Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Constant acceleration from accelerated observer's perspective

  1. Apr 27, 2013 #1
    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. jcsd
  3. Apr 27, 2013 #2


    User Avatar

    Staff: Mentor

    Google for "Rindler coordinates" - they were invented for exactly this situation.
  4. Apr 27, 2013 #3


    User Avatar
    Science Advisor

    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. :wink: 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
  5. Apr 27, 2013 #4
    Okay, thanks, guys!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook