Visions from the Atacama

atacamaFormsThe driest non-polar region in the world is Chile’s Atacama Desert. Some areas have gone 173 months without rain. At about a latitude of 20 degrees south of the equator, prevailing southeast trade winds carry moisture into the the eastern slopes of the Andes. But as the warm air ascends,  it encounters lower atmospheric pressure and expands at the expense of its own internal energy and cools. Precipitation ensues. As the remaining dry air descends over the other side of the Andes it compresses as it descends and heats up. Meanwhile the western shoreline is unusually deep, keeping the Pacific waters at that spot quite cold. There is also a cold current from the south, preventing cold onshore winds from delivering any moisture to the area. Dry air and no clouds for years on end are an astronomer’s fantasy, and it is why the Atacama Desert is the location of the European Southern Observatory’s Very Large Telescope facility (VLT). VLT consists of Antu, Kueyen, Melipal and Yepun, four telescopes that can be operated independently or in harmony to achieve a better resolution.

VLT-MilkyWaySet_6016-Crop-net
In the foreground VLT’s Antu, Kueyen and Melipal with the Milky Way in the sky.

VLT has been producing a large volume of sharp, beautiful images and noteworthy science. In 2009 VLT revealed that the star Betelgeuse has a vast plume of gas almost as wide as our Solar System and a gigantic bubble boiling on its surface. The star’s atmosphere is apparently constantly stretching out into space and then retracting, losing some material in the process.

Presently (late August, early September) from the southeastern part of North America, Betelgeuse is visible before dawn in the southeastern sky.  A star in Orion’s “right bicep”, it’s a red giant with an inconsistent peachy color.  A red giant that is massive enough to go supernova, Betelgeuse is within our galaxy, only 642 light years away. When it explodes some time within a thousand years, it will be bright enough to be visible on Earth in broad daylight. The plume observed from VLT reveals that Betelgeuse is asymmetric. All red giants shed material, but Betelgeuse is not spewing it out equally in all directions.

Two years later, again using the VLT,  astronomers discovered a surrounding nebula, bigger than Betelgeuse itself, stretching 60 billion kilometres beyond the star’s surface, or approximately 400 times the Earth-Sun distance.  The visible part of the nebula turns out to be made up of silicate and alumina dust. When we looked at the cosmic origins of the chemical elements, we learned that silicon is formed from a massive star capable of generating the necessary temperature and density. A red giant of Betelgeuse’s dimensions satisfies those requirements. With regard to the prominence of aluminum, it suggests that supernovae are not the only source of that element.eso1121a

Another telescope in the same desert is the Atacama Large Millimeter/submillimeter betelgeuse-captured-by-almaArray(ALMA). ALMA has just come up ( in 2017) with the highest-resolution image of Betelgeuse to date. It gives us a clearer look at its asymmetry.

A question that may arise is how can a star that’s approximately 20 times more massive than our sun reach the supernova stage in less than 10 million years, while our star, which although isn’t massive enough to go supernova, is still billions of years away from reaching the red giant stage? A very rough calculation will shed some light on this. Twenty times the mass means that there will also be 20 times the fuel available; however for stars of that size, the luminosity ratio is roughly equivalent to the mass ratio of the main sequence stars raised to a power of 3.5. In other words Betelgeuse has more fuel, but the added heat from the much stronger gravity makes the fuel at the core fuse at a prodigiously higher rate! The time that Betelgeuse spent on the main sequence was only 20 / ( 203.5) of the time that will be spent by our sun. Our sun’s life span on the main sequence is about 10 billion years, but Betelgeuse only spent 10×109[20 / ( 203.5) ] or about 6 million years, if we respect significant figures.

atacamaThe Atacama desert was named after a group of people who settled the northeastern border of the desert at least as far back as 500 AD. The Atacama people, known as atacameños or atacamas, used Rapé  smoking ceremonies, which they believed brought them closer to the gods. It’s fitting that their desert now hosts instruments and minds that try to get closer to distant secrets of the universe.

 

 

 

 

 

Other Sources:

http://www.eso.org/

http://www.eso.org/public/news/eso1121/

http://www.eso.org/public/news/eso0927/

Salaris, Maurizio; Santi Cassisi (2005). Evolution of stars and stellar populations. John Wiley & Sons. pp. 138–140

Astronomical Beauty of Hydrogen

If you enjoy chemistry, physics and math, it’s difficult not to love astronomy. Where else can you find an extraterrestrial interplay of this trio of intellectual endeavours?

Uncompounded, neutral hydrogen ( H2 ) occupies an extremely small part of the earth’s atmosphere. It is continuously produced naturally by tectonic and microbiological processes, thanks to iron-rich mafic rocks and iron-hydrogenase enzymes, respectively. Hydrogen gas is less dense than any other so it rises above the rest of the atmosphere’s components. Since hydrogen molecules have the same kinetic energy as air molecules, due to their lower molar mass, H2 ‘s molecules move 3.7 times faster than nitrogen and four times faster than oxygen. Although their molecules’ average speed of 1.8 km/s¹ at 10 °C is still inferior to Earth’s escape velocity of 11.2 km/s, the low mass of hydrogen’s charged products helps them accelerate to higher speeds by mechanisms in the ionosphere. Thus most of the lightweight gas leaves our planet.

Once hydrogen atoms join their own kind in space, they find plenty of company.  In the universe hydrogen is by far the most common element. It is the fuel of main sequence-stars; half of the galaxy’s hydrogen is in suns. The rest is found elsewhere in far lower densities in three forms:

(1) In molecular clouds the simplest element can be found bonded to its own kind as H2. For these clouds to exist, dust must shield them from ultraviolet rays, which are otherwise capable of breaking the single bond between atoms. But H2 is difficult to detect. Due to the perfectly symmetrical nature of the molecule, it cannot absorb infrared(IR) and change vibrational states. But luckily, His usually associated with the detectable and asymmetric carbon monoxide.

molecularcloud1
The above map of the Milky Way was made from the IR detection of carbon monoxide. It simultaneously gives us the location of molecular hydrogen and of molecular clouds in the galaxy. Source:Dame, Hartmann, Thaddeus/NASA

If a molecular cloud becomes large enough, thanks to gravity,  the cloud can become astronomically fertile and lead to the birth of a star. This is what’s happening in the so-called “Pillars of Creation”, part of Messier Object 16, commonly known as the Eagle Nebula.

Pillars
The Pillars of Creation, part of the Eagle Nebula, include molecular hydrogen.  https://en.wikipedia.org/wiki/Eagle_Nebula
nhi_allsky
Taken from http://www.astro.utu.fi

(2) In so-called HI regions, hydrogen can exist in free, atomic form (H). The spin of its lone proton (really a measurement of quantum angular momentum) is normally opposite that of its only electron. In a rare event, a collision between atoms can excite the hydrogen atom from the described spin state to one where the spin of the electron flips. When this state revert to its more common and stable one, energy in the ” twilight-zone” between microwaves and radio is emitted. Since there are so many free hydrogen atoms over large areas of space, despite the rarity of the events, there are enough of them to be observable with a radio telescope. Apparently small radio telescopes (1-3m) for observations of the 21 cm hydrogen
line can be built for about $300.

(3) An HII region is a thin plasma of charged hydrogen atoms and unbound electrons. It forms in the vicinity of certain stars whose radiation is capable of ionizing hydrogen gas. The star’s surface temperature and size will determine the size of the spherical HII region around it—the so-called Strömgren sphere. Some of the most beautiful sights of astronomy include HII regions. When the excited electron in each atom temporarily returns to a previously charged one, electromagnetic energy of select wavelengths is released. We don’t see the ultraviolet, but a transition from the third(n=3) to second(n=2) energy level leads to a prominent red emission at 656 nanometres, while n=4 to 2 and n=5 to 2 transitions create blue (486 nm) and blue-violet (486 nm), respectively. Each of the following familiar nebulae includes HII regions: Messier objects M8 (Lagoon Nebula), M16 (Eagle Nebula), M17 (Omega Nebula),  M42 (Orion Nebula), and M20 (Trifid Nebula), presented below in clockwise order.

One more : Heart and Soul Nebulas in Cassiopeia 

heartsoul_zauner_960

Additional Sources:
The International Encyclopedia of Astronomy. Patrick Moore, editor

https://mwmw.gsfc.nasa.gov/cgi-bin/mw/mw5_zoom2.pl?370,18

http://astronomy.swin.edu.au/cosmos/M/Molecular+Cloud

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/h21.html

http://www.spaceacademy.net.au/spacelab/projects/hlineobs/hlineobs.htm

https://phys.org/news/2016-07-oceans-large-overlooked-source-hydrogen.html

Footnote:

¹hydrogen’s average molecular speed can be calculated by equating 0.5 mv(kinetic energy) to 1.5 RT, where m is the molar mass in kg/mole while R = 8.31 J/mole K. Earth’s escape velocity in the absence of friction is obtained by equating kinetic energy to potential energy GMm/r, where G = gravitational constant and M and r are the mass and radius of the earth.

Mathematical Series Help Us Learn About the Moon

moonCan the moon’s sidereal month be calculated without using Kepler’s Laws? I’m sure one of Jean Meeus’ books has something to that effect, but since I haven’t seen anything of the sort online, here I go.

A lunar synodic month is the time it takes the moon to go through one cycle of phases. Its orbit, like any type of astronomical orbit, is never a perfect circle and is mostly controlled by the earth’s gravity. But the sun also plays a minor role in determining the lunar orbit, so the particular alignment of the Earth, Sun and moon will affect the length of the synodic month by as much as 13.5 hours.

If we measure the lunar month by how long it takes to complete one cycle of phases relative to the background of the stars, we are considering the lunar sidereal month. As the diagram reveals, the synodic month is longer than the sidereal month. To return to one of the 4 phases arbitrarily chosen, the moon has to complete one full turn(sidereal) around the earth, plus the equivalent angle covered by the earth in the time that it takes for the moon to be realigned with the sun’s centre and the earth.

siderealBlog
Two different months follow from two reference frames—the stars’ background for the sidereal month and the moon-earth sun’s centre alignment for the synodic month.  The diagram is mine and homemade with instruments from the 16th century known as the pencil and compass

Even though we will be neglecting the gravitational effects of the sun and even with the assumption that we are dealing with circular orbits (the latter is mostly corrected by the fact that we’re working with average values), we will calculate the average sidereal period and get within only 1.7 seconds of the accepted value. That’s a bit of a wow.

During the sidereal period, L, the earth moves at least (L days)/(365.2422 days)* (360 °). As a result the moon needs to cover part of that θ angle (see diagram) to get back to the full moon. Most of the extra time needed will be the above number of degrees divided by 360 and multiplied by the sidereal period L. Simplifying the expression, this added term becomes L²/365.2422. But the synodic period is not simply L + L²/365.2422. During the extra time that elapses while the moon covers L²/365.2422, the earth is not sitting still, waiting around with a coffee. It keeps revolving around the sun. So if we repeat the calculation we extend the total time needed with a third term, L³/365.2422². This continues and the times added on get progressively smaller. When we add them all up, we observe a geometric series:

Synodic period = L + L²/365.2422 + L³/365.2422² + ….

= L ( 1 + L /365.2422 + L²/365.2422² + L³/365.2422³+ ….)

A quick reminder on how to evaluate such a series:

We let the sum Sn = a + ar + ar² + ar³ + …. ar n-1. By simply multiplying the entire equation by r and by subtracting the two equations, if we then factor and solve for Sn, we will obtain the expression

Sn = a(1-rn)/(1- r)

=a/(1 – r) – arn/(1 – r).

If we then take the limit of the terms as n approaches infinity, as long as -1< r < 1, the second term approaches zero and a simple formula emerges:

Sn = a/(1 – r)

In our case a = L and r = L/ 365.2422.

The synodic period has been long been averaged out by the Babylonians and is also represented by the molad interval of the Hebrews. We now have one or two extra decimal places, so we’ll use 29.530556 days.

29.530556 = L/(1 – L/ 365.2422)

Solving for L, we get a sidereal period estimate of 27.32155 days instead of the accepted average of 27.32153 days. And an error of 0.00002 days, as mentioned at the onset, is equivalent to only 1.7 seconds.

Another geometric series emerges when we calculate the time between tides. A superficial analysis would lead us to the false conclusion that there are four tides every 24 hours or that 6 hours elapse between successive tides. Instead, while the earth rotates tthough the tidal cycle, the moon keeps orbiting our planet. A similar analysis leads to a total tidal period of

1 day + 1/ 29.29.530556 + 1/ 29.29.530556² + 1/1/ 29.29.530556³ + ….

Here a = 1 and r = 1/29.530556, so the average tidal period is 1/ ( 1- 1/29.530556) = 1.03505 days long.  or 24.84120 hours. That means we have, on average, a tide every 6.2103 hours, or one tide every 6 hours, 12.6 minutes.  The timing of tides of course is also influenced by varying lunar speeds, ocean-side landscape and sea floor features, so the length of the actual gap in tides varies over time and location.

Incidentally the cause of tides is often poorly explained in otherwise quality-college physics textbooks and on reputable websites. Contrary to the misconception,  the earth’s centrifugal force towards the moon-earth barycenter (center of mass) is not part of the reason we have two high tides. One can easily do the math and reveal that the centrifugal force is over 80 times weaker than the moon’s gravity even at a distanceΔ from where it presumably creates the second high tide. As explained by a physics professor(Donald E. Simanek who pondered the matter and researched it properly.,

The deformation of land and water of the earth due to the gravitational forces of the moon and sun acting on every part of the earth. This deformation results in two “tidal bulges” one on the side of the earth nearest the moon, and one on the opposite side. These “drive” all of the tidal phenomena with a periodicity synchronized with the moon’s position in the sky.

If you’re wondering why there are two bulges, you have to use vectors to calculate tidal force accelerations for a particle at various points inside and on the surface of our planet. When the gravitational equation is manipulated and part of the expression is represented by a McLaurin series, only one term with a ± in front of it contributes to the tidal forces, leading to following differential field shown in the illustration. The term has a  Δr in the numerator, which represents the difference in distance from the centre of Earth and also an R³ term in the denominator, where R is the distance between the earth and the body exerting tidal forces on it. The latter explains why the tidal forces depend more on the moon than on the sun. If one would naively apply the inverse-square law to rationalise tides, they would conclude that the tidal force due to the sun would be (Ms/Mm* Rm²/Rs²) or 177 times stronger than that of the moon. (Ms and  Mm represent the masses of the sun and moon respectively, and Rm and Rs symbolize the earth-moon and earth-sun distances. But the terms with the R³ term portrays the true tidal force relationship so that in fact the tidal force due to the sun’s influence is actually only 1.99 X 1030kg/7.35 X 1022kg*(3.84 X 108 m)³/(1.5 X 1011 m)³ or 0.45 of that of the moon.

Field_tidal.svg
The tidal force field from Wikipedia’s article on tidal forces

Once again a mathematical series leads to a serious result. 🙂

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