Re: OT: Relativity ,super-luminal communications and time travel (Was OT: Nuclear Waste (Was Re-Marketing Ada))

["Robert I. Eachus" <rieachus@comcast.net>]

Dmytry Lavrov wrote:

> Both could be original,IMO,if copyed very well.

That is where quantum mechanics comes in.  Heisenberg's Uncertainty 
Principle limits exact copying.  Quantum teleportation gets around that 
ALL of the QM state is copied exactly.  But the QM state of the orignial 
is destroyed and you can't use the process to make two different copies.

If a you think a copy of YOU without the QM data is still you, go for 
it.  I believe otherwise, but it will be a while before the experiment 
is tried.

> We send signal to "z-transmitter"(transmitter of super-luminal
> data,"emmits z-rays" :),it's sends to z-receiver A,then Mr. X routes
> it to z-transmitter B(Mr X can use normal light to route data,in facts
> X is not needed),then it comes to z-receiver B,and we can get it
> before we sends,and if time is too short,we routes it again to A to
> get necessary "negative time"...

The reason that the Nimtz process can't be used this way is that the 
superluminal transition takes the same (positive) time in all frames. 
You can calculate an equivalent velocity by taking the distance traveled 
with respect to some observer and divide by the (constant) transition 
time.  But since the time is constant and always positive there is no 
way to send data back in time.

Look at it this way.  We have three observers A and B a distance apart 
and C at A's location but travelling at 1/2 c.

    A,C ---------------------> B
        <---------------------

A sends a message to B, which takes k picoseconds.  B then sends a copy 
back to C again taking k picoseconds.  A and C see B recieve the message 
at different times in the future.  But both see the message from B 
arriving 2k picoseconds after A's message was sent.  Paradox averted. 
Notice that A and B compute the distances and (round trip) velocity the 
same, but C sees the distance (and speed) differently.

It is one of those elegant things that once your mind says "Aha!" you 
see that it is the only way to have superluminal communication without 
causality violations--the transmission time, not the speed must be the 
same for all observers.

> Also about name.Even without digging deep,"Closed timelike curve".If
> it's closed,it's mean that for particle there's no future.Particle
> stay on that loop.That's NOT the time mashine in any meaning,time
> mashine 1:don't have scientific meaning too;2:it have only sci-fi
> meaning that have nothing with looped particle or looped you.

Yes, and no.  This stuff gets pretty deep quickly, but some parts are 
simple.  Around any singuality (including black holes) in a steady state 
there must be a surface where any photon travelling perpendicular to the 
surface is on a closed timelike curve.  In other words it is in orbit 
around the black hole.  In the rest frame distant from the black hole, 
these photons have zero energy, but they have considerable energy 
relative to anything falling into the black hole.

Now rotate the black hole.  The radius at which a photon travels in a 
closed loop is different depending on whether it is going in the 
direction of the rotation or against it.  Orbits at an angle to the 
plane of rotation precess, an the number of orbits before a photon comes 
back to its starting point can be large.  (In fact it will pass "near" 
that point at a different angle at least once before it returns to 
approximately the same point with the same velocity vector.)  So for a 
rotating black hole there are many of these shells and they overlap.

Now let's look at two (non-rotating or with small angular momentum) 
black holes coming together.  There is a volume on the connecting axis 
where at one instant you are outside both sets of closed timelike 
curves, then you are inside another larger set around both black holes. 
  But how can you get out?  At a different matter infall, or the same 
one, you can exit if and only if you are moving backwards in time.  See 
how easy that is?  (Riiiggght!)

Constant mass infall doesn't create this situation, so just because a 
black hole has an accretion disk doesn't mean you can exit just anywhere 
in time.  For any black hole there are a finite number of potential 
entry and exit points, and the point of formation may or may not be one 
of them.  On the other hand even if a black hole that was formed in a 
supernova explosion did have the right characteristics at formation, you 
could come out in the middle of a supernova explosion.  (Ouch!)  Same 
for black holes formed during the first few milliseconds after the big 
bang, the exit point is going to be a little hot!

-- 
                                           Robert I. Eachus

100% Ada, no bugs--the only way to create software.