Re: Transverse paradox parallel question
- From: Darwin123 <drosen0000@xxxxxxxxx>
- Date: Thu, 24 Apr 2008 08:22:30 -0700 (PDT)
On Apr 23, 7:54 pm, Darwin123 <drosen0...@xxxxxxxxx> wrote:
On Apr 20, 9:31 pm, Peri of Pera <rie...@xxxxxxxxxxx> wrote:>Gravity on the spaceship has been engineered to be theCorrection. I made a mistake with the last line. I meant that
same as on earth.
Now here is the tricky part. How does that "gravity" on the space
ship work work? Look at it this way. Does this "gravity" work during
the deceleration of the space ship? To decelerate, that space ship has
to have a force exerted on it and the gravity mechanism.
Can he notice any difference in time watching the> two time pieces?
If the "gravity" on earth follows a Lorentz invariant rule, for
instance a Hamiltonian which is Lorentz invariant, then yes. Because
during the turn around, during the deceleration, the gravity in the
space ship can't be Lorentz invariant. The force on the space ship is
shifting the space ship and the clock it contains into different
inertial frames. The earth is staying in the same inertial frame. So
the gravity on the space ship is Lorentz invariant.
during the deceleration, the gravity on the space ship can't be
Lorentz invariant. The gravity on the space ship is Lorentz invariant
while it is moving at constant velocity relative to the earth.
However, in the (short?) time the train is turning around, the gravity
on the ship can't be Lorentz invariant.
Therefore, the gravity on the ship can't be the same as the
gravity on earth as seen from an inertial frame. During the turn
around, the Lorentz time dilation formula does not apply to the clock
on the train. During the turn around, even the Lorentz contraction of
distance does not apply for the rulers on the ship.
The observer on the ship makes inferences after he comes back.
He has to first collected all the data, including delays from speed of
light limitations. After he analyzes all the data and assuming he does
not accept the fact of his acceleration during turn around, he may
infer that the clock on earth ticked faster during his acceleration.
This does not violate SR. In fact he did accelerate during turn
around and so did not stay in one inertial frame. One can figure out
what he infers using SR, using the constraint that whatever the rocket
engines did to the clock was small on the order of aL/c^2 where a=F/m,
L-distance of turn around in the earth frame, and c is speed of light.
If the effects on the clock of the acceleration are bigger than this,
the ship has a broken clock.
The only issue here is whether analyzing the astronauts story can
be preformed entirely in SR, or is GR necessary. I take the view that
only SR is necessary. However, this is a fine semantic point.
Basically, the fact remains. The "twin paradox" is not a true paradox.
I vote we rename it the "twin ambiguity." Until the acceleration, the
"true" frame of reference is ambiguous. You can perform the
calculations in either the earth frame or the track frame. One also
has the ambiguity of "first order effects," a problem whenever one
uses Taylor expansion. How big is too big when it comes to being
"first order?" However, there is no physical or logical paradox in
relativity.
.
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- Transverse paradox parallel question
- From: Peri of Pera
- Re: Transverse paradox parallel question
- From: Darwin123
- Transverse paradox parallel question
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