Fusion in Stars
230-232

When does a protostar cease being a protostar and become a "real" star? To answer this question recall a number of ideas developed earlier:

1. As a protostar collapses its core temperature and density increase. This is a critical step because it tells you that the frequency of collisions between the atoms in the star (primarily hydrogen) goes up rapidly. As well, the increasing temperature means that the collisions are becoming increasingly violent.

2. There is a direct link between the temperature of a gas and velocity with which the particles in the gas move. The hotter the gas, the faster the particles move.
   
   
3. Because the atoms in the interior of a star are ionized and positively charged - they repel each other. Another name for this is Coulomb repulsion. Normally Coulomb repulsion pushes protons away from each other. However, if two protons are brought to within about 10^-15 m of each other a new force - The Strong Nuclear Force takes over. This force only operates at very close range but when it does it is about one thousand times stronger than the Coulomb force and is attractive. The protons fuse together.
   
   
4. When light elements such as Hydrogen and Helium fuse they release energy

All of these ideas help us understand how a star makes the transition from protostar to star. Use the following applet to explore what happens to two protons on a head-on collision course as you adjust the temperature of the gas (ignore the CNO cycle option for the moment). At what temperature do the Hydrogen nuclei begin to fuse together?

A "star is born" when the interior temperature and density are sufficient to produce atomic collisions vigorous enough to overcome Coulomb repulsion and cause fusion. The simplest form of fusion is called the pp-chain which requires temperatures of about 10 million K. We will explore the details of this fusion cycle ( as well as others) in the next section.

Example 9.4 If a protostar is not getting energy from fusion where does its energy come from?

Solution: Throughout the lifetime of a star there is another major energy source - gravitational potential energy. This is the same energy source that we use to run the turbines in a hydro-electric dam or that beaks our price-less china cup when we drop it. As matter "falls" in a gravitational field it speeds up. This means it gains kinetic energy. A collapsing protostar is able to convert gravitational energy into kinetic energy. Collisions between particles in the cloud becomes more energetic as it collapses and the cloud "heats up".

Paying the Power Bill

Our sun emits an incredible 4 x 10²⁶ Watts of energy. That's 400 trillion trillion W of power being radiated into space. How does our sun "pay" this enormous energy bill? Like all stars - the answer is nuclear fusion.

The reason fusion works to provide energy rests on two very important points:

1. When light nuclei fuse they form a nucleus with a slightly smaller mass than the combined mass of the "parent" nuclei
2. Einstein's theory of relativity tells us that there is a mass-energy equivalence expressed via which allows us to calculate how much energy is released when matter is converted to energy.

Example 9.5 How much energy is released by the fusion of 4 Hydrogen nuclei into Helium? The mass of a Hydrogen nucleus is 1.67 X 10^-27 kg and the mass of a Helium nucleus is 6.65 X 10^-27 kg .

Solution Four H nuclei have a combined mass of 6.69 X 10^-27 kg. A He nucleus has a mass of 6.65 X 10^-27 kg, so there is a mass deficit of 0.04 X 10^-27 kg - this tiny bit of mass has been turned into energy. Using you can conlude that the amount of energy released is:

While on the Main Sequence stars are busily turning Hydrogen into Helium. This occurs in the core of the star and proceeds until about 10% of the star's Hydrogen is converted to Helium. This defines the main sequence lifetime phase for a star.

The Proton-Proton Cycle The simplest way to convert Hydrogen into Helium is via one of three proton-proton reactions. The animation on the right shows one of these pp-cycles. Stars similar to and less massive than our sun utilize these fusion reactions. PP-chain fusion begins at tempertures of 10 million K.
	Figure 9.20 PP-chain reaction used by the sun and lower mass stars to derive energy from the fusion of Hydrogen inot Helium.
The CNO Cycle At remperatures above 15 million K a much more efficient fusion reaction occurs. This is the Carbon-Nitrogen-Oxygen cycle (CNO for short). Our sun is just not hot enough to use the CNO cycle effectively. But stars only 20% more massive have high enough core temperatures to use this cylce. For example, at a core temperature of 20 Million K the CNO cycle produces about 100 times as much energy as the PP-cycles.
	Figure 9.21 The CNO cycle is the dominant fusion reaction for stars more massive than the sun.

Example 9.6 Compare temperatures for the CNO cycle and PP cycles to operate. Why does the CNO cycle require higher temperatures than the PP-cycle does?

Solution Use the applet provided above to investigate when fusion begins to occur in the CNO cycle. You should see that it begins at tempertures greater than 16 million K while PP cycle fusion begins at around 10 million K. The CNO cycle requires a significantly higher temperature because the nuclei involved (Carbon, Nitrogen and Oxygen) have charges of +6, +7 and +8 electron charges (compared to +1 electron charge for the proton). So, the Coulomb repulsion is correspondingly greater for these nuclei. In order to over come this much higher velocities are required - hence higher temperatures.

The Stellar "Thermostat"

Throughout its "life" a star faces a continual struggle between two competing forces: 1) The inward crush of gravity produced by the mass of the star itself and 2) The outward pressure of the hot gases in the star. When the star is in equilibrium these two sets of forces balance each other. When a star is in this state it is said to be in hydrostatic equilibrium. But what happens if this balance is upset? Use the following applet to investigate this. In this applet a star is modeled by 5 concentric regions. Pull up or down on the red "radius widget" to change the size of the 2nd inner-most shell.

Applet: Hydrostatic Equilibrium which illustrates how star responds if its equillibrium is upset.

If you compress a layer in a star, that layer will heat which will increase the outward pressure in that layer. This causes the star to push back and lift the layer of material above it. Similarily, if you pull a layer outward, it will expand and the gas in that layer will cool. The gas no longer develops enough pressure to support the weight of material above it and the layer will begin to collapse. In either case, the star will after a few "wiggles" settle back down (or "damp out") into its previous equilibrium state. (In a later section we will investigate what drives certain kinds of variables stars and see there that the oscillation of the star does no damp out.)

This "self-correcting" ability of a star is analgous to how the thermostat in your house controls temperature. It also helps explain a very important relationship between the mass of a star and its overall structure. The mass of the star will determine the maximum compression and temperature that will occur in the core of the star. This in turn controls both the kind of fusion reaction that occurs and how much energy is being produced. The amount of energy released by nuclear fusion depends very sensitively on temperature. A more massive star will have a higher core temperature and a much higher energy generation rate. We will see how this affects the evolution of a star in the next section.

Practice

Rank order the stars numbered in the HR diagram shown on the right in order of: increasing mass increasing core temperature Which of the stars shown in the HR diagram use the CNO reaction to generate most of their energy? Explain why it is very unlikely that red dwarf stars would generate energy via the CNO process? True or False (and explain your answer!): Very massive stars can't use the PP-cycle fusion reaction.

To understand how stars produce energy

Chp 11.2

At very high temperatures atoms can lose their outer electrons. In the hot interior of a star a hydrogen atom will lose its electron and become a bare proton.

There is a lower limit for star formation. A protostar must have enough mass to compress its interior to 10 million K. If the mass of the protostare is less than 0.08Mo or about 8 times the mass of Jupiter it will not be able to become a star. It could, however, become a brown dwarf.