We all know about the ultimate speed limit... the speed of light.

If person A stands on Earth and shoots his pistol, he observes the bullet to fly directly away at 1000 mph. Person B is standing right next to him (not in front) and watches this event, and agrees that the bullet flies directly away at 1000 mph.

Let's change the situation and say that B is in a spaceship, and A is in a different (and very long) spaceship with lots of windows. B's ship is hovering in space (no thrusters/acceleration). A's ship is approaching from a distance and is going to pass B's ship (very close) but at incredible speed. Make careful note that A's ship is NOT thrusting or accelerating at all, it is "coasting". In fact, A's ship is moving, relative to B's ship at 10 mph less than the speed of light. WOW!

A stands in the middle of his ship and points his gun directly forward (in the direction of travel), and fires the same pistol at the exact moment that he is passing B.

The questions are: How fast does he observe the bullet leave the gun? How fast does B observe the bullet leave the gun?

How do your answers change (if at all) if A aims backwards when he fires?

Here is an Australian newbie giving it a shot:

(a)
Observer A will always observe the bullet going forwards (or backwards) at 10 mph.

(b)
From observer B’s point of view, the velocities will have to add relativistically. If they did not, then the bullet – a massive object – would be travelling at c, and this isn’t allowed. In order to add the velocities, we take the Lorentz transformation (which is a compact way of writing what B observers in terms of what A sees):

dx’ = ã(dx + v.dt), dt’ = ã(dt + v.dx/c²)

(The dashes refer to what B sees; ã is the Lorentz factor, and v is the relative velocity between A and B)

Divide the two gives
(dx + v.dt)
dx'/dt' = --------------
(dt + v.dx/c²)

u + v
So that u' = -------
u.v
1 + ---
c²

(where u' = dx'/dt', and u = dx/dt).

Now, for spaceship A travelling at v = c - 10 mph, and the bullet travelling at u = 10 mph = 16 km h^-1, we get:

16 km/h + c - 16 km/h
So that u' = -------------------------
1 + (1.48e-8)(0.99999994)

c
u' = --------------
1.000000014799

(Two points: (1) The c’s in the denominator cancel, (2) I am unable to work with anything besides metric.)

Thus, u' = 0.999999985c. That is, the bullet is travelling slightly faster, but the difference is less than 10 mph (or 16 km h^-1)

(c)
If the bullet is being shot backwards, we just change its sign in the equation above:

- 16 km/h + c - 16 km/h
u' = ------------------------- =
1 - (1.48e-8)(0.99999994)

0.99999988
u' = ----------- .c
0.999999985

This gives u' = 0.99999989c = 0.99999990 c. That is, B sees the bullet as travelling slower than A, but not by as much as 16 km h^-1.

Ta da! This is the work of a newbie, so apologies if it stinks (I haven't been able to see if the positioning of the fonts work).

Posted by Edward on 2003-12-04 08:17:24