Radiative transfer explains how light and other electromagnetic radiation change while travelling through matter. A measured beam can contain surviving background radiation, emission from material along the path and radiation scattered into the viewing direction. Understanding that mixture is essential when interpreting a spectrum, a thermal image, a cloudy atmosphere or a satellite observation.
The gold standard is to define the measured quantity, follow each gain and loss, and test whether the assumptions still hold when the wavelength, viewing angle or material changes. This guide develops that method through original numerical cases. All example values are hypothetical; they illustrate physical relationships rather than describe a particular instrument, atmosphere or material.
Use the Gold Standard master, the Education parent and the How X Works hub for the wider learning collection. The Radiation root explains radiation categories and measurement. Here the central question is more specific: how does a radiation field evolve between its source and its observer?
Choose your radiative transfer route
Chapters 1–3: Define quantities and mechanisms
Chapters 4–6: Solve rays and layers
Chapters 7–9: Resolve spectra and scattering
Chapters 10–12: Infer and diagnose
Chapter index
Define quantities and mechanisms · Chapters 1–3
1. Start with the quantity your detector measures
Solve rays and layers · Chapters 4–6
4. Derive the balance along one ray
Resolve spectra and scattering · Chapters 7–9
7. Keep spectral channels separate until integration
8. Connect thermal emission to temperature without skipping the path
Infer and diagnose · Chapters 10–12
10. Distinguish a forward prediction from an inverse inference
Practise and continue · Chapters 13–14
13. Choose a practical learning and research route
14. Questions that reveal whether the explanation is complete
CHAPTER 1 OF 14 · Define quantities and mechanisms
1. Start with the quantity your detector measures
Light has a direction as well as an energy and a wavelength. A camera looking toward a small patch of sky and a sensor collecting radiation from an entire hemisphere do not measure the same quantity. Before solving a transfer problem, describe the detector’s field of view, spectral band and collecting surface.
Spectral radiance, often called specific intensity in astrophysics, describes power per projected area, per solid angle and per spectral interval. Irradiance describes incoming power per receiving area after integration over relevant directions and wavelengths. Their units expose the difference: a radiance unit includes a steradian denominator; a broadband irradiance unit does not.
The University of Arizona radiometry reference defines radiance using projected area and solid angle. Those definitions are the basis of the geometry checks below.
An angular collection case
Suppose a small uniform source has band-integrated radiance 40 W/(m² sr), covers a solid angle of 0.02 sr and lies nearly along the receiving surface normal. The incoming irradiance from that patch is approximately 40 × 0.02 = 0.8 W/m². A receiving area of 0.1 m² intercepts approximately 0.08 W before instrumental losses.
Now move the same small patch to a direction making 60 degrees with the surface normal, retaining its radiance and solid angle. The projected-area factor is cos 60° = 0.5, so its contribution falls to approximately 0.4 W/m². The source did not become intrinsically dimmer; its geometric contribution to that surface changed.
The small-patch approximation matters. For a large angular field, integrate radiance times the cosine factor over its actual directions rather than multiply by one central angle. For a nonuniform spectrum, integrate with the instrument’s spectral response rather than multiply by an arbitrary bandwidth.
Why distance needs a geometry statement
A compact source can subtend a smaller solid angle at a more distant receiver. The received power then changes even if radiance along an unobstructed ray is conserved in the simple vacuum model. “Light falls with distance” is therefore incomplete until the quantity and source geometry are identified.
This distinction prevents a common double correction. If a calculation has already accounted for a source’s shrinking apparent solid angle, adding another inverse-square factor to its radiance can count geometric dilution twice. Draw the source, viewing cone and detector first; assign each factor to one physical operation.
CHAPTER 2 OF 14 · Define quantities and mechanisms
2. Separate absorption, emission and scattering
Absorption removes radiant energy from the tracked beam and transfers it to matter through an interaction. Emission adds radiation produced by matter. Scattering changes propagation direction and can also change frequency in some processes. These mechanisms enter the same transfer problem but do not have interchangeable consequences.
For a narrow direct beam, both absorption and scattering out of the beam reduce its intensity. Their combined removal coefficient is extinction. A detector collecting a wider angular range may recover some scattered light, so its measured attenuation can differ from the attenuation of the unscattered beam.
Two samples with the same direct transmission
Imagine two passive samples that each transmit 60 of every 100 units of incident direct-beam energy without interaction. In sample A, the remaining 40 units are absorbed. In sample B, 10 units are absorbed and 30 are scattered away from the original direction. A perfectly narrow direct-beam detector reports the same transmission for both.
An integrating collection system could distinguish them if it measures the scattered output appropriately. A heating measurement could also distinguish their absorbed energy, subject to other energy exchanges. The original direct-beam number alone cannot determine the energy deposited in the sample.
Suppose a second detector admits 12 units of B’s scattered radiation in addition to the 60 direct units. Its reading is 72. This does not violate the earlier transmission measurement; the two instruments selected different optical paths. State that selection before interpreting a discrepancy as instrument failure.
Keep the accounting boundary visible
For a steady passive sample with negligible emission in the measured band, incident energy is divided among transmitted, reflected or otherwise scattered outgoing energy, and absorbed energy. If there is luminescence, thermal emission, stored energy or a changing sample, extend the budget accordingly.
The specialist guide to LiDAR multiple scattering and waveform energy closure applies this path accounting to a canopy. A missing return is not automatically absorption: energy may leave the viewing cone, arrive later or fall below a processing threshold.
CHAPTER 3 OF 14 · Define quantities and mechanisms
3. Convert distance into optical depth
Optical depth measures accumulated interaction strength along a path. If the extinction coefficient χ has units m⁻¹, a uniform layer of thickness L has optical depth τ = χL. In a nonuniform medium, integrate χ along the path. Optical depth is dimensionless because inverse length is multiplied by length.
The direct unscattered beam in a passive medium follows I_out = I_in exp(−τ), assuming the appropriate extinction description and no addition to that selected beam. This exponential describes the surviving fraction after repeated opportunities for removal. It does not say that the same absolute amount disappears from every equal slice.
Build a layered example
Layer A has χ = 0.2 m⁻¹ and thickness 3 m, giving τ_A = 0.6. Layer B has χ = 0.5 m⁻¹ and thickness 0.8 m, giving τ_B = 0.4. Total optical depth is one, and direct transmission is exp(−1), approximately 0.368.
Starting with 100 units, layer A leaves about 54.9. Layer B then leaves about 36.8. Multiplying transmissions gives the same answer as adding optical depths and taking one exponential. This is a useful independent check on arithmetic and on a multilayer program.
Reversing these purely extinguishing layers leaves the same direct transmission. That conclusion depends on the restricted model: fixed coefficients, no source contribution and no feedback of the radiation on the material. Later we will see why emitting layers need not commute.
Optical thickness is not physical thickness
A long path through a weakly interacting medium and a short path through a strongly interacting medium can have equal optical depth. Conversely, a single material can be optically thin at one wavelength and optically thick at another. Always attach a wavelength or band and a path to the label.
The Potsdam radiative-transfer teaching notes introduce opacity, cross section and optical coordinates. In this guide, χ means extinction per length; other references may use a mass-specific opacity, which must first be multiplied by density. Matching symbols without checking units is unsafe algebra.
Consider a short distance ds along a ray. Let I be its spectral radiance in a chosen direction, and let j represent radiation added per unit path length in the same radiance units. To first order, the change is dI = (j − χI) ds. The loss is proportional to the radiation already present; the source term need not be.
The equation is a local balance, not an instruction to assume every medium is uniform. Both coefficients can vary with position and frequency. Scattering into the direction can be included in the source term, but then that source may depend on radiation arriving from other directions and require a coupled solution.
Check signs through a limiting case
Set j to zero and choose χ positive. The derivative must be negative whenever I is positive: the selected beam weakens along its direction of travel. If a chosen coordinate runs opposite to propagation, the derivative convention changes. The physics must still predict attenuation, not spontaneous gain.
Now set χ to zero while keeping constant j. The solution is I_out = I_in + jL. The source function defined below would involve division by zero, so the original distance equation is the better representation of this limit. A transformed equation can become inconvenient even when the underlying physical balance remains simple.
Define the source function
Where χ is positive, define S = j/χ and dτ = χ ds. The balance becomes dI/dτ = S − I, with τ increasing in the propagation direction used here. S has the same units as I. It indicates the intensity toward which a constant local source-and-removal balance drives the ray.
Czekala’s radiative-transfer course notes derive the integrating-factor solution and discuss the importance of coordinate direction. The notation here deliberately keeps one convention throughout. Before combining formulas from different references, redraw which boundary is zero and which way optical depth increases.
For constant source function S and total optical depth τ, the one-ray solution is I_out = I_in exp(−τ) + S[1 − exp(−τ)]. The first term is surviving background radiation. The second is the accumulated contribution generated along the path, including its subsequent attenuation.
Because exp(−τ) lies between zero and one for a passive layer, the result is a weighted average of I_in and S. That provides a fast check: the outgoing intensity must lie between those two values in this specific constant-source model.
A numerical solution with two checks
Take I_in = 12 units, S = 4 units and τ = ln 2. Transmission is one half, so I_out = 12/2 + 4/2 = 8 units. Six units come from the transmitted background and two from the slab’s source contribution.
If the layer becomes twice as optically thick, τ becomes ln 4 and transmission becomes one quarter. The result is 12/4 + 4 × 3/4 = 6 units. Greater thickness has moved the output closer to the source function, not toward zero.
For a different background, I_in = 1 with S still 4, the original half-transmitting slab gives 2.5 units. The same layer now brightens the ray. Whether a layer appears to absorb or emit relative to its background depends on the contrast between its source function and the incoming intensity.
Explain the limits before trusting a graph
As τ approaches zero, the outgoing intensity approaches the incoming value. As τ grows very large, the background contribution becomes negligible and the output approaches S. These limits can expose a sign error, an omitted source term or swapped boundary conditions.
For a varying S, the output is an integral of contributions weighted by their remaining attenuation to the observer. It is not generally obtained by inserting the arithmetic average of S into the constant-source formula. The position of a source matters because radiation generated farther away has more intervening material to traverse.
CHAPTER 6 OF 14 · Solve rays and layers
6. Test whether the order of layers matters
For one uniform layer, write the transformation as I_out = tI_in + (1−t)S, where t is transmission. Two layers apply two such transformations in sequence. Multiplication of the background transmission factors commutes, but the source contributions carry different foreground attenuation factors.
Reverse two layers
Let both layers transmit one half. Layer A has source function 2 and layer B has source function 10. Begin with I_in = 0. Passing through A first gives intensity 1; passing through B next gives 0.5 × 1 + 0.5 × 10 = 5.5.
Reverse the order. B first produces 5; A then produces 0.5 × 5 + 0.5 × 2 = 3.5. The total optical depth and the product of transmissions are unchanged, but the observed intensity changes by 2 units.
The explanation is physical rather than merely algebraic. Emission from the far layer must traverse the near layer. A bright foreground layer contributes without suffering the other layer’s attenuation afterward. Equal amounts of material can therefore yield different observations when their arrangement changes.
Derive the diagnostic difference
For arbitrary transmissions t_A and t_B, the difference between A-then-B and B-then-A is (1−t_A)(1−t_B)(S_B−S_A). The incoming background cancels from this difference. If either layer is transparent, or if their source functions are equal, this order effect vanishes within the stated model.
These conditions make useful software tests. A multilayer solver should reproduce equality for equal source functions and the calculated difference for unequal ones. Testing a special symmetric case alone can conceal an implementation that puts every source at the wrong depth.
A single measured output generally cannot reveal both source functions and their order. Several configurations may fit one number. Additional wavelengths, viewing directions or independent temperature information are needed to distinguish them. More precise measurement of the same number does not automatically resolve that ambiguity.
CHAPTER 7 OF 14 · Resolve spectra and scattering
7. Keep spectral channels separate until integration
Transfer coefficients can vary strongly with wavelength. A broad detector mixes the results from many spectral intervals according to its response. Averaging coefficients first and solving afterward is generally different from solving each channel and then averaging the outputs.
A two-channel counterexample
Imagine equal incoming power in two narrow channels, each carrying 50 units. The first channel has optical depth zero and the second optical depth two. With no emission or collected scattering, their outputs are 50 and 50 exp(−2), approximately 6.77. Total output is about 56.77 units.
Using the mean optical depth, one, would predict 100 exp(−1), approximately 36.79 units. The error is almost 20 units despite the apparently reasonable average. The exponential operation does not preserve arithmetic averaging of its input.
The effective optical depth inferred from the correct broadband transmission is −ln(0.5677), approximately 0.566. This is a property of the channel weights and this particular measurement, not the arithmetic mean optical depth of the material. Change the source spectrum or detector response and the effective value can change.
What a grey approximation does
A grey approximation removes spectral variation from selected optical properties. It can make a model easier to understand and compute, but its usefulness must be tested against the observable being predicted. A coefficient tuned to one broadband measurement need not reproduce a narrow absorption line.
The Climate Laboratory’s non-scattering transfer chapter explains the role of spectral dependence and the grey-gas simplification. Its atmospheric applications illustrate why a compact model needs a declared purpose. The two-channel case above is a reproducible way to test the averaging issue before studying a detailed atmospheric spectrum.
When comparing spectra expressed per wavelength and per frequency, transform the spectral density as well as the horizontal axis. Equal numerical widths in wavelength do not correspond to equal numerical widths in frequency. Integrating a density with the wrong interval produces an energy-accounting error even if the plotted curve looks plausible.
CHAPTER 8 OF 14 · Resolve spectra and scattering
8. Connect thermal emission to temperature without skipping the path
Under appropriate local thermodynamic equilibrium conditions, the thermal source function for true absorption and emission is the Planck function at the material temperature. This does not mean every observed ray equals the local Planck function: background radiation, optical depth and varying temperatures still enter the transfer solution.
A brightness temperature is the temperature a blackbody would need to reproduce a measured spectral radiance under the chosen convention. It is an interpretation of radiance. It need not equal a material’s kinetic temperature when emission is incomplete, reflection matters or intervening material modifies the signal.
A dimensionless thermal example
Work at one spectral channel and let the Planck radiance of a uniform layer be 10 units. Suppose background radiance is 2 and layer transmission is 0.3. The outgoing value is 0.3 × 2 + 0.7 × 10 = 7.6 units, below the layer’s blackbody radiance.
Now replace the background by 20 units without changing the layer. The output becomes 6 + 7 = 13 units, above the layer’s own blackbody value. An interpretation that assigns every arriving photon to the foreground layer would infer the wrong temperature in both cases.
The specialist article on emissivity and brightness-temperature bias develops a related surface-retrieval problem. A surface model and a volume-transfer model must be connected consistently; they are not interchangeable simply because both use the word emission.
Absence of contrast is not absence of material
If background intensity equals a constant source function, the slab solution gives the same intensity for every optical depth. There can be strongly interacting material along the path without contrast in that channel. Detectability depends on a contrast mechanism as well as on the amount of material.
This provides a useful experimental question: which controlled change would reveal the layer? Changing the background, looking in another spectral channel or measuring a different direction may help. Repeating the identical contrast-free measurement with a more sensitive instrument may still leave the structural ambiguity unresolved.
CHAPTER 9 OF 14 · Resolve spectra and scattering
9. Add scattering without inventing energy
Scattering couples directions. Radiation removed from one ray may become a source for another. A phase function describes the directional redistribution under its stated normalization. A strongly forward-scattering material and an isotropically scattering material can have the same total scattering coefficient yet produce different images and detector signals.
NASA’s aerosol-optics documentation distinguishes extinction, absorption and angular scattering properties. The following discrete example is an original conservation exercise, not a model fitted to an aerosol dataset.
Build a two-direction accounting model
Consider a deliberately simplified redistribution step with 80 energy units moving right and 20 moving left. Suppose each population retains 75% in its current direction and sends 25% into the opposite direction. The new right-moving amount is 0.75 × 80 + 0.25 × 20 = 65. The left-moving amount is 35.
The total remains 100, as required for this elastic redistribution model. The directional difference falls from 60 to 30. The model smooths directional imbalance while conserving total energy; conservation and directional change are separate tests.
If a program instead retains 75% and redistributes 35%, each input unit becomes 1.1 output units. The resulting brightness increase is a normalization error. No amount of spatial grid refinement fixes an interaction rule that creates energy in every step.
Why multiple scattering changes interpretation
After several scattering events, radiation reaching a detector can have followed paths much longer than the straight source-to-detector distance. Absorption along those paths and the detector’s angular acceptance both influence the signal. Treating all collected light as direct transmission can therefore misidentify material properties.
A two-direction model is a teaching abstraction, not a complete phase function. It has no continuous angular structure and cannot predict a detailed scattering halo. Its purpose is to make a conservation test transparent. More detailed models should retain that test while adding the geometry needed for their observations.
For a conservative scattering calculation, distinguish energy leaving through boundaries from energy disappearing numerically. Integrate all outgoing directions when testing global balance. A decrease in one detector’s signal can be physically correct even when the complete radiation field conserves energy exactly.
CHAPTER 10 OF 14 · Infer and diagnose
10. Distinguish a forward prediction from an inverse inference
A forward problem starts with a material model and boundary illumination, then predicts a measurement. An inverse problem begins with measurements and estimates material properties or sources. The second task is often harder because different parameter combinations can produce the same output.
Infer an optical depth with stated assumptions
For a pure attenuator, suppose the measured direct transmission is 0.4. Then τ = −ln(0.4), approximately 0.916. If the path length is independently known to be 2 m and the coefficient is uniform, χ is approximately 0.458 m⁻¹.
Without the path length, the measurement determines χL, not χ and L separately. A layer twice as thick with half the coefficient gives the same transmission. Calling the fitted coefficient a direct measurement of concentration would require another relationship between concentration and extinction.
Show a source ambiguity
Let I_in = 10 and I_out = 6. A source-free layer can explain this with transmission 0.6. But a layer with source function 2 can explain it with transmission 0.5, because 10 × 0.5 + 2 × 0.5 = 6. These are different optical depths with identical measured output.
The ambiguity is not resolved by reporting six decimal places in I_out. It is resolved by adding information that distinguishes the source term from the transmitted background, such as a controlled background change or an independent constraint on emission.
Design the second measurement
Keep the layer fixed and measure two backgrounds, I_a and I_b. In the constant-source model, the output difference is t(I_a−I_b). The common source contribution cancels. If the backgrounds differ by 8 units and the outputs differ by 4, transmission is 0.5.
This experiment works only if the layer remains unchanged between observations. If changing illumination heats the sample or alters its state, the cancellation no longer isolates one fixed t. A good inverse method states the experimental invariance it needs, not just the algebra that follows from it.
CHAPTER 11 OF 14 · Infer and diagnose
11. Diagnose saturation, noise and numerical error
At high optical depth, the direct transmitted signal becomes small. An additive background or detector offset can then dominate its relative error. Taking a logarithm of a nearly zero corrected signal amplifies uncertainty and can produce misleading optical-depth estimates.
An offset case
Suppose true transmission is 0.01 but an uncorrected additive contribution equals 0.005 of the incident reference. The instrument reports 0.015. The true optical depth is about 4.605; the naive estimate is −ln(0.015), about 4.200. A small absolute offset has produced a substantial change in the inferred depth.
The repair is to characterize the additive contribution and its uncertainty, not simply increase the number of digits in the logarithm. If corrected values become nonpositive because of noise, they cannot be treated as ordinary positive transmissions. Report the measurement limitation and use an analysis appropriate to detection limits.
Test a numerical step
For pure attenuation, an explicit first-order step gives I_next = I(1−Δτ). With Δτ = 1.5, it predicts a negative intensity from a positive input. The exact constant-coefficient update is I_next = I exp(−Δτ), which remains positive.
For a cell with constant S, the exact cell update is tI + (1−t)S. It reproduces the uniform-slab solution when repeated across any partition with the same source function and total optical depth. This is a stronger verification than checking whether a plotted curve looks smooth.
For spatially varying coefficients, refine the discretization and compare the quantity of interest. Convergence with respect to grid size is necessary for many calculations but does not establish that the physical model is correct. A perfectly converged solver can still omit relevant scattering or use a wrong boundary condition.
Check units, positivity where required, conservation, limiting cases and a reproducible analytic example before fitting extra parameters. Each test addresses a different failure mode; no single numerical score can substitute for all of them.
CHAPTER 12 OF 14 · Infer and diagnose
12. Transfer the model to a changed viewing geometry
A plane-parallel layer of vertical thickness H has a straight slant-path length H/cos θ for a ray making angle θ with the vertical, provided the layer is sufficiently broad and the flat-layer approximation applies. With uniform coefficients, its slant optical depth is the vertical optical depth divided by cos θ.
Same layer, oblique view
Take vertical optical depth 0.4. At normal viewing, direct transmission is exp(−0.4), approximately 0.670. At 60 degrees from the vertical, the path doubles and transmission becomes exp(−0.8), approximately 0.449.
Now halve the extinction coefficient while retaining the oblique view. The slant optical depth returns to 0.4. The direct transmission matches the original normal-view value. That equality does not imply the images are identical: projected areas, source variation and scattered contributions may still differ.
Know where the shortcut stops
Near grazing angles, a finite layer’s side boundaries or a planet’s curvature can matter. Extending H/cos θ without considering geometry can predict an unrealistic path. The first repair is a geometrically correct ray path, not an arbitrary cap placed on the exponential.
The changed-context test is to explain which quantity was held fixed and which dependence changed. “More oblique means darker” is not a universal conclusion for a glowing or strongly scattering scene. It follows directly only for the selected direct-beam attenuation model with the other assumptions retained.
This is the kind of transfer encouraged by the Education parent: rebuild the explanation from definitions when the situation changes. The learner should be able to calculate a new optical depth, identify the assumptions that survived and name one observation that the simplified calculation cannot predict.
CHAPTER 13 OF 14 · Practise and continue
13. Choose a practical learning and research route
Begin with a spreadsheet containing incident intensity, coefficient, path length and transmission. Plot intensity against distance for several positive coefficients. Verify that doubling length squares transmission rather than merely subtracting the same fixed percentage from the original input.
Next add a constant source function and reproduce the slab examples. Test a background below S, equal to S and above S. These three cases make brightening, unchanged intensity and dimming emerge from one equation. Record the boundary conditions alongside the graph so the plotted curve remains interpretable later.
Build an original investigation
Create a two-layer calculator with adjustable transmissions and source functions. Predict the sign of the layer-order difference before calculating it. Then test the equal-source and transparent-layer limits. This project develops a habit of designing checks that could expose a wrong implementation.
Extend the calculator to two spectral channels and reproduce the failure of mean-opacity averaging. Change the source weights while retaining the same material properties. Explain why a fitted broadband optical depth changes even though neither channel’s physical optical depth changed.
Follow the existing specialist routes
For remote observations, use the linked emissivity article to separate temperature from measured radiance. Use the LiDAR multiple-scattering article to follow delayed paths and energy budgets. Return to the Radiation root for the broader vocabulary of electromagnetic radiation and instruments.
For mathematical foundations, work through the source-linked university derivations with one consistent coordinate convention. For atmospheric modelling, compare a grey approximation with spectral channels before interpreting a complex output. For astronomy, begin with a uniform slab and then allow source function and opacity to vary along the ray.
Keep a model notebook with five entries for each calculation: observable, boundary illumination, material coefficients, neglected mechanisms and validation case. This is more useful than a collection of formulas detached from their assumptions. When a result disagrees with data, the notebook identifies which physical claim is actually being tested.
CHAPTER 14 OF 14 · Practise and continue
14. Questions that reveal whether the explanation is complete
Does optically thick mean completely black?
No. A thick absorbing and emitting layer can have a substantial outgoing intensity set by its source function. A scattering layer can redirect illumination. The phrase describes interaction strength along a specified path and spectral channel; appearance also requires illumination and source information.
Can the same material be transparent and opaque?
Yes, at different wavelengths or over different path lengths. A useful statement of transparency specifies both. Broadband labels can conceal narrow spectral features or a detector response that weights one region much more heavily than another.
Is less direct light evidence of more heating?
Not by itself. Scattering can remove light from a narrow viewing cone without depositing that energy as heat in the same way as absorption. A heating inference needs an absorbed-energy budget and any relevant nonradiative exchanges.
Does a successful fitted curve identify the true material?
Not necessarily. The inverse examples show exact ambiguities even before measurement noise is introduced. A fit becomes more informative when additional observations constrain source contributions, path lengths, spectral properties or directional behaviour independently.
What should be checked before using a sophisticated solver?
Reproduce direct attenuation, the constant-source slab, two-layer order reversal and a conservative redistribution example. Confirm that the quantities exported by the solver match the instrument quantities you intend to compare. A radiance prediction cannot be compared directly with an irradiance reading without the appropriate angular and spectral integration.
Radiative transfer becomes useful when the path from source to signal remains explicit. The equations connect that path to calculations, while changed backgrounds, wavelengths and viewing directions test the explanation. A trustworthy result shows both what the measurement supports and what further information is needed to distinguish the remaining possibilities.