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I've just begun learning of Astronomy and I can't figure out why any stars would begin their life with such elements if nuclear fusion hasn't created them.

Don't all stars begin life as Hydrogen?

I understand that the presence of these elements can change the luminosity of the given star. Is this also correct?

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    $\begingroup$ It would be true that all stars being life with just Hydrogen if only hydrogen gas clouds existed. However, we live in a universe that is 14 billion years old and many old and now dead stars have "polluted" the hydrogen clouds, filling them with trace amounts of heavier elements that then find their way into more recently formed stars. $\endgroup$
    – zephyr
    Mar 7, 2017 at 17:07
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    $\begingroup$ See astronomy.stackexchange.com/questions/16311/… $\endgroup$
    – ProfRob
    Mar 7, 2017 at 19:14
  • $\begingroup$ And see table 2 of weft.astro.washington.edu/courses/astro557/LODDERS.pdf $\endgroup$
    – ProfRob
    Mar 7, 2017 at 19:18

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Stars are mostly hydrogen and helium. The presence of small amounts of other elements does change their luminosity, and their spectra. The heavy elements were formed in other stars.

After the big bang there was a short period in which the conditions were right for hydrogen to fuse into helium. The conditions were not suitable for any heavier elements to form, the state of the universe right after this time was about 3/4 hydrogen, 1/4 helium.

After some time clouds of hydrogen and helium began to form (around clumps of dark matter) and inside some of these clumps, smaller clouds collapsed to form the first generation of stars. These were large and they had no elements other than hydrogen and helium. These stars produced in their cores heavier elements like Carbon and Oxygen, and when they had reached the end of their lives, being big stars, they exploded in supernovae, and these elements were returned to the interstellar space, along with other heavier elements that were formed by the supernova.

The next generation of stars formed from a mixture of hydrogen, helium and the heavier elements formed in the first generation. These stars still had less heavy elements than modern stars, but were more like normal stars. Some of these stars are still around, they are known as Population 2 stars. They are found mostly in globular clusters http://hyperphysics.phy-astr.gsu.edu/hbase/Starlog/pop12.html.

After the population 2 stars had lived their lives and died, either in a supernova or in a planetary nebula, elements that had been made in their cores was returned to the intersellar space, futher enriching it with heavy elements. It was from one of these clouds, seeded with elements from stars that had gone before it, that the sun and the solar system formed.

This is why Carl Sagan says We are star stuff.

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    $\begingroup$ Worth mentioning that Fe is the largest element which can be formed via "normal" fusion & all larger come from supernovas. $\endgroup$ Mar 8, 2017 at 16:31
  • $\begingroup$ @CarlWitthoft (and whomever gave it an uptick) this is partially incorrect. Approximately half of the elements heavier than iron (up to lead) are NOT from supernovae and are made by slow neutron capture in intermediate mass stars. $\endgroup$
    – ProfRob
    Mar 10, 2017 at 16:46
  • $\begingroup$ @RobJeffries you are correct - but only the s-process. TIL rather a lot about the different processes for burning/forming heavy elements, in particular that almost all those processes other than s-process occur during or just barely prior to the supernova shock wave. I should have said Fe was the limiting element produced via fusion rather than neutron (or putatively, proton) capture. $\endgroup$ Mar 10, 2017 at 18:07
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    $\begingroup$ @CarlWitthoft Furthermore it is now becoming apparent that supernovae may not even by responsible for creating most of the r-process elements. $\endgroup$
    – ProfRob
    Nov 4, 2017 at 0:53
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James' answer already covers the fact that since helium was formed in the Big Bang (see Big Bang Nucleosynthesis), stars always formed with some helium (about 1/4 by mass). Those first stars fused the hydrogen and helium into heavier elements, and some of them eventually spewed these products into their neighbourhoods, so the next generation of stars was born with those "metals" (all elements heavier than helium) from the start. Each subsequent "generation" of stars was born with more and more of the detritus of the previous generations, so the Sun was born with a wide range of metals. All that said, the metal content in your average stars is still low as a fraction of mass. e.g. around 2% in the Sun.

To answer the title question, however, I suggest starting with a widely-cited review article from 2009. Their Table 1 gives the logarithm of the number densities relative to hydrogen, plus 12. i.e. $$\log_{10}\epsilon_X=\log_{10}(n_X/n_\mathrm{H})+12$$ The values for carbon, sodium, silicon, and magnesium are 8.43 ± 0.05, 6.24 ± 0.04, 7.51 ± 0.03 and 7.60 ± 0.04, respectively. If you follow the paper trail, you'll find other estimates too.

Incidentally, the metal abundance of the Sun is actually a contemporary problem. Though this oversimplifies the details substantially, we generally measure the abundances by comparing spectral lines with models of the Sun's outer layers atmosphere. In the late 1990's and early 2000's, those models leaped forward from generally one-dimensional, hydrostatic models, to three-dimensional hydrodynamic models. Although the 3D models fit the spectral data better, they do so with lower abundances than we inferred with the 1D models. This revision isn't itself a problem but it means that our models of the Sun are quite a lot worse than we used to think (see Fig. 8 in the review), so there's a fair amount of effort going into trying to find what's missing from the models to explain the remaining discrepancy.

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Nuclear fusion is the process in which atomic nuclei are fused together. You may remember that like charges repel and opposite charges attract. In order for atomic nuclei to undergo nuclear fusion, they need to overcome the Coulomb Barrier, which is basically a minimum threshold of energy needed for like charges to overcome their repulsion. The only place this happens is inside the cores of stars due to the high temperature needed to overcome this threshold.

Although oversimplified, the main idea is that fusion of atomic nuclei combines tiny masses of particles and releases energy (via $\Delta E = \Delta m * c^2$). A very small amount of mass can be converted into a lot of energy. But the combined mass of fusing atomic nuclei is slightly less than the addition of the individual masses of atomic nuclei; it is this tiny mass difference that is converted into energy that we receive from the Sun in the form of heat and light.

Suppose you have a cup of water that you drop a rock into. Because the density of the rock is greater than the density of the water, the rock will sink to the bottom. Similarly, elements born via nuclear fusion are heavier than their "parents". In this way, nuclear fusion will keep building up heavier and heavier elements in the core of the star while the byproducts of the lighter elements "float" towards the top. But now that there are heavier elements in the core, the amount of energy required for nuclear fusion increases. Once the nuclear fusion chain reaches iron, the energy transfer is no longer cost-effective; the output energy from nuclear fusion will be less than the input energy needed for nuclear fusion, so the temperature is no longer sufficient to act as the input energy.

Once the star dies and goes nova, its elements (via fusion) are scattered in all directions, which can partially explain the galactic abundance of elements. You may notice from an HR-diagram that the characteristics of a star (like luminosity, mass, temperature, and size) change over time. If you look specifically at the sharp turns in the diagram throughout the lifespan of a star (like our Sun), you will notice that they occur when new fusion changes begin, such as the CNO cycle (wh

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    $\begingroup$ The edit cut off the end of my answer. If you edit my answer, the least you could do is post the continuation of it ... $\endgroup$
    – user15317
    Mar 12, 2017 at 0:47

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