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White dwarfs and neutron stars
White dwarfs and neutron stars

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4.3 The white dwarf mass-radius relationship

The mean density of a star is equal to its mass divided by its volume, rho macron equals three times cap m solidus left parenthesis four times pi times cap r cubed right parenthesis . It can be related to the core density if the density profile of the star is known. So-called polytropic stellar models are ones in which the pressure at some radius is proportional to the density at that radius to some power. For a star described by a polytropic model with cap p of r proportional to rho times left parenthesis r right parenthesis super five solidus three (a reasonable approximation), the mean density turns out to be one-sixth of the core density: rho macron equals rho sub c solidus six . It is therefore possible to write a relation between the radius, mass and core density of a star.

Where the core density is dictated by non-relativistic degenerate electrons, we have already seen how the core density and mass are related (Equation 15). We can eliminate the density from this equation and find an expression for the radius of the object as a function of its mass, as shown by the following activity.

Activity 6

  • a.Rearrange the definition of the mean density rho macron equals three times cap m solidus left parenthesis four times pi times cap r cubed right parenthesis to get an expression for the radius in terms of the mass and mean density.

  • b.Use the result for a polytropic model with cap p proportional to rho super five solidus three (i.e. rho macron equals rho sub c solidus six ) to re-express the radius in terms of mass and core density.

  • c.Substitute in the expression for the core density of non-relativistic degenerate electrons (Equation 15) to derive an expression for the radius of a white dwarf as a function of mass. Give your final answer in units normalised to the solar mass and solar radius. Note that cap g equals 6.67 multiplication 10 super negative 11 cap n m super two kg super negative two , h equals 6.63 multiplication 10 super negative 34 cap j s , m sub e equals 9.11 multiplication 10 super negative 31 kg and m sub cap h equals 1.67 multiplication 10 super negative 27 kg full stop

  • d.Re-express the final result in solar masses and Earth radii, where the Earth’s radius is cap r sub circled plus equals 6.37 multiplication 10 super six m.

Comment

  • a.The radius as a function of mass and average density is

    cap r equals left parenthesis three times cap m divided by four times pi times rho macron right parenthesis super one solidus three full stop
  • b.If the average density is 1/6 of the core density, then

    cap r equals left parenthesis 18 times cap m divided by four times pi times rho sub c right parenthesis super one solidus three full stop
  • c.Substituting in the core density from Equation 15 gives

    cap r equals left parenthesis 18 times cap m divided by four times pi right parenthesis super one solidus three times left parenthesis 16 times pi cubed divided by 81 right parenthesis super negative one solidus three times left parenthesis five times m sub e divided by h squared right parenthesis super negative one times cap g super negative one times left parenthesis m sub cap h divided by cap y sub e right parenthesis super negative five solidus three times cap m super negative two solidus three full stop

    Rearranging the terms with negative powers:

    cap r equals left parenthesis nine times cap m divided by two times pi right parenthesis super one solidus three times left parenthesis 81 divided by 16 times pi cubed right parenthesis super one solidus three times left parenthesis h squared divided by five times m sub e right parenthesis times cap y sub e super five solidus three divided by cap g times m sub cap h super five solidus three times cap m super negative two solidus three full stop

    Now consolidate the mass terms and the numerical factors:

    cap r equals left parenthesis 729 divided by 32 times pi super four right parenthesis super one solidus three times left parenthesis h squared divided by five times m sub e right parenthesis times cap y sub e super five solidus three divided by cap g times m sub cap h super five solidus three times cap m super negative one solidus three full stop

    Now, for convenience, express the mass in solar units:

    cap r equals left parenthesis 729 divided by 32 times pi super four right parenthesis super one solidus three times left parenthesis h squared divided by five times m sub e right parenthesis times cap y sub e super five solidus three divided by cap g times m sub cap h super five solidus three times cap m super negative one solidus three multiplication left parenthesis 1.99 multiplication 10 super 30 kg cap m sub circled dot operator super negative one right parenthesis super negative one solidus three full stop

    Then putting in the numbers:

    cap r equals left parenthesis 729 divided by 32 times pi super four right parenthesis super one solidus three multiplication left parenthesis 6.63 multiplication 10 super negative 34 cap j s right parenthesis squared divided by left parenthesis five multiplication 9.11 multiplication 10 super negative 31 kg right parenthesis multiplication left parenthesis 0.5 right parenthesis super five solidus three multiplication left parenthesis 1.99 multiplication 10 super 30 kg right parenthesis super negative one solidus three divided by left parenthesis 6.67 multiplication 10 super negative 11 cap n m super two times kg super negative two right parenthesis multiplication left parenthesis 1.67 multiplication 10 super negative 27 kg right parenthesis super five solidus three multiplication left parenthesis cap m divided by cap m sub circled dot operator right parenthesis super negative one solidus three full stop

    And so

    cap r equals 9.45 multiplication 10 super six times left parenthesis cap m divided by cap m sub circled dot operator right parenthesis super negative one solidus three m full stop

    Because cap r sub circled dot operator equals 6.96 multiplication 10 super eight m , the white dwarf radius is

    cap r sub WD equals left parenthesis 9.45 multiplication 10 super six m divided by 6.96 multiplication 10 super eight m cap r sub circled dot operator super negative one right parenthesis multiplication left parenthesis cap m sub WD divided by cap m sub circled dot operator right parenthesis super negative one solidus three
    cap r sub WD equals cap r sub circled dot operator divided by 74 multiplication left parenthesis cap m sub WD divided by cap m sub circled dot operator right parenthesis super negative one solidus three full stop
  • d.Because the radius of the Earth is cap r sub circled plus equals 6.37 multiplication 10 super six m ,

    cap r sub WD equals left parenthesis 9.45 multiplication 10 super six m divided by 6.37 multiplication 10 super six m cap r sub circled plus super negative one right parenthesis multiplication left parenthesis cap m sub WD divided by cap m sub circled dot operator right parenthesis super negative one solidus three
    cap r sub WD equals 1.5 times cap r sub circled plus multiplication left parenthesis cap m sub WD divided by cap m sub circled dot operator right parenthesis super negative one solidus three full stop

The result of the previous activity

cap r sub WD equals cap r sub circled dot operator divided by 74 multiplication left parenthesis cap m sub WD divided by cap m sub circled dot operator right parenthesis super negative one solidus three
Equation label: (18)

is remarkable for two reasons. First, it shows that the radius of a white dwarf with the same mass as the Sun is two orders of magnitude smaller, comparable in radius to the Earth, as illustrated in Figure 3.

Described image
Figure 3 An artist’s impression showing the relative sizes of the white dwarf Sirius B and the Earth.

Second, notice the mass-dependence of the radius: cap r sub WD proportional to cap m sub WD super negative one solidus three . Unlike low-mass main-sequence stars fusing hydrogen in helium, where cap r sub MS proportional to cap m sub MS super two solidus five , note the important result that the sign of the exponent is negative for white dwarfs!

While the more massive a main-sequence star is, the larger its radius, the more massive a white dwarf is, the smaller its radius!

The derivation above of the white dwarf mass-radius relationship was based on the assumption that the electrons are non-relativistic. We showed this to be incorrect as the mass increases and it would imply infinite density (zero radius) at the Chandrasekhar limit, 1.4 M. There is another formula which gives a more accurate picture of the radius of white dwarfs, but it is an empirical formula, simply fitted to observed data. This formula, derived by Michael Nauenberg in 1972, is

cap r sub WD almost equals 7.8 multiplication 10 super six m prefix multiplication of left square bracket left parenthesis 1.4 times cap m sub circled dot operator divided by cap m sub WD right parenthesis super two solidus three minus left parenthesis cap m sub WD divided by 1.4 times cap m sub circled dot operator right parenthesis super two solidus three right square bracket super one solidus two full stop
Equation label: (19)

The formula derived in Equation 18, while unable to correctly describe white dwarfs near the Chandrasekhar limit, does encapsulate real physical ideas that are valid for low-mass stars. There is a place in science for both formulae!