HELP! - Constant acceleration in SR and event horizon
Dean D
I'm looking at my lecture notes on four-vectors and there are some things I
have trouble understanding.
I remember the lecturer used invariance of s^2 to prove that an observer in
constant acceleration (hyperbolic motion) always have constant distance to
some point behind him (in his instantaneous inertial frame). The lecturer
then went on and proved that there is an event horizon at some distance
behind and that a photon sent to chase the observer can never catch up with
him. I can see that this is so by inspecting the space-time diagram but
unfortunately I don't have the proof written down ad I need the mathematical
proof (with focus on physics, not the rigorous mathematics).
Could someone help me out with this?
Thanks // Dean
Androcles
"Dean D" <Dean@fakeaddress_replay2NG.com> wrote in message news:A9jYh.39479$E02....@newsb.telia.net...
Mathematical proofs begin with axioms. Now, focus on the physics,
this as simple as it can be.
How far is it from A to A and how long will it take to get there?
If you don't get that right, all that is built on it will be ridiculous.
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
Your lecturer needs a psychiatrist. You've been helped, the rest is up
to you.
http://www.androcles01.pwp.blueyonder.co.uk/Vector/Vector.htm
http://www.androcles01.pwp.blueyonder.co.uk/Smart/Smart.htm
Dirk Van de moortel
"Dean D" <Dean@fakeaddress_replay2NG.com> wrote in message news:A9jYh.39479$E02....@newsb.telia.net...
See for instance at the end of
http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html
Using x and t as coordinate distance and ditto time, the covered distance
as seen in the (permanent) inertial frame is
x(t) = c^2/a ( sqrt( 1 + (a t/c)^2 ) -1 ) ,
provided
x(0) = 0
where a is the constant proper acceleration felt by the traveler.
You can easily verify that
limit { t -> inf ; x(t) / t } = c
and that
limit { t -> inf ; x(t) - c t } = -c^2/a
so the worldline x(t) has an oblique asymptote with equation
x = c t - c^2/a
which is the 'worldline' of a light signal sent ot at event
( t, x ) = ( 0, -c^2/a )
Since it is on the assymptote, the signal will not reach the traveller.
Dirk Vdm
karand...@yahoo.com
SperM.hotmail.com> wrote:
> "Dean D" <Dean@fakeaddress_replay2NG.com> wrote in messagenews:A9jYh.39479$E02....@newsb.telia.net...
>
> - Show quoted text -
Nice touch. I always like the page on accelerated motion in SR.
Dirk Van de moortel
<karand...@yahoo.com> wrote in message news:1177693371.0...@t39g2000prd.googlegroups.com...
[snip]
>
> Nice touch. I always like the page on accelerated motion in SR.
Thanks :-)
Dirk Vdm
karand...@yahoo.com
SperM.hotmail.com> wrote:
> <karandash2...@yahoo.com> wrote in messagenews:1177693371.0...@t39g2000prd.googlegroups.com...
> > On Apr 27, 9:33 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
> > SperM.hotmail.com> wrote:
>
> [snip]
>
>
>
> > Nice touch. I always like the page on accelerated motion in SR.
>
> Thanks :-)
>
> Dirk Vdm
I've got a present for you, I got Marcel to spout one of his biggest
"Luttgisms" ever:
Have a look:
Dean D
> as seen in the (permanent) inertial frame is
> x(t) = c^2/a ( sqrt( 1 + (a t/c)^2 ) -1 ) ,
> provided x(0) = 0
> where a is the constant proper acceleration felt by the traveler.
>
> You can easily verify that
> limit { t -> inf ; x(t) / t } = c
> and that
> limit { t -> inf ; x(t) - c t } = -c^2/a
> so the worldline x(t) has an oblique asymptote with equation
> x = c t - c^2/a
> which is the 'worldline' of a light signal sent ot at event
> ( t, x ) = ( 0, -c^2/a )
> Since it is on the assymptote, the signal will not reach the traveller.
But this is exactly what I can read of space-time diagram.
The proof I had in mind used invariance of s^2 (4-vector treatment)...
Also, what about prooving that an accelerated observer
always have constant distance to some point behind him in
his instantaneous inertial frame?
thanks in advance // Dean
Dirk Van de moortel
<karand...@yahoo.com> wrote in message news:1177699324....@s33g2000prh.googlegroups.com...
Uh :-)
Dirk Vdm
Androcles
"Dean D" <Dean@fakeaddress_replay2NG.com> wrote in message news:Q7tYh.39537$E02....@newsb.telia.net...
>> Using x and t as coordinate distance and ditto time, the covered distance
>> as seen in the (permanent) inertial frame
No such animal exists.
is
>> x(t) = c^2/a ( sqrt( 1 + (a t/c)^2 ) -1 ) ,
>> provided x(0) = 0
>> where a is the constant proper acceleration felt by the traveler.
>>
>> You can easily verify that
>> limit { t -> inf ; x(t) / t } = c
>> and that
>> limit { t -> inf ; x(t) - c t } = -c^2/a
>> so the worldline x(t) has an oblique asymptote with equation
>> x = c t - c^2/a
>> which is the 'worldline' of a light signal sent ot at event
>> ( t, x ) = ( 0, -c^2/a )
>> Since it is on the assymptote, the signal will not reach the traveller.
>
>
> But this is exactly what I can read of space-time diagram.
> The proof I had in mind used invariance of s^2 (4-vector treatment)...
>
> Also, what about prooving
The word is "proving". You seem to be a fuckwit. Go away.
Eric Gisse
It follows from the fact the observer is undergoing hyperbolic motion.
http://mathworld.wolfram.com/Hyperbola.html
Dirk Van de moortel
"Dean D" <Dean@fakeaddress_replay2NG.com> wrote in message news:Q7tYh.39537$E02....@newsb.telia.net...
>> Using x and t as coordinate distance and ditto time, the covered distance
>> as seen in the (permanent) inertial frame is
>> x(t) = c^2/a ( sqrt( 1 + (a t/c)^2 ) -1 ) ,
>> provided x(0) = 0
>> where a is the constant proper acceleration felt by the traveler.
>>
>> You can easily verify that
>> limit { t -> inf ; x(t) / t } = c
>> and that
>> limit { t -> inf ; x(t) - c t } = -c^2/a
>> so the worldline x(t) has an oblique asymptote with equation
>> x = c t - c^2/a
>> which is the 'worldline' of a light signal sent ot at event
>> ( t, x ) = ( 0, -c^2/a )
>> Since it is on the assymptote, the signal will not reach the traveller.
>
>
> But this is exactly what I can read of space-time diagram.
> The proof I had in mind used invariance of s^2 (4-vector treatment)...
Maybe this is what you're looking for?
Putting c = 1 and using accent notation for the derivative w.r.t.
proper time d/dT:
Velocity 4-vector V = ( t', x', 0, 0 ).
Acceleration 4-vector V' = ( t'', x'', 0, 0 )
with unknown functions t(T) and x(T).
You have V.V = 1, giving
t'^2 - x'^2 = 1
Since V'.V' is invariant, it must have the value it has in the comoving
inertial frame, giving
t''^2 - x''^2 = 0 - a^2
These 2 differential equations are pretty common and easiliy solved,
giving what you also find on
http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html :
t(T) = 1/a sinh( a T )
x(T) = 1/a cosh( a T ) - 1/a [ to make x(0) = 0 ]
from which you can eliminate the proper time T and find
the hyperbola
x(t) = 1/a ( sqrt( 1 + (a t)^2 ) -1 )
>
> Also, what about prooving that an accelerated observer
> always have constant distance to some point behind him in
> his instantaneous inertial frame?
not sure what you have in mind here...
Dirk Vdm
Eric Gisse
SperM.hotmail.com> wrote:
[...]
>
> > Also, what about prooving that an accelerated observer
> > always have constant distance to some point behind him in
> > his instantaneous inertial frame?
>
> not sure what you have in mind here...
Like how points on an ellipse or circle are always a fixed distance
away from something....
http://mathworld.wolfram.com/Hyperbola.html
>
> Dirk Vdm
Dirk Van de moortel
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:54QYh.146547$9G3.1...@phobos.telenet-ops.be...
Ah... of course. It was almost biting :-)
Okay, (back to c), take that event ( t, x ) = ( 0, -c^2/a ) again.
The coordinate distance to the point at x = -c^2/a is of course
xa(t) = x(t) + 1/a
= c^2/a ( sqrt( 1 + (a t/c)^2 ) -1 ) + c^2/a
= c^2/a sqrt( 1 + (a t/c)^2 )
So to get this distance as seen in the comoving inertial frame,
you just divide by gamma(t) = sqrt( 1 + (a t/c)^2 ), and you
trivially get
xa(t) / gamma(t) = c^2/a
which tells you that this distance is constant.
Dirk Vdm
Dean D
Thanks. That actually clarified a lot. (Sorry about the late replay)
Dirk Van de moortel
"Dean D" <Dean@fakeaddress_replay2NG.com> wrote in message news:Awp_h.40018$E02....@newsb.telia.net...
>
> Thanks. That actually clarified a lot. (Sorry about the late replay)
No problem.
Actually, thank *you* for having asked.
It took me a few hours to figure it out, and I have put the
little thing at the end of my page
http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html
It is *much* simpler and straightforward that the original
derivation :-)
Cheers,
Dirk Vdm