PX155 - H5 - the twin paradox

not a paradox: proof

alice and bobs are twins. alice travels to a distant star and returns to find bob many years older then her
but isn't time dilation reciprocal? shouldn't they both observe time running slowly for the other
use LTs to rectify:
- assume alice uses a spaceship that accelerates instantly to cruising speed
- frames :
  - $S =$ bob's rest frame
  - $S^{'} =$ alice's outbound rest frame, $+ u$ relative to bob
  - $S^{″} =$ alice's inbound rest frame, $- u$ relative to bob
- events::
  - $I$ : alice leaves earth, $x = x^{'} = x^{″} = 0$ , $t^{'} = t^{″} = 0$
  - $I I$ : alice reaches the star, $x = L$ , where, $L$ is the position of the star; $t = \frac{L}{u}$
  - $I I I$ : alice returns to earth, $x = 0$ , $t = 2 \frac{L}{u}$
- need $t^{'} = γ (t - \frac{u x}{c^{2}})$ , $t^{″} = γ (t + \frac{y x}{c^{2}})$

time	event 1, $x = 0$	event 2 $x = L$	event 3 $x = 0$
$t$	$0$	$\frac{L}{u}$	$2 \frac{L}{u}$
$t^{'}$	$0$	$γ (\frac{L}{u} - u \frac{L}{c^{2}})$	$2 γ \frac{L}{u}$
$t^{″}$	$0$	$γ (\frac{L}{u} + u \frac{L}{c^{2}})$	$2 γ \frac{L}{u}$

Δ t_{B} = (t_{2} - t_{1}) + (t_{3} - t_{2}) = 2 \frac{L}{u}

- alice:

Δ t_{A} = (t_{2}^{'} - t_{1}^{'}) + (t_{3}^{″} - t_{2}^{″}) = \frac{2 L γ}{u} (1 - \frac{u^{2}}{c^{2}}) = \frac{2 L}{γ u} = \frac{Δ t_{B}}{γ}

- $γ > 1 ⟹$ less times experienced by alice than by bob
- note: in frames $S^{'}$ and $S^{″}$ , $Δ t^{'} = γ Δ t_{B}$ ; $Δ t^{″} = γ Δ t_{B}$

alice and bob have identical clocks that they carry with them. each has a telescope with which they count ticks on each other's clock
bob's view of bob's clock:
- clock ticks with proper frequency $f_{0}$
- $N_{B} (B) = \frac{2 L_{B}}{u} f_{0}$
bob's view of alice's clock:
- subsequent ticks occur farther away for outbound journey
- for outbound frequency:

f_{1} = \frac{f_{0}}{γ (1 + \frac{u}{c})}

- for inbound frequency:

f_{2} = \frac{f_{0}}{γ (1 - \frac{u}{c})}

- in frame $S$ , bob receives last tick at $f_{1}$ at time, $\frac{L}{c}$ after it is emitted
- total ticks:

N_{B} (A) = (\frac{L}{u} + \frac{L}{c}) f_{1} + (\frac{L}{u} - \frac{L}{c}) f_{2}

N_{B} (A) = \frac{f_{0}}{γ} [\frac{\frac{L}{u} + \frac{L}{c}}{1 + \frac{u}{c}} + \frac{\frac{L}{u} - \frac{L}{c}}{1 - \frac{u}{c}}]

N_{B} (A) = \frac{f_{0}}{γ (1 - \frac{u^{2}}{c^{2}})} [(\frac{L}{u} + \frac{L}{c}) (1 - \frac{u}{c}) + (\frac{L}{u} - \frac{L}{c}) (1 + \frac{u}{c})]

N_{B} (A) = \frac{f_{0}}{γ (1 - \frac{u^{2}}{c^{2}})} [2 \frac{L}{u} - 2 \frac{L u}{c^{2}}]

N_{B} (A) = \frac{2 L}{u γ}

alice's view of alice's clock:
- for alice, length of he journey contracts to $\frac{L}{γ}$
- time taken $Δ t_{A}^{'} = 2 \frac{L}{γ u}$
- so ticks on her own clock:

N_{A} (A) = \frac{2 L}{γ u} f_{0}