M’ (the B one) before the arriving of the other (the A
> one). Which is your error? To suppose that lightning A and B,
> simultaneous for the embankment frame by definition, are simultaneous
> events also in the moving train frame [if the strikes occur
> simultaneously at equal distance] (your words between [ ]), which is
> precisely what you are trying to prove. Closing velocities (c-v) and (c
> +v), different ones for each frame that is moving with a non zero
> velocity v in the stationary frame, can be used without any problem to
> show that the simultaneous arriving of the two light rays at the
> centre point of the frame can occurs only in the stationary frame
> where v equals zero.> Ken Seto
>
> > > Ken Seto
>
> > > RVHG (Rafael Valls Hidalgo-Gati)- Hide quoted text -
>
> > > - Show quoted text -
>
> RVHG (Rafael Valls Hidalgo-Gato)- Hide quoted text -
papa...@gmail.com
>
> leer más »
Until your brainless mind understands that all your points (and your
so called "theory") are stupid and wrong.
The track observer is not blind like you...he is seeing what is going
on and sees how M', at t=0 is passing through his position and also he
will calculate, and later see, when the light signal A reaches M', and
that event, you moron, is exactly what M' will experience, so all
Einsteins calculations and drawings are dead right. So just deal with
it or indicate which part of the mathematical and graphical derivation
is wrong.
I will repeat it again and this time try to follow it line by line:
The following diagram shows the effect of the closing speed as
measured in the track frame.
< >
< * >
>< *
> / <
A>____________M’/______________<B
M C D ---> x
Where > is the A light signal moving away at c (in both frames as
shown below)
< is the B light signal moving away at c (in both frames as shown
below)
/ is the spacetime movement of the train observer following x = v*t
(this C is the point over the x axis, on the track frame, where the B
light signal reaches observer M’, according to the track observer M.
D is the point over the x axis on the track frame where the A light
signal reaches observer M’, according to the track observer M.
* are the intersection points in the spacetime.
v= 0.6c = 180000 km/sec
Equation, on frame K, for the back strike light is x_A(t) = ct –
100000
Equation, on frame K, for the front strike light is x_B(t) = 100000 –
ct
So at the track location we have x_A(t) = x_B(t) = 0 , which implies t
= 100000/c = 1/3 sec. So this is the time when the track observer sees
both strike light signals as simultaneous.
Now, according to the track observer M calculations (and later
observations), the front strike light signal will reach the train
observer location (which on the train frame K’ is x’=0) at point C,
while the back strike light signal will reach the train observer
location at point D (both as measured on frame K), which are given by
the following relations:
Point C: train observer M speed relation: x_M = vt = 180000t ; front
strike light signal equation x_B(t) = 100000 - ct. They intersect at
t_C = 100000 / (c + v) = 0.208 sec, at location x_C = vt_C = 37500 km.
On frame K’ of the train observer, point C space-time coordinates are
given by the Lorentz equations, resulting in:
x’_C = (x_C-vt) / (sqrt(1- v^2/c^2)) = (37500 – 180000 * 0.208)/0.8 =
0 (as it should be)
t’_C = (t_C-vx_C/c^2) / (sqrt(1- v^2/c^2)) = (0.208 – (180000*37500)/
300000^2)/0.8 = 0.167 sec. This is the time, on frame K' when the
front strike light signal reaches the train observer M'.
Point D: train observer M speed relation: x_M = vt = 180000t ; back
strike light signal equation x_B(t) = ct - 100000. They intersect at
t_D = 100000 / (c - v) = 0.833 sec, at location x_D = vt_D = 150000
km.
On frame K’, point D coordinates are given by Lorentz equations,
kenseto
> kenseto wrote:
> > "papar...@gmail.com" [Miguel Rios] wrote:
> >> You have been shown not one but numerous times that the closing speed
> >> is measured by the track observer and not by the train observer.
>
> > Hey idiot....I was the one who point this out to you. Closing speeds
> > between the light fronts and the train observer is observed only by
> > the track observer. However the track observer cannot use observed
> > closing speeds to determine the simultaneity of the strikes in the
> > train frame. [...]
>
> The track-fixed observer can determine that in the rest frame
> of the track, the two signals reach the train rider at different
> times. Ken Seto stated that the train rider sees the signals at
> the same time, which would be a contradictory result.
I said no such thing. Here's what I said:
From the track frame's point of view:
1. The strikes occur at equal distance of 0.5L m from him.
2. The speed of light in the track frame is isotropic.
3. Einstein stipulated that the track observer sees the flashes to be
simultaneous.
4. From items 1,2 and 3 the track observer determined that the
flashes
occur simultaneously to begin with.
5. The track observer says: The speed of light is isotropic in the
train frame.
6. The track observer said: The flashes occur at equal distance of
0.5L/gamma from the train observer.
7. The track observer said: The flashes occur simultaneously to begin
with.
8. Therefore the track observer must also said: The train observer
must also sees the strikes to be simultaneous at a time of 0.5L/
gamma*c second according to the train clock. This conclusion will
agree with the fact that the train clock is running slow.
Ken Seto
>
> To see why, let's say the train rider is actually a device
> consisting of two photo-detectors, an AND gate, a bomb. One
> photo-detector is aimed forward; the other backward. Each
> generate a voltage when the light from a lightning strike
> reaches it, coming from the direction in which it is aimed. The
> photo-detectors feed into the AND gate, which triggers the bomb.
> Thus the bomb detonates if and only if the light from the two
> lighting strikes reach the device simultaneously.
OK
>
> As we've seen, in the rest frame of the track, the light from
> the front strike reaches the device before light from the rear
> strike, so the bomb does not go off. According to Seto's stated
> theory, in the rest frame of the train the lights reach the
> device simultaneously, and thus the bomb would detonate.
Wrong. The track observer must also agree that the flashes arrive
simultaneously and thus the bomb goes off.
Ken Seto
PD
And please note the part of the video where the narrator says, "But
what does the passenger see?" It's about midway through the video.
Play it again. And watch carefully.
> This violates the isotropy of the speed of light in the
> train
It does no such thing. Note that in both frames, the rate at which the
light fronts move -- the speed at which those blue spheres grow -- is
identical.
> and violates the fact that the track observer determined that
> the flashes were simultaneous to begin with.
So not only can you not comprehend what you read, but you cannot
comprehend a cartoon of the same thing.
Your powers of comprehension are astoundingly small.
Spaceman
So he knew that RoS was completely an observational effect only.
:)
>
>> So Sheesh!
>> Simutenity.. Big ass boom, no matter what the stupid
>> ass observer says about "the observations"
>> This is also why Einstein was not in the final build team
>> of the damn bomb.
>
> No; I looked it up. Einstein was not on the project because
> the Army Intelligence Office denied him the needed security
> clearance. Manhattan Project scientists were not allowed to
> consult with Einstein or apprise him of their progress.
That might have been when he started his babble about
RoS.
:)
>> Get a clue.
>> What is seen is "observationally" is simply distance and motion
>> illusion and to find out about simultaneous stuff you need to
>> use absolute measurements and absolute timing to see if
>> it actually occured simultaneously.
>> What occurs can and does occur simulteaneously
>> if designed to, and does no matter about any stupid
>> ass observer frame.
>>
>> Now,
>> I have dropped the damn atomic bomb on this thread
>> Simulteneity DOES IN FACT OCCUR!
>> BBBBBBOOOOOOOOMMMMMMMM
>> You are all pancakes and dust and burned,
>> no matter if you saw the simultenous
>> actions to occur simultenous or not.
>> The END.
>> ELOL
>
> If you can take a break from congratulating yourself, you
> might check what real Manhattan Project scientists taught
> about relativity. They actually build the awesome weapons,
> which requires a lot of theory that works in reality. You
> convinced yourself that you understand such weapons, which
> requires a theory that works in your own head. Vastly
> different standards.
They did not use "relativity" they used many other theories
and not one of them had to do with relativity.
I suggest you learn more about it.
:)
Bryan Olson
> Bryan Olson wrote:
>> kenseto wrote:
>>> "papar...@gmail.com" [Miguel Rios] wrote:
>>>> You have been shown not one but numerous times that the closing speed
>>>> is measured by the track observer and not by the train observer.
>>> Hey idiot....I was the one who point this out to you. Closing speeds
>>> between the light fronts and the train observer is observed only by
>>> the track observer. However the track observer cannot use observed
>>> closing speeds to determine the simultaneity of the strikes in the
>>> train frame. [...]
>> The track-fixed observer can determine that in the rest frame
>> of the track, the two signals reach the train rider at different
>> times. Ken Seto stated that the train rider sees the signals at
>> the same time, which would be a contradictory result.
>
> I said no such thing.
Ken, I can cite you taking that (incorrect) position from
as far back as 2003 to as recently as the post I'm following.
> Here's what I said:
> From the track frame's point of view:
> 1. The strikes occur at equal distance of 0.5L m from him.
> 2. The speed of light in the track frame is isotropic.
> 3. Einstein stipulated that the track observer sees the flashes to be
> simultaneous.
> 4. From items 1,2 and 3 the track observer determined that the
> flashes
> occur simultaneously to begin with.
> 5. The track observer says: The speed of light is isotropic in the
> train frame.
> 6. The track observer said: The flashes occur at equal distance of
> 0.5L/gamma from the train observer.
> 7. The track observer said: The flashes occur simultaneously to begin
> with.
> 8. Therefore the track observer must also said: The train observer
> must also sees the strikes to be simultaneous at a time of 0.5L/
> gamma*c second according to the train clock. This conclusion will
> agree with the fact that the train clock is running slow.
>
>
>> To see why, let's say the train rider is actually a device
>> consisting of two photo-detectors, an AND gate, a bomb. One
>> photo-detector is aimed forward; the other backward. Each
>> generate a voltage when the light from a lightning strike
>> reaches it, coming from the direction in which it is aimed. The
>> photo-detectors feed into the AND gate, which triggers the bomb.
>> Thus the bomb detonates if and only if the light from the two
>> lighting strikes reach the device simultaneously.
>
> OK
>
>> As we've seen, in the rest frame of the track, the light from
>> the front strike reaches the device before light from the rear
>> strike, so the bomb does not go off. According to Seto's stated
>> theory, in the rest frame of the train the lights reach the
>> device simultaneously, and thus the bomb would detonate.
>
> Wrong. The track observer must also agree that the flashes arrive
> simultaneously and thus the bomb goes off.
He must and yet he does not; that's what makes your theory
contradictory, Ken. As we've seen, in the track-fixed
observer's rest frame the signals reach the device at
different times.
And Ken, you began the post denying you said that the
train-riding observer sees the signals at the same time.
So why do you now say the bomb goes off? The bomb is the
train-riding observer, and it detonates if and only if the
light from the two lighting strikes reach the device
simultaneously.
--
--Bryan
Bryan Olson
Do you even care whether what you write is true?
Which Manhattan Project scientists said that?
> I suggest you learn more about it.
Hypocrite.
--
--Bryan
Spaceman
When I use the term" might" does that imply I am saying it
is "definite"?
Do you even read what is stated?
When I type stuff, I know it is true unless I say words like
"might, probably, and etc..
I always use such words to designate whether I know it true
or if I only think such "could" be true..
:)
>> They did not use "relativity" they used many other theories
>> and not one of them had to do with relativity.
>
> Which Manhattan Project scientists said that?
None,
Just like none actually said they did use it.
So the statement stands.
kenseto
The train observer does not see what the narrator asserted because of
the following reasons:
1. The speed of light is isotropic in the train.
2. The strikes are at equal distance from the train observer.
3. It was determined by the track observer that the strikes were
simultaneous to begin with.
Your problem is that you (and the narrator) made the bogus assumption
that the light rays that hit the track observer simultaneously will
continue on to hit the train observer. This is wrong. The correct
interpretation is: Two light rays R1 and R2 arrive at the track
observer simultaneously and two different light rays R3 and R4 arrive
at the train observer simultaneously. If you watch the video carefully
you will note that the light spheres meet simultaneously at infinite
number of locations as they expand. The track observer is at one of
these locations and the train observer is at a different
location....this is due to the fact that they have a different state
of absolute motion wrt light.
Ken Seto
>
> > This violates the isotropy of the speed of light in the
> > train
>
> It does no such thing. Note that in both frames, the rate at which the
> light fronts move -- the speed at which those blue spheres grow -- is
> identical.
>
> > and violates the fact that the track observer determined that
> > the flashes were simultaneous to begin with.
>
> So not only can you not comprehend what you read, but you cannot
> comprehend a cartoon of the same thing.
>
> Your powers of comprehension are astoundingly small.
>
>
>
>
>
> > Ken Seto
>
> > - Hide quoted text -
>
> > > - Show quoted text -- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
kenseto
No the signals reaches the train observer simultaneously and the bomb
goes off....Boooom!!!! Whatyou asserted is based on your naive model
of light propagation. I suggest that you read up qed to get the
correct model of light propagation.
>
> And Ken, you began the post denying you said that the
> train-riding observer sees the signals at the same time.
I said from the track observer's point of view: he will see the
flashes to be simultaneous at an earlier time of 0.5L/c and the train
observer will see the flashes to be simultaneous at a later time of
0.5L/gamma*c.
> So why do you now say the bomb goes off? The bomb is the
> train-riding observer, and it detonates if and only if the
> light from the two lighting strikes reach the device
> simultaneously.
Because two different signals will arrive at the train observer
simultaneously.
I suggest that you watch the following video:
http://www.youtube.com/watch?v=wteiuxyqtoM
Please note that the light speres meet simultaneously at an infinite
number of locations. The track observer is at one of these locations
and the train observer is at a different location where these two
light spheres meet. Also toward the end of the video the narrator said
that even though the track observer disagrees with the train
observer's conclusion but they both conclusions are correct. This
statement is naive and self contradictory. Why? Take your example of
the bomb in the train....the bomb either goes off or not? That means
that only one of the conclusions is correct....not both as asserted by
the narrator.
Ken Seto
Spaceman
> Because two different signals will arrive at the train observer
> simultaneously.
> I suggest that you watch the following video:
> http://www.youtube.com/watch?v=wteiuxyqtoM
> Please note that the light speres meet simultaneously at an infinite
> number of locations. The track observer is at one of these locations
> and the train observer is at a different location where these two
> light spheres meet. Also toward the end of the video the narrator said
> that even though the track observer disagrees with the train
> observer's conclusion but they both conclusions are correct. This
> statement is naive and self contradictory. Why? Take your example of
> the bomb in the train....the bomb either goes off or not? That means
> that only one of the conclusions is correct....not both as asserted by
> the narrator.
Yes,
The passenger is ignoring thier motion to conclude
such bullshit.
If they calculated thier motion along with the
speed of light and the times the flashes occured,
they could determine the flashes did in fact, occur
simultaneously for real.
:)
Isn't it funny how they must ignore "relative" motion
to support the relative motion theory?
Bryan Olson
> Bryan Olson wrote:
>> Spaceman wrote:
>>> Bryan Olson wrote:
>>>> Spaceman wrote:
>>>>> Bryan Olson wrote:
>>>>> <snipped>
>>>>>> As we've seen, in the rest frame of the track, the light from
>>>>>> the front strike reaches the device before light from the rear
>>>>>> strike, so the bomb does not go off.
>>>>> I've got a better one for you,
>>>>> A group of men design a bomb that will only create an atomic
>>>>> explosion when a few simultaenous events all occur at the same
>>>>> time. A plane drops this bomb and you are watching it fall.
>>>>> Will the simultaneous events occur?
>>>>> Yes, no matter what you "the observer" sees.
>>>>>
>>>>> Hint: This is how an atomic bombs actually work.
>>>>> It is based upon a simultaneous explosion occuring
>>>>> simultanously.
>>>>> Because they needed simulteniety, Einstein was kicked
>>>>> out of the room.
>>>>> No simultaneity, no atomic boom.
>>>>> Einstein would have argued it would never work according
>>>>> to a moving observer and his RoS.
Here's an actual Manhattan Project scientists and Nobel
Prize winner lecturing on SR, including RoS:
http://youtube.com/watch?v=XuLNQP9n-so&watch_response
James, your claim that the bomb designers rejected Einstein's
relativity of simultaneity stands refuted. Not surprising, as
it was obviously a fabrication to begin with.
So why did you post all those unqualified falsehoods, such
as the reason Einstein was not on the "build team" for the
atomic bomb?
Have you checked my claim on that yet? I included enough
keywords that it should take you about 10 seconds to
Google up the facts, should you care even a little whether
what you wrote is true.
>>> They did not use "relativity" they used many other theories
>>> and not one of them had to do with relativity.
>> Which Manhattan Project scientists said that?
>
> None,
> Just like none actually said they did use it.
> So the statement stands.
It stands as a fabrication.
--
--Bryan
Spaceman
> Here's an actual Manhattan Project scientists and Nobel
> Prize winner lecturing on SR, including RoS:
>
> http://youtube.com/watch?v=XuLNQP9n-so&watch_response
>
> James, your claim that the bomb designers rejected Einstein's
> relativity of simultaneity stands refuted. Not surprising, as
> it was obviously a fabrication to begin with.
So you have proof Feynman was not in the "back room"
sweeping the floors with Einstein?
:)
What I stated stands,
You have no proof against my "theory".
And sadly you don't get that RoS is technically saying the
bomb will not work at all according to a relative motion observer.
The moving observer will argue that the simultenity did not
occur, yet it did, or the bomb would not work.
I feel sorry you are so brainwashed to not get that simultaneity
is possible for all frames if such frames actually include thier
relative motion in such calculations.
> So why did you post all those unqualified falsehoods, such
> as the reason Einstein was not on the "build team" for the
> atomic bomb?
I did not say he was not on the "team"
I joked that he was kicked into the back room to do the sweeping.
and still you have no proof such did not occur.
:)
And even more sad, you still don't get that without simultaniety
the bomb would not have worked at all.
PD
You said, Ken, that the gedanken does not declare what the train
observer actually sees. You said that this is only a conclusion based
on what the track observer says he SHOULD see. These two statements
that you made are shown in this video to be incorrect.
Now, Ken, you have two options:
1. You can acknowledge that you made an incorrect statement.
2. You can refuse to acknowledge any mistake and loop back around to
where you were several months ago, in an endless round-trip to
nowhere, forgetting everything on Wednesday what was told to you on
Tuesday.
If you can only do (2), then I've got precious little to say to you,
since I know it will be completely wasted, like talking to a table
leg. As you like to say, you can go ahead and die in denial of
reality.
PD
Bryan Olson
> Bryan Olson wrote:
>> Here's an actual Manhattan Project scientists and Nobel
>> Prize winner lecturing on SR, including RoS:
>>
>> http://youtube.com/watch?v=XuLNQP9n-so&watch_response
>>
>> James, your claim that the bomb designers rejected Einstein's
>> relativity of simultaneity stands refuted. Not surprising, as
>> it was obviously a fabrication to begin with.
>
> So you have proof Feynman was not in the "back room"
> sweeping the floors with Einstein?
> :)
Having cited Feynman on Einstein, from Feynman's own voice,
my case is made. If you prefer to believe the version you
made up rather than the reality in front of you, well, I
support your right to make that choice, stupid as it is.
> What I stated stands,
It stands as a fabrication.
> You have no proof against my "theory".
Ah, I never said I could prove it *to you*. I know many
facts that I can prove, but none that I can prove to a
potato.
> And sadly you don't get that RoS is technically saying the
> bomb will not work at all according to a relative motion observer.
For the same reason I do not get that 2 + 3 = 17.
> The moving observer will argue that the simultenity did not
> occur, yet it did, or the bomb would not work.
Have you considered learning what SR predicts? Your own theory
is clearly not working.
> I feel sorry you are so brainwashed to not get that simultaneity
> is possible for all frames if such frames actually include thier
> relative motion in such calculations.
You are, of course, free to feel as sorry for me as you like.
>> So why did you post all those unqualified falsehoods, such
>> as the reason Einstein was not on the "build team" for the
>> atomic bomb?
>
> I did not say he was not on the "team"
In post <0OidncZpANFgohzV...@comcast.com> James
M. "Spaceman" Driscoll Jr wrote:
This is also why Einstein was not in the final build team
of the damn bomb.
> I joked that he was kicked into the back room to do the sweeping.
> and still you have no proof such did not occur.
> :)
Except of course for what I cited: he wasn't in those rooms at
all, having been denied the needed security clearance. What,
you still haven't even bothered looking that up?
> And even more sad, you still don't get that without simultaniety
> the bomb would not have worked at all.
True: I do not get your theory there. Fortunately for the
Manhattan Project, the scientists knew to trust SR, not
nonsense like Driscoll-theory.
--
--Bryan
Bryan Olson
> Bryan Olson wrote:
>> kenseto wrote:
>>> Bryan Olson wrote:
>>>> kenseto wrote:
>>>>> "papar...@gmail.com" [Miguel Rios] wrote:
>>>>>> You have been shown not one but numerous times that the closing speed
>>>>>> is measured by the track observer and not by the train observer.
>>>>> Hey idiot....I was the one who point this out to you. Closing speeds
>>>>> between the light fronts and the train observer is observed only by
>>>>> the track observer. However the track observer cannot use observed
>>>>> closing speeds to determine the simultaneity of the strikes in the
>>>>> train frame. [...]
>>>> The track-fixed observer can determine that in the rest frame
>>>> of the track, the two signals reach the train rider at different
>>>> times. Ken Seto stated that the train rider sees the signals at
>>>> the same time, which would be a contradictory result.
>>> I said no such thing.
>> Ken, I can cite you taking that (incorrect) position from
>> as far back as 2003 to as recently as the post I'm following.
I make the same offer again, below.
>>> OK
No, it is based on clearly-established facts. In the rest frame
of the track, light from the flash at the front of the train
reaches the train-riding device/observer before reaching the
track-fixed observer, and light from the flash at the rear of
the train reaches the track-fixed observer before reaching the
train-riding device/observer.
Thus, in the rest frame of the track, the light from the strikes
reaches the train-riding device/observer at different times, and
the bomb does not go off. Thus Seto-theory cannot hold: in the
rest frame of the track, the train goes rolling along,
explosion-free; in the rest frame of the train, Seto-theory says,
"Boooom!!!!"
SR has no such contradiction. In SR, all frames of reference
agree on the simultaneity of co-located events. Relativity of
simultaneity pertains to spatially separated events, and is a
necessary consequence of the two postulates of SR.
> I suggest that you read up qed to get the
> correct model of light propagation.
Hypocrite
>> And Ken, you began the post denying you said that the
>> train-riding observer sees the signals at the same time.
>
> I said from the track observer's point of view: he will see the
> flashes to be simultaneous at an earlier time of 0.5L/c and the train
> observer will see the flashes to be simultaneous at a later time of
> 0.5L/gamma*c.
Yes Ken, that was among the various things you said; I never
claimed otherwise. My point is that you said what I claimed
you said. You also said other things, a few of them correct.
Ken, I put you on a specific (incorrect) position: that in
the familiar train-and-embankment gedanken, the train rider
sees the signals -- the light fronts -- at the same time.
You denied taking that position. I was telling the truth;
you, not so much. As I reported, you said it as far back as
2003, and it is still your stated position in this very
thread.
Want me to prove it, all the way to quotes and NNTP
message-id's? Ask and ye shall receive.
>> So why do you now say the bomb goes off? The bomb is the
>> train-riding observer, and it detonates if and only if the
>> light from the two lighting strikes reach the device
>> simultaneously.
>
> Because two different signals will arrive at the train observer
> simultaneously.
So why did you pretend I was wrong when I claimed, "Ken Seto
stated that the train rider sees the signals at the same time"?
Ken, you said it back in 2003, and you still say it today. Is a
little honesty too much to ask?
> I suggest that you watch the following video:
> http://www.youtube.com/watch?v=wteiuxyqtoM
Glad you like it, Ken. I watched that video within a minute of
reading PD pointing it out to you. It's good stuff. Fact is, Ken,
it flat-out refutes your theory.
> Please note that the light speres meet simultaneously at an infinite
> number of locations.
Let's be specific: In the rest frame of the track, the light
spheres meet simultaneously at all and only those locations that
are equally far from the front strike as from the rear strike. To
find these space-time points in other frames, we can use the use
the Lorentz transform
> The track observer is at one of these locations
> and the train observer is at a different location where these two
> light spheres meet.
Watch the video carefully: the train observer is *not* at
coordinates where the two light spheres meet simultaneously.
Ken, that means you are wrong on the points at issue in this
thread, and in dozens of previous threads.
> Also toward the end of the video the narrator said
> that even though the track observer disagrees with the train
> observer's conclusion but they both conclusions are correct. This
> statement is naive and self contradictory. Why?
Well exactly as next words of the narrator explain, of course:
"both are correct within their own frame of reference." This is
important, so let's go over it again, with more context:
Who's interpretation is correct: the observer on the platform
who claims that the strikes happened simultaneously, or the
observer on the train, who claims that the front strike
happened before the rear strike?
Einstein tells us that both are correct, within their own
frame of reference. This is a fundamental result of special
relativity.
Ken, this is what you've been missing for years: The strikes
were simultaneous in the rest frame of the "platform" (or,
equivalently, the "track", or the "embankment") which is the
frame of reference for the observer on the platform. The
strikes were *not* simultaneous in the rest frame of the
train, which is the frame of reference for the observer on
the train.
Ken, you've been appealing to the notion of the strikes being
"simultaneous to begin with", but that's your own
misunderstanding. Relativity says that the frames of reference
are equally valid; neither has special claim to being the one
true coordinate system. There is no "to begin with", because
we have equal justification for beginning with either.
> Take your example of
> the bomb in the train....the bomb either goes off or not? That means
> that only one of the conclusions is correct....not both as asserted by
> the narrator.
No Ken, the narrator asserted no such thing. It is *your* theory
that necessarily concludes both, even though, as you said, "the
bomb either goes off or not." Your theory, Ken, is contradictory.
SR has no such defect.
Sadly, I note now, as before:
Of course after Seto's failure to comprehend any of the
previous explanations, we do not expects him to grasp this
version either.
Ken, you *could* learn SR. Despite what others with more
experience than myself say, I believe it is *not* beyond your
ability. Nevertheless, I must face the reality that expectation
are low.
--
--Bryan
kenseto
wrote:
> kenseto wrote:
> > Because two different signals will arrive at the train observer
> > simultaneously.
> > I suggest that you watch the following video:
> >http://www.youtube.com/watch?v=wteiuxyqtoM
> > Please note that the light speres meet simultaneously at an infinite
> > number of locations. The track observer is at one of these locations
> > and the train observer is at a different location where these two
> > light spheres meet. Also toward the end of the video the narrator said
> > that even though the track observer disagrees with the train
> > observer's conclusion but they both conclusions are correct. This
> > statement is naive and self contradictory. Why? Take your example of
> > the bomb in the train....the bomb either goes off or not? That means
> > that only one of the conclusions is correct....not both as asserted by
> > the narrator.
>
> Yes,
> The passenger is ignoring thier motion to conclude
> such bullshit.
No James.....according to the RoS the passenger didn't reach any
conclusion by himself. The track observer forced the bogus conclusion
onto him. The track observer said that the train observer is moving
wrt the light fronts and that's why the track observer says that the
train observer will not see the strikes to be simultaneous.
The train observer determines the strikes to be simultaneous based on
the following reasons:
1. The speed of light in his frame is isotropic.
2. He is at equal distance from the strikes when they occur
simltaneously.
3. Therefore he see the strikes to be simultaneous.
Ken Seto
kenseto
That's right....the train observer does not see that he is moving wrt
the light fronts that give anisotropy of the speed of light. He (the
train observer) sees that the speed of light is isotropic in his
frame. That means that the gedanken as set up is wrong.
>You said that this is only a conclusion based
> on what the track observer says he SHOULD see.
That's right....the track observer forced the bogus claims that the
train observer is moving wrt the light fronts onto the train observer
and then insisted that the train observer will see the strikes to be
not simultaneous.
>These two statements
> that you made are shown in this video to be incorrect.
That's because the claims of the video are incorrect.....the train
observer didn't move wrt the light fronts in a way that will affect
the isotropy of the speed of light in the train.
Ken Seto
Spaceman
But here is the problem,
The train observer is at equal distance as the simultaneous strikes occur,
but since the train is in motion, the train observer will not be at equal
distance
when the flashes reach them
The light still travels at c wrt the track and thier source points but the
train
observer will be relative to that speed.
As I always say the speed of light is not relatively the same (c)
to all observers.
The train observer would have to use c+v and c-v for the speeds
of the light wrt the train observers frame.
The front flash has a closing speed of c+v and the rear flash
wrt the observer will have a closing speed of c-v.
And this is why the train observer in that example will not see
the lights as simultaenous even though they were in reality.
PD
No, Ken, the issue is not whether the gedanken is correct or not. That
is settled by experiment.
Your claim was that the observations made by the train observer are
NEVER MENTIONED AT ALL, right or wrong, in the gedanken.
Your claim is wrong. It IS mentioned in the gedanken. It IS mentioned
in the video. It is this simple fact that you will not acknowledge.
You LIED when you said it is not mentioned in the gedanken.
You now say it is mentioned (though you do not admit your previous
LIE) but that it is wrong, and you base this on your *thinking* about
it.
EXPERIMENT says that you are wrong about that as well. You are not
familiar with the experiments. You are only acquainted with the
gedanken, which is not an experiment of any kind, and you do not
understand the gedanken, either.
Please admit either that you lied or were wrong about the mention of
the train observer's actual observations.
If you cannot admit you are wrong about that, in the face of being
caught red-handed in the lie, then you have a serious personality
defect.
kenseto
Yes they will. The speed of light is indpendent of the motion of the
source or the observer. That's why the speed of light is a constant c
in all frames.
> The light still travels at c wrt the track and thier source points but the
> train
> observer will be relative to that speed.
> As I always say the speed of light is not relatively the same (c)
> to all observers.
NO....observed relative velocity by the track observer have no effect
on the isotropy of the speed of light in the train.
Ken Seto
> The train observer would have to use c+v and c-v for the speeds
> of the light wrt the train observers frame.
> The front flash has a closing speed of c+v and the rear flash
> wrt the observer will have a closing speed of c-v.
> And this is why the train observer in that example will not see
> the lights as simultaenous even though they were in reality.
>
> --
> James M Driscoll Jr
> Spaceman- Hide quoted text -
Spaceman
But, that is the problem, it is not constant to all frames simply
because it is a speed that is no more immune to relative motion
than any other speed would be.
Basically the motion of the observer proves the speed can not be the same
and actually is c+v if heading toward the light and c-v if heading away.
This is what causes the "supposed" non simulteniety.
It is observationally non simultaneous, but in reality, it is
simultaneous so the observer is proving the "speed of light" is not
constant to all frames or it would see the flashes simultaneously.
:)
>> The light still travels at c wrt the track and thier source points
>> but the train
>> observer will be relative to that speed.
>> As I always say the speed of light is not relatively the same (c)
>> to all observers.
>
> NO....observed relative velocity by the track observer have no effect
> on the isotropy of the speed of light in the train.
Actually yes, it does not have an effect on the constant speed of the light
itself
wrt the medium it travels in.
It only has a relative motion problem with the motion causing the
different times the light flash is seen.
In short, The light flashes are not seen simultaneously by the moving
observer simply because lightspeed is NOT the same to all frames.
kenseto
There is no experiment that settles the issue of RoS.
>
> Your claim was that the observations made by the train observer are
> NEVER MENTIONED AT ALL, right or wrong, in the gedanken.
>
> Your claim is wrong. It IS mentioned in the gedanken. It IS mentioned
> in the video. It is this simple fact that you will not acknowledge.
No it is not mentioned in the gedanken or the video. The gedanken and
the video assumed that the train observer will see the light from the
front of the train first and before he sees the light from the rear.
This bogus assumption is based on the initial bogus assumption that
the train observer is moving wrt the light fronts in such a way that
will affect the arrival time of the light fronts to the train
observer.
The true observations made by the train observer are as follows:
1. The speed of light is isotropic and independent of the motion of
the sources or his own motion wrt the light fronts. In other words he
is not moving in such a way that will affect the arrival speed of
light from different directions.
2. He (the train observer) measures that he is at equal distance from
the strikes.
3. The track observer determined that the strikes were simultaneous to
begin with.
4. From items 1, 2 and 3 the train observer concludes that the strikes
arrive at his location simultaneously.
Ken Seto
> > > > > > and- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -...
>
> read more »
PD
OK, so this is proof that something that was told to you YESTERDAY you
have forgotten today. In the video, starting at 57 seconds in, the
narrator says, "But what does the passenger really see?" and going on
for the next 15 seconds or so about what the passenger actually says.
And yet you have forgotten that you looked at it and listened to this,
and today you deny that it is mentioned in the video at all. Even this
morning you acknowledged that the statement was there, but that it's
wrong, and now you say the statement is not there at all.
Ken, you are not worth talking to, as your mind has gone to rot.
Ta-ta, Ken. You can continue to live in your tiny little sequestered
world in the confines of your children's house, and delude yourself
that what you saw yesterday was never there at all. Be my guest. I'm
done.
> ...
>
> read more »
kenseto
You are possibly the stupidest runt of the SRians in these NGs. After
the narrator says "But what does the passenger really see?" he follows
with a bunch of false assertions based on the naive notion that the
passager is moving wrt the light fronts in such a way that will
destroy the measured isotropy of the speed of light in the train.
Since the passenger measures the speed of light to be isotropic then
the narrator assertions are false.
The narrator should say:
1. The speed of light in the train is isotropic.
2. The flashes occur at equal distance from the strikes.
3. The strikes were determined to occur simultaneously.
4. Therefore the passenger will also see the flashes to be
simutlaneous.
Ken Seto
> > > You now say it is mentioned (though you do not admit your previous- Hide quoted text -
kenseto
The speed of light is a constant math ratio in all frames as follows:
Light path length of ruler (299.792,458 m long physically)/the
absolute time content for a clock second co-moving with the ruler.
The speed of light is not immune to relative motion. The observed
Doppler shift is evidence for the varying speed of incoming light from
a moving source as follows:
Speed of incoming light= c'= (Measured frequency of incoming light)
(Universal wavelength of that incoming light).
You ask: What is universal wavelength?
The answer:
For example: if the source of incoming light is sodium the universal
wavelength of sodium is 589 nm...the universal wavelength of sodium is
determined by the observer using his sodium source.
Then the arrival speed of sodium light from a moving sodium source is:
c'= (measured frequency of incoming sodium light)(589nm)
Ken Seto
> Basically the motion of the observer proves the speed can not be the same
> and actually is c+v if heading toward the light and c-v if heading away. This is what causes the "supposed" non simulteniety.
No this would destroy the isotropy of the speed of light in the train.
Since the train observer measures the speed of light in his frame to
be isotropic he does not see c+v or c-v. What this mean is that the
train observer does not move wrt the light fronts in such a way that
will destroy the isotropy of the speed of light in his frame. Please
watch the following video:
http://www.youtube.com/watch?v=wteiuxyqtoM
It shows that the two light spheres meets simultaneously at an
infinite number of locations. The track observer is at one of these
locations and the train observer is at a different location.
Ken Seto
> It is observationally non simultaneous, but in reality, it is
> simultaneous so the observer is proving the "speed of light" is not
> constant to all frames or it would see the flashes simultaneously.
> :)
>
> >> The light still travels at c wrt the track and thier source points
> >> but the train
> >> observer will be relative to that speed.
> >> As I always say the speed of light is not relatively the same (c)
> >> to all observers.
>
> > NO....observed relative velocity by the track observer have no effect
> > on the isotropy of the speed of light in the train.
>
> Actually yes, it does not have an effect on the constant speed of the light
> itself
> wrt the medium it travels in.
> It only has a relative motion problem with the motion causing the
> different times the light flash is seen.
>
> In short, The light flashes are not seen simultaneously by the moving
> observer simply because lightspeed is NOT the same to all frames.
>
> --
> James M Driscoll Jr
rval...@gmail.com
kenseto ha escrito:
> > > The train observer must make this determination by himself. In the
> > > train frame the speed of light is isotropic and therefore he does not
> > > observe any closing speed between him and the light fronts. Therefore
> > > if the strikes occur simultaneously at equal distance then he will
> > > observe them to be simultaneous.
> >
> > Read again Einstein’s text and be careful determining which are the
> > events considered simultaneous or not respect the two different
> > frames. The events are {two strokes of lightning A and B which are
> > simultaneous with reference to the railway embankment}(I put
> > Einstein’s word between { }). Observer M in the embankment will see
> > the two light rays arriving to him simultaneously by definition of the
> > experiment. But now we have a third event (the simultaneous arriving
> > of the two light ray to M) that must be present in all frames. This
> > excludes completely the possibility that the two light rays meet at
> > the centre point M’ in the moving train frame at the same time, as you
> > are suggesting.
>
> I suggested no such thing. Here's what I suggested.
Read your point 8. [The train observer must also sees the strikes to
be simultaneous]. Then you are saying that the rays arrive
simultaneously to M’.
> From the track frame's point of view:
> 1. The strikes occur at equal distance of 0.5L m from him.
> 2. The speed of light in the track frame is isotropic.
> 3. Einstein stipulated that the track observer sees the flashes to be
> simultaneous.
Einstein stipulated that the flashes ARE simultaneous for the track
frame (primary fact). And then, as a consequence of the stipulated
position M in the centre, and the constant and isotropic light speed,
the light rays arrive simultaneously to M, that is an event that must
be present in ANY frame.
> 4. From items 1,2 and 3 the track observer determined that the flashes
> occur simultaneously to begin with.
The flashes are simultaneous in the track frame by definition of the
experiment. Then, as a consequence, the observer at M MUST see the
light rays arriving simultaneously to him, an event that must be
present in ALL frames.
> 5. The track observer says: The speed of light is isotropic in the
> train frame.
The speed of light is isotropic in all frames by Einstein’s second
postulate.
> 6. The track observer said: The flashes occur at equal distance of
> 0.5L/gamma from the strikes.
False. The track observer doesn’t use any gamma. For him the light
rays are at equal distance from M’ ONLY at the starting instant of the
light rays (unique one when M’ coincides with M), and the equal
distance is 0.5L (not 0.5L/gamma). After this instant, the distances
from the light rays to M’ become different, owed to the different
closing velocities (c-v) and (c+v). Your deduction that the track
observer considers the light rays always at the same distance from M’
is incorrect.
> 7. The track observer said: The flashes occur simultaneously to begin
> with.
Yes, the flashes occur simultaneously IN THE TRACK FRAME by definition
of the experiment. You are assuming, erroneously, that they also occur
simultaneously in the train frame.
> 8. Therefore the track observer must also said: The train observer
> must also sees the strikes to be simultaneous at a time of 0.5L/
> gamma*c second according to the train clock. This conclusion will
> agree with the fact that the train clock is running slow.
>
Wrong deduction. You are forcing a simultaneous arriving of the light
rays to M’, that is consequence of your previous erroneous assumption
about the flashes being simultaneous also in the train frame. As I
said to you before, the simultaneous arriving of the light rays to
some point is an event that must be present in ALL frames. The rays
can’t arrive simultaneously to two different points M and M’. The
meeting of two light rays running with opposite senses is a SINGLE
event that must be present in all frames (which different space and
time co-ordinates in each frame, of course).
The flashes A and B are NOT simultaneous in the train frame. The B one
occurs before the A one, arriving first its ray to M’, even if both
rays run an equal distance at the same isotropic speed c (as seeing in
the train frame).
>
> Ken Seto
>
> >The same rays can’t be arriving at two different
> > points M and M’ at the same time in each of these points (no matter in
> > what frame are you observing the events), what implies that some ray
> > arrives to M’ (the B one) before the arriving of the other (the A
> > one). Which is your error? To suppose that lightning A and B,
> > simultaneous for the embankment frame by definition, are simultaneous
> > events also in the moving train frame [if the strikes occur
> > simultaneously at equal distance] (your words between [ ]), which is
> > precisely what you are trying to prove. Closing velocities (c-v) and (c
> > +v), different ones for each frame that is moving with a non zero
> > velocity v in the stationary frame, can be used without any problem to
> > show that the simultaneous arriving of the two light rays at the
> > centre point of the frame can occurs only in the stationary frame
> > where v equals zero.> Ken Seto
> >
> > > > Ken Seto
> >
> > > > RVHG (Rafael Valls Hidalgo-Gati)- Hide quoted text -
> >
> > > > - Show quoted text -
> >
> > RVHG (Rafael Valls Hidalgo-Gato)- Hide quoted text -
> >
> > - Show quoted text -
RVHG (Rafael Valls Hidalgo-Gato)
kenseto
These assertions are bogus. Why? Because it contradicts with the
measured isotropy of the speed of light in the train frame. M' cannot
claim the isotropy of the speed of light and at the same time claims
that he is moving wrt the light fronts at (c+v) and
(c-v).
>Your deduction that the track
> observer considers the light rays always at the same distance from M’
> is incorrect.
No it is not incorrect. Please watch the following video:
http://www.youtube.com/watch?v=wteiuxyqtoM
You will note that as the light spheres expand they will meet at an
infinite number of locations at equal distance from the origins of the
flashes. The track observer is located at one of these locations and
the train observer is at a different location but he will maintain the
same distance from the origins of the flashes.
Ken Seto
> - Show quoted text -- Hide quoted text -
kenseto
I am tired of argueing with an idiot like you. You seem to think that
it is OK for SR to make the following contradictory claims for the
train observer:
1. The train observer measures the speed of light to be isotropic.
2. The train observer is moving wrt the light fronts from the strikes
in such a way that will destroy the isotropy of the speed of light in
the train.
In case you are too stupid to understand the train observer can claim
one or the other but not both at the same time.
Ken Seto
PD
Ken is an idiot, and a senile idiot to boot.
1. He confuses closing speed with light speed, thinking that if
*closing speed* is anisotropic then this means that light speed is
anisotropic.
2. He thinks that if a velocity is anisotropic in the track frame,
then it is also anisotropic in the train frame.
Simple mistakes and easy to correct, but Ken is no longer capable of
correcting any of his mistakes. I'm afraid it's a hopeless case at
this point.
>
> Ken Seto
kenseto
Hey idiot....SR calims that closing speeds will affect the arrival
time of the light fronts to the observer in the train. This means that
the speed of light in the train is anisotropic.
> 2. He thinks that if a velocity is anisotropic in the track frame,
> then it is also anisotropic in the train frame.
No idiot....RoS said that the train observer sees the light from the
front before he sees the light from the rear. That can only mean that
the speed of light in the train is anisotropic. You are so stupid I
suggest that you go off to a corner and commit suicide.
Ken Seto
PD
No, it does NOT mean this. This is precisely what I just said. You
MISTAKENLY think that if closing speed is anisotropic, then light
speed is anisotropic. That is an error, which I just pointed out. You
chose to REPEAT the error, rather than correcting it. As I said, you
are incapable of correcting any mistake you make, even when the
mistake is pointed out to you. All you can do is repeat the same
mistake.
>
> > 2. He thinks that if a velocity is anisotropic in the track frame,
> > then it is also anisotropic in the train frame.
>
> No idiot....RoS said that the train observer sees the light from the
> front before he sees the light from the rear.
Yes.
> That can only mean that
> the speed of light in the train is anisotropic.
No.
Again, you REPEAT mistakes, rather than correct them. You are
incapable of correcting mistakes you make and instead can only repeat
them. As I said, Ken, there is no point in discussing anything with
you if you are incapable of correcting a mistake.
> You are so stupid I
> suggest that you go off to a corner and commit suicide.
>
> Ken Seto
>
> > Simple mistakes and easy to correct, but Ken is no longer capable of
> > correcting any of his mistakes. I'm afraid it's a hopeless case at
> > this point.
You see? You just proved that your case is hopeless.
>
> > > Ken Seto
Spaceman
Actually, yes it does mean that.
If it were not, the speed of light to the moving observer
would be the same relatively in both the front and rear.
But it can't be the same or the flash would still occur
simultenously to the observer.
The speed of light is therefore "relative" to the moving observer.
The speed of light is not constant to all frames of reference.
The End.
HA HA HA HA HA HA HA
!
PD
wrote:
Your laughter would mean something if it weren't coming from a
brainless clown.
Spaceman
Poor PD,
Instead of realizing the problem stated is a fact about lightspeed
not being constant to all frames.
He insults the messenger.
:)
kenseto
> On Jul 22, 8:54 am, kenseto <kens...@erinet.com> wrote:
>
>
>
>
>
> > On Jul 22, 8:39 am, PD <TheDraperFam...@gmail.com> wrote:
>
> > > On Jul 22, 7:29 am, kenseto <kens...@erinet.com> wrote:
>
> > > > I am tired of argueing with an idiot like you. You seem to think that
> > > > it is OK for SR to make the following contradictory claims for the
> > > > train observer:
> > > > 1. The train observer measures the speed of light to be isotropic.
> > > > 2. The train observer is moving wrt the light fronts from the strikes
> > > > in such a way that will destroy the isotropy of the speed of light in
> > > > the train.
>
> > > > In case you are too stupid to understand the train observer can claim
> > > > one or the other but not both at the same time.
>
> > > Ken is an idiot, and a senile idiot to boot.
> > > 1. He confuses closing speed with light speed, thinking that if
> > > *closing speed* is anisotropic then this means that light speed is
> > > anisotropic.
>
> > Hey idiot....SR calims that closing speeds will affect the arrival
> > time of the light fronts to the observer in the train. This means that
> > the speed of light in the train is anisotropic.
>
> No, it does NOT mean this.
Hey idiot....here's what RoS claims:
Closing speed as observed by the track observer makes the light from
the front of the train reaches the train observer before the light
from the rear. This sounds very anisotropic to me.
>This is precisely what I just said. You
> MISTAKENLY think that if closing speed is anisotropic, then light
> speed is anisotropic.
Hey idiot....there are two sources of light: one is at the front end
of the train and the other is at the rear end of the train. The train
obbserver is located at the middle of these two light sources. RoS
claims that the light front from the front reaches the train observer
before the light front from the rear. Why is this not mean that the
speed of light is anisotropic in the train?
>That is an error, which I just pointed out. You
> chose to REPEAT the error, rather than correcting it. As I said, you
> are incapable of correcting any mistake you make, even when the
> mistake is pointed out to you. All you can do is repeat the same
> mistake.
No the error is your failure to understand that different arrival
times of the light fronts means anisotropy of the speed of light.
>
>
>
> > > 2. He thinks that if a velocity is anisotropic in the track frame,
> > > then it is also anisotropic in the train frame.
>
> > No idiot....RoS said that the train observer sees the light from the
> > front before he sees the light from the rear.
>
> Yes.
>
> > That can only mean that
> > the speed of light in the train is anisotropic.
>
> No.
>
> Again, you REPEAT mistakes, rather than correct them. You are
> incapable of correcting mistakes you make and instead can only repeat
> them.
Hey idiot there is no mistake on my part. RoS does imply that the
speed of light in the train is anisotropic.
Ken Seto
>As I said, Ken, there is no point in discussing anything with
> you if you are incapable of correcting a mistake.
>
> > You are so stupid I
> > suggest that you go off to a corner and commit suicide.
>
> > Ken Seto
>
> > > Simple mistakes and easy to correct, but Ken is no longer capable of
> > > correcting any of his mistakes. I'm afraid it's a hopeless case at
> > > this point.
>
> You see? You just proved that your case is hopeless.
>
>
>
>
>
> > > > Ken Seto- Hide quoted text -
PD
That's because you don't know the difference between closing speed and
light speed. That's why you confuse anisotropic closing speed with
anisotropic light speed. They're the same thing to you. That's because
you are incapable of correcting any mistake and incapable of learning
the difference. That's why you are a hopeless case at this point.
>
> >This is precisely what I just said. You
> > MISTAKENLY think that if closing speed is anisotropic, then light
> > speed is anisotropic.
>
> Hey idiot....there are two sources of light: one is at the front end
> of the train and the other is at the rear end of the train. The train
> obbserver is located at the middle of these two light sources. RoS
> claims that the light front from the front reaches the train observer
> before the light front from the rear. Why is this not mean that the
> speed of light is anisotropic in the train?
As I said, you do not know the difference between anisotropic closing
speed and anisotropic light speed. You've made it plain three times in
a row that you do not know the difference. And you're calling me the
idiot.
>
> >That is an error, which I just pointed out. You
> > chose to REPEAT the error, rather than correcting it. As I said, you
> > are incapable of correcting any mistake you make, even when the
> > mistake is pointed out to you. All you can do is repeat the same
> > mistake.
>
> No the error is your failure to understand that different arrival
> times of the light fronts means anisotropy of the speed of light.
No, it does not, Ken. You do not know the difference between
anisotropic closing speed and anisotropic light speed.
>
>
>
>
>
> > > > 2. He thinks that if a velocity is anisotropic in the track frame,
> > > > then it is also anisotropic in the train frame.
>
> > > No idiot....RoS said that the train observer sees the light from the
> > > front before he sees the light from the rear.
>
> > Yes.
>
> > > That can only mean that
> > > the speed of light in the train is anisotropic.
>
> > No.
>
> > Again, you REPEAT mistakes, rather than correct them. You are
> > incapable of correcting mistakes you make and instead can only repeat
> > them.
>
> Hey idiot there is no mistake on my part. RoS does imply that the
> speed of light in the train is anisotropic.
I don't know how many times you want to repeat the same mistake. Every
time you do, it just reinforces the fact that you are incapable of
correcting a mistake. If you repeat the same mistake 1536 times, it
will have already been *abundantly* clear that you cannot correct a
mistake.
kenseto
Your are truly a fucking idiot. It is not closing speeds we are
talking about. RoS claims that the two light fronts from the opposite
directions take two different transit times to arrive at the train
observer. That's anisotropic of the speed of light in the train.
Ken Seto
gl...@aol.com
Ken Seto wrote,
< Therefore if the strikes occur simultaneously at equal distance then
he will observe them to be simultaneous. >
measured on the track.
Indeed, as per Einstein's definition, if strikes occur at equal
distances, the train's clocks have to be hand-set so that
tB - tA = tA' - tB.
By the way, if anyone wants to understand STR they should study
Einstein's 1905 paper instead of Einstein and Infeld's book to the
layman. And anyone who wants to really understand the equations in
that paper, and what's wrong with it, should study "A Flower for
Einstein" by Gerald Lebau.
e
Dono
>
> By the way, if anyone wants to understand STR they should study
> Einstein's 1905 paper instead of Einstein and Infeld's book to the
> layman. And anyone who wants to really understand the equations in
> that paper, and what's wrong with it, should study "A Flower for
> Einstein" by Gerald Lebau.
>
Why would anybody read your junk, Gerald?
PD
Yes, it is.
> RoS claims that the two light fronts from the opposite
> directions take two different transit times to arrive at the train
> observer.
Those are closing speeds. I cannot help it if you don't have the
foggiest idea what the difference between closing speed and light
speed is.
This isn't surprising. You don't know the meaning of half the terms
you use. You don't know what a vector component is, for instance, but
you use the term in a *definition* of relative velocity. Only a fool
would define a term using other terms he doesn't know the meaning of.
What does that make you, Ken? Have you no sense of shame? Does it not
bother you to further disgrace your family name?
kenseto
> Ken Seto wrote,
> < Therefore if the strikes occur simultaneously at equal distance then
> he will observe them to be simultaneous. >
> "Simultaneously" as measured by him is not the same as it is as
> measured on the track.
I agree...so what is your point? The track observer will see the
flashes to be simultaneous at time 0.5L/c. The track observer will
predict that the train observer sees the flashes to be simultaneous at
time 0.5L*gamma/c according to the track clock.
Ken Seto
Martin Hogbin
> Poor PD,
> Instead of realizing the problem stated is a fact about lightspeed
> not being constant to all frames.
> He insults the messenger.
> :)
>
Has Spaceman's program been updated? He seems to be
using more words than before.
Martin Hogbin
Spaceman
Yet another.
Nice insultation physics Martin.
:)
Glad to see your brain works a tiny bit beyond the
ROM that relativity has burned into your chips.
Now if you could just flush that ROM and re-configure it
for science instead.
Androcles
"Martin Hogbin" <goatN...@hogbin.org> wrote in message
news:o42dnbX1b-KWPBrV...@bt.com...
Obviously yours hasn't. The same old one line sniping drool it
has always been, you miserable shit.
rval...@gmail.com
Vacuum light speed is isotropic c in all frames, by postulate and
experiment. I don’t understand what contradiction are you talking
about. Observer at M’ considers himself at rest in the train frame
(v=0), and for him the closing velocities are c-0=c+0=c (the isotropic
c). Observer M (at rest in the embankment frame) is the one
considering M’ moving at some not zero v velocity, and different
closing velocities (c-v) and (c+v) with the light rays. You are mixing
(incorrectly) an M observation with an M’ one. Of course, you arrive
to a contradiction, but it is only your one.
>
> >Your deduction that the track
> > observer considers the light rays always at the same distance from M’
> > is incorrect.
>
> No it is not incorrect. Please watch the following video:
> http://www.youtube.com/watch?v=wteiuxyqtoM
> You will note that as the light spheres expand they will meet at an
> infinite number of locations at equal distance from the origins of the
> flashes. The track observer is located at one of these locations and
> the train observer is at a different location but he will maintain the
> same distance from the origins of the flashes.
>
It is totally incorrect. The distances that the track observer sees
between the light rays and the moving with v velocity M’ are inversely
proportional to the different closing velocities (c-v) and (c+v). As a
consequence, ray B arrives to M’ before ray A arrives to M’. As I
noted you more than one time before (without any comment from your
part yet), two light rays running in opposite senses over the same
right line can meet simultaneously only at a single point, no matter
what frame are you using to describe the movement. In our case this
point is M in the embankment frame (by definition of the experiment),
excluding totally that the rays can meet also simultaneously at a
different M’ point as you are (incorrectly) claiming. The event of the
light rays arriving simultaneously to M’ simply doesn’t exist at all
in any frame, including of course the train one.
Even if in the train frame the points A and B (where the flashes
originate) are at the same distance from M’, that doesn’t imply that
the flashes were simultaneous in the train frame (your primary error,
as I had notice to you more than one time).
I can’t see the video yet owed to local technical problems in my
computer. Anyway, a video shows only what his creator thinks it must
do. I am guessing that he shares with you the wrong idea about the two
flashes A and B being a single simultaneous event present in all
frames (or perhaps you are simply interpreting bad the video).
Absolute simultaneity can refer only to things occurring at the same
single space point. This is the more important concept introduced by
Einstein in his 1905 Relativity. Read again the corresponding Annalen
der Physik June 30 paper.
RVHG (Rafael Valls Hidalgo-Gato)
kenseto
Exactly M' measures the speed of light to be isotropic and thus he
does not see any closing velocity. But RoS insists that M' sees the
front flash before the rear flash. This means that the front flash
takes less transit time to reach the train observer than the transit
time that required for the rear flash to reach the train observer.
Since both flashes were generated simultaneously at equal distance
from the strikes, this means that the speed of light is anisotropic
according to RoS. Such RoS assertion is in direct conflict with the
measured isotropy of the speed of light in the train.
> Observer M (at rest in the embankment frame) is the one
> considering M’ moving at some not zero v velocity, and different
> closing velocities (c-v) and (c+v) with the light rays. You are mixing
> (incorrectly) an M observation with an M’ one. Of course, you arrive
> to a contradiction, but it is only your one.
No I didn't mix anything. The closing velocities as seen by M has no
effect on the isotropy of the speed of light in the train. But RoS use
them anyway to arrive at the bogus assertion that the train observer
actually sees the front flash before the rear flash. This bogus
assertion leads to the bogus concept of RoS.
>
> > >Your deduction that the track
> > > observer considers the light rays always at the same distance from M’
> > > is incorrect.
>
> > No it is not incorrect. Please watch the following video:
> >http://www.youtube.com/watch?v=wteiuxyqtoM
> > You will note that as the light spheres expand they will meet at an
> > infinite number of locations at equal distance from the origins of the
> > flashes. The track observer is located at one of these locations and
> > the train observer is at a different location but he will maintain the
> > same distance from the origins of the flashes.
>
> It is totally incorrect. The distances that the track observer sees
> between the light rays and the moving with v velocity M’ are inversely
> proportional to the different closing velocities (c-v) and (c+v).
Your statement is totally incorrect. What the track observer sees as
closing velocities has avbsolute no effect on the isotropy of the
speed of light in the train. Beside the train must make his own
determine whether the strikes are simultaneous as follows:
1. The speed of light in his frame is isotropic.
2. He is at equal distance from the strikes.
3. It was determined that the strikes were simultaneous to begin with.
4. Therefore the train observer must also sees the strikes to be
simutlaneous.
Ken Seto
> As a
> consequence, ray B arrives to M’ before ray A arrives to M’. As I
> noted you more than one time before (without any comment from your
> part yet), two light rays running in opposite senses over the same
> right line can meet simultaneously only at a single point, no matter
> what frame are you using to describe the movement. In our case this
> point is M in the embankment frame (by definition of the experiment),
> excluding totally that the rays can meet also simultaneously at a
> different M’ point as you are (incorrectly) claiming. The event of the
> light rays arriving simultaneously to M’ simply doesn’t exist at all
> in any frame, including of course the train one.
> Even if in the train frame the points A and B (where the flashes
> originate) are at the same distance from M’, that doesn’t imply that
> the flashes were simultaneous in the train frame (your primary error,
> as I had notice to you more than one time).
Ah....if the flashes were not simultaneous to begin with then the
track observer will not be able to see the flashes arriving at him
simultaneously. Why? Because the track observer is at equal distance
from the strikes and the speed of light in the track frame is
isotropic. It appear that it is you who made the error.
Ken Seto
> I can’t see the video yet owed to local technical problems in my
> computer. Anyway, a video shows only what his creator thinks it must
> do. I am guessing that he shares with you the wrong idea about the two
> flashes A and B being a single simultaneous event present in all
> frames (or perhaps you are simply interpreting bad the video).
> Absolute simultaneity can refer only to things occurring at the- Hide quoted text -
>
> - Show quoted text -...
>
> read more »
rval...@gmail.com
Wrong deduction. You have no reason at all to say that M’ [does not
see any closing velocity]. In the train frame simply the closing
velocities between the light rays and M’ and the isotropic c have the
same value, as I put to you very clear in my last comment.
> But RoS insists that M' sees the
> front flash before the rear flash. This means that the front flash
> takes less transit time to reach the train observer than the transit
> time that required for the rear flash to reach the train observer.
Wrong deduction again. In the train frame the transit time for both
light rays to reach M’ is exactly the same, running the same distance
at the same velocity c. Yes, M’ sees the front flash before the rear
flash, what implies that the front flash occurs BEFORE the rear flash.
The delay between the flashes is the same delay between the arriving
rays to M’.
> Since both flashes were generated simultaneously at equal distance
> from the strikes,
Yes, equal distance in both frames, BUT NOT SIMULTANEOSLY in both
frames. The flashes were generated simultaneously ONLY IN THE
EMBANKMENT FRAME.
> this means that the speed of light is anisotropic
> according to RoS. Such RoS assertion is in direct conflict with the
> measured isotropy of the speed of light in the train.
>
You are the one making the wrong deduction about light speed being
anisotropic according to RoS, owed to your primary error supposing the
flashes simultaneous in both frames.
> > Observer M (at rest in the embankment frame) is the one
> > considering M’ moving at some not zero v velocity, and different
> > closing velocities (c-v) and (c+v) with the light rays. You are mixing
> > (incorrectly) an M observation with an M’ one. Of course, you arrive
> > to a contradiction, but it is only your one.
>
> No I didn't mix anything.
Read your own words in the previous post [M' cannot claim the isotropy
of the speed of light and at the same time claims that he is moving
wrt the light fronts at (c+v) and (c-v)]. You are mixing (incorrectly)
the closing velocities (c+v) and (c-v) measured in the embankment
frame (an M observation) with the isotropy of the speed of light
measured in the train frame (an M’ observation).
> The closing velocities as seen by M has no
> effect on the isotropy of the speed of light in the train. But RoS use
> them anyway to arrive at the bogus assertion that the train observer
> actually sees the front flash before the rear flash. This bogus
> assertion leads to the bogus concept of RoS.
>
The train observer actually sees the front flash before the rear
flash, because in that frame the front flash occurs before the rear
one with the same delay, as I explained to you already in all detail
in a previous comment.
>
> >
> > > >Your deduction that the track
> > > > observer considers the light rays always at the same distance from M’
> > > > is incorrect.
> >
> > > No it is not incorrect. Please watch the following video:
> > >http://www.youtube.com/watch?v=wteiuxyqtoM
> > > You will note that as the light spheres expand they will meet at an
> > > infinite number of locations at equal distance from the origins of the
> > > flashes. The track observer is located at one of these locations and
> > > the train observer is at a different location but he will maintain the
> > > same distance from the origins of the flashes.
> >
> > It is totally incorrect. The distances that the track observer sees
> > between the light rays and the moving with v velocity M’ are inversely
> > proportional to the different closing velocities (c-v) and (c+v).
>
> Your statement is totally incorrect. What the track observer sees as
> closing velocities has avbsolute no effect on the isotropy of the
> speed of light in the train.
Isotropic light speed is the same c in all frames by postulate and
experience. Different closing velocities (c-v) and (c+v) are not
related with any anisotropy. I don’t understand what relationship are
you establishing between closing velocities in the embankment frame
and isotropy (or anisotropy) in the train frame. Are you rejecting
that (c-v) and (c+v) are the closing velocities between the light rays
and the moving with v velocity M’? Or maybe you are confusing the
light ray velocity in some frame with its closing velocity with
another moving entity?
> Beside the train must make his own
> determine whether the strikes are simultaneous as follows:
> 1. The speed of light in his frame is isotropic.
> 2. He is at equal distance from the strikes.
> 3. It was determined that the strikes were simultaneous to begin with.
False. This is your primary error. The flashes and the arriving of the
light rays to the observer are only simultaneous in the embankment
frame, not in the train frame.
> 4. Therefore the train observer must also sees the strikes to be
> simutlaneous.
>
False. In the train frame neither the flashes at A and B or the
arriving of the light rays to M’ are simultaneous, even if they have
the same delay, the front one occurring before the rear one.
> Ken Seto
>
> > As a
> > consequence, ray B arrives to M’ before ray A arrives to M’. As I
> > noted you more than one time before (without any comment from your
> > part yet), two light rays running in opposite senses over the same
> > right line can meet simultaneously only at a single point, no matter
> > what frame are you using to describe the movement. In our case this
> > point is M in the embankment frame (by definition of the experiment),
> > excluding totally that the rays can meet also simultaneously at a
> > different M’ point as you are (incorrectly) claiming. The event of the
> > light rays arriving simultaneously to M’ simply doesn’t exist at all
> > in any frame, including of course the train one.
> > Even if in the train frame the points A and B (where the flashes
> > originate) are at the same distance from M’, that doesn’t imply that
> > the flashes were simultaneous in the train frame (your primary error,
> > as I had notice to you more than one time).
>
> Ah....if the flashes were not simultaneous to begin with then the
> track observer will not be able to see the flashes arriving at him
> simultaneously.
Wrong deduction again. The flashes were not simultaneous in the train
frame, but simultaneous in the embankment one, where simultaneous
flashes implies simultaneous arriving of the light rays to the
observer M.
> Why? Because the track observer is at equal distance
> from the strikes and the speed of light in the track frame is
> isotropic. It appear that it is you who made the error.
>
Let me try to put it easy for you. We have equal distances from A and
B to the central observer and isotropic light speed c in all frames.
This implies that a possible delay between the flashes is preserved as
the same delay between the arriving of the light rays to the central
observer, and this is true for any frame.
If flashes are simultaneous (delay zero), the arriving of the light
ray to the observer is also simultaneous (same delay zero).
If flashes are not simultaneous (delay not zero), the arriving of the
light ray to the observer is also not simultaneous (same delay not
zero).
By definition of the experiment flashes are simultaneous (delay zero)
in the embankment frame. Then, the arriving of the light rays to the
central observer M is also simultaneous (delay zero). This
simultaneous arriving of the light rays to the point M constitutes an
event that must be present in any frame. But two light rays running in
the same right line with opposite senses can meet only in a single
point, being this valid in any frame, and this point must be always
the location of the observer M in the corresponding frame. Of course,
observer M is moving in all frames different from the embankment one
where he is at rest. As M is moving in the train frame, M’ is moving
in the embankment frame, being always M and M’ different points (they
are at the same point in the embankment frame only when the
simultaneous flashes A and B occur at the same distance). As the light
rays can meet in a single point (and this is M by definition), they
can’t meet at M’, what implies a non-zero delay in M’ that must be the
same non-zero delay of the flashes A and B in the train frame.
I avoided in this explanation to refer any closing velocity.
> Ken Seto
>
RVHG (Rafael Valls Hidalgo-Gato)
kenseto
No I didn't make any wrong deduction. I said that the track observer
concluded that the flashes were simultaneous to begin with. Thereefore
these same flashes will occur simultaneously TO BEGIN WITH for the
train observer. You and RoS then use the track as reference for motion
by the train observer wrt the light fronts. That's why you conclude
that the train observer sees the front flash before the rear flash.
Such conclusion is wrong and it violates the measured isotropy of the
speed of light made by the train observer. The train observer cannot
move wrt the light flashes in such a way that will violate the
measured isotropy.
Ken Seto
>
>
>
> > > Observer M (at rest in the embankment frame) is the one
> > > considering M’ moving at some not zero v velocity, and different
> > > closing velocities (c-v) and (c+v) with the light rays. You are mixing
> > > (incorrectly) an M observation with an M’ one. Of course, you arrive
> > > to a contradiction, but it is only your- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
papa...@hotmail.com
I will put here the complete mathematical derivation of the relativity
of simultaneity, following the same notation as Einstein in
http://www.bartleby.com/173/9.html
v= 0.6c = 180000 km/sec is the speed of the moving frame. Notice that,
according to SRT, you can’t differentiate on who is doing the move. So
we have to consider two cases: a) the track observer M is considered
at rest and the train observer M’ is moving at v, along the +x axis,
or b) the train observer M’ is considered at rest and the track
observer M is moving at v, along the -x axis.
c= 300000 km/sec
sqrt(1-v^2/c^2) = 0.8
Einstein proves that both observers disagree about the simultaneity of
the strikes.
1) Check for isotropy
Let K be the frame of the track observer M. Also let K’ be the frame
of the train observer M’. The Lorentz transformation equations between
these two frames are:
x = (x’+ vt’) / (sqrt(1- v^2/c^2)) ; t = (t’+ vx’/c^2) / (sqrt(1- v^2/
c^2))
x' = (x - vt) / (sqrt(1- v^2/c^2)) ; t’ = (t - vx/c^2) / (sqrt(1- v^2/
c^2))
When, on frame K, (x,t) = (0,0) it follows that, on frame K’, (x’,t’)
= (0,0) , meaning observers M and M’ coincide. At that instant of time
(t = t' =0), there are two lightning strikes at locations A, at x_A =
x’_A = -100000km, and at location B, at x_B = x’_B = +100000km
The strike light signal from A moves along the track frame K following
equation x = ct. Substituting this x in Lorentz equations, we obtain:
x' = (c-v)t / (sqrt(1-v^2/c^2)) ; t' = ((c-v)/c)t / (sqrt(1-v^2/
c^2)) , hence x'/t'=c. So in frame K’ the speed of the light signal is
again equal to c.
The strike light signal from B moves on the track frame K following
equation x = -ct. Substituting x in Lorentz equations, we obtain:
x' = - (c+v)t / (sqrt(1-v^2/c^2)) ; t' = ((c + v)/c)t / (sqrt(1-v^2/
c^2)) , hence x'/t' = - c. So both light signals, going in opposite
directions, move at c. Then it follows light speed is isotropic in the
train frame K’ and has the same value c as on the track frame K.
2) Point of view from the track observer M
Equation, on frame K, for the back strike light from A is x_A(t) = ct
– 100000
Equation, on frame K, for the front strike light from B is x_B(t) =
100000 – ct
So at the track observer M location we have x = x_A(t) = x_B(t) = 0 ,
which implies t = 100000/c = 1/3 sec. So this is the time when the
track observer sees both strike light signals from A and B as being
simultaneous.
Now, according to the track observer M calculations (see figure below
showing space time relations), the front strike light signal will
reach the train observer location (which on the train frame K’ is
x’=0) at point C, while the back strike light signal will reach the
train observer location at point D (both as measured on frame K),
which are given by the following relations:
< >
< * >
>< o
> / <
A>____________M’/_____________<B
M C D ---> x
Point C: train observer M’ speed relation on frame K: x_M’ = vt =
180000t ; strike light signal from B equation x_B(t) = 100000 - ct.
They intersect at t_C = 100000 / (c + v) = 0.208 sec, at location x_C
= vt_C = 37500 km. Now, on frame K’ of the train observer M’, point C
space-time coordinates are given by the Lorentz equations, resulting
in:
x’_C = (x_C-vt) / (sqrt(1- v^2/c^2)) = (37500 – 180000 * 0.208)/0.8 =
0 (as it should be), and t’_C = (t_C-vx_C/c^2) / (sqrt(1- v^2/c^2)) =
(0.208 – (180000*37500)/300000^2)/0.8 = 0.167 sec. This is the time,
on frame K' when the front strike light signal reaches the train
observer M'. Notice how M calculates a slower (longer) time for point
C.
Point D: train observer M’ speed relation on frame K: x_M’ = vt =
180000t ; back strike light signal equation x_B(t) = ct - 100000. They
intersect at t_D = 100000 / (c - v) = 0.833 sec, at location x_D =
vt_D = 150000 km. On frame K’, point D coordinates are given by
Lorentz equations, resulting in:
x’_D = (x_D-vt) / (sqrt(1- v^2/c^2)) = (150000 – 180000 * 0.803)/0.8 =
0 (as it should be), and t’_D = (t_D-vx_D/c^2) / (sqrt(1- v^2/c^2)) =
(0.803 – (180000*150000)/300000^2)/0.8 = 0.667 sec. This is the time,
on frame K' when the back strike light signal reaches the train
observer M'.
Conclusions from the track observer M point of view:
a) Track observer on frame K sees two simultaneous strikes at t = 1/3
sec, x = 0.
b) Train observer on frame K’ sees two non-simultaneous strikes, the
first at t’ = 0.167 sec, x’ = 0 and the second at t’ = 0.667 sec at x’
= 0.
3) Now we consider the point of view of the train observer M’
M’ in his own frame K’, considers the track observer, in frame K, is
moving towards the –x’ direction at a speed v = 0.6c
The strike light signal from A moves along the train frame K’ as x’ =
ct’. Substituting x’ in Lorentz equations, we obtain:
x = (c + v)t’ / (sqrt(1- v^2/c^2)) ; t = ((c + v)/c)t’ / (sqrt(1- v^2/
c^2)), hence x/t = c.
The strike light signal from B moves on the train frame K’ as x’ = -
ct’. Substituting x’ in Lorentz equations, we obtain:
x = - (c+v)t’ / (sqrt(1-v^2/c^2)) ; t = ((c + v)/c)t’ / (sqrt(1-v^2/
c^2)) , hence x/t = - c.
Then it follows light speed is isotropic in the track frame K and has
the same value c as on the train frame K.
< >
< * >
o ><
> \ <
A>____________\M_____________<B
D’C’ M’ ---> x'
Equation on train frame K’ for the back strike light signal is
x’_A(t’) = ct’ – 100000
Equation on train frame K’ for the front strike light signal is
x’_B(t’) = 100000 – ct’
So at the train location M’ we have x’_A(t’) = x’_B(t’) = 0 , which
implies t’ = 100000/c = 1/3 sec. So this is the time when the train
observer M’ sees both strike light signals as simultaneous.
Now, according to the train observer M’ calculations, the front strike
light signal will reach the track observer M location (which on the
track frame K is x=0) at point C’, while the back strike light signal
will reach the train observer location M at point D’ (both as measured
on train frame K’), which are given by the following relations:
Point C’: track observer M speed relation on frame K’: x’_M = -vt’ =
-180000t’ ; front strike light signal equation x’_B(t’) = 100000 –
ct’. They intersect at t’_C’ = 100000 / (c + v) = 0.208 sec, at
location x’_C’ = vt’_C’ = - 37500 km.
On frame K of the track observer, point C’ space-time coordinates are
given by the Lorentz equations, resulting in:
x_C’ = (x’_C’ + vt’) / (sqrt(1- v^2/c^2)) = (-37500 + 180000 * 0.208)/
0.8 = 0 (as it should be)
t_C’ = (t’_C’-vx’_C’/c^2) / (sqrt(1- v^2/c^2)) = (0.208 –
(180000*37500)/300000^2)/0.8 = 0.167 sec. This is the time, on frame K
when the front strike light signal reaches the track observer M.
Point D’: track observer M speed relation on frame K’: x’_M = -vt’ =
-180000t’ ; back strike light signal equation x’_B(t’) = ct’ - 100000.
They intersect at t’_D’ = 100000 / (c - v) = 0.833 sec, at location
x’_D’ = vt’_D = -150000 km.
On frame K, point D’ coordinates are given by Lorentz equations,
resulting in:
x_D’ = (x’_D’-vt’) / (sqrt(1- v^2/c^2)) = (-150000 + 180000 * 0.803)/
0.8 = 0 (as it should)
t_D’ = (t’_D’-vx’_D’/c^2) / (sqrt(1- v^2/c^2)) = (0.803 –
(180000*150000)/300000^2)/0.8 = 0.667 sec. This is the time, on frame
K when the back strike light signal reaches the track observer M.
Conclusions:
a) Train observer M’ on frame K’ sees two simultaneous strikes at t’ =
1/3 sec, x’ = 0.
b) Track observer M on frame K sees two non-simultaneous strikes, the
first at t = 0.167 sec, x = 0 and the second at t = 0.667 sec at x =
0.
We notice that both observer disagree about the strikes being or not
simultaneous and both are right in their observations (neither K nor
K’ frames are preferred frames).
This proves the relativity of simultaneity.
Miguel Rios
jem
> On Jul 22, 9:38 pm, gl...@aol.com wrote:
>> Ken Seto wrote,
>> < Therefore if the strikes occur simultaneously at equal distance then
>> he will observe them to be simultaneous. >
>> "Simultaneously" as measured by him is not the same as it is as
>> measured on the track.
>
> I agree...so what is your point? The track observer will see the
> flashes to be simultaneous at time 0.5L/c. The track observer will
> predict that the train observer sees the flashes to be simultaneous at
> time 0.5L*gamma/c according to the track clock.
Which means that whenever the train's speed exceeds c*sqrt(0.5), the
train observer won't see the light flash until AFTER he's passed the
place where the lightning struck the track. Setoland sure is a funny
place - I can see why you spend so much time ROTFLOL.
Sue...
> kenseto wrote:
> > On Jul 22, 9:38 pm, gl...@aol.com wrote:
> >> Ken Seto wrote,
> >> < Therefore if the strikes occur simultaneously at equal distance then
> >> he will observe them to be simultaneous. >
> >> "Simultaneously" as measured by him is not the same as it is as
> >> measured on the track.
>
> > I agree...so what is your point? The track observer will see the
> > flashes to be simultaneous at time 0.5L/c. The track observer will
> > predict that the train observer sees the flashes to be simultaneous at
> > time 0.5L*gamma/c according to the track clock.
>
> Which means that whenever the train's speed exceeds c*sqrt(0.5), the
> train observer won't see the light flash until AFTER he's passed the
> place where the lightning struck the track.
<< A light beam normally incident upon an uniformly
moving dielectric medium is, in general, subject to
bendings due to a transverse Fresnel-Fizeau light
drag effect. In most familiar dielectrics, the
magnitude of this bending effect is very small
and hard to detect. Yet, the effect can be
dramatically enhanced in strongly dispersive media
where slow group velocities in the m/s range have
been recently observed taking advantage of the
electromagnetically induced transparency effect.
In addition to the usual downstream drag that
takes place for positive group velocities, we
discuss a significant anomalous upstream drag
which is expected to occur for negative group
velocities. Furthermore, for sufficiently fast
speeds of the medium, higher-order dispersion
terms are found to play an important role and
to be responsible for light propagation along
curved paths or the restoration of the time and
space coherence of an incident noisy beam.
The physics underlying this class of slow-light
effects is thoroughly discussed. >>
http://prola.aps.org/abstract/PRA/v68/i6/e063819
See also:
http://en.wikipedia.org/wiki/Wave_impedance
http://en.wikipedia.org/wiki/Free_space
http://www-ssg.sr.unh.edu/ism/what.html
Sue...
[...]
kenseto
> On 25 jul, 09:01, kenseto <kens...@erinet.com> wrote:
>
> > On Jul 24, 5:20 pm, rvall...@gmail.com wrote:
>
> > No I didn't make any wrong deduction. I said that the track observer
> > concluded that the flashes were simultaneous to begin with. Thereefore
> > these same flashes will occur simultaneously TO BEGIN WITH for the
> > train observer. You and RoS then use the track as reference for motion
> > by the train observer wrt the light fronts. That's why you conclude
> > that the train observer sees the front flash before the rear flash.
> > Such conclusion is wrong and it violates the measured isotropy of the
> > speed of light made by the train observer. The train observer cannot
> > move wrt the light flashes in such a way that will violate the
> > measured isotropy.
>
> > Ken Seto
>
> I will put here the complete mathematical derivation of the relativity
Your math proof is based on a bogus model of the prpagation of the
light fronts. This bogus model is in turn based on the false
assumption that the same two light rays that arrive at M will continue
on to arrive at M' which leads you to make the bogus conclusion that
the train observer sees the front light front before he sees the rear
light front. This is wrong. The two light rays that hit M are absorbed
by M. Two different light rays from the light spheres will hit M'.
What this mean is that your math proof is useless. End of story.
Ken Seto
kenseto
> > On Jul 22, 9:38 pm, gl...@aol.com wrote:
> >> Ken Seto wrote,
> >> < Therefore if the strikes occur simultaneously at equal distance then
> >> he will observe them to be simultaneous. >
> >> "Simultaneously" as measured by him is not the same as it is as
> >> measured on the track.
>
> > I agree...so what is your point? The track observer will see the
> > flashes to be simultaneous at time 0.5L/c. The track observer will
> > predict that the train observer sees the flashes to be simultaneous at
> > time 0.5L*gamma/c according to the track clock.
>
> Which means that whenever the train's speed exceeds c*sqrt(0.5), the
> train observer won't see the light flash until AFTER he's passed the
> place where the lightning struck the track. Setoland sure is a funny
> place - I can see why you spend so much time ROTFLOL.
Indeed ROTFLOL....the two light rays that hits M were absorbed by M.
Two different light rays from the expanding light spheres will hit M'.
What this mean is that your are ROTFLOL at your own stupidity. LOL.
Ken Seto
>
>
>
>
>
> > Ken Seto
>
> >> Indeed, as per Einstein's definition, if strikes occur at equal
> >> distances, the train's clocks have to be hand-set so that
> >> tB - tA = tA' - tB.
>
> >> By the way, if anyone wants to understand STR they should study
> >> Einstein's 1905 paper instead of Einstein and Infeld's book to the
> >> layman. And anyone who wants to really understand the equations in
> >> that paper, and what's wrong with it, should study "A Flower for
> >> Einstein" by Gerald Lebau.
>
> >> e- Hide quoted text -
kenseto
>
> > On Jul 24, 5:20 pm, rvall...@gmail.com wrote:
>
> > No I didn't make any wrong deduction. I said that the track observer
> > concluded that the flashes were simultaneous to begin with. Thereefore
> > these same flashes will occur simultaneously TO BEGIN WITH for the
> > train observer. You and RoS then use the track as reference for motion
> > by the train observer wrt the light fronts. That's why you conclude
> > that the train observer sees the front flash before the rear flash.
> > Such conclusion is wrong and it violates the measured isotropy of the
> > speed of light made by the train observer. The train observer cannot
> > move wrt the light flashes in such a way that will violate the
> > measured isotropy.
>
> > Ken Seto
>
> I will put here the complete mathematical derivation of the relativity
BTW if you want to use the LT to see if the train observer will see
the light fronts to be simultaneous. You do as follows:
In the track frame: the transit time interval for the track observer
to see the flash to be simultaneous is Delta(t)=0.5L/c.
The LT equation for the transformation of this time interval into the
train frame is as follows:
Delta(t')= gamma[Delta(t)-vx/c^2]
When M and M' are coincide with each other x=0 and this equation
becomes:
Delta(t')= gamma[Delta(t)]
That means that the track observer will predict that the train
observer will see the flashes to be simultaneous at gamma[Delta(t)]
according to the track clock.
Ken Seto
>
> Miguel Rios
papa...@hotmail.com
> On Jul 25, 9:45 am, papa_r...@hotmail.com wrote:
>
> > < >
> > < * >
> > >< o
> > > / <
> > A>____________M’/_____________<B
> > M C D ---> x
>
>
> > < >
> > o ><
> > > \ <
> > A>____________\M_____________<B
> > D’C’ M’ ---> x'
>
> > 1/3 sec, x’ = 0.
> > b) Track observer M on frame K sees two non-simultaneous strikes, the
> > first at t = 0.167 sec, x = 0 and the second at t = 0.667 sec at x =
> > 0.
>
> > We notice that both observer disagree about the strikes being or not
> > simultaneous and both are right in their observations (neither K nor
> > K’ frames are preferred frames).
>
> > This proves the relativity of simultaneity.
>
> > Miguel Rios
>
> Your math proof is based on a bogus model of the prpagation of the
> light fronts. This bogus model is in turn based on the false
> assumption that the same two light rays that arrive at M will continue
> on to arrive at M' which leads you to make the bogus conclusion that
> the train observer sees the front light front before he sees the rear
> light front. This is wrong. The two light rays that hit M are absorbed
>
Your bogus answer is based in your inability to grasp and the
ignorance of the Einstein thought experiment.
Your assertion that strike light signal are lines is ridiculous, so is
your conclusion that follows.
You are also unable to understand that Einstein put the thought
experiment in simple words, and that it also works when the train
observer is moving towards the direction (x1, y1, z1), because the
light signal propagates in a sphere.
You do not understand the space-time diagrams either, which clearly
show what observers and light signals do.
Finally you do not grasp the concept of relativity, which is the main
subject of the thought experiment.
Miguel Rios
G. L. Bradford
"kenseto" wrote:
[.....]
Your math proof is based on a bogus model of the prpagation of the
light fronts. This bogus model is in turn based on the false
assumption that the same two light rays that arrive at M will continue
on to arrive at M' which leads you to make the bogus conclusion that
the train observer sees the front light front before he sees the rear
light front. This is wrong. The two light rays that hit M are absorbed
by M. Two different light rays from the light spheres will hit M'.
What this mean is that your math proof is useless. End of story.
Ken Seto
====================
"The two light rays that hit M are absorbed by M." Then it is your
contention that the light's wave front from position A never reaches B, and
that the light's wave front from B never reaches A? Since both have to pass
through M, simultaneously according to the track observer, in their
continuing expansions (though not simultaneously according to either
observation from point of view A or observation from point of view B
(singularly single-sided -- one sided only / no other side -- don't you
know? Thus there is no such thing as overtaking light from the rear;
light-TIME horizons of the universe, yes (recessionary time, multi...,
coordinate.....), but light itself, not a chance (again, it being
single-sided -- one sided or front-sided, so to speak -- only!))).
Or, aside, is it your contention that the light wave front from B will
reach point A simultaneously with the lightning strike at A? And that the
light wave front from A will reach point B simultaneously with the lightning
strike at B, thus the wave front from A reaching the farthest point forward
of the moving train simultaneously with the wave front from B reaching it,
and the wave front from B reaching the farthest point rearward of the moving
train simultaneously with the wave front from A reaching it (thus no matter
where train point M' is between track points A and B the light wave fronts
coming from the lightning strikes at A and B should reach it simultaneously,
right)? :-()
GLB
===================
Sue...
> Finally you do not grasp the concept of relativity, which is the main
> subject of the thought experiment.
<<
Einstein's relativity principle states that:
All inertial frames are totally equivalent
for the performance of all physical experiments.
In other words, it is impossible to perform a
physical experiment which differentiates in
any fundamental sense between different
inertial frames. By definition, Newton's
laws of motion take the same form in all
inertial frames. Einstein generalized this
result in his special theory of relativity by
asserting that all laws of physics take the
same form in all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
How does the ~thought experiment~ demonstrate
that principle ?
Sue...
>
> Miguel Rios
papa...@hotmail.com
You always put this quote which we all agree on, but you clearly do
not understand. Are you suggesting that Einstein train and track
thought experiment does not follow his own theory? Because that
thought experiment precisely indicates that both observers can detect
and agree on the simultaneity of two given events, but they will
disagree on that simultaneity if they are moving relative to the other
one.
Miguel Rios
Sue...
> On 26 jul, 12:48, "Sue..." <suzysewns...@yahoo.com.au> wrote:
>
>
>
> > On Jul 26, 10:53 am, papa_r...@hotmail.com wrote:
>
> > > Finally you do not grasp the concept of relativity, which is the main
> > > subject of the thought experiment.
>
> > <<
> > Einstein's relativity principle states that:
>
> > All inertial frames are totally equivalent
> > for the performance of all physical experiments.
>
> > In other words, it is impossible to perform a
> > physical experiment which differentiates in
> > any fundamental sense between different
> > inertial frames. By definition, Newton's
> > laws of motion take the same form in all
> > inertial frames. Einstein generalized this
> > result in his special theory of relativity by
> > asserting that all laws of physics take the
> > same form in all inertial frames. >>http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
>
> > How does the ~thought experiment~ demonstrate
> > that principle ?
>
> > Sue...
>
> > > Miguel Rios
>
> You always put this quote which we all agree on,
Yes... I repeat that question frequenty because
you never answer it.
Again:
<<
Einstein's relativity principle states that:
All inertial frames are totally equivalent
for the performance of all physical experiments.
In other words, it is impossible to perform a
physical experiment which differentiates in
any fundamental sense between different
inertial frames. By definition, Newton's
laws of motion take the same form in all
inertial frames. Einstein generalized this
result in his special theory of relativity by
asserting that all laws of physics take the
same form in all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
How does the ~thought experiment~ demonstrate
that principle ?
> but you clearly do
> not understand. Are you suggesting that Einstein train and track
> thought experiment does not follow his own theory? Because that
> thought experiment precisely indicates that both observers can detect
> and agree on the simultaneity of two given events, but they will
> disagree on that simultaneity if they are moving relative to the other
> one.
If they disagree when relative motion is introduced
then Einstein's theory must be wrong.
So the ~thought experiment~ demonstrates the theory
is wrong and you are presenting it because you are
a model railroader or some-such?
Sue...
>
> Miguel Rios
Androcles
<papa...@hotmail.com> wrote in message
news:16388d69-ed57-41d7...@m45g2000hsb.googlegroups.com...
| On 26 jul, 12:48, "Sue..." <suzysewns...@yahoo.com.au> wrote:
| > On Jul 26, 10:53 am, papa_r...@hotmail.com wrote:
| >
| > > Finally you do not grasp the concept of relativity, which is the main
| > > subject of the thought experiment.
| >
| > <<
| > Einstein's relativity principle states that:
| >
| > All inertial frames are totally equivalent
| > for the performance of all physical experiments.
| >
| > In other words, it is impossible to perform a
| > physical experiment which differentiates in
| > any fundamental sense between different
| > inertial frames. By definition, Newton's
| > laws of motion take the same form in all
| > inertial frames. Einstein generalized this
| > result in his special theory of relativity by
| > asserting that all laws of physics take the
| > same form in all inertial frames.
>>http://farside.ph.utexas.edu/teaching/em/lectures/node108.html
| >
| > How does the ~thought experiment~ demonstrate
| > that principle ?
| >
| > Sue...
| >
| >
| >
| > > Miguel Rios
|
| You always put this quote which we all agree on, but you clearly do
| not understand. Are you suggesting that Einstein train and track
| thought experiment does not follow his own theory?
Yes. He's the only idiot that can make lightning strike both ends
of a train simultaneously and say the strikes are not simultaneous.
Someone ought to make up their stupid mind.
papa...@hotmail.com
Well you are showing clearly now your agenda...the one of an
antirelativist. That is fine...you have the right to do what you want.
But it happens that you are wrong, regarding what Einstein and
Fitzpatrick say. And the reason is that you never read or understand
the rest of the quote (it reminds me of Seto...no matter how many
times he is corrected...there he goes again with the same mistakes).
So the answer to your question is in that very same page in the
following two sections:
"Consider a wave-like disturbance. In general, such a disturbance
propagates at a fixed velocity with respect to the medium in which the
disturbance takes place. For instance, sound waves (at S.T.P.)
propagate at 343 meters per second with respect to air. So, in the
inertial frame in which air is stationary, sound waves appear to
propagate at 343 meters per second. Sound waves appear to propagate at
a different velocity any inertial frame which is moving with respect
to the air. However, this does not violate the relativity principle,
since if the air were stationary in the second frame then sound waves
would appear to propagate at 343 meters per second in this frame as
well. In other words, exactly the same experiment (e.g., the
determination of the speed of sound relative to stationary air)
performed in two different inertial frames of reference yields exactly
the same result, in accordance with the relativity principle.
Consider, now, a wave-like disturbance which is self-regenerating, and
does not require a medium through which to propagate. The most well-
known example of such a disturbance is a light wave. Another example
is a gravity wave. According to electromagnetic theory, the speed of
propagation of a light wave through a vacuum is
c = 1 / sqrt{epsilon_0 * mu_0}=2.99729 * 10^8 meters per second
(1323)
where epsilon_0 and mu_0 are physical constants which can be evaluated
by performing two simple experiments which involve measuring the force
of attraction between two fixed changes and two fixed parallel current
carrying wires. According to the relativity principle, these
experiments must yield the same values for epsilon_0 and mu_0 in all
inertial frames. Thus, the speed of light must be the same in all
inertial frames. In fact, any disturbance which does not require a
medium to propagate through must appear to travel at the same velocity
in all inertial frames, otherwise we could differentiate inertial
frames using the apparent propagation speed of the disturbance, which
would violate the relativity principle."
Hence the key factor is that c is the same in every inertial frame,
and from that factor alone the relativity of simultaneity must appear.
If you do not understand this, then you are free to make your complain
to Drs. Einstein and Fitzpatrick.
Miguel Rios
Sue...
You are the one that continually offers a thought
experiment that you can not explain because the
model of light propagation is faulty.
It would seem that YOU are against Einstein's theory.
papa...@hotmail.com
You are at the same level as Seto....it appears you went together to
the same school as it seems.
Einstein thought experiment (not mine as your agenda puts) does not
"demonstrate" the principle of relativity. On the contrary, Einstein
thought experiment IS a direct consequence of the principle of
relativity. You do not understand that principle of relativity, so it
is clear you will not agree with Einstein thought experiment (and, of
course, neither you will with the twin "paradox" or the pole and barn
"paradox" as all good antirelativists do).
So just talk about your agenda with Seto and Spaceman...you all talk
the same language and understand each other...and remember that
whatever the stuff you are talking, the rest of us are not very
interested.
Miguel Rios