The Quantum Universe

by Brian Cox

  We’re now making progress because to balance the forces on the cube (which we remind you means that the cube doesn’t move and that means the star is not going to explode or collapse2) we require that

  where Pbottom and Ptop are the pressures of the electron gas at the upper and lower faces of the cube and A is the area of each side of the cube (remember, the force exerted by a pressure is equal to the pressure multiplied by the area). We have labelled this equation ‘(1)’ because it is very important and we will want to refer back to it.

  Step 3: Make a cup of tea and feel pleased with ourselves because, after carrying out step 1, we will have figured out the pressures, Pbottom and Ptop, and step 2 has made precise how to balance the forces. The real work is yet to come, though, because we still have to actually carry out step 1 and determine the pressure difference appearing on the left-hand side of equation (1). That is our next task.

  Imagine a star packed with electrons and other stuff. How are the electrons scattered about? Let’s focus our attention on a ‘typical’ electron. We know that electrons obey the Pauli Exclusion Principle, which means that no two electrons are likely to be found in the same region of space. What does that mean for the sea of electrons that we’ve been referring to as the ‘electron gas’ in our star? Because the electrons are necessarily separated from each other, we can suppose that each electron sits all alone inside a tiny imaginary cube within the star. Actually, that’s not quite right because we know that electrons come in two types – ‘spin up’ and ‘spin down’ – and the Pauli principle only forbids identical particles from getting too close, which means we can fit two electrons inside a cube. This should be contrasted with the situation that would arise if the electrons did not obey the Pauli principle. In that case the electrons would not be localized two-at-a-time inside ‘virtual containers’. Rather they could spread out and enjoy a much greater living space. In fact, if we were to ignore the various ways that the electrons can interact with each other and with the other particles in the star, there would be no limit to their living room.

  We know what happens when we confine a quantum particle: it hops about according to Heisenberg’s Uncertainty Principle, and the more it is confined the more it hops. That means that, as our would-be white dwarf collapses, so the electrons get increasingly confined and that makes them increasingly agitated. It is the pressure exerted by their agitation that will halt the gravitational collapse.

  We can do better than words, because we can use Heisenberg’s Uncertainty Principle to determine the typical momentum of an electron. In particular, if we confine the electron to a region of size Δx then it will hop around with a typical momentum p ∼ h/Δx. Actually, in Chapter 4 we argued that this is more like an upper limit on the momentum and that the typical momentum is somewhere between zero and this value; that piece of information is worth remembering for later. Knowing the momentum allows us, immediately, to learn two things. Firstly, if the electrons didn’t obey Pauli then they would not be confined to a region of size Δx but rather to some much larger size. That in turn would result in much less jiggling, and less jiggling means less pressure. So it is clear how the Pauli principle is entering the game; it is putting the squeeze on the electrons so that, via Heisenberg, they get a supercharged jiggle. In a moment we’ll convert this idea of a supercharged jiggle into a formula for the pressure, but first we should mention the second thing we can learn. Because the momentum p = mv, the speed of the jiggle also depends inversely on the mass, so the electrons are jumping around much more vigorously than the heavier nuclei that also make up the star, and that is why the pressure exerted by the nuclei is unimportant. So how do we go from knowing the momentum of an electron to computing the pressure a gas of similar electrons exerts?

  What we need to do first is to work out how big the little chunks containing the pairs of electrons must be. Our little chunks have volume (Δx)3, and because we have to fit all the electrons inside the star, we can express this in terms of the number of electrons within the star (N) divided by the volume of the star (V). We’ll 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 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 e
ffect 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 were to go ahead and calculate it, we would find that it is dwarfed by the much larger quantum pressure.

  Finally, we are ready to feed our equation for the quantum pressure into the key equation (1), which is worth repeating here:

  But this is not as easy as it sounds because we need to know the difference in the pressures at the upper and lower faces of the cube. We could re-write equation (1) entirely in terms of the density within the star, which is itself something that varies from place to place inside the star (it must be otherwise there would be no pressure difference across the cube) and then we could try to solve the equation to determine how the density varies with distance from the star’s centre. To do this is to solve a differential equation and we want to avoid that level of mathematics. Instead, we are going to be more resourceful and think harder (and calculate less) in order to exploit equation (1) to deduce a relationship between the mass and the radius of a white dwarf star.

  Obviously the size of our little cube and its location within the star are completely arbitrary, and none of the conclusions we are going to draw about the star as a whole can depend upon the details of the cube. Let’s start by doing something that might seem pointless. We are quite entitled to express the location and size of the cube in terms of the size of the star. If R is the radius of the star, then we can write the distance of the cube from the centre of the star as r = aR, where a is simply a dimensionless number between 0 and 1. By dimensionless, we mean that it is a pure number and carries no units. If a = 1, the cube is at the surface of the star and if a = ½ it is halfway out from the centre. Similarly, we can write the size of the cube in terms of the radius of the star. If L is the length of a side of the cube, then we can write L = bR where, again, b is a pure number, which will be very small if we want the cube to be small relative to the star. There is absolutely nothing deep about this and, at this stage, it should seem so obvious as to appear pointless. The only noteworthy point is that R is the natural distance to use because there are no other distances relevant to a white dwarf star that could have provided any sensible alternatives.

  Likewise, we can continue our strange obsession and express the density of the star at the position of the cube in terms of the average density of the star, i.e. we can write where f is, once again, a pure number and ρ is the average density of the star. As we have already pointed out, the density of the cube depends on its position inside the star – if it is closer to the centre, it will be more dense. Given that the average density does not depend on the position of the cube, then f must do so, i.e. f depends on the distance r, which obviously means it depends on the product aR. Now, here is the key piece of information that underpins the rest of our calculation: f is a pure number and R is not a pure number (because it is measuring a distance). This fact implies that f can only depend upon a and not on R at all. This is a very important result, because it is telling us that the density profile of a white dwarf star is ‘scale invariant’. This means the density varies with radius in the same way no matter what the radius of the star is. For example, the density at a point ¾ of the way out from the centre of the star will be the same fraction of the mean density in every white dwarf star, regardless of the star’s size. There are two ways of appreciating this crucial result and we thought we’d present them both. One of us explained it thus: ‘That’s because any dimensionless function of r (which is what f is) can only be dimensionless if it is a function of a dimensionless variable, and the only dimensionless variable we have is r/R = a, because R is the only quantity which carries the dimensions of distance that we have at our disposal.’

  The other author feels that the following is clearer: ‘f can in general depend in a complicated way on r, the distance of the little cube from the centre of the star. But let’s assume for the sake of this paragraph that it is directly proportional to it, i.e. f ∞ r2. In other words, f = Br, where B is a constant. Here, the key point is that we want f to be a pure number, whilst r is measured in (say) metres. That means that B must be measured in 1/metres, so that the units of distance cancel each other out. So what should we choose for B? We can’t just choose something arbitrary, like ‘1 inverse metres’, because this would be meaningless and has nothing to do with the star. Why not choose 1 inverse light years for example, and get a very different answer? The only distance we have to hand is R, the physical radius of the star, and so we are forced to use thi
s to ensure that f will always be a pure number. This means that f depends only on r/R. You should be able to see that the same conclusion can be drawn if we started out by assuming that f ∞ r2 say.’ Which is just what he said, only longer.

  This means that we can express the mass of our little cube, of size L and volume L3, sitting at a distance r from the centre of the star, as . We wrote f(a) instead of just f in order to remind us that f really only depends upon our choice of a = r/R and not on the large-scale properties of the star. The same argument can be used to say that we can write Min = g(a)M where g(a) is again only a function of a. For example, the function g(a) evaluated at a = 1/2 tells us the fraction of the star’s mass lying in a sphere of half the radius of the star itself, and that is the same for all white dwarf stars, regardless of their radius because of the argument in the previous paragraph.6 You might have noticed that we are steadily working our way through the various symbols which appear in equation (1), replacing them by dimensionless quantities (a, b, f and g) multiplied by quantities that depend only on the mass and radius of the star (the average density of the star is determined in terms of M and R because , the volume of a sphere). To complete the task, we just need to do the same for the pressure difference, which we can (by virtue of equation (4)) write as where h(a, b) is a dimensionless quantity. The fact that h(a, b) depends upon both of a and b is because the pressure difference not only depends on where the cube is (represented by a) but also on how big it is (represented by b): bigger cubes will have a larger pressure difference. The key point is that, just like f(a) and g(a), h(a, b) cannot depend upon the radius of the star.

  We can make use of the expressions we just derived to rewrite equation (1):