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need precisely N/2 containers to accommodate all of the electrons because we are allowed two electrons inside each container. This means that each container will occupy a volume of V divided by N/2, which is equal to 2(V/N). We’ll need the quantity N/V (the number of electrons per unit volume inside the star) quite a lot in what follows, so we’ll give it its own symbol n. We can now write down what the volume of the containers must be in order to contain all the electrons in the star, i.e. (Δx)3 = 2/n. Taking the cube root of the right hand side allows us to conclude that

We can now plug this into our expression from the Uncertainty Principle to get the typical momentum of the electrons due to their quantum jiggling:

where the ∼ sign means ‘something like’. Clearly this is a bit vague because the electrons will not all be jiggling in exactly the same way: some will move faster than the typical value and some will move more slowly. The Heisenberg Uncertainty Principle isn’t capable of telling us exactly how many electrons move at this speed and how many at that. Rather, it provides a more ‘broad brush’ statement and says if you squeeze an electron down then it will jiggle with a momentum something like h/Δx. We are going to take that typical momentum and assume it’s the same for all the electrons. In the process, we will lose a little precision in our calculation but gain a great deal of simplicity as a result, and we are certainly thinking about the physics in the right way.3

We now know the speed of the electrons and that is enough information to work out how much pressure they exert on the tiny cube. To see that, imagine a fleet of electrons all heading in the same direction at the same speed (v) towards a flat mirror. They hit the mirror and bounce back, again travelling at the same speed but in the opposite direction. Let us compute the force exerted by the electrons on the mirror. After that we can attempt the more realistic calculation, where the electrons are not all travelling in the same direction. This methodology is very common in physics – first think about a simpler version of the problem you want to solve. That way you get to learn about the physics without biting off more than you can chew and gain confidence before tackling the harder problem. Imagine that the electron fleet consists of n particles per cubic metre and that, for the sake of argument, it has a circular cross-section of area 1 square metre – as illustrated in Figure 12.4. In one second nv electrons will hit the mirror (if v is measured in metres per second). We know that because all of the electrons stretching from the mirror to a distance v × 1 second away will smash into the mirror every second, i.e. all of the electrons in the tube drawn in the figure. Since a cylinder has a volume equal to its cross-sectional area multiplied by its length, the tube has a volume of v cubic metres and because there are n electrons per cubic metre in the fleet it follows that nv electrons hit the mirror every second.

Figure 12.4. A fleet of electrons (the little dots) all heading in the same direction. All of the electrons in a tube this size will smash into the mirror every second.

When each electron bounces off the mirror it gets its momentum reversed, which means that each electron changes its momentum by an amount equal to 2mv. Now, just as it takes a force to halt a moving bus and send it travelling back in reverse, so it takes a force to reverse the momentum of an electron. This is Isaac Newton once again. In Chapter 1 we wrote his second law as F = ma, but this is a special case of a more general statement, which states that the force is equal to the rate at which momentum changes.4 So the whole fleet of electrons will impart a net force on the mirror F = 2mv × (nv), because this is the net change in momentum of the electrons every second. Due to the fact that the electron beam has an area of 1 square metre, this is also equal to the pressure exerted by the electron fleet on the mirror.

It is only a short step to go from a fleet of electrons