# Newbie relativity question

Hello everyone, great to meet you all.

My name is Alejandro Di Mare, a computer science bachelor from Costa Rica currently studying music in Pasadena, California. For the past few months physics have captured my attention so greatly I'm thinking about persuing a masters degree in the subject next year (I hope I'll eventually be able to help people in this forum instead of asking simple SR questions :P).

I've been reading a few posts on relativity trying to find if this has been addressed, but was unable to find it (I'm pretty sure someone must have asked this to you guys/girls, so I apologize if I force you to repeat yourselves).

I am observing 2 spaceships, A and B. Relative to me, A moves to my left at a speed of 0.75c, B moves to my right at a speed of 0.75c. I'm wondering if one of you guys/girls could explain me how a passenger in spaceship A perceives spaceship B and what results he would obtain in an attempt to meassure it's speed (in a reference frame where A is stationary.).

## Answers and Replies

yuiop
Hi Alejandro, welcome to PF :)

Have a read of this FAQ about relativistic velocity addition and see if it helps. Sorry for being lazy, but if you are still stuck, feel free to ask for further explanations/ clarifications ;)

Here is the link I left out :O --> http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

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Homework Helper
Welcome to PF!

Hi Alejandro! Welcome to PF!

In your example, we apply the rule for combining velocities: (u + v)/(1 + uv/c²).

With u = v = .75c, that gives 1.5c/(25/16), = 0.96c.

For more details, see wikipedia (which is very good on physics) …

Homework Helper
I'm also going to give you the short version of the answer hoping that you will ask additional questions if you aren't completely satisfied:
from the point of view of A (in which A is stationary), B is moving past it at high velocity. To find that velocity, one should add both speeds, using the relativistic formula,
[itex]w = \frac{u + v}{1 + u v / c^2}[/tex]
which gives back the standard formula w = u + v when u and v are much smaller than c. You can just use this formula to calculate the velocity of B, according to A (and will get something around 0.96c, rather than 1.5c).

tiny-tim beat me... by a lot even (did it take me so long to type that short answer?)[/edit]

ah, beautiful, just what I was looking for, for some reason I'm able to picture the addition of velocities with frames that are approaching each other (or the typical example of a guy walking on a train, it's not difficult to invision the length contractions that make sense of the adddition of velocities) but I'm still having trouble seeing it with reference frames moving away from each other :P. Thanks for the help

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yuiop
ah, beautiful, just what I was looking for, for some reason I'm able to picture the addition of velocities with frames that are approaching each other (or the typical example of a guy walking on a train, it's not difficult to invision the length contractions that make sense of the adddition of velocities) but I'm still having trouble seeing it with reference frames moving away from each other :P. Thanks for the help

If it helps any, the velocity that A sees B aproaching at, is exactly the same as the velocity of B that he measures when they are going away from each other (except for the sign).

Is it safe for me (the observer that sees each ship moving in opposite directions at 0.75c) to say that they're moving away from each other at a speed greater than c?
I made the space-time diagram and see no problems with this affirmation, hope you can check it out and tell me if there's anything wrong with it :P

The ships are moving away from each other with an angle of 60° (greater than 45°, which to my understanding would mean they're moving away from each other at c.). Of course I understand that if we rotate the diagram using Lorentz transformations to see how the same situation looks using spaceship A as the reference frame spaceship B's line would not be at an angle greater than 45° from spaceship A's line.

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