Sea ice covers an ocean-sized region with a shifting mixture of level floes, ridges, leads, snow, and melt ponds. A satellite has to separate those features even through long winter darkness and frequent cloud. It can do that only by measuring how each surface emits or reflects electromagnetic radiation.

A passive microwave radiometer can map most of the Arctic each day, but it blends small features into a coarse footprint. Imaging radar resolves individual floes and leads over a smaller area. An altimeter measures surface elevation along a narrow track, and ice thickness must then be calculated from that elevation. Every final map combines a measured signal with assumptions used to interpret it.

Why photographs are not enough

Winter sea ice covers roughly $14\text{ million km}^2$, far too much to survey from the surface each day. A camera in orbit still faces darkness and cloud. At $80^\circ\mathrm{N}$, the lecture gives about 131 days without sunrise; at the pole, about 183 days. These are examples of polar night, not counts of complete darkness, because twilight and atmospheric refraction soften the change between day and night. During the brighter part of the year, cloud remains common, especially when summer open water adds moisture to the air.

A visible-light camera records sunlight reflected from the surface, so it needs both daylight and a clear path through the atmosphere. Under those conditions, it can show metre-scale features, natural colour, floe boundaries, and melt ponds. During polar night there is no sunlight to form the reflected image. Cloud can also hide the surface or resemble the bright ice below.

Thermal infrared measures radiation emitted by the surface, so it does not need sunlight. It can show temperature contrasts, such as a warm lead cutting through colder ice, but most cloud blocks the view of the surface. A complete observing system has to do several things:

1. locate the ice and estimate its concentration;
2. distinguish ice types such as first-year and multi-year ice;
3. estimate thickness, the hardest quantity;
4. follow drift and deformation;
5. describe the snow, melt ponds, and melt state on top.

An instrument that maps the whole Arctic each day usually cannot resolve a narrow lead or pressure ridge. An instrument that resolves those features covers less area or samples only a narrow track. Its signal must also respond to the property being measured, such as liquid water, surface roughness, or elevation.

One electromagnetic toolbox

Every sensor in this lecture measures electromagnetic radiation. Two properties determine what it can detect: wavelength and measurement mode. Wavelength affects how radiation interacts with cloud droplets, snow grains, brine, bubbles, surface roughness, and boundaries between materials. Measurement mode determines whether the instrument receives radiation already leaving Earth or sends out a pulse and records its echo.

Microwave wavelengths are much larger than ordinary cloud droplets. A representative cloud droplet is about $10\,\mu\mathrm{m}$ across, visible green light has a wavelength near $0.5\,\mu\mathrm{m}$, and radiation at $19\,\mathrm{GHz}$ has a wavelength of about $15.8\,\mathrm{mm}$. Ordinary cloud droplets scatter these microwaves much less than they scatter visible or infrared radiation. Microwave instruments can usually observe the surface through cloud, and they do not need sunlight.

Cloud is not completely transparent to every microwave. Atmospheric water vapour, cloud liquid water, and precipitation can still change the signal, with different effects at different frequencies.

Passive and active microwave

A passive microwave radiometer receives the weak microwave radiation emitted naturally by the surface and atmosphere. It sends out no pulse. Its wide swath allows frequent mapping across the Arctic, but each measurement averages a relatively large area.

An active microwave radar sends a microwave pulse toward the surface and measures the echo. Because it supplies its own illumination, it also works in darkness. Radar can resolve much finer structures than a radiometer, but each image covers a more limited swath, and the echo changes with viewing geometry and surface conditions.

The continuous passive-microwave climate record is conventionally reported from 1979. By 2026, 47 years have elapsed since that start. This gives scientists a nearly five-decade daily record that continues through polar night and most cloudy weather.

Brightness temperature separates water and ice

A radiometer reports brightness temperature, $T_b$. This is the temperature that a perfect emitter would need to produce the measured microwave radiance. In a simple surface-only picture,

$$T_b\approx \epsilon T,$$

where $T$ is physical temperature and $\epsilon$ is microwave emissivity. This is why $T_b$ is reported in kelvin even though it is not automatically the surface's physical temperature. A surface with low emissivity emits less microwave radiation than a perfect emitter at the same temperature. Polarization, the atmosphere, and scattering within snow and ice also change the reading.

Open water and sea ice can have nearly the same physical temperature but very different brightness temperatures. At the lecture's representative frequencies, open water has relatively low emissivity, so it appears microwave-dark. First-year ice has high emissivity and appears brighter. Multi-year ice is dimmer than first-year ice because summer drainage leaves a fresher, bubble-rich interior that scatters some of the radiation.

These values are classroom tie points, not constants for every place and time. Weather, season, polarization, ice temperature, snow, salinity, and melt state can all shift them.

At $19\,\mathrm{GHz}$, the first-year-ice tie point is $102\,\mathrm{K}$ above the open-water value, so that channel separates water from ice well in this example. The change between frequencies helps separate ice types. From 19 to $37\,\mathrm{GHz}$, the representative multi-year value falls by $25\,\mathrm{K}$, while the first-year value falls by only $8\,\mathrm{K}$. Operational algorithms combine several frequencies and polarizations because no single channel separates every possible surface.

The ice-type contrast comes from the structure described in §4, Why brine matters. Salty first-year ice and fresher, air-rich multi-year ice have different emissivities and internal scattering, which produces different brightness temperatures.

From brightness temperature to concentration

As a simple teaching model, treat one pixel as a mixture of open water and solid first-year ice. Let

• $T_b$ be the measured brightness temperature;
• $T_w=136\,\mathrm{K}$ be the open-water tie point;
• $T_i=238\,\mathrm{K}$ be the first-year-ice tie point.

If $f_{\mathrm{ice}}$ is the fraction of the pixel covered by ice, then

$$T_b=(1-f_{\mathrm{ice}})T_w+f_{\mathrm{ice}}T_i.$$

Solving for the unknown fraction gives

$$f_{\mathrm{ice}} =\frac{T_b-T_w}{T_i-T_w} =\frac{T_b-136\,\mathrm{K}}{102\,\mathrm{K}}.$$

For $T_b=187\,\mathrm{K}$,

$$f_{\mathrm{ice}} =\frac{187-136}{238-136} =\frac{51}{102} =0.50.$$

In this simplified model, the pixel is 50% ice and 50% water. For $T_b=217\,\mathrm{K}$,

$$f_{\mathrm{ice}} =\frac{217-136}{238-136} =\frac{81}{102} \approx0.794,$$

so the estimated ice concentration is about 79%, or roughly 80%.

Real retrieval algorithms combine several channels and polarizations with weather filters and tie points that change with surface conditions. The one-channel mixture is especially unreliable during melt. Liquid water in a melt pond changes the microwave emission from an ice-covered floe and can make part of the floe resemble open water. The algorithm may then report a lower ice concentration even though solid ice remains beneath the pond. §3, Melt onset and melt ponds describes how the ponds form and alter the surface energy balance.

Concentration, extent, and area

Sea-ice concentration, $f_i$, is the estimated fraction of grid cell $i$ covered by ice. Ice extent counts the full area of every cell whose concentration reaches a chosen threshold. Ice area instead multiplies every cell's area by its estimated ice fraction.

With the common 15% threshold, ice extent, $E$, is

$$E=\sum_i A_i\,\mathbf{1}\!\left(f_i\ge 0.15\right),$$

where $A_i$ is cell area and the indicator equals 1 when the threshold is met and 0 otherwise. A cell at 15% and a cell at 100% both contribute their full area to extent.

Ice area, $A$, keeps the concentration inside each cell:

$$A=\sum_i A_i f_i.$$

The lecture's worked example uses four neighbouring cells. Each is $25\,\mathrm{km}\times25\,\mathrm{km}=625\,\mathrm{km}^2$, and their concentrations are 95%, 80%, 40%, and 10%. Extent includes the first three cells because their concentrations are at least 15%, but it excludes the fourth:

$$E=3(625)=1875\,\mathrm{km}^2.$$

Area weights each cell by its concentration, including the 10% in the fourth cell:

$$\begin{aligned} A&=(0.95+0.80+0.40+0.10)(625)\\ &=2.25(625)\\ &=1406.25\,\mathrm{km}^2. \end{aligned}$$

The same four cells have an extent of $1875\,\mathrm{km}^2$ and an ice area of $1406.25\,\mathrm{km}^2$. Extent is one third larger here because each of the first three cells contributes all $625\,\mathrm{km}^2$, including the open water within those cells. Some operational products remove concentrations below 15% before computing area as well, so the product definition must be checked. The basic distinction remains full-cell counting for extent and concentration weighting for area.

A trend in extent cannot be compared directly with a trend in area. Melt-pond errors often affect area more strongly because area uses every concentration value. Extent is still affected whenever a concentration error moves a cell from one side of the 15% threshold to the other.

The 25 km blind spot

Frequent Arctic-wide coverage requires coarse spatial averaging. Many long-record concentration products use a nominal $25\,\mathrm{km}$ grid, so each grid cell covers about $625\,\mathrm{km}^2$. The sensor's actual footprint depends on its frequency and design and may be larger than the grid spacing, especially for a 19 GHz channel. A grid cell is not a camera pixel with sharp $25\,\mathrm{km}$ edges.

Suppose a cell reports 96% concentration. The remaining 4% is $25\,\mathrm{km}^2$ of water within a $625\,\mathrm{km}^2$ cell, but the measurement does not show how that water is arranged. That 4% could fill many narrow cracks or one lead wide enough for navigation. Pressure ridges only tens of metres wide also disappear into the cell average. The reported concentration can be accurate even though the locations, widths, and shapes of the openings remain unknown.

That missing geometry changes the local heat exchange. A narrow opening can release far more winter heat per square metre than thick, snow-covered ice, while a ridge can obstruct travel even though it occupies little area. The physical processes are described in §4, Divergence: leads and polynyas and §4, Convergence: rafting and ridging.

Imaging radar reveals structure

Imaging radar sends out its own pulse, so it works in darkness and through most cloud. It usually points sideways rather than straight down. An echo from the near side of the swath returns before an echo from the far side, which gives the across-track, or range, coordinate.

Resolving fine detail along the flight path would normally require an impractically long physical antenna. Synthetic aperture radar (SAR) instead combines echoes coherently as the satellite moves through a sequence of positions. Those measurements act like one much longer antenna. Depending on its operating mode, SAR can produce pixels from metre scale to tens of metres, fine enough to resolve floes, leads, and deformation that passive microwave averages together.

Two broad sets of properties control the strength of a SAR echo:

• surface roughness relative to the radar wavelength and viewing direction;
• electrical properties, which change with salinity, brine volume, temperature, wetness, snow, and internal air inclusions.

Under cold-season conditions, these properties often produce the following patterns:

Brightness alone does not identify the surface or the ice type. Wind can roughen open water enough to produce a bright echo, while a smooth multi-year floe can look dark. Wet snow and melt ponds can also change the return even though the underlying ice remains. SAR brightness is backscatter strength, not visible whiteness or albedo. Incidence angle, polarization, wavelength, and image processing all affect it as well.

Repeated SAR images can show where the ice moved. If the same floe is displaced by $\Delta\mathbf{x}$ in time $\Delta t$, its mean velocity is

$$\mathbf{v}\approx\frac{\Delta\mathbf{x}}{\Delta t}.$$

Tracking many floes produces a motion field. Differences across that field show floes translating, rotating, converging, and separating, rather than giving only the path of one buoy.

Wavelength changes what radar can see

Radar bands respond to different roughness scales and penetrate to different interfaces:

Short wavelengths respond strongly to small surface features. Longer wavelengths can often travel farther through cold, dry snow or ice that absorbs little microwave energy, allowing them to interact with larger or deeper structures. They do not penetrate without limit. Wet snow, saline first-year ice, liquid brine, absorption, and strong internal boundaries can absorb or scatter the signal before it travels far. Spatial resolution also depends on antenna design, bandwidth, viewing geometry, and processing, not just wavelength.

Thickness comes from freeboard

An imaging radar image does not directly show total ice thickness, and neither does an altimeter. An altimeter measures the distance from the satellite to a reflecting surface. Scientists use nearby leads to estimate local sea level, then compare that level with the height of the floe. The height of the ice surface above the water is ice freeboard, $f_i$.

Converting freeboard to thickness requires hydrostatic balance. Let

• $H_i$ be ice thickness;
• $h_s$ be snow depth;
• $\rho_w$, $\rho_i$, and $\rho_s$ be seawater, ice, and snow densities.

The weight of the ice and snow must equal the weight of the seawater displaced by the submerged ice:

$$\rho_i H_i+\rho_s h_s=\rho_w(H_i-f_i),$$

so

$$H_i=\frac{\rho_w f_i+\rho_s h_s}{\rho_w-\rho_i}.$$

Using the stated example densities

$$\rho_w=1024,\qquad \rho_i=915,\qquad \rho_s=320 \quad \mathrm{kg\,m^{-3}},$$

a snow-free floe has

$$H_i=\frac{1024}{1024-915}f_i\approx9.39f_i.$$

This gives the rough ten-to-one rule. With these densities and no snow, a $20\,\mathrm{cm}$ ice freeboard corresponds to about $1.88\,\mathrm{m}$ of ice. If the same measured ice freeboard carries $10\,\mathrm{cm}$ of snow, the full equation gives

$$H_i =\frac{1024(0.20)+320(0.10)}{1024-915} \approx2.17\,\mathrm{m}.$$

Snow adds weight and pushes the floe farther into the water. To maintain the same measured ice freeboard while carrying snow, the floe must be thicker. Any precise conversion from freeboard to thickness needs estimates of snow depth and of the snow, ice, and water densities.

Laser and radar altimeters also reflect from different effective surfaces:

• a laser usually reflects from the top of the snow, measuring snow freeboard;
• a Ku-band radar return is often interpreted as coming from near the snow-ice interface, giving an estimate of ice freeboard after propagation corrections.

In an ideal cold, dry, co-located case,

$$h_s\approx h_{\mathrm{laser}}-h_{\mathrm{radar}}.$$

Actual snow does not always provide one clean radar-reflecting boundary. If it is layered, saline, or wet, the strongest return may come from somewhere within the snow rather than from the snow-ice interface. The laser and radar satellites may also pass over separate areas of a drifting floe at different times. These uncertainties in snow depth and radar penetration can cause large errors in the thickness estimate.

The hydrostatic equation also amplifies small elevation errors. With the example densities and all other inputs held fixed,

$$\frac{\partial H_i}{\partial f_i} =\frac{1024}{109} \approx9.39, \qquad \frac{\partial H_i}{\partial h_s} =\frac{320}{109} \approx2.94.$$

For this density set:

• a $1\,\mathrm{cm}$ absolute freeboard error contributes about $9.4\,\mathrm{cm}$ of thickness error;
• a $10\,\mathrm{cm}$ absolute snow-depth error contributes about $29\,\mathrm{cm}$ of thickness error.

These are dimensional sensitivities, not relative-error percentages that apply to every retrieval. Uncertain ice and snow density, surface roughness, sea-surface height, and departures from hydrostatic balance add more error. Thin ice has such a small freeboard that separating its surface from local sea level is especially difficult.

Snow and ice also slow the conductive loss of ocean heat. The insulation they provide is described in §2, The surface energy balance and §4, Snow increases the insulation.

Match the instrument to the question

Resolution and coverage pull in opposite directions

• Imaging
• Broad coverage
• Profile altimeter

Approximate positions used in the resolution-versus-coverage plot

Instrument	Approximate placement
High-resolution SAR	about 10 m resolution and roughly 80 km swath
Wide-swath SAR	about 40 m resolution and several hundred kilometres of swath
Laser altimeter	about 15–20 m resolution along a narrow profile
Radar altimeter	about 300 m resolution along a profile a few kilometres wide
Optical / thermal	about 500 m resolution and coverage of a few thousand kilometres
Passive microwave	near 25 km resolution with daily, broad Arctic coverage
Scatterometer	near 25 km resolution and roughly 1,000 km coverage

Observers would like metre-scale detail across the entire Arctic every day, represented by the empty upper-left part of the diagram. No current instrument provides both that coverage and that detail. Broad-swath radiometers, high-resolution radar images, and narrow altimeter tracks are used together instead.

Passive microwave supplies a long, frequent record of concentration but averages away narrow leads and ridges. SAR recovers much of that geometry and can track floe motion, though it covers less area at once. Altimetry measures elevation along a track; thickness is then inferred from freeboard, snow, density, and hydrostatic balance. Because each product depends on different measurements and assumptions, its uncertainty and spatial scale limit the conclusions that can be drawn from the final map.