On May 24, 2:04 pm, altergnostic <
altergnos...@gmail.com> wrote:
> I completely agree that this does not violate causality.
> BUT
> This is not the explanation i found around.
Lets see the toher explanation...
> What i found (and which was
> answered around here as well) is that the stretching doesn't have to travel
> from the pulling ship through the wire until the wire breaks,
yes it does.
> but that the
> whole length of the wire changes "simultaneously" in spacetime.
No it doesnt. Thast description is a *simplified* and perhaps naive
description of the process. The laws of physics must apply; there are
no 'rigid' things according to SR; the wire is stretchable.
> Meaning, the increase in proper distance doesn't
> take time to travel from the front ship to the breakage point.
The 'naive' description is that eventually, when the ships have
attained their final speed and so as the wire, the wire is Lorentz
contracted. But since we posit that we ant to 'force' it to maintain
the full length, then those two effects are incompatible: eventually
the wire breaks. You dont need the intermolecular cohesion description
to conclude that the wire will break. But if you want a full analysis,
you do need to consider it.
> Again, your description seems rather classical,
It might seem that to you but it is a fully relativistic description.
Actually, in classical mechanics the wire will stretch and break
too...but thats a different topic.
> and the breakage is always
> observed after both engines started.
> If that is really the correct answer, then i understand why there is no
> causality violation.
There you go.
> But vertical seems to believe what i have concluded as
> well: that there are frames in which the breakage occurs before the second
> ship fires up.
The wire breaks always after the pulling ship, in all frames. In S
frame:
E1:Front ship starts pulling @ t1
E2: wire breaks @ t2
dt' = g(dt - dx*v/c2). But dt >= dx/c (causal, it breaks after
pulling in S), thus
dt = g(dt - c*dt*v/c2) = d*dt*(1 - v/c) >= 0 for all v.
Note_1: For E3 = back ship starts 'pushing' @ t3 = t1 (it is
simultaneous in S).
Thus dt for these two events is 0 thus dt' can be < 0. But this is not
what you are looking for. You are concerned with the breakage which
is not simul to t1 = 0 in S.
Note_2: But again, even pushing (squishing and destroying) will break
the wire if one assumes its rigid, which is no longer part of our
discussion since nothing is rigid.
> To point out another, lesser issue i see in your explanation, think about
> the frame where the rear ship fires first: the wire is compressed a bit
> before the front ship ignites.
Yes, as hinted in Note_1.
> In the frame where the front ship pulls first, the wire surely CAN break
> before the rear ship is seen to ignite,
yes.
> hence there is no compression in this frame.
None yet, and is, a priori, not needed.
> The only way i see to deal with this disagreement
I see no disagreement here. All works out fine.
> is to take this compression as a regular LC (as in inertial frames) where no physical
> process is not altered. But since this is a non-inertial frame, i remain
> puzzled.
>
> Could you clarify further?
Just did. But perhaps a note_3: "this" is a non-frame? Which one? S is
inertial so as S', the final steadystate frame. You can describe the
EM intemolecular stretching/compression via any iframe. You cant
readily describe the effects via non iframe, but thats a different
subject.