<p>In astrophysics, being &quot;optically thick&quot; means that light (and it is applied to photons of any frequency, not just visible light) cannot travel across or out of a body (which may be the source of the radiation) without being absorbed or scattered.</p>
<p>Mathematically this is often described in terms of an intensity traversing across a slab of material of thickness <span class="math-container">$x$</span> as
<span class="math-container">$$I = I_0 \exp(-\rho \sigma x)\ , $$</span>
where <span class="math-container">$I_0$</span> is the intensity entering the slab, <span class="math-container">$\rho$</span> is the density and <span class="math-container">$\sigma$</span> is the absorption/scattering cross-section in units of area per mass.</p>
<p>If the product <span class="math-container">$\rho \sigma x &gt; 1$</span> then little radiation emerges on the other side of the slab and it would be said to be &quot;optically thick&quot;. Clearly a high density, a high cross-section or a large physical thickness, will make something optically thick.</p>
<p>In the context of <em>this</em> question, &quot;optically thick to pair production&quot; means that the (gamma ray) photons produced inside a source cannot escape before interacting with a nucleus (or proton) and producing an electron-positron pair.</p>
<p>The cross-section for the pair-production interaction of gamma rays with matter increases (roughly) as the square of the atomic number of the nuclei present and with photon energy. This means that a source of a given size and physical composition will become &quot;optically thick&quot; to gamma rays above some threshold energy.</p>

CC BY-SA 4.0
2022-05-22T09:02:16.003Z
Excellent response, thanks very much.
georgeamccarthy
CC BY-SA 4.0
2022-05-22T07:03:06.687Z
2022-05-22T07:03:06.687Z
ProfRob
<p>I have seen several references in the high energy astrophysics literature (including the example below) which make reference to an astronomical source being &quot;thick to pair production&quot;.</p>
<p>What does this mean? Why would something be thick to pair production?</p>
<hr />
<p>from Broderick and Tchekhovskoy (2015) <a href="https://arxiv.org/abs/1506.04754" rel="nofollow noreferrer">Horizon-Scale Lepton Acceleration in Jets: Explaining the Compact Radio Emission in M87</a>:</p>
<blockquote>
<p>4.3. Post-Gap Cascade</p>
<p>Given the threshold seed photon energy of 1.2 meV, the minimum gamma-ray energy for which the seed photons bath is optically-thick to pair production is <span class="math-container">$\epsilon_{\gamma, min}$</span> = 640 GeV, hence the asymptotic Lorentz factor for M87 is (see Equation 27)</p>
<p><span class="math-container">$$\gamma_{\gamma \gamma} = 2.6 \times 10^6, \tag{59}$$</span></p>
<p>implying enhancement in the number density due to the postgap cascade of roughly <span class="math-container">$n_{\infty} / n_g$</span> = 670 (see Equation 28).</p>
</blockquote>

<binary-star><gamma-ray-bursts><high-energy-astrophysics>
What does it mean for something to be optically thick to pair production?
georgeamccarthy
# What does it mean for something to be optically thick to pair production?

(Asked by: username_0 on 2022-05-19)

I have seen several references in the high energy astrophysics literature (including the example below) which make reference to an astronomical source being "thick to pair production".

What does this mean? Why would something be thick to pair production?

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from Broderick and Tchekhovskoy (2015) [Horizon-Scale Lepton Acceleration in Jets: Explaining the Compact Radio Emission in M87](https://arxiv.org/abs/1506.04754):

> 4.3. Post-Gap Cascade
>
> Given the threshold seed photon energy of 1.2 meV, the minimum gamma-ray energy for which the seed photons bath is optically-thick to pair production is $\epsilon\_{\gamma, min}$ = 640 GeV, hence the asymptotic Lorentz factor for M87 is (see Equation 27)
>
> $$\gamma\_{\gamma \gamma} = 2.6 \times 10^6, \tag{59}$$
>
> implying enhancement in the number density due to the postgap cascade of roughly $n\_{\infty} / n\_g$ = 670 (see Equation 28).

**Tags:** binary-star, gamma-ray-bursts, high-energy-astrophysics


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## 1 answers

### Answer by username_1 on 2022-05-22. Score: 6 | Accepted answer

In astrophysics, being "optically thick" means that light (and it is applied to photons of any frequency, not just visible light) cannot travel across or out of a body (which may be the source of the radiation) without being absorbed or scattered.

Mathematically this is often described in terms of an intensity traversing across a slab of material of thickness $x$ as
$$I = I\_0 \exp(-\rho \sigma x)\ , $$
where $I\_0$ is the intensity entering the slab, $\rho$ is the density and $\sigma$ is the absorption/scattering cross-section in units of area per mass.

If the product $\rho \sigma x > 1$ then little radiation emerges on the other side of the slab and it would be said to be "optically thick". Clearly a high density, a high cross-section or a large physical thickness, will make something optically thick.

In the context of *this* question, "optically thick to pair production" means that the (gamma ray) photons produced inside a source cannot escape before interacting with a nucleus (or proton) and producing an electron-positron pair.

The cross-section for the pair-production interaction of gamma rays with matter increases (roughly) as the square of the atomic number of the nuclei present and with photon energy. This means that a source of a given size and physical composition will become "optically thick" to gamma rays above some threshold energy.

> **Comments:**
> **username_0**: Excellent response, thanks very much.  

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