Velocity transformations - light at an angle

[Phynos]

Homework Statement

(a) A light signal is fired at $60^o$ North of West. Calculate the west-east velocity component of the signal according to an observer traveling due East at 0.5c. State your answer as a multiple of c.

(b) Calculate the North-South velocity component of the light signal according to an observer traveling due East at 0.5c. State your answer as a multiple of c.

(c) Calculate the West-East velocity component of the light source according to an observer traveling due West at 0.5c. State your answer as a multiple of c.

(d) Calculate the North-South velocity components of this light source according to an observer traveling due West at 0.5c. State your answer as a multiple of c.

Homework Equations

$U'_x = \Large \frac{U_x - V}{1 - \frac{U_x V}{c^2}}$

$U'_y = \Large \frac{U_y}{\gamma(1 - \frac{U_y V}{c^2})}$

$\gamma = \Large \frac{1}{ \sqrt{1 - \frac{v^2}{c^2}}}$

The Attempt at a Solution

I'm getting mixed up about what sign each term should have. There must be a easier way to figure this out than trial and error?

For the cases where the observer is moving east at 0.5c...

(a) $U'_x = \Large \frac{0.5c + cos60*c}{1 + \frac{0.5c*cos60*c}{c^2}} \normalsize = 0.8c$

Here I have v negative because the observer is moving in the opposite direction as the east-west component of the light.

(b) $\gamma = \Large \frac{1}{ \sqrt{1 - \frac{(0.5c)^2}{c^2}}} \normalsize = 1.1547$

$U'_y = \Large \frac{sin60*c}{1.1547*(1 - \frac{sin60*c*0.5*c}{c^2})} \normalsize = 1.322c$

Which is nonsense so I assume that sign in the denominator should be positive?

$U'_y = \Large \frac{sin60*c}{1.1547*(1 + \frac{sin60*c*0.5*c}{c^2})} \normalsize = 0.5234c$

Assuming that's correct, what's the reason for the sign swap? That's not the equation I was given...

For the cases where the observer is moving west at 0.5c...

(c) $U'_x = \Large \frac{0.5c - cos60*c}{1 + \frac{0.5c*cos60*c}{c^2}} \normalsize = 0$

(d) $\gamma = 1.1547$

$U'_y = \Large \frac{sin60*c}{1.1547*(1 + \frac{sin60*c*0.5*c}{c^2})} \normalsize = 0.5234c$

What kind of result is that!? No velocity along the x axis, all along the y axis, except the y component of the velocity is less than c. This is nonsense.

Also, hopefully the Latex is more readable, it took forever to get it right.

 

[andrewkirk]

Check your formulas with the source. I'm pretty sure the $U_y$ in the denominator of your second formula (for $U_y'$) should be $U_x$.

 

[Phynos]

Ahhh, yes. Good catch. But still I think I need the minus in the denominator to be addition, or else the answer is 1. I can't understand conceptually why this should be the case though.

$U'_y = \Large \frac{U_y}{\gamma(1 + \frac{U_x V}{c^2})}$

(b) $U'_y = \Large \frac{sin60*c}{1.1547*(1 + \frac{cos60*c*0.5*c}{c^2})} \normalsize = 0.6c$

(d) For this one it only makes sense if I have the minus in the denominator now, then I get 1c. Which is great because the x-component is zero. But the problem again is that I can't get my head around why I must use that sign (Apart from just getting the right answer). If I had a strategy for figuring it out I would feel better about these problems.

$U'_y = \Large \frac{sin60*c}{1.1547*(1 - \frac{cos60*c*0.5*c}{c^2})} \normalsize = 1c$

 

[andrewkirk]

But still I think I need the minus in the denominator to be addition,

Yes, it is a plus. The formula is

$$
U'_y = \Large \frac{U_y}{\gamma(1 + \frac{U_x V}{c^2})}
$$
where $V$ is the velocity of the 'stationary' frame relative to the observer's frame. Since both $U_x$ and $V$ are heading Westward relative to the observer, they have the same sign in that formula, so that the denominator is an addition.

In (d), the 'stationary' frame is now moving Eastward relative to the observer, so $U_x$ and $V$ now have opposite signs, and the denominator becomes a subtraction.

Well done with the latex by the way. It makes it much easier to help with a problem if one can read the formulas clearly.

 

[Phynos]

That makes sense. Thanks for your help!

I'll be sure to use Latex from now on then.