Reaching the Rindler horizon in a finite proper time

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Hi,

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:

https://www.physicsforums.com/showpost.php?p=3316839&postcount=375

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.
 

Answers and Replies

  • #2
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I also don't understand how to work out that the worldine is given by r=(t,sech(t),0,0).
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:
[itex]T=atanh(t)[/itex]
[itex]t=tanh(T)[/itex]

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]

But don't understand why you have to take T=∞ in order to find X=0?
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]
 
  • #3
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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] ?

Now, to find the proper time we use formula 1.97 in Sean Carroll's lecture notes
[tex]\tau = \int \sqrt{-g_{\mu\nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}} \, dt = \int sech^{2}(t) \, dt = tanh(t)[/tex]
And how do you get from the square root expression to the sech-squared term?


Thanks for your patience!
 
  • #4
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5,317
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] ?
This follows directly from the coordinate transformation equations (remember their convention is different from yours):
http://en.wikipedia.org/wiki/Rindler_coordinates#Relation_to_Cartesian_chart

Just substitute in x=1.

And how do you get from the square root expression to the sech-squared term?
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(
\begin{array}{cccc}
-x^2 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}
\right)= \left(
\begin{array}{cccc}
-sech(t)^2 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}
\right)[/tex]
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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