Most of what we know about stars we have learned from observations of our closest star, the Sun. It is reasonable to assume that more distant stars work in roughly the same way as the Sun and we have no evidence to make us think otherwise. Indeed, current evidence suggests the Sun is very much like any Main Sequence star.
You will recall that our Sun is a yellow star (G2-type). Because it is slightly on the small side, some astronomers classify it as a yellow dwarf. Regardless, the Sun provides
us with a wealth of information (plus a lot of light and heat )!
Throughout this lesson we will be using the Sun as an example
but, unless I say otherwise, you can assume that what goes for
the Sun also goes for other stars on the Main Sequence.
So. How does a star work?
Gravity pulls matter towards the center of a star and continues to do so until the internal pressure balances the gravitational pressure. If you have ever studied physics or chemistry you will recall that as the internal pressure of a gas rises, its temperature also increases. With a ball of gas as big as a star this pressure and heat can become substantial! When the center of the ball reaches a temperature of about 10 millionoC (or oK, it doesn't really matter) the ball of gas "ignites" as nuclear fusion takes hold. A star is born and it moves to the Main Sequence. It's position on the Main Sequence depends upon the star's mass.
Is this like coal burning?
No, not at all. If the entire Sun was made of coal it would "burn" for only 5,000 years (assuming it had oxygen to help it burn). Stars are not powered by mere chemical energy, the way coal burns. Chemical energy, and chemical reactions have to do with moving electrons from one place to another, usually between atoms. Chemical energy is nonsense in the center of a star because it is so hot in there that all the atoms are stripped of all their electrons. The core of a star is made of nothing but "raw and naked" atomic nuclei - no electrons are associated. Atoms stripped of electrons have a positive charge and are called cations.
Hydrogen is the most abundant atom (element) in a star. That's
because hydrogen is the most abundant atom (element) in the universe.
It was produced in the "Big Bang", but that's another
story. Hydrogen is also the simplest of atoms. Most hydrogen
is just a proton, which we abbreviate "p". Because all
the hydrogens in a star's core are stripped of electrons, all
the hydrogens are cations and are abbreviated "p+".
A star gets most of its energy by fusing these protons (p+). It
is not easy to fuse cations because their positive charge causes
them to repel each other but inside the hot, high pressure of
a star's core this fusion can occur.
The first step is the fusion of two hydrogen nuclei. These hydrogen
nuclei are nothing more than single protons so they each have a charge
of +1. When they fuse they produce a new hydrogen nucleus called
deuteron. We abbreviate that "D". Deuteron consists
of one proton, with a charge of +1, and one neutron that has no
charge at all!. The leftover positive charge, released when one of
the protons became a neutron, is emitted as a peculiar particle
called a positron. A positron is the antimatter version
of an electron, so it has a positive charge instead of the negative
charge of an electron. A normal electron is abbreviated "e-"
but because the positron has a positive charge it's abbreviated
"e+". Also emitted in this event is a weird particle
called a neutrino, abbreviated "". Neutrinos
have no charge and very, very little mass). If you think neutrinos are hard to detect, you're right! Neutrinos interact with matter very poorly but they
are becoming an increasingly important clue in our
studies of stars and the universe. But let's not go to far too
quickly. (We'll come back to neutrinos in a future lesson.)
Let's summarize step one with an equation. I'll show the mass as superscripts to remind us how much mass makes up each nucleus (this mass is called its "nucleon count") and I
will also place a superscript after the ion to remind us of its
charge.
1p+ + 1p+ ---fuse----> 2D+ + e+ +
This reaction is called the "slow step" because, well,
it's so slow! It takes about 10 billion years for half of the
hydrogen in the core of an average-sized star to go through this process. That's because this fusion, of these two small, positively charged particles, is not permitted by classical physics! The protons will repel each other - and the closer they get the more they will repel. That's a basic law of electrostatics. For those two protons to fuse you would need much higher temperatures (more energy) or more pressure (density) than that found in the center of most stars. Fortunately, quantum physics provides a "loophole". There is a possibility - a probability - that these two protons will get close enough to each other to fuse. That's a very small probability but, given enough time, it will happen. This quantum event is called the tunnel effect because, through the strange probabilistic world of quantum mechanics, particles are able to get over the "energy hump" that seperates these two particles through a quantum mechanical "tunnel". Strange but true. If it were not for the quantum tunnel effect, protons would not fuse and stars would not shine.
Is this step when all the nuclear energy is created?
No. Only a small amount of energy is given off at this step, about 1.4 MeV. [Let's not get hung up on the definition of a MeV. Instead, simply use it as a way to keep track of the energy.] This energy output is insignificant compared to what comes later but it's an important step because it creates the important ingredient - deuteron (D).
Within a few seconds of being formed the deuteron (2D)
fuses with another proton (p) to produce a new element ( ) called
helium-3 and abbreviated "3He". The creation of new
elements from old ones is called transmutation and it is
through transmutation that our universe has become populated with
over 90 different elements. Some transmutations, such as this
one, occur as a normal process of a star's nuclear fusion. Other
transmutations occur when unstable elements breakdown or when
a star explodes, but I'll leave that to a future lesson. [Helium-3 is present in relatively large quantities on the Moon
and some folks have considered harvesting the Moon's helium-3.]
This second step (the transmutation of hydrogen to helium) can
be written
1p+ + 2D+ ---fuse----> 3He++ +
Lambda () is a standard abbreviation in physics meaning "light".
So this second step produces helium from hydrogen and gives off
a ray of light. Actually, a "ray of light" is not a very scientific way to express it. We tend to think of light rays as beams of light and associate it with the light from a flashlight, the ray of light we see enter a dark room, or the path of light as it passes through a lens. It would be more accurate to describe this single unit of light energy as a photon. The photon released in this particular reaction is
a very powerful, high frequency (short wavelength) kind of light called
a gamma ray.
And that produces the star's light?
This reaction produces 5.5MeVs and accounts for about a third of the star's energy.
However, this isn't the end of the story. Two helium-3s (3He)
are fused together to create helium-4 (4He) and two protons are
released like this
3He++ + 3He++ ---fuse----> 4He++ + 1p+ + 1p+
This reaction releases 12.9 MeVs of energy mostly as kinetic energy. Kinetic energy is the "energy of motion" and in this particular reaction it refers to the motion caused when these three cations (the He++ and two p+) are repelled from each other due to their charges. As these particles collide with other particles, this kinetic energy is quickly turned to high energy light rays - gamma
rays again.
If you are keeping track of all this you can summarize these nuclear
reactions as
1p+ + 1p+ ---fuse----> 2D+ + e+ + + (1.4 MeVs) X 2 (because
you'll need two helium-3s in the last step) = 2.8 MeVs
1p+ + 2D+ ---fuse----> 3He++ + (5.5 MeVs) X 2 (because, again, you'll
need two helium-3s in the last step) = 11 MeVs
3He++ + 3He++ ---fuse----> 4He++ + 1p+ + 1p+ + (12.9 MeVs) = 12.9
MeVs
That's a total of 6 p+ ---fuse----> 4He++ + 21p+ + 2e+ + 2
+ 2
+ 26.7 MeVs
So, as you can see, all three steps produce some energy. The first step, the slow step, produces only about 10% of the output but without the first step you cannot get the other two steps.
There are alternative pathways. For example a helium-3
can fuse with a proton to make helium-4 like this
3He++ + 1p+ ---fuse----> 4He++ + e+ +
This step releases 19.8 MeVs. Notice that you only need one helium-3
for this step. That means you only need one each of steps #1 and
#2 to create the required helium-3 and if you work though the
math you'll see that you still get a total of 26.7 MeVs that way
too. [The laws of thermodynamics always apply - even in the center of a star!]
Your math is "dazzling" but where does the ENERGY come from!?
It comes from the difference in mass between what goes into the reaction and what comes out.
Let's look at some of those steps again but now focus on the masses
involved. We can ignore gamma rays because they have no mass.
We can also ignore the neutrinos for much the same reason. The positrons
have some mass but not much so we will ignore them too. That leaves
us only with the atomic nuclei to think about. We measure the mass of atomic nuclei in "amu"s which stands for "atomic mass unit". Don't confuse it with an AU which is an Astronomical Unit.
A proton (1p+) has a mass of 1.0078 amu.
A deuteron (2D+) has a mass of 2.0141 amu.
When you put two protons together you should have a mass of 2.0156
amu. (That's 2 X 1.0078 amu.) However the deuteron formed in the
slow step has a mass of 2.0141. That means 0.0015 amu is missing!
It isn't really missing. Einstein's equation shows us that energy
and mass can be converted from one to the other. That missing
0.0015 amu is converted into the 1.4 Mev of energy!
[I'll spare you the details of working through the equation to
convert mass and energy but you could do it yourself if you learned
a little about the units we use for energy.]
Before leaving this subject let's take a look at the mass change
in the last reaction
3He++ + 3He++ ---fuse----> 4He++ + 1p+ + 1p+
Helium-3 has a mass of 3.016 amu so the two helium-3 fused should
have a mass of 6.032. The products are helium-4, which has a
mass of 4.0026, and two protons, which have a mass of 1.0078 each.
The total mass of all these products (one helium-4 and two protons)
equals 6.0182. That means 0.0138 amu are missing. Notice that
is more missing mass than lost in the slow step. It's about 9.2
times as much. That's why this step produces more energy than
the first step. That 0.0138 amu is converted to 12.9 MeVs, which
is exactly 9.2 times as much energy as released in
the slow step!
Cool huh?!
Is it important to know this?
Well, not really but some folks like to know the details that power a star. Don't worry if you cannot understand these detailed equations. (If you can understand them, then - good for you!) However, I want you to have read through them, at least once, to see that we have a very good understanding of how a star works and to realize that astrophysicists can use these equations in order to make predictions and test theories.
I don't expect you to become another Arthur Eddington.
Who?
Arthur S. Eddington (1882-1944) was, in my opinion, the world's most brilliant astrophysicist. (Yet he never won a Noble Prize! ) He made many contributions to the field of astronomy and astrophysics - perhaps the most important of which is the Eddington limit. Every amateur astronomer should understand the idea behind the Eddington limit because it a fundamental concept in understanding how bright (luminous) a stable star can be. So, let me teach you about the Eddington limit.
Generally speaking, a star's luminosity is determined by the amount of energy it puts out. You can imagine, from what you have read above, that a star with a lot of mass will have a lot of fusion going on in its core and it will produce a lot of energy which translates into a star of high luminosity. As energy wells up from the core it pushes the overlaying materials (upwards) away from the core. Gravity pulls those materials back (downwards). Therefore a star has a kind of balancing act - energy pushes the materials away from the star's core but the gravitational attraction of the mass in the star's core draws the materials back. However, if the energy output is very high the star will, literally, blow itself apart!
The Eddington limit is that upper limit to the luminosity of a star of any given mass. A star at the Eddington limit just barely balances its energy output with its inward gravitational force. Any star that exceeds its Eddington limit will be blown apart. So the Eddington limit provides an upper limit to the luminosity : mass ratio for a stable star.
We measure luminosity in watts and the Eddington limit can be expressed as a ratio using the mass of our Sun as point of comparison. The equation (you don't have to know it) is ...
Let's use our own star, the Sun, as an example. The Sun has a mass of approximately 2X1027(kilo)tons.
[By the way, the best way to determine the mass of any big object in space is to measure how fast something orbits it. I won't go into the details of how that's done but you should understand that we need not, indeed cannot, "weigh" a star to determine its mass. Instead we use the laws of gravity to figure out its mass by measuring how it interacts with something else - usually a satellite.]
We can use the above equation to determine the Sun's Eddington limit. In fact, this is very easy to do because the mass of this "star" is the mass of the Sun so we divide the same number by itself to get the value of 1X1031 watts. That means the Sun cannot produce more than 1031 watts otherwise it would blow up!
How many watts does the Sun produce?
Roughly 4X1026 watts. So the Sun has enough mass to keep itself stable. Indeed, it could increase its luminosity another 10,000 times before we would need to worry about it exploding. (However, if the Sun increased its luminosity by only a factor of 2 we would find ourselves living on an uncomfortably hot planet! )
Of course, working out the Eddington limit for our Sun is fun but not particularly useful. The Eddington limit can be used another way. Take, for example, Rigel. I don't know the mass of Rigel but I know its luminosity is about 60,000 times that of our Sun. Astronomers determined the luminosity of Rigel by comparing its relative magnitude to its distance.
First let's find Rigel's luminosity in watts.
If Rigel is 60,000 times more luminous than our Sun and
the Sun's luminosity is 4X1026 watts then
Rigel's luminosity is 60,000 times 4X1026 watts
which equals 2.4X1031 watts.
(Again, if you are uncomfortable with the math, just go with the flow and trust me. )
Hey, that's more than the Eddington limit!
Well, yes and no. It's more than the Sun's Eddington limit (1031 watts) but we don't know, from this information, the Eddington limit of Rigel. However, we do know that Rigel is a stable star. So, what do you know about the mass of Rigel?
Ah, it must be more massive than the Sun.
Right! As a matter of fact, Rigel must be AT LEAST 2.4 times more massive than our Sun, otherwise its energy output would have blown it apart. The reason Rigel can get away with being so luminous is because it has enough mass to hold itself together against its tremendous fusion power!
Off hand, I don't know the mass of Rigel but, using the Eddington limit, I can give it a lowest estimate of
2.4 (Rigel's minimum mass compared to the Sun) X 2X1027(kilo)tons (the mass of the Sun) = 4.8X1027(kilo)tons.
I don't really expect you to calculate Eddington limits but I want you to understand how astrophysicists go about figuring out certain things. The Eddington limit is one of many equations that astronomers use to understand stars.
So, nothing is brighter than its Eddington limit?
Well, certainly no STABLE STAR exceeds its Eddington limit. Later in this course I will tell you about some very energetic objects like nova and quasars. They exceed the Eddington limit but that's OK because those things are NOT stable stars.
Why do we use the Sun's mass for comparisons?
It's convenient!
There's nothing special about the Sun but it's very special to us Earthlings and astronomers have decided to use certain characteristics of the Sun as a standard. Luminosity, for example, is based upon the Sun's absolute magnitude. (Remember that the Sun's luminosity equals 1 and acts as unit 1 on the H-R diagrams.) Astronomers also use the Sun's mass as a "standard mass".
One solar mass is the mass of our Sun = 2X1027(kilo)tons.
If we use solar masses in our calculations, the math becomes easier. For example, Rigel is at least 2.4 solar masses.
Eddington understood that a stable star is in balance between its mass and energy output. He went on to predict that a star must have a mass over 0.5 solar masses (half the Sun's mass) to burn its nuclear fuel in a stable manner. Recall I said a protostar's mass is less than 6.7% the Sun's mass. In fact, Eddington was slightly off in his estimate but that isn't bad considering that he developed his theories between 1916 and 1926 when no one had a clue about nuclear fusion. (Einstein explained that E = mc2 in 1905 but it wasn't until the 1930s that nuclear fusion in stars was beginning to be understood.)
On the other hand, Eddington also predicted that a star with a mass greater than 50 solar masses would be too energetic!
His predictions, based upon his theories of star structure and power, involve a level of detail that we will not go into. Regardless, his predictions seem to be in line with reality. He may have been a little off on his ideas about small stars but Eddington certainly understood the upper limits to a stable star. No one has found a stable star with a mass of more than 50 solar masses.
In this lesson you have learn (or at least read about) some very important and fundamental concepts in astronomy. I could have gone into further details on the nuclear physics and more complicated math but there really is no need in our course. Instead, I hope you understand how important math and physics are to astronomers. (Anyone thinking about becoming a professional astronomer must be comfortable with those subjects.) Arthur Eddington is a fine example of an astronomer who used his understanding of math and physics to better visualize, in his mind's eye, how a star works.
Now that you understand how a star works you will be well prepared to understand our next lesson about the structure of a star. (Don't worry. There are no equations or math in the next lesson! )
If you are ready to learn about the structure of a stable star, please continue on. Otherwise, take a break and come back after you're refreshed.