<p>Imagine the X-ray pulse from the GRB travelling outwards like a spherical shell. If it then encounters a slab of dust on its way into our galaxy, then the intersection of the X-ray shell and the slab will be a ring drawn on the slab.</p>
<p>Some of the X-rays will then be scattered and change their original direction and head towards the Earth. We then see those scattered X-rays as a ring, coming from the slab, surrounding the GRB. Because the scattered X-rays have travelled further than the direct photons from the GRB, then the X-ray ring will appear after the GRB.</p>
<p><a href="https://i.sstatic.net/ff6uN.png" rel="noreferrer"><img src="https://i.sstatic.net/ff6uN.png" alt="GRB dust ring geometry" /></a></p>
<p>Although the scattering is not equal in all directions there is no preferred angle - it is a form of Rayleigh scattering (see <a href="https://iopscience.iop.org/article/10.1086/306018/fulltext/37392.text.html" rel="noreferrer">Smith &amp; Dwek 1997</a>). In that respect the Physics is not the same as visible light scattering from ice crystals in the Earth's atmosphere. What determines the deflection angle is how close the dust slab is.</p>
<p>As a formula
<span class="math-container">$$
\theta (t) = \bigg [ \frac{2c}{d} \frac{(1-x)}{x} (t-{T}_0) \bigg ]^{0.5},
$$</span>
where <span class="math-container">$T_0$</span> is the time of the GRB, <span class="math-container">$c$</span> is the speed of light, <span class="math-container">$t$</span> is the time the ring is observed <span class="math-container">$d$</span> is the distance to the GRB and <span class="math-container">$x$</span> is the ratio of the distance to the dust slab and <span class="math-container">$d$</span>. Usually <span class="math-container">$x\ll 1$</span> and this simplifies to (<a href="https://academic.oup.com/mnras/article/472/2/1465/4082223" rel="noreferrer">Pintore et al. 2017</a>)
<span class="math-container">$$\theta (t) = \bigg [ \frac{2c (t-{\rm T}_0)}{d_{\rm dust}} \bigg ]^{0.5}.
$$</span></p>

CC BY-SA 4.0
2022-10-18T15:33:10.963Z
Usually I wait longer than this to accept an answer but this seems unlikely to be surpassed. At least one thing I read mentioned "slabs" of dust, and I was familiar with the idea that a pulse of radiation creates a spherical shell. But I didn't have that thought in my head because I was so entranced by the spurious comparison to sun halos. Good answer!
Mark Foskey
CC BY-SA 4.0
2022-10-18T07:34:58.110Z
2022-10-18T08:37:41.980Z
2022-10-18T08:37:41.980Z
ProfRob
<p>This <a href="https://apod.nasa.gov/apod/ap221017.html" rel="noreferrer">APOD image</a> shows two nested rings in the X-ray spectrum surrounding a gamma ray burst. Or maybe I should consider it to be a single ring around a disc. The APOD description says,</p>
<blockquote>
<p>...the X-rays also bounced off regions high in dust right here in our
Milky Way Galaxy, creating the unusual reflections. The greater the
angle between reflecting Milky Way dust and the GRB, the greater the
radius of the X-ray rings, and, typically, the longer it takes for
these light-echoes to arrive.</p>
</blockquote>
<p>This seems reminiscent of how sun halos form, where there is a preferential angle for sunlight to reflect off of ice crystals in the atmosphere, leading to a ring at a certain angular radius around the sun. But the second sentence quoted above seems to suggest that the angle could be different, although it says nothing about what could make it different. Is the analogy with sun halos valid? Is it known what determines the deflection angle?</p>
<p><a href="https://i.sstatic.net/3g7fk.png" rel="noreferrer"><img src="https://i.sstatic.net/3g7fk.png" alt="Bright spot in disk with gray ring surrounding" /></a></p>

CC BY-SA 4.0
2022-10-18T05:25:42.550Z
It can't be a perfect analogy, since halos are formed by *internal* reflections in crystals Perhaps a corona would be a better analogue, those are formed by diffraction rather than reflection.
James K
<x-ray><gamma-ray-bursts>
Are X-ray rings around GRBs analogous to sun halos?
Mark Foskey
# Are X-ray rings around GRBs analogous to sun halos?

(Asked by: username_0 on 2022-10-18)

This [APOD image](https://apod.nasa.gov/apod/ap221017.html) shows two nested rings in the X-ray spectrum surrounding a gamma ray burst. Or maybe I should consider it to be a single ring around a disc. The APOD description says,

> ...the X-rays also bounced off regions high in dust right here in our
> Milky Way Galaxy, creating the unusual reflections. The greater the
> angle between reflecting Milky Way dust and the GRB, the greater the
> radius of the X-ray rings, and, typically, the longer it takes for
> these light-echoes to arrive.

This seems reminiscent of how sun halos form, where there is a preferential angle for sunlight to reflect off of ice crystals in the atmosphere, leading to a ring at a certain angular radius around the sun. But the second sentence quoted above seems to suggest that the angle could be different, although it says nothing about what could make it different. Is the analogy with sun halos valid? Is it known what determines the deflection angle?

[![Bright spot in disk with gray ring surrounding](https://i.sstatic.net/3g7fk.png)](https://i.sstatic.net/3g7fk.png)

**Tags:** x-ray, gamma-ray-bursts
> **Comments:**
> **username_1**: It can't be a perfect analogy, since halos are formed by *internal* reflections in crystals Perhaps a corona would be a better analogue, those are formed by diffraction rather than reflection.  

---

## 1 answers

### Answer by username_2 on 2022-10-18. Score: 6 | Accepted answer

Imagine the X-ray pulse from the GRB travelling outwards like a spherical shell. If it then encounters a slab of dust on its way into our galaxy, then the intersection of the X-ray shell and the slab will be a ring drawn on the slab.

Some of the X-rays will then be scattered and change their original direction and head towards the Earth. We then see those scattered X-rays as a ring, coming from the slab, surrounding the GRB. Because the scattered X-rays have travelled further than the direct photons from the GRB, then the X-ray ring will appear after the GRB.

[![GRB dust ring geometry](https://i.sstatic.net/ff6uN.png)](https://i.sstatic.net/ff6uN.png)

Although the scattering is not equal in all directions there is no preferred angle - it is a form of Rayleigh scattering (see [Smith & Dwek 1997](https://iopscience.iop.org/article/10.1086/306018/fulltext/37392.text.html)). In that respect the Physics is not the same as visible light scattering from ice crystals in the Earth's atmosphere. What determines the deflection angle is how close the dust slab is.

As a formula
$$
\theta (t) = \bigg [ \frac{2c}{d} \frac{(1-x)}{x} (t-{T}\_0) \bigg ]^{0.5},
$$
where $T\_0$ is the time of the GRB, $c$ is the speed of light, $t$ is the time the ring is observed $d$ is the distance to the GRB and $x$ is the ratio of the distance to the dust slab and $d$. Usually $x\ll 1$ and this simplifies to ([Pintore et al. 2017](https://academic.oup.com/mnras/article/472/2/1465/4082223))
$$\theta (t) = \bigg [ \frac{2c (t-{\rm T}\_0)}{d\_{\rm dust}} \bigg ]^{0.5}.
$$

> **Comments:**
> **username_0**: Usually I wait longer than this to accept an answer but this seems unlikely to be surpassed. At least one thing I read mentioned "slabs" of dust, and I was familiar with the idea that a pulse of radiation creates a spherical shell. But I didn't have that thought in my head because I was so entranced by the spurious comparison to sun halos. Good answer!  

---

