The Virtuosi (Posts about betelgeuse)https://thephysicsvirtuosi.com/enContents © 2019 <a href="mailto:thephysicsvirtuosi@gmail.com">The Virtuosi</a> Thu, 24 Jan 2019 15:05:00 GMTNikola (getnikola.com)http://blogs.law.harvard.edu/tech/rss- Betelgeuse, Betelgeuse, Betelgeuse!https://thephysicsvirtuosi.com/posts/old/betelgeuse-betelgeuse-betelgeuse-/Corky<div><hr>
<p><a href="http://3.bp.blogspot.com/-6s6rSrItOdk/TrS6st3tdVI/AAAAAAAAAPk/GpGfjY5KZ3Y/s1600/betelgeuse.png"><img alt="image" src="http://3.bp.blogspot.com/-6s6rSrItOdk/TrS6st3tdVI/AAAAAAAAAPk/GpGfjY5KZ3Y/s320/betelgeuse.png"></a>
A very cold person points out Betelgeuse</p>
<hr>
<p><em>Betelgeuse is a massive star at the very end of its life and could
explode any second now!</em> Every time I hear that I get really <em>really</em>
excited. Like a kid in a candy store that's about to see a star blow up
like nobody's business. This giddiness will last for a solid minute
before I realize that "any second now" is taken on astronomical
timescales and roughly translates to "sometime in the next million years
maybe possibly." Then I feel sad. But you know what always cheers me up?
Calculating things! Hooray! So let's take a look at the ways Betelgeuse
could end its life (even if it's not going to happen tomorrow) and how
these would affect Earth. First, a little background. Betelgeuse is the
bright orangey-red star that sits at the head/armpit of Orion. It is one
of the <a href="http://en.wikipedia.org/wiki/List_of_brightest_stars">brightest</a>
stars in the night sky. Its distance has been measured by the
<em><a href="http://en.wikipedia.org/wiki/Hipparcos">Hipparcos</a></em>satellite to be
about 200 parsecs [1] from Earth (about 600 light years). Betelgeuse is
at least 10 times as massive as our Sun and has a diameter that would
easily accomodate the orbit of Mars. In fact, the star is big enough and
close enough that it can actually be spatially resolved by the Hubble
Space Telescope! Being so big and bright, Betelgeuse is destined to die
young, going out with a bang as a
<a href="http://en.wikipedia.org/wiki/Supernova#Core_collapse">core-collapse</a>
supernova. This massive explosion ejects a good deal of "star stuff"
into interstellar space [2] and leaves behind either a <a href="http://en.wikipedia.org/wiki/Neutron_star">neutron
star</a> or a <a href="http://en.wikipedia.org/wiki/Black_hole">black
hole</a>. Alright, now that we're
all caught up, let's turn our focus on this "massive explosion" bit.
What kind of energy scale are we talking about if Betelgeuse blows up?
Well, a pretty good upper bound would be if all of the star's mass (10
solar masses worth!) were converted directly to energy, so $$ E_{max} =
mc^2 =
10M_{\odot}\times\left(\frac{2\times10^{30}\~\mbox{kg}}{1\~M_{\odot}}\right)\times
\left(3\times10^8\~\mbox{m/s}\right)^2 $$ which is about $$
E_{max} \sim 10^{48}\~\mbox{J} $$ and that's nothing to shake a
stick at. But remember, this is if the <em>entire star</em> were converted
directly to energy, and that would be hard to do. Typical fusion
efficiencies are about \~1% [3], so let's say a reasonable estimate for
the total nuclear energy available is $$ E_{nuc} \sim \eta_{f}
\times E_{max} \sim 10^{-2} \times 10^{48}\~\mbox{J} \sim
10^{46}\~\mbox{J}. $$ This is the total energy released by a typical
supernova. As it turns out though, 99% of this energy is carried away in
the form of neutrinos and only about 1% is carried away in photons.
Since we are mainly concerned with how this explosion will affect Earth,
and the neutrinos will just pass on by, we will only consider the 1% of
energy released in photons that would reasonably interact with Earth.
That gives us $$ E_{ph} \sim 0.01 \times E_{nuc} \sim
10^{44}\~\mbox{J}. $$ Neato, so that's the total amount of energy
released in a supernova in the form of photons. How much of this energy
would be deposited at the Earth if Betelgeuse exploded? Well, if the
energy is deposited isotropically (that is, the same in all directions),
then the fluence (or time integrated energy flux) is given by $$ F_{ph}
= \frac{E_{ph}}{4\pi d^2}. $$ All this is saying is that the total
energy release by the supernova spreads out uniformly over a sphere of
radius d, so the fluence will give us the amount of energy deposited in
each square meter of that sphere (the units of fluence here are J/m^2).
The total energy deposited on Earth is then $$ E_{\oplus} = F_{ph}
\times \pi R^2_{\oplus}. $$ Hot dog! Let's plug in some numbers,
already. The total energy deposited on the Earth by a symmetrically
exploding Betelgeuse at a distance of d = 200 pc (where 1 pc = 3 <em>
10^16 m) is $$E_{\oplus}=\frac{E_{ph}}{4\pi d^2}\times\pi
R^2_{\oplus}\sim
10^{19}\~\mbox{J}\left(\frac{E_{ph}}{10^{44}\~\mbox{J}}\right)\left(\frac{d}{200\~\mbox{pc}}\right)^{-2}.$$
Well, 10^19 J certainly </em>seems* like a lot of energy. In fact, it is
roughly the amount of energy contained in the entire nuclear arsenal of
the United States [4]. But it is spread over the entire atmosphere. Is
there a way to gauge how this would affect life on Earth? We could see
how much it would heat up the atmosphere using specific heats: $$ E =
m_{atm}c_{air}\Delta T $$ where c is the specific heat of air
(\~10^3 J per kg per K). Oops, looks like we need to know the mass of
the atmosphere. But we can figure this out, the answer is pushing right
down on our heads! We know the pressure at the surface of the Earth (1
atm = 101 kPa) and that pressure is just the result of the weight of the
atmosphere pushing down on us. Since pressure is just force / area, we
have $$ P = F/A = m_{atm}g / A_{\oplus} $$ So $$ m_{atm} =
\frac{P\times4\pi
R^2_{\oplus}}{g}=\frac{10^5\~\mbox{Pa}\times4\pi
(6\times10^6\~\mbox{m})^2}{9.8\~\mbox{m/s}^2}\approx4\times10^{18}\~\mbox{kg}.$$
Neato, gang. So we could see a temperature rise of about $$ \Delta T =
\frac{E_{ph}}{m_{atm}c_{air}}=\frac{10^{19}\~\mbox{J}}{4\times10^{18}\~\mbox{kg}\times10^3\~\mbox{J/
kg K}}\approx0.003\~\mbox{K}, $$ or three one-thousandths of a degree.
Remember, too, that this will be an upper bound since we are assuming
that all this energy is deposited into the atmosphere before it has a
chance to cool. In fact, if the energy is deposited over the course of
hours or days, this value will be much less. So it looks like we've
wrapped this thing up: Betelgeuse exploding will most certainly not put
the Earth in any danger. Or did we? We have considered the case of a
symmetric supernova, but there's more than one way to blow up a star.
Massive stars can also end their lives in a fantastic explosion called a
<a href="http://en.wikipedia.org/wiki/Gamma-ray_burst">gamma-ray burst</a> (GRBs to
the hep cats that study them, some fun facts relegated to [5]). GRBs are
still an intense area of current study, but the current picture (for one
type of GRB, at least) is that they are the result of a star blowing up
with the energy of the explosion focussed into two narrow beams (see
picture below). Since the flux isn't distributed over the whole sphere,
GRBs can be seen at much greater distances than a typical supernova.</p>
<hr>
<p><a href="http://2.bp.blogspot.com/-siXXiFnQWHY/TrITJA8z_sI/AAAAAAAAAPU/qStD4_9qLBI/s1600/grb.jpg"><img alt="image" src="http://2.bp.blogspot.com/-siXXiFnQWHY/TrITJA8z_sI/AAAAAAAAAPU/qStD4_9qLBI/s320/grb.jpg"></a>
Example of a gamma-ray burst, with the explosion in two beams.</p>
<hr>
<p>So how will this change our answer? Well, it's going to change the
fluence we calculated above. Instead of spreading the energy out over
the whole sphere, it's only going to go to some fraction of the 4pi
steradians. So we get $$ F_{ph} = \frac{E_{ph}}{4\pi f_{\Omega}
d^2}, $$ where f_{omega} is called the "beaming fraction" and tells us
what fraction of the sphere the energy goes through. Typical GRB beams
range from 1 to 10 degrees in radius. Converting this to radians, we can
find the beaming fraction as $$ f_{\Omega} = \frac{2 \times \pi
\theta^2}{4\pi} \approx
10^{-4}\left(\frac{\theta}{1^\circ}\right)^2,$$ so the beaming
fraction is 10^-4 and 10^-2 for a beam angle of 1 degree and 10
degrees, respectively. Alright, so now we can redo the calculations we
did for the supernova case, but keeping this beaming fraction around.
The total amount of energy that would hit Earth is then about
$$E_{\oplus}=\frac{E_{ph}}{4\pi f_{\Omega} d^2}\times\pi
R^2_{\oplus}\sim
10^{23}\~\mbox{J}\left(\frac{E_{ph}}{10^{44}\~\mbox{J}}\right)\left(\frac{d}{200\~\mbox{pc}}\right)^{-2}\left(\frac{\theta}{1^\circ}\right)^{-2}.$$
Holy sixth-of-a-moley! Continuing as we did above, we find that this
could potentially heat up the atmosphere by $$ \Delta T =
\frac{E_{ph}}{m_{atm}c_{air}}=\frac{10^{23}\~\mbox{J}}{4\times10^{18}\~\mbox{kg}\times10^3\~\mbox{J/
kg
K}}\approx3\~\mbox{K}\left(\frac{\theta}{1^\circ}\right)^{-2},
$$ which is certainly non-negligible. Now, this won't destroy the planet
[6], but it could make things really uncomfortable. This will be
especially true when you realize that a fair amount of the energy
carried away from a gamma-ray burst is in the form of (wait for it...)
gamma-rays, which will wreck havoc on your DNA. Remember, though, that
this is an absolute worst-case scenario since we have assumed the
smallest beaming angle. But this may still make us a little nervous, so
is there anyway to figure out if Betelgeuse could, in fact, beam a
gamma-ray burst towards Earth? Yes, yes there is. Jets and beams like
those in GRBs typically point along the rotation axis of the star [7].
If we could determine the rotational axis of Betelgeuse, then we could
say whether or not there's a chance it's pointed towards us. It just so
happens that Betelgeuse is the only star (aside from our Sun) that is
spatially resolved. If you could measure spectra along the star, you
could look for Doppler shifting of absorption lines and say something
about the velocity at the surface of the star. Luckily, this has already
been done for us (see, for example <a href="http://adsabs.harvard.edu/abs/1998AJ....116.2501U">Uitenbroek et al.
1998</a>). These
measurements are hard to do since the star is only a few pixels wide,
but it appears as though the rotation axis is inclined to the
line-of-sight by about 20 degrees (see figure below). That means this
would require a beam with at least a 20 degree radius to hit the Earth.
This appears to be outside the typical ranges observed. So even if
Betelgeuse were to explode in a gamma-ray burst, the beam would miss
Earth and hit some dumb other planet nobody cares about.</p>
<hr>
<p><a href="http://2.bp.blogspot.com/-xYILwh0eLEM/TrSTBxflbxI/AAAAAAAAAPc/Amma1aQtmPU/s1600/rotation.png"><img alt="image" src="http://2.bp.blogspot.com/-xYILwh0eLEM/TrSTBxflbxI/AAAAAAAAAPc/Amma1aQtmPU/s400/rotation.png"></a>
Figure reproduced from Uitenbroek et al. (1998)</p>
<hr>
<p>Alright, so the moral of the story is that Betelgeuse is completely
harmless to people on Earth. When it does explode, it will be a
brilliant supernova that would likely be visible at least a little bit
during the day. It will be the coolest thing that anyone alive (if there
are people...) will ever see. Sadly, this explosion could take place at
just about any time during the next million years. Assuming a uniform
distribution over this time period and a human lifetime of order 100
years, there is something like a 1 in 10,000 chance you'll see this in
your life. Feel free to hope for a spectacular astronomical sight, but
don't lose sleep worrying about being hurt by Betelgeuse!
<strong>Semi-excessive Footnotes:</strong> [1] This has nothing to do with the Kessel
Run. For a description of the actual distance unit see
<a href="http://en.wikipedia.org/wiki/Parsec">Wikipedia</a>. For a circuitous
retconning to correct for one throwaway line in <em>Star Wars</em>, see
<a href="http://starwars.wikia.com/wiki/Kessel_Run">Wookieepedia</a>. [2] This is
how anything heavier than helium gets distributed throughout the
universe. The hydrogen and helium formed after the Big Bang gets fused
into heavier elements in stars and then dispersed out through
supernovae. In fact, most things heavier than Iron actually <em>require</em>
supernovae to even exist. If you have any gold on you right now (I'm
looking at you Mr. T), that only exists <em>because a star exploded!</em> [3]
Let's consider the case of turning 4 protons into a Helium nucleus.
<a href="http://en.wikipedia.org/wiki/Helium-4">Helium-4</a> has a binding energy
of about 28 MeV, which means that the total energy of a bound He-4
nucleus is 28 MeV less than its free protons and neutrons (in other
words, we need to put in 28 MeV to break it up). So the process of
turning 4 protons into a Helium nucleus gives off 28 MeV worth of
energy. But we had a total of 4 times 1000 MeV worth of matter we could
have turned into energy. Thus, the process was 28 MeV/4000 MeV \~ 0.7%
efficient at turning matter into energy. [4] Sometime last year, the
United States
<a href="http://www.defense.gov/npr/docs/10-05-03_Fact_Sheet_US_Nuclear_Transparency__FINAL_w_Date.pdf">disclosed</a>
that its nuclear arsenal as of Sept 2009 was something like 5000
warheads. Assume these to be Megaton warheads. A Megaton is about 4 <em>
10^15 J, so the total energy in the US arsenal is about 5000 * 4 </em>
10^15 J = 2 * 10^19 J. [5] A fun fact about GRBs: They were
discovered by a <a href="http://en.wikipedia.org/wiki/Vela_(satellite)">military
satellite</a> looking for
illegal nuclear tests, which would emit some gamma-rays. Instead of
seeing a signal on Earth, they saw bursts coming from space. I really
really hope that someone's first thought was that the Russians were
testing nukes on the Moon or something. <em>We must not allow a moon-nuke
gap!</em> [6] We here at the Virtuosi are contractually obligated to only
destroy the Earth in our posts on Earth day. I apologize for any
inconvenience this may cause. [7] I am not exactly sure why this is the
case. It is certainly observed to be the case and I thought there was a
straightforward explanation for why this was the case, but I don't
really have a good explanation. Although, maybe there just <a href="http://en.wikipedia.org/wiki/Polar_jet">isn't a good
one yet.</a> [8] For comparison,
there is about a 1 in 3000
<a href="http://www.lightningsafety.com/nlsi_pls/probability.html">chance</a>
you'll be struck by lightning in your lifetime.</p></div>betelgeusescott bakulathings that probs wont kill youhttps://thephysicsvirtuosi.com/posts/old/betelgeuse-betelgeuse-betelgeuse-/Sat, 05 Nov 2011 01:25:00 GMT