For an H-R diagram, we usually plot surface temperature on the x-axis and luminosity on the y-axis. For individual stars in the galaxy, we cannot determine the luminosity of the star just by observing it - we also need to know its distance. However, all of the stars in a star cluster are about the same distance away from us. That means that any difference in the apparent brightness between stars is caused by them having different luminosities, not because they are different distances away from us. That means that we can observe the stars in a star cluster and plot an H-R diagram with apparent brightness instead of luminosity on the y-axis and still get a valid H-R diagram.

 The rate of nuclear fusion depends strongly upon the temperature. High temperatures mean fast motions and when the particles move fast, they have a

larger chance to “penetrate the electrostatic shield” provided by the repulsive electric

force between the two nuclei. Once they penetrate deep enough, the attractive strong force will overwhelm the electric repulsion, and fusion takes place.  When a star becomes a red giant, there is no longer any fusion going on in the core of

the star; so no heat generated; so core contracts and heats up (conversion of gravitational potential energy into heat); contraction stops when core becomes degenerate. Thus the outside of the contracted red-giant core is hotter than the core of the hydrogen-burning MS star. The hydrogen that burns in the shell just outside the core is thus hotter, which means a higher rate of burning and hence larger energy production.

If we suppose that the proton-proton chain of fusion is happening in the Sun, then there should be a specific amount of light produced. That amount of light matches up with what we observe from the Sun.

Neutrinos are a by-product of the proton-proton chain. We have observed neutrinos coming from the Sun, and in the correct amount to match with the proton-proton chain being the source.

Sirius, in the constellation Canis Major, is emitting energy at a rate about 24 times

larger than the Sun. Sirius’s mass is about 2 solar masses. Do you think that Sirius will have fusion inside its core for a longer or shorter amount of time than the Sun? Explain your reasoning. (You should think about the entire lifetime of the Sun and Sirius.)

Sirius has ∼2 times more mass than the Sun, so the first impression might be that

Sirius will have fusion for longer than the Sun because it has more “fuel” to fuse. However, Sirius is using up its mass at a much faster rate than the Sun: 24x faster. That means that even though Sirius has more mass than the Sun, it will use up that mass much more quickly. Consequently, Sirius overall lifetime during which it has fusion in its core will be shorter than

the Sun’s.

FYI1: According the the mass-luminosity relation (L = M3.5), Sirius’ energy output should

be 11.3x larger than that of the Sun. This is smaller than Sirius’ actual (observed) luminosity (24x). There are at least 3 reasons for this: 1) its mass is slightly larger than 2.0 [luminosityis 1.03x larger], 2) the mass-luminosity relation is only approximate: at larger masses the

exponent is larger than 3.5 [ [luminosity is 1.31x larger] and 3) Sirius is slightly evolved, and so it produces more energy as when it first arrived on the MS. It’s age is ∼150 Myr, while the MS lifetime is 780 Myr [luminosity is 1.06x larger]. Applying all corrections brings the luminosity up to 11.3 x 1.03 x 1.31 x 1.06 = 16.2x LSUN . We are still a factor 24/16.2=1.48 short.

FYI 2: This is seriously mysterious. This extra luminosity is due to the fact that Sirius has

about 2x as much heavy elements as the Sun, and this is know to increase the luminosity. That factor 1.48 is about in the range as to what I would expect.

FYI 3: On the other hand, the mass-luminosity relation is more valid for stars as they first reach the MS. The Sun did so 4.5 Gyr ago. Since the Sun evolves at about 10% per Gyr, it was 1.1^4*.5 = 1.54x fainter than it is today. This is tantalizingly close to the missing factor 1.48x from above. So what about the abundance of heavy elements?

FYI 4: Well, Sirius has a white dwarf as a companion (Sirius B). This means that Sirius B was originally heavier than Sirius A, hence it evolved faster than Sirius A. During it’s evolution, Sirius B has shed a lot of material, part of which has been captured by Sirius A. This could also have changed the “metal” content of Sirius A. So maybe the high “metal” content of Sirius A is just a surface feature. Evidently, it’s all pretty complicated as soon as you start digging into a particular example.

Clusters of stars have been instrumental in helping astronomers understand the various

phases of stellar evolution. Which three critical properties of star clusters made this possible, and why are these factors indeed critical.

There are a number of possible answers. You need to pick three at two

points per correct answer.

1) All stars are at practically at the same distance. Even if we don’t know the distance very accurately, we can still compute how much brighter one star is as compared to the other stars. This allows us to create an HR diagram that looks pretty much like how

it would look like if we would know the exact distances. That is to say, the relative

positions of the various groups of stars (MS, giants, white dwarfs etc.) have the same relative locations, and so you will still be able to tell which star is a MS, WD, giant etc.

FYI: if we make an error in the distance by a factor of 2, the luminosity is off by a factor

of 4 (inverse-square law). BUT, all stars are affected by this error in the same way and

the HR diagram shifts up/down by this factor while preserving its shape.

2) All stars have approximately the same age (born at the same time) and so the HR

diagram of a cluster gives a good indication as to where a star of a given mass is at the age of the cluster. By comparing different clusters with different ages, one can put together a reasonable picture of the evolution of stars.

3) The composition of all stars in the cluster is pretty much the same, and so you don’t

have to worry about how composition affects stellar evolution.

4) They are all in about the same part of the galaxy, and so interstellar extinction towards the cluster is pretty uniform (because the cluster is pretty small as compared to the Galaxy). Thus, we don’t have to worry about different stars changing color and brightness

by different amounts of extinction.

5) See question #4c)

Clusters of stars have been instrumental in helping astronomers understand the various

phases of stellar evolution. How many stars are, typically, present in open clusters?

There are 100s to 1000s of stars per open cluster.

Explain why the number of stars in a cluster is also important, but not “critical’.”

Hint: consider the extreme case that the cluster contains just a single star.

Since there are many stars in a cluster, many phases of the evolution of

the cluster “will be sampled” by the cluster. That is to say, the various regions of the HR diagram (MS, giant, WD, ...) will be populated: if there are enough stars in the cluster, one might be lucky to find one or a few stars in rare stages of stellar evolution such as planetary nebula or tip of the red-giant branch (He flash). As an extreme case, if the cluster would have just one star, it wouldn’t tell us very much: just one dot on the HR diagram, and only one phase would be populated, and we would have no idea as to how that star might change over time or how it would be different if its mass were different, etc

Explain what hydrostatic equilibrium is, in the context of the Sun and fusion.

In the Sun, hydrostatic equilibrium is a balance between pressure and gravity. The

gravitational force of the Sun upon itself is trying to pull all of the Sun’s mass inward and compress it. This is counteracted by the thermal pressure of the Sun, which is pushing outward. The thermal pressure is caused by the thermal energy of the Sun’s particles in its interior. The Sun’s particles have a lot of thermal energy because they are densely packed and because thermonuclear fusion is happening in the core, which heats them up. Because these two processes are balanced, the Sun remains at the same size (in terms of

diameter).

If we could somehow turn a dial and turn up the Sun’s core temperature, how would the

Sun’s fusion rate and diameter be affected? Explain your reasoning.

Describe the five phases a star like our own Sun will go through from the moment

fusion started on the main sequence through it’s end as a white dwarf. Describe/Name: 1) each of these phases, 2) where it is generating energy, 3) which elements are being produced by nuclear fusion, 4) the luminosity of the star as compared to as the main-sequence phase, 5) same for radius, 6) the importance of mass loss (stellar wind) and it’s effect on the total mass of the star, and 7) does the wind contain a significant amount of “nuclear ashes” (helium, beryllium, carbon, oxygen, ...) that were previously generated by this star’s fusion processes? [hint: is the star convective?]

[image]

NOTES TO TABLE:

1) MS phase: mass loss does occur, but at a really small rate (1/10 15 M SUN per year)

2) Red giant phase: you may also call this simply “giant” or luminosity class III. Luminosities during the RGB phase are 10 – 100x larger than on the MS. Significant mass loss occurs during the RGB phase, but much less as during the AGB phase (#4).

3) Horizontal branch: structure of star is like on MS, except that it burns helium in its center (no shell burning). Luminosity is 10s – 100x larger, radius about 10x larger. Small amount of mass loss.

4) 2nd ascend to giant branch, or asymptotic giant branch (AGB). Have a helium burning shell making carbon (and oxygen) and a hydrogen burning shell creating helium. NO nuclear fusion. Luminosities are 100s – 10,000x larger than on MS, and radii several 100x larger. The

interior becomes convective (needs to transport lost of energy outwards), and the nuclear ashes are mixed throughout the star. The stellar wind is very strong containing the waste products of nuclear fusion. The star looses a large fraction of its mass while on the AGB.

5) Central star of planetary nebula: when the central star of the PN is still very hot and luminous, some mass loss occurs. Not very much, so if you answered “yes” or “no,” either is more-or-less correct. And since the central star of a planetary nebula is essentially the exposed core of a star, its wind naturally contains nuclear ashes. Radius is about that of a white dwarf

(Earth-sized (5,000–10,000 km). But, because it extremely high temperature (exposed core of star), it is still very luminous. It has run out of fissionable material, so no more energy production.

6) White dwarf: like #5, except that is has cooled down dramatically and so the stellar wind has all but disappeared. So no mass loss is the most correct answer, but of course there is always some mass loss (just like on the Sun), and this material would be enhanced with nuclear waste products.

Emission nebulae are comprised of gas (and dust) and are found near hot (and

bright) stars. These stars are hot enough that they emit lots of photons that are energetic enough to ionize the gas in the nebula. In this ionization process, electrons absorb a photon that is energetic enough to separated it from the nucleus. Eventually, these electrons recombine with the nuclei, and when doing so, they emit radiation of the same wavelength, but in an arbitrary direction If you include something like this in your ANSWER, you’ll earn 2 pts: You might wonder as to why there is any net effect: photons get absorbed and re-emitted. How comes one sees a nebula? The answer lies in the fact that these bright stars emit light in all directions, not just towards us. In the nebula, the light that was absorbed at positions not directly between us and the star ionizes the gas as well. And, when the electrons recombine, part of those photons emitted will be directed towards us: hence the extended nebula.

Describe two ways we can observe some of the properties of dark nebulae in the Milky Way.

Dark nebulae absorb light from the stars “behind” them from our point of view. When the starlight is only partially absorbed, the blue light is absorbed more, so the stars appear redder than they should. We can learn about the density of a dark nebula by studying this effect.

• If the dark nebulae only partially absorb light from the stars behind them, then when we look at the stars’ spectra, there will be extra absorption lines from the nebulae. The absorption lines can tell us about the composition and velocity of the nebulae.

• Dark nebulae are too cold to emit visible light. However, at these low temperatures they emit radio light, either from the atoms or from the molecules in the nebulae. When we observe the nebulae in radio light, we can learn about the shape of the nebulae and the specific atoms or molecules in them.

• The dust in the nebulae is too cold to emit visible light, but it is warm enough to emit infrared light. When we observe the  nebulae in infrared light, we can learn about the amount of dust in them and the shape of the nebulae.

In the core of the Sun, hydrogen is continuously “transmuted” into helium.

(a) What is Einstein’s most famous equation, and: 1) explain this equation in words, 2) what are the units of the terms that make up the equation

Assuming that all the energy that the Sun radiates is generated through fusion in the core, how many kilograms of hydrogen are converted into helium every second. HINT: it may be easier if you first assume that mass is converted to energy with 100% efficiency. Then, take into account the actual conversion efficiency of 0.7% (7 parts per 1,000).

the luminosity is LSun = 3.8 10^26 J/s, so every second, the amount of energy produced equals 3.8 10^26 J. This energy comes from the mass-to-energy conversion: E = m c^2. Solving for the mass, and assuming 100% efficiency, we get: m = E/c^2 = 3.8 10^26 J/(3 10^8 m/s)

2 = 4.2 10^9 kg. Now, the Sun is only 0.7% efficient, so it actually converts (4.2 10^9) × 1000/7 = 6 10^11kg (600 million tons) of hydrogen into helium, every second.

Astronomers often say that the Sun converts hydrogen into helium.

(a) At a recent party, a physicist told me that that’s actually not correct, but that instead “it converts hydrogen nuclei into helium nuclei.” Was she right, wrong or both?, Explain your answer.

That is right indeed. In the core of the Sun, the material is fully ionized: it is a plasma where all electrons are stripped off the nuclei. And so it is really the nuclei that

are doing the “fusing.” And of course she’s also wrong since it is not the whole story

Another party-goer chimes in and say: “you are both wrong: in reality it turn protons into neutrons!” Was he right, wrong or both?, Explain your answer

What is the mass-luminosity relation for main-sequence stars.

EQN: How does the luminosity of main-sequence stars depend on mass?

ANSWER [5 pts]: The mass-luminosity relation tells us how luminous mains sequence stars are, given their mass. Alternatively, if you know the absolute magnitude (i.e., Luminosity) of a main-sequence star, you can compute its mass:

EQN: LMS = M3.5 MS

• What is the Stefan-Boltzmann law

EQN: How does the radius of a star depend on stellar luminosity and temperature?

ANSWER [5 pts]: The Stefan-Boltzmann law tells us how much energy is radiated away per second per unit area as a function of its temperature [if you include the area term (4π R2), that’s

OK too]. EQN: L = σT^4×4π R^2) and in solar units: L∗/LSUN = (T∗/TSUN )^4

( ×(R∗/RSUN )^2)

For the longest time, our life form used silver and gold as “money.” Where/how were these

elements formed.

These elements are formed through (a special form of) nuclear fusion during

supernova explosions.

What is special about the main-sequence?

Briefly describe the other three groups of stars (temperatures, masses, radii, luminosities, ...).

Since this galaxy is reported to be quite far away, an individual star would have to be pretty bright intrinsically for it to be visible with your backyard telescope. Thus, such

a star would be most likely a super giant, the brightest among the stars.

per correctly filled out column. To compute the luminosity in the 5th column,

you need use the mass-luminosity relation for main-sequence stars: L = M3.5

(see question #5). To compute the radius, we use the Stefan-Boltzmann relation with area: L = 4π R^2σ T^4. However, when computing things in solar units, we do not care about factors 4π and σ because these have the same value for the Sun as for the star.