<p>Suppose civilizations are randomly distributed in the galaxy (not necessarily evenly, but independent of each other). That means that there is a CDF <span class="math-container">$\Pr[d&lt;x]=F(x)$</span>, the probability for the distance <span class="math-container">$d$</span> to one of them being less than <span class="math-container">$x$</span>. If there are <span class="math-container">$N$</span> civilizations, then the probability that <em>all</em> of them are inside distance <span class="math-container">$x$</span> is <span class="math-container">$\Pr[d_1,d_2,\ldots d_N&lt;x]=F(x)^N$</span>. So if you know <span class="math-container">$F(x)$</span> you can easily find the median distance for a random civ by solving <span class="math-container">$F(x)=1/2$</span>, and the median distance to the remotest by solving <span class="math-container">$F(x)=2^{-1/N}$</span>. You can calculate the mean too <a href="https://stats.stackexchange.com/questions/10159/find-expected-value-using-cdf">by cleverly integrating the CDF</a>. Note that the same trick can be used to find the distribution of the closest civilization distance: it is <span class="math-container">$\Pr[d_1,d_2,\ldots d_N&gt;x]=1-(1-F(x))^N$</span>.</p>
<p>The above is nice and general, but not exactly what the question is. In fact, trying to reverse engineer the calculator formula is likely wasted effort: it doesn't make much sense (more civilizations should make it give a more remote remotest civ).</p>
<p>My simplest approach would be to say that civilizations are uniformly distributed in a spherical volume corresponding to the milky way (of radius <span class="math-container">$R$</span>) with Earth in the center (yes, this is wrong but not entirely wrong). Then their distance CDF will be <span class="math-container">$F(x) = (x/R)^3$</span> up to <span class="math-container">$x=R$</span>, and 1 beyond. So I get a CDF for the remotest that is <span class="math-container">$(x/R)^{3N}$</span> and a median distance <span class="math-container">$x=R \cdot 2^{-1/3N}$</span>. This behaves right: as you get more and more civs, the outermost will be almost exactly at the edge <span class="math-container">$x=R$</span>.</p>
<p><a href="https://i.sstatic.net/13yKW.png" rel="noreferrer"><img src="https://i.sstatic.net/13yKW.png" alt="enter image description here" /></a></p>
<p>A better model is to assume the Milky Way is a cylinder, and that the sun is in an offset position from the center. Unfortunately I don't know enough stochastic geometry to get <span class="math-container">$F(x)$</span> in a closed form. Basically, <span class="math-container">$F(x)=$</span>[galactic volume inside x]/[total galactic volume]. This can be solved numerically. But if you are solving things numerically you can in principle go full hog and use realistic star densities or even galactic habitable zones.</p>
<p>Addendum: I couldn't resist actually doing it. I took an approximate stellar density model (from Carrol &amp; Ostlie) that I assumed is equivalent to a civilization probability density, did a numerical integration in Matlab and calculated the approximate distance.
<a href="https://i.sstatic.net/0I57d.png" rel="noreferrer"><img src="https://i.sstatic.net/0I57d.png" alt="enter image description here" /></a></p>
<p>This doesn't give an explicit formula, but shows the overall behavior. <span class="math-container">$F(x)$</span> is basically a sigmoid with a turning point near the distance to the galactic center. The furthest distance grows slowly since there are rare stars far away from the galactic plane.</p>
<p>Adding a ring-shaped galactic habitable zone (not shown) makes it more of a linear rise, producing a much flatter distance distribution (furthest civilization will be at the far end of the ring).</p>

CC BY-SA 4.0
2022-04-30T11:30:08.360Z
2022-04-30T12:26:58.310Z
2022-04-30T12:26:58.310Z
Anders Sandberg
<p>I'm planning on making a video related to Fermi Paradox, and I've used <a href="https://www.omnicalculator.com/physics/alien-civilization" rel="nofollow noreferrer">Omni Calculator's Alien Civilization calculator</a> as a reference. I noticed they have a field that calculates the &quot;maximum&quot; distance to an alien civilization based on the amount of civilization in the Milky Way.</p>
<p>I've attempted to recreate the equation used for calculating this as their calculator doesn't explain how this is calculated, but my number doesn't exactly match theirs. I used the following equation:</p>
<p><span class="math-container">$$d_{max} = \sqrt[3]{\frac{6\left(\frac{1}{N_{civilizations}}V_{galaxy}\right)}{\pi}}$$</span></p>
<p>What it does is take the number of civilizations in the galaxy, then find how large a sphere volume one civilization would take up assuming they were evenly distributed, and then finds the diameter of such a sphere, and that diameter is the maximum distance to an alien civilization. So, by definition the Earth would be at the center of our volume, and to get to the next civilization you would have to leave our volume, so the distance to the next volume is the radius of the volume, and to get to the planet at the center of the next volume you'd have to cross another radius of distance, meaning the total distance is the diameter of the volume. To be clear, I'm not sure if this equation even calculates a meaningful number even assuming the civilizations are evenly distributed, which they are not.</p>
<p>So, if I take the number of civilizations in a galaxy as 928, and the volume of the Milky Way as 8.333 trillion cubic light years:</p>
<p><span class="math-container">$$d_{max} = \sqrt[3]{\frac{6\left(\frac{1}{928}\times8.333\times10^9\hspace{1mm}\mathrm{ly^3}\right)}{\pi}} = 2579\hspace{1mm}\mathrm{ly}$$</span></p>
<p>the result of 2579 light years is about 1.29 times lower than the Omni Calculator's result of 3320 light years.</p>
<p>Does this equation even work assuming the civilizations are evenly distributed across the galaxy, and does anyone have a better equation for calculating the maximum distance to an alien civilization?</p>
<p><strong>EDIT 1:</strong> I might have found the equation in the scientific paper
The Astrobiological Copernican Weak and Strong Limits for Intelligent Life, but I'll have to read it thoroughly to actually understand it, so I'll update here when I do that.</p>
<p><strong>EDIT 2:</strong> I believe I have found the exact equation used in the scientific paper <a href="https://arxiv.org/abs/2004.03968" rel="nofollow noreferrer">The Astrobiological Copernican Weak and Strong Limits for Intelligent Life</a>, and on the Omni Calculator's page, the equation is (I slightly modified it to add the civilization volume, which was in the paper a separate equation) as follows:
<span class="math-container">$$d_{max} = 2\times\sqrt{\frac{\frac{1}{N_{civilizations}}V_{galaxy}}{0.3 kpc\times\pi}}$$</span></p>
<p>where <strong>N</strong> is the number of civilizations in the galaxy (you can derive that number either from <a href="https://en.wikipedia.org/wiki/Drake_equation" rel="nofollow noreferrer">the Drake Equation</a> or from the equation used in the mentioned paper), the <strong>V</strong> is the volume of a galaxy (Milky Way's is approximately 8 trillion cubic light-years) and <strong>kpc</strong> is the distance unit (1 kpc = 3261.56 light years).</p>
<p>The results match almost 1:1 to Omni Calculator's. I believe the very slight difference is due to the different galaxy volumes used. I used 8 trillion cubic light-years when calculating, and Omni Calculator might have used a more specific number), or potentially due to Omni Calculator more precisely calculating the civilization volume based on the given parameters (the latter is likely correct).</p>
<p><strong>Note:</strong> The number given by the equation has a very large uncertainty.</p>

CC BY-SA 4.0
2022-04-29T23:09:29.380Z
How is that the 'maximum distance'? That's the 'average' distance to the six closest neighbours or something around that.
planetmaker
CC BY-SA 4.0
2022-04-29T23:32:16.880Z
@planetmaker No, it isn't. This equation gets you a diameter of a sphere with the volume in the brackets. That volume is the volume one civilization would occupy if they were evenly distributed.
OptimusePrime
CC BY-SA 4.0
2022-04-29T23:33:52.553Z
exactly. And that certainly is not "the maximum distance.", but the average or median or something like that. The maximum distance is the diameter of the Galaxy: all civilizations except one in one globular cluster and one in another the whole Galaxy diameter away. If you found a reference for the equation, please link it.
planetmaker
CC BY-SA 4.0
2022-04-29T23:49:50.247Z
@planetmaker Well yes it's something like that because the equation assumes the civilizations are evenly distributed across the galaxy. I'll link the reference once I actually find the equation within the scientific paper because it has something like 45 pages.
OptimusePrime
CC BY-SA 4.0
2022-04-29T23:51:56.853Z
Maybe the paper says something about why they use it?
planetmaker
CC BY-SA 4.0
2022-04-29T23:54:42.727Z
@planetmaker Not sure. I will have to read it tomorrow, and I'll update the question then. Although it likely does.
OptimusePrime
CC BY-SA 4.0
2022-04-29T23:59:07.933Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/135934/discussion-between-optimuseprime-and-planetmaker).
OptimusePrime
CC BY-SA 4.0
2022-04-30T11:36:46.280Z
@OptimusePrime I have converted your images of equations to [MathJax](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference), please check that I have done so correctly.
GrapefruitIsAwesome
CC BY-SA 4.0
2022-04-30T11:53:21.167Z
@GrapefruitIsAwesome I checked, and the equations are correct.
OptimusePrime
CC BY-SA 4.0
2022-04-30T12:19:59.567Z
@OptimusePrime Thanks!
GrapefruitIsAwesome
<galaxy><life><extra-terrestrial><astrobiology>
Calculating "maximum" distance to an alien civilization
OptimusePrime
# Calculating "maximum" distance to an alien civilization

(Asked by: username_0 on 2022-04-29)

I'm planning on making a video related to Fermi Paradox, and I've used [Omni Calculator's Alien Civilization calculator](https://www.omnicalculator.com/physics/alien-civilization) as a reference. I noticed they have a field that calculates the "maximum" distance to an alien civilization based on the amount of civilization in the Milky Way.

I've attempted to recreate the equation used for calculating this as their calculator doesn't explain how this is calculated, but my number doesn't exactly match theirs. I used the following equation:

$$d\_{max} = \sqrt[3]{\frac{6\left(\frac{1}{N\_{civilizations}}V\_{galaxy}\right)}{\pi}}$$

What it does is take the number of civilizations in the galaxy, then find how large a sphere volume one civilization would take up assuming they were evenly distributed, and then finds the diameter of such a sphere, and that diameter is the maximum distance to an alien civilization. So, by definition the Earth would be at the center of our volume, and to get to the next civilization you would have to leave our volume, so the distance to the next volume is the radius of the volume, and to get to the planet at the center of the next volume you'd have to cross another radius of distance, meaning the total distance is the diameter of the volume. To be clear, I'm not sure if this equation even calculates a meaningful number even assuming the civilizations are evenly distributed, which they are not.

So, if I take the number of civilizations in a galaxy as 928, and the volume of the Milky Way as 8.333 trillion cubic light years:

$$d\_{max} = \sqrt[3]{\frac{6\left(\frac{1}{928}\times8.333\times10^9\hspace{1mm}\mathrm{ly^3}\right)}{\pi}} = 2579\hspace{1mm}\mathrm{ly}$$

the result of 2579 light years is about 1.29 times lower than the Omni Calculator's result of 3320 light years.

Does this equation even work assuming the civilizations are evenly distributed across the galaxy, and does anyone have a better equation for calculating the maximum distance to an alien civilization?

**EDIT 1:** I might have found the equation in the scientific paper
The Astrobiological Copernican Weak and Strong Limits for Intelligent Life, but I'll have to read it thoroughly to actually understand it, so I'll update here when I do that.

**EDIT 2:** I believe I have found the exact equation used in the scientific paper [The Astrobiological Copernican Weak and Strong Limits for Intelligent Life](https://arxiv.org/abs/2004.03968), and on the Omni Calculator's page, the equation is (I slightly modified it to add the civilization volume, which was in the paper a separate equation) as follows:
$$d\_{max} = 2\times\sqrt{\frac{\frac{1}{N\_{civilizations}}V\_{galaxy}}{0.3 kpc\times\pi}}$$

where **N** is the number of civilizations in the galaxy (you can derive that number either from [the Drake Equation](https://en.wikipedia.org/wiki/Drake_equation) or from the equation used in the mentioned paper), the **V** is the volume of a galaxy (Milky Way's is approximately 8 trillion cubic light-years) and **kpc** is the distance unit (1 kpc = 3261.56 light years).

The results match almost 1:1 to Omni Calculator's. I believe the very slight difference is due to the different galaxy volumes used. I used 8 trillion cubic light-years when calculating, and Omni Calculator might have used a more specific number), or potentially due to Omni Calculator more precisely calculating the civilization volume based on the given parameters (the latter is likely correct).

**Note:** The number given by the equation has a very large uncertainty.

**Tags:** galaxy, life, extra-terrestrial, astrobiology

> **Comments:**
> **username_1**: How is that the 'maximum distance'? That's the 'average' distance to the six closest neighbours or something around that.  
> **username_0**: @username_1 No, it isn't. This equation gets you a diameter of a sphere with the volume in the brackets. That volume is the volume one civilization would occupy if they were evenly distributed.  
> **username_1**: exactly. And that certainly is not "the maximum distance.", but the average or median or something like that. The maximum distance is the diameter of the Galaxy: all civilizations except one in one globular cluster and one in another the whole Galaxy diameter away. If you found a reference for the equation, please link it.  
> **username_0**: @username_1 Well yes it's something like that because the equation assumes the civilizations are evenly distributed across the galaxy. I'll link the reference once I actually find the equation within the scientific paper because it has something like 45 pages.  
> **username_1**: Maybe the paper says something about why they use it?  
> **username_0**: @username_1 Not sure. I will have to read it tomorrow, and I'll update the question then. Although it likely does.  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/135934/discussion-between-username_0-and-username_1).  
> **username_2**: @username_0 I have converted your images of equations to [MathJax](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference), please check that I have done so correctly.  
> **username_0**: @username_2 I checked, and the equations are correct.  
> **username_2**: @username_0 Thanks!  

---

## 1 answers

### Answer by username_3 on 2022-04-30. Score: 5

Suppose civilizations are randomly distributed in the galaxy (not necessarily evenly, but independent of each other). That means that there is a CDF $\Pr[d<x]=F(x)$, the probability for the distance $d$ to one of them being less than $x$. If there are $N$ civilizations, then the probability that *all* of them are inside distance $x$ is $\Pr[d\_1,d\_2,\ldots d\_N<x]=F(x)^N$. So if you know $F(x)$ you can easily find the median distance for a random civ by solving $F(x)=1/2$, and the median distance to the remotest by solving $F(x)=2^{-1/N}$. You can calculate the mean too [by cleverly integrating the CDF](https://stats.stackexchange.com/questions/10159/find-expected-value-using-cdf). Note that the same trick can be used to find the distribution of the closest civilization distance: it is $\Pr[d\_1,d\_2,\ldots d\_N>x]=1-(1-F(x))^N$.

The above is nice and general, but not exactly what the question is. In fact, trying to reverse engineer the calculator formula is likely wasted effort: it doesn't make much sense (more civilizations should make it give a more remote remotest civ).

My simplest approach would be to say that civilizations are uniformly distributed in a spherical volume corresponding to the milky way (of radius $R$) with Earth in the center (yes, this is wrong but not entirely wrong). Then their distance CDF will be $F(x) = (x/R)^3$ up to $x=R$, and 1 beyond. So I get a CDF for the remotest that is $(x/R)^{3N}$ and a median distance $x=R \cdot 2^{-1/3N}$. This behaves right: as you get more and more civs, the outermost will be almost exactly at the edge $x=R$.

[![enter image description here](https://i.sstatic.net/13yKW.png)](https://i.sstatic.net/13yKW.png)

A better model is to assume the Milky Way is a cylinder, and that the sun is in an offset position from the center. Unfortunately I don't know enough stochastic geometry to get $F(x)$ in a closed form. Basically, $F(x)=$[galactic volume inside x]/[total galactic volume]. This can be solved numerically. But if you are solving things numerically you can in principle go full hog and use realistic star densities or even galactic habitable zones.

Addendum: I couldn't resist actually doing it. I took an approximate stellar density model (from Carrol & Ostlie) that I assumed is equivalent to a civilization probability density, did a numerical integration in Matlab and calculated the approximate distance.
[![enter image description here](https://i.sstatic.net/0I57d.png)](https://i.sstatic.net/0I57d.png)

This doesn't give an explicit formula, but shows the overall behavior. $F(x)$ is basically a sigmoid with a turning point near the distance to the galactic center. The furthest distance grows slowly since there are rare stars far away from the galactic plane.

Adding a ring-shaped galactic habitable zone (not shown) makes it more of a linear rise, producing a much flatter distance distribution (furthest civilization will be at the far end of the ring).


---

