Release of trapped light during supernova

[jml53]

TL;DR Summary

Does the release of light transiting from the core of a star to its surface contribute to brilliance of a supernova? Q

I was talking with my 14 year old son and he asked me a great question. He said he was thinking about the fact that it takes 100,000-1,000,000 years for energy released in the core of a star to reach the surface and be released as a photon. The journey is a random walk of microsteps of emission and reabsorbtion.

With that in mind, he asked if all of those in transit photons are released as a star goes supernova, and if so, how much does that contribute to the brightness of the supernova itself? I have a B.S. in Applied Physics (from 30 years ago) and still follow developments closely. I'd never heard anyone raise or address this question. I was pretty impressed by his insight. Does anyone know if this has been studied and if there is anything to his question?

 

[Drakkith]

I was talking with my 14 year old son and he asked me a great question. He said he was thinking about the fact that it takes 100,000-1,000,000 years for energy released in the core of a star to reach the surface and be released as a photon. The journey is a random walk of microsteps of emission and reabsorbtion.

This is half-accurate. It is true that photons are being emitted and absorbed all the time, but energy transfer inside a plasma, especially a plasma that is stratified and under compression by pressures in a range from 0.1 billion to 100 billion atmospheres, is extremely complicated. There are multiple distance scales to take into account and radiation transfer vs convection transfer can dominate at different distances from the core for different stars. Plus, energy transfer from one particle to another can take place without one emitting a photon and another absorbing it. Plain old collisions can and do occur.

With that in mind, he asked if all of those in transit photons are released as a star goes supernova, and if so, how much does that contribute to the brightness of the supernova itself?

Even though the energy transfer is complex, if we take a snapshot of every photon in the star at the time the collapse begins, we can say that no, these photons, on the whole, do not make it to our eyes. First, remember that the collapse process takes some non-zero amount of time, during which many of these photons will be absorbed by the plasma and many more will be emitted. Even after the collapse is 'complete', the stellar remnant (neutron star or black hole) is surrounded by an opaque cloud of gas and dust that is still extremely hot near the center and cooler near the outside. Only the photons emitted by the outer layers will be able to escape to reach out eyes. Essentially, even in death, we only see the light emitted from the equivalent of a photosphere. Virtually all other photons will be absorbed before they get out of the cloud.

Only once this cloud of plasma has expanded enough will photons emitted from the inner layers be able to start making it through the cloud to get to us. Remember that the Sun's photosphere is only about 100 km thick, about 0.014% of the Sun's radius. And it is amongst the least dense layers of the Sun. The inner layers are MUCH denser and it takes some time for these layers to be blown off and expanded enough for light to transmit through.

 

[Drakkith]

To clarify a bit, we can see the Sun, or any star, only because light is emitted from the upper layer, which we call the photosphere, and the reason the photosphere emits light is because it is hot. A supernova is much the same way. The outermost layers of the expanding shell of gas is initially still very hot. Hot enough to emit the same visible light it was previously emitting before the collapse. Some portion of the vast amount of energy released during the collapse of the core is absorbed by the rest of the star, heating it up. This increased temperature combined with the increasing size of the expanding shell is why a supernova brightens so drastically. Hotter gas means more light emitted and a larger size means more surface area from which to emit light. This light is NOT coming from inside the supernova, that light is absorbed almost as soon as it is emitted. It is coming directly from the 'surface' of the expanding shell.

There are some other facts to consider when talking about brightness, like opaqueness from the ionization of hydrogen and the energy released by the decay of radioactive isotopes, but the main thing to take away here is that a supernova is so bright because the expanding shell is very hot and very large.

At least, that's my understanding of the process. I'm not an astrophysicist or anything so I could be mistaken.
@Ken G have I misunderstood anything here?

 

[Vanadium 50]

I would not use the word "photon" and I certainly wouldn't use the phrase "trapped light" as it will surely lead to incorrect conclusions.

The term astronomers use is "opacity". The interior of the sun is opaque and light produced gets absorbed. As it does, the plasma heats up and as it does so, emits light. The process continues - and yes, energy transport via this process is slow. (Although this statement is somewhat academic, as it is not the only energy transport proess, and we can't sensibly talk about the transport of this erg and that erg)

To answer your question, yes, this is part of the energy in a supernova, but it is normally considered just as part of the plasma temperature.

 

[snorkack]

How much of Sun´s heat content in any time is in the energy of photons inside, and how much in kinetic energy of electrons and nuclei?

 

[jml53]

Drakkith and Vanadium 50 thanks for your feedback. You've articulated a lot of the ideas that my instincts were pointing me to.

Based on your posts, I think a good way for me explain it to him is that the bulk of the star has a very high energy density, largely described by the kinetic energy (temperature) of the plasma at any given depth. The energy is constantly in motion through convection, as well as through absorption and emission of photons (thermal radiation).

As you've pointed out, we normally only see photons emitted from the very outer layer of the photosphere; the rest of the star is opaque, preventing photons from traveling any appreciable distance.

When the supernova occurs a large portion of its mass is expelled, leaving the remainder of the core to collapse. The plasma density in the portion that is expelled drops considerably as a function of Volume over time and the pressure drops accordingly. Conservation of energy says that the energy that was captured in the bulk is still there, but it is likely that the volume of the transparent outer layer will go up quickly after the explosion, allowing the release of the stored energy density at a much faster rate. Similarly, the energy of the supernova itself, including things like "the energy released by the decay of radioactive isotopes" will further heat this expanding plasma, adding considerably to its initial energy content.

I'm just pleased that he's asking good questions. Thanks for helping me get him a good answer.

 

[snorkack]

Drakkith and Vanadium 50 thanks for your feedback. You've articulated a lot of the ideas that my instincts were pointing me to.

Based on your posts, I think a good way for me explain it to him is that the bulk of the star has a very high energy density, largely described by the kinetic energy (temperature) of the plasma at any given depth.

Directly tied to temperature and composition.
Some energy is carried by the kinetic energy of nuclei and electrons, some by the photons.
In stars, generally, the total energy of photons is a minority of standing stock of energy - that ties to the stability conditions.

The energy is constantly in motion through convection, as well as through absorption and emission of photons (thermal radiation).

And conduction.
Although photons carry a small share of energy present at any time, they move at speed of light, for small distances, while electrons and nuclei move at a small fraction of speed of light. Therefore energy moved by photons is a bigger fraction of energy moved than the energy held by photons is of total energy held.
But both photons, electrons and nuclei have short free paths in dense gas. Thus when convection is possible, it carries energy along with it.

 

[Ken G]

To clarify a bit, we can see the Sun, or any star, only because light is emitted from the upper layer, which we call the photosphere, and the reason the photosphere emits light is because it is hot. A supernova is much the same way. The outermost layers of the expanding shell of gas is initially still very hot. Hot enough to emit the same visible light it was previously emitting before the collapse. Some portion of the vast amount of energy released during the collapse of the core is absorbed by the rest of the star, heating it up. This increased temperature combined with the increasing size of the expanding shell is why a supernova brightens so drastically. Hotter gas means more light emitted and a larger size means more surface area from which to emit light. This light is NOT coming from inside the supernova, that light is absorbed almost as soon as it is emitted. It is coming directly from the 'surface' of the expanding shell.

There are some other facts to consider when talking about brightness, like opaqueness from the ionization of hydrogen and the energy released by the decay of radioactive isotopes, but the main thing to take away here is that a supernova is so bright because the expanding shell is very hot and very large.

At least, that's my understanding of the process. I'm not an astrophysicist or anything so I could be mistaken.
@Ken G have I misunderstood anything here?

Yes, that all sounds good to me. The OP is correct that the 14 year old is showing great insight in asking that question, it shows they understand the concept of a conservation law. As pointed out above, it's not necessarily conservation of photons, because the photons can get absorbed during the process, but it is conservation of energy. It happens to turn out, however, that the energy contained inside a star prior to supernova is much less than the energy of the supernova itself (which here comes from the gravitational collapse of the core of the star).

I also think the 14 year old is ready to be told the gist of the virial theorem, which states that the total kinetic energy (which basically includes the photon energy, we don't need to bother the kid with relativity yet) is half the total gravitational potential energy. For many of the kinds of stars that undergo core collapse (a lot more massive than the Sun), the photon energy is not that much less than the gas kinetic energy, and indeed there is a similar number of photons in the star as gas particles. But since all that energy is characterized by the gravitational energy, and the latter is about to go rocketing up when the core collapses, the initial energy in the star, including the photons, is not terribly important.

I suspect the 14 year old will also appreciate the fact that most of the energy released in the supernova is not in the photons, but in neutrinos. Then the next most important energy pathway is in the kinetic energy of the gas that flies off at speeds that would cross the Earth in just a second. The amount of light energy released is a distant third!

 

[jml53]

Thanks again for your insights. I'll share this all with my son, he really appreciated your initial feedback and this is even better. I'd forgotten to mention the neutrino energy. I've talked to him about them in other context previously and this will help tie it all together.