# Constant acceleration from accelerated observer's perspective

1. Apr 27, 2013

### Whovian

First, an unnecessary short introduction of why I'm asking this question in the first place. Due to some doubts about a certain plot device in a science-fiction novel, I'm trying to figure out how long it would take to accelerate someone to, say, 50% of the speed of light under constant acceleration from their own reference frame (for various reasons, I'm assuming about 10*g.) Due to this being a small ameteur project, I'll probably be using Wolfram Alpha.

In any case, I got the outrageously small answer of about 2.5 weeks, which doesn't sound even close to right to me. I think I've isolated my problem to how I formulate this in terms of acceleration as measured from Earth. If this bit's correct, yes, I'll post the rest of my work in a hope of others finding a minor mistake.

So, now for a formulation of the problem. Say Bob's in a spaceship and accelerating with acceleration $100\ \dfrac{\mathrm{m}}{\mathrm{s}^2}\approx 10\cdot g$ in his own reference frame. How long, measured in seconds from Earth, will it take for him to be moving at .5c with respect to Earth?

Now, after a large amount of squandering about with exhaust-based engines in which the exhaust has mass << the spaceship which resulted in very ugly expressions for momentum ($\dfrac{1\ \mathrm{m}^3\cdot\delta\cdot\frac{v-r}{1-\frac{v\cdot r}{c^2}}}{\sqrt{1-\left(\frac{\frac{v-r}{1-\frac{v\cdot r}{c^2}}}{c}\right)^2}}$, for instance,) I decided to stop and just come out with what my intuition told me the acceleration as measured from Earth is, which is $100\ \dfrac{\mathrm{m}}{\mathrm{s}^2}\cdot\sqrt{1-\left(\dfrac vc\right)^2}$. Solving this differential equation yields the nutters result of a periodic function. So, clearly, something's wrong with this expression. But what, exactly?

2. Apr 27, 2013

### Staff: Mentor

Google for "Rindler coordinates" - they were invented for exactly this situation.

3. Apr 27, 2013

### Bill_K

The trajectory of an object undergoing constant proper acceleration is called hyperbolic motion. Because, I guess, it can be expressed in terms of hyperbolic functions. An equivalent but simpler result is

x = (1/a)√(1 + a2t2) - 1
v = dx/dt = at/√(1 + a2t2)

EDIT: Should point out that this is with c = 1, and sure enough, v → 1 as t → ∞.

Last edited: Apr 27, 2013
4. Apr 27, 2013

### Whovian

Okay, thanks, guys!