PX155 - I9 - how far could you go

depart earth and accelerate with $1 g$ . how far can you travel in $10 y e a r s$ ?

outline derivation:

let $S$ be the rest frame of the earth, $S^{'}$ be your instantaneous rest frame, which is moving at $v$ in frame $S$
after $δ t^{'}$ in $S^{'}$ , you are moving at speed, $δ v^{'}$ , in $S^{'}$ , which is no longer your rest frame
- in $S$ ,

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

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

- time dilation:

δ t = γ δ t^{'}

\frac{d v}{d t} = lim_{δ t \to 0} \frac{δ v}{δ t} = \frac{1}{γ^{3}} \frac{d v^{'}}{d t} = \frac{1}{γ^{3}} a^{'} = \frac{g}{γ^{3}}

- this is a velocity dependent acceleration problem, because $γ = γ (v)$ :

\int_{0}^{v} d v = \int_{0}^{t} \frac{g}{γ^{3}} . d t . . .

- we get $v (t)$ . rearrange for $γ (t)$
- integrate $v (t)$ from $0$ to $T$ to find the distance. need $T$ in terms of $T^{'}$ , where $T^{'} = 10 y e a r s$ :

δ t = γ δ t^{'}

where, $γ = γ (t)$
$$\int_{0}^{T} \frac{1}{\gamma(t)}dt = \int_{0}^{T_{0}'} dt'$$
- get $T (T^{'})$
- distance travelled in frame $S$ :

x (T^{'}) = \int_{0}^{T (T^{'})} v (t) . d t

- for $a^{'} = 9$ , $T^{'} = 10 /, y e a r s$ , $x (T^{'}) = 14, 900 l y$