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From: rotchm on 8 Nov 2009 13:12
On Nov 8, 12:55 pm, DSeppala <dsepp...(a)austin.rr.com> wrote:
> To clarify your confusion caused by my phrasing.
>    The two identical rockets are positioned to travel along lines
> parallel to the x-axis (as if they were racing side by side once they
> start accelerating).  

SIDE by SIDE...ok


>The tip of one rocket is at x = 0, and the tip
> of the other is at x = L'.  


Thus they are NOT side by side???

Make up your mind. Try again.

Eg. Let S be an ( 2 dim) i-frame.

Two rockets ( or their tips) initially ar rest at locations ( x=0, y
= 1) and ( x=0, y = -1).
They are thus SIDE by SIDE, understand ???

From: rotchm on 8 Nov 2009 13:26
Ok trying to understand what you meant... You are just citing a
different variant of the Boughn identical accelerated twins. Read it
and similar material. We have explained those situations to you
befeore..for the past many years!

We explained to you how to s olve "1+2"
We explained to you how to s olve "1+3"
We explained to you how to s olve "1+4"
We explained to you how to s olve "1+5"

Now you are asking to explain "1+6". No. Figure it out yourself or
learn how to read...its all explained somewhere on the net and books.


On Nov 8, 12:55 pm, DSeppala <dsepp...(a)austin.rr.com> wrote:
> To clarify your confusion caused by my phrasing.
>    The two identical rockets are positioned to travel along lines
> parallel to the x-axis (as if they were racing side by side once they
> start accelerating).  The tip of one rocket is at x = 0, and the tip
> of the other is at x = L'.  In another inertial reference frame that
> is traveling at velocity V with respect to the inertial reference
> frame the two rockets are initially in, the distance between the tips
> of the rockets is measured as L.  At time t0 in this moving reference
> frame, observers in this frame turn the the thrusters on both rockets
> simultaneously.  The accelerometers on board each rocket show a
> constant and identical acceleration. As measured in the moving frame
> (where the thrusters were simultaneously turned on) at any instant of
> time, the tips of the two rockets are always L meters apart since both
> rockets simultaneously go through identical motions (but at different
> points along the x-axis).
>    Now in the original rocket inertial frame, the thrusters weren't
> turned on simultaneously. The thrusters of one rocket was turned on
> before the thrusters of the other rocket.  From the point of view an
> observer in the first rocket, how does he describe what happens during
> the constant acceleration?  First he starts accelerating toward the
> other rocket.  Then at some point in time as measured by the first
> rocket's clocks, the other rocket starts to accelerate.  At time t1
> (as measured by the first rocket's clocks) when the other rocket
> begins its acceleration, the two rockets were approaching each other
> with some velocity. As they both continue to accelerate along the x-
> axis, why doesn't the tip of the first rocket ever reach the same x
> position in space as the tip of the second rocket and eventually pass
> the second rocket?
>    Hope that clarification removes your confusion.
> David
> On Nov 8, 9:54 am, rotchm <rot...(a)gmail.com> wrote:
>
>
>
> > Whoah... so many errors in the formulation of your question.
>
> > On Nov 8, 8:25 am, DSeppala <dsepp...(a)austin.rr.com> wrote:
>
> > > The following seems contradictory.
>
> > Yes, "seems".
>
> > > There are  two identical rockets separated a distance L as measured on
> > > the x-axis in an inertial reference frame.  
>
> > Ok, so both are ON the x axis. The usual config.
>
> > >Let the two rockets be on
> > > parallel lines to the x-axis.  
>
> > ? Now you mean that they are both NOT on the x axis but on parallel
> > lines *to* the x axis?
>
> > >An observer in a frame moving at V
> > > relative to the x-axis turns on the thrusters of both rockets
> > > simultaneously as measured in his frame. As viewed in this frame the
> > > tip of each rocket remains a distance L away
>
> > ? They are a distance L in which frame, the "initial" frame  you
> > referenced or this new frame  V ? make up your  mind.
>
> > At this point, you must completely rephrase your problem because it is
> > too badly posed.- Hide quoted text -
>
> - Show quoted text -

From: xxein on 8 Nov 2009 22:09
On Nov 8, 8:25 am, DSeppala <dsepp...(a)austin.rr.com> wrote:
> The following seems contradictory.
> There are  two identical rockets separated a distance L as measured on
> the x-axis in an inertial reference frame.  Let the two rockets be on
> parallel lines to the x-axis.  An observer in a frame moving at V
> relative to the x-axis turns on the thrusters of both rockets
> simultaneously as measured in his frame. As viewed in this frame the
> tip of each rocket remains a distance L away from the tip of the other
> rocket even as the rockets accelerate. This occurs because  both
> rockets undergo identical simultaneous accelerations, so as one rocket
> changes position the other does the identical motion at a different
> location in space as measured in this frame.
>     How do people on board the rockets view things? In the inertial
> reference frame the rockets were initially in, the turning on of the
> thrusters of each of the two rockets were not simultaneous events.
> The thrusters of one rocket were turned on before the thrusters of the
> other rocket. Let's say you are in the rocket where the thruster was
> turned on first, and you start accelerating toward the other rocket at
> some constant acceleration rate.  After T seconds as measured  in your
> accelerating rocket, the thrusters of the other rocket are turned on
> so that both of you are now accelerating in the same direction.  You
> note that at this time you had a closing velocity of V (you and the
> other rocket were approaching each other).  You know the other rocket
> is identical to yours, and you are still accelerating at the same
> constant rate.  If the closing rate continues  at least at V or
> greater, after some point in time the tip of your rocket will pass the
> tip of the other rocket.  But we've already established that as
> measured in the inertial reference frame where turning on the
> thrusters were simultaneous events, the tips of the two rockets never
> are at the same point in space at the same time, and hence can never
> pass each other.  It seems to me at some point in time the rocket that
> started accelerating first must pass the other rocket.
>    Thanks for explaining this from the point of view of someone in the
> first rocket.
> David Seppala
> Bastrop Texas

xxein: The rockets proceed at distance L from one another. What you
'see' is the front rocket appearing closer because you are in waiting
on the light to see it move as you move toward it. Just the opposite
occurs for the front rocket to 'see' you. It is already moving away
from you before it sees your light arrive when you appear stationary.

But that is only the beginning of the explanation. Suppose you both
pass through the same instantaneous velocity of .2c during this
acceleration. The light travel time from front rocket to rear rocket
diminishes. But light travel time will increase for light to get to
the front rocket from the rear one. Iow, the front rocket will appear
increasingly closer to the rear rocket while the rear rocket appears
increasingly further away as your speeds increase. C is a finite
velocity.

If they were point particles able to observe each other, they would
show limits to this observability. As they got to c in velocity, the
rear one would view the front one as being 1/2 L away (if both on the
same axis instead of parallel axes) because the time for light to get
there would be c/2. Light has doubled it's speed to get to you
because you are approaching that light at c velocity (2c for the
effect). Otoh, for the front to view the rear, the rear would
disappear because the light would never get to it. This is the basis
for a physical explanation for velocity addition.

This is not nearly the end of the explanation either. This is
reflected light between observers with co-moving clocks. Not
generated light by the movers (that gives a different measurement and
remain as L because of time dilation).

The end of the explanation goes seamlessly to the Pioneer (so-called)
anomaly. Yeah. Right through gravity and cosmology.

I could only wish that there is a non-brainwashed person who could
persue the physic instead of the taught physics.
From: BURT on 8 Nov 2009 22:16
On Nov 8, 5:25 am, DSeppala <dsepp...(a)austin.rr.com> wrote:
> The following seems contradictory.
> There are  two identical rockets separated a distance L as measured on
> the x-axis in an inertial reference frame.  Let the two rockets be on
> parallel lines to the x-axis.  An observer in a frame moving at V
> relative to the x-axis turns on the thrusters of both rockets
> simultaneously as measured in his frame. As viewed in this frame the
> tip of each rocket remains a distance L away from the tip of the other
> rocket even as the rockets accelerate. This occurs because  both
> rockets undergo identical simultaneous accelerations, so as one rocket
> changes position the other does the identical motion at a different
> location in space as measured in this frame.
>     How do people on board the rockets view things? In the inertial
> reference frame the rockets were initially in, the turning on of the
> thrusters of each of the two rockets were not simultaneous events.
> The thrusters of one rocket were turned on before the thrusters of the
> other rocket. Let's say you are in the rocket where the thruster was
> turned on first, and you start accelerating toward the other rocket at
> some constant acceleration rate.  After T seconds as measured  in your
> accelerating rocket, the thrusters of the other rocket are turned on
> so that both of you are now accelerating in the same direction.  You
> note that at this time you had a closing velocity of V (you and the
> other rocket were approaching each other).  You know the other rocket
> is identical to yours, and you are still accelerating at the same
> constant rate.  If the closing rate continues  at least at V or
> greater, after some point in time the tip of your rocket will pass the
> tip of the other rocket.  But we've already established that as
> measured in the inertial reference frame where turning on the
> thrusters were simultaneous events, the tips of the two rockets never
> are at the same point in space at the same time, and hence can never
> pass each other.  It seems to me at some point in time the rocket that
> started accelerating first must pass the other rocket.
>    Thanks for explaining this from the point of view of someone in the
> first rocket.
> David Seppala
> Bastrop Texas

You are saying that contraction would effect simultaneity by changing
distances for light.

Mitch Raemsch

From: DSeppala on 8 Nov 2009 22:33
On Nov 8, 12:12 pm, rotchm <rot...(a)gmail.com> wrote:
> On Nov 8, 12:55 pm, DSeppala <dsepp...(a)austin.rr.com> wrote:
>
> > To clarify your confusion caused by my phrasing.
> >    The two identical rockets are positioned to travel along lines
> > parallel to the x-axis (as if they were racing side by side once they
> > start accelerating).  
>
> SIDE by SIDE...ok
>
> >The tip of one rocket is at x = 0, and the tip
> > of the other is at x = L'.  
>
> Thus they are NOT side by side???
>
> Make up your mind. Try again.
>
> Eg. Let S be an ( 2 dim) i-frame.
>
> Two rockets ( or their  tips) initially ar rest at locations ( x=0, y
> = 1) and ( x=0, y = -1).
> They are thus SIDE by SIDE, understand ???

okay, I will be more pedantic for you.
Initially one rocket tip is at x = 0, y = 0. The other rocket tip is
at x = L' and y = D where D is greater than the diameter of the rocket
so that they never can crash into each other if they travel at
different velocities along those lines parallel to the x-axis.
In this frame at time t0, the rocket at x = 0, y = 0 starts
accelerating in the positive x direction at some constant acceleration
rate, which I'll call g. At some time t1 where t1 > t0 the second
rocket starts accelerating in an identical fashion in the positive x
direction. As viewed from the first rocket, why do the observers on
board the first rocket say that they can never catch up to the second
rocket even though their speed along the x-axis is always greater than
the second rocket's speed along the x-axis?
David
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