The Chandrasekhar limit: a simplified approach

This article outlines a simplified approach to approximating the Chandrasekhar limit for white dwarf stars at a level appropriate for advanced high school students, beginning undergraduate students, and high school teachers. Using a combination of introductory quantum mechanics and Einstein’s theory of special relativity, the electron degeneracy pressure is calculated in the non-relativistic and ultra-relativistic limits. By combining the electron degeneracy energy with the gravitational energy for a constant density star, an approximation to the Chandrasekhar mass is derived.

Due to a typesetting error the limits of the integral in equation (15)

Introduction
In his pioneering work of 1931, the Indian physicist Subrahmanyan Chandrasekhar used a combination of quantum mechanics and relativity to show that it is not possible for a white dwarf star to exist with a mass exceeding 1.4 solar masses [1]. The detailed calculations underpinning Chandrasekhar's original work are complex and usually first encountered in advanced undergraduate or postgraduate degree courses. There have been several excellent articles and books written at an introductory level that aim to make the work of Chandrasekhar accessible to a wider audience [2][3][4]. The goal of the current article is to outline a simplified approach to approximating the Chandrasekhar mass that only requires knowledge of high school calculus and introductorylevel quantum mechanics. This article is aimed at students and teachers who wish to understand the logic behind Chandrasekhar's famous result.

Electron degeneracy energy
Our starting point is to understand what happens when an electron is confined to a region of space [5][6][7][8][9]. We will begin by considering a simplified model in which an electron is confined to a onedimensional box. According to classical mechanics, a freely moving electron of mass m e and momentum p has kinetic energy equal to E = p 2 /2m. From a quantum mechanical perspective, an electron is described by a wave function that provides information about the probability of finding the electron within a particular region of space. Assuming that the electron is confined to the onedimensional box, it follows that the probability of finding the electron outside the box must be equal to zero, and therefore the wave function must have zero amplitude at the edges of the box. This constrains the allowed wavelengths of the electron wavefunction as depicted in figure 1.
If L is the size of the box, then the allowed wavelengths are given by λ = 2L/n where n is a Figure 1. For an electron confined to a onedimensional box, the electron wavelength can only take certain discrete values. These values determine the allowed energy levels of the system. positive integer representing the number of halfwavelengths that fit inside the box. According to quantum mechanics the momentum of an electron is related to the wavelength by the de Broglie relation: where we have introduced the reduced Planck constant ℏ = h/2π and the wave number k = 2π /λ. If we combine the de Broglie relation (1) with the wavelength constraint, λ = 2L/n, and the classical energy, E = p 2 /2m e , we find that: This expression is telling us that when you confine an electron to a region of space, the energy of the electron is quantised and can only take certain discrete values characterised by the quantum number n. We see that if the size of the box, L, is made smaller, then the energy of all the possible states increases. We can easily generalise this result to three dimensions by confining the electron to a three dimensional box of volume V = L 3 . In this case the electron wavelength is constrained in each of the three coordinate directions and the energy becomes: Since n x , n y and n z can only take positive integer values, the 'allowed states' are represented by the positive octant of a sphere in n-space. It follows that the volume of available states in nspace will be equal to one-eighth the volume of a sphere with radius n: To find the number of energy states corresponding to a given value of n, we need to divide the volume of the positive octant of a sphere with radius n by the volume of a single state in n-space, which is simply 1, and we find 1 : What if we have multiple electrons inside our box? Naively you might expect all of the electrons to occupy the lowest energy ground state as this will minimise the overall energy of the system. However, according to Pauli's Exclusion principle, only two electrons can occupy each quantum state, one with spin up, and one with spin down. This means that if there are N states, then the maximum number of electrons that can occupy these states is given by N e = 2N. If we wish to find the total energy of our system, we will need to allocate two electrons to each energy state and then sum over all of the different energy states.
Next, consider one octant of a spherical shell in n-space of infinitesimal thickness dn. This shell contains a volume equal to one-eighth the surface area of the sphere of radius n, multiplied by the thickness of the shell, dn:

The Chandrasekhar limit: a simplified approach
Following our previous discussion, the number of electrons in the shell will be equal to: The energy of the electrons represented by states that lie within the shell segment can then be calculated by multiplying (3) and (7): To find the total energy, we simply need to integrate this from 0 to n, where n labels the highest energy state: Our final step is to express the total energy in terms of the number of electrons contained within the volume, V. Since two electrons can occupy each state, we can use (5) to write: If we substitute this into (9) we find: This represents the total electron degeneracy energy that results from squeezing N e electrons into a volume, V, of space. The degeneracy energy plays a similar role to the internal energy of an ordinary gas in the sense that it causes a pressure to be exerted. It is this electron degeneracy pressure that can support a white dwarf star and stop it from completely collapsing under the force of its own gravity. To see how this balancing act is possible, we first need to derive an expression for the gravitational energy of a star.

Gravitational energy
Imagine building a spherical star layer by layer 2 . Assuming the star has acquired mass, m, and has radius, r, the work necessary to bring in the next increment of mass, dm, is equal to: where G is Newton's gravitational constant. The minus sign simply indicates that the potential energy of an in-falling mass will decrease as it approaches the centre of the star. Next, if we rewrite the mass, m, in terms of the star density, ρ, and radius, r, we have we find that the mass segment, dm, can be written as: where dr represents the increase in radius resulting from the additional mass that has been added to the star 3 . If we substitute (13) into (12) and write m in terms of ρ and r we find: It follows that the total gravitational energy of a sphere of radius R and density ρ is given by the integral of d E grav from 0 to R: Next, if we assume that the mass of the star, M, is roughly equal to the number of nucleons contained within it multiplied by the mass of a single proton, then we find that the star density, ρ, can be written as: 2 See, for example [7][8][9]. 3 To simplify our analysis we have assumed that the star has a constant density throughout its interior.

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where N n is the number of nucleons, and M p is the mass of a proton. It follows that the gravitational energy (15) can then be written as: This equation tells us how the gravitational energy of a star varies with the radius of the star.

Energy of a white dwarf star
Now that we have expressions for both the electron degeneracy energy and gravitational energy of the star, we can express the total energy as the sum of both 4 : where: and: Note that in order to write the total energy in this form, we first had to re-express the electron degeneracy energy in terms of the star radius, R.
Next, we want to find the radius of the star that corresponds to the minimum total energy (figure 2), and to do this we simply need to differentiate the total energy and then set the answer equal to zero: And therefore we see that: where R 0 is the radius corresponding to minimum energy. If we substitute (19) and (20) into (22) we find that the energy of the star is minimised when: Figure 2. If we plot a graph of energy vs radius, we see that there exists a minimum energy value where the gradient is zero.
Since the star is assumed to be electrically neutral, the number of protons in the star will be equal to the number of electrons, and we will further assume that half of the nucleons are protons, and therefore the number of electrons is approximately equal to half the number of nucleons N e ∼ 1 2 N n . 5 If we sub this into (23) we find: To get a sense of what this equation is telling us, we will calculate the radius of a white dwarf star that has a mass equal to the mass of our Sun. In that case, the number of nucleons can be estimated by dividing the mass of the Sun by the mass of a single proton: If we substitute this into (24) we find: If we compare this with the radius of the Earth, which is R E ∼ 6.4 × 10 6 m, we see that a white dwarf with a mass equal to our Sun would be roughly the same size as the Earth. It is also worth noting that if we re-express the radius equation in terms of the mass of the star, we see that: And therefore it follows that: This is telling us that as the mass of a white dwarf star increases, the volume of the star will actually decrease. This equation implies that no matter how massive the star, it will always be possible to find a value of R 0 for which the degeneracy pressure can stop the gravitational collapse. So, is this true?

Relativistic degeneracy pressure
Our analysis so far has been based on the assumption that the velocity of an electron is unbounded, and therefore there is no limit to the electron degeneracy pressure that can, in principle, be exerted. But in reality, this is not true. According to Einstein's theory of special relativity, nothing can travel faster than the speed of light. This inherent restriction on the maximum velocity of the electrons places an upper limit on the amount of electron degeneracy pressure a white dwarf star can exert, which in turn places a limit on the maximum mass of a white dwarf star. So, how do we account for this?
Firstly, we need to replace the classical energy momentum relation with the relativistic version: As the velocity of an object approaches the speed of light, the first term inside the bracket will dominate the second term, and therefore in the 'ultra-relativistic' limit we can assume that p 2 c 2 ≫ m 2 c 4 , in which case the total energy will be approximated by E ≈ pc. If we then combine the de Broglie relation, p = ℏk, with the ultrarelativistic energy momentum relation, E ≈ pc, we find: where E r refers to the ultra-relativistic energy. We can then repeat the same analysis as before to calculate the total electron degeneracy energy, only this time we will use the ultra-relativistic energy momentum relation instead. In this case we simply make the replacement in (8): We then use the fact that: to rewrite (31) as: Following the same logic that lead to (9), the total electron degeneracy energy can be calculated by integrating (33) from 0 to n: Next, if we substitute (10) into (34) and rewrite in the energy in terms of the radius, R, we find that: If we then combine the ultra-relativistic electron degeneracy energy (35) with the gravitational energy (17) we find: where: July 2023 A M Low Figure 3. When A > B, a white dwarf star can reduce its energy by increasing its size. When A < B the total energy is negative and unbounded, suggesting that the star could reduce its energy by reducing its radius to zero. and: We see that when A > B, the total energy will be positive for all values of R, and the white dwarf star will be able to reduce its energy by increasing its radius ( figure 3). If the stars radius increases, then so will the volume of the star. According to Heisenberg's uncertainty principle this will cause the average velocity uncertainty of the electrons to decrease, and the ultra-relativistic energy equation will no longer apply. We would then simply use our earlier analysis (23) to determine the stable radius of the star.
When A < B, the total energy will be negative for all values of the radius R, and the white dwarf star will be able to reduce its energy by reducing it is radius. Furthermore, we see from figure 4 that this energy reduction is unbounded and therefore the white dwarf star will rapidly collapse down to zero radius unless some new kind of pressure kicks in to stop the collapse. In this case, it would appear that the mass of the star is simply too large, and a stable white dwarf star cannot exist 6 .

The Chandrasekhar mass
A natural question arises: what is the maximum possible mass of a white dwarf star? As you might have guessed, to answer this question we simply 6 The ultimate fate of the star depends on the mass of the star. See, for example [10]. need to look at the limiting case in which A = B (since this represents the tipping point between the two different scenarios). Setting A = B and rearranging for N n we find: To estimate the corresponding maximum mass, we simply multiply the number of nucleons, N n , by the mass of a proton: . (40) where M c is known as the Chandrasekhar mass. This equation is telling us the maximum possible mass of a white dwarf star. If we sub in the numbers, we find a value for the Chandrasekhar mass of: Considering the crude approximations that we have used in our analysis, it is remarkable that we have found a result so close to Chandrasekhar's result 7 : The main reason for the difference in numerical factor can be traced back to our constantdensity approximation. If we repeat the analysis and factor in the density variations within the star, we will obtain an answer closer to the famous 1.4 solar mass result. A more detailed analysis would also take into account the fact that, although small, the temperature of a star does play some role in determining its structure. And finally, we should take account of the fact that the electrons experience an electrostatic force of attraction from the charged atomic nuclei. However, despite these added complexities, the remarkable thing is that we have been able to arrive at a 7 See for example [7].