Relativistic Addition of Velocities

[Piamedes]

Homework Statement

A rocket is traveling at speed V along the x-axis of frame S. It emits a signal (for example, a pulse of light) that travels with speed c along the y prime axis of the rocket's rest frame S prime. What is the speed of the signal as measured in S?

Homework Equations

$$v_{y} = v_{y}^{'} \gamma (1-v_{x}V/c^2)$$

The Attempt at a Solution

I know the answer is C. That's readily apparent because the speed of light is constant in all inertial reference frames. But when I actually plug in the value C to this equation it doesn't give C as the answer.
From my understanding, the variables represent:

$$v_{y}$$ is the velocity of the signal relative to the S frame

$$v_{y}^{'}$$ is the velocity of the signal relative to the S prime frame

V is the relative speed between the S and S prime frames.

and $$v_{x}$$ is the velocity of the rocket relative to the S frame

So $$v_{y}$$ is what I'm solving for,

$$v_{y}^{'} = c$$

and V is just some arbitrary speed v

Plugging in:

$$v_{y} = c \gamma (1-v_{x}V/c^2)$$

But the rocket is at rest in S prime, so $$v_{x}=v$$

$$v_{y} = c \gamma (1-v^2/c^2)$$

$$v_{y} = c \sqrt{1-v^2/c^2}$$

Where exactly am I going wrong? Am I misunderstanding the meaning behind each of the variables, or did I make some algebra error somewhere?

Thanks for the help.

 

[turin]

Does the signal travel in exactly the y-direction in S?

 

[Piamedes]

Thanks, I think I have it:

That is the velocity of the signal in the y direction, but it also has a velocity component in the x direction because it was emitted by a moving source. The velocity in the x direction is just v. So the speed V is just:

$$V^2 = v_{y}^2 + v_{x}^2$$

$$V^2 = c^2 (1-v^2/c^2) + v^2$$

$$V^2 = c^2$$

$$V = c$$

Which gives the speed as c

Does that still work in relativity?

 

[turin]

Does that still work in relativity?

What do you mean? How did you get that without relativity?

 

[Piamedes]

I was just referring to if the magnitude of a velocity vector was still the square root of the sum of squares of its components.

 

[turin]

Ah. Excellent question! The short answer is, "No." However, the more detailed answer specific to this case is actually, "Yes, the magnitude of a 3-vector can still be obtained in the 'usual' way." By "usual way", I mean the Pythagorean way (as you suggest). This is a consequence of the flat Euclidean nature of the 3-D space.

However, take caution, for future reference! This is not true of 4-vectors, which live (in Special Relativity) in flat so-called Minkowski space (named after one of Einstein's mathematical mentors). A 4-vector is a vector that includes all 3 space components as well as the time component (for a total of 4 components). In this case, you must modify the Pythagorean idea, and there are two cases. If you want to consider length, the relativistically invariant magnitude of length is the "proper length", s, and is given by:

s² = x² + y² + z² - c²t²

If you want to consider time, the relativistically invariant magnitude of time is the "proper time", τ, and is given by:

τ² = t² - ( x² + y² + z² )/c²

Heed the very important minus signs.