What is the relationship between an event happening and “seeing” an event happen?

In relativity, I often hear that an event is in our "future" or "past" because its light cannot reach us.

What is the relationship between seeing something happen and something happening? For instance, we never see a monkey fall into a black hole, so it must never have happened. Isn't this misleading?

As a related question, I think, we should not see the monkey permanently etched on the event horizon. The fact that we are "seeing" the monkey (extremely red-shifted) implies that the image is losing photons and sooner or later, there won't be any more photons representing the monkey's image.

• (1) In Special Relativity, simultaneity is relative, so we must synchronize clocks. However in General Relativity simultaneity is local. Generally there is no way to synchronize clocks. We "see" the monkey at the event horizon forever, but the monkey sees itself falling through with no delay. (2) You are correct. We can only see the photons reflected by the monkey in its frame before the monkey passes the event horizon. After that last photon we see nothing. In our frame, the monkey is still on this side forever, but it's time is just too slow for any photons to get reflected. – safesphere Dec 5 '17 at 16:43

Your first question is one of philosophy, not of physics. Of course things can happen without us seeing them; physics tells you what you will see.

For your second question, your misunderstanding is in how time dilation works. The photons that you see at two different moments in time are not part of the same image; they were emitted by/reflected off of the object when it was at two different positions relative to the event horizon. The photons from when it was at one position take longer to reach you than photons from another position; this can be true even in Euclidean space.

But near a black hole, space is more and more intensely curved; general relativity states that the more extreme the curvature of space, the slower time will appear to move to an outside observer. Because of this, the time it will be perceived to take to fall through the event horizon will be infinite to an outside observer. So at any finite time, the observer will be able to see some of the photons from the object when it was a certain distance to the event horizon. If the observer waits a while, then looks again, he/she/it will see photons from when the object was closer to the event horizon.

EDIT:

We have to be a bit careful in what exactly we mean by "you will see the object forever." The object, after all, emits a finite number of photons before crossing the event horizon, so there should in every case be some finite time at which the object emits its very last photon. Now we can ask the question: When, on average, does this occur?

Let's assume that we have an object falling into a Schwarzschild black hole, so that the math will be as easy as possible. In the object's frame of reference, there is a probability density $\rho$ for photon emission that is independent of proper time $\tau$, so we can write the probability $P$ of the object emitting a photon in a given proper time interval $d\tau$ as $$P=\rho d\tau$$.

The Schwarzschild metric, with $G=c=1$, is defined as

$$d\tau^2=\left(1-\frac{2M}{r}\right)dt^2-\left(1-\frac{2M}{r}\right)^{-1}dr^2-r^2(d\theta^2+\sin^2\theta d\phi^2)$$

Let's assume the object is falling straight in, so that $d\theta=d\phi=0$ and we can ignore the last term of the metric. The remaining terms can be combined with the basic tenets of general relativity to yield

$$\frac{d\tau}{dt}=\frac{m}{E}\left(1-\frac{2M}{r}\right)$$ $$\frac{dr}{dt}=-\left(1-\frac{2M}{r}\right)\sqrt{\frac{2M}{r}}$$

where $m$ is the mass of the infalling object and $E$ is the total energy of the system, both of which are constant here.

Now we can find the probability that a faraway observer will see a photon emitted in a given time interval $dt$ as measured by the observer. Using the fact that $d\tau=\frac{d\tau}{dt}dt$, we can write

$$P=\frac{\rho m}{E}\left(1-\frac{2M}{r}\right)dt$$

so we can define a probability density per observer's unit time $p(t)$ as

$$p(t)=\frac{\rho m}{E}\left(1-\frac{2M}{r(t)}\right)$$

We can obtain $r(t)$ from the above ODE for $\frac{dr}{dt}$. While there is no elementary solution for this ODE, it is reasonably easy to numerically integrate via Mathematica or your favorite software. The expectation value of the time at which you will see a photon emitted is

$$\langle t\rangle=\int tp(t)dt$$

Using Mathematica, it is easy to see that this expected value diverges (you can play around to convince yourself here: https://sandbox.open.wolframcloud.com/app/objects/29cae95d-423d-444b-88ad-d5fdaf316673). Therefore, on average, you will always expect to see another photon, and we can reasonably expect that the object persists forever, on average.

• "They" refers to the photons, not to the object. – probably_someone Dec 5 '17 at 16:41
• @safesphere Oh, give me a break. "They" has been used as singular shorthand for "he/she/it" since the days of Shakespeare (en.wikipedia.org/wiki/Singular_they). But whatever, I'll edit anyway, since you seem to think that it's wrong. – probably_someone Dec 5 '17 at 16:50
• @safesphere The observer is far away from the event horizon. – probably_someone Dec 5 '17 at 17:05
• It seems to me that "simultaneous" is defined in terms of "seeing" something. – Jus12 Dec 5 '17 at 17:12
• @safesphere I was not aware that objects at finite temperature did not emit a blackbody spectrum of photons. Since the OP does not assume a monkey at absolute zero, the object will continue to emit photons arbitrarily close to the event horizon. – probably_someone Dec 5 '17 at 17:55

The key point is that within the context of relativity, there is no clear ordering in time of events that are outside each other's lightcone. Which event is first will depend on the frame of reference of the observer. This effect is known as relativity of simultaneity. The only situation in which we can unambiguously say that an event has happened, is if the event is on our causal past, i.e. our past lightcone. This has nothing with being able to see it, but everything the with the ambiguity of time-ordering in relativity.

• The second answer based on misunderstanding the question. Also, why do you say that the monkey is outside of our light cone? We can see the monkey, so it is in our light cone. Finally, relativity of simultaneity is a concept from Special Relativity that is not applicable in General Relativity where simultaneity generally is not relative, but local. – safesphere Dec 5 '17 at 16:54
• Can we not rephrase it as "the only situation in which we can unambiguously say that an event has happened, is if we can in theory see that event"? Also, if there is no universal "now", how did the two ends of our observable universe reach the same temperature? This may mean that there is a universal now, just not defined via "seeing". – Jus12 Dec 5 '17 at 17:21
• @Jus12 Yes, you can say this, but the opposite is not true, of course. For example, if the monkey falls, we see it frozen, but its mass would increase the Hawking radiation, so, if Hawking is right, we could tell the monkey is already inside while we still see it outside. Although "already" is not a well defined term in General Relativity in this case. On your other question, the "universal now" (comiving time, see Wiki) is defined globally (asymptotically), but not necessarily locally. For example, we can define it between the Earth and a planet on the other side, but not around a black hole. – safesphere Dec 5 '17 at 19:07

It might be helpful to think of terms of 4 dimensions rather than 3 + 1. In spacetime there are world lines, objects on them have different 4 velocities at different points on thier worldline, but they dont "traverse" thier worldline, it simply is. In this model light (and other massless stuff) has a "line" that consists of a point, ie the length of the worldline is zero. If you can get to that point you might interact with it, but there are lots of other events at that point and it may interact with one of those events instead. Some events are a negative distance away and your worldline wont intersect that point, pretty straightforward.