I was trying to calculate the gravitational binding energy of a white dwarf just before it went on a type Ia supernova in order to calculate the kinetic energy of the ejecta, but I wasn't able to get the radius since the radius is zero at the Chandrasekhar limit.

So how to calculate the binding energy of a WD just before a type Ia SN ?

EDIT : Found this which gives a number for the binding energy of a WD at Chandrasekhar limit, but still I don't know how it was calculated.

EDIT 2 : For Rob Jeffries, please take a look at page 5 of this, it clarifies somethings I think. Any knowledge on how this internal energy was calculated ?

  • $\begingroup$ Typically, $R_{wd}\sim R_{earth}$. $\endgroup$
    – Kyle Kanos
    Commented Sep 2, 2014 at 14:03
  • $\begingroup$ Very good question. Clearly one needs a better theory than the basic Chandrasekhar arguments. Kippenhahn and Weigert mention that pycnonuclear reactions and inverse $\beta$ decay become important at the high-mass end, and they refer to Hamada and Salpeter for more details. Perhaps someone can pick up here and make a full answer... $\endgroup$
    – user10851
    Commented Sep 2, 2014 at 14:07
  • $\begingroup$ @KyleKanos, this seems to be the value used in the link I just edited the question for, but still I don't understand why this value is used. $\endgroup$ Commented Sep 2, 2014 at 14:12
  • $\begingroup$ @AbanobEbrahim: It depends on using non-relativistic electron degeneracy, which doesn't correctly apply here; using relativistic electron degeneracy gives the correct $\mathcal R\to0$ you've found in the first link. $\endgroup$
    – Kyle Kanos
    Commented Sep 2, 2014 at 14:18
  • 2
    $\begingroup$ @akrasia I don't know where they get that number from, because every supernova researcher has a different opinion on Type Ia progenitors. It may be that most are double degenerate, it may be that none are. A lot more work needs to be done. $\endgroup$
    – user10851
    Commented Sep 2, 2014 at 15:32

1 Answer 1


The gravitational binding energy is the sum of the gravitational potential energy, $\Omega$, and the total internal kinetic energy, $U$.

If you calculate $\Omega + U$ for a star governed solely by ideal ultra-relativistic electron degeneracy pressure, the net binding energy is zero. This corresponds to the "traditional" Chandrasekhar limit for infinite density and zero radius, which occurs at a mass of $1.44 M_{\odot}$ for a carbon or oxygen white dwarf.

In truth, this situation does not occur in nature. There are a number of small corrections to the equation of state - e.g. electrostatic interactions, but more importantly there are at least two reasons why the white dwarf would become unstable at a mass lower than the canonical Chandrasekhar mass and at finite radius. (i) Neutronisation (aka electron capture) may occur, leading to the removal of degenerate electrons and instability; (ii) If one uses the appropriate Tolman-Oppenheimer-Volkhoff (TOV) general relativistic expression for hydrostatic equilibrium, then the WD becomes unstable (for a carbon WD) at around $1.397M_{\odot}$ and at a small, but finite radius of about 1000 km (see Mathew & Nandy 2017 ).

An approximation of $\Omega \sim -1.5GM^2/R$ (which is valid for a gas governed by relativistic degeneracy pressure - i.e.for a $n=3$ polytrope, http://www.astro.princeton.edu/~gk/A403/polytrop.pdf) gives $\Omega= -6\times 10^{44}$ J.

It is somewhat harder to calculate $U$ on the back of an envelope - you really need to integrate a numerical model in spherical shells, solving the TOV hydrostatic equilibrium equation in GR. However, here goes. Let's get an estimate by using the energy density of gas at the average density of the WD ($6.6\times10^{11}$ kg/m$^3$).

For a carbon gas at this density, the Fermi momentum is $p_F=1.9\times10^{-21}$ kg m/s and the relativity parameter, $p_F/m_e c \simeq 7$. Then, approximating this as ultra-relativistic, the average kinetic energy per electron is $(3/4)p_{F}c$ and the kinetic energy density $u=8.4\times10^{25}$ kg/m$^3$. Multiplying by the stellar volume gives $U=3.5\times10^{44}$ J.

Hence binding energy $\Omega + U = -2.5\times10^{44}$ J.

This is five times the $-5\times 10^{43}$ J quoted in the references you dug out. This could easily be due to my crude approximations in the calculations of $\Omega$ and $U$ (subtracting one big, uncertain number from another), but I also note that in your references they talk about a central density of $2\times10^{12}$ kg/m$^3$, whereas the central density of a WD at the GR Chandrasekhar limit is actually $2.35\times10^{13}$ kg/m$^3$. So I guess their WD is also factor of 2-3 bigger and so their GPE is a factor of 2-3 smaller because of this.

I am puzzled by where this central density comes from (if indeed that's what it is) and would appreciate any comments on this (rather than a downvote).


The OP raises the question of rotation. This might change things. Boshkayev et al. (2013) finds a GR Chandrasekhar limit of 1.386$M_{\odot}$ for non-rotating WDs and a central density of $2.12\times10^{13}$ kg/m$^3$ (consistent with what I use above). The rotating models (shown in Fig.2) show that a WD with 1.38$M_{\odot}$ and central density of 2-3$\times10^{12}$ kg/m$^3$ is possible, but these should be stable - the Chandrasekhar limit is increased by rotation and occurs at lower central densities in these cases, but always I think $\geq 7\times10^{12}$ kg/m$^3$.

Further footnote

After correspondence with one of the authors of the original SN Type 1A progenitor papers, it turns out they are using WD structures that do not use GR in the calculation. Hence the lower central densities at a given mass.

  • $\begingroup$ I edited the main question, please look at this report I put the link for and tell me if that solves the problem. It also seems the central density is not correct, but the radius and GPE is, so what do you think ? $\endgroup$ Commented Sep 3, 2014 at 2:05
  • $\begingroup$ Also, could you please explain why we multiplied but that factor of 3 in calculating the GPE ? And does a non rotating WD change anything ? $\endgroup$ Commented Sep 3, 2014 at 2:27
  • $\begingroup$ @Abanob Ebrahim - you mean the factor of 3/2 ? This is because the star gas a centrally condensed density profile. The factor 3/2 is appropriate for an n=3 polytrope, which is approximately correct for a star supported by relativistic degeneracy pressure. $\endgroup$
    – ProfRob
    Commented Sep 3, 2014 at 7:04
  • $\begingroup$ @Abanob Ebrahim - see p.4 of astro.princeton.edu/~gk/A403/polytrop.pdf $\endgroup$
    – ProfRob
    Commented Sep 3, 2014 at 11:17
  • $\begingroup$ Thanks for going through all these calculations. One thing I found while trying to figure out how the internal energy was calculated was that at one model where the net binding energy was calculated to be 0.54$\times$$10^{44}$ J, the radius for the WD they used was 1500 km, so can using this correct the difference in your calculations ? $\endgroup$ Commented Sep 3, 2014 at 14:55

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