I know there is a proportionality between the luminosity and $\frac{dN}{dt}$, but I am not sure why they are proportional and why there is a need for an initial mass to compute the luminosity if the half-life equation is:

$N(t) = N_{0} \cdot (\frac{1}{2})^{t/t_{half-life}^{1/2}}$

and how does this apply to supernovae?

I tried to use Cobalt decay to help me understand, but I do not see how to progress from:

enter image description here

$\frac{dN}{dt} = \frac{N_{0}}{N(t)} \cdot (\frac{1}{2})^{t/t_{half-life}^{1/2}}$

  • $\begingroup$ The luminosity is proportional to the number of atoms decaying per unit time, and obviously depends on how many atoms there are at the start (the $N_{0}$ in your equation). $\endgroup$ Dec 9, 2022 at 12:26

1 Answer 1


You need an indication of how much mass is decaying, in order to have an idea of how much energy will be released. That $N_0$ indicates quantity of substance, which you can express in grams, moles, number of atoms or whatever units you prefer.

How does decay in general apply to supernovae? In thermonuclear reactions in stellar cores new molecules are being formed all the time, but they are often unstable and decay.

In supernovae explosions, conditions are so extreme that radiation itself can break apart heavy nuclei, releasing tons of energy and enriching the environment with heavy elements. Many of these processes will produces free neutrons, which are highly unstable and cannot exist outside of a nucleus for more than about 800 seconds.

  • 1
    $\begingroup$ The OP is almost certainly referring to type Ia supernovae where the light curve/luminosity output is determined by the decay of Ni to Fe via the Cobalt decay mentioned. $\endgroup$
    – ProfRob
    Dec 9, 2022 at 10:29

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