Boltzmann constant). Its value is 1.38 10 16 erg/K.

depends strongly on T. If we double the temperature

(The quantity kT is proportional to the kinetic

of an object, the rate at which it gives off energy

energy per particle in the gas.) The Rayleigh“Jeans

goes up by a factor of 16. If we change the tempera-

law agrees with experimental results at low fre-

ture by a factor of ten (say from 300 K to 3000 K), the

energy radiated goes up by a factor of 104. quencies (long wavelengths), but disagrees at high

frequencies. In fact, you can see from equation

For a star, we are interested in the total lumi-

(2.8) that, as we go to higher and higher frequen-

nosity. The luminosity is the total energy per sec-

cies, the energy given off becomes arbitrarily large.

ond (i.e. the power) given off by the star. The

16 PART I PROPERTIES OF ORDINARY STARS

However, in looking at the blackbody curves (as in I( , T) d I( , T) d

Fig. 2.5) you see that there is a peak and then there

Solving for I( , T) gives

is less energy at higher frequencies. (The classical

prediction of arbitrarily large energies at high fre- I( , T) I( , T) (d /d )

quencies was sometimes referred to as the “ultra-

To find I( , T) we must be able to evaluate

violet catastrophe”.)

c/ , so

d /d . We do this by remembering that

The first step in solving the problem of theo-

that

retically predicting blackbody curves was to

deduce an empirical formula for the observed 2

c/

d /d (2.10a)

spectra. By “empirical” we mean a formula that is

We don™t care about the minus sign, which

arbitrarily put together to describe the observa-

just tells us that frequency increases when wave-

tions, but which is not derived from any theory.

length decreases. Using this result gives

Once that is done, then we try to find a theory

that can be used to derive the formula. I( , T) (c/ 2)

I( , T) (2.10b)

In 1900, Max Planck, a German physicist, pro-

Now we can substitute c/ for to obtain the

duced an empirical formula that accurately

final result:

describes the experimental blackbody spectra:

2h 3>c2 2hc2 5

I1 , T 2

I1 , T 2 (2.10c)

(2.9)

>kT ehc kT

1

eh 1

Remember, the Planck function accurately

In this equation, h is called Planck™s constant,

10 27 erg s. describes blackbody spectra, but it was origi-

and has the numerical value 6.63

nally presented as an empirical formula. There

This value was determined to provide the best

was still no theoretical understanding of the ori-

agreement with observed blackbody spectra.

gin of the formula. Planck continued his work

Since the Rayleigh“Jeans law adequately

in an effort to derive the formula from some

describes blackbody spectra at low frequencies,

theory. Planck found that he could derive the

the Planck law must reduce to the Rayleigh“Jeans

formula from classical physics if he inserted a

law in the limit of low frequencies. We can see

mathematical trick. The trick amounted to tak-

this if we take low frequencies to mean that h

ing a sum rather than an integral. The trick cor-

V kT (or equivalently h /kT V 1). In that case, we

responds to the physical statement that a black-

can take advantage of the fact that for x V 1,

ex 1 x. The Planck function then becomes body can only emit radiation at a frequency in

multiples of the quantity h . That is, the energy

2h 3 kT

I1 , T 2 could only be emitted in small bundles or quanta

c2 h (singular quantum). The quanta have energy h .

Even though Planck was able to derive the black-

2

2kT

body formula correctly, he was still not satisfied.

c2

There was no justification for the restriction

which is the Rayleigh“Jeans law. that energy must be quantized.

Equation (2.9) gives the Planck function in

terms of frequency. How do we find it as a func- 2.4.2 Photons

tion of wavelength? Your first guess might be sim- An explanation for why energy must be quan-

ply to substitute c/ for each occurrence of in tized was proposed by Albert Einstein, in 1905. (It

equation (2.9). However, we must remember that was for this explanation that Einstein was later

I( , T) gives the energy per second per frequency awarded the Nobel Prize in physics.) Einstein

interval, whereas I( , T) gives the energy per sec- was trying to explain a phenomenon known as

ond per wavelength interval. The functions must the photoelectric effect, in which electrons can be

reflect that difference (especially since they will ejected from a metal surface if light falls on the

need different units). We therefore require that surface. (This is the basis for photocells, which

2 CONTINUOUS RADIATION FROM STARS 17

are used in many applications.) Laboratory stud-

2.5 Stellar colors

ies had shown that increasing the intensity of

the light falling on the surface increased the

We have seen that the color of a star can tell us

number of electrons ejected from the surface,

about the star™s temperature. However, we now

but not their energy. Einstein said that all radia-

need a way of quantifying a color, rather than

tion (whether from a blackbody or otherwise)

just saying something is red, green or blue. For

must come in small bundles, called photons. The

example, if we compare two blue stars, how do we

energy, E, of a photon with a frequency is given

decide which one is bluer?

by

We define two standard wavelength ranges,

E h (2.11)

centered at 1 and 2 and take the ratio of the

observed brightnesses, b( 1)/b( 2). We then con-

This explains the observed properties of the

vert that brightness ratio into a magnitude dif-

photoelectric effect by stating that each electron

ference (using equation (2.3)), giving

is ejected by a single photon striking the surface.

Increasing the intensity of light increases the

m2 m1 2.5 log10[b( 1)/b( 2)] (2.12)

number of photons striking the surface per sec-

ond, and therefore increases the rate at which We define the quantity m2 m1 as the color,

electrons are ejected. Increasing the frequency of measured in magnitudes, corresponding to the

the light increases the energy at which the elec- wavelength pair, 1, 2. For definiteness, let™s

trons are ejected. (This latter prediction was assume that 2 1. As we increase the tempera-

finally tested by Robert Millikan in 1916.) A further ture, b( 1)/b( 2) increases. This means that the

quantity m2 m1 decreases, since the magnitude

test of the photon hypothesis came in an analysis

of collisions between light (photons) and elec- scale runs backwards. If we know that an object is

trons, by A. H. Compton. The fact that light is com- radiating exactly like a blackbody, we need only

posed of photons explains why Planck had to take the ratio of brightnesses at any two wave-

assume that energy is quantized in deriving the lengths to determine the temperature.

formula for blackbody spectra. As we have said, we don™t really measure the

The assertion that light is essentially a parti- intensity of radiation at a wavelength. Instead, we

cle went against the then accepted ideas about measure the amount of energy received in some

light. The question of whether light is a particle wavelength interval. We can control that wave-

length interval by using a filter that only passes

or a wave had been going on for centuries. For

example, Newton believed that it is a particle, light in that wavelength range. When we use a

and he worked out a theory of refraction “ the filter, we are actually measuring the integral of

I( , T) over some wavelength range. Actually, the

bending of light when it passes, for example,

from air to glass “ on the basis of light speeding situation is more complicated. The transmission of

up when it enters the glass. (We now know, how- any real filter is not 100% over the selected range,

ever, that as a wave it slows down.) The wave the- and this must be factored in (see Problem 2.23).

ory became dominant with the demonstration of Another complication is that continuous spec-

interference effects by Young and the explana- tra of stars do not exactly follow blackbody

tion of electromagnetic waves by Maxwell. In curves. Therefore, observations through two fil-

explaining the photoelectric effect, Einstein was ters are not generally sufficient to tell us the tem-

saying that the particle picture must be revived. perature of the star. Over the years, a system of

The explanation that, somehow, light can exhibit standard filters has been developed, so that

both wave and particle properties is referred to as astronomers at various observatories can com-

the wave“particle duality. This concept is the pare their results. The wavelength ranges of the

foundation of what we refer to as the quantum rev- various filters are shown in Table 2.2. The most

olution, since it was such a radical departure from commonly discussed filters are U (for ultraviolet)

B (for blue) and V (for visible, meaning the center

previous theories. We will discuss this point far-

ther in the next chapter. of the visible part of the spectrum). More recently,

18 PART I PROPERTIES OF ORDINARY STARS

still L, but it is spread over an area of 4 d2 so the

Filter systems.

Table 2.2.

energy flux, f, is

Width (nm)a

Filter Peak wavelength (nm)

L/4 d2

f (2.13)

U 350 70

The received flux falls off inversely as the

B 435 100

square of the distance.

V 555 80

Unfortunately, distances to astronomical

R 680 150

objects are generally hard to determine. There is

I 800 150

a direct method for determining distances to

a

Full width at half maximum. nearby stars. It is called trigonometric parallax, and

amounts to triangulation from two different

observing points. You can demonstrate parallax

R (for red), and I (for infrared) have been added.

for yourself by holding out a finger at arm™s

(There are actually a couple of filters in different

length and viewing it against a distant back-

parts of the infrared.)

ground. Look at the finger alternately using your

For example, the B V color is defined by

left and right eye. The finger appears to shift

B V 2.5 log10 [I( V) / I( B)] constant

against the distant background. Bring the finger

where I( V) and I( B) are the intensities averaged closer and repeat the experiment. The shift now

over the filter ranges. (The constant is adjusted so appears larger. If you could move your eyes far-

that B V is zero for a particular temperature ther apart, the effect would be even greater.

star, designated A0. These designations will be Even the closest stars are too far away to

discussed in the next chapter.) As the tempera- demonstrate parallax when we just use our eyes.

ture of an object increases, the ratio of blue to vis- However, we can take advantage of the fact that

ible increases. This means that the B V color the Earth orbits the Sun at a distance defined to be

decreases (again because the magnitude scale runs one astronomical unit (AU). Therefore, if we observe

backwards.) a star and then observe it again six months later,

we have viewing points separated by 2 AU. The sit-

uation is illustrated in Fig. 2.6. We note the posi-

2.6 Stellar distances

tion of the star against the background of distant

stars, and then six months later we note the angle

So far we have discussed how bright stars appear by which the position has shifted. If we take half

as seen from Earth. However, the apparent bright- of the value of this angle, we have the parallax

ness depends on two quantities: the intrinsic angle, p.

luminosity of the star and its distance from us. Once we know the value of p, we can construct

(As we will see in Chapter 14, starlight is also a right triangle with a base of 1 AU and the other

dimmed when it passes through clouds of inter- leg being the length, d, the unknown distance to

stellar dust.) Two identical stars at different dis- the star. From the right triangle, we can see that

tances will have different apparent brightnesses.

tan p 1 AU/d (2.14)

If we want to understand how stars work, we

must know their total luminosities. This requires Since p is small, tan(p) p (rad), which is the

correcting the apparent brightness for the dis- value of p, measured in radians. Equation (2.14)

tance to the star. then gives us

If we have a star of luminosity L, we can cal-

p(rad) 1 AU/d (2.15)

culate the observed energy flux at a distance d. If

no radiation is absorbed along the way, all the

It is not very convenient measuring such

energy per second leaving the surface of the star

small angles in radians, so we convert to arc sec-

will cross a sphere at a distance d in the same

onds (see Box 2.1):

time. It will just be spread over a larger area.

105p(rad)

Therefore, the energy per second reaching d is p( ) 2.06

2 CONTINUOUS RADIATION FROM STARS 19

Box 2.1. Angular measure

The natural unit for measuring angles is the radian. If we

have a circle of radius R, and two lines from the center

making an angle with each other, then the length of the

arc bounded by the two lines is

p

L (rad)R

where (rad) is the value of measured in radians. Since

d

the full circumference of a circle is 2 R, the angle corre-

sponding to a full circle must be 2 radians. This tells us

that a full circle, that is 360 , is equal to 2 radians, or

180 is equal to radians. In astronomy, we often deal

with very small angles, and measurements in arc seconds

( ) are convenient. We can convert measurements by

saying

1“ 2 1rad2

180° 60¿ 60“

rad 1° 1¿

1 AU

S E

105 (rad)

2.06

When we take the derivatives of trigonometric func-

Fig 2.6. Geometry for parallax measurements.The ¬gure

tions (for example, d(sin )/d cos ), it is assumed

is not to scale. In reality the distance to the star, d, is much

that the angles are in radians. If not, a conversion factor

greater than 1 AU, so the parallax angle, p, would normally

must be carried through the differentiation.

be very small.

When angles expressed in radians have a value that

is much less than unity, we can use a Taylor series to

approximate them:

d1sin ¿ 2

where p( ) is the parallax angle measured in arc

sin1 2 sin 10 2

seconds. Substituting this into equation (2.15) 0

d¿

gives

cos(0)

105/p( )

d/1 AU 2.06 (2.16)

d1tan ¿ 2

This gives the distance to the star in AU (1 AU

tan1 2 tan 10 2

1.50 108 km). 0

d¿

This method suggests a convenient unit for

sec2(0)

measuring distances. We define the parsec (abbre-

viated pc) as the distance of a star that produces

d1cos ¿ 2

a parallax angle p of 1 arc sec. From equation

cos1 2 cos 10 2

105 AU (or

(2.16), we can see that 1 pc 2.06 0

d¿

1013 km, or 3.26 light years). We rewrite

3.09

equation (2.16) as 1 sin(0)

1

d(pc) 1/p( ) (2.17)

Note that for small , sin and tan are both approxi-

Remember, as an object moves farther away,

mately equal to , so they must be equal to each other.

the parallax angle decreases. Therefore, a star at

(Remember, in each of the above expressions, must be

a distance of 2 pc will have a parallax angle of

expressed in radians.)

0.5 arc sec.

20 PART I PROPERTIES OF ORDINARY STARS

Using the fact that log(x2 ) 2 log(x) gives

Example 2.3 Distance to the nearest star

The nearest star (Proxima Centauri) has a parallax

m M 5 log10(d/10 pc) (2.18)

p 0.76 arc sec. Find its distance from Earth in

The quantity 5 log10(d/10 pc), which is equal to

parsecs.

(m M), is called the distance modulus of the star.

It indicates the amount (in magnitudes) by which

SOLUTION

distance has dimmed the starlight. If you know

We use equation (2.17) to give

any two of the quantities (m, M or d) you can use

d(pc) 1/(0.76)

equation (2.18) to find the third. For any star that

d 1.32 pc we can observe, we can always measure m, its

apparent magnitude. Therefore, we are generally

With current ground-based equipment, we can

faced with knowing M and finding d or knowing

measure parallax to within a few hundredths of

d and finding M. In the next chapter we will look

an arc second. Parallax measurements are there-

at some ways of determining M.

fore useful for the few thousand nearest stars.

They are a starting point for a very complex system Example 2.4 Absolute magnitude

of determining distances to astronomical objects. A star is at a distance of 100 pc, and its apparent

We will encounter a variety of distance determi- magnitude is 5. What is its absolute magnitude?

nation methods throughout this book. The

trigonometric parallax method is the only one SOLUTION

that is direct and free of any assumptions. For this We use equation (2.18) to find

reason, astronomers would like to extend their

M m 5 log (d/10 pc)

capability for measuring parallax. The Hipparcos

satellite measures parallaxes to 10 3 arc sec. 5 5 log (100 pc/10 pc)

5 5 log (10)

2.7 Absolute magnitudes

0

We should note that changing the distance of

The magnitudes discussed in Section 2.1, based on

a star changes its apparent magnitude, but it

observed energy fluxes, are called apparent magni-

does not change any of its colors. Because colors

tudes. In order to compare intrinsic luminosities of

are defined to be differences in magnitudes, each

stars, we define a system of absolute magnitudes. The

is changed by the distance modulus. For example,

absolute magnitude of a star is that magnitude

using equation (2.18)

that it would appear to have as viewed from a stan-

dard distance, d0. This standard distance is chosen mB MB 5 log (d/10 pc)

to be 10 pc. From this definition, you can see that if

mV MV 5 log (d/10 pc)

a star is actually at a distance of 10 pc, the absolute

and apparent magnitudes will be the same. Taking the difference gives

To see how this system works, consider two

mB mV MB MV

identical stars, one at a distance d and the other

at the standard distance d0. We let m be the appar-

Therefore, the distance modulus never appears in

ent magnitude of the star at distance d, and M

the colors.

that of the star at distance d0. (Of course, M will

When we talk about determining an absolute

be the absolute magnitude for both stars.) The

magnitude, we are really only determining it over

energy flux falls off inversely as the square of the

some wavelength range, corresponding to the

distance, therefore the ratio of the flux of the star

wavelength range of the observations. We would

at d to that from the star at d0 is (d0/d)2 . Equation

like to have an absolute magnitude that corre-

(2.3) then gives us

sponds to the total luminosity of the star. This

2.5 log10(d/d0)2 magnitude is called the bolometric magnitude of

m M

2 CONTINUOUS RADIATION FROM STARS 21

the star. (As we will see in Chapter 4, a bolometer ated BC), which relates the bolometric magnitude

to the absolute visual magnitude MV. Therefore

is a device for measuring the total energy received

from an object.) For any type of star, we can define

MBOL MV BC (2.19)

a number, called the bolometric correction (abbrevi-

Chapter summary

We saw in this chapter what can be learned from face area than cooler ones (as described by the

the brightness and spectrum of the continuous Stefan“Boltzmann law), and also have their spec-

radiation from stars. tra peaking at shorter wavelengths (as described

We introduced a logarithmic scale, the mag- by Wien™s displacement law). We saw how

nitude scale, for keeping track of brightness. attempts to understand the details of blackbody

Apparent magnitude is related to the observed spectra (Planck™s law) contributed to the idea of

energy flux from the star, and the absolute mag- light coming in bundles, called photons, with

nitude is related to the intrinsic luminosity of specific energies. With a knowledge of blackbody

the star. spectra, we saw how stellar colors can be used to

We saw how, even though stars are obvious to deduce stellar temperatures.

us in the visible part of the spectrum, they, and We saw how finding the distances to astro-

other astronomical objects, give off radiation in nomical objects is very important, but can be

other parts of the spectrum. The richness of infor- quite difficult. If we don™t know the distance to

mation in other parts of the spectrum is a theme an object, we cannot convert its apparent bright-

that we will come back to throughout the book. ness into a luminosity. We introduced one

We introduced the concept of a blackbody, method of measuring distances “ trigonometric

which is useful because the continuous spectrum parallax. It is the most direct method, but only

of a star closely resembles that of a blackbody. works for nearby stars. The problem of distance

Hotter bodies give off more power per unit sur- determination will come up throughout the book.

Questions

2.1. Why is the magnitude scale logarithmic? 2.8. What are the different ways in which the

2.2. Are there any other types of measurements word “spectrum” is used in this chapter?

that we encounter in the everyday world that 2.9. Give some examples of objects whose spectra

are logarithmic? (Hint: Think of sound.) are close to that of blackbodies.

2.3. Why are astronomical observations potentially 2.10. How can we determine the temperature of a

useful in measuring the speed of light? blackbody?

2.4. What are the factors that have resulted in 2.11. If the peak of a blackbody spectrum shifts to

early astronomical observations being in the shorter wavelengths as we reach higher tem-

“visible” part of the spectrum? peratures, how can it be that a hotter black-

2.5. What do we mean by “atmospheric window”? body gives off more energy at all wavelengths

2.6. Why was Maxwell™s realization that a varying than a cooler one?

electric field can create a magnetic field 2.12. What is the evidence for the existence of

important in understanding electromagnetic photons?

waves? 2.13. Explain how we quantify the concept of color.

2.7. (a) Estimate the number of people on Earth 2.14. What is the value of using standard filters in

who are exactly 2 m tall. (By “exactly” we looking at stellar spectra?

mean to an arbitrary number of decimal 2.15. Suppose you could communicate with an

places.) (b) How does this relate to the way we astronomer on a planet orbiting a nearby

define the intensity function I( )? star. (The astronomer is native to that planet,

22 PART I PROPERTIES OF ORDINARY STARS

2.16. How would parallax measurement improve if

rather than having traveled from Earth.) You

we could do our observations from Mars?

determine the distance to the star (by

2.17. As we determine the astronomical unit more

trigonometric parallax) to be 2 pc. The distant

accurately, how does the relationship

astronomer says that you are wrong; the dis-

between the AU and the parsec change?

tance is only 1 pc. What is the problem?

Problems

2.1. What magnitude difference corresponds to a wavelength being 501.00 nm and 510.00 nm.

factor of ten change in energy flux? What do you conclude?

2.2. One star is observed to have m 1 and *2.10. (a) Use equation (2.9) to derive max, the fre-

another has m quency at which I( , T) peaks. Convert this

1. What is the ratio of

energy fluxes from the two stars? max into a wavelength max. (b) Use equation

2.3. The apparent magnitude of the Sun is 26.8. (2.10c) to find the wavelength at which it

How much brighter does the Sun appear than peaks. (c) How do the results in (a) and (b)

the brightest star, which has m 1? compare?

2.4. (a) What is the distance modulus of the Sun? 2.11. For a 300 K blackbody, over what wavelength

(b) What is the Sun™s absolute magnitude? range would you expect the Rayleigh“Jeans

2.5. Suppose two objects have energy fluxes, f and law to be a good approximation?

f f, where f V f. Derive an approximate 2.12. Derive an approximation for the Planck func-

expression for the magnitude difference m tion valid for high frequencies (h W kT).

between these objects. Your expression should 2.13. As we will see in Chapter 21, the universe is

have m proportional to f. (Hint: Use the filled with blackbody radiation at a tempera-

fact that ln (1 x) … x when x V 1.) ture of 2.7 K. (a) At what wavelength does the

2.6. Show that our definition of magnitudes has spectrum of that radiation peak? (b) What

the following property: If we have three stars part of the electromagnetic spectrum is this?

with energy fluxes, f1 , f2 and f3 , and we 2.14. (a) We observe the blackbody spectrum from

define a star to peak at 400 nm. What is the temper-

ature of the star? (b) What about one that

m2 m1 2.5 log10( f1/f2)

peaks at 450 nm?

m3 m2 2.5 log10( f2/f3) 2.15. Derive an expression for the shift in the

peak wavelength of the Planck function for a

then

blackbody of temperature T, corresponding to

m3 m1 2.5 log10(f1/f3) a small shift in temperature, T.

2.16. Calculate the energy per square centimeter

2.7. Suppose we measure the speed of light in a

per second reaching the Earth from the Sun.

laboratory, with the light traveling a path of

2.17. How does the absolute magnitude of a star

10 m. How accurately do you have to time the

vary with the size of the star (assuming the

light travel time to measure c to eight signifi-

temperature stays constant)?

cant figures?

2.18. (a) What is the energy of a photon in the

2.8. Let 1 and 2( 1, 2) be the wavelength (fre-

middle of the visible spectrum ( 550 nm)?

quency) limits of the visible part of the spec-

(b) Approximately how many photons per sec-

trum. Compare ( 1 2)/( 1 2) with ( 1

ond are emitted by (i) a 100 W light bulb,

2)/( 1 2). Comment on the significance.

(ii) the Sun?

2.9. (a) Calculate the frequencies corresponding

2.19. If we double the temperature of a blackbody,

to the wavelengths 500.00 nm and 500.10 nm.

by how much must we decrease the surface

Use these to check the accuracy of equation

area to keep the luminosity constant?

(2.10a). (b) Repeat the process for the second

*An asterisk denotes a harder Problem or Question. The convention continues throughout the book.

2 CONTINUOUS RADIATION FROM STARS 23

2.20. (a) How does the absolute bolometric magni- can be measured using the method of

tude vary with the temperature of a star trigonometric parallaxes? (b) By what factor

(assuming the radius stays constant)? (b) Does will the volume of space over which we can

the absolute visual magnitude vary in the measure parallax change if we can measure

to 0.001 arc sec? (c) Why is the volume of

same way?

*2.21. For a star of radius R, whose radiation follows space important?

a blackbody spectrum at temperature T, derive 2.27. If we lived on Mars instead of the Earth, how

an expression for the bolometric correction. large would the parsec be?

2.22. Suppose we observe the intensity of a black- 2.28. Suppose we discover a planet orbiting a

body, I0, in a narrow frequency range cen- nearby star. The distance to the star is 3 pc.

tered at 0. Find an expression for T, the tem- We observe the angular radius of the planet™s

orbit to be 0.1 arc sec. How many AU from

perature of the blackbody in terms of I0 and

0. (a) First do it in the Rayleigh“Jeans limit the star is the planet? (Hint: You can solve

and (b) in the general case. this problem by “brute force”, converting all

*2.23.Suppose we receive light from a star for the units. For an easier solution, think about

which the received energy flux is given by the what the answer would be if the star were 1

function f( ). Suppose we observe the star pc from us and the angular radius of the

orbit were 1 arc sec, and then scale the result

through a filter for which the fraction of

light transmitted is t( ). Derive an expression accordingly.)

for the total energy detected from the star. 2.29. Derive an expression for the distance to a star

(Hint: Start by thinking of the energy in terms of its distance modulus.

detected in a small wavelength range.) 2.30. If we make a 0.05 magnitude error in measur-

2.24. What is the distance to a star whose parallax ing the apparent magnitude of a star, what

is 0.1 arc sec? error does that introduce in our distance

2.25. Derive an expression for the distance of an determination (assuming its absolute magni-

object as a function of the parallax angle seen tude is known exactly)?

by your eyes?

2.26. (a) If we can measure parallaxes as small as

0.1 arc sec, what is the greatest distance that

Computer problems

2.1. Make a fourth column for Table 2.1, showing the lation you may assume that magnitudes are deter-

range of photon frequencies for each part of the mined in a narrow range of wavelengths around

spectrum. Make a fifth column showing the range the peak of each filter.

of photon energies for each part of the spectrum. 2.3. For the Sun, plot the difference between the

Make a sixth column showing the temperatures Rayleigh“Jeans approximation and the Planck for-

that blackbodies would have to peak at the wave- mula, as a function of wavelength, for wavelengths

lengths corresponding to the boundaries between in the visible part of the spectrum.

the parts of the spectrum 2.4. For the Sun, calculate the energy given off over the

2.2. Make a graph of the magnitude difference MB MV wavelength bands that correspond to the U, B and

V filters. Use this to estimate the colors U B and

as a function of temperature for a temperature

B V.

range of 3000 K to 30 000 K. To simplify the calcu-

Chapter 3

Spectral lines in stars

In Chapter 2 we discussed the continuous spectra gave the strongest lines letter designations that

of stars and saw that they could be closely we still use today.

described by blackbody spectra. In this chapter, The origin of these lines was a mystery for

some time. In 1859, the German chemist Gustav

we will discuss the situations in which the spec-

Robert Kirchhoff noticed a similar phenomenon in

trum shows an increase or decrease in intensity

over a very narrow wavelength range. the laboratory. He found that when a beam of

white light was passed through a tube containing

some gas, the spectrum showed dark lines. The

gas was absorbing energy in a few specific narrow

3.1 Spectral lines wavelength bands. In this situation, we refer to

the lines as absorption lines. When the white light

We know that if we pass white light through a was removed, the spectrum showed bright lines,

or emission lines, the wavelengths where absorp-

prism, light of different colors (wavelengths) will

emerge at different angles with respect to the ini- tion lines had previously appeared. The gas could

tial beam of light. If we pass white light through emit or absorb energy only in certain wavelength

a slit before it strikes the prism (Fig. 3.1), and bands.

then let the spread-out light fall on the screen, at Kirchhoff found that the wavelengths of the

each position on the screen we get the image of emission or absorption lines depend only on the

the slit at a particular wavelength. type of gas that is used. Each element or com-

Both William Hyde Wollaston (1804) and Josef von pound has it own set of special wavelengths. If

Fraunhofer (1811) used this method to examine two elements which don™t react chemically are

sunlight. They found that the normal spectrum mixed, the spectrum shows the lines of both ele-

was crossed by dark lines. These lines represent ments. Thus, the emission or absorption spec-

wavelengths where there is less radiation than at trum of an element identifies that element as

nearby wavelengths. (The lines are only dark in uniquely as fingerprints identify a person. This

comparison with the nearby bright regions.) The identification can be carried out without under-

linelike appearance comes from the fact that, at standing why it works.

each wavelength, we are seeing the image of the Whether we see absorption or emission depends

slit. It is this linelike appearance that leads us to in part on whether or not there is a strong enough

call these features spectral lines. If we were to make background source providing energy to be absorbed

a graph of intensity vs. wavelength, we would (Fig. 3.2). The strength of the spectral lines also

find narrow dips superimposed on the continuum. depends on how much gas is present and on the

The solar spectrum with dark lines is sometimes temperature of the gas. Sample emission and

referred to as the Fraunhofer spectrum. Fraunhofer absorption spectra of stars are shown in Fig. 3.3.

26 PART I PROPERTIES OF ORDINARY STARS

Screen

Intensity

Prism

Wavelength

Gas with

Slit

Atoms & Molecules

(a)

t

igh

nl

Su

Fig 3.1. If we allow white light to fall all over a prism, the

Intensity

red from one part will overlap the blue from another part,

and we can™t see a clear spectrum. Instead, we pass white

Continuum

light through a slit ¬rst.The beam of light is then spread out Source

as it passes through the prism. On the screen, we are seeing

Wavelength

a succession of images of the slit in different colors. If there

is a color missing from the white light, this will show up as a

gap on the screen in the shape of the slit.

(b)

Fig 3.2. Conditions for the formation of emission and

absorption lines. (a) We look at a cloud of gas with the

3.2 Spectral types atoms or molecules capable of producing spectral lines. Since

there is no continuum radiation to absorb, we can only have

When spectra were taken of stars other than emission. (b) We now look through the gas at a background

continuum source.This can produce absorption lines.

the Sun, they also showed absorption spectra.

Presumably, the continuous radiation produced

in a star passes through an atmosphere in which

the absorption lines are produced. Not all stars B stars the next strongest, and so on. These letter

have absorption lines at the same wavelength. designations were called spectral classes or spectral

Astronomers began to classify and catalog the types. We now know that the different spectral

spectra, even though they still did not understand types correspond to different surface tempera-

the mechanism for producing the lines. This points tures. However, the sequence A, B, . . . is not a

out an important general technique in astronomy “ temperature-ordered sequence. For reasons we

studying large numbers of objects to look for gen- will discuss below, hydrogen lines are strongest in

eral trends. In one very important study, over intermediate temperature stars.

200 000 stars were classified by Annie Jump Cannon The spectral classes we use, in order of decreas-

at the Harvard College Observatory. The benefactor ing temperature, are O, B, A, F, G, K, M. We break

of that study was Henry Draper, and the catalog of each of these classes into ten subclasses, identified

stellar spectra was named after him. The stars in by a number from zero to nine; for example, the

this catalog are still known by their HD numbers. sequence O7, O8, O9, B0, B1, B2, . . . , B9, A0, A1, . . . .

One set of spectral lines common to many (For O stars the few hottest subclasses are not

stars was recognized as belonging to the element used.) For some of the hotter spectral types, we

hydrogen. The stars were classified according to even use half subclasses, for example, B1.5. It was

the strongest hydrogen absorption lines. In this originally thought that stars became cooler as

system, A stars have the strongest hydrogen lines, they evolved, so that the temperature sequence

3 SPECTRAL LINES IN STARS 27

Fig 3.3. Samples of stellar spec-

tra.These are high resolution

spectra, with the visible part of

the spectrum (400 to 700 nm)

broken into 50 slices.Wavelength

increases from left to right along

each strip and from bottom to

top. (a) Procyon, also known as

Alpha Canis Majoris (the brightest

star in Canis Major). It has spec-

tral type F5 (see Section 3.2),

making it a little warmer than the

Sun. (b) Arcturus, also known as

Alpha Bootes. It is spectral type

K1, being cooler than the Sun.

[NOAO/AURA/NSF]

(a)

(b)

was really an evolutionary sequence. Therefore, are tied to the nature of matter and light. In

the hotter spectral types were called early and the Chapter 2, we saw the beginnings of the quantum

cooler spectral types were called late. We now revolution with the realization that light exhibits

know that these evolutionary ideas are not cor- both particle and wave properties. We now see

rect. However, the nomenclature still remains. We how the ideas of quantization apply to the struc-

even talk about a B0 or B1 star being ˜early B™ and ture of the atom.

a B8 or B9 as being a ˜late B™. The modern picture of the atom begins with

the experiments of Ernest Rutherford, who studied

the scattering of alpha particles (helium nuclei)

3.3 The origin of spectral lines off gold atoms. Most of the alpha particles passed

through the gold atoms without being deflected,

The processes that result in atoms being able to suggesting that most of the atom is empty space!

emit or absorb radiation at certain wavelengths Some alpha particles were deflected through

28 PART I PROPERTIES OF ORDINARY STARS

index of refraction of air, 1.000 29, giving

large angles, suggesting a concentration of posi-

656.28 nm. (It is interesting to note that when spec-

tive charge at the center of each atom. This con-

centration is called the nucleus. A sufficient num- troscopists tabulate wavelengths, those longer than

200 nm are given as they would be in air, since that

ber of electrons orbit the nucleus to keep the

is how they usually will be measured. Radiation

atom electrically neutral.

with wavelengths less than 200 nm doesn™t pene-

There were still some problems with this pic-

trate through air, and its wavelengths are usually

ture. It did not explain why electron orbits were

measured in a vacuum, so the vacuum values are

stable. Classical electricity and magnetism tells

tabulated.)

us that an accelerating charge gives off radiation.

An electron going in a circular orbit is accelerat-

3.3.1 The Bohr atom

ing, since its direction of motion is always chang-

The next advancement was by the Danish physi-

ing. Therefore, as the electrons orbit, they should

cist Neils Bohr who tried to understand hydrogen

give off radiation, lose energy and spiral into the

(the simplest atom), illustrated in Fig. 3.4. He pos-

nucleus. This is obviously not happening. The sec-

tulated the existence of certain stationary states. If

ond problem concerns the origin of spectral

the electron is orbiting in one of these states, the

lines. There is nothing in the Rutherford model

atom is stable. Each of these states has a particu-

of the atom that allows for spectral lines.

lar energy. We can let the energy of the nth state

The arrangement of spectral lines in a partic-

be En and the energy of the mth state be Em. For

ular element is not random. For example, in 1885,

definiteness, let En Em.

Johann Jakob Balmer, a Swiss teacher, realized that

Under the right conditions, transitions between

there was a regularity in the wavelengths of the

states can take place. If the electron is in the

spectral lines of hydrogen. They obeyed a simple

higher energy state, it can drop down to the lower

relationship which became known as the Balmer

energy state, as long as a photon is emitted with

formula:

an energy equal to the energy difference between

R(1/22 1/n2)

1/

The constant R is called the Rydberg constant,

and its value is given by 1/R 91.17636 nm. The

4th Orbit

quantity n is any integer greater than two. By set-

ting n to 3, 4, . . . , we obtain the wavelengths for

the visible hydrogen lines (also known as the

Balmer series). Of course, this was just an empirical 3rd Orbit

formula, with no theoretical justification.

Example 3.1 First Balmer line 2nd Orbit

Calculate the wavelength of the longest wavelength

Balmer line. This line is known as the Balmer- 1st Orbit

alpha, or simply H .

Nucleus

SOLUTION

We let n 3 in equation (3.1) to obtain

R(1/22 1/32)

1/

Substituting for R and inverting gives

656.47 nm

This is the wavelength as measured in a vacuum.

Fig 3.4. The Bohr atom. Electrons orbit the nucleus in

We generally refer to the wavelength in air, since

allowed orbits.The relative sizes of the orbits are correct,

that is how we measure it at a telescope. The wave-

but on this scale the nucleus should be much smaller.

length in air is that in vacuum divided by the

3 SPECTRAL LINES IN STARS 29

the two states. If the frequency of the photon is , We put this into equation (3.2) to find the total

energy in terms of r:

this means that

(1/2)e2/r

E (3.4)

h En Em

The minus sign indicates that the total energy is

If the electron is in the lower energy state, it

negative. To see what this means, remember that

can make a transition to the upper state if the

we have defined the potential energy such that it

atom absorbs a photon with exactly the right

is zero when the electron and proton are infi-

energy. This explanation incorporated Einstein™s

nitely far apart. The electron and proton being far

idea of photons.

apart with no motion is the minimal condition

Bohr pointed out that one could calculate the

for the electron being free of the proton. So, if the

energies of the allowed states by assuming that

electron is barely free of the proton, the total

the angular momentum J of the orbiting elec-

energy would be zero. So, if the total energy is

trons is quantized in integer multiples of h/2 .

negative, as in equation (3.4), then we must add

The combination h/2 appears so often we give it

energy [(1/2)e2/r] if we want to bring it up to zero,

its own symbol, h (spoken as ˜h-bar™). We apply

which would free the electron. So the negative

this to a hydrogen atom, with an electron, with

energy means that the electron is not free. In this

charge e, orbiting a distance r from a nucleus

case we say that the system is bound.

with charge e. We assume that the nucleus is

We still have to find the allowed values of r.

much more massive than the electron, so we can

The angular momentum is J mvr. The quantiza-

ignore the small motion of the nucleus, since

tion condition becomes

both the nucleus and electron orbit their com-

mon center of mass. mvr nh/2

We first look at the kinetic energy

Solving for v gives

(1/2)mv2

KE

v nh/2 mr

The potential energy, relative to the potential

energy being zero when the electron is infinitely Squaring and multiplying by m gives

far from the nucleus, is given by mv2 n2h2/4 2 mr2

e2/r

PE By equation (3.2), we have

(By writing the potential energy in this form, e2/r n2h2/4 2

mr2

rather than with a factor of 1/4 0, we are using

We now solve for r, giving the radius of the nth

cgs units. This means that charges are expressed

orbit:

in electrostatic units, esu, with the charge on the

electron being 4.8 10 10 esu.) The total energy n2h2/4 2

me2

rn (3.5)

is the sum of kinetic and potential:

Substituting into equation (3.4) gives the energy

(1/2)mv2 e2/r

E (3.2) of the nth state:

We can relate v and r by noting that the elec- En (1/2)e4m(4 2)/n2h2 (3.6)

trical force between the electron and the nucleus, 2

e2/r2, must provide the acceleration to keep the Note that this has the 1/n dependence that we

electron in the circular orbit, v2/r. This tells us would expect from the Balmer formula.

One modification that we should make is to

that

account for the motion of the nucleus (since it is

2 22

mv /r e /r not infinitely massive). We should replace the

mass of the electron, m, in equations (3.5) and

Multiplying both sides of the equation by r

(3.6) by the reduced mass of the electron and pro-

gives

ton. The reduced mass, mr, is defined such that

2 2

(3.3) the motion of the electron, as viewed from the

mv e /r

30 PART I PROPERTIES OF ORDINARY STARS

(moving) proton, is as if the proton were fixed It is then very easy to calculate the energies of

and the electron™s mass is reduced to mr. An the other levels. For example, E2 3.4 eV.

expression for mr is (see Problem 3.2) Therefore the energy difference, E2 E1 is equal

to 10.2 eV. The energy levels are shown in Fig. 3.5.

memp

This diagram is a convenient graphical representa-

mr

me mp

tion of energy levels, called an energy level diagram.

In this diagram the levels are plotted as horizon-

0.9995 me

tal lines with vertical locations proportional to

where me and mp are the masses of the electron the energy. We can draw vertical arrows indicat-

and proton, respectively. ing possible transitions between the levels. The

length of the arrow would then indicate the

Example 3.2 Hydrogen atom energy

energy change associated with that transition.

Compute the energy of the lowest (ground) energy

Note that the levels are closer together as one

level in a hydrogen atom. Also, find the radius of

goes to higher values of n.

the orbit of the electron in that state.

Since the zero of potential energy is arbi-

trarily defined, we sometimes choose to shift

SOLUTION

the energy scale by the binding energy (13.6 eV

We use equation (3.6) with n 1 to give

for hydrogen). This would make the energy of

11>22 14.8 esu2 4 19.11 g 2 14 2

10 28 2

10 10 the ground state (n 1) zero, and for the n 2

11 2 2 16.63 erg s 2 2

E1 27

10 state, 10.2 eV. A free electron would then have

11

2.2 10 erg

n

An erg is not a convenient unit to use to keep

track of such small energies, so we convert to 0 5

electron volts, eV (1 eV 1.6 10 12 erg, is the 4

3

energy acquired by an electron in being acceler-

ated through a potential difference of 1 volt),

2

giving

Balmer

Energy (eV)

E1 13.6 eV

The radius is given by equation (3.5):

16.23 erg s 2 2

27

10

19.11 g 2 14.8 esu2 2 14 2

r1 28 10 2

10 10

9

5.25 10 cm

0.0525 nm

Note that if we take n in equation (3.6) we

get E 0. However, n corresponds to a free 1

13.6

electron. Therefore, to move the electron far from Lyman

the nucleus, we must add 13.6 eV. The energy that

Fig 3.5. Hydrogen energy levels.The right hand column

we must add to an atom to break it apart is called

gives the principal quantum number, n.The energies are rela-

the binding energy. The energy goes to do work

tive to the state in which the electron and proton are in¬-

against the electrical attraction between the elec- nitely far apart, so the ground state energy is 13.6 eV.

tron and the nucleus as you try to pull the elec- Transitions (which can be either emission or absorption) are

tron away. grouped according to the lower level of the transition. For

Now that we have evaluated E1, we can rewrite example, the Balmer series consists of emissions with the

equation (3.6) as electrons ending in state n 2, and absorptions starting in

the n 2 state.

13.6 eV/n2

En (3.7a)

3 SPECTRAL LINES IN STARS 31

an alpha ( ) transition; if n m 2, we have a

an energy of 13.6 eV, or greater. The values of

beta ( ) transition. The first Balmer line is then

the energy differences between these states are

designated H2 . (Note that for the Balmer series

unaffected by this shift in the zero point of the

of hydrogen only, we sometimes drop the 2 and

energy.

just say H , H , etc.)

We can use equation (3.6) to derive the Balmer

formula. First, we rewrite the equation as

3.3.2 Quantum mechanics

hcR/n2

En (3.7b)

The Bohr model of the atom allowed physicists to

where understand the organization of energy levels.

However, it was far from a complete theory. One

(1/2)e4m(4 2

)/ch3

R (3.7c)

shortcoming was that it did not explain why

The energy of an emitted or absorbed photon some spectral lines are stronger than others.

must equal the energy difference between the More fundamentally, it was an ad hoc theory. Bohr

two states: had no explanation of why stationary states exist,

or why angular momentum must be quantized in

En Em h

some particular way. These were just postulates.

hc/ A much deeper understanding was needed.

An important step was made by Louis de

Taking the energies from equation (3.7b) gives

Broglie, who proposed the revolutionary idea that

R(1/m2 1/n2)

1/ (3.8)

if light could exhibit a wave“particle duality,

then maybe all matter could. That is, an electron

which looks very similar to the Balmer formula,

orbiting a nucleus has certain wavelike proper-

except that the Balmer formula has a 2 instead of

the m. This means that the Balmer series all have ties, and it is those properties that determine the

states that are stable. One could think of the elec-

the second energy level as their lower level.

tron as having a certain wavelength. Stationary

We can use equation (3.8) to divide the hydro-

states could be those whose circumference con-

gen spectrum into different series. A given series

tained an integral number of wavelengths, pro-

is characterized by having the same lower energy

ducing a pattern that reinforced during each

state. For example, the Balmer series consists of

orbit (like a standing wave). It was necessary to

absorptions accompanying transitions from level

have expressions for the wavelength and fre-

2 to any higher levels, and emissions accompany-

quency of a particle, and de Broglie noted that if

ing transitions from higher levels down to level 2.

the wavelength was taken as h/p (where p is the

The first Balmer transition (involving levels 2 and

momentum of the particle) and the frequency as

3) has the smallest energy difference of the series.

E/h, then the orbits allowed by the standing wave

(Clearly the energy difference between levels 2

idea were the same as the orbits that Bohr found

and 3 is less than the energy difference between

from his postulates (see Problem 3.8).

levels 2 and 4, or between levels 2 and 5, and so

This is clearly a departure from our normal

on.) The Balmer series is important because the

way of looking at matter around us, and we can-

first few transitions fall in the visible part of the

not go through all of the ramifications here. To

spectrum. The series with the lower energy level

being level 1 is called the Lyman series. Even the this point, we have gone far enough to under-

stand stellar spectra. The picture as presented by

lowest transition in the Lyman series is in the

Bohr and de Broglie is quantum theory in its

ultraviolet.

most naive form. It was realized that if particles

We have developed a labeling system for vari-

behave, in some fashion, like waves then the

ous transitions. First we give the chemical symbol

description of particle motions (mechanics) must

for the element (e.g. H for hydrogen). Then we

give the m for the lowest level that characterizes be changed from Newton™s laws of motion to laws

of motion involving waves. (Of course, in the

the series (1 for Lyman, 2 for Balmer, etc.). Finally,

limit of large objects, such as apples falling to

we give a Greek letter denoting the number of

levels jumped. For example, if n m 1, we have Earth, these new laws of motion must reduce to

32 PART I PROPERTIES OF ORDINARY STARS

Newton™s laws, because we know that Newton™s tion. The number of atoms per unit volume in a

given state is called the population of that state. In

laws work quite well for apples and planets.)

Theories that describe the mechanics of waves this section we look at the factors that determine

are called wave mechanics or quantum mechanics. the populations of the various states. We refer to

processes that can alter the populations as excita-

One such theory was presented in 1925 by the

German physicist Erwin Schrödinger. In his theory tion processes. We have already seen one type of

the information about the motion of a particle is excitation process “ the emission and absorption

contained in a function, called a wave function. of photons. Electrons can jump to a higher level

Schrodinger™s interpretation of the wave func- when a photon is absorbed or they can jump to a

tion was that it is related to the probability of find- lower level when a photon is emitted.

ing a particle in a particular place with a partic- Populations can also be changed by collisions

ular momentum. This replaced the absolute with other atoms, as illustrated in Fig. 3.6. For

example, atom 1 can be in state i. It then under-

determinism of classical physics, with the state-

ment that we can only predict where a particle is goes a collision with atom 2, and makes a transi-

likely to be, but not exactly where it will be. tion to a higher state, j. In the process the kinetic

However, we can predict the average positions energy of atom 2 is decreased by the difference

and momenta of a large group of particles, and it between the energies of the two states in atom 1,

Ej Ei. The reverse process is also possible, with

is these average properties that we see (and meas-

ure) in our everyday world. Many physicists atom 2 gaining kinetic energy and atom 1 drop-

ping from state j to state i.

(including Einstein) were not comfortable with

this probabilistic interpretation, but quantum

theory has been very successful in predicting the

outcome of a wide variety of experiments. We (a)

will pick up on some of the threads of the quan-

tum revolution later in this book.

3.4 Formation of spectral lines

Now that we have some idea of how atoms can

emit or absorb radiation, we can return to stellar

spectra. The first point to realize is that in a star

Before

we are not talking about the radiation from a sin-

gle hydrogen atom, but from a large number of (b)

them. We see a strong H absorption line in stars

because many photons are removed from the con-

tinuum by this process. It is clear, however, that

having a lot of hydrogen does not assure us of a

strong H absorption. In order for such absorp-

tion to take place, a significant number of atoms

must be in level 2, ready to absorb a photon. If all

the hydrogen is in level 1, you will not see the

Balmer series, no matter how much hydrogen is Fig 3.6. Collisional excitation. In each case, the left frame

present. shows the atoms before the collision and the right frame

shows them after. In each frame, the occupied level is indi-

3.4.1 Excitation cated by a heavier line. (a) To a lower state. After the colli-

sion, atom 1 is in a lower state and atom 2 is moving faster.

In general, the strength of a particular transition

(b) To a higher state. After the collision atom, 1 is in a higher

(emission or absorption) will depend on the num-

state and atom 2 is moving slower.

ber of atoms in the initial state for that transi-

3 SPECTRAL LINES IN STARS 33

The collisional excitation rates will depend on the atoms are in the ground state, so the ratio is

the kinetic temperature of the gas. The higher zero. As the temperature increases, the quantity

the temperature the faster the atoms are moving. in square brackets gets smaller, so the exponent

For atoms of kinetic temperature Tk the average becomes less negative, and the ratio increases. If

we let Tk go to infinity the ratio of populations

kinetic energy per atom is (3/2)kTk. As the tem-

perature increases more energy is available for approaches the ratio of statistical weights. For a

collisions. This makes higher energy states easier given temperature, increasing the energy separa-

to reach. Also, since the particles are moving tion between the two levels makes the exponent

faster, they spend less time between collisions. more negative, lowering the ratio. This makes

There are more collisions per second. sense, since the greater the energy separation, the

When a gas is in thermodynamic equilibrium harder it is to excite the atom to the higher level.

(which we discussed in the previous chapter), The Boltzmann distribution provides us with a

with a kinetic temperature Tk, the ratios of the convenient reference point, even for a system that

level populations are given by a Boltzmann distri- is not in thermodynamic equilibrium. For any

bution. If we let ni and nj be the populations of lev- given population ratio nj/ni, we can always find

els i and j, respectively, their ratio is given by some value of T to plug into equation (3.9) to make

the equation correct. We call such a temperature

nj gj

31Ej Ei 2>kTk 4 the excitation temperature. When they are not in

e (3.9)

ni gi

equilibrium, each pair of levels can have a differ-

ent excitation temperature. In thermodynamic

In this equation gi and gj are called statistical

equilibrium all excitation temperatures are equal

weights. They are needed because certain energy

to each other and to the kinetic temperature.

levels are actually groupings of sublevels that

have the same energy. The statistical weight of a

3.4.2 Ionization

level is just a count of the number of sublevels in

If we know the temperature in the atmosphere of

that level. Typically, g are small integers.

a star, we can use the Boltzmann equation to pre-

To help us understand the Boltzmann distri-

dict how many atoms will be in each state, i, and

bution, Fig. 3.7 shows how the ratio of popula-

predict the strengths of various spectral lines.

tions for an atom with just two levels depends on

However, there is still an additional effect that we

temperature. When the temperature is zero, all

have not taken into account “ ionization. If the tem-

perature is very high, some of the colliding parti-

2

cles will have kinetic energies greater than the

ionization energy of the atom, so the electron

Population Ratio

1.5 will be torn away in the collision. Once a hydro-

gen atom is ionized, it can no longer participate

in line emission or absorption.

1

When the gas is ionized, electrons and posi-

tive ions will sometimes collide and recombine.

When the total rate of ionizations is equal to the

0.5

total rate of recombinations, we say that the gas

is in ionization equilibrium. If the gas is in thermal

0 equilibrium and ionization equilibrium, then the

1 — 105 2 — 105

0

Saha equation tells us the relative abundances of

various ions. We let n(Xr) and n(Xr 1) be the densi-

Temperature (K)

ties of the r and r 1 ionization states, respec-

tively, of element X. (For example, if r 0, then

Fig 3.7. Level populations as a function of temperature for

a two-level system. In this case we have put in energies and we are comparing the neutral species and the

statistical weights (3, 5) for the n 2 and n 3 states of first ionized state.) The ionization energy to go

hydrogen (¬rst Balmer transition).

from r to r 1 is Eion. The electron density is ne,

34 PART I PROPERTIES OF ORDINARY STARS

and the kinetic temperature is Tk. Finally, gr and

Ionization energies (eV).

Table 3.1.

gr 1 are the statistical weights of the ground elec-

Atom Singly ionized Doubly ionized

tronic states of Xr and Xr 1 (assuming that most

of each species is in the ground electronic state).

H 13.6 “

The Saha equation tells us that

He 24.6 54.4

12 2 gr 2 mr kTk 3>2

a be

ne n1X r 3Ei>kTk 4 C 11.3 24.4

n1X r 2

(3.10)

h2

gr 1 N 14.5 29.6

O 13.6 35.1

The Saha equation has the same exponential

Na 5.1 47.3

energy dependence as the Boltzmann distribution.

K 4.3 31.8

3>2

However, there is an additional factor of T k . This

Ca 6.1 11.9

comes from the fact that a free electron has more

Fe 7.9 16.2

states available to it at higher Tk than at lower Tk.

In addition there is a factor of ne on the left. This

is because a higher abundance of electrons leads to

so the left side of equation (3.10) simplifies to

a higher rate of recombinations, driving down the

n2>n0, where n0 is the number of neutrals. This

fraction of atoms that are ionized. Just as we did e

extra factor of ne makes even this simpler form of

with the excitation temperature in the Boltzmann

the Saha equation harder to solve for (ne/n0 ) than

equation, we can define an ionization temperature Ti,

the Boltzmann equation is to solve for the ratio of

which makes the Saha equation correct, even if

level populations. In Fig. 3.8, we show the ratio

the gas is not in thermodynamic equilibrium.

ne/n0 as a function of temperature, for a value of

In this equation ne is the number of electrons

ne reasonable for stars like the Sun.

from all sources, since any electron can combine

The ionization energies of some common

with a hydrogen ion (for example) no matter

atoms are given in Table 3.1. This table is useful in

where that electron came from (hydrogen,

deciding which ions you are likely to encounter

helium, etc.) In many situations, virtually all of

at various temperatures. In designating ionized

the ions are hydrogen. That is because hydrogen

atoms, there is a shorthand that has been adopted.

is by far the most abundant element, and because

The roman numeral I is used to designate the

the next most abundant element, helium, is very

neutral species, II the singly ionized species, III

hard to ionize. In that case, the number of elec-

the doubly ionized species, and so on. For exam-

trons is equal to the number of positive ions, n ,

ple, neutral hydrogen is H(I), ionized hydrogen

(H ) is H(II), doubly ionized carbon is C(III).

1

Ionization Fraction

3.4.3 Intensities of spectral lines

We are now in a position to discuss the intensities

of various absorption lines in stars. We will take

H as an example to see the combined effects of

0.5 excitation and ionization. At low temperatures,

essentially all of hydrogen is neutral, and most of

it is in the ground state. Since little H will be in

the second state, there will be few chances for H

absorption. The H line will be weak.

0

As we go to moderate temperatures, most of

1 — 104 2 — 104

0

the hydrogen is still neutral. However, more of

Temperature (K) the hydrogen is in excited states, meaning that a

Fig 3.8. The ratio of electrons to the total number of reasonable amount will be in level 2. H absorp-

hydrogen atoms (neutral plus ion), for an electron density tion is possible. As the temperature increases, the

appropriate to stars like the Sun.

H absorption becomes stronger.

3 SPECTRAL LINES IN STARS 35

At very high temperatures, the hydrogen

becomes ionized. Since there is less neutral hydro-

gen, the H line becomes weaker. This explains

why the H line is strongest in middle-temperature

stars, and why the original scheme of classifying

by hydrogen line strengths did not produce a

sequence ordered in temperature.

We can apply a similar analysis to other ele-

ments. The details will differ because of different

energy level structures and different ionization

Fig 3.10. The relative strengths of spectral lines from

energies. It should be noted that, after hydrogen

important species as a function of spectral type. Each species

and helium, the abundances of the elements fall

shows the effects of excitation and ionization. For example,

off drastically (see Appendix F for the abundances

the increase in H line strengths from K to A stars occurs

of the elements). In fact, astronomers often refer

because the increasing temperature results in more hydro-

to hydrogen, helium and ˜everything else™. The gen in the n 2 (and higher) levels. However, the higher

˜everything else™ are collectively called metals, temperatures of the B and O stars ionize much of the

even though many of the elements don™t fit our hydrogen and the lines get much weaker.

common definition of a metal.

We now look at the properties of different star has strong ZrO (zirconium oxide) lines

spectral types, in order of increasing tempera- as opposed to TiO lines, we call it an S-type.

ture. Sample spectra are shown in Fig. 3.9, and K Temperatures range from 3500 to 5000 K. There

the behaviors of a few spectral lines are shown in are many lines from neutral metals. The H

Fig. 3.10. lines are stronger than in M stars but most

of the H is still in the ground state.

M Temperatures in M stars are below 3500 K,

G Temperatures in the range 5000“6000 K. The

explaining their red color. The temperature

Sun is a G2 star. The H lines are stronger

is not high enough to produce strong H

than in K stars, as more atoms are in excited

absorption, but some lines from neutral

states. The temperature is high enough for

metals are seen. The stars are cool enough

metals with low ionization energies to be

for simple molecules to form, and many

partially ionized. Two prominent lines are

lines are seen from molecules such as CN

from Ca(II). When Fraunhofer studied the

(cyanogen) and TiO (titanium oxide). If cool

solar spectrum, he gave the strongest lines

stars show strong CH lines, we designate

letter designations. These Ca(II) lines are the

them as C-type or ˜carbon stars™. If any M

H and K lines in his sequence.

F Temperatures range from 6000 to 7500 K. The H

lines are a little stronger than in G stars. The

ionized metal lines are also stronger.

A Temperatures range from 7500 to 10 000 K.

These stars are white“blue in color. They

have the strongest H lines. Lines of ionized

metals are still present.

B Temperatures are in the range 10 000“30 000 K,

and the stars appear blue. The H lines are

beginning to weaken because the tempera-

tures are high enough to ionize a significant

Fig 3.9. Samples of spectra from stars of different spectral

fraction of the hydrogen. The lines of neutral

types.The name of the star appears on the right of each

spectrum, and the spectral type appears on the left. In each and singly ionized helium begin to appear.

spectrum, the wavelength increases from left to right. Hotter Otherwise there are relatively few lines in

stars are at the top. [NOAO/AURA/NSF]

the spectrum.

36 PART I PROPERTIES OF ORDINARY STARS

O Temperatures range from 30 000 to over magnitude. However, if we find a group of stars

60 000 K, and the stars appear blue. The ear- all at the same distance, we can plot their appar-

liest spectral types that have been seen are ent magnitudes, since the distance modulus