<p>Nickel 56  decays to Cobalt 56 via electron capture decay, with a half-life of 6.1 days and a decay constant of <span class="math-container">$\lambda = 1.31\times 10^{-6}$</span> s<span class="math-container">$^{-1}$</span>.</p>
<p>About 1.75 MeV of energy is lost as gamma rays and a further 0.41 MeV in the form of an electron neutrino (<a href="http://articles.adsabs.harvard.edu//full/1994ApJS...92..527N/0000527.000.html" rel="nofollow noreferrer">Nadyozhin 1994)</a></p>
<p>Let's assume that we are talking about the period of time after the initial detonation, where the fusion energy of carbon and oxygen is released and is sufficient to unbind the white dwarf.</p>
<p>Let's assume that apart from the very first few seconds of the supernova, that the envelope is transparent to neutrinos, so that energy is lost.</p>
<p>Let's further assume that the gamma rays are able to thermalise rapidy in the envelope, that the energy is able to diffuse to the &quot;photosphere&quot; of the expanding fireball in <span class="math-container">$&lt; 6 $</span> days and that the work done in expanding any ejecta is also negligible. This latter may not be justified in a type Ia supernova.</p>
<p>The decay equation <span class="math-container">$N= N_0 \exp(-\lambda t)$</span>, means the rate of gamma ray energy deposition will be
<span class="math-container">$$ \frac{dE}{dt} = 1.75\ \lambda N_0 \exp(-\lambda t) \ {\rm MeV/s}, $$</span>
where <span class="math-container">$N_0$</span> is the number of nickel nuclei you begin with.</p>
<p>Let's assume about <span class="math-container">$0.5M_{\odot}$</span> of Nickel is produced (see <a href="https://arxiv.org/abs/1507.02501" rel="nofollow noreferrer">Childlress et al. 2015</a>, incidentally, this paper illustrates how complex this question really is). This means <span class="math-container">$N_{0} \simeq 1.1\times 10^{55}$</span>.</p>
<p>Thus I make <span class="math-container">$dE/dt$</span> (assumed to result in the emergent luminosity), when <span class="math-container">$t=0$</span>, to be
<span class="math-container">$2.5\times10^{49}$</span> MeV/s, or <span class="math-container">$10^{10} L_{\odot}$</span>.</p>
<p>This value is about a factor of two higher than the average type Ia supernova peak bolometric luminosity measured by <a href="https://arxiv.org/abs/1402.6842" rel="nofollow noreferrer">Scalzo et al. (2014)</a>, which were well modelled with nickel masses of around <span class="math-container">$0.5 M_{\odot}$</span>. So I would suggest that one or more of the assumptions above (probably the one about no work being done on the ejecta) is a bit flaky. However, I also notice that the peak luminosity occurs about 2 weeks after the initial rise, so I would suggest that the &quot;instant&quot; reprocessing assumption is also incorrect and that the released energy is smoothed out somewhat so that a lot of the nickel has already decayed by the time you get to peak luminosity.</p>

CC BY-SA 4.0
2020-08-11T20:31:32.310Z
This is excellent, thanks very much -- I had been looking for some literature on the subject so your references are appreciated. I'll be doing some reading!
notadoctor
CC BY-SA 4.0
2020-08-11T17:20:05.927Z
2020-08-11T17:20:05.927Z
ProfRob
<p>Okay, yes, this was the way I went about it, and it seems satisfactory enough. Just use a nuclear physics reference (or Google) to get the isobar masses:</p>
<p><span class="math-container">$M_{^{56}Ni} = 55.9421u$</span></p>
<p><span class="math-container">$M_{^{56}Co} = 55.9398u$</span></p>
<p>Then you have <span class="math-container">$\Delta m=0.0023u=(0.0023)(931.5\frac{MeV}{c^2})=2.14245\frac{MeV}{c^2}$</span> and thus the energy per decay is <span class="math-container">$2.14245MeV$</span>.</p>
<p>From here, just figure out how many Ni-56 nuclei you started with via your initial mass and straight division through by the Ni-56 mass above, then multiply by your energy released per decay and convert to the units you need.</p>
<p>In my case, I ended up with about <span class="math-container">$2\times10^{10}L_{sun}$</span>. Hope this helps someone else one day!</p>

CC BY-SA 4.0
2020-08-11T07:30:04.280Z
This makes a number of very big assumptions, the main one being that the luminosity is in the form of the emitted gamma rays. Clearly that isn't true, most of the luminosity emerges at UV to IR wavelengths. That's ok, so long as the timescale for the gamma ray energy to diffuse to the "photosphere" of the fireball is less than 6 days and that none of the released energy is being absorbed and doing work.
ProfRob
CC BY-SA 4.0
2020-08-11T16:19:33.153Z
This is a very good point. The initial gamma ray burst would likely only last a few hours, not 6.1 days. I wonder if I should be considering the UV/IR luminosities instead -- but then I am not sure how to determine those from just the initial mass.
notadoctor
CC BY-SA 4.0
2020-08-11T16:48:02.153Z
No, the gamma rays will be released on a timescale given by the decay half life. My point is that you are assuming that the observable luminosity follows this timescale too. In order to get from (a) to (b) the gamma rays have to be reprocessed into UV-optical-IR light at the "photosphere" of the fireball. In addition, *some* of that energy may do work on any ejecta and be "lost". I am not advocating any change to your simple method; just making explicit your impicit assumptions. However I don't see where you calculated the number of decays per second, which is what you need.
ProfRob
CC BY-SA 4.0
2020-08-11T06:27:11.987Z
2020-08-11T06:27:11.987Z
notadoctor
<p>The initial portion (and peak) of a SNIa's light curve is powered by the <span class="math-container">$\beta$</span>-decay reaction:</p>
<p><span class="math-container">$$ ^{56}Ni \rightarrow ^{56}Co + e^+ + \nu_e + \gamma $$</span></p>
<p>Supposing we know how much Ni-56 is created in the supernova (i.e., an initial mass of Ni-56), how can I go about estimating the luminosity peak? The half-life of Ni-56 is 6.1 days, if that's relevant.</p>
<p>Does it make sense if I just look up/calculate the energy released in this reaction, multiply it by the number of Ni-56 atoms required to comprise the given initial mass, and call that the luminosity? Any other better ideas?</p>

CC BY-SA 4.0
2020-08-11T07:44:53.357Z
YOu need to know how many Ni56 atoms are decaying per second. This will initially be a bit more than half of them divided by 6.1 days. I think the extra factor is 1/ln(2), but I'm not sure and I haven't time to look it up just now.
Steve Linton
CC BY-SA 4.0
2020-08-11T16:17:16.850Z
Thanks Steve, this sounds familiar. I will look into it.
notadoctor
<supernova><stellar-astrophysics><nucleosynthesis><light-curve>
How can I estimate the initial luminosity of a SNIa given initial mass of Ni-56?
notadoctor
# How can I estimate the initial luminosity of a SNIa given initial mass of Ni-56?

(Asked by: username_0 on 2020-08-11)

The initial portion (and peak) of a SNIa's light curve is powered by the $\beta$-decay reaction:

$$ ^{56}Ni \rightarrow ^{56}Co + e^+ + \nu\_e + \gamma $$

Supposing we know how much Ni-56 is created in the supernova (i.e., an initial mass of Ni-56), how can I go about estimating the luminosity peak? The half-life of Ni-56 is 6.1 days, if that's relevant.

Does it make sense if I just look up/calculate the energy released in this reaction, multiply it by the number of Ni-56 atoms required to comprise the given initial mass, and call that the luminosity? Any other better ideas?

**Tags:** supernova, stellar-astrophysics, nucleosynthesis, light-curve

> **Comments:**
> **username_1**: YOu need to know how many Ni56 atoms are decaying per second. This will initially be a bit more than half of them divided by 6.1 days. I think the extra factor is 1/ln(2), but I'm not sure and I haven't time to look it up just now.  
> **username_0**: Thanks username_1, this sounds familiar. I will look into it.  

---

## 2 answers

### Answer by username_2 on 2020-08-11. Score: 1 | Accepted answer

Nickel 56 decays to Cobalt 56 via electron capture decay, with a half-life of 6.1 days and a decay constant of $\lambda = 1.31\times 10^{-6}$ s$^{-1}$.

About 1.75 MeV of energy is lost as gamma rays and a further 0.41 MeV in the form of an electron neutrino ([Nadyozhin 1994)](http://articles.adsabs.harvard.edu//full/1994ApJS...92..527N/0000527.000.html)

Let's assume that we are talking about the period of time after the initial detonation, where the fusion energy of carbon and oxygen is released and is sufficient to unbind the white dwarf.

Let's assume that apart from the very first few seconds of the supernova, that the envelope is transparent to neutrinos, so that energy is lost.

Let's further assume that the gamma rays are able to thermalise rapidy in the envelope, that the energy is able to diffuse to the "photosphere" of the expanding fireball in $< 6 $ days and that the work done in expanding any ejecta is also negligible. This latter may not be justified in a type Ia supernova.

The decay equation $N= N\_0 \exp(-\lambda t)$, means the rate of gamma ray energy deposition will be
$$ \frac{dE}{dt} = 1.75\ \lambda N\_0 \exp(-\lambda t) \ {\rm MeV/s}, $$
where $N\_0$ is the number of nickel nuclei you begin with.

Let's assume about $0.5M\_{\odot}$ of Nickel is produced (see [Childlress et al. 2015](https://arxiv.org/abs/1507.02501), incidentally, this paper illustrates how complex this question really is). This means $N\_{0} \simeq 1.1\times 10^{55}$.

Thus I make $dE/dt$ (assumed to result in the emergent luminosity), when $t=0$, to be
$2.5\times10^{49}$ MeV/s, or $10^{10} L\_{\odot}$.

This value is about a factor of two higher than the average type Ia supernova peak bolometric luminosity measured by [Scalzo et al. (2014)](https://arxiv.org/abs/1402.6842), which were well modelled with nickel masses of around $0.5 M\_{\odot}$. So I would suggest that one or more of the assumptions above (probably the one about no work being done on the ejecta) is a bit flaky. However, I also notice that the peak luminosity occurs about 2 weeks after the initial rise, so I would suggest that the "instant" reprocessing assumption is also incorrect and that the released energy is smoothed out somewhat so that a lot of the nickel has already decayed by the time you get to peak luminosity.

> **Comments:**
> **username_0**: This is excellent, thanks very much -- I had been looking for some literature on the subject so your references are appreciated. I'll be doing some reading!  

---

### Answer by username_0 on 2020-08-11. Score: 0

Okay, yes, this was the way I went about it, and it seems satisfactory enough. Just use a nuclear physics reference (or Google) to get the isobar masses:

$M\_{^{56}Ni} = 55.9421u$

$M\_{^{56}Co} = 55.9398u$

Then you have $\Delta m=0.0023u=(0.0023)(931.5\frac{MeV}{c^2})=2.14245\frac{MeV}{c^2}$ and thus the energy per decay is $2.14245MeV$.

From here, just figure out how many Ni-56 nuclei you started with via your initial mass and straight division through by the Ni-56 mass above, then multiply by your energy released per decay and convert to the units you need.

In my case, I ended up with about $2\times10^{10}L\_{sun}$. Hope this helps someone else one day!

> **Comments:**
> **username_2**: This makes a number of very big assumptions, the main one being that the luminosity is in the form of the emitted gamma rays. Clearly that isn't true, most of the luminosity emerges at UV to IR wavelengths. That's ok, so long as the timescale for the gamma ray energy to diffuse to the "photosphere" of the fireball is less than 6 days and that none of the released energy is being absorbed and doing work.  
> **username_0**: This is a very good point. The initial gamma ray burst would likely only last a few hours, not 6.1 days. I wonder if I should be considering the UV/IR luminosities instead -- but then I am not sure how to determine those from just the initial mass.  
> **username_2**: No, the gamma rays will be released on a timescale given by the decay half life. My point is that you are assuming that the observable luminosity follows this timescale too. In order to get from (a) to (b) the gamma rays have to be reprocessed into UV-optical-IR light at the "photosphere" of the fireball. In addition, *some* of that energy may do work on any ejecta and be "lost". I am not advocating any change to your simple method; just making explicit your impicit assumptions. However I don't see where you calculated the number of decays per second, which is what you need.  

---

