How big will the Betelgeuse supernova appear?

[swampwiz]

I was looking at the seemingly very bright full Moon this early morning, and it got me thinking about how Betelgeuse will be as bright as the full Moon when it goes supernova (i.e., at its brightest). I suppose that first there is the question of the size of the bright exploding matter (but I suspect that it will still be virtually a point in the sky at its brightest), and then the question of how much optical dispersion (that's probably not the correct term; perhaps scintillation is the correct term?) such an immensely bright point object would have. If it were to still be a point object with the same amount of dispersion as a regular star like Sirius, perhaps it would be so bright that it could be dangerous to look at?

 

[Drakkith]

The explosion will be essentially a point source for our eyes and, while very bright for an object in the night sky, will still be roughly 500,000 times dimmer than the Sun. I doubt it will be dangerous to look at even after accounting for smaller apparent size compared to the Sun and Moon.

 

[mathman]

Unless it happens soon (very unlikely) you won't see it. Estimates seem to be sometime within one million years.

 

[swampwiz]

The explosion will be essentially a point source for our eyes and, while very bright for an object in the night sky, will still be roughly 500,000 times dimmer than the Sun. I doubt it will be dangerous to look at even after accounting for smaller apparent size compared to the Sun and Moon.

But what I was getting at is that if it is a "point", then it essentially has infinite brightness at that point (which obviously isn't the case). The Sun is 500K times brighter, but it has a certain steradial measure (I think that's the term) over which the brightness is distributed. If the steradial measure for Betelgeuse were 500K times smaller, than the retinal cones that happen to map to Betelgeuse would get burned at the same level that such cones that map to the Sun would.

 

[Drakkith]

But what I was getting at is that if it is a "point", then it essentially has infinite brightness at that point (which obviously isn't the case).

If you look at a star, you'll notice that the brighter the star, the larger it appears to be. It's not a single bright dot in your vision, it's a slightly spread out spiky mess that gets larger the brighter the star. This is because the light cannot be focused down to a perfect spot due to diffraction and other wave properties. The apparent size, or apparent angular diameter is larger than the actual angular diameter. When we say that an object is a 'point source' we mean that its angular diameter is smaller than than our ability to resolve it.

In short, your eye doesn't focus the light down to a perfect point, it gets spread out a little bit, keeping your retinal cells nice and safe.

 

[DrStupid]

It's not a single bright dot in your vision, it's a slightly spread out spiky mess that gets larger the brighter the star.

Maybe that can be quantified. The diameter of the inner disc is

$d = 2.44 \cdot \frac{{\lambda \cdot f}}{D}$

With the focal length $f \approx 2cm$ of the human eye, the diameter $D \approx 6 \ldots 8mm$ of the pupils and waveleghts in the range of $\lambda \approx 400 \ldots 750nm$ I get

$d \approx 2.5 \ldots 6\mu m$

The intensity would be amplified by a factor of

$\frac{{D^2 }}{{d^2 }} \approx 10^6 \ldots 10^7$

Now we need an estimation of the intensity of the supernova as seen from Earth and check if the resulting irradiance on the retina could be a problem.

 

[hutchphd]

Now we need an estimation of the intensity of the supernova as seen from Earth and check if the resulting irradiance on the retina could be a problem.

If you work out the diameter of the solar image on the retina it is approximately 0.2 mm. So this area is roughly $10^4$ bigger than the Betelgeuse diffraction spot. If sunlight is 5x $10^5$ brighter than Betelgeuse the relative retinal solar flux is 50 times as much.
The recommended optical density for solar filters is bigger than 3.5 so I believe this could represent a problem.

 

[swampwiz]

If you work out the diameter of the solar image on the retina it is approximately 0.2 mm. So this area is roughly $10^4$ bigger than the Betelgeuse diffraction spot. If sunlight is 5x $10^5$ brighter than Betelgeuse the relative retinal solar flux is 50 times as much.
The recommended optical density for solar filters is bigger than 3.5 so I believe this could represent a problem.

So you're saying that Betelgeuse would appear on a retinal-unit-area basis to be 50X dimmer than the Sun? What does the solar filter having an optical density bigger than 3.5 mean?

 

[hutchphd]

In this circumstance the transmissivity of $10^{-3.5}$ which less than 1/50 by a good bit.