Prev: Riedt vs Einstein
Next: LA RACINE DU MAL EN PHYSIQUE, In a dozen of days CERN’s LHC will kill all of you
From: DSeppala on 8 Nov 2009 22:40
On Nov 8, 12:26 pm, rotchm <rot...(a)gmail.com> wrote:
> 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!
In the Boughn identical accelerated twins variant the twins eventually
return to the same frame. In my question, the first rocket is
continually is approaching the second rocket but can never catch up to
the second rocket, even if they accelerate for all eternity (by that I
mean as t becomes infinite). I don't understand why.
David
>
> 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 -- Hide quoted text -
>
> - Show quoted text -

From: eric gisse on 8 Nov 2009 23:57
DSeppala wrote:

> On Nov 8, 12:26 pm, rotchm <rot...(a)gmail.com> wrote:
>> 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!
> In the Boughn identical accelerated twins variant the twins eventually
> return to the same frame. In my question, the first rocket is
> continually is approaching the second rocket but can never catch up to
> the second rocket, even if they accelerate for all eternity (by that I
> mean as t becomes infinite). I don't understand why.

Of COURSE you don't understand. The real mystery is why you persist.

[...]

From: Inertial on 9 Nov 2009 03:17
"DSeppala" <dseppala(a)austin.rr.com> wrote in message
news:34874b17-57b4-4cd7-a898-fa7ea37f66bd(a)r5g2000yqb.googlegroups.com...
> On Nov 8, 12:26 pm, rotchm <rot...(a)gmail.com> wrote:
>> 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!
> In the Boughn identical accelerated twins variant the twins eventually
> return to the same frame.

Depends on the version of it .. some have them eventually stopping
accelerating, some do not

> In my question, the first rocket is
> continually is approaching the second rocket

the rockets get further apart, not closer together

> but can never catch up to
> the second rocket, even if they accelerate for all eternity (by that I
> mean as t becomes infinite). I don't understand why.

Do some more reading on borne rigid motion, and bells spaceship 'paradox'


From: BURT on 9 Nov 2009 05:36
On Nov 9, 12:17 am, "Inertial" <relativ...(a)rest.com> wrote:
> "DSeppala" <dsepp...(a)austin.rr.com> wrote in message
>
> news:34874b17-57b4-4cd7-a898-fa7ea37f66bd(a)r5g2000yqb.googlegroups.com...
>
> > On Nov 8, 12:26 pm, rotchm <rot...(a)gmail.com> wrote:
> >> 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!
> > In the Boughn identical accelerated twins variant the twins eventually
> > return to the same frame.
>
> Depends on the version of it .. some have them eventually stopping
> accelerating, some do not
>
> > In my question, the first rocket is
> > continually is approaching the second rocket
>
> the rockets get further apart, not closer together
>
> > but can never catch up to
> > the second rocket, even if they accelerate for all eternity (by that I
> > mean as t becomes infinite).  I don't understand why.
>
> Do some more reading on borne rigid motion, and bells spaceship 'paradox'

Whats the speed of light for a rocket moving directly behind it?

Mitch Raemsch
From: rotchm on 9 Nov 2009 10:21

> 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.

So R1 cannot coincide ( touch, catch up ?) with R2 ? ( R = rockets)
The diameters are irrelevant. And since the rockets travel along
parallel paths, they cannot conincide ( crash). Your velocity argument
seems irrelevant.

Start over.

> 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?


They cannot catch up because you posed the scenario that way !! R1 and
R2 travel on ( non-touching) parallel paths !! DuH!


You mean catch up as in, to have the same x coordinate? Then,

??? Observers onboard the first rocket would not say that they can
never catch up.
In fact, they can catch up to the second rocket.

Let R2 be at L', = 1 cm. Let t0 = 0 and t1 = 3600 sec ( 1 hour btw!)..
Let g = 1.
Within a fraction of a second, R1 would catch up with R2 and pass it.
Only an hour later will R2 start to move...

You description is again totally confusing and does not make sense.
Go read, ALOT. See how the authors write and explain scenarios. Once
you learnt how to be fully clear in your descriptions, then you can
start to worry about the math and physics.

First  |  Prev  |  Next  |  Last
Pages: 1 2 3 4
Prev: Riedt vs Einstein
Next: LA RACINE DU MAL EN PHYSIQUE, In a dozen of days CERN’s LHC will kill all of you