# Two objects are thrown into a black hole. The first crosses the event horizon at time's end, so when does the second one cross?

An observer throws an object towards a black hole, and then an arbitrary amount of time later, throws a second object towards the black hole. Disregarding Hawking radiation and assuming the Black Hole will last forever, it will take an infinite amount of time from the point of the view of the observer for the first object to reach the event horizon. In other words, the first object crosses the event horizon right at the very end of time. But if this is true, then when does the second object reach the event horizon? After the first object reaches the event horizon, there is theoretically no more time that can be elapsed, yet we know that the second object must arrive at the event horizon after the first object.

Update: Perhaps it would be better to rephrase some of the above. Just for clarification, I am not thinking of infinity as a number, but more like indexes in infinite set theory. Infinity is not a number, but there is a concept denoting the last index in an infinite set, omega. In this case, the state of the object corresponds to an index of the time set, and the state where the object crosses the event horizon is defined to correspond to an index of omega. Two sets with last indices omega and omega + 2 have the same cardinality, but are just indexed differently. My question was that the state of the first object when it passes the horizon corresponds to index omega, and the index corresponding to the event when the second object crosses the event horizon must come after omega. However, time is defined with a last index of omega, so my question is: what is the index CORRESPONDING to (not equal to, as with numbers) the event when the second object crosses? Ie, if an index of omega is ascribed to the event when the first object reaches the event horizon, what index do we ascribe to the event when the second object reaches the event horizon?

• Will time ever end? – QuIcKmAtHs Jan 14 '18 at 5:00
• I was only commenting based on my limited understanding of quantum mechanics – QuIcKmAtHs Jan 14 '18 at 6:17
• Depends on your definition. You can define the end of time as the point in time when an object reaches the event horizon of a black hole. – Zaif Jan 14 '18 at 6:21
• @Zaif, it does not take until the "very end of time" to reach the black hole - a spaceship would be crushed and incinerated promptly whilst all fundamental activity slowed to a crawl (so for an observer going into the black hole, the crushing and incinerating would appear to happen increasingly quickly relative to their sense of time, as would activity in the rest of the universe which would become overpoweringly fast and bright). For the remote observer, local activity aboard the object would appear to slow and approach stasis, whilst the image faded to red and then eventually to black. – Steve Jan 14 '18 at 16:35
• @Steve right, but I was asking about it from a purely mathematical point of view, disregarding Hawking radiation, red-shift, and assuming the objects, the blackhole, and the observer could last forever. – Zaif Jan 14 '18 at 20:24

## 5 Answers

(It's important to note that this is, of course, talking about an idealized mathematical setup - in real life a real black hole does not last for infinite time and it's not certain the universe will either.)

Both reach it at infinite time. It's for the same reason the graphs of both $f(x) = \frac{1}{x + 1}$ and $g(x) = \frac{1}{x + 2}$ have limit 0 at infinite $x$ despite starting at different points when $x = 0$.

ADD (2018-01-16): The other answer here mentions how that "infinity" is "not a number", and "should not be regarded as one". I'd say this depends on your point of view. Whether you call "infinity" a "number" or not depends on what objects you choose to admit under the label of "number" (and also, what objects you choose to label with the word "infinity") which is, admittedly something that is rather not admitting of a precise formal definition (that is, there is no precise formal mathematical definition of what is a "number", except given that some kinds of mathematical objects are called as such.). The relevant concept of "infinity" here is that of the "extended real number line" - I note the original questioner mentioned something about infinities in his post saying they were like infinite set cardinalities. This is not correct - the relevant notion is the "infinity" as used in calculus which is formally a member of this extended set, and a suitably continuous function can be extended to it by taking the limit.

If one objects to this formalism (though I don't see why one should - it is perfectly sensible as long as one plays by the rules that govern it, which is required for all mathematics), instead of saying "that it is 'reached at infinite time'" one can say both "approach arbitrarily close to the horizon at suitably large times", or that "both reach the horizon in the limit of arbitrarily large times". In any case a limit is required because the relevant functions are not defined directly at $t = \infty$ (on the ERL) - it is much the same case as for a "removable singularity" like of $f(x) = x^0$.

Now as for the physical world, it is not something we can empirically test or confirm whether or not time goes on arbitrarily far, much less if the terminal point $t = \infty$ that would be added in an extended real number setup for convenience actually existed. It is rather a feature of our models, and admittedly quite idealized ones at that, that can be left off without affecting any prediction that is actually testable. (Indeed one could argue that any claim as to a putative "end of time" is not empirically testable at all because once it hits, we cease to exist and thus cannot register it as a truth.) In reality, a real black hole is almost certainly limited by the Hawking radiative evaporation which cuts off its lifespan at a very high but finite time (about $10^{100}$ s for the biggest black holes, versus the age of the current Universe at about 435 Ps ($435 \times 10^{15}$ s) or 13.8 Ga.). The rule for the time lapse in that case is that any object in free-fall will just reach the horizon at the instant the black hole vanishes. Thus also, both objects reach the horizon at the same time, only now it is at a finite and mathematically uncontroversial "number".

• As interesting as the question was, this is a really simple answer to it. Thanks! – Zaif Jan 16 '18 at 22:34

It is tempting to regard infinity as a number,i.e. there is a time $t=\infty$, but infinity is not a number. Instead it is a limit, and it is a limit that can never be reached. If we graph the radial distance of our infalling object against time we'd get something like:

But the $t$ axis never ends and the red line never meets it. So there is no end of time that you can label $\infty$ and the infalling object never meets the event horizon.

This is the problem with your question. It makes no sense to ask how much later the second object meets the horizon because neither object ever meets the horizon.

Admittedly physicists have a habit of putting infinity on their spacetime diagrams - these are known as Penrose diagrams - but it is understood that this shows a limit not an actual time or distance.

• Sorry, I wasn't thinking about infinity as a number, more like indexes in infinite set theory. Infinity is not a number, but there is a concept denoting the last index in an infinite set, omega. In this case, the state of the object corresponds to an index of the time set, and the state where the object crosses the event horizon corresponds to the index of omega. Two sets with last indices omega and omega + 2 have the same cardinality, but are just indexed differently. My question was that the state of object1 when it passes the horizon corresponds to index omega, and the index... – Zaif Jan 14 '18 at 20:15
• ...corresponding to the event when the second object crosses the event horizon must come after omega. However, time is defined with a last index of omega, so what is the index corresponding to the event when the second object crosses? – Zaif Jan 14 '18 at 20:18

Contrary to the current votes, this is an excellent question!

First of all read the answer of John Rennie, his reserves are perfectly justified, and any answer daring to go beyond his answer must be considered with very much caution.

Secondly, your question and the current answers are based on the point of view of an observer outside the black hole. You have to distinguish a) the perception of the observer (what he could see) and b) the spacetime diagram of an observer, which is not the same (in his spacetime diagram you can read what is simultaneous for the observer etc.). It is important to notice that the phenomenon is the same in both points of view: a) the observer observes an infalling object to be remaining eternally outside of the event horizon, and equally b) according to his relative concept of simultaneity (spacetime diagram) this applies to all infalling objects.

So we could close the file saying that both objects are never reaching the event horizon. However, we should be aware of the fact that this answer is not as free of doubt as it seems: We know that infalling observers are not perceiving the infinite time in front of the event horizon. From their point of view they are reaching (and crossing?) the event horizon within finite time. However, their spacetime diagram (see above b)) would tell them that simultaneously with their crossing of the event horizon, the outside of the event horizon has come to its end.

It is curious to notice that all other observers of the universe (and all other objects and particles if we consider them as observers) are experiencing the same thing: all objects of the universe are reaching all event horizons of the universe simultaneously with the end of time (according to their respective relative spacetime diagram, see above b)).

So, as a result, we could say: an infinite time is not a mathematically defined moment. But we should be aware of the fact that infinity could play a certain role within the logic of the universe.

• How can anyone go beyond JohnRennie's answer? His answer consists (correctly) of pointing out that the OP's question is misconceived! Just a small point about your answer though, the "event horizon" only works one way - nothing can communicate out, but the world outside does not "come to its end" and can still communicate in (that is logically necessary, for how otherwise would the object itself get into the hole!). – Steve Jan 14 '18 at 21:10

In other words, the first object crosses the event horizon right at the very end of time.

Assuming the Schwarzschild black hole solution, the Schwarzschild coordinates $r,t$ do not map the whole of the event horizon, i.e., a single event (suppressing the angular coordinates for simplicity) is mapped to $r = 2M, -\infty \lt t \lt \infty$.

So it isn't the case that a test particle in this geometry crosses the horizon at the "very end of time". In the Schwarzschild coordinates, the coordinates of 'the observer at $r = \infty$', there is no value of the time coordinate $t$ that maps to the event of the particle crossing the horizon.

The idea of "at infinite time" here is an artifact of using Schwarzschild coordinates to describe the black hole. The Schwarzschild coordinates are convenient for many purposes -- first and foremost that they are stationary outside the black hole -- but they only describe a proper subset of the entire ideal spacetime of the black-hole solution to the GR equations.

As a simple analogy, imagine that we had some good reason to describe points in the plane not by ordinary coordinates but by the logarithms of those coordinates. This description could make us think all there exists is the first quadrant of the space -- a point that moves towards the $x$- or $y$-axis will see one of its coordinates go towards negative infinity. So by the same reasoning as in your question, we could say that we cannot possibly cross those axes because we're going to "run out of numbers" first.

We can re-coordinatize the Schwarzschild such that a particle that approaches the event horizon doen't run out of numbers. Kruskal–Szekeres coordinates is one such recoordinatization, where the outer Schwarzschild space fits as an open subset.

In Kruskal–Szekeres coordinates we can see that two particles that fall into the black hole from the same direction, but starting at the same time, will actually cross the event horizon at different events that are connected by a light-like curve (i.e. a null geodesic). In this coordinate system the crossing events have ordinary finite numbers as coordinates. Generally the particles will then continue along separate paths through the inner spacetime and meet the singularity at different points.

## protected by Qmechanic♦Jan 14 '18 at 21:08

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