<p>Neutrinos wouldn't pass through <a href="http://en.wikipedia.org/wiki/White_dwarf" rel="noreferrer">white dwarfs</a> or <a href="http://en.wikipedia.org/wiki/Neutron_star" rel="noreferrer">neutron stars</a> as easily as through Earth, measured by the scattering rate per distance, since these types of stars are very dense: white dwarfs about one million times denser than water, neutron stars about as dense as atomic nuclei.</p>

<p>To scatter a neutrino with a chance of > 50%, you "just" need the inner of a neutron star along 10 times the diameter of Earth's orbit around the sun:</p>

<p>The neutrino-nucleon scattering cross section is about $10^{-38}cm^2=10^{-42}m^2$ for neutrinos with energies above about 1 GeV, according to <a href="http://pdg.lbl.gov/2013/reviews/rpp2012-rev-nu-cross-sections.pdf" rel="noreferrer">this paper</a>. A <a href="http://en.wikipedia.org/wiki/Atomic_nucleus" rel="noreferrer">proton</a> has a <a href="http://en.wikipedia.org/wiki/Neutron" rel="noreferrer">diameter</a> of about $1.75\cdot 10^{-15}m$.
With $$1.75\cdot 10^{-15}m/10^{-42}m^2=1.75 \cdot 10^{27}$$ we need about $1.75\cdot 10^{27}$ nucleons in a row to scatter a neutrino. That's about $$1.75\cdot 10^{27}\cdot 1.75\cdot 10^{-15}m=3\cdot 10^{12}m=20\mbox{ a.u.}$$ as a rough estimate. (For simplicity I've neglected the exponentially decreasing intensity of the neutrino beam in the scattering medium.) Neutrinos will probably scatter a little more easily in a neutron star than in the same number of single traversed neutrons, but that's a guess based on the scattering cross section for heavier nuclei.</p>

CC BY-SA 3.0
2014-01-21T21:23:12.550Z
2014-01-21T21:35:48.363Z
2014-01-21T21:35:48.363Z
Gerald
<p>One impressive thing about neutrinos is that you would need about one light year of lead to stop a neutrino about <a href="http://www.sciencedaily.com/articles/n/neutrino.htm" rel="nofollow">50% of the time</a>. Which is due to neutrinos interacting only weakly with matter.</p>

<p>So, to answer your question, it would make $X$ time difference in neutrino opacity if you take you planet $X$ times larger in radius or $X$ times denser. However, the net effect will be still vanishingly small.</p>

<p>Also, molecular-level structures do not affect the interaction with neutrinos, but rather the nuclear structure.</p>

CC BY-SA 3.0
2014-01-17T14:49:40.503Z
Ok, so is the nuclear density of atoms in red or brown dwarf stars high enough to interact with neutrinos?
Cymatical
CC BY-SA 3.0
2014-01-17T15:24:20.970Z
It depends on what do you mean by interact. The Earth also interacts with neitrinos. However, for both the Earth and brown dwarfs neutrino opacity is small and close to zero.
Alexey Bobrick
CC BY-SA 3.0
2014-01-19T05:44:50.327Z
@AlexeyBobrick It might help to talk about interaction cross-section. You could be a bit more precise with your answer when you say "still vanishingly small".
astromax
CC BY-SA 3.0
2014-01-19T12:15:33.293Z
@astromax, given that the question is "interacts or not", the answer is simply "not", on the argument that $10^{4}-10^{6} \textrm{km}$ of ordinary matter is by about ten orders of magnitude smaller than a light year.
Alexey Bobrick
CC BY-SA 3.0
2014-01-19T16:02:52.900Z
Well, hold up a second. You're "impressive thing" in retrospect is incorrect. It's a light year of lead for it to interact 50% of the time. I'm editing your answer to reflect that. Secondly, I would still find it appropriate and informative to include a general discussion about interaction cross-section. Take it or leave it - no need to debate with me regarding this.
astromax
CC BY-SA 3.0
2014-01-19T16:16:55.870Z
@astromax: As is with any opacities, of course one light year is the damping length in an exponent. If I do add any further quantitative discussion, I will edit the question first. In the current state it seems rather unclear.
Alexey Bobrick
CC BY-SA 3.0
2014-01-17T13:50:46.240Z
2014-01-19T16:05:09.140Z
2014-01-19T16:05:09.140Z
Alexey Bobrick
<p>Would dense molecular structures on heavier planets (crystalline or other compounds which are generally unknown to us) allow neutrinos to pass through as easily as on Earth?</p>

<matter><neutrinos>
Neutrino interaction with dense molecular structures
Cymatical
# Neutrino interaction with dense molecular structures

(Asked by: username_0 on 2014-01-17)

Would dense molecular structures on heavier planets (crystalline or other compounds which are generally unknown to us) allow neutrinos to pass through as easily as on Earth?

**Tags:** matter, neutrinos


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## 2 answers

### Answer by username_1 on 2014-01-21. Score: 5

Neutrinos wouldn't pass through [white dwarfs](http://en.wikipedia.org/wiki/White_dwarf) or [neutron stars](http://en.wikipedia.org/wiki/Neutron_star) as easily as through Earth, measured by the scattering rate per distance, since these types of stars are very dense: white dwarfs about one million times denser than water, neutron stars about as dense as atomic nuclei.

To scatter a neutrino with a chance of > 50%, you "just" need the inner of a neutron star along 10 times the diameter of Earth's orbit around the sun:

The neutrino-nucleon scattering cross section is about $10^{-38}cm^2=10^{-42}m^2$ for neutrinos with energies above about 1 GeV, according to [this paper](http://pdg.lbl.gov/2013/reviews/rpp2012-rev-nu-cross-sections.pdf). A [proton](http://en.wikipedia.org/wiki/Atomic_nucleus) has a [diameter](http://en.wikipedia.org/wiki/Neutron) of about $1.75\cdot 10^{-15}m$.
With $$1.75\cdot 10^{-15}m/10^{-42}m^2=1.75 \cdot 10^{27}$$ we need about $1.75\cdot 10^{27}$ nucleons in a row to scatter a neutrino. That's about $$1.75\cdot 10^{27}\cdot 1.75\cdot 10^{-15}m=3\cdot 10^{12}m=20\mbox{ a.u.}$$ as a rough estimate. (For simplicity I've neglected the exponentially decreasing intensity of the neutrino beam in the scattering medium.) Neutrinos will probably scatter a little more easily in a neutron star than in the same number of single traversed neutrons, but that's a guess based on the scattering cross section for heavier nuclei.


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### Answer by username_2 on 2014-01-17. Score: 4

One impressive thing about neutrinos is that you would need about one light year of lead to stop a neutrino about [50% of the time](http://www.sciencedaily.com/articles/n/neutrino.htm). Which is due to neutrinos interacting only weakly with matter.

So, to answer your question, it would make $X$ time difference in neutrino opacity if you take you planet $X$ times larger in radius or $X$ times denser. However, the net effect will be still vanishingly small.

Also, molecular-level structures do not affect the interaction with neutrinos, but rather the nuclear structure.

> **Comments:**
> **username_0**: Ok, so is the nuclear density of atoms in red or brown dwarf stars high enough to interact with neutrinos?  
> **username_2**: It depends on what do you mean by interact. The Earth also interacts with neitrinos. However, for both the Earth and brown dwarfs neutrino opacity is small and close to zero.  
> **username_3**: @username_2 It might help to talk about interaction cross-section. You could be a bit more precise with your answer when you say "still vanishingly small".  
> **username_2**: @username_3, given that the question is "interacts or not", the answer is simply "not", on the argument that $10^{4}-10^{6} \textrm{km}$ of ordinary matter is by about ten orders of magnitude smaller than a light year.  
> **username_3**: Well, hold up a second. You're "impressive thing" in retrospect is incorrect. It's a light year of lead for it to interact 50% of the time. I'm editing your answer to reflect that. Secondly, I would still find it appropriate and informative to include a general discussion about interaction cross-section. Take it or leave it - no need to debate with me regarding this.  
> **username_2**: @username_3: As is with any opacities, of course one light year is the damping length in an exponent. If I do add any further quantitative discussion, I will edit the question first. In the current state it seems rather unclear.  

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