sci.physics.relativity

"David" wrote in message
news:ajn7e3dsff7rlah716s5rh6md692tko8i6@ ...
> 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.

In the brick frame .. ok

> 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.

Yes

> 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.

If it were the same (or VERY close to) velocity, then you have multiple
bricks overlapping (at the same place and time) .. so they won't fit