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Hyperbolic motions

  1. Feb 12, 2010 #1
    Hi everybody

    I've been lately a little bit concerned over the hyperbolic motions that have the following equations in (ct,x)-space:

    [tex]\frac{x^2}{(c^2/a)^2}-\frac{(ct)^2}{(c^2/a)^2}=1[/tex].

    We know that events horizons are the lines that form a 45-degree angle by both ct- and x-axis. So what does actually assure us that here, for instance, for t=0, [tex]x=\pm c^2/a[/tex] lie inside events horizens? Is this just because [tex]a[/tex] can't in magnitude gets higher than [tex]c[/tex]?

    AB
     
  2. jcsd
  3. Feb 12, 2010 #2

    George Jones

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    No.

    [tex]x^2 - \left(ct\right)^2 = \left( \frac{c^2}{a} \right)^2,[/tex]

    so [itex]a \rightarrow \infty[/itex] gives the horizons. For [itex]t=0[/itex], any value of [itex]x[/itex] except [itex]x = 0[/itex] lies inside the horizons.
     
  4. Feb 12, 2010 #3
    Yeah, I got it!

    Thanks
     
  5. Feb 12, 2010 #4

    George Jones

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    Also, differentiating

    [tex]x^2 - \left(ct\right)^2 = \left( \frac{c^2}{a} \right)^2,[/tex]

    gives

    [tex]\frac{dx}{dt} = c \frac{ct}{x}.[/itex]

    Consequently,

    [tex]-c < \frac{dx}{dt} < c[/tex]

    gives that [itex]\left(ct , x \right)[/itex] lies inside the horizons.
     
  6. Feb 12, 2010 #5
    Could you explain this a little bit more?
     
  7. Feb 12, 2010 #6

    DrGreg

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    Combine the following and what do you get?
     
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