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Relativistic Bullet - perpendicular (Posted on 2005-01-17) Difficulty: 3 of 5
Suppose you're traveling on a space ship at 9/10 the speed of light (.9c). You have a high-powered rifle that shoots bullets at the same speed. Suppose you shoot the bullet perpendicular to your direction of travel.

It appears that the bullet would travel at a 45-degree angle (northeast, if the ship is traveling north and the bullet is shot eastward), at about 1.2728c which is faster than light. Why is this wrong, and what would the actual speed and direction be?

See The Solution Submitted by Ken Haley    
Rating: 4.5000 (6 votes)

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Solution Another (simpler) solution | Comment 9 of 15 |
There's actually another, simpler, way to think about this problem.  If we consider the time dialation phenomenon, then an observer on Earth (to whom the rocket ship is moving at 0.9c) sees event on the ship happening at β≈0.4359 or 43.59% of there normal rate (See previous post as to why this number).  Thus, according to the Earth observer, the bullet is only shot at a perpendicular speed of 0.4359 * 0.9c = 0.3923c.  Thus when the ship and bullet travel 0.9 units in the direction of the ship's motion, the bullet only moves 0.3923 units away from the ship.  Therefore the bullet's angle of travel is arctan(0.3923/0.9) ≈ 23.55 degrees, and its total speed is √(0.9² + 0.3923²) ≈ 0.9818c.

Most special relativity problems can be solved in multiple ways - using either length contraction in on frame or time dialation in the other.  I always found this nice, since if one approach seemed to difficult, I could always try the other :)

  Posted by Jay Schamel on 2005-01-20 09:57:48
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