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Would that person go blind?

Neutrino detectors and the abundance of Neutrinos would detect the upcoming visible show about 3 hours before any visible signs, so there would be time to point certain telescopes that could handle the brightness towards it.

I'm curious if an individual with a telescope pointed in that direction would have an unpleasant surprise. Would the scientific community be wise to not announce the massive stellar explosion until after it's visible to avoid potential negative effects from over-eager amateur astronomers.

I realize this is a kind of silly question and it might depend too much on the telescope, but I'm curious.

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    $\begingroup$ "When it goes ..." - It's 642.5 light years away, so it would need to have already gone supernova over 550 years ago ... But we know what you meant, and Mark's answer is OK, as is the other. $\endgroup$ – Rob Dec 8 '18 at 23:35
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    $\begingroup$ @Rob Depends on your reference frame ;) ( en.wikipedia.org/wiki/Relativity_of_simultaneity ) $\endgroup$ – jpa Dec 9 '18 at 13:09
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No, it would not be a problem. Supernovae are not at all like flashbulbs – they brighten over a period of many days and dim again even more slowly. Here are a number of different light curves taken from Wikipedia: Luminous output of different types of Supernovae

The rise is fast on an astronomical scale – several orders of magnitude over a period of roughly ten days – but very slow on a human scale. An amateur looking at it would not notice any significant change in brightness, but if the same person came back a few hours later or the next night, the change would be very evident.

As far as we can tell, the reason is that the light at peak brightness is caused by emissions from material blown off by the explosion. For example, in Type 1a SNe, most of the light is from the radioactive decay of the huge mass of ejected nickel-56 (half life 6 days).

The Wikipedia article on supernovae is quite good and covers this all in more detail.

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    $\begingroup$ This answer explains that the brightness increases gradually, but doesn't seem to answer the question as to whether or not a person watching Betelgeuse would go blind (unless we count "no, it would not be a problem" as the answer). $\endgroup$ – JBentley Dec 10 '18 at 13:10
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    $\begingroup$ Note that the OP asked if there'd be an "unpleasant surprise" -- the answer is, "No, there would be no surprise." You would not be blinded unless you really wanted to be. Obviously if you stare into a bright enough light with a big enough telescope for a long enough time you can blind yourself in one eye -- you'd have to do it all over again with the other, of course. I don't feel that constitutes a "surprise." $\endgroup$ – Mark Olson Dec 10 '18 at 13:44
  • $\begingroup$ @JBentley It's not fast enough and it's not bright enough to blind you. $\endgroup$ – Florin Andrei Dec 10 '18 at 19:03
  • $\begingroup$ @Mark Olson : Yep. OP is almost surely imagining it being like a camera flashbulb (or perhaps, a nuke?) "suddenly" going off and you are looking into said flashbulb with your telescope. But it's not. $\endgroup$ – The_Sympathizer Dec 27 '18 at 10:46
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If you insist on observing the exploding Betelgeuse at peak brightness, you could potentially damage your eye. The complete answer enters the realm of physiology. Here I'll discuss the astronomical parts:

Betelgeuse will explode as a type II supernova, the typical brightness of which is around $M \sim -17$. With a distance of $d\simeq200\,\mathrm{pc}$, its distance modulus is $$ \mu = 5\log(d/\mathrm{pc}) - 5 \simeq 6.5, $$ so its apparent magnitude will be $$ m = M + \mu \simeq -10.5. $$

For these calculations I assume that the Sun is the threshold for damaging your eye (a brief look at the Sun is okay, a longer look will cause permanent damage. But… physiology…). The Sun has an apparent magnitude of $m_\odot = -26.7$, i.e. it is $\Delta m = 16.2$ magnitudes brighter. In other words, Betelgeuse will be $$ f = 10^{\Delta m/2.5} \simeq 3\times10^6 $$ times dimmer than the Sun.

However, the Sun is an extended source, spanning an angle of roughly $\theta_\mathrm{Sun} = 32$ arcminutes across. In contrast, Betelgeuse is a point source, which when transferred through the atmosphere and the telescope, is spread out over $\theta_\mathrm{Bet} \sim$ a few square arcseconds. Thus its light will be more concentrated; i.e. it will be much brighter, but it will hit a much smaller area of your retina. However, your eye will also move around, smearing out the light. Not being a physiologist, for the sake of this calculation I assume that the light is smeared out over a disk 1 arcminute across (about the size of a planet seen from Earth).

Thus, the factor $f$ will itself be a factor $(\theta_\mathrm{Sun} / \theta_\mathrm{Bet})^2 \simeq 1000$ times larger — that is, Betelgeuse is only $\sim 3\,000$ times dimmer than the Sun.

Hence, for our assumptions your eye will be damaged if you observe exploding Betelgeuse through a telescope with an area $\sim 3\,000$ larger — or roughly 55 times wider — than your pupil. In bright light, the pupil contracts to roughly 3 mm in diameter, so if observing through a telescope of 16 cm or larger, you could damage your eye.

Based on evolutionary models of Betelgeuse, Dolan et al. (2016) estimate an apparent magnitude of $m=-12.4$, i.e. roughly 6 times brighter than our estimate. This would mean that you only need a 7 cm telescope to damage your eye.

However, as Mark writes in his answer, supernovae don't increase to their peak brightness in matters of seconds, but rather in matters days (roughly half a mag per day), so you have plenty of time to look away.

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    $\begingroup$ I'm not sure this is the right calculation. Betelgeuse subtends only a tiny angle, meaning that when it's in focus all its light is focused on effectively a single point on the retina, whereas the sun's light is spread over an area. I'd be surprised if that didn't make it dangerous when using a much smaller telescope. $\endgroup$ – Nathaniel Dec 10 '18 at 8:07
  • $\begingroup$ Additionally, this answer assumes that the brightness of the sun is the minimum required to go blind. For a complete answer, we should probably consider how much brightness the human eye can withstand without being blinded. It could be the case that a much smaller telescope than the one you propose is sufficient. $\endgroup$ – JBentley Dec 10 '18 at 13:12
  • $\begingroup$ @Nathaniel Yes, you're right. I'm not sure what the completely correct approach would be. A point source — which when seen through the atmosphere would span a few square arcsec — would be much brighter, but damage only a tiny fraction of your eye. $\endgroup$ – pela Dec 10 '18 at 14:08
  • $\begingroup$ @JBentley That's true, that's an assumption. But considering that you can actually look at the Sun for a brief period of time without permanently damaging your eye, I think makes this threshold okay, at least for an order-of-magnitude estimate. $\endgroup$ – pela Dec 10 '18 at 14:13
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    $\begingroup$ I edited to take into account these very relevant comments. $\endgroup$ – pela Dec 10 '18 at 15:17
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Brightness varies inversely with the square of the distance. Betelgeuse is about 642.5 light years away and has an apparent magnitude of 0.42. My grasp of apparent magnitude concepts is a bit wobbly, but I believe if it grew a million times as bright, it might have an apparent magnitude of -14.5 or so, which is a lot more like the brightness of the moon than the sun.

Given the great distance, the decrease in brightness due to distance, and the countless amounts of dust & gas between earth and Betelgeuse, I think you'd probably be fine. You might be dazzled by its brightness -- a bit like looking at a light bulb, I imagine -- but I doubt it would cause any physical harm.

EDIT: I hope a real astronomer sounds off here. I'm not sure what kind of supernova we might expect from Betelgeuse, but apparently supernovas (supernovae?) can achieve a theoretical brightness equal to 5 trillion suns!

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    $\begingroup$ Does it matter that the brightness comparable to the moon will be concentrated in a star-sized point? $\endgroup$ – Barmar Dec 10 '18 at 5:25
  • $\begingroup$ I am not qualified to say. I will speculate that its intensity will be diffused quite a bit by all the dust & gas between us and Betelgeuse -- it's 640 light-years away. On the other hand, the moon looks extremely bright in my own telescope. Bright enough to dazzle the eyes quite a bit. $\endgroup$ – S. Imp Dec 10 '18 at 18:15
  • $\begingroup$ I assumed you'd already included the diffusion in your estimate of the apparent magnitude. $\endgroup$ – Barmar Dec 10 '18 at 19:15
  • $\begingroup$ My calculation was a simple one. I simply assumed that Betelgeuse might increase in brightness by a million, and converted that increase to apparent magnitude. Check the third paragraph here. This sort of sidesteps complicated diffusion calculations by just assuming diffusion will not change if the thing gets brighter. $\endgroup$ – S. Imp Dec 10 '18 at 20:09
  • $\begingroup$ That seems like a reasonable assumption for broad estimates like this. In reality, diffusion is probably dependent on light wavelength, which I assume changes during supernova. But if we're just dealing with orders of magnitude, it's probably not significant. $\endgroup$ – Barmar Dec 10 '18 at 20:13

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