<p>The supernova rate is only estimated/known to a factor of 2-3, so high levels of accuracy are not possible or warranted.</p>
<p>If we assume high mass stars are born uniformly in a disc of radius  <span class="math-container">$r$</span>  light years, then you can work out a &quot;surface density rate&quot; of supernovae of <span class="math-container">$1/100 \pi r^2$</span> per square light year, per year (adopting your rate of one per century over the Galaxy).</p>
<p>Then, consider a thin annulus of radius <span class="math-container">$x$</span> and width <span class="math-container">$dx$</span> around the Earth. The number of supernovae that explode per year will be
<span class="math-container">$$dN = \frac{2\pi x}{100 \pi r^2} dx$$</span>
and the number that have exploded in the last <span class="math-container">$x$</span> years, who's light is yet to reach is just this multiplied by another factor of <span class="math-container">$x$</span>.</p>
<p>If the Sun was at the centre of the Galaxy the calculation is then simple
<span class="math-container">$$N = \frac{1}{50r^2}\int_{0}^{r} x^2\ dx = \frac{r}{150},$$</span>
with <span class="math-container">$r$</span> measured in light years. If <span class="math-container">$r\sim 30000$</span> light years (I think 50,000 is a bit big), then the number is 200.</p>
<p>Unfortunately, that isn't the geometry. Instead of being a circular annulus around the Sun, you have to work with an annulus that is truncated where it reaches the &quot;edge&quot; of the Galactic disc. I may add to this answer later, but I doubt this will change the number above by very much.</p>

CC BY-SA 4.0
2020-10-07T15:19:16.000Z
Thank you.
I guess the difference between being in the center of the galaxy and halfway towards the outer edge would be the timing of the received light and of course the distribution in the sky. I don't see that that would change the estimate.
JRig
CC BY-SA 4.0
2020-10-07T06:37:41.587Z
2020-10-07T11:36:55.723Z
2020-10-07T11:36:55.723Z
ProfRob
<p>The Milky Way has a diameter of around 100.000 light years with the Solar system placed some 25,000 light years from the center.</p>
<p>If we assume a supernova rate of 1 per century, and also assume that supernovae are distributed somewhat uniformly in our galaxy, then it seems to me that light and neutrinos from hundreds maybe even thousands of these events are propagating through the Milky Way towards us at this very moment.</p>
<p>Can this be estimated more accurately? Does the supernovae distribution even affect this estimate, given the age of the galaxy?</p>

<milky-way><supernova>
How many supernovae events are on their way from within the Milky Way?
JRig
# How many supernovae events are on their way from within the Milky Way?

(Asked by: username_0 on 2020-10-06)

The Milky Way has a diameter of around 100.000 light years with the Solar system placed some 25,000 light years from the center.

If we assume a supernova rate of 1 per century, and also assume that supernovae are distributed somewhat uniformly in our galaxy, then it seems to me that light and neutrinos from hundreds maybe even thousands of these events are propagating through the Milky Way towards us at this very moment.

Can this be estimated more accurately? Does the supernovae distribution even affect this estimate, given the age of the galaxy?

**Tags:** milky-way, supernova

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## 1 answers

### Answer by username_1 on 2020-10-07. Score: 2 | Accepted answer

The supernova rate is only estimated/known to a factor of 2-3, so high levels of accuracy are not possible or warranted.

If we assume high mass stars are born uniformly in a disc of radius $r$ light years, then you can work out a "surface density rate" of supernovae of $1/100 \pi r^2$ per square light year, per year (adopting your rate of one per century over the Galaxy).

Then, consider a thin annulus of radius $x$ and width $dx$ around the Earth. The number of supernovae that explode per year will be
$$dN = \frac{2\pi x}{100 \pi r^2} dx$$
and the number that have exploded in the last $x$ years, who's light is yet to reach is just this multiplied by another factor of $x$.

If the Sun was at the centre of the Galaxy the calculation is then simple
$$N = \frac{1}{50r^2}\int\_{0}^{r} x^2\ dx = \frac{r}{150},$$
with $r$ measured in light years. If $r\sim 30000$ light years (I think 50,000 is a bit big), then the number is 200.

Unfortunately, that isn't the geometry. Instead of being a circular annulus around the Sun, you have to work with an annulus that is truncated where it reaches the "edge" of the Galactic disc. I may add to this answer later, but I doubt this will change the number above by very much.

> **Comments:**
> **username_0**: Thank you. I guess the difference between being in the center of the galaxy and halfway towards the outer edge would be the timing of the received light and of course the distribution in the sky. I don't see that that would change the estimate.  

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