<p>Simple answer: the matter density is incredibly high. The energy density may permit coherent scattering of the neutrinos among the nucleons. There is a lot of matter, and it is not transparent to neutrinos. The mean free path for scattering is very small, and the neutrinos are bouncing back and forth.</p>

<p><a href="http://www.int.washington.edu/talks/WorkShops/RIA/People/Liebendoerfer_M/liebendoerfer.RIA.pdf" rel="nofollow noreferrer">Neutrino Transport in Core Collapse Supernovae</a></p>

<p><a href="https://arxiv.org/pdf/nucl-th/0211003.pdf" rel="nofollow noreferrer">Neutrino mean free path and in-medium nuclear interaction</a></p>

CC BY-SA 3.0
2017-04-24T17:31:28.283Z
2017-04-24T18:08:44.457Z
2017-04-24T18:08:44.457Z
dantopa
<p>You are barking up the wrong tree. The passage you quote is talking about how neutron stars cool <em>after</em> they have formed. Neutrinos are emitted in a burst lasting a few seconds after core collapse but then continue to be produced whilst the neutron star is hot.</p>

<p>Neutrino production is indeed fearsomely high in the first fraction of a second after core collapse. However, the hot, dense material is opaque to neutrinos (mean free paths of 10-100 m) and they diffuse outward in much the same way as photons from the Sun. But the timescale for this is only about 10 seconds. Once the neutron star has cooled to about $10^{10}$ K, the fermions (neutrons, protons, electrons) in the neutron star become degenerate and the neutron star becomes transparent to neutrinos because only the small fraction of fermions within $kT$ of their Fermi surfaces can interact with thermal neutrinos and the neutrino mean free paths increase as $\sim T^{-2}$ and the production rate of neutrinos falls as $T^6$.</p>

<p>Still, the neutron star would then radiate away all of its thermal energy within a further 10 seconds, but the main neutrino-generating reactions - cycles of beta and inverse beta decay (aka the <a href="https://en.m.wikipedia.org/wiki/Urca_process" rel="noreferrer">Urca process</a>) become blocked (except perhaps right in the core at the highest densities) by an inability to simultaneously conserve energy and momentum in the degenerate gas. Instead, neutrinos continue to be generated by the modified Urca process at a <em>much</em> slower rate, using "bystander" baryons to conserve momentum.
$$ n + n \rightarrow n + p + e + \bar{\nu}_e$$
$$ n + p + e \rightarrow n + n + \nu_e$$</p>

<p>Though the efficiency of this process is low, so is the thermal energy contained within degenerate gases. Thus rapid cooling still occurs. I am not clear where Wikipedia gets its information from, but cooling to even a surface temperature of $10^{6}$K (the interiors are 1-2 orders of magnitude hotter) is likely to take longer than a few years - somewhere between 100 and $10^4$ years, unless direct URCA processes are somehow permitted (e.g. in quark matter, see <a href="https://arxiv.org/abs/astro-ph/0402143" rel="noreferrer">Yakovlev &amp; Pethick 2004, for a thorough review of neutron star cooling</a>).</p>

CC BY-SA 3.0
2017-04-24T19:49:20.367Z
2017-04-24T20:18:23.607Z
2017-04-24T20:18:23.607Z
ProfRob
<p>After a supernova neutrinos get released by electron capture,and they escape in a flood.</p>

<p>My guess is because they have tiny masses and get momentum from momentum conservation.</p>

<blockquote>
  <p>However, the huge number of neutrinos it emits carry away so much energy that the temperature of an isolated neutron star falls within a few years to around $10^6$ kelvin.</p>
</blockquote>

<p>Reference : <a href="https://en.wikipedia.org/wiki/Neutron_star#Mass_and_temperature" rel="nofollow noreferrer">Wikipedia</a>,<a href="http://www.astro.umd.edu/~miller/nstar.html" rel="nofollow noreferrer">Original</a>(wikipedia reference)
Wouldn't most of the neutrinos scatter after supernova almost instantaneously before a formation of a neutron star?</p>

CC BY-SA 3.0
2017-04-24T11:56:48.307Z
You might want to look up the estimates for how long it takes a plain old photon in the interior of our sun to reach the surface and "escape" into space.   I think the mechanism is similar here: neutrinos have a much smaller cross-section but the neutron star is kinda dense.
Carl Witthoft
CC BY-SA 3.0
2017-04-24T12:11:00.173Z
So in general it takes at minimum around 4000 years for a photon to travel from interior of the Sun to the [surface](https://image.gsfc.nasa.gov/poetry/ask/a11354.html).Anyway according to [this](https://physics.stackexchange.com/questions/277212/how-do-neutrinos-pass-through-the-sun-so-quickly) they shouldn't slow down that much because they don't interact that much as photons.
kingW3
CC BY-SA 3.0
2017-04-24T13:09:23.673Z
FYI, I think 4,000 years is a gross underestimate. It should be on the order of a million years plus or minus (and your source even says this is a possible timescale).
zephyr
CC BY-SA 3.0
2017-04-24T13:12:52.643Z
Can you provide a source for your (presumably) quoted text?
zephyr
CC BY-SA 3.0
2017-04-24T18:42:27.043Z
@zephyr Agreed about 4,000 years,that's why I said minimum.Anyway I edited in the references
kingW3
<supernova><neutron-star><neutrinos>
Why do neutrinos take time to scatter?
kingW3
# Why do neutrinos take time to scatter?

(Asked by: username_0 on 2017-04-24)

After a supernova neutrinos get released by electron capture,and they escape in a flood.

My guess is because they have tiny masses and get momentum from momentum conservation.

> However, the huge number of neutrinos it emits carry away so much energy that the temperature of an isolated neutron star falls within a few years to around $10^6$ kelvin.

Reference : [Wikipedia](https://en.wikipedia.org/wiki/Neutron_star#Mass_and_temperature),[Original](http://www.astro.umd.edu/~miller/nstar.html)(wikipedia reference)
Wouldn't most of the neutrinos scatter after supernova almost instantaneously before a formation of a neutron star?

**Tags:** supernova, neutron-star, neutrinos

> **Comments:**
> **username_1**: You might want to look up the estimates for how long it takes a plain old photon in the interior of our sun to reach the surface and "escape" into space.   I think the mechanism is similar here: neutrinos have a much smaller cross-section but the neutron star is kinda dense.  
> **username_0**: So in general it takes at minimum around 4000 years for a photon to travel from interior of the Sun to the [surface](https://image.gsfc.nasa.gov/poetry/ask/a11354.html).Anyway according to [this](https://physics.stackexchange.com/questions/277212/how-do-neutrinos-pass-through-the-sun-so-quickly) they shouldn't slow down that much because they don't interact that much as photons.  
> **username_2**: FYI, I think 4,000 years is a gross underestimate. It should be on the order of a million years plus or minus (and your source even says this is a possible timescale).  
> **username_2**: Can you provide a source for your (presumably) quoted text?  
> **username_0**: @username_2 Agreed about 4,000 years,that's why I said minimum.Anyway I edited in the references  

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## 2 answers

### Answer by username_4 on 2017-04-24. Score: 6 | Accepted answer

You are barking up the wrong tree. The passage you quote is talking about how neutron stars cool *after* they have formed. Neutrinos are emitted in a burst lasting a few seconds after core collapse but then continue to be produced whilst the neutron star is hot.

Neutrino production is indeed fearsomely high in the first fraction of a second after core collapse. However, the hot, dense material is opaque to neutrinos (mean free paths of 10-100 m) and they diffuse outward in much the same way as photons from the Sun. But the timescale for this is only about 10 seconds. Once the neutron star has cooled to about $10^{10}$ K, the fermions (neutrons, protons, electrons) in the neutron star become degenerate and the neutron star becomes transparent to neutrinos because only the small fraction of fermions within $kT$ of their Fermi surfaces can interact with thermal neutrinos and the neutrino mean free paths increase as $\sim T^{-2}$ and the production rate of neutrinos falls as $T^6$.

Still, the neutron star would then radiate away all of its thermal energy within a further 10 seconds, but the main neutrino-generating reactions - cycles of beta and inverse beta decay (aka the [Urca process](https://en.m.wikipedia.org/wiki/Urca_process)) become blocked (except perhaps right in the core at the highest densities) by an inability to simultaneously conserve energy and momentum in the degenerate gas. Instead, neutrinos continue to be generated by the modified Urca process at a *much* slower rate, using "bystander" baryons to conserve momentum.
$$ n + n \rightarrow n + p + e + \bar{\nu}\_e$$
$$ n + p + e \rightarrow n + n + \nu\_e$$

Though the efficiency of this process is low, so is the thermal energy contained within degenerate gases. Thus rapid cooling still occurs. I am not clear where Wikipedia gets its information from, but cooling to even a surface temperature of $10^{6}$K (the interiors are 1-2 orders of magnitude hotter) is likely to take longer than a few years - somewhere between 100 and $10^4$ years, unless direct URCA processes are somehow permitted (e.g. in quark matter, see [Yakovlev & Pethick 2004, for a thorough review of neutron star cooling](https://arxiv.org/abs/astro-ph/0402143)).


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### Answer by username_3 on 2017-04-24. Score: 1

Simple answer: the matter density is incredibly high. The energy density may permit coherent scattering of the neutrinos among the nucleons. There is a lot of matter, and it is not transparent to neutrinos. The mean free path for scattering is very small, and the neutrinos are bouncing back and forth.

[Neutrino Transport in Core Collapse Supernovae](http://www.int.washington.edu/talks/WorkShops/RIA/People/Liebendoerfer_M/liebendoerfer.RIA.pdf)

[Neutrino mean free path and in-medium nuclear interaction](https://arxiv.org/pdf/nucl-th/0211003.pdf)


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