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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