sci.physics.relativity

Let there be a line of N planets, all at rest, spanning a length of
many light-years along the x-axis. Let N be a very large number. Call
this the rest frame.

Let there be a moving frame with velocity V along the x-axis. Let V
almost be equal to c, such that the span of N planets, as measured in
the moving frame, measures less than the length of a brick that has
zero velocity as measured in the moving frame. Now let that brick
move very slowly in the y-direction along the y-axis (as measured in
the moving frame) so that it crosses the x-axis at the same time this
line of planets is also spanning the same x-coordinates. In this
moving frame, this brick will simultaneously hit each of the N
planets.

From the point of view of rest frame of the planets, this brick will
sequentially hit planet after planet as it travels nearly at the speed
of light. Now let each point on the planet that this brick hits have
a second brick on the surface at that same impact point that has
almost the same velocity as the first brick. In other words, these
two bricks have almost zero relative velocity. We have N planets, so
we have N bricks that get hit. From the point of view on the surface
of each planet, there is a collision on each of the planets of two
almost identical objects (the high speed bricks).

From the point of view of the moving frame, these N collisions happen
simultaneously (the given information). The moving frame brick hits
each of these N bricks on the line of planets. The N bricks all have
almost zero relative velocity with respect to the brick that hits
them, hence they are not length contracted as measured in the moving
frame. From the point of view of the moving frame, how can N bricks
(where N is a very large number) simultaneous span the same length as
a single brick that has virtually zero relative velocity with respect
to each of the bricks it hits?
David Seppala