A friend in a spaceship travels past you at a high speed. He tells you that his ship is 20m long and that the identical ship that you are sitting in is 19m long. According to your observations, (a) how long is your ship, (b) how long is his ship, and (c) what is the speed of your friend's ship (relative to yours)?
(a) Since your ship is identical to your friend's, it must have the same proper length. The proper length of his ship is 20m, so that must be the proper length of your ship, and this is what you would measure while in the rest frame of your ship.
Your friend, moving with speed v relative to you, measures your 20m ship to be 19m long. You are moving with speed v relative to your friend, so you will measure his 20m long ship to be 19m long as well.
To determine the relative speed of the ships, use the length contraction formula
L = Lp Sqrt{1 - v²/c²},
so (1 - v²/c²) = (L/Lp)².
Move things around a bit more to get that
v²/c² = 1 - L²/Lp²
or v/c = Sqrt{1 - L²/Lp² = Sqrt{1 - (19/20)²} = 0.22.
Therefore, we find that v = 0.22c.
Note in the last example that it is perfectly reasonable to leave the answer for a speed as a fraction times the speed of light.
Recall that in "classical" mechanics, momentum was useful because it is conserved in many situations. We want to modify the definition of momentum for special relativity so that it is also conserved in situations, such as in collisions. We also want this new definition to reduce to the old definition of momentum when the speed approaches zero. The relativistic quantity that satisfies these conditions is
An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is 2.50×10-28kg (a pion), and that of the heavier fragment is 1.67×10-27kg (a proton). If the lighter fragment has a speed of 0.893c after the breakup, what is the speed of the heavier fragment?
Relativistic momentum is conserved, just as in the non-relativistic case.
The initial momentum of the system is zero since the unstable particle is at rest.
For the final momentum to be zero, then the momenta of the two particles must have equal magnitude and be in opposite directions (sketch).
gl = 1 / Sqrt{1 - 0.893²} = 2.22
The momentum of the light particle is
pl = gl ml vl
and this must equal the momentum of the heavy particle,
ph = gh mh vh.
Therefore,
gh vh = gl vl (ml / mh) = (2.22)(0.893c)(2.50×10-28kg / 1.67×10-27kg) = 0.297c
Now, it takes a bit of algebra to solve this for v, since v appears in a square root in the g term.
The solution works out to be
vh = c / Sqrt{(1/0.297)² + 1} = 0.285c.
Earlier we discussed that if two objects (spaceship 1 and spaceship 2) are moving with respect to each other, then the velocity of 2 measured with respect to 1, v21, must equal the velocity of 1 measured with respect to 2, v12.
As you might now expect, the result is not a simple addition of velocities. Instead, we "add" velocities according to the following equation:
A space vehicle is moving at a speed of 0.75c with respect to Earth. An atomic particle is projected at 0.90c in the same direction as the spaceship's velocity with respect to an observer inside the vehicle. What is the speed of the projectile as seen by an observer on Earth?
Let's identify the Earth as object 1, the vehicle as object 2, and the projectile as object 3.
Therefore, v12 = 0.75c, and v23 = 0.90c.
Note that v23 is positive since the particle moves in the same direction as the vehicle.
Adding the velocities as learned previously, we would get the result that the speed of the projectile as seen from Earth is v13 = v12 + v23 = 0.75c + 0.90c = 1.65c!
This is greater than the speed of light which is not possible!
Using the relativistic equation we get:
v13 = (0.75c + 0.90c) / (1 + (0.75c)(0.90c)/c²) = 1.65c / 1.675 = 0.985c
This value is close to the speed of light, but not greater!