# If I sliced the universe in half, would the slice go through a star?

This question is based on a discussion with a 10-year old. So if it is not clear how to interpret certain details, imagine how a 10-year old would interpret them.

This 10-year old does not know about relativistic issues, so assume that we are living in a Newtonian universe.
In this model, our universe is homogenous and isotropic, with properties such as we see around us. Specifically, the density and size distribution of stars is what the current models say they are.

This universe has the same size as our observable universe, around 45 billion light years.

If we froze time, and took a plane through this universe, would this plane go through a star?

I cannot figure out if the chance of this happening is close to zero or close to one. I know that distances between stars are very big, so the plane is much more likely to be outside a star than inside a star, so my intuition wants to say that the chance is very small. But on the other hand, this plane will be very big... So based on that, my intuition says that the chance is close to one. I expect the chance to be one of these extremes, I would be very surprised if the chance were close to 50%...

Clearly, my intuition fails here. And I don't know how to approach this problem better (generating entire universes of stars and calculating if a plane intersects one of the stars takes too much time...).

Rough estimates are perfectly acceptable, I only want to know if the chance is close to zero or close to one!

Edit: Reading the comments/answers, I noticed that my reference to the 10-year old did not have the intended effect.

Some of the answers/comments focussed on how an answer to the title question could be explained to a 10-year old. That was not my question, and I was a bit surprised to see several people interpreting it that way. My question is the one summarized in the title.

And some of the comments were about the definition of observable universe, and that it necessarily would slice through earth because earth is in the center of our observable universe. I added the reference of the 10-year old to avoid such loopholes...

Rob Jeffries' and Accumulation's interpretation of the question was exactly what I meant, so their answers satisfied me.

• Are you asking the probability that a randomly chosen plane through the earth passes through at least one star? Mar 2 '18 at 8:14
• Comments are not for extended discussion; this conversation has been moved to chat. Mar 4 '18 at 3:33

There are about $10^{23}$ stars in the observable universe. Thanks to the expansion of the universe, those stars are currently spread over a sphere that is about $d=2.8\times 10^{10}$ parsecs across.

Of course some stars will have died whilst their light has been travelling towards us, but others will have been born, so I am going to ignore that complication.

If we imagine the stars uniformly spread through this volume$^{*}$, they have a number density of $n=3 \times 10^{-58}$ m$^{-3}$ (or $\sim 10^{-8}$ pc$^{-3}$). If we then define an average radius for a star $R$ we can ask how many stars lie within $R$ of a plane that goes through the Earth. The volume occupied by this slice is $2\pi d^2 R/4$ and the number of stars within that volume is $$N = \pi d^2 R n/2.$$

If $R \sim 1 R_{\odot}$ (many stars are much bigger, most stars are a bit smaller), then $N \sim 2\times 10^5$. So my surprising conclusion (to me anyway) is that many stars would be "sliced" by a plane going through the entire observable universe.

$*$ NB: Stars are not distributed uniformly - they are concentrated in galaxies and those galaxies are organised into groups, clusters and filamentary superstructures. However, on the largest scales the universe is rather homogeneous (see the cosmic microwave background) and so to first order the smaller-scale non-uniformity will not affect an estimate of the average total number of "sliced" stars across the observable universe, but may mean there is a larger variance in the answer than simple Poissonian statistics would suggest.

Could the clustering of stars affect the conclusion? It could if the clustering is strong enough that the median number of stars within $R$ of the plane becomes $<1$, but with the mean number unchanged. As an example consider an extreme bimodal model where all stars are found in galaxies of $N_*$ stars, where the average density is $n_*$. The "structure" of the universe could then be characterised by uniformly distributed galactic "cubes" of side $L = (N_*/n_*)^{1/3}$ and of voids with side $(n_*/n)^{1/3} L = (N_g/n)^{1/3}$. The number density of galaxies is the number of galaxies divided by the volume of the observable universe $n_g = (10^{23}/N_*)/(\pi d^3/6)$

The number of galaxies intersected by the plane will be $$N_g \sim \left(\frac{6\times 10^{23}}{\pi d^3 N_*}\right)\left(\frac{\pi d^2}{4}\right) L = 1.5 \times 10^{23} \left(\frac{L}{N_* d}\right)$$ and in each of those galaxies there will be $\sim L^2 R n_* = R N_*/L$ intersections with a star.

If we let $n_*= 0.1$ pc$^{-3}$ (the local stellar density in our Galaxy) and $N_* =10^{11}$ (the size of our Galaxy), then $L= 10^4$ pc, $N_g = 5\times 10^{5}$ and the number of stellar intersections per galaxy will be about 0.25. thus the average number of intersections will be about the same (by design) but the variance won't be much different either.

I think the only way density contrasts could give an appreciable chance of no intersection is if $N_g<1$, and thus $L/N_* < 2 \times 10^{-13}$ - i.e. if galaxies/structures contain lots more stars and are very dense so that there is a good chance that the plane will not intersect a single "galaxy". For example if $N_* = 10^{21}$ and $n_* = 10^3$ pc$^{-3}$, then $L= 10^6$ pc and $N_g \sim 0.05$. In this circumstance (which looks nothing like our universe) there is a high chance that the plane would not intersect one of the 100 big "galaxies", but if it did there would be about $10^7$ stellar intersections.

• Comments are not for extended discussion; this conversation has been moved to chat.
– rob
Feb 26 '18 at 16:09
• Interestingly, a single line will intersect approximately $N=\pi R^2 d n\approx 4\cdot10^{-13}$ stars, which falls squarely in the 'probably not' category. Apr 4 '19 at 13:14

As a rough approximation that is easy to try out with a child, you could try this:

1. Find or print out a big star chart or sky photo. Something like this.

2. Throw a long, narrow stick on top of it. See if it goes on top of any stars.

This will not be very accurate because not all stars are visible, and brightness hides the real sizes of the stars. But it should quite clearly demonstrate that the chances of a random plane hitting a star is pretty good.

--

Update: This seems to be rather popular answer. However I agree with the comments that the accuracy of this example is very poor, and it may actually be harmfully misleading. So it might be a good idea to follow it up with a discussion about its limitations, which, if nothing else, serves to illustrate the actual complexity in getting an accurate answer to the simple question.

Some points to consider:

• How large a paper would you need to accurately represent the size of the stars? A map of the whole sky would have to be about 1000 kilometers in size to have most of the visible stars be 1 millimeter in diameter. More distant stars would be ever smaller in perspective projection.

• How many stars are invisible? You can see about 5 000 stars with naked eye, but there are 10¹⁹ stars in the universe.

• How thick is the stick? Even a hair would be wider than a distant star, so ideally you would need an infinitely thin edge for accurate results.

And then the biggest uncertainty, which was already mentioned in the question:

• How big is the universe really? Limiting oneself to the observable universe is one possibility, but that's probably not the way the question was originally phrased.
• A very appropriate method for a 10-year old, although as you comment, the stars are (a lot?) smaller than they appear on a photograph which is of course offset by all the stars visible. It would be cool to know if this over or underestimation could be quantified. Feb 26 '18 at 19:42
• When I was somewhere between 5-7 years old, I remember getting the concept that the number of stars in the sky was uncountable, in the sense that if we could see all of them and started counting, we'd spend our whole life counting and still die before we finished. So I think it should be possible to explain to a 10 year old that there's a lot of stars that look so dim/small that we can't see them, and the sheer magnitude of how many there are. Feb 26 '18 at 22:28
• A good quick extension that can build educationally on this answer would be to then take the Hubble Ultra Deep Field images, and explain that (roughly speaking) every tiny spot on the night sky that we see with the naked eye as black actually contains about the amount of stuff that we can see in the HUDF pictures (the filaments vs. voids thing, and even galaxies vs inter-galactic space distinction, can be glossed over with a general "some spots have more, some have less", if the kid is unfamiliar with the difference already and you don't want to sidetrack). Feb 26 '18 at 22:32
• It's worth noting, though probably not worth worrying about, that one subtle (and I mean really subtle) natural human thinking fallacy that this way of teaching encourages is the idea that concepts that are intuitively true at n-dimensions are always intuitively generalizeable to n+1 dimensions. Intersection problems in particular do not always generalize to higher-dimensional systems, but I think in this case it's a reasonably step of intuition. Feb 26 '18 at 22:42
• Plus, what you see in that photo are mostly galaxies, not stars, and galaxies are mostly empty, and appear much larger/substantial due to how the camera picks it up than it would if you took points to scale. The exercise you propose would be so far in the inaccurate realm that it's worth skipping. Feb 27 '18 at 17:03

This is the Hubble Ultra-Deep Field, a long exposure photography taken by Hubble space telescope.

• It contains an estimated 10 000 galaxies. Each one of those contains an average of 100 billion stars.
• It shows a very small portion ($\frac{1}{13\ 000\ 000})$ of the whole sky. The diagonal is a tenth of a full moon's diameter.
• It was choosen because it has a low density of bright stars in the near-field. It looks completely black to the naked eye or to common telescopes.
• It looks very similar to other parts of the sky and the galaxies are very far away. The distribution would look the same anywhere else in the universe.

I couldn't manage to find a single line avoiding the galaxies. According to @RobJeffries' great answer, a line will cross an average of 100 galaxies on this picture alone, and will slice roughly 25 stars.

• The size of a star on a CCD image is to do with the optics and pixel size. The actual angular size of the stars is many orders of magnitude smaller than any line you could conceivably draw on the image Feb 26 '18 at 23:27
• @RobJeffries. That's no star. Feb 27 '18 at 4:56
• Even if the plane would slice through a million galaxies, that does not mean it would slice through a star. Galaxies are surprisingly empty!
– user78574
Feb 27 '18 at 7:50
• @Pakk: They're empty in 3D, yes, so your argument would surely be valid if we were talking about a line. I'm not sure galaxies still look mostly empty when projected onto a plane, though. Anyway, I find this picture simply jaw-dropping, I couldn't resist including it. :D Feb 27 '18 at 7:53
• Your point about the galaxies is fine. As I show in my answers you expect to interesect $\sim 10^5$ (large) galaxies along a great arc. The HUDF is 2.4 arcminutes across, so you might naively expect a line to intersect ~100 galaxies. However that doesn't answer the question (at all) until you work out how many stars are intersected by the line through each galaxy. That turns out to be 0.25 (see my calculation), so your image features roughly 25 stellar intersections. Feb 27 '18 at 9:34

This question is just a two dimensions version of Olber's paradox. In an infinite homogeneous universe, any line of sight ends in the surface of a star. Therefore, in such a universe, every line in a given plane goes through an star. In consequence, every plane in such universe slices a lot of stars.

• The question postulates a finite universe. Feb 26 '18 at 15:16
• @EmilioMBumachar You have a point, here.
– Pere
Feb 26 '18 at 15:48
• @Dronz I suppose there's an assumption that the stars are distributed randomly over the large scale. But one could imagine a universe where the stars are in a perfect lattice -- a plane between and parallel to the grid would not intersect any stars. Feb 26 '18 at 18:17
• @Edouard Yes, Emilio M Bumachar pointed it
– Pere
Feb 27 '18 at 19:06
• If the question involved a line rather than a plane, then I think the equivalence to Olbers' paradox would be correct. A classic analysis of Olbers' paradox is that if the universe were infinite, every line through the earth would intersect a star, so the night sky would be bright. It isn't, so the universe must be finite. You can tinker with this reasoning just slightly and conclude that since the night sky is not bright, a randomly chosen line seldom hits a star. A plane is just different from a line.
– user4552
Mar 2 '18 at 2:54

First of all, note that if something has a 1 in X chance of happening in each “try”, and there are Y “tries”, the formula exp(-Y/X) gives a rough estimate for the probability of it not happening. So, for instance, if you roll a die six times, the probability (to two decimal places) of not getting any ones is 33.49%, while the approximation I gave above yields 36.78%. As X and Y increase, this approximation gets better. If Y is significantly larger than X, then the probability is pretty much zero. So how many “tries” are there, and what’s the probability for each “try”?

Suppose we have a unit system in which the radius of stars is 1. (Note that my calculations are rough estimates, so I will not worry about the variation in such quantities as the radii of stars, and unlike @Rob Jeffries, I will not be keeping track of such “small” constants such $\pi$. In the calculations that follow, there are likely multiple points at which a mathematically minded person might find that I’m off by such a factor, but that shouldn’t alter the final answer.)

Now let’s say that the distance between stars is D, and the radius of the universe is U (again, these are measured in star radius units). Then each plane has U2 “tries” at getting a star, so our Y is U2, and our X is D3. So our probability is exp(-U2/D3). The radius of the universe is around 1011 light years, so if a star radius is about 10-4 light years, then U= 1015. If the distance between stars is, say, 101 light years, then D = 105. So that gives (1015)2/(105)3 = 1030/1015 = 1015. exp(10-15) is pretty much zero for practical purposes; it’s literally an astronomically small number.

• "Then each plane has $U^2$ “tries” at getting a star" - I don't follow, why?
– user78574
Feb 27 '18 at 20:04
• @Pakk Imagine splitting the universe into boxes the size of a star. The plane will cut through U<sup>2</sup> such boxes, and each box either has a star or doesn't. Feb 27 '18 at 20:19
• @RobJeffries "Star radius is orders of magnitude out. " Yes, I explicitly stated that my calculations are rough estimates. "U2/D3 is not dimensionless," Yes, it is. U is the radius of the universe divided by the radius of stars. D is the distance between stars divided by the radius of stars. I said that twice. Feb 28 '18 at 15:52
• Star radius is more than 3 orders of magnitude out, not even a rough approximation. $D=10^8$ and $U=10^{18}$. $U^2/D^3=10^{12}$. Still 7 orders of magnitude different to my estimate somewhere, probably because the average distance between stars in the universe is not 10 light years it is more like 1000 light years. Feb 28 '18 at 16:56
• I think this answer is underappreciated. @RobJeffries' answer is more explicit and has better estimates, but this answer takes a slightly different approach and arrives at the same conclusion ($D\approx 10^{10}$ and $U\approx 10^{18}$ are better approximations, to arrive at $exp(-U^2/D^3)=exp(-10^{6})$.) +1 from me.
– user78574
Mar 5 '18 at 14:00