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Imagine I pointed a laser into the night sky. How likely is it that a particular photon will ever hit anything?

This question bothered me for a while. I know very little about astronomy. When I asked my roommate he laughed and said that it must be very little, since the universe is mostly empty. I know it is true, but it seems to me that volume of the objects in the universe does not matter. Rather, it should be the area of objects when projected on a sphere that matters.

On one hand, universe is pretty empty, so hitting something outside the Solar System might be hard. But on the other hand, there are a lot of possible targets. These targets, however, appear smaller when the distance increases.

Which one of these, or other, forces will dominate? Would it make a difference if we ignored gravity?

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    $\begingroup$ I agree with @zephyr's answer. Also see Olber's Paradox. This question may also be a repeat of: astronomy.stackexchange.com/questions/11737 $\endgroup$
    – user21
    Commented Oct 31, 2017 at 17:23
  • $\begingroup$ I agree with the above. There are points of contact with Olber paradox treatment $\endgroup$
    – Alchimista
    Commented Nov 2, 2017 at 11:41

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I can't give you an exact number, but the chances your photon will hit anything within the next several billion years is exceedingly small. Nay, infinitesimally small. Almost no chance that it will happen.

What you're really asking, stated as a physics question is, what is the mean free path of a photon in the present universe?

If we consider an object (be it an electron, photon, or baseball) traveling through some medium (be it the Sun, space, or Earth's atmosphere), we can calculate how far, on average, it can travel before hitting something. This distance is known as the mean free path. You want to know how far a photon can travel through our universe before it hits something, in other words, what is the photon's mean free path?

The answer is that a photon's mean free path in our universe is larger than the observable universe. What that means is your photon can travel from Earth to the edge of the observable universe and still not hit anything (on average).

To support this fact, I point you to the Cosmic Microwave Background (CMB) radiation. This is radiation which was produced more than 13 billion years ago in the early universe and has been traveling through the universe since, almost completely unimpeded. In fact, if you tally up all the photons in the universe, you'll find that a very large fraction are CMB photons. This tells you that these photons, despite being produced 13 billion years ago and traveling through the universe since then, have never hit anything since and most are still traveling through space, unimpeded.

You are correct when you say that its the projected area of objects on our sky that matters, but you have to take into account the extremely minuscule projected area the matter in the rest of the universe has, primarily due to how far away it is.

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  • $\begingroup$ I recommend rewording the last paragraph to "... miniscule projected area of the matter in the rest of the universe...." rather than jsut "the universe" $\endgroup$
    – Carl Witthoft
    Commented Oct 31, 2017 at 18:25
  • $\begingroup$ I slightly object: the mean free path does not answer the question asked. It does not tell you the probability of collision. It only tells you the expected value (or: average over many trials). So, on average a given photon escapes the known universe, but that does not tell you the probability for a single one. But agreed overall that the probability would be in the "extremely low" category. $\endgroup$
    – jwd
    Commented Mar 28 at 16:37
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xkcd worked this out!

It was part of a related question: what if you pointed an infinite power laser in random directions, how much damage would happen? I'll try to summarize the parts of his analysis relevant to this question.

If you choose a truly random direction, you'd have about a 50% chance of hitting the Earth. For the times you don't hit Earth, 89,999 times out of 90,000 your beam will pass out of the Milky Way without hitting anything. The objects you have the highest chance of hitting are the Sun or Moon, but the odds of hitting either of them is about 1 in 180,000. He goes on to show that your odds of hitting one of Jupiter's moons is about 1 in 1 trillion. Hitting a star is even harder, "even if you aim for the core [of the Milky Way]."

He doesn't seem to have worked it out to the level of individual photons hitting individual hydrogen atoms, but given that space is mostly empty anyway the angular size method he uses should be a reasonable approximation.

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    $\begingroup$ Good find! This brings up an interesting point. In my answer, I assumed (and I think the question hinted it was intended) that the laser was purposefully pointed to avoid collision with nearby objects such as Earth, the Sun, or the Moon. But obviously, if you're just randomly picking a direction, half the time you'll hit the Earth! $\endgroup$
    – zephyr
    Commented Nov 2, 2017 at 13:22
  • $\begingroup$ Even though I believe the computations, I still find it surprising how hard it is to it the Moon. It looks pretty big! $\endgroup$
    – ElChorro
    Commented Nov 7, 2017 at 0:24

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