5.6 Relativistic velocity transformation  (Page 2/4)

Velocity transformation equations for light

Suppose a spaceship heading directly toward Earth at half the speed of light sends a signal to us on a laser-produced beam of light ( [link] ). Given that the light leaves the ship at speed c as observed from the ship, calculate the speed at which it approaches Earth.

Strategy

Because the light and the spaceship are moving at relativistic speeds, we cannot use simple velocity addition. Instead, we determine the speed at which the light approaches Earth using relativistic velocity addition.

Solution

1. Identify the knowns: $v=0.500c;u\prime =c.$
2. Identify the unknown: u .
3. Express the answer as an equation: $u=\frac{v+u\prime }{1+\frac{vu\prime }{c^2}}.$
4. Do the calculation:
   
   $\begin{array}{cc}\hfill u& =\frac{v+u\prime }{1+\frac{vu\prime }{c^2}}\hfill \\ & =\frac{0.500c+c}{1+\frac{(0.500c)(c)}{c^2}}\hfill \\ & =\frac{(0.500+1)c}{\left(\frac{c^2+0.500c^2}{c^2}\right)}\hfill \\ & =c.\hfill \end{array}$

Significance

Relativistic velocity addition gives the correct result. Light leaves the ship at speed c and approaches Earth at speed c . The speed of light is independent of the relative motion of source and observer, whether the observer is on the ship or earthbound.

Got questions? Get instant answers now!

Relativistic package delivery

Suppose the spaceship in the previous example approaches Earth at half the speed of light and shoots a canister at a speed of 0.750 c ( [link] ). (a) At what velocity does an earthbound observer see the canister if it is shot directly toward Earth? (b) If it is shot directly away from Earth?

Strategy

Because the canister and the spaceship are moving at relativistic speeds, we must determine the speed of the canister by an earthbound observer using relativistic velocity addition instead of simple velocity addition.

Solution for (a)

1. Identify the knowns: $v=0.500c;u\prime =0.750c.$
2. Identify the unknown: u .
3. Express the answer as an equation: $u=\frac{v+u\prime }{1+\frac{vu\prime }{c^2}}.$
4. Do the calculation:
   
   $\begin{array}{cc}\hfill u& =\frac{v+u\prime }{1+\frac{vu\prime }{c^2}}\hfill \\ & =\frac{0.500c+0.750c}{1+\frac{(0.500c)(0.750c)}{c^2}}\hfill \\ & =0.909c.\hfill \end{array}$

Solution for (b)

1. Identify the knowns: $v=0.500c;u\prime =\mathrm{-0.750}c.$
2. Identify the unknown: u .
3. Express the answer as an equation: $u=\frac{v+u\prime }{1+\frac{vu\prime }{c^2}}.$
4. Do the calculation:
   
   $\begin{array}{cc}\hfill u& =\frac{v+u\prime }{1+\frac{vu\prime }{c^2}}\hfill \\ & =\frac{0.500c+(\mathrm{-0.750}c)}{1+\frac{(0.500c)(\mathrm{-0.750}c)}{c^2}}\hfill \\ & =\mathrm{-0.400}c.\hfill \end{array}$

Significance

The minus sign indicates a velocity away from Earth (in the opposite direction from v ), which means the canister is heading toward Earth in part (a) and away in part (b), as expected. But relativistic velocities do not add as simply as they do classically. In part (a), the canister does approach Earth faster, but at less than the vector sum of the velocities, which would give 1.250 c . In part (b), the canister moves away from Earth at a velocity of $\mathrm{-0.400}c,$ which is faster than the −0.250 c expected classically. The differences in velocities are not even symmetric: In part (a), an observer on Earth sees the canister and the ship moving apart at a speed of 0.409 c , and at a speed of 0.900 c in part (b).

Got questions? Get instant answers now!