by Brian Cox
The gas in these first proto-stars became hotter and hotter as they collapsed in on themselves, as anyone who has used a bicycle pump will know, because compressing a gas makes it heat up. When the gas reaches temperatures of around 100,000 degrees, the electrons can no longer be held in orbit around the hydrogen and helium nuclei and the atoms get ripped apart, leaving a hot plasma of bare nuclei and electrons. The hot gas tries to expand outwards and resist further collapse but, for sufficiently massive clumps, gravity wins out. Because protons have positive electric charge they will repel each other but, as the gravitational collapse proceeds and the temperature continues to rise, the protons move faster and faster. Eventually, at a temperature of several million degrees, the protons are moving so fast that they get close enough for the weak nuclear force to take over. When that happens, two protons can react with one another; one of them spontaneously changes into a neutron with the simultaneous emission of a positron and a neutrino (exactly as illustrated in Figure 11.3 on page 202). Freed from the electrical repulsion, the proton and the neutron fuse under the action of the strong nuclear force to make a deuteron. This process releases huge amounts of energy because, just as in the formation of a hydrogen molecule, binding things together releases energy.
The energy release in a single fusion event isn’t large by everyday standards. One million proton–proton fusion reactions generate roughly the same amount of energy as the kinetic energy of a mosquito in flight or a 100 watt light-bulb radiates in a nanosecond. But that is huge on atomic scales and, remember, we are talking about the dense heart of a collapsing gas cloud in which there are around 1026 protons per cubic centimetre. If all the protons in a cubic centimetre were to fuse into deuterons, 1013 joules of energy would be liberated, which is enough to power a small town for one year.
The fusion of two protons into a deuteron is the start of a fusion jamboree. The deuteron itself is eager to fuse with a third proton to make a light version of helium (called helium-3) with the emission of a photon, and those helium nuclei then pair up and fuse into regular helium (called helium-4) with the emission of two protons. At each stage, the fusing together liberates more and more energy. And, just for good measure, the positron, which was emitted right back at the start of the chain, also rapidly fuses with an electron in the surrounding plasma to produce a pair of photons. All of this liberated energy makes for a hot gas of photons, electrons and nuclei that pushes against the in-falling matter and halts any further gravitational collapse. This is a star: nuclear fusion burns up nuclear fuel in the core, and that generates an outward pressure that stabilizes the star against gravitational collapse.
There is, of course, only a finite amount of hydrogen fuel available to burn and, eventually, it will run out. With no more energy released there is no more outward pressure; gravity once again takes control and the star resumes its postponed collapse. If the star is massive enough, the core will heat up to temperatures of around 100 million degrees. At that stage, the helium produced as waste in the hydrogen-burning phase ignites, fusing together to produce carbon and oxygen, and once again the gravitational collapse is temporarily halted.
But what happens if the star is not massive enough to initiate helium fusion? For stars less than about half the mass of our Sun, this is the case, and for them something very dramatic happens. The star heats up as it contracts, but, before the core reaches 100 million degrees, something else halts the collapse. That something is the pressure exerted by electrons due to the fact that they are in the grip of the Pauli Exclusion Principle. As we have learnt, the Pauli principle is crucial to understanding how atoms remain stable, and it underpins the properties of matter. Here is another string to its bow: it explains the existence of compact stars that survive despite the fact that they no longer burn up any nuclear fuel. How does this work?
As the star gets squashed, so the electrons within it get confined to a smaller volume. We can think of an electron in the star in terms of its momentum p and hence its associated de Broglie wavelength, h/p. In particular, the particle can only ever be described by a wave packet that is at least as big as its associated wavelength.1 This means that, when the star is dense enough, the electrons must be overlapping each other, i.e. we cannot imagine them as being described by isolated wave packets. This in turn means that quantum mechanical effects, and the Pauli principle in particular, are important in describing the electrons. Specifically, they are being squashed together to the point where two electrons are attempting to occupy the same region of space, and we know from the Pauli principle that they resist this. In a dying star, therefore, the electrons avoid each other and this provides a rigidity that resists any further gravitational collapse.
This is the fate of the lightest stars, but what of stars like our Sun? We left them a couple of paragraphs ago, burning helium into carbon and oxygen. What happens when they run out of helium? They too must then start to collapse under their own gravity, which means they will have their electrons squashed together. And, just as for the lighter stars, the Pauli principle can eventually kick in and halt the collapse. But, for the most massive of stars, even the Pauli Exclusion Principle has its limits. As the star collapses and the electrons get squashed closer together, so the core heats up and the electrons move faster. For heavy enough stars, the electrons will eventually be moving so fast that they approach the speed of light, and that is when something new happens. When they close in on light-speed, the pressure the electrons are able to exert to resist gravity is reduced to such an extent that they aren’t up to the job. They simply cannot beat gravity any more and halt the collapse. Our task in this chapter is to calculate when this happens, and we’ve already given away the punchline. For stars with masses greater than 1.4 times the mass of the Sun, the electrons lose and gravity wins.
That completes the overview that will provide the basis for our calculation. We can now go ahead and forget all about nuclear fusion, because stars that are burning are not where our interest lies. Rather, we are keen to understand what happens inside dead stars. We want to see just how the quantum pressure from the squashed electrons balances the force of gravity, and how that pressure becomes diminished if the electrons are moving too fast. The heart of our study is therefore a balancing game: gravity versus quantum pressure. If we can make them balance we have a white dwarf star, but if gravity wins we have catastrophe.
Although not relevant for our calculation, we can’t leave things on such a cliff-hanger. As a massive star implodes, two further options remain open to it. If it is not too heavy then it will keep squashing the protons and electrons until they too can fuse together to make neutrons. In particular, one proton and one electron convert spontaneously into a neutron with the emission of a neutrino, again via the weak nuclear force. In this way the star relentlessly converts into a tiny ball of neutrons. In the words of Russian physicist Lev Landau, the star converts into ‘one gigantic nucleus’. Landau wrote those words in his 1932 work ‘On the Theory of Stars’, which appeared in print in the very same month that the neutron was discovered by James Chadwick. It is probably going too far to say that Landau predicted the existence of neutron stars but, with great prescience, he certainly anticipated something like them. Perhaps the credit should go to Walter Baade and Fritz Zwicky, who wrote in the following year: ‘With all reserve we advance the view that supernovae represent the transitions from ordinary stars into neutron stars, which in their final stages consist of extremely closely packed neutrons.’ The idea was considered so outlandish that it was parodied in the Los Angeles Times (see Figure 12.1), and neutron stars remained a theoretical curiosity until the mid 1960s.
Figure 12.1. A cartoon from the 19 January 1934 edition of the Los Angeles Times.
In 1965, Anthony Hewish and Samuel Okoye found ‘evidence for an unusual source of high radio brightness temperature in the crab nebula’, although they failed to identify it as a neutron star. The positive ID came in 1967 by Iosif Shklovsky and, shortly afterwards, after more deta
iled measurements, by Jocelyn Bell and Hewish himself. This first example of one of the most exotic objects in the Universe was subsequently named the ‘Hewish Okoye Pulsar’. Interestingly, the very same supernova that created the Hewish Okoye Pulsar was also observed by astronomers, a thousand years earlier. The great supernova of 1054, the brightest in recorded history, was observed by Chinese astronomers and, as shown by a famous drawing on an overhanging cliff edge, by the peoples of Chaco Canyon in the south-western United States.
We haven’t yet said how those neutrons manage to fend off gravity and prevent further collapse, but you can probably guess how it works. The neutrons (just like electrons) are slaves to the Pauli principle. They too can halt further collapse and so, just like white dwarves, neutron stars represent a possible end-point in the life of stars. Neutron stars are a detour as far as our story goes, but we can’t leave them without remarking that these are very special objects in our wonderful Universe: they are stars the size of cities, so dense that a teaspoonful weights as much as a mountain, held up by nothing more than the natural aversion to one another of spin-half particles.
There is only one option remaining for the most massive stars in the Universe – stars in which even the neutrons are moving close to light-speed. For such giants, disaster awaits, because the neutrons are no longer able to generate sufficient pressure to resist gravity. There is no known physical mechanism to stop a stellar core with a mass of greater than around three times the mass of our Sun falling in on itself, and the result is a black hole: a place where the laws of physics as we know them break down. Presumably Nature’s laws don’t cease to operate, but a proper understanding of the inner workings of a black hole requires a quantum theory of gravity, and no such theory exists today.
It is time to get back on message and to focus on our twin goals of proving the existence of white dwarf stars and calculating the Chandrasekhar mass. We know how to proceed: we must balance the electron pressure with gravity. This is not going to be a calculation we can do in our heads, so it will pay to make a plan of action. Here’s the plan; it’s quite lengthy because we want to clear up some background detail first and prepare the ground for the actual calculation.
Step 1: We need to determine what the pressure inside the star is due to those highly compressed electrons. You might be wondering why we are not worrying about the other stuff inside the star – what about the nuclei and the photons? Photons are not subject to the Pauli principle and, given enough time, they’ll leave the star in any case. They have no hope of fighting gravity. As for the nuclei, the half-integer spin nuclei are subject to Pauli’s rule but (as we shall see) their larger mass means they exert a smaller pressure than do the electrons and we can safely ignore their contribution to the balancing game. That simplifies matters hugely – the electron pressure is all we need, and that is where we should set our sights.
Step 2: After we’ve figured out the electron pressure, we’ll need to do the balancing game. It might not be obvious how we should go about things. It’s one thing to say ‘gravity pulls in and the electrons push out’ but it is quite another thing to put a number on it.
The pressure is going to vary inside the star; it will be larger in the centre and smaller at the surface. The fact that there is a pressure gradient is crucial. Imagine a cube of star matter sitting somewhere inside the star, as illustrated in Figure 12.2. Gravity will act to draw the cube towards the centre of the star and we want to know how the pressure from the electrons goes about countering it. The pressure in the electron gas exerts a force on each of the six faces of the cube, and the force is equal to the pressure at that face multiplied by the area of the face. That statement is precise; until now we have been using the word ‘pressure’ assuming that we all have sufficient intuitive understanding that a gas at high pressure ‘pushes more’ than a gas at low pressure. Anyone who has had to pump air into a flat car tyre knows that.
Figure 12.2. A small cube somewhere within the heart of a star. The arrows indicate the pressure exerted on the cube by the electrons within the star.
Since we are going to need to understand pressure properly, a brief diversion into more familiar territory is in order. Sticking with the tyre example, a physicist would say that a tyre is flat because the air pressure inside is insufficient to support the weight of the car without deforming the tyre: that’s why we get to go to all the best parties. We can go ahead and calculate what the correct tyre pressure should be for a car with a mass of 1,500 kg if we want 5 centimetres of tyre to be in contact with the ground, as illustrated in Figure 12.3: it’s chalk dust time again.
If the tyre is 20 cm wide and we want a 5 cm length of the tyre to be touching the road, then the area of tyre in contact with the ground will be 20 × 5 = 100 square centimetres. We don’t know the requisite tyre pressure yet – this is what we want to calculate – so let’s represent it by the symbol P. We need to know the downward force on the ground exerted by the air within the tyre. This is equal to the pressure multiplied by the area of tyre in contact with the floor, i.e. P × 100 square centimetres. We should multiply that by four, because our car has four tyres: P × 400 square centimetres. That is the total force exerted on the ground by the air within the tyres. Think of it like this: the air molecules inside the tyre are pounding the ground (they are, to be pedantic, pounding the rubber in the tyre in contact with the ground, but that isn’t important). The ground doesn’t usually give way, in which case it pushes back with an equal but opposite force (so we did use Newton’s third law after all). The car is being pushed up by the ground and pulled down by gravity and since it doesn’t sink into the ground or leap into the air, we know that these two forces must balance each other. We can therefore equate the P × 400 square centimetres of force pushing up with the downward force of gravity. That force is just the weight of the car and we know how to work that out using Newton’s second law, F = ma, where a is the acceleration due to gravity at the Earth’s surface, which is 9.81 m/s2. So the weight is 1,500 kg × 9.8 m/s2 = 14,700 Newtons (1 Newton is equal to 1 kg m/s2 and it is roughly the weight of an apple). Equating the two forces implies that
Figure 12.3. A tyre deforming slightly as it supports the weight of a car.
This is an easy equation to solve: P = (14,700/400) N/cm2 = 36.75 N/cm2. A pressure of 36.75 Newtons per square centimetre is probably not a very familiar way of stating a tyre pressure, but we can convert it into the more familiar ‘bar’. 1 bar is standard air pressure, and is equal to 101,000 Newtons per square metre. There are 10,000 square centimetres in a square metre, so 101,000 Newtons per square metre is equivalent to 10.1 Newtons per square centimetre. Our desired tyre pressure is therefore 36.75/10.1 = 3.6 bar (or 52 psi – you can work that one out for yourself). We can also use our equation to deduce that, if the tyre pressure decreases by 50% to 1.8 bar, then we’ll double the area of tyre in contact with the ground, which makes for a flatter tyre. After that refresher course on pressure we are ready to return to the little cube of star matter illustrated in Figure 12.2.
If the bottom face of the cube is closer to the centre of the star then the pressure on it should be a little bit bigger than the pressure pressing on the top face. That pressure difference gives rise to a force on the cube that wants to push the cube away from the centre of the star (‘up’ in the figure) and that is just what we want, because the cube will, at the same time, be pulled towards the centre of the star by gravity (‘down’ in the figure). If we could work out how to balance those two forces then we’d have developed some understanding of the star. But that is easier said than done because, although step 1 will allow us to work out how much the cube is pushed out by the electron pressure, we still have to figure out by how much gravity pulls in the opposite direction. By the way, we do not need to worry about the pressure pushing against the sides of our cube because the sides are equidistant from the centre of the star, so the pressure on the left side will balance the pressure on the right side and that ensures the cube does
not move to the left or right.
To work out the force of gravity on the cube we need to make use of Newton’s law of gravity, which tells us that every single piece of matter within the star pulls on our little cube by an amount that decreases in strength the farther the piece is from our cube. So more distant pieces pull less than closer ones. To deal with the fact that the gravitational pull on our cube is different for different pieces of star matter, depending on their distance away, looks like a tricky problem but we can see how to do it, in principle at least – we should chop the star into lots of pieces and then work out the force on the cube for each and every such piece. Fortunately, we do not need to imagine chopping the star up because we can exploit a very beautiful result. Gauss’ law (named after the legendary German mathematician Carl Friedrich Gauss) informs us that: (a) we can totally ignore the gravity from all the pieces sitting further out from the centre of the star than our little cube; (b) the net gravitational effect of all of the pieces that sit closer to the centre is exactly as if all of those pieces were squashed together at the exact centre of the star. Using Gauss’ law in conjunction with Newton’s law of gravity we can say that the cube experiences a force that pulls it towards the centre of the star and that force is equal to
where Min is the mass of the star lying within a sphere whose radius reaches only as far out as the cube, Mcube is the mass of the cube and r is the distance of the cube from the star’s centre (and G is Newton’s constant). For example, if the cube sits on the surface of the star then Min is the total mass of the star. For all other locations, Min is smaller than that.