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to a gas of electrons. Rather than all the electrons ploughing along in the same direction, we have to take into account that some travel up, some down, some to the left and so on. The net effect is to reduce the pressure in any one direction by a factor of 6 (think of the six faces on a cube) to (2mv) × (nv)/6 = nmv2/3. We can replace v in this equation by our Heisenberg-informed estimate of the typical speeds at which the electrons are zipping about (i.e. the previous equation (2)) to get the final result for the pressure exerted by the electrons in a white dwarf star:5

If you recall, we said that this was only an estimate. The full result, using a lot more mathematics, is

This is a nice result. It tells us that the pressure at some place in the star varies in proportion to the number of electrons per unit volume at that place raised to the power of . You should not be concerned that we did not get the constant of proportionality correct in our approximate treatment – the fact that we got everything else right is what matters. In fact, we did already say that our estimate of the momentum of the electrons is probably a little too big and this explains why our estimate of the pressure is bigger than the true value.

Knowing the pressure in terms of the density of electrons is a good start but it will suit our purposes better to express it in terms of the actual mass density in the star. We can do this under the very safe assumption that the vast majority of the star’s mass comes from the nuclei and not the electrons (a single proton has a mass nearly 2,000 times greater than that of an electron). We also know that the number of electrons must be equal to the number of protons in the star because the star is electrically neutral. To get the mass density we need to know how many protons and neutrons there are per cubic metre within the star and we should not forget the neutrons because they are a by-product of the fusion process. For lighter white dwarfs, the core will be predominantly helium-4, the end product of hydrogen fusion, and this means that there will be equal numbers of protons and neutrons. Now for a little notation. The atomic mass number, A, is conventionally used to count the number of protons + neutrons inside a nucleus and A = 4 for helium-4. The number of protons in a nucleus is given the symbol Z and for helium Z = 2. We can now write down a relationship between the electron density, n, and the mass density, ρ:

and we’ve assumed that the mass of the proton, mp, is the same as the mass of the neutron, which is plenty good enough for our purposes. The quantity mpA is the mass of each nucleus; ρ/mpA is then the number of nuclei per unit volume, and Z times this is the number of protons per unit volume, which must be the same as the number of electrons – and that’s what the equation says.

We can use this equation to replace n in equation (3), and because n is proportional to ρ the upshot is that the pressure varies in proportion to the density to the power of . The salient physics we have just discovered is that

and we should not be worrying too much about the pure numbers that set the overall scale of the pressure, which is why we just bundled them all up in the symbol κ. It’s worth noting that κ depends on the ratio of Z and A, and so will be different for different kinds of white dwarf star. Bundling some numbers together into one symbol helps us to ‘see’ what is important. In this case the symbols could distract us from the important point, which is the relationship between the pressure and the density in the star.

Before we move on, notice that the pressure from quantum jiggling doesn’t depend upon the temperature of the star. It only cares about how much we squeeze the star. There will also be an additional contribution to the electron pressure that comes about simply because the electrons are whizzing around ‘normally’ due to their temperature, and the hotter the star, the more they whizz around. We have not bothered to talk about this source of pressure because time is short and, if we