<p>The peak absolute magnitude of a Type Ia supernova is -19.5. The relation between absolute and apparent magnitude for nearby objects (<span class="math-container">$z \approx 0)$</span> is
<span class="math-container">$$m = M + 5\,(\log_{10}(d) - 1)$$</span>
where <span class="math-container">$d$</span> is the distance to the object in parsecs, <span class="math-container">$M$</span> is the absolute magnitude, and <span class="math-container">$m$</span> is the apparent magnitude. The -1 on the right hand side is because absolute magnitude is defined as being magnitude at a distance of 10 parsecs. The factor of 5 and the log base 10, that's also a part of the definition. The definition is more or less consistent with how human eyes work.</p>
<p>Difference sources provide different distances to the Crab Nebula, ranging from 6500 light years to 6700 light years, or 1993 to 2053 parsecs. I'll use 2000 parsecs as a nice round number. Doing the math, <span class="math-container">$-19.5 + 5\,(\log_{10}(2000) - 1) \approx -8$</span>. So even a Type Ia supernova at 2000 parsecs distance would not appear anywhere as close to bright as would a full moon. It would however appear to be brighter than Venus at its brightest (maximum apparent magnitude = -4.6).</p>
<p>The Crab Nebula was not a Type Ia supernova. It was instead a Type II supernova, with various sources stating its peak absolute luminosity magnitude being -18 rather than -19.5. That still means that SN 1054 was brighter than is Venus at its brightest.</p>

CC BY-SA 4.0
2022-06-12T04:26:36.147Z
And I thought they are super...
Felicia
CC BY-SA 4.0
2022-06-13T20:00:33.843Z
An object so distant that light emitted at a time before proto Indoeuropean was spoken.  So old that when Stonehenge was built, it was already half way.... and it is bright enough to cast a shadow! Isn't that "super" enough for you!?
James K
CC BY-SA 4.0
2022-06-11T21:42:26.713Z
2022-06-11T21:42:26.713Z
David Hammen
<blockquote>
<p>The peak luminosity of all type Ia supernovae is 1.60×10<sup>36</sup> Watts. If we see a star go supernova from earth, and we observe its intensity to be 1.0 W/m2. The intensity of the Sun is about 6.33x10<sup>7</sup> W/m2.</p>
</blockquote>
<p>Now, I have read that a supernova explosion intensity can exceed that of the whole Milky Way (or even <a href="https://www.science.org/content/article/universe-s-most-luminous-supernova-was-50-times-brighter-milky-way" rel="nofollow noreferrer">50 times</a>!. The 1054 supernova, of which  the Crab Nebula is a remnant (see pictures below), was seen to be about four times the intensity of Venus. But shouldn't it at 6700 lightyears distance, have looked  brighter than the Sun for a moment?</p>
<p>The Crab Nebula supernova remnant:</p>
<p><a href="https://i.sstatic.net/sCp1s.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/sCp1s.jpg" alt="The Crab Nebula supernova remnant" /></a></p>
<p>The neutron star in the nebula, spinning about 30 times per second!</p>
<p><a href="https://i.sstatic.net/eAAn0.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/eAAn0.jpg" alt="The neutron star in the nebula, spinning about 30 times per second!" /></a></p>

CC BY-SA 4.0
2022-06-11T17:43:48.133Z
Can you please indicate what source you are quoting in your question?
GrapefruitIsAwesome
CC BY-SA 4.0
2022-06-11T19:22:01.023Z
The numbers you quote are for a type Ia supernova, which [is not the type of supernova that the Crab Nebula is](https://earthsky.org/clusters-nebulae-galaxies/crab-nebula-was-an-exploding-star/)
Justin T
CC BY-SA 4.0
2022-06-11T21:06:45.150Z
There are $5\cdot 10^{40} m^2$ on the surface of a sphere radius 6700 light years, so the expected wattage per m^2 would be significantly lower than the sun.  9.4*10^15 meters/light year.
Sheldon
CC BY-SA 4.0
2022-06-11T23:17:43.860Z
Welcome, Felicia. :) As Grapefruit indicated, you *need* to give some form of attribution when you use material you didn't create. That applies to those images, too. Please see https://astronomy.stackexchange.com/help/referencing
PM 2Ring
CC BY-SA 4.0
2022-06-12T04:41:05.830Z
Images are particularly important because they are often covered by copyright. Copying verbatim a few dozen words, maybe even a hundred words, from some source falls under "fair use". It's okay, but you should show that you are copying (we have the '>' markdown for just that) and you should give proper attribution. A thousand words? That's not "fair use". The problem therein is that "an image is worth a thousand words". Using a copyright-protected image without attribution and without the permission of the owner is verboten at this site.
David Hammen
CC BY-SA 4.0
2022-06-12T04:43:11.363Z
Fortunately, these two images are not copyrighted. They were taken by the Hubble prior to 2008 and hence are public domain. You should be giving credit to NASA/ESA/STScI.
David Hammen
<the-sun><supernova><nebula>
Wouldn't the 1054 supernova have temporarily enlightened the Earth like the Sun?
Felicia
# Wouldn't the 1054 supernova have temporarily enlightened the Earth like the Sun?

(Asked by: username_0 on 2022-06-11)

> The peak luminosity of all type Ia supernovae is 1.60×1036 Watts. If we see a star go supernova from earth, and we observe its intensity to be 1.0 W/m2. The intensity of the Sun is about 6.33x107 W/m2.

Now, I have read that a supernova explosion intensity can exceed that of the whole Milky Way (or even [50 times](https://www.science.org/content/article/universe-s-most-luminous-supernova-was-50-times-brighter-milky-way)!. The 1054 supernova, of which the Crab Nebula is a remnant (see pictures below), was seen to be about four times the intensity of Venus. But shouldn't it at 6700 lightyears distance, have looked brighter than the Sun for a moment?

The Crab Nebula supernova remnant:

[![The Crab Nebula supernova remnant](https://i.sstatic.net/sCp1s.jpg)](https://i.sstatic.net/sCp1s.jpg)

The neutron star in the nebula, spinning about 30 times per second!

[![The neutron star in the nebula, spinning about 30 times per second!](https://i.sstatic.net/eAAn0.jpg)](https://i.sstatic.net/eAAn0.jpg)

**Tags:** the-sun, supernova, nebula
> **Comments:**
> **username_1**: Can you please indicate what source you are quoting in your question?  
> **username_2**: The numbers you quote are for a type Ia supernova, which [is not the type of supernova that the Crab Nebula is](https://earthsky.org/clusters-nebulae-galaxies/crab-nebula-was-an-exploding-star/)  
> **username_3**: There are $5\cdot 10^{40} m^2$ on the surface of a sphere radius 6700 light years, so the expected wattage per m^2 would be significantly lower than the sun.  9.4*10^15 meters/light year.  
> **username_4**: Welcome, Felicia. :) As Grapefruit indicated, you *need* to give some form of attribution when you use material you didn't create. That applies to those images, too. Please see https://astronomy.stackexchange.com/help/referencing  
> **username_5**: Images are particularly important because they are often covered by copyright. Copying verbatim a few dozen words, maybe even a hundred words, from some source falls under "fair use". It's okay, but you should show that you are copying (we have the '>' markdown for just that) and you should give proper attribution. A thousand words? That's not "fair use". The problem therein is that "an image is worth a thousand words". Using a copyright-protected image without attribution and without the permission of the owner is verboten at this site.  
> **username_5**: Fortunately, these two images are not copyrighted. They were taken by the Hubble prior to 2008 and hence are public domain. You should be giving credit to NASA/ESA/STScI.  

---

## 1 answers

### Answer by username_5 on 2022-06-11. Score: 7

The peak absolute magnitude of a Type Ia supernova is -19.5. The relation between absolute and apparent magnitude for nearby objects ($z \approx 0)$ is
$$m = M + 5\,(\log\_{10}(d) - 1)$$
where $d$ is the distance to the object in parsecs, $M$ is the absolute magnitude, and $m$ is the apparent magnitude. The -1 on the right hand side is because absolute magnitude is defined as being magnitude at a distance of 10 parsecs. The factor of 5 and the log base 10, that's also a part of the definition. The definition is more or less consistent with how human eyes work.

Difference sources provide different distances to the Crab Nebula, ranging from 6500 light years to 6700 light years, or 1993 to 2053 parsecs. I'll use 2000 parsecs as a nice round number. Doing the math, $-19.5 + 5\,(\log\_{10}(2000) - 1) \approx -8$. So even a Type Ia supernova at 2000 parsecs distance would not appear anywhere as close to bright as would a full moon. It would however appear to be brighter than Venus at its brightest (maximum apparent magnitude = -4.6).

The Crab Nebula was not a Type Ia supernova. It was instead a Type II supernova, with various sources stating its peak absolute luminosity magnitude being -18 rather than -19.5. That still means that SN 1054 was brighter than is Venus at its brightest.

> **Comments:**
> **username_0**: And I thought they are super...  
> **username_6**: An object so distant that light emitted at a time before proto Indoeuropean was spoken.  So old that when Stonehenge was built, it was already half way.... and it is bright enough to cast a shadow! Isn't that "super" enough for you!?  

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