The challenge: During the space wars, you were standing on the Moon, watching two fast spaceships passing horizontally overhead, line astern, one chasing the other at identical constant speeds relative to you. When the midpoint between them was directly overhead, you observed them to simultaneously fire a single shot at one another, but the two bullets spectacularly collided and exploded. If the muzzle velocities were identical, did that collision happen exactly halfway between the two spaceships?

The answer is no. The bullets collided closer to the rear ship due to the fact that the front ship fired earlier than the rear ship. This is all due to the relativity of simultaneity. When observers are in relative motion, they don't agree on what's simultaneous and what's not, because simultaneity is relative. The 'man on the moon' observed those shots as simultaneous, but the crew of both ships agreed that the pursuer has been 'beaten to the draw' by the pursued.

The easiest way to visualize the situation is by means of a Loedel spacetime diagram. The blue coordinates X,cT represent the 'man on the moon', while the red coordinates X,cT represent the inertial frame of the two passing spaceships, flying line astern, d meters apart as measured by their on-board radars.

The blue dots marked x_r (rear) and x_f (front) represent the spatial positions of the simultaneous firing of the two guns in the blue (moon) frame. In the red (spaceship) frame the guns were not fired simultaneously, because its lines of simultaneity are parallel to the red x-axis. In that frame, the front ship fired dv/c² seconds before the rear ship fired, where v is the speed of the two ships relative to the man on the moon and c the speed of light.^[1]

If the front ship fired first, the two bullets must have collided aft of their moving midpoint. Now let's say the muzzle speeds of both guns were u m/s relative to the firing ship. Neglecting the Moon's weak gravity for the moment, we need to solve a simple Newtonian equation in the spaceship frame to obtain the time to collision (t) for the rear ship's bullet.

ut + u(t + dv/c²) = d, where

t + dv/c² is the time to collision for the front ship's bullet. This gives:

t = (d/2u) (1-uv/c²), from where it is easy to determine the position of the collision relative to whoever. The collision distance x_c from the rear ship, in the ships frame of reference:

x_c = ut = (d/2) (1-uv/c²), which is obviously less than halfway.

Let's plug in some 'realistic' values for the era of "Space War I". Say the ships were streaking past the Moon at 10^-4 c (30 km/s) and the bullet muzzle velocity was 10 times that. The value uv/c² works out to 10^-4 x 10^-3 = 10^-7, or 1 part per 10 million. If the ships were separated by 2 km, the deviation from midpoint would have been 0.1 mm, probably a tad too small to notice. By the time of "Space War II", technology should have pushed both speeds up by a factor 10, so the deviation from the midpoint might become 10^-3 x 10^-2 = 10³ x 10^-5 m, or 1 cm, easily detectable by the technology of that time.

In conclusion, a few words on the effect of the Moon's gravity on the collision point. In the ship-frame, the Moon is moving 'backwards', in the same direction as the leading ship's bullet and against the chasing ship's bullet. This will cause a very slight asymmetry in the gravitational accelerations of the two bullets, but for this scenario (bullet travel time is some 3.33 ms) it is negligible when compared to the non-simultaneity effect. For example, the 'vertical drop' of the horizontally fired bullets will be in the order of 0.01 mm and the asymmetry effect is another order of magnitude less.

Utterly useless and sometimes controversial information, but then, after the moon dust has settled, you will probably never view simultaneity in the same light again!

Jorrie

[1] This free download from Relativity 4 Engineers deals with synchronization of clocks in depth. (eBook page 33+)