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Let's say I have a particle in a uniformly acclerated motion such that

$$ a_\mu a^\mu = -g^2, \quad g > 0\ {\rm and\ constant} \tag1 $$

I want to prove that any signal coming from a apace-time point $(t > 0, x < t - 1/g)$ cannot ever reach the particle. From Eq. (1) I know that the equations for the particle's motion are

$$ t = \frac{1}{g}\sinh(gs), \quad x = \frac{1}{g}(\cosh(gs) - 1) $$

Where $s$ is the proper time. Then, the trajectory of the particle is:

$$ x_p(t) = \frac{1}{g}\Big(\sqrt{1 + (gt)^2} - 1\Big) $$

So, in order to prove that it cannot be reached by any signal coming from $(t > 0, x)$ that travels a finite time $\Delta t$, I tried proving that the interval was time-like and see if I got the condition $x < t - 1/g$. So,

$$ \int_t^{t + \Delta t}\ dt^2 - \int_x^{x_p(t + \Delta t)} \ dx^2 < 0 $$

And solving the inequation I got that it was fulfilled for,

$$ x > \frac{1}{g}\Big(\sqrt{1 + (g(t + \Delta t))^2} - 1\Big) + \Delta t\\ x < \frac{1}{g}\Big(\sqrt{1 + (g(t + \Delta t))^2} - 1\Big) - \Delta t $$

This has nothing to do with $x < t - 1/g$, so clearly I'm misunderstanding something. How would you do it?

$\bf{COMMENT}$

The answer provided by Eli is an alternative way to solve the problem but it doesn't explain why mine is wrong. So if anyone can say the reason to that, please comment because I'd love to know it

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  • $\begingroup$ But when and where was the signal sent from? And what do you mean by reach? Is an unknown observer trying to bounce a signal off the particle? Or by reach, is this a rest-line issue? And I have no idea why you're integrating. Let $\gamma$ be the signal, is it true $\gamma(0)=x_{p}(0)$? Also, what is the invariant of the motion of the particle in this problem? $\endgroup$
    – Cinaed Simson
    Commented Sep 20, 2019 at 19:32
  • $\begingroup$ Oops - I just realized $\gamma$ was a very poor choice for the signal variable - so call it $signal$ instead of $\gamma$. Is it true $signal(0)=x_{p}(0)$? $\endgroup$
    – Cinaed Simson
    Commented Sep 20, 2019 at 20:03
  • $\begingroup$ @CinaedSimson As the post says, the signal comes from $(t, x)$ such that $t > 0,\ x < t - 1/g$ so it is not necessary that $signal(0) = x_p(0)$. And by reach the particle I mean that the signal (another particle or light) at some time $T$ satisfies that $signal(T) = x_p(T)$ $\endgroup$
    – Vicky
    Commented Sep 20, 2019 at 21:00
  • $\begingroup$ The $0$ in $signal(0) = x_p(0)$ is $\tau=0$ - where $\tau$ is the proper time. It's unlikely you'll even see a rest-line - assuming you were an observer riding on the particle. In any case, Eli answered your question - never - the two lines don't intersect. $\endgroup$
    – Cinaed Simson
    Commented Sep 21, 2019 at 22:53

1 Answer 1

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enter image description here

the red one is

$x_p=\frac{1}{g}\,(\sqrt{(g\,t)^2+1}-1)$

the blue equation is

$x_l=t-\frac{1}{g}$

so if you don't have intersection point between $x_p(t)$ and $x_l(t)$ you can't reach the space time domain.

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  • $\begingroup$ The blue line represents a signal at light speed, right? So any other worldline with velocity less than $c$ won't be able to reach the red one if the blue cannot. Is that the idea? Nevertheless, about my idea with the interval what do you think is failing? The integral limits maybe? $\endgroup$
    – Vicky
    Commented Sep 18, 2019 at 16:09
  • $\begingroup$ @Vicky yes that is idea, about your integral limits I have to think about it $\endgroup$
    – Eli
    Commented Sep 18, 2019 at 16:59

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