I think this is a special relativity question

[psych2012]

Homework Statement

Two spaceships are traveling on the interstellar freeway. Ship A is traveling at 0.5 c and Ship B is traveling at 0.8 c. Ship B passes Ship A. How fast does the pilot of Ship A see Ship B moving away from her? (c is speed of light)

Homework Equations

I think the velocity-addition formula is relevant.
V = (v1 + v2) / (1 + ((v1/c) x (v2/c)))

The Attempt at a Solution

V = (0.5c + 0.8c) / (1 + ((0.5c/c) x (0.8c/c))) = 0.93c
This answer just doesn't make sense to me which leads me to believe that one of the values should be negative, or I need to alter the formula.

I'm wondering if maybe it should be V = (0.8c - 0.5c) / (1 + ((0.8c/c) x (0.5c/c))) = 0.21c. I really don't know.

 

[Filip Larsen]

I would say you are correct, that one of the speeds should be negative.

As I read the problem, both ships have measured their speed relative to the freeway, whereas the velocity addition equation, as it is usually derived, is meant to find the speed w of a something with speed w₁ measured in one frame that moves relative to another frame with speed w₂, both speed being positive in the same direction. Translating this to your problem, one possibility is to map the speeds as w = v₁, w₁ = v and w₂ = v₂, with the other possibility being to swap v₁ and v₂. Inserting this into the velocity addition equation and isolate for v you will find that is transforms into the same velocity addition equation again, now only with a minus for either v₁ or v₂ depending on which of the mappings you chose.

 

[psych2012]

Any other opinions? I'm wondering now if it would just be as simple as 0.8c - 0.5c = 0.3c, since the velocity-addition formula is usually used if there is a third stationary object observing the other two objects in motion, and there is no third object in this problem.

 

[Filip Larsen]

Any other opinions?

You are of course welcome to ask others to comment, but my reply was not really an opinion even if it did started out sounding like one.

I'm wondering now if it would just be as simple as 0.8c - 0.5c = 0.3c, since the velocity-addition formula is usually used if there is a third stationary object observing the other two objects in motion, and there is no third object in this problem.

Direct addition of velocities, while valid in Newtonian mechanics, is never valid in relativistic mechanics and this is true no matter if it involves physical objects or not. The essential aspect is that when you have three different inertial frames of reference that move (on a line) relative to each other you can from the knowledge of two of the speeds calculate the third using the velocity addition equation.

 

[psych2012]

You are of course welcome to ask others to comment, but my reply was not really an opinion even if it did started out sounding like one.

I apologize for suggesting that your reply was an opinion. I used your reasoning to arrive at an answer that I understand and am satisfied with. Thanks for your help.

(0.8c - 0.5c) / (1 + (0.8c/c)(-0.5c/c)) = 0.5c