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Interpreting results from constant-acceleration/distance/subjective-time calculator

  1. Oct 14, 2012 #1
    Hello, first time poster here!

    I am trying to understand some results I am getting from this "Relativistic Star Ship Calculator". This site sets up the question I'm trying to answer exactly like I imagined it should (thanks to The Forever War). From the description:

    The star ship accelerates continuously from the origin to the midpoint of the mission.
    At the midpoint, the ship turns its thrusters to face the destination.
    The ship decelerates continuously from the midpoint to the destination.

    The way I understood it, if you are accelerating at 1G and hope to come arbitrarily close to the speed of light, it should take (3*10^8 m/s)/(9.8 m/s^2)= ~354 days. Similarly, if you hope to decelerate from such a speed, it will require the same amount of time. So say about 2 years of combined acceleration/deceleration.

    Further, I thought that at a velocity which is arbitrarily close to c, your Lorentz Factor becomes arbitrarily high, meaning that light years can be traveled in seconds (or less). So the conclusion I reached is that you can travel virtually any distance in ~2 years if you had a ship that accelerates at 1G and can reach a very high fraction of c.

    When inputting large distances in the calculator linked above, I get results that don't fit with my rationale. Example:

    Input: Distance: 10 LY
    Acceleration: 1G
    Output: Time: 4.9 Y
    Max Speed: 0.987 c

    So this calculator assumes your trip takes 4.9 Y (not 2) and you only reach 0.987 c (not .99999....c)

    Can someone explain the fault in my reasoning?

    Thanks so much!
     
    Last edited: Oct 14, 2012
  2. jcsd
  3. Oct 14, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Interpreting results from constant-acceleration/distance/subjective-time calculat

    Velocities do not add that way. That would be true under Galilean relativity, but not Special relativity. (See: How Do You Add Velocities in Special Relativity?)

    The longer the acceleration is applied, the less affect it has on the speed with respect to the original frame. (Note that the acceleration is measured from an inertial frame co-moving with the rocket at any moment--not with respect to the original frame.)

    After one year (in rocket time) of accelerating at g, your speed would be only about .77c. (See: The Relativistic Rocket)

    That's true.
    The problem is that you are not going as fast as you think. It takes a while to reach a really high speed at that acceleration.
     
  4. Oct 14, 2012 #3
    Re: Interpreting results from constant-acceleration/distance/subjective-time calculat

    Thanks Doc Al, I'll read those links. I'm kind of miffed that I made it out of my Physics course (which covered special relativity) without understanding that bit. Anyway, good to know!
     
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