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Topic: This Week's Finds in Mathematical Physics (Week 276)
Replies: 10   Last Post: Jul 14, 2009 10:56 PM

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Phillip Helbig---remove CLOTHES to reply

Posts: 33
Registered: 12/13/04
Re: This Week's Finds in Mathematical Physics (Week 276)
Posted: Jun 21, 2009 4:42 PM
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In article <h1jgpm$ikn$1@glue.ucr.edu>,
baez@math.removethis.ucr.andthis.edu (John Baez) writes:
> It would be nice to see some calculations of just how much power
> we'd get from a supernova at that distance. I must admit that
> "brighter than a million moons" doesn't really do it for me. Does
> anyone out there have what it takes to crunch the numbers?


We don't know enough about how Betelgeuse will behave exactly to do
better than an order of magnitude, but that's good enough.

A rule of thumb is that a supernova is about as bright as a galaxy. A
galaxy has about 10**11 stars, but on average they are not as luminous
as the Sun. Say a galaxy is a few times 10**10 times as bright as the
Sun. 5 magnitudes is a factor of 100. (The absolute magnitude of an L*
galaxy (a "typical" galaxy, luminosity-wise) is about -21.5, that of the
Sun about 5.5. That makes 27 magnitudes, or a few times 10**10. So
that looks consistent.) 10**10 is 25 magnitudes. Say a typical star
has an absolute magnitude of 6. Increase that by 27 magnitudes and we
arrive at -21. The absolute magnitude of Betelgeuse is about -5. So
that means that as a supernova Betelgeuse will be about 16 magnitudes
brighter than it is now. Its apparent magnitude now is about 0.5 (it's
variable), so as a supernova its apparent magnitude would be about
-16.5.

The Sun has an apparent magnitude of about -27. So we are looking at
something about 10.5 magnitudes fainter than the Sun, say roughly
one-ten-thousandth the brightness of the Sun.

At night, though, it will seem even brighter than that, since a) the eye
is dark-adapted and b) the contrast to the dark nighttime sky is greater
than that of the Sun to the blue daytime sky. (Of course, at its
brightest phase as a supernova Begelgeuse might not be in the nighttime
sky.)

The apparent magnitude of the full moon is about -12.5, so Betelgeuse as
a supernova would be about 4 magnitudes brighter. That's less than a
factor of 100, not "brighter than a million moons". (One could arrive
at "brighter than a million moons" if one assumes that Betelgeuse as a
supernova would be 27 magnitudes brighter than it is now, rather than 27
magnitudes brighter than a typical star. A factor of a million
corresponds to a difference of 15 magnitudes. That would be -27.5.
That is brighter than the Sun! Betelgeuse as a supernova will be
bright, but not that bright!)



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