<p>I <em>think</em> I understand this now. I was trying to imagine how you could take a snapshot of a supernova and infer from it the half-life some some element. In fact, I think the measurement comes from watching the radioactive decay happen over time, in the aftermath of a supernova.</p>
<p><a href="https://openstax.org/books/astronomy/pages/23-3-supernova-observations" rel="nofollow noreferrer">Openstax Astronomy 23.3 &quot;Supernova Observations&quot;</a> shows a graph entitled &quot;Figure 23.13 Change in the Brightness of SN 1987A over Time.&quot; It shows brightness increasing during the days right after the supernova, and the rest of the curve not following a consistent fading pattern. It explains:</p>
<blockquote>
<p>One of the elements formed in a supernova explosion is radioactive nickel, with an atomic mass of 56 (that is, the total number of protons plus neutrons in its nucleus is 56). Nickel-56 is unstable and changes spontaneously (with a half-life of about 6 days) to cobalt-56. (Recall that a half-life is the time it takes for half the nuclei in a sample to undergo radioactive decay.) Cobalt-56 in turn decays with a half-life of about 77 days to iron-56, which is stable. Energetic gamma rays are emitted when these radioactive nuclei decay. Those gamma rays then serve as a new source of energy for the expanding layers of the supernova. The gamma rays are absorbed in the overlying gas and re-emitted at visible wavelengths, keeping the remains of the star bright.</p>
<p>As you can see in Figure 23.13, astronomers did observe brightening due to radioactive nuclei in the first few months following the supernova’s outburst and then saw the extra light die away as more and more of the radioactive nuclei decayed to stable iron. The gamma-ray heating was responsible for virtually all of the radiation detected from SN 1987A after day 40. Some gamma rays also escaped directly without being absorbed. These were detected by Earth-orbiting telescopes at the wavelengths expected for the decay of radioactive nickel and cobalt, clearly confirming our understanding that new elements were indeed formed in the crucible of the supernova.</p>
</blockquote>
<p>So if I understand correctly, the logic would be:</p>
<ul>
<li>SN 1987A occurred 168,000 light years from earth</li>
<li>Therefore, humans observed it 168,000 years after it happened</li>
<li>As they watched the supernova unfold, astronomers saw radiation that appeared to be generated by the decay of known, short-lived isotopes, and that decay happened at the same rates we observe today</li>
<li>Therefore, the decay rates of these isotopes were the same 168,000 years ago as they are today</li>
</ul>

CC BY-SA 4.0
2020-05-26T10:13:00.163Z
FWIW, SN1987A was relatively close to us. We've observed some *much* more distant [type II supernovae](https://en.m.wikipedia.org/wiki/Type_II_supernova), some notable examples are given in [this list](https://en.m.wikipedia.org/wiki/List_of_supernovae). At very large distances, we have to take into account the expansion of space, which causes a Doppler shift in the light from the supernova. The time dilation also stretches the duration of the supernova light curve, but the amount of stretching is consistent with the Doppler shift.
PM 2Ring
CC BY-SA 4.0
2020-05-26T02:40:07.637Z
2020-05-26T02:40:07.637Z
2020-06-17T09:47:40.653Z
Nathan Long
<p>When an unstable atom decays, it releases energy. At an earlier time, there are more unstable atoms, so the released energy is larger. At different times, there are different atoms left, so the energy released is different. You can measure the decay rate by following the time-series of this energy.</p>

<p><span class="math-container">$$ E(t) = E_0 \exp[-\lambda t]$$</span></p>

<p>where <span class="math-container">$E$</span> is the energy, <span class="math-container">$t$</span> is time, <span class="math-container">$E_0$</span> is energy at <span class="math-container">$t=0$</span>, <span class="math-container">$\lambda$</span> is the decay constant. This is a very simple picture. In reality, it will be a bit more complicated because of other effects causing the deviation of the observed energy, such as scattering.</p>

CC BY-SA 4.0
2019-01-13T23:05:28.997Z
This leaves a lot unanswered for me. By looking at a glow of a specific brightness, how would you know that it was caused by a the decay of a specific amount of a specific element, or by radioactive decay at all?
Nathan Long
CC BY-SA 4.0
2019-01-14T01:34:16.420Z
Because other power sources would have the light curve behave differently. For example, fallback accretion onto a blackhole would be $\propto t^{-5/3}$, or $\propto t^{-2}$ for a magnetar spindown engine, etc.
Kornpob Bhirombhakdi
CC BY-SA 4.0
2019-01-14T16:08:21.870Z
I had to Google "light curve". :) Wikipedia says it's a graph of light intensity over time. Assuming you somehow could prove that some light from a supernova was being generated by decay of element X, wouldn't you have to measure the light for a very long time in order to draw that curve?
Nathan Long
CC BY-SA 4.0
2019-01-14T16:48:42.757Z
Define long  ???
Kornpob Bhirombhakdi
CC BY-SA 4.0
2019-01-13T18:39:36.307Z
2019-01-13T18:39:36.307Z
Kornpob Bhirombhakdi
<p>Radiometric dating relies on accurate knowledge of the half-lives of radioactive elements. Skeptics have argued that since many elements' half-lives are too long for anyone to have observed them, we can't know what they were in the distant past.</p>

<p>There are various counterarguments based on terrestrial evidence, like agreement between elements in the same sample, and corroboration from varves, tree rings, and ice layers.</p>

<p>Another counterargument I've seen is by looking at supernova, we get a window "back in time" to when their light began its long journey to us, and that much of this light was generated by radioactive decay, confirming that decay rates were the same then as now.</p>

<p>I don't quite follow this explanation, and would like some help. Distant starlight doesn't give us a rock sample in which to analyze proportions of parent and daughter elements. Does it still somehow give evidence of specific radioactive decay rates? If so, how?</p>

<p>I know next to nothing about astronomy, so please explain this to me like I'm a child.</p>

<supernova>
Can radioactive decay rates be confirmed by looking at supernova?
Nathan Long
# Can radioactive decay rates be confirmed by looking at supernova?

(Asked by: username_0 on 2019-01-13)

Radiometric dating relies on accurate knowledge of the half-lives of radioactive elements. Skeptics have argued that since many elements' half-lives are too username_0 for anyone to have observed them, we can't know what they were in the distant past.

There are various counterarguments based on terrestrial evidence, like agreement between elements in the same sample, and corroboration from varves, tree rings, and ice layers.

Another counterargument I've seen is by looking at supernova, we get a window "back in time" to when their light began its username_0 journey to us, and that much of this light was generated by radioactive decay, confirming that decay rates were the same then as now.

I don't quite follow this explanation, and would like some help. Distant starlight doesn't give us a rock sample in which to analyze proportions of parent and daughter elements. Does it still somehow give evidence of specific radioactive decay rates? If so, how?

I know next to nothing about astronomy, so please explain this to me like I'm a child.

**Tags:** supernova


---

## 2 answers

### Answer by username_0 on 2020-05-26. Score: 3

I *think* I understand this now. I was trying to imagine how you could take a snapshot of a supernova and infer from it the half-life some some element. In fact, I think the measurement comes from watching the radioactive decay happen over time, in the aftermath of a supernova.

[Openstax Astronomy 23.3 "Supernova Observations"](https://openstax.org/books/astronomy/pages/23-3-supernova-observations) shows a graph entitled "Figure 23.13 Change in the Brightness of SN 1987A over Time." It shows brightness increasing during the days right after the supernova, and the rest of the curve not following a consistent fading pattern. It explains:

> One of the elements formed in a supernova explosion is radioactive nickel, with an atomic mass of 56 (that is, the total number of protons plus neutrons in its nucleus is 56). Nickel-56 is unstable and changes spontaneously (with a half-life of about 6 days) to cobalt-56. (Recall that a half-life is the time it takes for half the nuclei in a sample to undergo radioactive decay.) Cobalt-56 in turn decays with a half-life of about 77 days to iron-56, which is stable. Energetic gamma rays are emitted when these radioactive nuclei decay. Those gamma rays then serve as a new source of energy for the expanding layers of the supernova. The gamma rays are absorbed in the overlying gas and re-emitted at visible wavelengths, keeping the remains of the star bright.
>
> As you can see in Figure 23.13, astronomers did observe brightening due to radioactive nuclei in the first few months following the supernova’s outburst and then saw the extra light die away as more and more of the radioactive nuclei decayed to stable iron. The gamma-ray heating was responsible for virtually all of the radiation detected from SN 1987A after day 40. Some gamma rays also escaped directly without being absorbed. These were detected by Earth-orbiting telescopes at the wavelengths expected for the decay of radioactive nickel and cobalt, clearly confirming our understanding that new elements were indeed formed in the crucible of the supernova.

So if I understand correctly, the logic would be:

* SN 1987A occurred 168,000 light years from earth
* Therefore, humans observed it 168,000 years after it happened
* As they watched the supernova unfold, astronomers saw radiation that appeared to be generated by the decay of known, short-lived isotopes, and that decay happened at the same rates we observe today
* Therefore, the decay rates of these isotopes were the same 168,000 years ago as they are today

> **Comments:**
> **username_1**: FWIW, SN1987A was relatively close to us. We've observed some *much* more distant [type II supernovae](https://en.m.wikipedia.org/wiki/Type_II_supernova), some notable examples are given in [this list](https://en.m.wikipedia.org/wiki/List_of_supernovae). At very large distances, we have to take into account the expansion of space, which causes a Doppler shift in the light from the supernova. The time dilation also stretches the duration of the supernova light curve, but the amount of stretching is consistent with the Doppler shift.  

---

### Answer by username_2 on 2019-01-13. Score: -1

When an unstable atom decays, it releases energy. At an earlier time, there are more unstable atoms, so the released energy is larger. At different times, there are different atoms left, so the energy released is different. You can measure the decay rate by following the time-series of this energy.

$$ E(t) = E\_0 \exp[-\lambda t]$$

where $E$ is the energy, $t$ is time, $E\_0$ is energy at $t=0$, $\lambda$ is the decay constant. This is a very simple picture. In reality, it will be a bit more complicated because of other effects causing the deviation of the observed energy, such as scattering.

> **Comments:**
> **username_0**: This leaves a lot unanswered for me. By looking at a glow of a specific brightness, how would you know that it was caused by a the decay of a specific amount of a specific element, or by radioactive decay at all?  
> **username_2**: Because other power sources would have the light curve behave differently. For example, fallback accretion onto a blackhole would be $\propto t^{-5/3}$, or $\propto t^{-2}$ for a magnetar spindown engine, etc.  
> **username_0**: I had to Google "light curve". :) Wikipedia says it's a graph of light intensity over time. Assuming you somehow could prove that some light from a supernova was being generated by decay of element X, wouldn't you have to measure the light for a very username_0 time in order to draw that curve?  
> **username_2**: Define username_0  ???  

---

