A rocket is moving to the right at 1/2 the speed of light relative to the Earth. A light bulb in the center of a room inside the rocket suddenly turns on Call the light hitting the front (right) side of the room event A and the light hitting the back (left) side of the room event B. Which event occurs first, A or B, or are they simultaneous as viewed by... a) the astronaut in the rocket ship and b) a person at rest on Earth?

a) The two events occur simultaneously as viewed by an astronaut on the rocket.

b) The two events so occur simultaneously as viewed by a stationary person on earth.

Explanation:

a) For the astronaut on the rocket.

The astronaut is stationary with respect to the rocketship and he sees the light rays move with the speed of light relative to the rocket. Since the light is at the centre of the room and the two days move with the same speed, they both hit the front and back side of the room at the same times.

b) A person at rest on earth.

For relative velocities for velocities close to the speed of light, the mode of treatment is different from the normal route.

u = {(u' + v)/[1 + (u'v/c²)]}

where u = velocity of the light rays relative to earth = ?

u' = velocity of the light rays relative to the rocket = c for the front of the room, -c for the back of the room

v = velocity of the rocket relative to Earth = 0.5c

For the light hitting the front of the room

u = {(c + 0.5c)/[1 + (0.5c²/c²)]}

u = (1.5c/1.5) = c

For the back of the room

u = {(c - 0.5c)/[1 + (-0.5c²/c²)]}

u = (0.5c/0.5) = c

The two answers are equal, indicating that the light rays hit the front and back of the room at the same time and the two events occur simultaneously too.

If this relativity calculation is done for the case of the astronaut too, the answer obtained would be c and -c respectively, indicating that the light rays hit the front and back of the room simultaneously.

Hope this Helps!!!