Specific absorption rate: measurement and pitfalls
Understand what SAR (W/kg) measures, when it applies, how it is measured or modelled, and how it is most often misused.
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ContinueSpecific absorption rate (SAR) is RF power absorbed per tissue mass, in W/kg. It is an internal absorption rate, not a field strength in air and not accumulated energy. A meaningful result states the operating state, body-coupling geometry and both mass and time averaging rules.
Use the field foundations to distinguish internal and external quantities, and duty-cycle averaging to distinguish squared-field averages from amplitude averages.
Define the quantity before calculating
Pointwise SAR is SAR = dP_abs/dm = (1/ρ)dP_abs/dV. For a linear isotropic conductive-loss model with a sinusoidal internal RMS field, SAR = (σ |E_(rms,int)|²)/ρ. Use tissue conductivity σ appropriate to the frequency and model, and density ρ. A sinusoid’s peak amplitude is square root of 2 times its RMS value. Other waveforms need their own mean square; do not apply the sinusoidal factor indiscriminately.
An air-field probe does not measure E_(rms,int). Coupling depends on body and source geometry, polarisation, frequency and nearby structures. Substituting an air reading in this internal-field formula cannot establish SAR without a validated coupling argument.
Average absorbed power with the correct mass
For a specified tissue region Vₘ of mass m = ∫_Vₘρ dV, P_abs = ∫_Vₘρ SAR dV, SARₘ = P_abs/m. Average by mass, not by the number of sampled points or regions. Whole-body SAR uses the body’s total absorbed power and mass; a local average uses its defined tissue region. A small high-absorption region can have little influence on the whole-body average and still control a required local comparison.
A compliance calculation must also implement the prescribed region shape, mass, search procedure and time window. A conveniently chosen 10 g region is not automatically the required maximum local SAR. For example, the ICNIRP 2020 guidelines, Table 2 and its notes, define separate whole-body and local comparisons, with the local heating metric changing above 6 GHz. A high-frequency superficial-absorption question may therefore require absorbed power density, rather than treating SAR as the answer to every RF question.
Average power over the required time window
For fixed on-state spatial absorption and zero between bursts, the average over complete identical periods is SAR bar = D SARₒₙ. The assumptions matter: changes in transmitted power, position or coupling can change the on-state distribution, and an arbitrary short window need not contain the same duty fraction.
The RMS field corresponding to this on/off mean square scales as square root of D, not D. SAR and absorbed power scale as D for that fixed on-state model. Also check any required short-duration energy restrictions; a passing long-window average cannot close every exposure condition.
Guided practice 1
Combine mass and time averages in order
Problem
In a supplied model, a specified 10 g tissue region contains 4.0 g at on-state SAR 5.0 W/kg and 6.0 g at 1.0 W/kg. The spatial pattern stays fixed, with an on-state duty fraction of 0.40 and zero absorption between bursts. Average over complete periods. Find the on-state absorbed power in the region, its on-state mass-averaged SAR and its time-averaged SAR. Does this region alone establish the maximum local or whole-body SAR?
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
Add absorbed powers first. Taking the arithmetic mean of the two SAR values would weight unequal masses equally.Show solution step by step
Use the mass weights
Method
Calculate P = ∑ᵢ mᵢ SARᵢ.Reason
Mass-averaged SAR is total absorbed power divided by the averaging mass.Working
Answer
Pₒₙ = (0.0040)(5.0) + (0.0060)(1.0) = 0.026 W. The region’s mass is 0.010 kg, so SAR_(m,on) = 0.026/0.010 = 2.6 W/kg. The unweighted arithmetic mean of 3.0 W/kg is wrong for these unequal masses.
Apply the time weight and delimit the result
Method
Multiply by 0.40 for the specified complete-period average.Reason
The calculation proves only the average of the supplied region and exposure history.Working
Answer
SAR bar ₘ = 0.40(2.6) = 1.04 W/kg, and average absorbed power in the region is 0.0104 W. No other regions or whole-body absorbed power were supplied, so neither the maximum local value nor the whole-body average is established. No limit comparison was supplied either. Check your conclusion as well as the three numerical entries.
Connect absorption to temperature without equating them
For a uniform element with constant SAR and specific heat capacity c, neglecting heat exchange gives Δ T ≈ SAR t/c. If SAR varies, integrate its history over time. Real conduction, perfusion and geometry change the temperature response. This short-time energy estimate is not a compliance method or a universal prediction of local temperature from a mass-averaged value.
SAR can be obtained from calibrated tissue-equivalent measurements or validated numerical models. Reports need the device configuration, position, material properties, resolution and convergence or measurement uncertainty, plus the prescribed peak search and averaging. A published device test value belongs to those test conditions; it is not a live personal absorbed-dose reading.
Calculate SAR and distinguish local from whole-body averages
Calculate SAR from a peak electric field
Worked example 2
Calculate SAR from a peak electric field
Problem
A sinusoidal internal field has peak amplitude Eₚₑₐₖ = 100 V/m:
Use σ = 1.0 S/m and ρ = 1000 kg/m³.
Find the RMS field and SAR, and explain the error from treating the peak field as RMS.
Show full solution
Convert peak field to RMS
Method
Divide the peak field by the square root of two.Reason
The stated SAR calculation uses RMS field rather than peak amplitude.Working
Answer
Eᵣₘₛ = Eₚₑₐₖ/(square root of 2) = 70.7 V/m.Calculate SAR
Method
Use conductivity times the squared RMS field divided by density.Reason
SAR depends on the field squared, so the amplitude convention affects the result.Working
Answer
With σ = 1.0 S/m and ρ = 1000 kg/m³:
SAR = (σ Eᵣₘₛ²)/ρ ≈ 5.0 W/kg.State the peak-field error
Working
Answer
The RMS field is 70.7 V/m and SAR is 5.0 W/kg. Treating 100 V/m as RMS would give 10 W/kg, exactly twice the correct value for this sinusoid. The peak-to-RMS factor is waveform-dependent.
Convert local SAR to absorbed power
Worked example 3
Convert local SAR to absorbed power
Problem
SAR = 2.0 W/kg over m = 10 g = 0.01 kg:
Find the absorbed power in the stated mass.
Show full solution
Select the averaging mass
Method
Use the stated local mass in kilograms.Reason
The mass-averaged absorption rate becomes power when multiplied by the averaging mass.Working
Answer
SAR = 2.0 W/kg over m = 10 g = 0.01 kg:
State the absorbed power
Working
Answer
P_abs ≈ SAR · m = 0.02 W = 20 mW.
Average SAR from repeated transmit bursts
Worked example 4
Average SAR from repeated transmit bursts
Problem
Constant SARₒₙ = 6 W/kg occurs for tₒₙ = 1 ms every Tᵣₑₚ = 10 ms, with zero SAR between bursts. Average over complete periods:
Find the time-averaged SAR.
Show full solution
Identify the burst duty fraction
Method
Use the ratio of burst duration to repetition period.Reason
The time average weights the on-state SAR by the fraction of time spent transmitting.Working
Answer
SARₒₙ = 6 W/kg, with tₒₙ = 1 ms every Tᵣₑₚ = 10 ms (D = 0.10):
State the time-averaged SAR
Working
Answer
SAR bar ≈ D SARₒₙ = 0.6 W/kg.
Compare local and whole-body SAR averages
Worked example 5
Compare local and whole-body SAR averages
Problem
Find the whole-body average for a 10 g region at 4 W/kg and the remaining 70 kg at 0.02 W/kg; interpret the effect of the local peak.
Show full solution
Weight each region by its mass
Method
Combine each region’s absorption rate with its mass before averaging over the body.Reason
A small local region and the rest of the body contribute in proportion to their masses.Working
Answer
P_abs = 4 W/kg × 0.010 kg + 0.02 W/kg × 70 kg = 1.44 W.The total mass is 70.010 kg, so SAR_WB = 1.44/70.010 W/kg.
State the whole-body interpretation
Working
Answer
The whole-body average is 0.0206 W/kg, while the 10 g region remains at 4 W/kg. The small region has little effect on the whole-body average; its local criterion must still be evaluated separately. A low whole-body average cannot establish that every local average passes.
Check a SAR claim
Ask whether the field is internal, whether its amplitude is RMS or peak, and whether tissue properties match the frequency. Reconstruct absorbed power and mass weighting before time averaging. State the averaging region, search, time-window placement and operating states. Compare all applicable local, whole-body and short-duration criteria using the adopted definitions; do not let a favourable whole-body average replace the local check.
For external surveys, use field and power-density metrics where the governing method permits them. An external reference-level comparison and a direct internal-dosimetry comparison are different assessment routes; a failed reference-level comparison does not by itself quantify internal SAR.
Knowledge check
Check your understanding 1
Why is a device’s published compliance SAR not a live measure of its user’s absorbed dose?
Show answer
The published value comes from defined test states, positions and averaging rules. Actual use can have different power, duty and coupling to the body. SAR is a physical absorption rate, but that particular test result is not a live personal dosimeter reading.
Check your understanding 2
What two averaging definitions must always accompany a local SAR value?
Show answer
Mass averaging (e.g., the defined averaging mass and search procedure) and time averaging (the defined time window and how it was applied).
Check your understanding 3
Why is a “one-line conversion” from air E to SAR generally not defensible?
Show answer
Because internal absorption depends on coupling geometry, posture, polarization, and near-field behavior. Air E does not uniquely determine internal tissue fields without a model.
Exercises
Check your understanding 4
Exercise 1 — RMS vs peak. If Eₚₑₐₖ = 80 V/m in tissue (sinusoid), what is Eᵣₘₛ?
Show answer
Eᵣₘₛ = Eₚₑₐₖ/square root of 2 ≈ 56.6 V/m.
Check your understanding 5
Exercise 2 — SAR scaling sanity. If internal Eᵣₘₛ increases by 20%, how does SAR change (all else equal)?
Show answer
SAR scales with Eᵣₘₛ², so it changes by about (1.2)² = 1.44 (a 44% increase).
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Syllabus and review details
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