# Distribution of spectral types of “giant” stars

I'm looking for the distribution of spectral types in the "giant" star category. This category is defined as basically B and O main sequence stars and any star that is larger than main sequences outside of it.

My first thought was that the M, K, G, F, and A type distribution were the same as they are in the main sequence but then I realized M has to be dropped, and then i'm not quite sure how K, G, F, and A types evolve and usually only Red Giants are mentioned and I couldn't find numbers for the giants so I'm stuck.

To be clear, I want to know the percentage of giant stars that are in each of the spectral types. So if there are 1000 stars in a given area and 100 giant stars in that areas I only want to know the percentage of each spectral type of the stars in that 100.

OK, I see what you want.

What I did was take the revised Hipparcos catalogue and select all stars closer than 50pc. At this distance there should be a complete sample of the types of stars you were interested in. I can plot an absolute magnitude versus colour diagram for these 7096 stars (see below) and use this to select giants and OB main sequence stars.

I do this by using data from chapter 2 of Zombeck (1992), I define O and B main sequence stars to have $B-V<0$ and $M_V<5$ (to exclude hot white dwarfs). Evolved giants are selected (a bit arbitrarily) to lie significantly above the main sequence.

Then I can plot a frequency histogram of the B-V values of those stars and label them with spectral types, again according to tables in Zombeck (1992). The result is seen below.

Of these 235 stars: None are O stars, 29 are B stars, about 120 are G giants, about 80 are K giants and only 2-3 are M giants

## Colors, Temperatures, and Spectral Types of Stars

If you study this plot, or one of the interactive blackbody radiation demonstrators we used in the last lesson, you can prove to yourself that the color of a star provides a fairly accurate measurement of its surface temperature. For example, a 4500 K blackbody peaks in the red part of the spectrum, a 6000 K blackbody in the green part of the spectrum, and a 7500 K blackbody in the blue part of the spectrum.

Measuring a star’s spectrum is not always easy, but astronomers can often measure a star’s color reasonably easily. To do this, they put a blue filter (B) on the telescope and observe the star. They then re-observe the same star with a visual (V), or yellow, filter. The B filter measures the star’s brightness in blue light, and the V filter measures the star’s brightness in yellow light. The difference between these two, B-V, is the star’s color. The animation below shows a plot of Frequency vs. Intensity, There is a yellow band showing the frequency range that corresponds to the V filter, and a blue band that illustrates the frequency range for the B filter. When you click the play button, you see an animated curve representing blackbodies of different temperatures, and it marks the B and V measurements through these two filters for the different blackbodies. Note how for the three different objects with three different temperatures, that for the coolest object: its B intensity is smaller than its V intensity, for the warmer object: they are roughly the same, and for the hottest object: its B intensity is larger than its V intensity.

In addition, look at this image: Hubble Space Telescope image of star cluster 47 Tucanae. Astronomers took images through different colored filters (in this case, near-infrared, I, visual, V, and ultraviolet, U), and added the three images together to produce a close approximation of the colors we would see of these stars with our own eyes. You can tell that many of the stars are similar in color however some stand out as being much redder than the others. These red stars have the coolest temperatures among the stars in the cluster.

Another good example is this color image of Albireo taken by students at the University of California, Berkeley. They adopted the double star system Albireo as the “Cal Star,” because the two stars (one blue and one yellow) match the school’s colors. Once again, we know from the colors of these stars that the blue star is hotter than the yellow star, because its apparent color indicates that the peak of its emission is in the blue, while the other star’s peak is in the yellow part of the spectrum.

At Hubblesite, they have an extended tutorial on the "Meaning of Color in Hubble Images." which includes a discussion of the filters used by astronomers to determine the color of astronomical objects.

Recall from Lesson 3 that the spectrum of a star is not a true blackbody spectrum because of the presence of absorption lines. The absorption lines visible in the spectra of different stars are different, and we can classify stars into different groups based on the appearance of their spectral lines. In the early 1900s, an astronomer named Annie Jump Cannon took photographic spectra of hundreds of thousands of stars and began to classify them based on their spectral lines. Originally, she started out using the letters of the alphabet to designate different classes of stars (A, B, C…). However, some classes were eventually merged with others, and not all letters were used. The original classification scheme used the strength of the lines of hydrogen to order the spectral types. That is, spectral type A had the strongest lines, B slightly weaker than A, C slightly weaker than B, and so on.

For more information on her life and work, visit the homepage for Annie Jump Cannon at Wellesley College.

Recall from Lesson 3 that the electrons in a gas are the cause of absorption lines—all the photons with the correct amount of energy to cause an electron to jump from one energy level to a higher energy level get absorbed as they pass through the gas. The absorption lines from hydrogen observed in the visible part of the spectrum are called the Balmer series, and they arise when the electron in a hydrogen atom jumps from level 2 to level 3, level 2 to level 4, level 2 to level 5, and so on. The strength of the Balmer lines (that is, how much absorption they cause) depends on the temperature of the cloud. If the cloud is too hot, the electrons in hydrogen have absorbed so much energy that they can break free from the atom. This is called “ionization,” and ionized hydrogen cannot create absorption lines because it no longer has an electron left to absorb any photons. So, very hot stars will have weak Balmer series hydrogen lines because most of their hydrogen has been ionized. Recall also that it takes energy to raise an electron from a lower level to a higher level. So, if the cloud of gas is too cool, the electrons will all be in the lowest energy level (the ground state, level 1). Since the Balmer series lines require electrons to already be in level 2, if there are no hydrogen electrons in level 2 in the gas, there will not be any Balmer series hydrogen lines created by that gas. So, very cool stars will have weak Balmer series hydrogen lines, too. Thus, the stars with the strongest hydrogen lines must be in the middle of the temperature sequence, since their atmospheres are hot enough that hydrogen will have its electrons in level 2, but not so hot that hydrogen becomes ionized.

This theory for the absorption by hydrogen was not understood until after much of the work on stellar classification had been completed. So, after the origin of the strengths of the lines was understood to have some dependence on temperature, the spectral classes for stars were reordered with the hottest stars at the beginning of the sequence and the coolest stars at the end of the sequence. The current order of spectral types is:

For decades, astronomy students have been taught a mnemonic to remember this order: O Be A Fine Girl (or Guy), Kiss Me!

Astronomers divide each class into 10 subclasses—so for example, a G star can be a G0, G1, G2. G9. Our Sun is a G2 star.

In the figure above, spectra of thirteen stars with normal spectral types and three special spectral types observed by the Kitt Peak / WIYN 0.9 meter telescope are presented. You can see two prominent trends in the spectral lines visible in the stars:

• O stars have few lines at all, while M stars have many
• The Hydrogen lines (the four most prominent lines in the A1 star) are strongest in the B6 - F0 stars

One summary comment about this discussion is that stars can be roughly classified by their colors, since the spectral types are arranged by temperature. Also, the apparent color of a star gives you a measurement of its temperature, but more accurate classification usually requires a high quality spectrum.

## 1 Answer 1

The color of a star depends on its mass and temperature. The distribution of these also depends on the age of the galaxy. When the galaxy is very young, there are large amounts of gas still available for star formation, and many young, heavy, hot stars mean that the galaxy is very bright and shines in bluer light. This can be seen as an analogy of the quick flare just when you strike a match. These stars have a short life span and soon die, and as they quickly spend a large part of the available gas in the galaxy, the star formation activity decreases.

With lower star formation and the most of the very massive stars already gone in a fireworks of supernovas, the older galaxy appears redder. This partly due to the fewer blue stars, but also because many more of the low-mass red dwarfs - which form slower but live much longer than the massive, bright blue ones - have accumulated. The galaxy also gets a growing contribution from the heavier main-sequence stars turning into red giants, but this contribution is quite small not always so small, it turns out (*).

How this evolution is going is quite complicated. There are some pretty well-trusted models for the mass distributions of stars in young, star-forming galaxies, called the initial mass function. So the one simplified approach could be to draw a random sample from this distribution and from their masses determine their spectral type and hence their color. But this distribution is only valid at age zero the mass and hence color distribution is subject to significant evolution over time. Unfortunately, this evolution is not simple, and much of it is still unknown. We don't know, for example, if young galaxies have a period of steady star formation, form most of their massive stars in a single or a series of bursts, or something else.

Also, the evolution is not smooth, as collisions and mergers between galaxies have a heavy influence on Star Formation Rates and hence mass and color distribution. Such collisions and mergers have been found to be very common in the local Universe, and probably much more so in the early Universe.

We can say one thing, though: the Milky Way stellar mass distribution seems to be relatively typical for galaxies at our stage of evolution that is, a medium-large spiral galaxy with a steady but not strong star formation activity. Later, the Milky Way is likely to merge with first Andromeda and later other galaxies, which will probably trigger more star bursts and later deprive the new supergalaxy of almost all its gas. At this point, it will become an elliptical galaxy with few blue stars and many more red stars, and much less dust and gas than it has today.

So, to answer the question:

You first need to determine what type of galaxy you want. If you want a young, star forming galaxy, one of the initial mass function mentioned on Wikipedia should be just fine, and this will also be the simplest solution. If you want smaller starburst galaxies, gas-rich Milky Way-type galaxies or large, red, gas-poor elliptical galaxies, you should probably try searching the literature for mass functions to use - see for example

A. Tamm, E. Tempel, P. Tenjes, O. Tihhonova, T. Tuvikene. Stellar mass map and dark matter distribution in M31. Astron. Astrophys. (accepted). arXiv:1208.5712 [astro-ph.CO].

Once you have found your mass function, I'd draw a random sample from this, and from the masses determine the spectral type. This is actually not possible to do without a full-fledged numerical stellar model, but if you limit yourself to main-sequence stars, there are some approximations you can use here and in the Wikipedia entry for the mass-luminosity relation. This should (although I haven't had the time to double check) give you the temperature of the star from the mass, from which you can infer the color by using Wien's displacement law.

In short, there is no easy way to do what you want, but I hope I have sketched a way to do it not-too-wrong.

(*) From a remark from a professor at our department during a Ph. D. Thesis defense today.

This paper seems to have something that could be practically useful as to the evolution of the Mass Function (i.e. the mass distribution) over time. This is for galaxy clusters, not for individual galaxies, so the result is, you could say, "unrealistically typical", but should be a good start:

Guido De Marchi, Francesco Paresce, Simon Portegies Zwart. The stellar IMF of Galactic clusters and its evolution. arXiv:astro-ph/0409601

## Contents

Cool stars, particularly class M, show molecular bands, with titanium(II) oxide (TiO) especially strong. A small proportion of these cool stars also show correspondingly strong bands of zirconium oxide (ZrO). The existence of clearly detectable ZrO bands in visual spectra is the definition of an S-type star. [1]

• α series, in the blue at 464.06 nm, 462.61 nm, and 461.98 nm
• β series, in the yellow at 555.17 nm and 571.81 nm
• γ series, in the red at 647.4 nm, 634.5 nm, and 622.9 nm [2]

The original definition of an S star was that the ZrO bands should be easily detectable on low dispersion photographic spectral plates, but more modern spectra allow identification of many stars with much weaker ZrO. MS stars, intermediate with normal class M stars, have barely detectable ZrO but otherwise normal class M spectra. SC stars, intermediate with carbon stars, have weak or undetectable ZrO, but strong sodium D lines and detectable but weak C2 bands. [3]

S star spectra also show other differences to those of normal M class giants. The characteristic TiO bands of cool giants are weakened in most S stars, compared to M stars of similar temperature, and completely absent in some. Features related to s-process isotopes such as YO bands, Sr I lines, Ba II lines, and LaO bands, and also sodium D lines are all much stronger. However, VO bands are absent or very weak. [4] The existence of spectral lines from the period 5 element Technetium (Tc) is also expected as a result of the s-process neutron capture, but a substantial fraction of S stars show no sign of Tc. Stars with strong Tc lines are sometimes referred to as Technetium stars, and they can be of class M, S, C, or the intermediate MS and SC. [5]

Some S stars, especially Mira variables, show strong hydrogen emission lines. The Hβ emission is often unusually strong compared to other lines of the Balmer series in a normal M star, but this is due to the weakness of the TiO band that would otherwise dilute the Hβ emission. [1]

The spectral class S was first defined in 1922 to represent a number of long-period variables (meaning Mira variables) and stars with similar peculiar spectra. Many of the absorption lines in the spectra were recognised as unusual, but their associated elements were not known. The absorption bands now recognised as due to ZrO are clearly listed as major features of the S-type spectra. At that time, class M was not divided into numeric sub-classes, but into Ma, Mb, Mc, and Md. The new class S was simply left as either S or Se depending on the existence of emission lines. It was considered that the Se stars were all LPVs and the S stars were non-variable, [6] but exceptions have since been found. For example, π 1 Gruis is now known to be a semiregular variable. [7]

The classification of S stars has been revised several times since its first introduction, to reflect advances in the resolution of available spectra, the discovery of greater numbers of S-type stars, and better understanding of the relationships between the various cool luminous giant spectral types.

### Comma notation Edit

The formalisation of S star classification in 1954 introduced a two-dimensional scheme of the form SX,Y. For example, R Andromedae is listed as S6,6e. [1]

X is the temperature class. It is a digit between 1 (although the smallest type actually listed is S1.5) and 9, intended to represent a temperature scale corresponding approximately to the sequence of M1 to M9. The temperature class is actually calculated by estimating intensities for the ZrO and TiO bands, then summing the larger intensity with half the smaller intensity. [1]

Y is the abundance class. It is also a digit between 1 and 9, assigned by multiplying the ratio of ZrO and TiO bands by the temperature class. This calculation generally yields a number which can be rounded down to give the abundance class digit, but this is modified for higher values: [1]

In practice, spectral types for new stars would be assigned by referencing to the standard stars, since the intensity values are subjective and would be impossible to reproduce from spectra taken under different conditions. [1]

A number of drawbacks came to light as S stars were studied more closely and the mechanisms behind the spectra came to be understood. The strengths of the ZrO and TiO are influenced both by temperature and by actual abundances. The S stars represent a continuum from having oxygen slightly more abundant than carbon to carbon being slightly more abundant than oxygen. When carbon becomes more abundant than oxygen, the free oxygen is rapidly bound into CO and abundances of ZrO and TiO drop dramatically, making them a poor indicator in some stars. The abundance class also becomes unusable for stars with more carbon than oxygen in their atmospheres. [8]

This form of spectral type is a common type seen for S stars, possibly still the most common form. [9]

### Elemental intensities Edit

The first major revision of the classification for S stars completely abandons the single-digit abundance class in favour of explicit abundance intensities for Zr and Ti. [10] So R And is listed, at a normal maximum, with a spectral type of S5e Zr5 Ti2. [9]

In 1979 Ake defined an abundance index based on the ZrO, TiO, and YO band intensities. This single digit between 1 and 7 was intended to represent the transition from MS stars through increasing C/O ratios to SC stars. Spectral types were still listed with explicit Zr and Ti intensity values, and the abundance index was included separately in the list of standard stars. [8]

### Slash notation Edit

The abundance index was immediately adopted and extended to run from 1 to 10, differentiating abundances in SC stars. It was now quoted as part of the spectral type in preference to separate Zr and Ti abundances. To distinguish it from the earlier abandoned abundance class it was used with a slash character after the temperature class, so that the spectral class for R And became S5/4.5e. [3]

The new abundance index is not calculated directly, but is assigned from the relative strengths of a number of spectral features. It is designed to closely indicate the sequence of C/O ratios from below 0.95 to about 1.1. Primarily the relative strength of ZrO and TiO bands forms a sequence from MS stars to abundance index 1 through 6. Abundance indices 7 to 10 are the SC stars and ZrO is weak or absent so the relative strength of the sodium D lines and Cs bands is used. Abundance index 0 is not used, and abundance index 10 is equivalent to a carbon star Cx,2 so it is also never seen. [4]

The derivation of the temperature class is also refined, to use line ratios in addition to the total ZrO and TiO strength. For MS stars and those with abundance index 1 or 2, the same TiO band strength criteria as for M stars can be applied. Ratios of different ZrO bands at 530.5 nm and 555.1 nm are useful with abundance indices 3 and 4, and the sudden appearance of LaO bands at cooler temperatures. The ratio of Ba II and Sr I lines is also useful at the same indices and for carbon-rich stars with abundance index 7 to 9. Where ZrO and TiO are weak or absent the ratio of the blended features at 645.6 nm and 645.0 nm can be used to assign the temperature class. [4]

### Asterisk notation Edit

With the different classification schemes and the difficulties of assigning a consistent class across the whole range of MS, S, and SC stars, other schemes are sometimes used. For example, one survey of new S/MS, carbon, and SC stars uses a two-dimensional scheme indicated by an asterisk, for example S5*3. The first digit is based on TiO strength to approximate the class M sequence, and the second is based solely on ZrO strength. [2]

### Standard stars Edit

This table shows the spectral types of a number of well-known S stars as they were classified at various times. Most of the stars are variable, usually of the Mira type. Where possible the table shows the type at maximum brightness, but several of the Ake types in particular are not at maximum brightness and so have a later type. ZrO and TiO band intensities are also shown if they are published (an x indicates that no bands were found). If the abundances are part of the formal spectral type then the abundance index is shown.

Comparison of spectral types under different classification schemes
Star Keenan
(1954) [1]
Keenan et al.
(1974) [11]
Ake
(1979) [8]
Keenan-Boeshaar
(1980) [4]
R Andromedae S6,6e: Zr4 Ti3 S4,6e S8e Zr6 4 S5/4.5e Zr5 Ti2
X Andromedae S3,9e Zr3 Ti0 S2,9e: S5.5e Zr4 5 S5/4.5e Zr2.5 Tix
RR Andromedae S7,2e: Zr2 Ti6.5 S6,2e: S6.5e Zr3 Ti6 2 S6/3.5e Zr4+ Ti4
W Aquilae S4,9: Zr4 Ti0 S3,9e: S6/6e Zr6 Ti0
BD Camelopardalis S5,3 Zr2.5 Ti4 S3.5 Zr2.5 Ti3 2 S3.5/2 Zr2+ Ti3
BH Crucis SC8,6: [12] SC4.5/8-e Zr0 Tix Na10:
Chi Cygni S7,1e: Zr0-2 Ti7 S7,2e S9.5 Zr3 Ti9 1 S6+/1e = Ms6+ Zr2 Ti6
R Cygni S3.5,9e: Zr3.5 Ti0 S3,9e S8e Zr7 Ti3: 4 S5/6e Zr4 Tix
R Geminorum S3,9e: Zr3 Ti0 S3,9e S8e Zr5 5 S4/6e Zr3.5 Tix

There are two distinct classes of S-type stars: intrinsic S stars and extrinsic S stars. The presence of Technetium is used to distinguish the two classes, only being found in the intrinsic S-type stars.

### Intrinsic S stars Edit

Intrinsic S-type stars are thermal pulsing asymptotic giant branch (TP-AGB) stars. AGB stars have inert carbon-oxygen cores and undergo fusion both in an inner helium shell and an outer hydrogen shell. They are large cool M class giants. The thermal pulses, created by flashes from the helium shell, cause strong convection within the upper layers of the star. These pulses become stronger as the star evolves and in sufficiently massive stars the convection becomes deep enough to dredge up fusion products from the region between the two shells to the surface. These fusion products include carbon and s-process elements. [14] The s-process elements include zirconium (Zr), yttrium (Y), lanthanum (La), technetium (Tc), barium (Ba), and strontium (Sr), which form the characteristic S class spectrum with ZrO, YO, and LaO bands, as well as Tc, Sr, and Ba lines. The atmosphere of S stars has a carbon to oxygen ratio in the range 0.5 to < 1. [15] Carbon enrichment continues with subsequent thermal pulses until the carbon abundance exceeds the oxygen abundance, at which point the oxygen in the atmosphere is rapidly locked into CO and formation of the oxides diminishes. These stars show intermediate SC spectra and further carbon enrichment leads to a carbon star. [16]

### Extrinsic S stars Edit

The Technetium isotope produced by neutron capture in the s-process is 99 Tc and it has a half life of around 200,000 years in a stellar atmosphere. Any of the isotope present when a star formed would have completely decayed by the time it became a giant, and any newly formed 99 Tc dredged up in an AGB star would survive until the end of the AGB phase, making it difficult for a red giant to have other s-process elements in its atmosphere without technetium. S-type stars without technetium form by the transfer of technetium-rich matter, as well as other dredged-up elements, from an intrinsic S star in a binary system onto a smaller less-evolved companion. After a few hundred thousand years, the 99 Tc will have decayed and a technetium-free star enriched with carbon and other s-process elements will remain. When this star is, or becomes, a G or K type red giant, it will be classified as a Barium star. When it evolves to temperatures cool enough for ZrO absorption bands to show in the spectrum, approximately M class, it will be classified as an S-type star. These stars are called extrinsic S stars. [16] [17]

Stars with a spectral class of S only form under a narrow range of conditions and they are uncommon. The distributions and properties of intrinsic and extrinsic S stars are different, reflecting their different modes of formation.

TP-AGB stars are difficult to identify reliably in large surveys, but counts of normal M-class luminous AGB stars and similar S-type and carbon stars have shown different distributions in the galaxy. S stars are distributed in a similar way to carbon stars, but there are only around a third as many as the carbon stars. Both types of carbon-rich star are very rare near to the galactic centre, but make up 10% – 20% of all the luminous AGB stars in the solar neighbourhood, so that S stars are around 5% of the AGB stars. The carbon-rich stars are also concentrated more closely in the galactic plane. S-type stars make up a disproportionate number of Mira variables, 7% in one survey compared to 3% of all AGB stars. [18]

Extrinsic S stars are not on the TP-AGB, but are red giant branch stars or early AGB stars. Their numbers and distribution are uncertain. They have been estimated to make up between 30% and 70% of all S-type stars, although only a tiny fraction of all red giant branch stars. They are less strongly concentrated in the galactic disc, indicating that they are from an older population of stars than the intrinsic group. [16]

Very few intrinsic S stars have had their mass directly measured using a binary orbit, although their masses have been estimated using Mira period-mass relations or pulsations properties. The observed masses were found to be around 1.5 – 5 M [16] until very recently when Gaia parallaxes helped discover intrinsic S stars with solar-like masses and metallicities. [15] Models of TP-AGB evolution show that the third dredge-up becomes larger as the shells move towards the surface, and that less massive stars experience fewer dredge-ups before leaving the AGB. Stars with masses of 1.5 – 2.0 M will experience enough dredge-ups to become carbon stars, but they will be large events and the star will usually skip straight past the crucial C/O ratio near 1 without becoming an S-type star. More massive stars reach equal levels of carbon and oxygen gradually during several small dredge-ups. Stars more than about 4 M experience hot bottom burning (the burning of carbon at the base of the convective envelope) which prevents them becoming carbon stars, but they may still become S-type stars before reverting to an oxygen-rich state. [19] Extrinsic S stars are always in binary systems and their calculated masses are around 1.6 – 2.0 M . This is consistent with RGB stars or early AGB stars. [17]

Intrinsic S stars have luminosities around 5,000 – 10,000 L , [20] [21] although they are usually variable. [16] Their temperatures average about 2,300 K for the Mira S stars and 3,100 K for the non-Mira S stars, a few hundred K warmer than oxygen-rich AGB stars and a few hundred K cooler than carbon stars. Their radii average about 526 R for the Miras and 270 R for the non-miras, larger than oxygen-rich stars and smaller than carbon stars. [22] Extrinsic S stars have luminosities typically around 2,000 L , temperatures between 3,150 and 4,000 K, and radii less than 150 R . This means they lie below the red giant tip and will typically be RGB stars rather than AGB stars. [23]

Extrinsic S stars lose considerable mass through their stellar winds, similar to oxygen-rich TP-AGB stars and carbon stars. Typically the rates are around 1/10,000,000th the mass of the sun per year, although in extreme cases such as W Aquilae they can be more than ten times higher. [20]

It is expected that the existence of dust drives the mass loss in cool stars, but it is unclear what type of dust can form in the atmosphere of an S star with most carbon and oxygen locked into CO gas. The stellar winds of S stars are comparable to oxygen-rich and carbon-rich stars with similar physical properties. There is about 300 times more gas than dust observed in the circumstellar material around S stars. It is believed to be made up of metallic iron, FeSi, silicon carbide, and forsterite. Without silicates and carbon, it is believed that nucleation is triggered by TiC, ZrC, and TiO2. [21]

Detached dust shells are seen around a number of carbon stars, but not S-type stars. Infrared excesses indicate that there is dust around most intrinsic S stars, but the outflow has not been sufficient and longlasting enough to form a visible detached shell. The shells are thought to form during a superwind phase very late in the AGB evolution. [20]

BD Camelopardalis is a naked-eye example of an extrinsic S star. It is a slow irregular variable in a symbiotic binary system with a hotter companion which may also be variable. [24]

The Mira variable Chi Cygni is an intrinsic S star. When near maximum light, it is the sky's brightest S-type star. [25] It has a variable late type spectrum about S6 to S10, with features of zirconium, titanium and vanadium oxides, sometimes bordering on the intermediate MS type. [4] A number of other prominent Mira variables such as R Andromedae and R Cygni are also S-type stars, as well as the peculiar semiregular variable π 1 Gruis. [25]

The naked-eye star ο 1 Ori is an intermediate MS star and small amplitude semiregular variable [7] with a DA3 white dwarf companion. [26] The spectral type has been given as S3.5/1-, [4] M3III(BaII), [27] or M3.2IIIaS. [7]

## 6.2: Blackbody Radiation

• Contributed by OpenStax
• General Physics at OpenStax CNX

By the end of this section you will be able to:

• Apply Wien&rsquos and Stefan&rsquos laws to analyze radiation emitted by a blackbody
• Explain Planck&rsquos hypothesis of energy quanta

All bodies emit electromagnetic radiation over a range of wavelengths. In an earlier chapter, we learned that a cooler body radiates less energy than a warmer body. We also know by observation that when a body is heated and its temperature rises, the perceived wavelength of its emitted radiation changes from infrared to red, and then from red to orange, and so forth. As its temperature rises, the body glows with the colors corresponding to ever-smaller wavelengths of the electromagnetic spectrum. This is the underlying principle of the incandescent light bulb: A hot metal filament glows red, and when heating continues, its glow eventually covers the entire visible portion of the electromagnetic spectrum. The temperature (T) of the object that emits radiation, or the emitter, determines the wavelength at which the radiated energy is at its maximum. For example, the Sun, whose surface temperature is in the range between 5000 K and 6000 K, radiates most strongly in a range of wavelengths about 560 nm in the visible part of the electromagnetic spectrum. Your body, when at its normal temperature of about 300 K, radiates most strongly in the infrared part of the spectrum.

Radiation that is incident on an object is partially absorbed and partially reflected. At thermodynamic equilibrium, the rate at which an object absorbs radiation is the same as the rate at which it emits it. Therefore, a good absorber of radiation (any object that absorbs radiation) is also a good emitter. A perfect absorber absorbs all electromagnetic radiation incident on it such an object is called a blackbody.

Figure (PageIndex<1>): A blackbody is physically realized by a small hole in the wall of a cavity radiator.

Although the blackbody is an idealization, because no physical object absorbs 100% of incident radiation, we can construct a close realization of a blackbody in the form of a small hole in the wall of a sealed enclosure known as a cavity radiator, as shown in Figure (PageIndex<1>). The inside walls of a cavity radiator are rough and blackened so that any radiation that enters through a tiny hole in the cavity wall becomes trapped inside the cavity. At thermodynamic equilibrium (at temperature T), the cavity walls absorb exactly as much radiation as they emit. Furthermore, inside the cavity, the radiation entering the hole is balanced by the radiation leaving it. The emission spectrum of a blackbody can be obtained by analyzing the light radiating from the hole. Electromagnetic waves emitted by a blackbody are called blackbody radiation.

Figure (PageIndex<2>): The intensity of blackbody radiation versus the wavelength of the emitted radiation. Each curve corresponds to a different blackbody temperature, starting with a low temperature (the lowest curve) to a high temperature (the highest curve).

The intensity (I(lambda, T)) of blackbody radiation depends on the wavelength (lambda) of the emitted radiation and on the temperature T of the blackbody (Figure (PageIndex<2>)). The function (I(lambda, T)) is the power intensity that is radiated per unit wavelength in other words, it is the power radiated per unit area of the hole in a cavity radiator per unit wavelength. According to this definition, (I(lambda, T)dlambda) is the power per unit area that is emitted in the wavelength interval from (lambda) to (lambda + dlambda). The intensity distribution among wavelengths of radiation emitted by cavities was studied experimentally at the end of the nineteenth century. Generally, radiation emitted by materials only approximately follows the blackbody radiation curve (Figure (PageIndex<3>)) however, spectra of common stars do follow the blackbody radiation curve very closely.

Figure (PageIndex<3>): The spectrum of radiation emitted from a quartz surface (blue curve) and the blackbody radiation curve (black curve) at 600 K.

Two important laws summarize the experimental findings of blackbody radiation: Wien&rsquos displacement law and Stefan&rsquos law. Wien&rsquos displacement law is illustrated in Figure (PageIndex<2>) by the curve connecting the maxima on the intensity curves. In these curves, we see that the hotter the body, the shorter the wavelength corresponding to the emission peak in the radiation curve. Quantitatively, Wien&rsquos law reads

[lambda_T = 2.898 imes 10^ <-3>m cdot K label]

where (lambda_) is the position of the maximum in the radiation curve. In other words, (lambda_) is the wavelength at which a blackbody radiates most strongly at a given temperature T. Note that in Equation ef, the temperature is in kelvins. Wien&rsquos displacement law allows us to estimate the temperatures of distant stars by measuring the wavelength of radiation they emit.

Example (PageIndex<1>): Temperatures of Distant Stars

On a clear evening during the winter months, if you happen to be in the Northern Hemisphere and look up at the sky, you can see the constellation Orion (The Hunter). One star in this constellation, Rigel, flickers in a blue color and another star, Betelgeuse, has a reddish color, as shown in Figure (PageIndex<4>). Which of these two stars is cooler, Betelgeuse or Rigel?

Figure (PageIndex<4>): In the Orion constellation, the red star Betelgeuse, which usually takes on a yellowish tint, appears as the figure&rsquos right shoulder (in the upper left). The giant blue star on the bottom right is Rigel, which appears as the hunter&rsquos left foot. (credit left: modification of work by NASA c/o Matthew Spinelli)

We treat each star as a blackbody. Then according to Wien&rsquos law, its temperature is inversely proportional to the wavelength of its peak intensity. The wavelength (lambda_^<(blue)>) of blue light is shorter than the wavelength (lambda_^<(red)>) of red light. Even if we do not know the precise wavelengths, we can still set up a proportion.

Writing Wien&rsquos law for the blue star and for the red star, we have

When simplified, this gives

Therefore, Betelgeuse is cooler than Rigel.

Significance

Note that Wien&rsquos displacement law tells us that the higher the temperature of an emitting body, the shorter the wavelength of the radiation it emits. The qualitative analysis presented in this example is generally valid for any emitting body, whether it is a big object such as a star or a small object such as the glowing filament in an incandescent lightbulb.

The flame of a peach-scented candle has a yellowish color and the flame of a Bunsen&rsquos burner in a chemistry lab has a bluish color. Which flame has a higher temperature?

The second experimental relation is Stefan&rsquos law, which concerns the total power of blackbody radiation emitted across the entire spectrum of wavelengths at a given temperature. In (PageIndex<2>) , this total power is represented by the area under the blackbody radiation curve for a given T. As the temperature of a blackbody increases, the total emitted power also increases. Quantitatively, Stefan&rsquos law expresses this relation as

where (A) is the surface area of a blackbody, (T) is its temperature (in kelvins), and (&sigma) is the Stefan&ndashBoltzmann constant, (sigma = 5.670 imes 10^ <-8>W/(m^2 cdot K^4)). Stefan&rsquos law enables us to estimate how much energy a star is radiating by remotely measuring its temperature.

Example (PageIndex<2>): Power Radiated by Stars

A star such as our Sun will eventually evolve to a &ldquored giant&rdquo star and then to a &ldquowhite dwarf&rdquo star. A typical white dwarf is approximately the size of Earth, and its surface temperature is about (2.5 imes 10^4 K). A typical red giant has a surface temperature of (3.0 imes 10^3 K) and a radius

100,000 times larger than that of a white dwarf. What is the average radiated power per unit area and the total power radiated by each of these types of stars? How do they compare?

If we treat the star as a blackbody, then according to Stefan&rsquos law, the total power that the star radiates is proportional to the fourth power of its temperature. To find the power radiated per unit area of the surface, we do not need to make any assumptions about the shape of the star because P/A depends only on temperature. However, to compute the total power, we need to make an assumption that the energy radiates through a spherical surface enclosing the star, so that the surface area is (A = 4pi R^2), where R is its radius.

A simple proportion based on Stefan&rsquos law gives

The power emitted per unit area by a white dwarf is about 5000 times that the power emitted by a red giant. Denoting this ratio by (a=4.8×10^3), Equation ef <6.5>gives

We see that the total power emitted by a white dwarf is a tiny fraction of the total power emitted by a red giant. Despite its relatively lower temperature, the overall power radiated by a red giant far exceeds that of the white dwarf because the red giant has a much larger surface area. To estimate the absolute value of the emitted power per unit area, we again use Stefan&rsquos law. For the white dwarf, we obtain

The analogous result for the red giant is obtained by scaling the result for a white dwarf:

Significance

To estimate the total power emitted by a white dwarf, in principle, we could use Equation ef<6.7>. However, to find its surface area, we need to know the average radius, which is not given in this example. Therefore, the solution stops here. The same is also true for the red giant star.

An iron poker is being heated. As its temperature rises, the poker begins to glow&mdashfirst dull red, then bright red, then orange, and then yellow. Use either the blackbody radiation curve or Wien&rsquos law to explain these changes in the color of the glow.

The wavelength of the radiation maximum decreases with increasing temperature.

Suppose that two stars, (&alpha) and (&beta), radiate exactly the same total power. If the radius of star (&alpha) is three times that of star (&beta), what is the ratio of the surface temperatures of these stars? Which one is hotter?

(T_/T_ <eta>= 1/sqrt <3>cong 0.58), so the star (eta) is hotter.

The term &ldquoblackbody&rdquo was coined by Gustav R. Kirchhoff in 1862. The blackbody radiation curve was known experimentally, but its shape eluded physical explanation until the year 1900. The physical model of a blackbody at temperature T is that of the electromagnetic waves enclosed in a cavity (Figure (PageIndex<1>)) and at thermodynamic equilibrium with the cavity walls. The waves can exchange energy with the walls. The objective here is to find the energy density distribution among various modes of vibration at various wavelengths (or frequencies). In other words, we want to know how much energy is carried by a single wavelength or a band of wavelengths. Once we know the energy distribution, we can use standard statistical methods (similar to those studied in a previous chapter) to obtain the blackbody radiation curve, Stefan&rsquos law, and Wien&rsquos displacement law. When the physical model is correct, the theoretical predictions should be the same as the experimental curves.

In a classical approach to the blackbody radiation problem, in which radiation is treated as waves (as you have studied in previous chapters), the modes of electromagnetic waves trapped in the cavity are in equilibrium and continually exchange their energies with the cavity walls. There is no physical reason why a wave should do otherwise: Any amount of energy can be exchanged, either by being transferred from the wave to the material in the wall or by being received by the wave from the material in the wall. This classical picture is the basis of the model developed by Lord Rayleigh and, independently, by Sir James Jeans. The result of this classical model for blackbody radiation curves is known as the Rayleigh&ndashJeans law. However, as shown in Figure (PageIndex<5>), the Rayleigh&ndashJeans law fails to correctly reproduce experimental results. In the limit of short wavelengths, the Rayleigh&ndashJeans law predicts infinite radiation intensity, which is inconsistent with the experimental results in which radiation intensity has finite values in the ultraviolet region of the spectrum. This divergence between the results of classical theory and experiments, which came to be called the ultraviolet catastrophe, shows how classical physics fails to explain the mechanism of blackbody radiation.

Figure (PageIndex<5>): The ultraviolet catastrophe: The Rayleigh&ndashJeans law does not explain the observed blackbody emission spectrum.

The blackbody radiation problem was solved in 1900 by Max Planck. Planck used the same idea as the Rayleigh&ndashJeans model in the sense that he treated the electromagnetic waves between the walls inside the cavity classically, and assumed that the radiation is in equilibrium with the cavity walls. The innovative idea that Planck introduced in his model is the assumption that the cavity radiation originates from atomic oscillations inside the cavity walls, and that these oscillations can have only discrete values of energy. Therefore, the radiation trapped inside the cavity walls can exchange energy with the walls only in discrete amounts. Planck&rsquos hypothesis of discrete energy values, which he called quanta, assumes that the oscillators inside the cavity walls have quantized energies. This was a brand new idea that went beyond the classical physics of the nineteenth century because, as you learned in a previous chapter, in the classical picture, the energy of an oscillator can take on any continuous value. Planck assumed that the energy of an oscillator ((E_n)) can have only discrete, or quantized, values:

[E_n = nhf, , where , n = 1,2,3, ldots label]

In Equation ef, (f) is the frequency of Planck&rsquos oscillator. The natural number (n) that enumerates these discrete energies is called a quantum number. The physical constant (h) is called Planck&rsquos constant:

[h = 6.626 imes 10^ <-34>J cdot s = 4.136 imes 10^ <-15>eV cdot s label<6.10>]

Each discrete energy value corresponds to a quantum state of a Planck oscillator. Quantum states are enumerated by quantum numbers. For example, when Planck&rsquos oscillator is in its first (n 1) quantum state, its energy is (E_1 = hf) when it is in the (n = 2) quantum state, its energy is (E_2 = 2hf) when it is in the (n = 3) quantum state, (E_3 = 3hf) and so on.

Note that Equation ef shows that there are infinitely many quantum states, which can be represented as a sequence <hf, 2hf, 3hf,&hellip, (n &ndash 1)hf, nhf, (n + 1)hf,&hellip>. Each two consecutive quantum states in this sequence are separated by an energy jump, (delta E = hf). An oscillator in the wall can receive energy from the radiation in the cavity (absorption), or it can give away energy to the radiation in the cavity (emission). The absorption process sends the oscillator to a higher quantum state, and the emission process sends the oscillator to a lower quantum state. Whichever way this exchange of energy goes, the smallest amount of energy that can be exchanged is hf. There is no upper limit to how much energy can be exchanged, but whatever is exchanged must be an integer multiple of hf. If the energy packet does not have this exact amount, it is neither absorbed nor emitted at the wall of the blackbody.

PLANCK&rsquoS QUANTUM HYPOTHESIS

Planck&rsquos hypothesis of energy quanta states that the amount of energy emitted by the oscillator is carried by the quantum of radiation, (Delta E):

Recall that the frequency of electromagnetic radiation is related to its wavelength and to the speed of light by the fundamental relation (flambda = c). This means that we can express Equation ef <6.10>equivalently in terms of wavelength (lambda). When included in the computation of the energy density of a blackbody, Planck&rsquos hypothesis gives the following theoretical expression for the power intensity of emitted radiation per unit wavelength:

where c is the speed of light in vacuum and kBkB is Boltzmann&rsquos constant, (k_B = 1.380 imes 10^ <-23>J/K). The theoretical formula expressed in Equation ef <6.11>is called Planck&rsquos blackbody radiation law. This law is in agreement with the experimental blackbody radiation curve (Figure (PageIndex<2>)). In addition, Wien&rsquos displacement law and Stefan&rsquos law can both be derived from Equation ef<6.11>. To derive Wien&rsquos displacement law, we use differential calculus to find the maximum of the radiation intensity curve (I(lambda, T)). To derive Stefan&rsquos law and find the value of the Stefan&ndashBoltzmann constant, we use integral calculus and integrate (I(lambda, T)) to find the total power radiated by a blackbody at one temperature in the entire spectrum of wavelengths from (lambda = 0) to (lambda = infty). This derivation is left as an exercise later in this chapter.

Figure (PageIndex<6>): Planck&rsquos theoretical result (continuous curve) and the experimental blackbody radiation curve (dots).

Example (PageIndex<3>): Planck&rsquos Quantum Oscillator

A quantum oscillator in the cavity wall in Figure (PageIndex<1>) is vibrating at a frequency of (5.0 imes 10^ <14>Hz). Calculate the spacing between its energy levels.

Energy states of a quantum oscillator are given by Equation ef. The energy spacing (Delta E) is obtained by finding the energy difference between two adjacent quantum states for quantum numbers n + 1 and n.

We can substitute the given frequency and Planck&rsquos constant directly into the equation:

[egin Delta E &= E_ &minus En = (n + 1)hf &minus nhf [5pt] &= hf [5pt] &= (6.626 imes 10^ <&minus34>, J cdot s)(5.0 imes 10^ <14>, Hz) [5pt] &= 3.3 imes 10^ <&minus 19>, J end]

Significance

Note that we do not specify what kind of material was used to build the cavity. Here, a quantum oscillator is a theoretical model of an atom or molecule of material in the wall.

A molecule is vibrating at a frequency of (5.0 imes 10^<14>, Hz). What is the smallest spacing between its vibrational energy levels?

Example (PageIndex<4>): Quantum Theory Applied to a Classical Oscillator

A 1.0-kg mass oscillates at the end of a spring with a spring constant of 1000 N/m. The amplitude of these oscillations is 0.10 m. Use the concept of quantization to find the energy spacing for this classical oscillator. Is the energy quantization significant for macroscopic systems, such as this oscillator?

We use Equation ef as though the system were a quantum oscillator, but with the frequency f of the mass vibrating on a spring. To evaluate whether or not quantization has a significant effect, we compare the quantum energy spacing with the macroscopic total energy of this classical oscillator.

For the spring constant, (k = 1.0 imes 10^3 N/m), the frequency f of the mass, (m = 1.0 , kg), is

The energy quantum that corresponds to this frequency is

[Delta E = hf = (6.626 imes 10^ <-34>J cdot s)(5.0 , Hz) = 3.3 imes 10^ <-33>J onumber]

When vibrations have amplitude (A = 0.10 , m), the energy of oscillations is

[E = dfrac<1> <2>kA^2 = dfrac<1><2>(1000 , N/m)(0.1 , m)^2 = 5.0 , J onumber]

Significance

Thus, for a classical oscillator, we have (Delta E/E approx 10^<-34>). We see that the separation of the energy levels is immeasurably small. Therefore, for all practical purposes, the energy of a classical oscillator takes on continuous values. This is why classical principles may be applied to macroscopic systems encountered in everyday life without loss of accuracy.

Would the result in Example (PageIndex<4>) be different if the mass were not 1.0 kg but a tiny mass of 1.0 µg, and the amplitude of vibrations were 0.10 µm?

No, because then (Delta E /E approx 10^<-21>)

When Planck first published his result, the hypothesis of energy quanta was not taken seriously by the physics community because it did not follow from any established physics theory at that time. It was perceived, even by Planck himself, as a useful mathematical trick that led to a good theoretical &ldquofit&rdquo to the experimental curve. This perception was changed in 1905 when Einstein published his explanation of the photoelectric effect, in which he gave Planck&rsquos energy quantum a new meaning: that of a particle of light.

## After the Collapse.

After the great collapse of the core corporation where millions of electrons were employees, conditions in the once envied Micro-Estates deteriorated. Electrons from all around were forced to move into the estates when their suburban lifestyle became untenable. Overcrowding began to run rampant, and the electrons degenerated into survival mode. Each electron was forced to hold the little ground he or she had left at any cost forgotten were the days where electrons were bound by society to stay the customary distance away from one another.

In the story this last piece of ground that an electron has is its micro-estate. In reality, the analog to these are called microstates and no two electrons, or any other fermion for that matter, can occupy the same state. A microstate is single discrete environment an electron can exist in, defined by several quantum mechanical properties (quantum numbers), and is a unique realization of an energy that a single electron can have. This fundamental property of electrons leads to an outward pressure that stops the star from collapsing further due to its own gravity. Under these conditions the force of repulsion that two electrons would normally experience (like charges repel!)and the thermal properties that usually determine pressure, are negligible compared to this degeneracy condition.

### Frequency and Wavelength

Electromagnetic radiation is the flow of energy in the form of periodic oscillations of electric and magnetic fields that can propagate through a vacuum at the speed of light or through any medium that is transparent to them at a speed less than the speed of light. For such periodic waves, the frequency f has an inverse relationship to the wavelength λ:

where v is the phase velocity of the wave in a media. Note that in a dispersive media, for example, in optical glass, the speed depends on frequency and therefore the wavelength is not completely inversely proportional to the frequency. In a vacuum, v = c, where c is the speed of light in a vacuum, the expression above becomes:

This formula is used in our calculator.

### Photon Energy

The amount of energy carried by a single photon is directly proportional to the photon’s frequency and inversely proportional to its wavelength. The higher the photon’s frequency, the greater its energy, and the higher the photon’s energy, the larger its frequency. Photon energy depends only on its frequency or wavelength. The intensity of radiation does not affect the photon’s energy. The relationship between photon energy E and its wavelength λ or frequency f is determined by the following equations, which are used in this calculator:

where h = 6.62607015×10 -34 J·s is the Plank’s constant and c is the speed of light in a vacuum. As c and h are both constant, photon energy is inverse proportional to its wavelength and directly proportional to its frequency.

Photon energy is measured in any unit of energy, for example in joules or electronvolts (eV), 1 eV being equal to 1.602176565×10⁻¹⁹ J.

### Black Body Temperature

A black body is an idealized physical body that absorbs all the electromagnetic radiation (light, radio waves, gamma rays, etc.) that strikes it. To stay in thermal equilibrium, it must emit radiation at the same rate as it absorbs the radiation. The energy that was absorbed by a black body, is re-emitted in a wide spectrum and is called black-body radiation.

Black bodies are so important in physics because they do not reflect radiation, they only emit it like all other bodies. The spectrum of light produced by a hot object depends on its temperature. At room temperature, the peak in this radiation is mostly in the infrared (thermal) region. Our bodies, when at their normal temperature of about 310 K (37 °C), radiate only in the infrared part of the spectrum. Hotter bodies, emit radiation with its spectrum shifted into higher and higher frequencies, into the visible part of the spectrum and start glowing red or even white when they are heated even more.

Stars are often modeled as black bodies in astronomy. The temperature of a star can be determined from the frequency of the peak of its radiation spectrum. Stars that are hotter than our nearest star — the Sun, produce most of their light in the blue and violet regions of the spectrum, that is the light with higher frequencies. Stars that are cooler than the Sun, produce most of their light in red or infrared regions of the spectrum. That means their light has lower frequencies. The Sun with its surface temperature of about 6000 K radiates most strongly in the visible part of the electromagnetic spectrum.

In this converter, we calculate the temperature using Wien’s displacement law, which states that the wavelength of the peak of the emission fmax of black-body radiation is inversely proportional to the absolute temperature (Т) or the peak frequency increases linearly with the absolute temperature:

where c is the speed of light in a vacuum and b = 2.8977729×10 -3 m·K is a proportionality constant called Wien’s displacement constant.

## Star Facts: Pollux

Image Credit: scienceblogs.com

Pollux is the brightest star in the constellation Gemini, and the 17th brightest in the entire night sky. However, contrary to convention, Pollux bears the designation Beta Geminorum, which by rights should apply to Castor (Alpha Geminorum), which is the second brightest star in Gemini, and only the 23rd brightest star in the night sky. At a distance of just 33.78 light years away, Pollux also has the distinction of being the closest giant star to the Sun.

Since 1943, this orange giant has been one of the stable standards by which other stars spectra are classified, and along with Castor, Pollux features prominently in literature throughout history, including in a romance-era poem by Shelley called ‘Homer’s Hymn to Castor and Pollux’ in which Pollux is referred to thus:

“Ye wild-eyed Muses, sing the Twins of Jove,
Whom the fair-ankled Leda, mixed in love
With mighty Saturn’s Heaven-obscuring Child,
On Taygetus, that lofty mountain wild,
Brought forth in joy: mild Pollux, void of blame,
And steed-subduing Castor, heirs of fame.”

Quick Facts

• Constellation: Gemini
• Coordinates: RA 7h 45m 19s | Dec +28° 1′ 35?
• Distance: 33.78 light years
• Star Type: Orange Giant (K0 III)
• Mass: 2.04 sol
• Radius: 8.8 sol
• Apparent Magnitude: +1.14
• Luminosity: 43 sol
• Surface Temperature: 4,666
• Rotational Velocity: 2.8 km/s (One rotation = 558 days)
• Age: 724 million years
• Other Designations: Beta Geminorum, 78 Geminorum, BD+28°1463, GCTP 1826.00, Gliese 286, HD 62509, HIP 37826,

Pollux is the brightest star in Gemini, a northern constellation that can be seen from latitudes of between +90° and -60°. It is a first magnitude star (+1.14), which together with the star Castor (+1.58), mark the heads of the legendary twins from Greek mythology. The rest of the stars in the constellation represents their bodies, with Gemini easily located northeast of Orion, with other nearby constellations including Leo to its east, Taurus to its west, and Canis Minor to its south.

Since there are no other bright stars immediately around either Pollux or Castor when observing from low to mid-northern latitudes, the two stars are conspicuous and therefore easily spotted. However, since the two stars are relatively close to each other, it is easy to confuse one with the other, so to make sure the right star is observed bear in mind that Pollux appears as pale yellow, whereas Castor is white, with a pale blue tinge at times. Pollux is also the brighter star of the pair, which makes its identification easier.

Pollux becomes visible for most northern observers from about mid-October, when it rises above the north-eastern horizon before midnight, with the actual time depending on the observer’s location. It can then be seen throughout winter and spring in the northern hemisphere, with its best time for viewing during its midnight culmination around the 15th of January each year when the star is visible right through the night. On that day, Pollux crosses the meridian at around local midnight, which means the star crosses an imaginary line drawn from due north through the zenith, to a position due south, during which it can be seen high in the sky from much of the planet’s surface. Bear in mind also that Pollux is a circumpolar star from central Alaska, the northern reaches of Canada, and northward of central Scandinavia.

Physical Properties

Size, Mass, Luminosity

Apart from being more than forty times as bright as the Sun, Pollux has just more than twice the Sun’s mass, and almost nine times its size. Having consumed all, or most of its hydrogen fuel, Pollux, has now evolved off of the main sequence into a giant star with an effective surface temperature of 4,666K, which is typical of K-type, orange giant stars.

Metallicity and Magnetic Field

Pollux’s metalicity (which refers to elements in a star other than hydrogen and helium), is still somewhat uncertain, with some investigators estimating the total metal abundance to be no more than about 85% of that of the Sun, while others have shown the total metal abundance to be as high as 155% of that of the Sun. Investigations are continuing.

Recent studies have revealed the presence of weak magnetic activity on Pollux, based on measurements of X-ray emissions from the star that were made with the ROSAT orbiting telescope. Interestingly, the detected 1027 erg s-1 X-ray emission from Pollux is about the same as that produced by the Sun. Nonetheless, the magnetic field of Pollux turns out to be well below 1 Gauss in strength, which is the weakest magnetic field ever to have been discovered on any star. By comparison, the magnetic field of Jupiter is on average about 10 Gauss, which is roughly 20,000 times stronger than Earth’s 0.5 Gauss-strength magnetic field.

The presence (and strength) of Pollux’s magnetic field suggests that Pollux was once an Ap-type main sequence star, which is a chemically-peculiar class of main sequence stars (hence the “p” in the classification) that display large overabundance of metals such as strontium, chromium and europium in their spectra. Together with Bp-class stars, Ap-type stars typically have very low rotational velocities, which in the case of Pollux rotates at 2.8 km/sec, seeming to confirm Pollux’s past life as a chemically peculiar Ap-class star.

In June of 2006, the presence of a planet with an orbital period of 589.64 ± 0.81 days was confirmed. Named Thestias, the planet has just more than twice the mass of Jupiter, and orbits Pollux in a nearly circular orbit at a distance of 1.64 astronomical units. Interestingly, Thestias was the grandfather of Pollux in Greek mythology.

Image Credit: Greg Parker

According to Greek Mythology, Polydeukes and Kastor, whose Latin names are Pollux and Castor, were the identical twin son’s of Queen Leda of Sparta, whose other children included Helen of Troy, and Clytemnestra, who later became the wife of King Agamemnon of Mycenae. While Castor was the son of Leda’s husband King Tyndareus, Pollux was the offspring of the adulterous Zeus, and was therefore immortal.

Castor was a skilled warrior horseman, while Pollux was a formidable boxer, and throughout their lives, Pollux and Castor were inseparable, and enjoyed many adventures together, including helping Jason and the Argonauts steal the Golden Fleece. In terms of etymology, Polydeukes means “very sweet” or even “charming” (according to some sources) in Greek, which is difficult to reconcile with the brutal and bloody sport that boxing was in ancient Greece.

On account of their striking appearance together, the stars Pollux and Castor have symbolized twins in many cultures throughout history. In ancient India, for instance, they represented the Ashwins, or twin horsemen of the dawn in Persia they were Du Paikar, the two figures in Phoenicia they depicted two gazelles or kid-goats while in China they symbolized Yin and Yang, the opposite but interconnected forces of nature.

## Centre for Astrophysics and Supercomputing

Study or work with the largest astronomical research group in Victoria and reach for the stars.

The Centre for Astrophysics and Supercomputing (CAS) is dedicated to inspiring a fascination of the universe through research and education. We aim to understand how the universe around us came to be and determine the nature of the physical laws making it work. By pursuing this work, we will enable technology development, train the next generation of physical scientists and inspire the public.

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Professor Jean Brodie , Director of the Centre for Astrophysics and Supercomputing

## Diffraction Grating

When there is a need to separate light of different wavelengths with high resolution, then a diffraction grating is most often the tool of choice. This "super prism" aspect of the diffraction grating leads to application for measuring atomic spectra in both laboratory instruments and telescopes. A large number of parallel, closely spaced slits constitutes a diffraction grating. The condition for maximum intensity is the same as that for the double slit or multiple slits, but with a large number of slits the intensity maximum is very sharp and narrow, providing the high resolution for spectroscopic applications. The peak intensities are also much higher for the grating than for the double slit.

When light of a single wavelength , like the 632.8nm red light from a helium-neon laser at left, strikes a diffraction grating it is diffracted to each side in multiple orders. Orders 1 and 2 are shown to each side of the direct beam. Different wavelengths are diffracted at different angles, according to the grating relationship.