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Reaching the Rindler horizon in a finite proper time

  1. Nov 13, 2011 #1

    I am trying to show that timelike geodesics reach the Rindler horizon (X=0) in a finite proper time.

    The spacetime line element is

    [itex]ds^{2} = -\frac{g^{2}}{c^{2}}X^{2}dT^{2}+dX^{2}+dY^{2}+dZ^{2}[/itex]

    Ive found something helpful here:


    But don't understand why you have to take T=∞ in order to find X=0?

    I also don't understand how to work out that the worldine is given by r=(t,sech(t),0,0).

    Any help will be greatly appreciated.
  2. jcsd
  3. Nov 13, 2011 #2


    Staff: Mentor

    OK, to be consistent with your notation above I will use (T,X,Y,Z) for Rindler coordinates and I will use (t,x,y,z) for Minkowski coordinates. (This is the opposite of the convention used in the thread you linked to and also the opposite of the convention in the Wikipedia page http://en.wikipedia.org/wiki/Rindler_coordinates).

    In Minkowski coordinates we can easily see that the worldline [itex]r=(t,1,0,0)[/itex] is a geodesic. Transforming this to Rindler coordinates gives [itex]R=(atanh(t),\sqrt{1-t^2},0,0)[/itex]. Now, this parameterizes the worldline by the Minkowski time coordinate which, although it is not wrong, it is not very aesthetically pleasing. So we solve for t in terms of T:

    Substituting that in we get
    [itex]R=(atanh(tanh(T)), \sqrt{1-(tanh(T))^2},0,0)[/itex]
    [itex]R=(T, \sqrt{(sech(T))^2},0,0)[/itex]
    [itex]R=(T, sech(T),0,0)[/itex]

    We want to find when it crosses the Rindler horizon, which is located at X=0. So, we solve [itex]0=sech(T)[/itex] and get [itex]T=\infty[/itex]
  4. Nov 13, 2011 #3
    Thanks for your quick reply.

    Still a couple of things I don't follow...

    How exactly do you transform from
    [itex]\textbf{r}=(t,1,0,0)[/itex] to [itex]\textbf{R}=(atanh(t), \sqrt{1-t^{2}},0,0)[/itex] ?

    And how do you get from the square root expression to the sech-squared term?

    Thanks for your patience!
  5. Nov 13, 2011 #4


    Staff: Mentor

    This follows directly from the coordinate transformation equations (remember their convention is different from yours):

    Just substitute in x=1.

    We have
    [tex]\frac{d\mathbf x}{dt} = \frac{d\mathbf r}{dt} = \frac{d}{dt}(t,sech(t),0,0) = (1,-sech(t) \, tanh(t),0,0)[/tex]
    [tex]\mathbf g = \left(
    -x^2 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1
    \right)= \left(
    -sech(t)^2 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1
    Substituting that in we obtain
    [tex]\tau = \int \sqrt{-g_{\mu\nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}} \, dt [/tex]
    [tex]\tau = \int \sqrt{sech(t)^2-sech(t)^2 tanh(t)^2} \, dt [/tex]
    [tex]\tau = \int \sqrt{sech(t)^4} \, dt [/tex]
    [tex]\tau = \int sech(t)^2 \, dt [/tex]
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