PX155 - H6 - relativistic velocity addition

derivation and examples

light travels at $c$ in all initial frames
what about objects moving at $v^{'} < c$ in frame $S^{'}$ along the x axis? what is $v$ for that object in frame $S$ ?
use LTs: $$t=\gamma(t' + \frac{ux'}{c})$$

x = γ (x^{'} + u t^{'})

differentiating wrt $t^{'}$ : $$\frac{dt}{dt'} = \gamma\left(1+ \frac{u}{c^{2}} \frac{dx'}{dt'} \right) = \gamma \left(1 + \frac{uv'}{c^{2}}\right)$$

\frac{d x}{d t^{'}} = γ (\frac{d x^{'}}{d t^{'}} + u) = γ (v^{'} + u)

v = \frac{d x}{d t} = \frac{d x}{d t^{'}} \frac{d t^{'}}{d t} = \frac{d x}{d t^{'}} {(\frac{d t}{d t^{'}})}^{- 1}

the relativistic velocity addition is: $$\therefore v = \frac{v'+u}{1+ \frac{uv'}{c^{2}}}$$

∴ v^{'} = \frac{v - u}{1 - \frac{u v}{c^{2}}}

eg: two particles approach each other at $0.8 c$ - "head on" - in the rest frame of some observer. what is the speed of one particle in the rest frame of the other?
- let observer at rest in $S^{'}$
- $S^{'}$ moves at $+ v^{'}$ in the rest frame of left-going particle. so, taking that frame as $S$ with $u = v^{'}$
$v = \frac{v^{'} + u}{1 + \frac{u v^{'}}{c^{2}}} = \frac{v^{'} + u^{'}}{1 + \frac{u v^{'}}{c^{2}}} = \frac{0.8 c + 0.8 c}{1 + {0.8}^{2}} \approx 0.9756 c$
eg: two aeroplanes approach each other head on at $500 m p h$ . find the speed of one in the rest frame of other, as a correction to the galilean result
- we want to find:

δ v = \frac{v^{'} + u}{1 + \frac{u v^{'}}{c^{2}}} - (v^{'} + u)

δ v = \frac{v^{'} + u - (1 + \frac{u v^{'}}{c^{2}}) (v^{'} + u)}{1 + \frac{u v^{'}}{c^{2}}} = \frac{- (v^{'} + u) u v^{'}}{(1 + \frac{u v^{'}}{c^{2}}) c^{2}}

δ v \approx \frac{- u v^{'}}{c^{2}} (v^{'} + u)

- we have, $c = 186, 000 m p s = 186, 000 \times 3600 m p h$
- therefore,

δ v \approx - 5.6 \times 10^{- 10}

what about a 3D velocity vector with components $(v_{x}, v_{y}, v_{z})$ ?
in $S^{'}$ : $v_{y}^{'} = \frac{d y^{'}}{d t^{'}} = \frac{d y^{'}}{d t} {(\frac{d t^{'}}{d t})}^{- 1}$
$\frac{d t^{'}}{d t} = γ (\frac{1 - u}{c^{2}} v_{x})$
since $y^{'} = y$ , $\frac{d y^{'}}{d t} = \frac{d y}{d t} = v_{y}$

v_{y}^{'} = \frac{v_{y}}{γ (1 - \frac{u v_{x}}{c^{2}})}

v_{z}^{'} = \frac{v_{z}}{γ (1 - \frac{u v_{x}}{c^{2}})}

v_{x}^{'} = \frac{v_{x} - u}{(1 - \frac{u v_{x}}{c^{2}})}

aberration of light

eg: a beam of light travels down the y-axis toward origin of frame $S$ . what does $S^{'}$ measure?
- in $S$ , beam has velocity, $(0, - c, 0)$
- in $S^{'}$

v_{x}^{'} = \frac{v_{x} - u}{1 - \frac{u v_{x}}{c^{2}}} = - u

v_{y}^{'} = \frac{v_{y}}{1 - \frac{u v_{x}}{c^{2}}} = - \frac{c}{γ}

- in frame $S^{'}$ :

{\vec{v}}^{'} = (- u, - \frac{c}{γ}, 0)

- sanity check:

| {\vec{v}}^{'} |^{2} = u^{2} + \frac{c^{2}}{γ^{2}} = u^{2} + c^{2} (\frac{1 - u^{2}}{c^{2}}) = c^{2}