What does it mean for a proto-neutron star to be opaque to neutrinos?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $$ f_\mathrm{esc} \equiv \frac{I(\tau)}{I_0} = e^{-\tau}. $$