<p><strong>tl;dr</strong>: On Earth, atmospheric molecules scatter blue light more than red. In contrast, on Mars atmospheric, <span class="math-container">$\mu\mathrm{m}$</span>-sized hematite dust particles happen to give the opposite effect.</p>
<hr />
<h1>Scattering theory</h1>
<p>First a bit of theory. Pretty pictures later.</p>
<h3>Scattering cross section</h3>
<p>Scattering of light of wavelength <span class="math-container">$\lambda$</span> on particles of radius <span class="math-container">$r$</span> may be described in general using <a href="https://en.wikipedia.org/wiki/Mie_scattering" rel="nofollow noreferrer">Mie theory</a>. Using the variable <span class="math-container">$x \equiv 2\pi r/\lambda$</span>, the scattering cross section <span class="math-container">$\sigma$</span> increases steeply with <span class="math-container">$x$</span> until <span class="math-container">$x\sim2\pi$</span>, after which it decreases again and starts to oscillate, as seen in this figure:</p>
<p><a href="https://i.sstatic.net/6HWkrjdB.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6HWkrjdB.png" alt="xsec" /></a></p>
<p><sup>Scattering cross section (normalised to the geometrical cross section) as a function of particle size-to-wavelength ratio (from <a href="https://en.wikipedia.org/wiki/Mie_scattering" rel="nofollow noreferrer">Wikipedia</a> with my own annotations).</sup></p>
<p>In other words, when a particle is much smaller than the wavelength of the light (<span class="math-container">$x\ll 1$</span>), the probability of scattering the light increases steeply with decreasing wavelength — in fact the cross section goes as <span class="math-container">$\lambda^{-4}$</span>. This regime is called Rayleigh scattering. The scattering efficiency reaches a maximum for particles roughly equal to the wavelength of the light (<span class="math-container">$x\simeq2\pi$</span>; green arrow in the figure). Hereafter, it falls again, exhibits some oscillations that have to do with interference in the particle, slowly converging to the &quot;geometric&quot; value <span class="math-container">$\sigma = \pi r^2$</span> (i.e. what would be expected classically).</p>
<p>In many (astrophysical) situations dust tends to be a distribution of sizes with more small than large particles (typically <span class="math-container">$n\propto r^{-3.5}$</span>), meaning that an ensemble of dust has much more particles that scatter short wavelengths (blue) than long wavelengths (red). Thus, blue light is filtered out and red light light tends to be transmitted, in effect reddening the light.</p>
<p>On the other hand, if the dust consists of particles of the same size, then the extinction may rise and fall with wavelength, as seen in the figure. For example, optical light with <span class="math-container">$\lambda \simeq [0.4\text{–}0.7] \,\mu\mathrm{m}$</span>, scattered by particles of size <span class="math-container">$0.7\,\mu\mathrm{m}$</span>, is scattered according to the orange region in the figure — i.e. red light is scattered more efficiently than blue light. The exact form depend on the refractive index of the scatterer.</p>
<h3>Phase function</h3>
<p>When a photon is scattered, it doesn't happen in a completely random direction, but follows a direction distribution given by the <em>phase function</em>. In the Rayleigh regime, there is a slight preference for forward and backward scattering, but for large <span class="math-container">$x$</span> there is a strong preference for forward scattering, as seen in this diagram (from <a href="http://pages.mtu.edu/%7Escarn/teaching/GE4250/scattering_lecture_slides.pdf" rel="nofollow noreferrer">these slides</a>):</p>
<p><a href="https://i.sstatic.net/joXoa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/joXoa.png" alt="phase" /></a></p>
<h1>Sunsets on Earth</h1>
<p>On Earth the sunlight is scattered primarily by molecules which are much smaller than the wavelength of the light, i.e. we're in the Rayleigh regime. Hence, blue light tends to be scattered a lot, creating the blue sky. As the blue light is filtered out of the line of sight (LOS) toward the Sun, its light — white to begin with — becomes more yellow. And as the Sun sets, its light has to go through a thicker layer of atmosphere, enhancing this <em>reddening effect</em> and giving a red sunset.</p>
<h1>Sunsets on Mars</h1>
<h3>Blue Sun</h3>
<p>On Mars, there is almost no atmosphere, only 1% of Earth's. Instead, the Sun's light is scattered by the fine, red dust swirled up in the (thin) air. Martian dust contains <a href="https://en.wikipedia.org/wiki/Hematite" rel="nofollow noreferrer">hematite</a> (<span class="math-container">$\alpha\text{-}\mathrm{Fe}_2\mathrm{O}_3$</span>), the mineral responsible for Mars' red color. This iron oxide has a very wavelength-dependent refractive index, and the grain sizes lie rather tightly around a few <span class="math-container">$\mu\mathrm{m}$</span>, scattering red light more than blue as in the (ideal) example given above (e.g. <a href="http://adsabs.harvard.edu/abs/1997JGR...102.9039O" rel="nofollow noreferrer">Ockert-Bell et al. 1997</a>; <a href="http://adsabs.harvard.edu/abs/1999ficm.conf.6085L" rel="nofollow noreferrer">Lane et al. 1999</a>). This has nothing to do with the fact that the dust is red, but is merely a consequence of the size of the particles.</p>
<p>Hence, in this case it is the <em>red</em> light which is filtered away, making the Sun appear blue, especially as it sets.</p>
<h3>Blue halo and red sky</h3>
<p>The unusual size distribution of the dust particles described above is the reason for the blue disk of the Sun, but cannot explain the blue halo around it (<a href="https://www.osapublishing.org/ao/abstract.cfm?uri=ao-53-9-1808" rel="nofollow noreferrer">Ehlers et al. 2014</a>). This spectacular phenomenon is caused by the difference in the phase function described above: Whereas Rayleigh scattering is rather isotropic, Mie scattering tends to scatter in the forward direction, and  Martian dust in particular has a large asymmetry parameter (e.g. <a href="http://adsabs.harvard.edu/abs/2007JGRE..112.8S13V" rel="nofollow noreferrer">Vincendon et al. 2007</a>), meaning that light that <em>is</em> scattered has a preference for being scattered along the LOS, i.e. toward the observer of the sunset.</p>
<p>Moreover, the asymmetry parameter is dependent on the wavelength of the light, with the forward scattering of blue light being much more pronounced than for red light, as seen in this figure:</p>
<p><a href="https://i.sstatic.net/nxnIH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/nxnIH.png" alt="asym" /></a>
<sup>Probability of scattering at a given angle for blue light (425 nm, left) and red light (694 nm, right). The three line types correspond to different models (from <a href="https://www.osapublishing.org/ao/abstract.cfm?uri=ao-53-9-1808" rel="nofollow noreferrer">Ehlers et al. (2014)</a>).</sup></p>
<p>Hence, the red light has a larger tendency to be scattered out of the LOS, creating the reddish sky.</p>
<h1>Pretty picture</h1>
<p>I promised a pretty picture. Here's a Martian sunset:</p>
<p><a href="https://i.sstatic.net/Hy7dC.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Hy7dC.jpg" alt="MartianSunset" /></a>
<sup>Credit: <a href="https://mars.nasa.gov/resources/5300/a-martian-sunset/?site=insight" rel="nofollow noreferrer">NASA/JPL-Caltech/Texas A&amp;M/Cornell</a></sup></p>
<h1>Blue Sun and Moon on Earth</h1>
<p>On rare occasions, aerosols in Earth's atmosphere can make the Sun and Moon  appear blue. When the volcano Krakatoa erupted in 1853, the Moon looked blue in some places on Earth for almost a month (<a href="http://www.mv.helsinki.fi/home/asmi/Aitken/" rel="nofollow noreferrer">Aitken 1892</a>).</p>

CC BY-SA 4.0
2019-01-20T00:07:32.153Z
Interesting! But regular Rayleigh scattering probably plays a role, doesn't it? It is just sub-dominant in the martian sky, as the total amount of Rayleigh happens (on Earth) in the Troposphere and is thus linked to the total atmospheric mass, which is lower on Mars?
AtmosphericPrisonEscape
CC BY-SA 4.0
2019-01-20T02:16:55.047Z
re asymmetry parameter: https://i.sstatic.net/ES3c1.png from http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html However I don't understand your argument how using the term means the forward scattered light is a different color. It may be in that reference but it's pretty deep. *Why exactly is forward-scattering from red dust bluer (less red) than large angle scattering??*
uhoh
CC BY-SA 4.0
2019-01-20T14:26:57.883Z
@uhoh That was bad phrasing from my side. Forward scattering makes the halo more _pronounced_, because the light is not scattered so far from the LOS to the source, not more blue. The blueness is caused by the preferential scattering of red photons, i.e. they may be scattered multiple times and are hence lost, whereas blue photons are scattered perhaps ~1 time, and hence are still seen roughly along the direction toward the Sun. Thanks, I'll edit.
pela
CC BY-SA 4.0
2019-02-01T10:55:41.160Z
@uhoh It looked a bit more into this, and in fact the dust _both_ preferentially scatters red photons _and_ does indeed forward-scatter blue photons more than red photons. The preferential scattering of red photons doesn't have anything to do with the dust being red, but with the fact that it's mostly micron-sized particles. I updated the answer.
pela
CC BY-SA 4.0
2019-02-01T11:01:32.903Z
okay that's super! thanks for digging in there, I will give these a read. I'm also struggling with Mie scattering [in good old Terran air](https://physics.stackexchange.com/q/458041/83380) and so the more I can read about scattering in general, the better.
uhoh
CC BY-SA 4.0
2015-05-23T19:54:48.403Z
2024-05-21T08:54:53.100Z
2024-05-21T08:54:53.100Z
pela
<p>I understand why our sky is blue and turns red when the sun sets or rises. I know why the sky on mars is red, but why does it turn blue during sunsets and rises? Shouldn't this actually not work due to the shorter wavelenght blue has compared to red?</p>

<light><mars>
Sunsets: Mars/Earth
Astrony
# Sunsets: Mars/Earth

(Asked by: username_0 on 2015-05-23)

I understand why our sky is blue and turns red when the sun sets or rises. I know why the sky on mars is red, but why does it turn blue during sunsets and rises? Shouldn't this actually not work due to the shorter wavelenght blue has compared to red?

**Tags:** light, mars


---

## 1 answers

### Answer by username_1 on 2015-05-23. Score: 20 | Accepted answer

**tl;dr**: On Earth, atmospheric molecules scatter blue light more than red. In contrast, on Mars atmospheric, $\mu\mathrm{m}$-sized hematite dust particles happen to give the opposite effect.

---

# Scattering theory

First a bit of theory. Pretty pictures later.

### Scattering cross section

Scattering of light of wavelength $\lambda$ on particles of radius $r$ may be described in general using [Mie theory](https://en.wikipedia.org/wiki/Mie_scattering). Using the variable $x \equiv 2\pi r/\lambda$, the scattering cross section $\sigma$ increases steeply with $x$ until $x\sim2\pi$, after which it decreases again and starts to oscillate, as seen in this figure:

[![xsec](https://i.sstatic.net/6HWkrjdB.png)](https://i.sstatic.net/6HWkrjdB.png)

Scattering cross section (normalised to the geometrical cross section) as a function of particle size-to-wavelength ratio (from [Wikipedia](https://en.wikipedia.org/wiki/Mie_scattering) with my own annotations).

In other words, when a particle is much smaller than the wavelength of the light ($x\ll 1$), the probability of scattering the light increases steeply with decreasing wavelength — in fact the cross section goes as $\lambda^{-4}$. This regime is called Rayleigh scattering. The scattering efficiency reaches a maximum for particles roughly equal to the wavelength of the light ($x\simeq2\pi$; green arrow in the figure). Hereafter, it falls again, exhibits some oscillations that have to do with interference in the particle, slowly converging to the "geometric" value $\sigma = \pi r^2$ (i.e. what would be expected classically).

In many (astrophysical) situations dust tends to be a distribution of sizes with more small than large particles (typically $n\propto r^{-3.5}$), meaning that an ensemble of dust has much more particles that scatter short wavelengths (blue) than long wavelengths (red). Thus, blue light is filtered out and red light light tends to be transmitted, in effect reddening the light.

On the other hand, if the dust consists of particles of the same size, then the extinction may rise and fall with wavelength, as seen in the figure. For example, optical light with $\lambda \simeq [0.4\text{–}0.7] \,\mu\mathrm{m}$, scattered by particles of size $0.7\,\mu\mathrm{m}$, is scattered according to the orange region in the figure — i.e. red light is scattered more efficiently than blue light. The exact form depend on the refractive index of the scatterer.

### Phase function

When a photon is scattered, it doesn't happen in a completely random direction, but follows a direction distribution given by the *phase function*. In the Rayleigh regime, there is a slight preference for forward and backward scattering, but for large $x$ there is a strong preference for forward scattering, as seen in this diagram (from [these slides](http://pages.mtu.edu/%7Escarn/teaching/GE4250/scattering_lecture_slides.pdf)):

[![phase](https://i.sstatic.net/joXoa.png)](https://i.sstatic.net/joXoa.png)

# Sunsets on Earth

On Earth the sunlight is scattered primarily by molecules which are much smaller than the wavelength of the light, i.e. we're in the Rayleigh regime. Hence, blue light tends to be scattered a lot, creating the blue sky. As the blue light is filtered out of the line of sight (LOS) toward the Sun, its light — white to begin with — becomes more yellow. And as the Sun sets, its light has to go through a thicker layer of atmosphere, enhancing this *reddening effect* and giving a red sunset.

# Sunsets on Mars

### Blue Sun

On Mars, there is almost no atmosphere, only 1% of Earth's. Instead, the Sun's light is scattered by the fine, red dust swirled up in the (thin) air. Martian dust contains [hematite](https://en.wikipedia.org/wiki/Hematite) ($\alpha\text{-}\mathrm{Fe}\_2\mathrm{O}\_3$), the mineral responsible for Mars' red color. This iron oxide has a very wavelength-dependent refractive index, and the grain sizes lie rather tightly around a few $\mu\mathrm{m}$, scattering red light more than blue as in the (ideal) example given above (e.g. [Ockert-Bell et al. 1997](http://adsabs.harvard.edu/abs/1997JGR...102.9039O); [Lane et al. 1999](http://adsabs.harvard.edu/abs/1999ficm.conf.6085L)). This has nothing to do with the fact that the dust is red, but is merely a consequence of the size of the particles.

Hence, in this case it is the *red* light which is filtered away, making the Sun appear blue, especially as it sets.

### Blue halo and red sky

The unusual size distribution of the dust particles described above is the reason for the blue disk of the Sun, but cannot explain the blue halo around it ([Ehlers et al. 2014](https://www.osapublishing.org/ao/abstract.cfm?uri=ao-53-9-1808)). This spectacular phenomenon is caused by the difference in the phase function described above: Whereas Rayleigh scattering is rather isotropic, Mie scattering tends to scatter in the forward direction, and Martian dust in particular has a large asymmetry parameter (e.g. [Vincendon et al. 2007](http://adsabs.harvard.edu/abs/2007JGRE..112.8S13V)), meaning that light that *is* scattered has a preference for being scattered along the LOS, i.e. toward the observer of the sunset.

Moreover, the asymmetry parameter is dependent on the wavelength of the light, with the forward scattering of blue light being much more pronounced than for red light, as seen in this figure:

[![asym](https://i.sstatic.net/nxnIH.png)](https://i.sstatic.net/nxnIH.png)
Probability of scattering at a given angle for blue light (425 nm, left) and red light (694 nm, right). The three line types correspond to different models (from [Ehlers et al. (2014)](https://www.osapublishing.org/ao/abstract.cfm?uri=ao-53-9-1808)).

Hence, the red light has a larger tendency to be scattered out of the LOS, creating the reddish sky.

# Pretty picture

I promised a pretty picture. Here's a Martian sunset:

[![MartianSunset](https://i.sstatic.net/Hy7dC.jpg)](https://i.sstatic.net/Hy7dC.jpg)
Credit: [NASA/JPL-Caltech/Texas A&M/Cornell](https://mars.nasa.gov/resources/5300/a-martian-sunset/?site=insight)

# Blue Sun and Moon on Earth

On rare occasions, aerosols in Earth's atmosphere can make the Sun and Moon appear blue. When the volcano Krakatoa erupted in 1853, the Moon looked blue in some places on Earth for almost a month ([Aitken 1892](http://www.mv.helsinki.fi/home/asmi/Aitken/)).

> **Comments:**
> **username_2**: Interesting! But regular Rayleigh scattering probably plays a role, doesn't it? It is just sub-dominant in the martian sky, as the total amount of Rayleigh happens (on Earth) in the Troposphere and is thus linked to the total atmospheric mass, which is lower on Mars?  
> **username_3**: re asymmetry parameter: https://i.sstatic.net/ES3c1.png from http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html However I don't understand your argument how using the term means the forward scattered light is a different color. It may be in that reference but it's pretty deep. *Why exactly is forward-scattering from red dust bluer (less red) than large angle scattering??*  
> **username_1**: @username_3 That was bad phrasing from my side. Forward scattering makes the halo more _pronounced_, because the light is not scattered so far from the LOS to the source, not more blue. The blueness is caused by the preferential scattering of red photons, i.e. they may be scattered multiple times and are hence lost, whereas blue photons are scattered perhaps ~1 time, and hence are still seen roughly along the direction toward the Sun. Thanks, I'll edit.  
> **username_1**: @username_3 It looked a bit more into this, and in fact the dust _both_ preferentially scatters red photons _and_ does indeed forward-scatter blue photons more than red photons. The preferential scattering of red photons doesn't have anything to do with the dust being red, but with the fact that it's mostly micron-sized particles. I updated the answer.  
> **username_3**: okay that's super! thanks for digging in there, I will give these a read. I'm also struggling with Mie scattering [in good old Terran air](https://physics.stackexchange.com/q/458041/83380) and so the more I can read about scattering in general, the better.  

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