The definition of Event Horizon I took from Wikipedia:

An Event Horizon is a boundary in spacetime beyond which events cannot affect an outside observer.

From the definition, I can abstractly define such boundary without involving Gravity. Now, I want to prove it by construction.

Is it possible without involving Gravity?

  • $\begingroup$ Are you thinking of a Rindler horizon? en.wikipedia.org/wiki/Rindler_coordinates#The_Rindler_horizon $\endgroup$ Commented Mar 11, 2014 at 2:13
  • $\begingroup$ Aha, this setup is great, but it's also exploiting the same thing to achieve that. And, I'd like to stick with normal inertial reference frame. $\endgroup$ Commented Mar 11, 2014 at 2:42
  • $\begingroup$ What do you mean when you say that you can define such a boundary? You cannot just declare a surface to be an event horizon. Whether it is or not depends on the casual structure of the spacetime. How would you do that without gravity! Also the wiki definition is not very accurate, an event horizon is a connected component of the boundary of the casual past of the future null infinity. By the definition you've cited the future null cone in Minkowski spacetime would be an event horizon since events within it cannot influence events outside it. $\endgroup$
    – MBN
    Commented Mar 11, 2014 at 11:18
  • $\begingroup$ Nothing can stop me to declare such surface abstractly. If it's wrong, I won't be able to prove it (by construction or whatever). That's it. But, if base definition is wrong, then I won't say anything. :) $\endgroup$ Commented Mar 11, 2014 at 16:13
  • $\begingroup$ What I meant was that by calling it something doesn't make it that kind of a thing. $\endgroup$
    – MBN
    Commented Mar 11, 2014 at 17:45

1 Answer 1


I am not sure if I understand the question correctly, but from the way I understand it I would assume that the question itself is already ill posed.

in GR gravity is curvature of spacetime. So anything not involving Gravity would mean having a flat (Minkowski) spacetime.

And there are no event horizons in flat spacetime. (There can be coordinate singularities though, see the discussion in the comments.)

  • 1
    $\begingroup$ That last statement isn't true. The universe is (approximately) flat yet there is a horizon due to the accelerated expansion. Likewise an accelerating observer will see a horizon even in Minkowski spacetime - this is the Rindler horizon Alfred Centauri refers to in his comment. $\endgroup$ Commented Mar 11, 2014 at 9:59
  • $\begingroup$ @JohnRennie: The universe may be flat but the spacetime is not. And these horizons you mention are not event horizons. $\endgroup$
    – MBN
    Commented Mar 11, 2014 at 10:02
  • $\begingroup$ @MBN: it's true that it would require infinite time to form a true horizon for an accelerating observer, but then it would take an infinite time to form a true horizon round a black hole. In both cases observers would only see an apparent horizon. $\endgroup$ Commented Mar 11, 2014 at 10:17
  • $\begingroup$ @JohnRennie: Yes, but to me there is a fundamental difference between an event horizon and other types of horisons. One is a property of the spacetime, the other of an observer. $\endgroup$
    – MBN
    Commented Mar 11, 2014 at 10:24
  • $\begingroup$ @MBN: true, but I still don't like that last sentence (though not enough to downvote :-). $\endgroup$ Commented Mar 11, 2014 at 10:31

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