RF dosimetry and thermal modelling

How RF exposure links to absorbed dose and heating: SAR, absorbed power density, averaging rules, and thermal sanity checks.

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An external field reading does not tell you directly how much power a tissue region absorbs, or how warm that region becomes. Dosimetry connects fields to absorbed power; thermal modelling then accounts for heat storage and transfer. This lesson develops those two steps and shows why an average absorption rate cannot describe every local temperature or exposure history.

Think in a chain:

Source state → EM field distribution → absorbed power → heat transfer → temperature rise

The field-strength lesson explains what an external reading can establish. Here, keep internal and external fields distinct, and label whether a value is instantaneous, on-state or averaged over the full interval. The numerical tissue properties and thermal times below are specified training assumptions, not a clinical prediction or an exposure-limit specification.

Choose the evaluation this situation needs

Use this when: you need to evaluate internal exposure in situations where external measurements alone are insufficient (close-to-body devices, near-field geometries, complex coupling, high spatial gradients).
Core outputs: spatial distributions of SAR (W/kg) and/or absorbed power density (W/m²), plus optional temperature rise Δ T(t,r).
What “good” looks like: clearly defined geometry, operating states, averaging rules, uncertainty drivers, and a validation story (measurement, benchmarking, or standard-aligned method).
Big idea: heating depends on absorption and heat transport over time and space. Apply every criterion required by the adopted framework, including separate brief-exposure conditions where specified; an acceptable long-window average does not automatically resolve a short exposure.

Connect absorption, power flow and heat storage

Absorption: SAR and absorbed power

Specific absorption rate (SAR) is absorbed power per mass:

SAR = dP_abs/dm = (1/ρ)dP_abs/dV [W kg⁻¹].

A commonly used conductive-loss relationship is:

SAR = (σ |E_(rms,int)|²)/ρ,

where σ is tissue conductivity (S/m), ρ is tissue density (kg/m³), and E_(rms,int) is the internal RMS electric field (V/m).

Absorbed power in a volume:

P_abs = ∫V ρ SAR dV.

Mass-averaged SAR over a defined averaging mass m:

SARₘ = 1/m∫_Vₘ ρ SAR dV.

Here Vₘ is the region containing mass m = ∫_Vₘρ dV. This is a mass-weighted average: for uniform subregions, sum mᵢSARᵢ to find absorbed power, then divide by their total mass. Averaging the listed SAR values equally works only when the subregions have equal masses. A volume average is also different when densities differ.

Standards define how the averaging mass is selected and searched—do not invent your own procedure. ICNIRP 2020, Table 2 notes and Appendix A, specify the mass, time and area conventions for that framework. The calculations here illustrate weighting; they do not implement its spatial search or prove compliance.

From fields to power flow

Instantaneous Poynting vector:

S(t) = E(t) × H(t).

For time-harmonic fields using peak-amplitude phasors:

⟨S⟩ = (1/2)Re{E × H*}.

In a plane wave (far field), with η₀ ≈ 377 Ω:

S ≈ Eᵣₘₛ²/η₀ ≈ Hᵣₘₛ² η₀.

In the near field, these plane-wave relations can fail; you may need both E and H, or direct dosimetry.

Temperature rise: a working thermal model

Pennes bioheat equation:

ρ c (∂ T)/(∂ t) = ∇ · (k∇ T) + ρ SAR + Qₘₑₜ-ρb cb ωb (T-Tb).

Here T is tissue temperature; c is specific heat capacity in J/(kg·K), k thermal conductivity in W/(m·K), and Qₘₑₜ metabolic heat generation per volume in W/m³. The blood density ρb, specific heat cb and perfusion rate ωb describe exchange with blood at temperature Tb; in this convention ωb has units s⁻¹. Every term has units W/m³. The perfusion term removes heat when tissue is warmer than blood and supplies heat when it is cooler. Conduction can likewise carry heat into or out of a region.

For a uniformly heated element, the short-time adiabatic estimate of the RF-induced rise (neglecting heat transfer and perfusion) is:

Δ T ≈ SAR/c t.

To obtain this estimate, write the added RF energy as m SAR t and the element’s heat-capacity change as mcΔ T. Equating them cancels the mass. For a time-varying SAR, replace SAR t by ∫SAR(t) dt. This estimates the RF-induced change relative to the initial thermal baseline, assuming constant heat capacity and no added heat exchange.

Use this for sanity checks; temperature is boundary-condition sensitive. A spatially averaged SAR does not bound a smaller local hotspot’s temperature. Applying the average to every point silently replaces a nonuniform absorption distribution with a uniform one.

Select and report the required metric

Start by deciding whether you need an internal metric (SAR / absorbed PD) or whether an external metric (E/H/incident PD) is sufficient.

When external metrics are usually sufficient

External metrics are most defensible when:

You are in a clear far-field environment (plane-wave-like),
Your framework provides reference levels in terms of E/H/S for that frequency range,
Your instrumentation supports the required averaging.

When dosimetry becomes the right tool

Dosimetry is indicated when:

Near-field coupling dominates (close distances, strong gradients, reactive fields),
Geometry is complex (body near conductors/enclosures/vehicles; wearable/handset positioning),
Surface-localized absorption is the endpoint,
Time variability can’t be confidently captured by measurement (duty, scheduling, beam behavior).

What to report

Scenario definition: bands, bandwidth, max/typical states, time behavior assumptions.
Quantity and averaging: SAR or absorbed PD with explicit time window and mass/area averaging definitions.
Peak location and mechanism: where the maximum occurs and why.
Validation evidence: standard-aligned method, benchmarking, and/or measurement cross-checks.
Uncertainty drivers: tissue properties, positioning tolerances, numerical discretization, and source variability.

State the averaging region and time window

Time averaging: energy over a window

For a power-like quantity y(t):

y bar = 1/T∫_t₀^(t₀ + T) y(t) dt.

Two practical questions:

What averaging window does the framework define?
Where is the worst-case window placement (window position that yields the highest average)?

Spatial, mass and area averaging

Pointwise maxima can be dominated by tiny geometric details, mesh/probe artifacts, or sensitive placement. Averaging over a defined mass/area reduces sensitivity and better reflects heat diffusion and repeatability.

Calculate absorbed power, averaged SAR and temperature rise

Find absorbed power in an averaging mass

Problem

Local mass-averaged SAR is 1.5 W/kg over 10 g = 0.01 kg:

Find the absorbed power.

Show full solution

Select the averaging mass
Method
Use the mass in kilograms with the mass-averaged SAR.
Reason
Multiplying the energy absorption rate per unit mass by that mass gives the absorbed power.
Working
Answer
Local mass-averaged SAR is 1.5 W/kg over 10 g = 0.01 kg:
State the absorbed power
Working
Answer
P_abs ≈ SAR · m = 1.5 × 0.01 = 0.015 W = 15 mW.

Average SAR over repeated bursts

Problem

SARₒₙ = 6 W/kg during bursts; tₒₙ = 2 ms every Tᵣₑₚ = 20 ms:

Assume constant SAR during each burst, zero between bursts and an averaging interval containing complete periods. Find the duty cycle and time-averaged SAR.

Show full solution

Apply the duty cycle
Method
Divide the burst duration by the repetition period and multiply the on-state SAR by that fraction.
Reason
The on-state absorption contributes only during the transmitting fraction of each period.
Working
Answer
D = tₒₙ/Tᵣₑₚ = 0.10, SAR bar ≈ D SARₒₙ = 0.6 W/kg.
State the averaging cautions
Working
Answer
The duty cycle is 0.10, and the time-averaged SAR is 0.6 W/kg. SAR is power per mass, so it scales with D. The corresponding RMS field over the full interval scales with square root of D, not D. Peak-hold readings do not perform this averaging.

Estimate SAR in a superficial absorbing layer

Problem

Uniform, normally incident power density S_inc = 10 W/m² over A = 100 cm² = 0.01 m²:

Assume 50% absorption, layer thickness δ = 2 mm and ρ ≈ 1000 kg/m³.

Estimate the incident power, absorbed power, layer mass and SAR; this is not a compliance conversion.

Show full solution

Find incident power
Method
Multiply the incident power density by the exposed area.
Reason
Power density integrated over the stated area gives the incident power.
Working
Answer
P_inc = S_incA = 0.10 W.
Specify absorption and layer geometry
Method
Apply the stated absorbed fraction and identify the layer thickness and density.
Reason
The absorbed power and absorbing mass are both needed for the SAR estimate.
Working
Answer
Assume 50% absorbed: P_abs ≈ 0.05 W. Concentrated in layer thickness δ = 2 mm with ρ ≈ 1000 kg/m³:
State the layer mass and SAR
Working
Answer
m = ρ Aδ = 0.02 kg, SAR ≈ P_abs/m = 2.5 W/kg.

Check the estimated temperature rise

Problem

Uniform SAR 2 W/kg sustained for t = 60 s in a tissue element with c ≈ 3600 J/(kg · K). Ignore heat transfer and perfusion and estimate only the RF-induced rise:

Estimate the temperature rise and state what to check if a model predicts much larger rises.

Show full solution

Estimate the temperature rise
Method
Divide SAR by the specific heat and multiply by the exposure duration.
Reason
The stated thermal estimate relates absorbed energy per unit mass to temperature change.
Working
Answer
Δ T ≈ (SAR/c)t = (2 W/kg × 60 s)/(3600 J/(kg · K)) ≈ 0.033 K.
State the modelling check
Working
Answer
The RF-induced rise is 0.033 K under these assumptions. A much larger prediction for the same element and input needs investigation: check the SAR value and averaging, heat-transfer boundary conditions and perfusion. A smaller hotspot may have higher local SAR, so this calculation does not bound every temperature in a nonuniform tissue volume.

Resolve the absorption distribution before estimating heating

Problem

A training volume contains two uniform subregions: 2.0 g with SAR 6.0 W/kg and 8.0 g with SAR 0.50 W/kg. Both values are constant for 30 s. Each subregion has specific heat c = 3600 J/(kg · K). They start at the same temperature; neglect all heat exchange, including exchange between the subregions, and estimate only the added RF heating.

Find the total absorbed power and the mass-averaged SAR over the whole 10 g. Then find the temperature rise in the higher-SAR subregion and the mass-averaged rise over the whole volume. Explain whether the averaged SAR can be assigned to both subregions or used to bound the hotter subregion’s rise. This volume is a weighting exercise, not a standard-defined peak-search result.

Account for each subregion before averaging

Total absorbed power

Unit: mW

Mass-averaged SAR over the complete volume

Unit: W/kg

Temperature rise in the higher-SAR subregion

Unit: K

Mass-averaged temperature rise over the complete volume

Unit: K

Interpret the average and its limits (self-review)

Hints

Hint 1

Start with conserved power

Each contribution is mass times SAR. Adding those powers preserves the energy account; averaging the two SAR values equally does not.

Hint 2

Keep local and averaged results separate

Under the stated thermal assumptions, each subregion’s rise follows its own SAR. Their mean is weighted by mass because the heat capacity is the same per kilogram.

Show solution step by step

Recover absorbed power and average SAR
Method
Sum the power contributions, then divide by the complete mass.
Reason
SAR is absorbed power per mass; the average must preserve total absorbed power.
Working
Answer
The subregions absorb (0.0020)(6.0) = 0.012 W and (0.0080)(0.50) = 0.0040 W. Thus P_abs = 0.016 W = 16 mW and SAR_10g = 0.016/0.010 = 1.6 W/kg. Multiplying the average by 0.010 kg recovers the total power.
Estimate local and average rises
Method
Apply the adiabatic energy balance to each subregion, then weight the rises.
Reason
The local absorption is nonuniform, but the stipulated heat capacity and exposure duration are common.
Working
Answer
The higher-SAR subregion rises by 6.0(30)/3600 = 0.050 K; the other rises by 0.50(30)/3600 ≈ 0.00417 K. The mass fractions are 0.20 and 0.80, giving 0.20(0.050) + 0.80(0.00417) ≈ 0.0133 K. Equivalently, 1.6(30)/3600 ≈ 0.0133 K.
Interpret the distinction
Working
Answer
The volume absorbs 16 mW and averages 1.6 W/kg. Its mean rise is about 0.013 K, while the hotter subregion rises by 0.050 K under this model. Assigning the mean SAR everywhere preserves the total power but loses the hotspot; it cannot bound the local rise. The equivalence between mean SAR and mean rise here depends on the common heat capacity and neglected heat exchange. Real temperature prediction requires local absorption, properties, conduction, perfusion and boundary conditions. Compare your explanation against those distinctions; the checker evaluates the four numerical quantities only.

Include heat loss when following temperature over time

An adiabatic calculation stores all added RF energy. To see what changes when heat leaves, use a deliberately simple one-element model. Let θ be the element’s added temperature above a fixed baseline, and suppose its incremental heat loss is proportional to θ. A positive time constant τ describes that loss. With constant specific heat c, dθ/dt = SAR(t)/c-θ/τ. The first term supplies heating in K/s; the second removes it. This replaces spatial heat transfer with one stipulated response time. It does not model a particular body region, thermoregulation or changing surroundings.

For constant SAR and initial θ(0) = 0, the solution is θ(t) = θ_∞(1-e^(-t/τ)), θ_∞ = (SAR τ)/c. Differentiate the expression: its rate is (SAR/c)e^(-t/τ), equal to the heating term minus the loss term. At times much shorter than τ, 1-e^(-t/τ) ≈ t/τ, recovering the adiabatic result. At long times the temperature approaches a finite rise, where input and loss balance. After one time constant it reaches about 63% of that rise.

If RF absorption then stops, the added temperature remains initially positive and decays as θ(t_off + u) = θ(t_off)e^(-u/τ), where u is elapsed time after switch-off. Zero SAR stops new RF heating; it does not instantly remove stored heat. ICNIRP 2020, Appendix A, equation 22, discusses a first-order lag under simplified conditions. The numerical time constant below is supplied for this exercise, rather than inferred from that discussion.

Check your understanding 1

Two training histories act on separate identical one-element models, initially at their fixed baseline, with c = 3600 J/(kg · K) and τ = 30 s:

History A: SAR is 6.0 W/kg for the first 60 s, then zero for 30 s.
History B: SAR is 4.0 W/kg throughout the complete 90 s.

Find the absorbed RF energy per mass in each history. For A, find the added temperature at 60 s and 90 s. For B, find it at 90 s. Both histories have the same 90 s time-averaged SAR: does that require equal final temperatures in this model? Explain the decisive difference, and state why none of these results establishes exposure compliance or predicts a person’s temperature.

Absorbed RF energy per mass in either history

Unit: J/kg

History A: added temperature at 60 s

Unit: K

History A: added temperature at 90 s

Unit: K

History B: added temperature at 90 s

Unit: K

Compare the histories and delimit the inference (self-review)

Show answer

Both absorb 6.0(60) = 4.0(90) = 360 J/kg, averaging 4.0 W/kg over 90 s. For A, the on-period equilibrium rise would be 6.0(30)/3600 = 0.050 K. At 60 s, θA = 0.050(1-e⁻²) ≈ 0.0432 K. During the next 30 s it falls by a factor e⁻¹, giving θA(90) ≈ 0.0159 K.

For B, the equilibrium rise is 4.0(30)/3600 ≈ 0.0333 K, and θB(90) = (4.0)(30)(1-e⁻³)/3600 ≈ 0.0317 K. B finishes warmer despite equal absorbed energy: it continues to supply heat while A is cooling. Each result is below the 0.10 K that the equal energies would produce without any heat loss. The energy histories have the same integral, but their timing relative to heat loss differs.

Compare your explanation for energy equality, the two-stage history A, continuing absorption in B and the stipulated thermal assumptions. The checker evaluates the energy and three temperature rises only. The exercise supplies no frequency, adopted exposure category, spatial evaluation, applicable limit or validated tissue model. It illustrates temperature dynamics; it does not replace standard-defined averaging or establish a compliance or clinical conclusion.

Check the model and its inputs

Treating an air-field reading as the internal field in SAR = σ E²/ρ.
Mixing peak, RMS, and time-average quantities (especially with peak-hold meters or high PAPR waveforms).
Reporting “maximum SAR” without stating averaging mass and peak-search procedure.
Using plane-wave conversions in the near field.
Thermal modelling with undocumented boundary conditions (convection/contact/ambient) or unrealistic perfusion assumptions.
Numerical artifacts: mesh-dependent hotspots, insufficient convergence checks, unvalidated material properties.

Knowledge check

Check your understanding 2

What is the “chain” that a dosimetry + thermal argument should follow?

Show answer

Source state → EM field distribution → absorption (SAR / absorbed power density) → heat transfer → temperature rise. A defensible argument states assumptions and averaging at each step.

Check your understanding 3

Why do standards use averaging windows and averaging masses/areas?

Show answer

Because pointwise peaks can be dominated by tiny geometry details and measurement/model artifacts, and heating depends on energy deposition over time and over a region. Averaging improves repeatability and ties the metric to physiological relevance.

Check your understanding 4

When is dosimetry typically indicated rather than external E/H/S measurements?

Show answer

When near-field coupling dominates, geometry is complex, or the framework requires internal/surface dose metrics (close-to-body devices, steep gradients, refined evaluation after screening exceedance).

Exercises

Check your understanding 5

If SAR = 1.8 W/kg averaged over 10 g, estimate absorbed power in that averaging mass.

Show answer

m = 10 g = 0.01 kg. P_abs ≈ SAR · m = 1.8 × 0.01 = 0.018 W = 18 mW.

Check your understanding 6

Estimate the RF-induced Δ T in a uniformly heated element for SAR = 2 W/kg sustained for t = 30 s with c = 3600 J/(kg · K). Ignore heat transfer and perfusion.

Show answer

Δ T ≈ (SAR/c) t = (2/3600) × 30 ≈ 0.0167 K.

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

Last reviewed: 2026-02-13

Choose the evaluation this situation needs

Connect absorption, power flow and heat storage

Absorption: SAR and absorbed power

From fields to power flow

Temperature rise: a working thermal model

Select and report the required metric

When external metrics are usually sufficient

When dosimetry becomes the right tool

What to report

State the averaging region and time window

Time averaging: energy over a window

Spatial, mass and area averaging

Calculate absorbed power, averaged SAR and temperature rise

Find absorbed power in an averaging mass

Average SAR over repeated bursts

Estimate SAR in a superficial absorbing layer

Check the estimated temperature rise

Resolve the absorption distribution before estimating heating

Include heat loss when following temperature over time

Check your understanding 1

Check the model and its inputs

Further reading

Knowledge check

Check your understanding 2

Check your understanding 3

Check your understanding 4

Exercises

Check your understanding 5

Check your understanding 6