120° Parhelia & Multiple Internal Reflection Dynamics: How Horizontally Oriented Plate Crystals and Double Internal Reflections Forge Luminous White Mock Suns

1. Opening Scene: The Sub-Zero Mirage

At twenty-eight degrees below zero on a windless morning across the high plateau, the atmosphere ceases to behave like empty air and begins to behave like an optical laboratory. Step outside, and the cold strikes your trachea with the dry, sharp bite of aerosolized glass. Every exhalation crystallizes instantly, hanging suspended in the breathless air before drifting earthward. The barometer on your wrist reads a towering 1,028 hectopascals—the unmistakable signature of a deep Arctic anticyclone settling over frozen terrain.

Look toward the horizon, and you will find that the sky is not clear blue, nor is it shrouded in conventional cloud. Instead, a pale, luminescent gauze of diamond dust—a ground-level precipitation of pristine, microscopic ice crystals—permeates the boundary layer. The low sun, perched scarcely fifteen degrees above the southern snowfields, blazes with an unearthly intensity. Flanking it to the left and right, at roughly twenty-two degrees of angular separation, burn the familiar twin beacons of parhelia (the common sun dogs), flashing with vivid chromatic fire: ruby red along their inner flanks, melting into emerald green, and trailing off into faint cerulean tails.

       [ 22° Sun Dog ] -------- ( Sun ) -------- [ 22° Sun Dog ]
              \                                         /
               \---------------[ PARHELIC CIRCLE ]-----/
                                      |
                     [ 120° Parhelion: Pure White ]

Yet if you turn your back on the blinding dazzle of the sun and trace your gaze along the faint, milk-white ribbon of the parhelic circle—a razor-sharp band of light girdling the sky at the exact elevation of the solar disk—something far rarer and stranger reveals itself. Far down the circle, located deep in your peripheral vision at an azimuth of exactly one hundred and twenty degrees from the sun, floats a pair of brilliant, ghostly spots.

Unlike the fiery sun dogs flanking the solar disk, these distant apparitions possess no spectral color whatsoever. They are dead, pristine white—radiant patches of pure sunlight that appear to have detached from the solar core and drifted around the dome of the sky. To the uninitiated observer, they seem impossible: mock suns burning where no prism should be able to bend light. Yet their presence is governed by a breathtakingly elegant conspiracy of crystalline geometry, fluid dynamics, and multiple internal reflections.

2. What's Actually Happening — Plain English First

To understand why these ghostly white sun dogs appear so far from the sun, we must shrink down to the scale of cloud microphysics and observe the billions of ice crystals drifting silently through the frozen air.

Think of an ice cloud not as a uniform fog, but as a vast ballroom filled with trillions of microscopic, six-sided glass coins. These are hexagonal plate crystals, sculpted by cold vapor into flat, six-sided prisms with perfectly smooth top and bottom faces (called basal planes) and six vertical side faces (called prism facets).

         Top Basal Face (Face 1)
              /--------\
             /          \
  Prism     |            | Prism Face (Face 4)
  Face (6)  |            |
             \          /
              \--------/
        Bottom Basal Face (Face 2)

In calm air, these miniature crystal plates do not tumble erratically. Much like a leaf falling through a still autumn afternoon or a coin settling through water, aerodynamic drag acts on their broad, flat surfaces. The rushing air beneath them forces them to settle with their wide top and bottom faces oriented almost perfectly level with the ground.

Now, imagine a single beam of sunlight striking one of these horizontally leveled crystals:

1. The Entry: The sunbeam enters through the flat, horizontal top roof of the crystal. As it crosses from the air into the denser ice, the ray bends downward.
2. The First Bounce: Instead of shooting straight through, the light strikes one of the six vertical side walls inside the crystal. Because the angle is steep, the polished ice wall acts like an internal mirror, bouncing the light across the crystal interior.
3. The Second Bounce: The reflected beam travels across to a non-adjacent vertical side wall, striking it and bouncing a second time.
4. The Exit: Finally, having ricocheted twice off the vertical walls, the beam hits the horizontal bottom floor of the crystal and refracts back out into the open sky.

Here is the crucial insight that explains both the position and the color of the 120° parhelion:

• Why it turns exactly 120 degrees: Because the side walls of a regular hexagon are locked into fixed 60-degree and 120-degree angles, two successive bounces off alternating side walls act like a precise corner reflector. No matter how the crystal is rotated horizontally, that dual bounce always deflects the beam's horizontal compass heading by precisely 120 degrees.
• Why it is pure white: When sunlight enters the top flat roof and exits the parallel bottom floor, the crystal behaves like a flat sheet of window glass rather than a triangular prism. The bending that splits colors upon entry is exactly reversed and undone upon exit. Every wavelength of the rainbow—red, green, and violet—emerges travelling in the exact same direction. The colors recombine perfectly, leaving a brilliant, achromatic white spot in the sky.

3. The Science (For Those Who Want to Go Deeper)

To formalize the optics of the 120° parhelion, we employ ray-tracing analysis through hexagonal crystalline lattices, governed by Snell's Law, spatial coordinate transformations, and classical boundary-layer aerodynamics.

TOP-DOWN RAY PATH (Hexagonal Plate):
              Face 4
            /--------\
  Face 5   /          \  Face 3
          |    (1)     |  <-- Ray enters Face 1 (top)
          |   /  \     |  --> Hits Face 3 (internal reflection)
  Face 6  |  /    \    |  --> Hits Face 5 (internal reflection)
           \/______\  /   --> Exits Face 2 (bottom)
            \--------/
              Face 2     Net Azimuthal Deflection = 120°

The 1-3-5-2 Ray Path and Chromatic Cancellation

In standard halo nomenclature established by atmospheric optical physicists, the basal faces of a hexagonal prism are designated as Face 1 (top) and Face 2 (bottom), while the six vertical prism facets are numbered sequentially from 3 to 8 around the perimeter.

The formation of the 120° parhelion follows the deterministic path designated as Ray Path 1–3–5–2 (or symmetrically, 1–4–6–2, 1–5–7–2, etc.).

Let us trace a ray of light entering a plate crystal whose basal planes are aligned parallel to the horizontal $xy$-plane:

1. Entrance Refraction (Basal Face 1)

Let the solar elevation angle above the horizon be $h$. The solar ray vector in Cartesian coordinates (where $\hat{z}$ points toward the zenith) is:

$$\mathbf{s}_0 = (\cos h \cos \phi_0, \, \cos h \sin \phi_0, \, -\sin h)$$

The angle of incidence relative to the surface normal of the upper basal face ($\mathbf{n}_1 = (0, 0, 1)$) is:

$$\theta_0 = 90^\circ - h$$

Applying Snell's law at the air-ice interface, where $n_{\text{air}} \approx 1.000$ and $n_{\text{ice}}(\lambda)$ is the wavelength-dependent refractive index of solid water ice $I_h$:

$$\sin \theta_0 = n_{\text{ice}}(\lambda) \sin \theta_r$$

$$\theta_r(\lambda) = \arcsin\left(\frac{\cos h}{n_{\text{ice}}(\lambda)}\right)$$

Inside the crystal, the unit vector of the refracted ray becomes:

$$\mathbf{s}_{\text{int}} = (\sin \theta_r \cos \phi_0, \, \sin \theta_r \sin \phi_0, \, -\cos \theta_r)$$

Notice that the vertical component of the ray velocity inside the ice is governed strictly by $-\cos \theta_r$.

2. Successive Internal Reflections (Prism Faces 3 and 5)

The internal ray encounters a vertical prism face (Face 3). Because the prism facets are perpendicular to the basal planes, their surface normal vectors $\mathbf{n}_v$ lie entirely within the horizontal $xy$-plane:

$$\mathbf{n}_v = (\cos \psi_k, \, \sin \psi_k, \, 0)$$

By the classical vector law of specular reflection, the reflected ray direction $\mathbf{s}'$ is:

$$\mathbf{s}' = \mathbf{s}{\text{int}} - 2(\mathbf{s}{\text{int}} \cdot \mathbf{n}_v)\mathbf{n}_v$$

Because $\mathbf{n}_v \cdot \hat{z} = 0$, the dot product evaluates purely in the horizontal coordinates:

$$s'z = s{\text{int}, z} - 2(0) = -\cos \theta_r$$

The vertical trajectory of the ray is completely unaffected by reflection off vertical walls. The ray continues downward at the identical angle $\theta_r$ relative to the vertical axis.

The ray then propagates across the crystal to strike a non-adjacent prism face (Face 5), whose normal vector $\mathbf{n}{v2}$ is inclined at an angle of $\alpha = 60^\circ$ relative to $\mathbf{n}{v1}$. A second reflection occurs:

$$\mathbf{s}'' = \mathbf{s}' - 2(\mathbf{s}' \cdot \mathbf{n}{v2})\mathbf{n}{v2}$$

Once again, $s''_z = s'_z = -\cos \theta_r$.

3. Exit Refraction (Basal Face 2)

The ray arrives at the lower horizontal basal face (Face 2), whose surface normal is $\mathbf{n}_2 = (0, 0, -1)$. The angle of incidence on this exit face is precisely $\theta_r(\lambda)$.

Applying Snell's law at the exit boundary from ice back into ambient air:

$$n_{\text{ice}}(\lambda) \sin \theta_r(\lambda) = 1.000 \cdot \sin \theta_{\text{exit}}$$

Substituting $\sin \theta_r(\lambda) = \frac{\sin \theta_0}{n_{\text{ice}}(\lambda)}$:

$$n_{\text{ice}}(\lambda) \left( \frac{\sin \theta_0}{n_{\text{ice}}(\lambda)} \right) = \sin \theta_{\text{exit}}$$

$$\sin \theta_{\text{exit}} = \sin \theta_0 \implies \theta_{\text{exit}} = \theta_0 = 90^\circ - h$$

$$\frac{d\theta_{\text{exit}}}{d\lambda} = 0$$

Geometric Proof of the 120° Azimuthal Deflection

Why does this double reflection result in a net azimuth change of exactly 120°? We can prove this using the planar geometry of the hexagonal crystal.

In the horizontal plane, let the angle of the ray before any internal reflection be $\Phi_0$. The two reflecting vertical faces of the regular hexagon have normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$ separated by an angle $\alpha = 60^\circ$ (since non-adjacent prism faces meet at an interior angle of $120^\circ$, their outward normal vectors form an angle of $60^\circ$).

SIDE CROSS-SECTION (Basal Parallel Planes):
              Air (n = 1.000)
    ----------------------------------- Face 1 (Top)
              Ice (n ≈ 1.31)
                 \
                  \ (Internal reflections preserve vertical angle)
                   \
    ----------------------------------- Face 2 (Bottom)
              Air (n = 1.000)

The deflection angle $\delta_1$ produced by the first mirror plane whose normal is at angle $\psi_1$ is:

$$\Phi_1 = 2\psi_1 - \Phi_0 + 180^\circ$$

The second reflection off a plane with normal at $\psi_2 = \psi_1 + 60^\circ$ yields:

$$\Phi_2 = 2\psi_2 - \Phi_1 + 180^\circ = 2(\psi_1 + 60^\circ) - (2\psi_1 - \Phi_0 + 180^\circ) + 180^\circ$$

$$\Phi_2 = 2\psi_1 + 120^\circ - 2\psi_1 + \Phi_0 - 180^\circ + 180^\circ = \Phi_0 + 120^\circ$$

$$\Delta \Phi = |\Phi_2 - \Phi_0| = 120^\circ$$

The total azimuthal rotation $\Delta \Phi$ is independent of the initial orientation of the crystal ($\psi_1$). Every oriented plate crystal undergoing the 1-3-5-2 ray path contributes light to the exact same azimuthal position: $\pm 120^\circ$ relative to the sun.

Critical Angle and Solar Elevation Limits

Internal reflection at the vertical prism faces is governed by the critical angle $\theta_c$ for the ice-air interface:

$$\theta_c = \arcsin\left(\frac{1}{n_{\text{ice}}}\right)$$

For solid ice with $n_{\text{ice}} \approx 1.309$ at $\lambda = 589.3\text{ nm}$:

$$\theta_c = \arcsin\left(\frac{1}{1.309}\right) \approx 49.88^\circ$$

WORKED EXAMPLE: Total Internal Reflection Threshold
Target: Calculate the critical angle and maximum solar elevation for optimal 120° parhelia.

1. Refractive Index of Ice: n = 1.309
2. Critical Angle: θ_c = arcsin(1 / 1.309) = 49.88°
3. Solar Elevation Limit:
   For rays entering at solar elevation h, the internal ray angle θ_r = arcsin(cos(h) / n).
   Total internal reflection requires the grazing angle on the vertical wall to exceed θ_c.
   When h > 32°, internal rays strike vertical faces at angles steeper than the critical threshold,
   causing light leakage and dramatic loss of spot luminance.

When the solar elevation $h$ is low ($h < 30^\circ$), internal reflection off the vertical walls is total (100% reflectance) for a broad swath of crystal azimuths. As the sun climbs above approximately $32^\circ$, the angle of incidence on the vertical faces drops below $\theta_c$. The reflection shifts from total to partial Fresnel reflection, causing substantial light leakage through the side facets and extinguishing the 120° parhelia.

Aerodynamic Stability of Horizontally Oriented Plates

The optical coherence of the 120° parhelion relies entirely on the horizontal alignment of the plate crystals. If crystals tumble randomly, light is scattered diffusely, washing out the feature.

When a flat hexagonal plate with diameter $D$ and thickness $L$ (where $D/L \gg 1$) falls through viscous air at low to moderate Reynolds numbers ($0.5 < Re < 100$), hydrodynamic forces create a powerful stabilizing torque:

1. Center of Pressure Shift: When a falling plate tilts by a small angle $\theta_{\text{tilt}}$, the stagnation point and center of aerodynamic pressure shift toward the leading edge.
2. Restoring Moment: This pressure imbalance generates a restorative torque that drives $\theta_{\text{tilt}} \to 0$, forcing the basal plane perpendicular to the relative upward airflow.
3. Laminar Damping: In non-turbulent atmospheric layers, crystal wobble variance is constrained to $\sigma_\theta < 0.5^\circ$, maintaining the crisp, point-like geometry of the 120° mock sun.

Distinguishing the Parhelic Family

To avoid misidentification in the field, atmospheric scientists categorize features along the parhelic circle based on their distinct optical signatures and ray paths:

4. Practical Outdoor Guidance

Observing a 120° parhelion requires patience, keen spatial awareness, and specific meteorological conditions. Because they appear two-thirds of the way around the sky from the sun, most observers look right past them.

WHERE TO LOOK:
                  ( Solar Horizon )
                         Sun
                          |
             [22°]        |        [22°]
          Sun Dog <--- 22° ---> Sun Dog
                          |
             ===========================  Parhelic Circle
            /                           \
           /                             \
     [ 120° Parhelion ]           [ 120° Parhelion ]
     (Look 120° Away)             (Look 120° Away)

1. What to Look for in the Sky

• Locate the Anchors First: Confirm the presence of bright 22° parhelia and a visible parhelic circle. If the parhelic circle is strong and extends past the zenith, follow the white band around to the left or right.
• Measure the Angle: Extend your arm fully. Spread your thumb and little finger wide (spanning roughly $20^\circ$). Six hand-spans along the parhelic circle from the sun brings your eye directly to the $120^\circ$ position.
• Scan for the White Pillow: Look for a soft, brilliant white patch resting directly on the parhelic circle. It will look like a miniature, cloud-like condensation of pure light.

2. Meteorological Instrument Checklist

• Barometric Pressure: Watch for high, steady pressure ($>1,020\text{ hPa}$) associated with continental polar air masses or Siberian/Arctic high-pressure systems.
• Ambient Temperature: Surface temperatures typically fall below $-15^\circ\text{C}$ ($5^\circ\text{F}$) during ground-level diamond dust events, or high cirrostratus decks at $-40^\circ\text{C}$ aloft during warm-front overrunning.
• Surface Winds: Calm or light breezes ($< 5\text{ knots}$). Turbulent shear destroys the horizontal aerodynamic leveling of the plate crystals, dispersing the 120° parhelion into diffuse haze.

3. Field Observer's Rule of Thumb

• Polarized Sunglasses Trick: Wear linearly polarized sunglasses and tilt your head. Because internal reflections at near-critical angles heavily polarize light, the 120° parhelion will dramatically brighten and dim as you rotate your head, making faint displays pop against cirrus background clouds.

For deeper technical exploration of crystal ray tracing, consult the NOAA National Weather Service Glossary and the World Meteorological Organization International Cloud Atlas.

5. Today's Meteorological Rule of Thumb

When brilliant sun dogs ignite the morning frost, never keep your eyes locked on the sun: trace the white ribbon of the parhelic circle two-thirds of the way around the horizon, where double-reflecting ice plates spin sunlight into ghostly, achromatic twins at exactly 120 degrees.