The vast majority of energy from a supernova is emitted as neutrinos and high-energy radiation, both of which being invisible to the naked eye. Yet a supernova can outshine its galaxy in the visible spectrum for weeks.

How is a supernova generating so much visible light that nuclear fusion pales in comparison? What is the ratio of visible and invisible energy emitted by a supernova, and how does it compare to stars' normal operation?

  • 1
    $\begingroup$ Just curious, do you have a reason for specifically asking about Type II supernovae? $\endgroup$
    – krismath
    Sep 20, 2014 at 13:05
  • $\begingroup$ I was researching neutron stars and type II supernovae and wanted to make the question as specific as possible. But I expect the answers for the other types to be similar, that's why I put it in parentheses. $\endgroup$
    – Cephalopod
    Sep 20, 2014 at 19:48
  • $\begingroup$ Well, Type Ia supernovae are due to white dwarfs, so that explanation would have a different answer. You also might want to include Type Ib/c in your question, because they are also due to gravitational collapse. The main difference between Type I and Type II is the presence (or lack thereof) of hydrogen in spectral lines. $\endgroup$
    – HDE 226868
    Sep 20, 2014 at 20:23

1 Answer 1


The difference is based on the different efficiency of the processes. We can describe the luminosity by:

$$L = \eta m c^2$$

where $\eta$ is the conversion efficiency, and describes how much matter can be converted into luminosity (photons). Main sequence stars (if you mean this by "normal operation") extract energy from matter by nuclear fusion. The efficiency of nuclear fusion in stars is of the order of $\eta = 1\%$.

The other competitive mechanism, much more efficient than nuclear fusion, is conversion from gravitational potential. In core-collapse SNe, this is the mechanism that brings to the explosion. Gravitational potential energy is defined as $U=-\frac{GMm}{r}$ where m is the mass of a particle, at a distance r, from a large object of mass M; G is the gravitation constant.

If you consider the SN scenario, $r$ is the distance of a particle that is falling from the external layer towards the center (center at $r=0$). So the particle goes from $r_1$ to $r_2$ with $r_2 < r_1$. Then the system lost part of the potential energy, converted to kinetic energy. All the external matter collapse in a very brief time (order of seconds), from the external shell to the compressed nucleus, bringing more compression and then generating energetic shock in the explosion.

The virial theorem says that half the change in gravitational energy stays with the star, while the other half is emitted as luminosity. The efficiency coefficient for gravitational conversion is $\eta\sim10\%$, i.e., 10 times larger than the nuclear fusion efficiency.

From this reference, you can check that the ratio between the invisible against visible emission is of the order $\sim10^7$ (I don't know what kind of reference is that, but the authors are totally trusty).

Something more here and here.


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