Supernova Questions

Hi!

Please don't spare me any juicy details. I can take it! I'm specifically interested in type II supernovae and their stellar precursors in my following questions.

Stars above the mass of 8 times the size of the sun go through successive burning phases in their cores, resulting in a cores of iron eventually being formed shortly before collapse. I've often heard it said that "iron can't be fused and so the process [supporting the core from collapse] halts", but wouldn't it be more correct to say that the fusion of iron takes energy away instead? Iron can be fused - it just detracts from the energy supporting the core from collapse, right?

During this phase of the life of the star, light is emitted with sufficient energy to break nuclei apart within the core, allowing for the precursor elements to be produced again. How much extra support does this add during the final phase of the life of the star? Does it buy any extra time to help support the star from collapse? My guess would be that the energy that went into splitting the nuclei apart is going to be more than would be produced by fusing the elements again, though I haven't run the numbers to confirm this.

My third question relates to the fate of any companion stars and planets that may be close to the star. During the collapse of the core a truly vast amount of neutrinos are emitted with extreme energies, enough that they have been detected as a precursor to the light of an explosion from significant distances. I would assume that the neutrinos are emitted in such quantities, and with such energies, that they may result in the destruction of anything within the solar neighbourhood? Let alone the explosion of the superheated radioactive gas blasted outwards by the supernova itself. I imagine anything nearby would struggle during the latter phases of the stars life, without even considering the supernova.

How far would be considered as being safe for this sort of supernova?

RyanJ said:

How far would be considered as being safe for this sort of supernova?

Depends on what you mean by “safe”….
Is a safe distance one at which you won’t be instantly vaporized, or one where it’s safe to watch the explosion without protective eyewear, or something in between?

But you’ll have way more fun with this problem if you work it out yourself. Figure a supernova releases ##10^{44}## joules. You can imagine all that energy flowing through a sphere of radius ##r## and area ##4\pi r^2## with the supernova at its center, and calculate the number of joules per square meter.

Try the calculation, make reasonable assumptions about how long it takes the supernova to release that energy. (For comparison, the sun delivers something like 1000 joules per square meter per second to the surface of the earth, so that’s “protective eyewear” territory).

You will conclude that no double-digit number of light years is going to safe. But set up a spreadsheet, play with the ranges of values you find with Google, just for fun search Wikipedia for “hypernova”…

This thread reminds me of the xkcd: What If article, "LETHAL NEUTRINOS."



https://what-if.xkcd.com/73/

Nugatory said:

Try the calculation, make reasonable assumptions about how long it takes the supernova to release that energy. (For comparison, the sun delivers something like 1000 joules per square meter per second to the surface of the earth, so that’s “protective eyewear” territory).

You will conclude that no double-digit number of light years is going to safe. But set up a spreadsheet, play with the ranges of values you find with Google, just for fun search Wikipedia for “hypernova”…

Will you? I seem to make a different conclusion.
For Sun, note that Sun is 3% bigger and 6% brighter in winter. We do not notice. We notice Sun being lower in sky, rising and setting further south, short days - but not that Sun is brighter.
It follows that if the sky contained a star 6% the brightness of Sun, we would notice it - such a star high in sky would flood the landscape with light brighter than that of Sun low above horizon! - but it would not actually affect us with its heat.
The 1054 supernova was 6500 ly away, and it shone at -6... at peak. It was visible in daytime for just 23 days.
Sun is -26,8. 6% solar luminosity is -23,7.
SN1054 at 4,3 ly should be something like -22, right? Just over 1% of solar luminosity.
What forms of energy from a nearby supernova would harm us the way ordinary perihelion does not?

RyanJ said:

I've often heard it said that "iron can't be fused and so the process [supporting the core from collapse] halts", but wouldn't it be more correct to say that the fusion of iron takes energy away instead? Iron can be fused - it just detracts from the energy supporting the core from collapse, right?

Close. The final chain of fusion that occurs in stars is the silicon burning chain, where alpha particles are fused with nuclei to create successively more massive nuclei. The binding energy per nucleon of each element in the chain is as follows:

Si-28: 8.447 (MeV)
S-32: 8.493
Ar-36: 8.519
Ca-40: 8.551
Ti-44: 8.533
Fe-52: 8.609
Ni-56: 8.642

As you can see, Ni-56 has more binding energy per nucleon than Fe-52, so there is a net energy gain from the chain. Note that the binding energy per nucleon of Ti-44 is less than that of Ca-40. Luckily the next few steps after titanium all release energy too, so that a quick series of fusion events can proceed past calcium and titanium. Nickel is actually the final element to be produced in large quantities from fusion inside massive stars. When the supernova happens, much of this nickel is ejected into space where it decays to cobalt-56 and then to iron-56, which is stable. I believe a good portion of the afterglow of a supernova is from the energy emitted from the decay of nickel-56 and perhaps cobalt-56 (half-lives of 6 days and 77 days respectfully).

The issue with Ni-56 is two-fold:

1. While Ni-56 doesn't have the highest binding energy per nucleon out of all isotopes/elements, the binding energy per nucleon of any possible isotope in this chain is lower than that of Ni-56. Here are the next three isotopes in the chain:
Zn-60: 8.583
Ge-64: 8.530
Se-68: 8.476
So no energy can be gained from further fusion.*

2. Photodisintegration increases with temperature, and the temperature is so high towards the end of this chain that fusion can't keep up with it and the chain effectively stops.

*Note that the fusion chain is actually fusing alpha particles to nuclei, which continues to release energy up until tin-100 or so. But those alphas originally came from nuclei split during photodisintegration and we can just look at the energy per nucleon of the nuclei to determine where the chain stops producing energy.

What would be the binding energies of:
Cu-60 vs. Co-56?
Ni-60 vs. Fe-56?

RyanJ said:

During this phase of the life of the star, light is emitted with sufficient energy to break nuclei apart within the core, allowing for the precursor elements to be produced again. How much extra support does this add during the final phase of the life of the star?

I'm not sure about stars in the range of 10ish stellar masses, but in certain stars of 130-250 solar masses the absorption of too many photons inside the core leads to a pair-instability supernova, as it leads to an overall reduction in core pressure.

snorkack said:

What would be the binding energies of:
Cu-60 vs. Co-56?
Ni-60 vs. Fe-56?

This site should have everything you want: http://barwinski.net/isotopes/query_select.php

Drakkith said:

This site should have everything you want: http://barwinski.net/isotopes/query_select.php

Ah, thanks!
Turns out that the daughters of Ni-56 are also resistant to just adding α when alternative for that α is more Ni-56:
Co-56 8,695 Cu-60 8,666
Fe-56 8,790 Ni-60 8,781
Yet Fe-56 is not the most strongly bound nucleus, by that measure:
Fe-58 8,792
Ni-62 8,794.

RyanJ said:

During this phase of the life of the star, light is emitted with sufficient energy to break nuclei apart within the core, allowing for the precursor elements to be produced again. How much extra support does this add during the final phase of the life of the star? Does it buy any extra time to help support the star from collapse?

You have it backward, the fact that photodisintegrating nuclei removes energy means that support is taken away by that process, the opposite of providing "extra" support. Indeed this is the crucial aspect of the core collapse, at high enough energy scales there appear all kinds of new "endothermic" processes (processes like photodisintegration that remove energy from the core and reduce its support). These processes produce a kind of endothermic runaway, more or less the opposite of the exothermic runaway in a thermonuclear explosion. The result is core collapse, which produces an explosion somewhat ironically by releasing gravitational energy.

Ken G said:

core collapse (...) ironically

*chuckle*

Yeah, that irony could be the basic reason why supernova simulations have such a tough time actually getting explosions to happen! Nature may "abhor a vacuum", but it seems to like irony even less...

Ken G said:

You have it backward, the fact that photodisintegrating nuclei removes energy means that support is taken away by that process, the opposite of providing "extra" support. Indeed this is the crucial aspect of the core collapse, at high enough energy scales there appear all kinds of new "endothermic" processes (processes like photodisintegration that remove energy from the core and reduce its support). These processes produce a kind of endothermic runaway, more or less the opposite of the exothermic runaway in a thermonuclear explosion. The result is core collapse, which produces an explosion somewhat ironically by releasing gravitational energy.

Ahhh. That does make sense, thank you for clarifying that segment.