I have read that a proto-neutron star is opaque to neutrinos but I do not really know what does means. Could someone explain this to me? I am a mathematician, not a physicist so I found it hard to understand these things.


3 Answers 3


Opaque in this context means the same thing as it would optically, for light. The neutrinos do not pass through and are either scattered or absorbed due to the density of the matter in the star.

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    $\begingroup$ "the density of the matter in the start" touches on the complete answer, I think. Some particles interact with the electromagnetic force, but neutrinos only interact with the weak force (and gravity). The reason this matters is that the range of the EM force is much greater than the range of the weak force. Therefore less dense matter leaves significant gaps between the weak force generated by constituent particles, and the neutrinos will go right through those gaps, and are statistically less likely to encounter a weak force that blocks them. $\endgroup$ Jan 12, 2022 at 17:33

You say you're a mathematician, so let's put this in terms of math. Let's say you have a source of neutrinos with an initial intensity $I_0$, in units of $\text{neutrinos/sec/m}^2$, (that is, the number of neutrinos passing through a surface area per second. This beam of neutrinos is passing through a dense medium of density $\rho$, in units of $\text{kg/m}^3$. After traveling a distance $x$, in units of $m$, the initial intensity will have been attenuated (via absorption or scattering) to an intensity of

$$I(x) = I_0 e^{-\kappa\rho x}$$

In this equation, $\kappa$ is the opacity of the medium, in units of $m^2/kg$ that the neutrinos are passing through. Hopefully you can see that $I(x)$ decreases as the neutrinos pass through the medium (i.e., as $x$ increases). This implies the intensity is dying off as the neutrino beam gets absorbed and reflected while traveling through the medium.

By increasing the opacity, the intensity falloff increases, meaning the neutrinos have a harder and harder time penetrating through the medium. If the opacity is very low (or even zero) such that most of the neutrinos (or all of them) can pass through the medium, we would say it is transparent. If the opacity is very high such that most of the neutrinos cannot pass through the medium, we would say it is opaque.

  • $\begingroup$ Nice answer, but isn't "have a harder and harder time penetrating through the medium" slightly imprecise? $\endgroup$
    – shamalaia
    Jan 12, 2022 at 10:16
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    $\begingroup$ @shamalaia : This is why mathematicians (and physicists and others) write the equation down -- so that minor imprecisions and ambiguities in natural language can be resolved by glancing at the equation rather than by writing unreadable, tortuous prose. For an example of the latter, go read Fermat in the original. $\endgroup$ Jan 12, 2022 at 13:50

An opaque medium is a medium which is not transparent.

Typically, the term refers to electromagnetic radiation, but in principle it may refer to any entity, also neutrinos. A medium can be opaque to one kind of particles, but transparent to another. For instance, Earth's atmosphere is largely opaque to UV radiation, but transparent to visible light.

Optical depth

"Not being transparent" is a somewhat… opaque, or at least imprecise term, since particles always have an ever-so-small probability of penetrating the medium. In the theory of radiative transfer, we also use the terms "optically thin" for transparent, and "optically thick" for opaque, and what we mean is that the probability of escaping the medium is "very large" and "very small" respectively.

This probability is given by the "optical depth" $\tau$, which increases with 1) the number density $n$ of the medium's particles, 2) the particles' cross section $\sigma$, and 3) the distance $r$ traveled through the medium. That is, $$ \tau \equiv n\,\sigma \,r. $$

Quantitatively, a medium is optically thin/transparent if $\tau$ is $\ll1$, and optically thick/opaque if $\tau\gg1$.

Note that, although the cross section does have units of an area, it shouldn't be thought of as an actual area; rather it is the probability of interaction. Note also that for radiation, the cross section typically depends on the wavelength of the incoming photon, so the optical depth also becomes wavelength-dependent (cf. the discussion above on UV vs. optical light).

In other words, a medium is more opaque the more particles it contains, the "larger" those particles are, and the more there is.

Absorption vs. scattering

If you consider a beam of particles traveling through a medium, its intensity $I$ may be reduced by either being absorbed and disappearing completely, or being scattered, in which case they continue their journey, but in another direction. The total, effective cross section can then be written as $\sigma_\mathrm{tot} = \sigma_\mathrm{abs} + \sigma_\mathrm{sca}$.

In some cases — e.g. if you're observing a star through a layer of dust — you may not care which of the two processes reduced the star's intensity; you just care how much light you're missing. In other cases — such as neutrinos escaping a neutron star — scattered particles still escape, just in another direction, and for every particles scattered out of your line of sight, another one is scattered into the line of sight (on average). In this case, you may use $\sigma_\mathrm{tot} = \sigma_\mathrm{abs}$.

(I don't know if neutrinos actually undergo both processes; maybe someone else can chime in here.)

Escape fraction

For a small distance $dr$, the particles of the medium cover a fractional area $n\sigma$, so a beam of initial intensity $I_0$ is reduced by $$ dI = -I_0 n\sigma dr = -I_0 d\tau. $$ Integrating along the path of the beam, covering a total optical depth $\tau$ reduces the intensity to $I(\tau) = I_0 e^{-\tau}$. The escape fraction is thus $$ f_\mathrm{esc} \equiv \frac{I(\tau)}{I_0} = e^{-\tau}. $$

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    $\begingroup$ I started writing this answer when no-one had answered, then got distracted, then finished. Only after posting were I able to see that others had answered in the meantime. I'll let my answer stand, but the other answers are also good :) $\endgroup$
    – pela
    Jan 12, 2022 at 11:01

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