# Transformation of angles in special relativity

My question is about the problem below

Depicted are two space ships (the USS Voyager and the USS Enterprise), each with velocity $v=c/2$ relative to the space station (Babylon 5). At the exact moment the two space ships are closest together (at a distance d) the USS Enterprise fires off a shuttle with velocity $u=3c/4$ relative to the space station.

The question is: Under what angle $\alpha'$ (as measured by USS Enterprise) must the shuttle be fired off in order to meet the USS Voyager?

I see two possible ways to approach this:

1. Calculate the angle $\alpha$ measured by the space station Babylon 5 and figure out how angles transform when we change our frame of reference.
2. Figure out the velocity in x-/y-direction in the frame of reference of the USS Enterprise and deduce the angle $\alpha'$ from that.

Now, if both ways are correct they should give the same answer. But they don't seem to, so what is wrong?

Here is what I have tried:

Let $u_x$, $u_y$ be the velocities (in the corresponding directions) the shuttle needs to have - as observed by Babylon 5. Let $u_x'$, $u_y'$ be the velocities of the shuttle observed by USS Enterprise.

Then by the rules for adding velocities:

\begin{eqnarray*} u_y' &=& \frac{u_y + v}{1 + \frac{vu_y}{c^2}} \\ u_x' &=& \frac{u_x }{1 + \frac{vu_y}{c^2}}\cdot \sqrt{1-\frac{v^2}{c^2}} \end{eqnarray*}

So this gives

$$\tan(\alpha') = \frac{u_y'}{u_x'} = \frac{u_y + v}{u_x} \cdot\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

But on the other hand, we could imagine the guys on Babylon 5 drawing a big triangle to show the crew on USS Enterprise the trajectory the shuttle needs to take.

The people on Babylon 5 will then draw a triangle with angle $\alpha$ satisfying $$\tan(\alpha) = \frac {u_y}{u_x}$$

So, if side $a$ of this triangle is parallel to the x-axis and $b$ is parallel to the y-axis, then we have $$\frac ba = \tan(\alpha) = \frac{u_y}{u_x}$$

Since for the USS Enterprise $a$ is the same, but side $b$ is contracted, they will see a traingle with side $b' = b/\gamma$ and $a' = a$. Therefore

$$\tan(\alpha') = \frac{b'}{a'} = \frac ba \sqrt{1-\frac{v^2}{c^2}} = \frac{u_y}{u_x}\sqrt{1-\frac{v^2}{c^2}}$$

These are two different results in general. I cannot figure out what is wrong with either of them...

Thanks for reading, help will be greatly appreciated! :)

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O.k. I think that the second result is wrong: If $\alpha=0$ (i.e. $u_y=0$) then the shuttle must be fired off with a component in the direction opposite to the direction the USS Enterprise is moving. This is exactly what the first result says, but it is not at all what the second result says. I'm still unsure as to what exactly is the problem with the approach leading to the second formula but I think it has something to do with not taking the movement of USS Enterprise into account somehow. –  Sam Aug 16 '11 at 17:52

Let's look at the Gallilean case first (to which this problem must reduce in non-relativistic limit anyway). Your first approach works the same way but we can get rid of unnecessarily complicated terms. In particular, we have $\tan(\alpha) = \frac{v}{u_x}$ and $\tan(\alpha') = \frac{2v}{u_x}$ (I've used that $u_y = v$ which is true also relativistically). You should convince yourself that this is the correct transformation.

Now, for the second approach we get that $\alpha = \alpha'$ since there are no contractions in Gallilean case. The conflict with the first approach rests on the fact that since the triangle is formed using the velocity vector, it will change when boosted. Therefore it's not valid to deduce that the triangle only transforms by contraction in one direction (which reduces to identity in non-relativistic limit). In fact, this contracting effect is quite negligible when compared to the deformation of the triangle caused by the boost.

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"... since the triangle is formed using the velocity vector, it will change when boosted." I think this is exactly right. Thanks! =) –  Sam Aug 16 '11 at 20:10

There's an inconsistency with the way the velocity composition formulas are consistent with the axis labeled as in the diagram, but your $\tan$ equations are consistent with the axis labeled the other way around. I'll stick with the unconventional way the vertical axis is labeled $x$, the horizontal $y$ in your diagram, so

$\tan(\alpha') = \frac{u_y'}{u_x'} = \frac{u_y + v}{u_x} \cdot\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

should be

$\tan(\alpha') = \frac{u_x'}{u_y'} = \frac{u_x}{u_y + v} \cdot\sqrt{1-\frac{v^2}{c^2}}$

Now let's look at your second method. The Lorentz contraction (LC) formula is a relationship between space intervals for the same pair of events in different frames, but simultaneous in one. I'll leave you to prove their time interval in the other frame $$\Delta t' = \gamma \frac V {c^2} \Delta x$$ Therefore, you need to be very clear in how you're using the LC formula to relate the space intervals of which pair of events.

A better way would be to find the coordinates of the event of the shuttle reaching USS voyager in the USS Enterprise frame using the Lorentz transformation$$y'=\gamma(y + Vt) = \gamma(b + Vt) = \gamma(u_yt + Vt),\qquad x'= x = a = u_xt$$ giving $$tan\alpha' = \frac {x'} {y'} = \frac{u_x}{\gamma(u_y + v)} \cdot$$ So the two methods are consistent with one another, as expected.

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