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Hyperbolic path in Minkowski space

  1. Aug 27, 2013 #1
    The path described by a constantly accelerating particle is given by:

    [itex]x=c\sqrt{c^2/a'^2+t^2}[/itex]

    where a prime denotes an observer traveling with the particle and a letter without a prime a resting observer.

    If we leave the [itex]c^2/a'^2[/itex] out it reduces to x=ct, which makes sense. The distance traveled by a photon is simply it's speed times the travel time. If we think of a constantly accelerating particle, a particle which would first need to accelerate to a speed c (or at least close to it), we would expect that it would travel a shorter distance in the same amount of time, since it would first need to accelerate and hence travel at a smaller average speed. However, when you look at the equation above, you can see the travel time gets a little bonus, [itex]c^2/a'^2[/itex]:

    [itex]x=c\left[t+f\left(c^2/a'^2\right)\right][/itex]

    According to the equation, the slower one accelerates, the larger distance one will be able to travel. To me that does not make any sense at all, so I wondered whether I made any mistakes and decided to verify the answer on the internet. However, without blinking an eye, every single source I find online gives me the same answer.

    So where do I go wrong in all this?

    Sources:
    http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity [Broken])
    http://www.physik.uni-leipzig.de/~schiller/ed10/Uniform relativistic acceleration.pdf
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 27, 2013 #2
    Careful: where is the particle at t = 0?
     
  4. Aug 27, 2013 #3
    [itex]x\left(t=0\right)=c^2/a'[/itex]

    So I should define a new axis x'' then?

    [itex]x''=x-c^2/a'=c\sqrt{c^2/a'^2+t^2}-c^2/a'=c\left(\sqrt{c^2/a'^2+t^2}-c/a'\right)[/itex]

    If a' goes to infinity:

    [itex]x''=ct[/itex]

    If a' goes to zero:

    [itex]x''=c\left(\sqrt{c^2/a'^2}-c/a'\right)=0[/itex]

    Let's say a'=0.5c/yr, t=5 yr:

    [itex]x''=c\left(\sqrt{4+25}-2\right)=2.39ly[/itex]

    t=infinity:

    [itex]x''=c\left(\sqrt{c^2/a'^2+t^2}-c/a'\right)=c\left(\sqrt{t^2}-c/a'\right)=c\left(t-c/a'\right)=ct[/itex]

    This makes all sense to me. Thank you.
     
  5. Aug 27, 2013 #4
    This source http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] gives the distance travelled as:

    [tex]d =\frac{c^2}{a}\left(\sqrt{1+\left(\frac{at}{c}\right)^2}-1\right)[/tex]

    d in this case increases with increasing proper acceleration a.

    For your example in your second post, when a=0.5c/yr, t=5 yr, d=3.385 lyr.
     
    Last edited by a moderator: May 6, 2017
  6. Aug 27, 2013 #5
    It's the same, but I made a calculation error. I did [itex]\sqrt{25+4}-3[/itex] instead of [itex]\sqrt{25+4}-2[/itex]

    I prefer my version though, since it's in the form of c times a time.

    In my formula it's also much more obvious that d increases with increasing a'.
     
  7. Aug 27, 2013 #6
    Here is how to get from the equation given in the Baez link I posted, to the equations in your links.

    Starting with:

    [tex]d =\frac{c^2}{a}\left(\sqrt{1+\left(\frac{at}{c}\right)^2}-1\right)[/tex]

    which can be rewritten as:

    [tex]d =\sqrt{\frac{c^4}{a^2}+c^2t^2}-\frac{c^2}{a}[/tex]

    where d=0 when time t=0.

    In the Rindler coordinate system (X,T), the distance from the origin is c^2/a at time T=0 and the proper acceleration (α) of a given particle is proportional to c^2/X also at T=0. This means that the above equation can be written in terms of X as:

    [tex]X = d+\frac{c^2}{\alpha} = \sqrt{\frac{c^4}{\alpha^2}+c^2T^2}[/tex]

    which is essentially your equation at the start of this thread. Note that α is constrained to be the value of c^2/X at time T=0.

    P.S. I see I missed your post while posting this and that you had already figured this out. I was thrown by the fact you got a different result but that was just a numerical error.
     
    Last edited by a moderator: May 6, 2017
  8. Sep 18, 2013 #7
    If we rearrange that equation in terms of the co-ordinate time we get this:


    [tex]t =\sqrt{\left(\frac{2d}{a}\right)+\left(\frac{d}{c}\right)^2}[/tex]

    ...so the co-ordinate time can be thought of as having two time components - the Newtonian time to travel a displacement d at a constant acceleration a, and the light travel time. I'm not sure if that makes sense, but that's how it looks to me from the equations.
     
    Last edited by a moderator: May 6, 2017
  9. Sep 18, 2013 #8
    Thanks, that makes total sense.
     
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