Re: Evil Twin Paradox
- From: "Todd" <td@xxxxxxx>
- Date: Wed, 8 Aug 2007 10:29:42 -0500
"David" <dseppala@xxxxxxxxxxxxx> wrote in message news:u87eb31kdqll2atvpp1bpqtja2i9n3718n@xxxxxxxxxx
This problem is a variation of the barn-pole paradox. It is a simple
application of the length contraction concept, but the results don't
seem to make sense. I would appreciate it if someone would point out
the error.
Evil Twin Paradox
An evil twin and a good twin are intially in the same reference frame
which I'll call the rest frame. The feuding twins are separated by an
extremely long steel wall that extends along the x-axis. One twin is
on the postivie y side of the steel wall and the other twin is on the
negative y side of the steel wall. Each twin has a weapon that can
shoot a 100 millimeter projectile in the y direction perpendicular to
the x-axis. The steel wall prevents the twins from shooting at each
other, but there is a section of this wall that has a long gap
extending along the x-direction. The length of this gap is L=10
light-years.
The good twin happens to be at rest at the mid-point of this 10
light-year gap. The evil twin learns of this and decides to leave the
rest frame and travel at velocity V parallel to the x-axis and wall,
and fire his weapon (which is always pointed perpendicular to the
x-axis as measured in his moving reference frame) at the good twin as
he passes the good twin who is at the mid-point of this 10 light-year
wide gap in the steel wall. The good twin learns of the evil twin's
plan well before the evil twin takes off. So the good twin prepares
his weapon so that his 100 mm projectile can hit the evil twin as he
passes the good twin's position as the evil twin moves with velocity V
just parallel to the x-axis.
Both twins know the exact initial positions, and time that the evil
twin starts moving parallel to the x-axis, and they each know the
exact relative velocity V. But the evil twin has an ultra-fast
propulsion system that makes his velocity V relative to the rest frame
almost equal to c. The good twin has zero velocity with respect to
the rest frame (and the steel wall).
When I apply Einstein's length contraction formula
L' = L * SQRT(1-(V**2/c**2))
in this situation, since V is abitrarily close to c, L' gets
arbitrarily close to zero. In other words, this 10 light-year gap in
the steel wall as measured in the moving frame of the evil twin will
be smaller than the size of a proton if V is almost equal to c. If
the moving twin attempts to fire his 100 millimeter projectile through
this proton sized gap (he fires in the y-direction as the wall and
good twin pass), the 100 millimeter projectile will hit the steel wall
and cannot pass through this proton sized gap.
The good twin is firing his 100 millimeter projectile in the y
direction through the 10 light-year wide gap so his projectile can
easily hit the evil twin as he passes.
If the steel wall were not there, either twin could shoot the other
twin as they passed each other. Or if the steel wall prior to the gap
was removed, then the evil twin could shoot the passing good twin. Or
if the steel wall following the gap was removed, the evil twin could
hit the good twin. But if both sections of the steel wall are
present, then using this length contraction formula, the evil twin's
100 millimeter projectile cannot pass through the proton sized gap.
That makes no sense to me.
Thanks ahead for your physics insights (and also for the usual barrage
of comments about what an idiot I am).
David Seppala
As for so many SR "paradoxes", the key to the resolution is the relativity of simultaneity. To see this, let's replace the bullet of width 100 mm by two point particles separated by 100 mm. Thus, let the rocket observer (evil twin) place two point particles on his x' axis separated by 100 mm. He then simultaneously fires the particles so that from his point of view they travel perpendicularly to the x' axis toward the wall with the gap.
From his point of view the wall is moving and the gap is contracted to muchless than 100 mm width. Also, from his point of view the two particles arrive at the wall simultaneously. Clearly, he must conclude that both particles will not make it through the gap.
If SR is to survive, the observer in the wall-frame must also conclude that both particles fail to make it through the gap. How can this be so if in the wall frame the gap is much wider than 100 mm? The answer, of course, is that in the wall frame the two particles were not fired simultaneously and do not arrive at the wall simultaneously. Moreover, due to this non-simultaneous firing, the values of x where the two particles cross the x-axis of the wall will differ by more than the width of the gap even though according to the wall frame the two particles were separated by much less than 100 mm before they were fired. So, the wall observer agrees that both particles cannot make it through the gap.
The same argument works for the bullet. Just consider the two point particles in the above discussion as representing two points of the bullet.
Todd
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