Unlike other supernova, which can vary in their size, Type Ia supernova are all about the same size. This is due to the fact that they are caused when a white dwarf star gains enough mass from its binary companion to reach the Chandrasekhar limit of 1.44 solar masses.

The energy released in a Type Ia supernova is estimated to be $10^{44}$ joules. Since the explosions are all the same size, the luminosity is also the same, and for this reason Type Ia supernova are used as standard candles to gauge distances of objects from our reference point across the universe.

I was doing some rough calculations using the percentage of mass that is actually converted into energy during nuclear fusion and then plugging this figure into $E=mc^2$ and I keep getting a deficit even when I plug in extra-realistic figures for the portion of the 1.44 solar mass that can be expected to fuse.

As a hobby physicist, I could be making some errors. What would be a reasonable calculation of the total energy output from nuclear fusion for a 1.44 Solar mass white dwarf which fuses all its fusion fuel all at once?

  • $\begingroup$ By the way, I'm pretty sure that "supernova" is singular and "supernovae" is the plural. Or "supernovas", of course, which is always valid in English. $\endgroup$
    – Ryan Reich
    Commented Jul 17, 2017 at 21:54
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    $\begingroup$ A good fraction of the light released by a Type Ia supernova is not directly produced by fusion. The fusion reaction produces about half a solar mass of nickel-56, which is unstable to beta decay to cobalt-56 (half life of a week), which then beta-decays again to iron-56 (half life of a couple of months). These fission reactions are responsible for maintaining the brightness of the supernova for weeks to months after the initial explosion. $\endgroup$
    – Calchas
    Commented Jul 17, 2017 at 23:43
  • $\begingroup$ @calchas But not responsible for unbinding the star. $\endgroup$
    – ProfRob
    Commented Jul 18, 2017 at 5:46

1 Answer 1


A back of the envelope calculation (and that is all this is) would go along the lines of assuming that the white dwarf is made entirely of $^{12}$C (it isn't) and is entirely converted into $^{56}$Ni (it isn't).

The appropriate mass to use would be $\sim 1.4M_{\odot}$ (it is actually a touch lower - the real "Chandrasekhar mass" at which instability sets in is determined by GR collapse; or by inverse beta decay; or by the onset of pyconuclear reactions, all of which take place at $\rho \sim 3 \times 10^{13}$ kg/m$^3$ when the white dwarf has a mass of about 1.37-1.38$M_{\odot}$).

If the star is entirely $^{12}$C, then this means $1.40 \times 10^{56}$ carbon nuclei, containing $1.68\times 10^{57}$ baryons. To conserve the baryon number, the number of $^{56}$Ni nuclei produced is smaller by a factor of 12/56.

The mass of each carbon nucleus (by definition) is $12m_u$, where $m_u$ is the atomic mass unit. The mass of each nickel nucleus is $55.94m_u$.

Thus the change in mass converting all the carbon into nickel is $$ \Delta M \simeq 1.40\times10^{56}\times 12m_u - 1.40\times10^{56}\times (12/56)*55.94m_u$$ $$\Delta M \simeq 1.8\times 10^{54} m_u = 3.0\times10^{27}\ {\rm kg}$$

Converting this to energy gives $2.7\times 10^{44}$J, which is indeed roughly the energy involved in a type Ia supernova. This is what is responsible for "exploding" the star, since with an initial radius of $\sim 1000$ km, it has a gravitational binding energy, $\sim -3GM^2/5R = -3\times 10^{44}$ J.

A slightly less back of the envelope calculation would include the internal energy of the relativistic electrons, which shrinks the magnitude of the binding energy considerably (it would be exactly zero for a star entirely governed by ideal ultra-relativistic degeneracy pressure and halved for non-relativistic degeneracy pressure), so that a large fraction of the energy released can actually go into photons, neutrinos and the kinetic energy of the ejecta.

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    $\begingroup$ That makes type Ia supernova surprisingly efficient, if it is powered by fusion; I'd naively (not-a-physcist) expect that the runaway fusion would shove the vast, vast majority of the mass out of the gravitational well before it could fuse. Instead, your (back-of-envelope) estimate is that ~1/3 of the entire star undergoes fusion! That is one steep cliff. $\endgroup$
    – Yakk
    Commented Jul 17, 2017 at 18:03
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    $\begingroup$ @Yakk There's some evidence that the fusion proceeds as a supersonic detonation rather than a gradual flame spreading "slowly" through the remnant. In this case, loosely speaking, the material ahead of the front would not be "informed" about the fusion going on elsewhere in the remnant until the detonation front passed through it. $\endgroup$
    – Calchas
    Commented Jul 17, 2017 at 18:46
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    $\begingroup$ Rob, I'm too lazy to do the computation without asking ... do you happen to have a figure on the gravitational binding energy released? Just at the level of "similar", "smaller but comparable", "negligible", or "silly rabbit it isn't even close" would be fine. $\endgroup$ Commented Jul 17, 2017 at 22:49
  • $\begingroup$ @dmckee $R\sim 1000$ km, so $3GM^2/5R \sim 3\times 10^{44}$ J ... $\endgroup$
    – ProfRob
    Commented Jul 18, 2017 at 5:44
  • $\begingroup$ @Calchas now, how to do that artificially... $\endgroup$
    – ikrase
    Commented Jan 9, 2020 at 6:17

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