2 A spacecraft is launched from the surface of the Earth with a velocity
3 of $0.600\U{c}$ at an angle of $50.0\dg$ above the horizontal positive
4 $x$ axis. Another spacecraft is moving past with a velocity of
5 $0.700c$ in the negative $x$ direction. Determine the magnitude and
6 direction of the velocity of the first spacecraft as measured by the
7 pilot of the second spacecraft.
8 \end{problem*} % problem 9.22
11 The Lorentz transformations
13 t' &= \gamma \p({t - v x/c^2}) \\
14 x' &= \gamma \p({x - v t}) \\
17 yield the velocity transformations
19 u_x' &= \pderiv{t'}{x'}
20 = \frac{\gamma\p({\partial x - v \partial t})}{\gamma\p({\partial t - v \partial x/c^2})}
21 = \frac{\pderiv{x}{t} - v}{1 - v \pderiv{t}{x}/c^2}
22 = \frac{u_x - v}{1 - v u_x/c^2} \\
23 u_y' &= \pderiv{t'}{y'}
24 = \frac{\partial y}{\gamma\p({\partial t - v \partial x/c^2})}
25 = \frac{\pderiv{t}{y}}{\gamma\p({1 - v \pderiv{t}{x}/c^2})}
26 = \frac{u_y}{\gamma\p({1 - v u_x/c^2})} \\
29 We can use these velocity transformations on our spacecraft's velocity.
31 \gamma &= \frac{1}{\sqrt{1-v^2/c^2}}
32 = \frac{1}{\sqrt{1-(-0.700)^2}} = 1.40 \\
33 u_x &= 0.600c\cos(50.0\dg) = 0.386c \\
34 u_y &= 0.600c\sin(50.0\dg) = 0.460c \\
35 u_x' &= \frac{u_x - v}{1 - vu_x/c^2}
36 = \frac{0.321c + 0.700c}{1 + 0.700\cdot0.321}
38 u_y' &= \frac{u_y}{\gamma(1 - vu_x/c^2)}
39 = \frac{0.383c}{1.40\cdot(1 + 0.700\cdot0.321)}
41 u' &= \sqrt{u_x'^2 + u_y'^2}
43 \theta' &= \arctan(u_y'/u_x')