# Simple Relativity Question

1. Oct 18, 2005

### stallion

This question is concerned with special relativity and relative velocity.
If a spaceship is traveling at 2x10 eighth m/s and someone shines a laser straight ahead out the front of the ship. Suppose someone in front of the ship
records the velocity of the light. Since the speed of light is the same in all frames of reference the speed should be 3x18 to the eighth and not 5x10 to the eighth(adding the two velocities).

Here is my question...since the ship is moving with a force which allows it
to travel at 2x10 to the eighth, doesn't this same force push on the photons in
the light and force them to travel faster?

2. Oct 18, 2005

### JesseM

Leaving aside relativity, I think you're misunderstanding something about Newtonian physics here--if a ship is moving at a constant velocity through empty space with no resistance, it doesn't need a force pushing on it to maintain that speed, in Newtonian physics all objects naturally move in straight lines at constant velocity unless a force is acting on them (forces produce acceleration, which cause objects to change speed, direction, or both).

Also, there's no such thing as absolute speed. Even if the ship is moving at 2*10^8 m/s in your frame, in the ship's own frame it is at rest and you are moving at 2*10^8 m/s in the opposite direction, and both frames are equally valid.

Aside from that, it's just a property of the coordinate systems used in relativity that insure that something moving at c in one frame must be moving at c in other frames (the reason for this choice of coordinate systems is that it insures the laws of physics obey the same equations in all coordinate systems). If frame A uses coordinates (x,y,z,t) and B uses (x',y',z',t'), and the origin of B's coordinate system is moving at velocity v along the x-axis of A's coordinate system, then the transformation is:

$$x' = \gamma (x - vt)$$
$$y' = y$$
$$z' = z$$
$$t' = \gamma (t - vx/c^2)$$
where $$\gamma = 1/\sqrt{1 - v^2/c^2}$$

So say you have a light beam moving at speed c along the x-axis of A's coordinate system. At t=0 minutes, the light beam is at position x=0 light-minutes, and at t=1 minutes, the light beam is at position x=1 light-minute. So if you plug x=0, t=0 into the above transformation, you get x'=0, t'=0; if you plug in x=1, t=1 into it, you get $$x' = \gamma (1 - v)$$ and $$t' = \gamma (1 - v)$$. So, using speed = (change in position)/(change in time), you see that in both frames the speed is 1c.