# What stellar number density would two galaxies have to have for another star to collide with the Sun during a galactic merger?

Note: This sprung from some annoyance I've had with the tag, and the goal of this question is to attract some really high-quality answers that any question using the tag deserves. An answer that meets the tough criteria of the tag will receive a bounty, and my gratitude.

The Milky Way and Andromeda will collide a few billion years in the future. Stellar collisions will be rare because - as Douglas Adams put it - "Space is big. Really, really big." In the galactic disk, the number density of stars is quite low. Chances are, the Solar System will not be ejected from the galaxy or collide with another star.

I'd like the star in my planetary system - similar to the Solar System, for all intents and purposes - in a spiral galaxy like the Milky Way to collide with another star in a spiral galaxy Andromeda, during the collision.

Obviously, there's no way for there to be a 100% chance of this happening. I'd settle for a 90% chance, give or take.

What stellar number density would both galaxies have to have?

Or, as Ayelis put it,

So basically, how many stars (of a typical stellar class distribution) would the Andromeda galaxy need to contain within its current bounds in order to pose a reasonable (90%+) threat of a stellar collision with our Sun when our galaxies collide?

I am aware, by the way, of this paper by Cox and Loeb.

Remember, this question has the tag. Make sure you understand what kind of answers are expected. I don't want to scare anyone off, but I really do want awesome answers here. Good work will absolutely be rewarded.

Note: Yes, I know that the resulting number will be quite large. However, it will be finite and calculable. Simply saying, "It's too large because [x,y,z reasons]" is not enough. The equations do not lie.

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

• This is along the lines of worldbuilding.stackexchange.com/questions/19000/…, FYI, in that it's designed to encourage great, well-thought-out answers. Like that, I'm happy to clarify anything that's unclear (or based on inaccurate notions). – HDE 226868 Jun 12 '15 at 15:27
• The thing is, you are conflating a chance of any 2 stars colliding between two galaxy's and a specific star colliding with one from another galaxy. It's like the chances of 2 people having the same birthday in a room of 30 is almost a guarantee, vs. some else in the same room having MY birthday. – bowlturner Jun 12 '15 at 15:42
• This is a probability calculation (as you likely know). The probability of at least one star collision during a typical galaxy merger is probably high but not 100%. However, to get the probability to 90% for a specific star (the Sun) colliding with another, would require ridiculous stellar densities. Putting it another way, in order to get the Sun up to a 90% chance of collision, this would have to be the average collision rate for the galaxy merger so 90% of the galaxy stars would have to collide with other stars. – Jim2B Jun 12 '15 at 17:11
• The best method of doing this calculation would be particle path approach (making sure that paths don't intersect) but I haven't done that sort of calculation since the 80's. To [hard-science] calculate this, you would need supercomputing time and a team of scientists to write the code. I'm guessing at best these stellar densities are rare and may not actually exist. Consider that our Sun passing through a 3 or 4 star system leads to >> 99% chance of missing all the stars. – Jim2B Jun 12 '15 at 17:14
• Then you'd use the mean free path calculation (en.wikipedia.org/wiki/…) and then plug in galaxy disk thickness. But even this provides an incorrect answer because it doesn't account for the attraction the "particles" (stars) have for each other. But as a first order analysis, this might work. – Jim2B Jun 12 '15 at 21:06

Well, this is easily answered, but first we need to get a few assumptions stated. First, let's assume that all stars are the same size as our sun, about 5 light-seconds across (I'll use ls often in the following, as a contraction for light-seconds). Second, we'll assume that our system will pass through the Andromeda pretty much on a diameter through the galactic disk, for a path length of 260,000 light years. Third, the Andromeda will be modeled as possessing a uniform density of N stars/cubic light-year. This is clearly not true, but we need to start somewhere. Finally, two stars will be said to collide if they pass within the Roche limit for the sun, 2.5 times the radius, or 6.25 ls.

The first thing to realize is that, at the current approach velocity, 110 kps, it's going to take a looong time to make the first pass through — about 700 million years. Second, the Milky way and Andromeda will continue to interact after they pass each other, eventually forming the Milkomeda galaxy, so looking at the first pass-thru is not nearly the whole story. Finally, the density of the Milky Way is irrelevant, since we are only dealing with the behavior of a single star.

So. Let's start by modeling the path through Andromeda as a series of volumes 1 light-year on a side. That will be 260,000 of these volumes. It is straightforward to determine the collision probability within any given volume. Actually, we start by determining the miss probability; that is, the probability that the Sun will miss all the stars in a volume, and raise that to the 260,000 power to get the probability that the Sun will miss all the stars in its path. As it happens $.9999911^{260000} = .0988$. Close enough. So the probability of collision in a single volume must be $1 - .9999911$, or $8.9 \times 10^{-6}$.

Now let's think about one target volume. Its area is 1 light year by 1 light year, or (if you do the seconds to years conversion) $9.94 \times 10^{14}\ {ls}^2$. The critical area around each star is $30.7 \ ls^2$. The miss probability is high enough that we can simply ignore the possibility of one star "hiding behind" another.

Then the number of stars required to block $8.9 \times 10^{-6}$ of the volume's area is given by

$$8.9 \times 10^{-6} = N \times 30.7 ls^2 / 9.94 \times 10^{14} ls^2$$

and $$N = 8.9 \times 10^{-6} \times 9.94 \times 10^{14} / 30.7$$

so $N = 2.9 \times 10^8$, or in rough numbers 300 million stars / cubic light year.

In other words, assuming uniform spacing, the stars in the Andromeda galaxy would need to be spaced at 800 light-minute intervals, assuming cubic packing. Another way to look at it is that an equivalent spacing closer to home would result in 2900 stars between us and Alpha Centauri. Or, even better, this density corresponds to packing the entire Milky Way into a volume 10 light years on a side.

Can you say, "supermassive black hole"? I knew you could.

That's not to say such an assembly of stars would constitute a black hole - only that it's just a matter of time.

• Nice. I think this seems to be pretty accurate. It does assume that the galaxies hit each other completely perpendicularly, but that's not a horrible assumption. – HDE 226868 Jun 12 '15 at 21:58
• No, it just assumes that the path of the Sun (or whatever the subject star is) does so. And, of course, any other assumption decreases the path length of the subject star within Andromeda, and requires an increased density to compensate. – WhatRoughBeast Jun 12 '15 at 22:01
• Ah, right, I visualized that incorrectly. – HDE 226868 Jun 12 '15 at 22:02
• I would think that gravity would increase the odds of a collision, although it is a bit hard to quantify that effect, particularly if the interaction is a slow one. – ohwilleke Nov 1 '16 at 4:42

It's not that much about density

I'm not an astrophysicist, I love reading about space and would love also to have a real awesome hard-science answer to this question, but while reading up on it this is what I came to conclude.

To answer this question properly (which I doubt I will entirely), you first need to determine two things. 1 - how star collisions happen, and 2 - how dense is your average galaxy.

1 - I'm basing this off of this page (which I think is plausible but does not make it hard proof) Can Stars Collide?. The probability no matter what the event of 2 stars directly crashing into one another is ridiculously low. If you consider the speeds at which stars travel through space, the vastness of space etc... Even if 10 stars were to pass through our solar system right now, the chances of any of them directly hitting the sun would be very slim. (That being said they would still completely mess up everything). What is much more likely to happen, is depending on the speed at which they meet the stars would either start orbiting themselves and eventually merge, or if meeting at ridiculously high speeds and passing close enough slingshot into one another and blow everything up. The latter however is unlikely to happen. Scholz's star passed through our outer Solar system a couple of thousands of years ago and its trajectory was only slightly perturbed (and it is much smaller than the sun).

2 - Galaxies are fancy to look at from afar and seem to house ridiculously huge amounts of stars (and look rather dense). Once again that's just because the human eye cannot even begin to represent the vastness of space. From Wikipedia I found that the density of stars in space surrounding the Sun (which is far out from the center of our Galaxy which is denser) is of one star per 284 cubic light years. (That's not a lot — Scholz's star passed within .8 ly of the Sun) The core is thought to have about 500 times that density (or 1 star per 1.75 cubic ly) and that's ignoring the fact that dust, gases, and the gigantic black hole at the center of it accounts for most of that mass. Still even if it were 1/1.75, again probabilities are still ridiculously low, although making huge amounts of merges and orbiting stars much more likely.

All of that being said. When Andromeda merges with the Milky Way (because if you've read all of the above I trust everyone understands there won't be any explosions on a galactic level) a few dozens of stars will most likely merge or collide with one another closer to the cores. The galaxy will change, tidal forces reshaping it, but with the relatively low density of stars all around, most of that will occur with very little effect on the stars of the outer parts themselves. Many more stars will likely collide and merge near the core in the aftermath but not that much more than whats already taking place anyway.

What would be needed for the Sun to be hit.

• First off if we merge into Andromeda near its core, that would obviously up the chances of the Sun merging/colliding with another star by a lot, yet were still talking about low probabilities and that probably won't happen because I assume the cores will attract one another. Seeing as we're far from the center that leaves us relatively untouched, but then who knows how the two galaxies will merge. Maybe Andromeda will be coming at an angle and its core will pass through the arms of the Milky Way before reaching our core.

• Second, if that were to happen, we'd either end up sucked into Andromeda's core (which I don't count as a collision, even though it's just as world-ending) or get trapped in it which means that sooner or later the sun WOULD merge or collide with another star but only simply as a result of being near the core of the new formed galaxy. So the only way it seems (to me) to have a high probability of hitting one of Andromeda's stars is to pass very near its core.

Most of everything I've written comes from searches on Wikipedia (which is not absolute truth or devoid of errors) and the links I posted. Hope it helps.

• . . . But collision rates, no matter how low, are related to density, right? I recently did a project on stellar collisions in globular clusters, and the formula for collision rate per units volume was given by $\gamma=\frac{1}{2}n^2\langle\sigma V\rangle$, where $n$ is number density. You would have to have ridiculously high densities for something like what I'm talking about to happen, but there would still be a finite density, right? – HDE 226868 Jun 12 '15 at 16:56
• Yes but I don't think a Galaxy exists that has a density of stars other than at it's core that will come even close to ensuring a collision. – Spacemonkey Jun 12 '15 at 16:59
• I understand that, but the density could still be determined. – HDE 226868 Jun 12 '15 at 17:00
• The way I see and understand it because of the way Galaxies are, a Galaxy with that density of stars would not possibly exist, there would always be a core sucking in stars from is outer reaches or sub cores forming in its outer reaches and then mutually attracting one another and we end up with one core again. Making the outer reaches always low in density of stars. What you seem to be talking about is more like passing through a hypothetical dense cloud of stars. In which case yes, that could have a finite density to ensure a 90%+ collision rate. – Spacemonkey Jun 12 '15 at 17:10
• Which means we need to find out how close a star has to pass to the sun for there to be a 90%+ chance of it colliding/merging into it. – Spacemonkey Jun 12 '15 at 17:11

The criteria can't be met.

Lets look at the problem a bit differently—lets cast a ray randomly (but on the correct angle of approach) through Andromeda and see if it hits something. The ray needs to be a somewhat thicker than the sun because gravity will take over and turn a near miss into a hit. To get your 90% chance of a hit 90% of such rays must intersect at least one star.

Now, lets take one of those random rays and put an observer at its midpoint (relative to its passage through Andromeda.) He looks along the ray—there's a 69% chance he sees a star looking either way. I can think of no path for the ray that doesn't produce the same effect anywhere on a plane that contains the ray—thus anywhere I look along that plane I have at least a 69% chance of finding a star. If this is the plane of maximum density we would find fewer stars as we looked above or below it but galaxies aren't paper-thin, there will still be quite a few stars as you look some degrees away from this optimum.

Now, the sun occupies about 5 millionths of the sky and yet keeps us warm. What's going to happen when you have a band several degrees wide that's 2/3 stellar surface (and it tapers off beyond that, not an abrupt ending)? You will have an incredible amount of absorbed heat. Our observer is hypothetical and doesn't matter but the stars around certainly do—they're going to be substantially heated by the glow from other stars. That's not going to be one bit good for their lifespan. I don't think there's going to be a galaxy left by the time it gets here if it's that dense.

Note, also, that this math assumes the ray is on the path of maximum density. Since we see a glorious spiral in the sky the actual path is much closer to the one of minimum density—thus most spots in the sky will be star to our observer. With no ability shed heat stars would soon be destroyed. Even without any speedup to their life consider the sun: Energy output: 1.2E34 J/year. Binding energy: 6.9E41 J. It's going to produce enough energy to disassemble itself in 57 million years.

• Interesting. Where did you get the numbers from? – HDE 226868 Jun 12 '15 at 21:00
• Yup, and the densest star clusters only go up to like 1.5-2 / cubic ly which still isn't enough because stellar dynamics will make sure they keep mostly to their initial trajectory when passing through the Sun's neighborhood which has like 0.003/ cubic ly and 'history' has already proven that a star passing within 0.5 ly from the sun won't affect it much. – Spacemonkey Jun 12 '15 at 21:01
• @HDE226868 the 69% is basic math as is the 57 million years. The energy numbers are from Wikipedia (they are both on a table of ener, the sky are of the sun I found in several places, always as steradians, I ended up tracking down the conversion factor. – Loren Pechtel Jun 12 '15 at 21:50
• @LorenPechtel I can't quite follow the derivation from 90% to 69%, nor where in the world you get 57 million years. I also don't know which Wikipedia article you're talking about. :-) – HDE 226868 Jun 12 '15 at 21:52
• @HDE226868 Two events with a 69% probability each have a combined probability of at least one happening of 90%. The 57 million is simple division of the two numbers I gave. – Loren Pechtel Jun 12 '15 at 22:03