Combining uncertainties: absolute uncertainties for sums and differences

Percentage uncertainties for products and quotients, and multiplying by the power

Percentage difference from an accepted value, compared with the percentage uncertainty, to judge agreement

Error bars, steepest and shallowest gradient lines, and the uncertainty in a gradient and an intercept

Natural and base-10 logarithms, the rules you need, and significant figures in logs

Plotting ln y against x to test y = Aeᵏˣ (capacitor discharge, radioactive decay, absorption)

Plotting log y against log x to test y = kxⁿ and find n and k

Units and significant figures for log columns, axes, gradients and intercepts

1. At A2 you carry uncertainties all the way into the final answer: add absolute uncertainties for sums and differences, add percentage uncertainties for products and quotients, and use the result to judge whether the answer agrees with an accepted value.

2. A quantity raised to a power n contributes n times its percentage uncertainty (a square root is the power ½). A small difference between two large readings can have a large percentage uncertainty.

3. Percentage difference = |experimental − accepted| ÷ accepted × 100%. If it is smaller than the percentage uncertainty, the result agrees with the accepted value within its uncertainty.

4. Draw error bars, then the steepest and shallowest lines that still pass through them. Uncertainty in the gradient ≈ (steepest − shallowest) ÷ 2.

5. y = Aeᵏˣ → plot ln y against x: gradient k, intercept ln A. y = kxⁿ → plot log y against log x: gradient n, intercept log k.

6. A log has no unit: label axes ln (V / V).

Give a log to as many decimal places as the number has significant figures.

At AS you found the uncertainty of one reading at a time. At A2 most results are calculated from several readings, so you need rules for combining uncertainties. These rules give a cautious, worst-case estimate. Start with a result that is a sum or a difference of two measured values. Each reading could be out by its full uncertainty, and the two errors could happen to push the result the same way. That is true even for a subtraction: if the first reading is slightly too high and the second slightly too low, the difference is too large by both amounts together. So the absolute uncertainties add.

Typical cases: a temperature rise, an extension from two scale readings, and a count rate corrected for background. Work out the absolute uncertainty first; turn it into a percentage only if the result goes on into a product or quotient.

The danger is a small difference between two large readings: the absolute uncertainty stays the same size but becomes a much larger fraction of the result. Design the experiment so the difference is as large as practical.

WORKED EXAMPLE · QUESTION

A block is heated electrically. A thermometer with 1 °C divisions reads 18.5 °C at the start and 46.0 °C at the end; each reading is ± 0.5 °C. Calculate the temperature rise, its absolute uncertainty and its percentage uncertainty, and compare the last with the percentage uncertainty of the final reading alone.

SOLUTION

Step 1 The rise is a difference: Δθ = 46.0 − 18.5 = 27.5 °C.

Step 2 Add the absolute uncertainties: 0.5 + 0.5 = ± 1.0 °C.

Step 3 Percentage uncertainty: 1.0 ÷ 27.5 × 100 = 3.6%. The 46.0 °C reading alone is only 0.5 ÷ 46.0 × 100 = 1.1%.

Δθ = 27.5 °C ± 1.0 °C (3.6%), over three times the 1.1% of one reading. Heating for longer would cut the percentage, because the ± 1.0 °C stays the same.

For products and quotients the percentage uncertainties combine. If a is 2% too big and b is 3% too big, ab is about 5% too big. If b is 3% too small and on the bottom of a fraction, a/b is again about 5% too big: uncertainties never cancel.

Powers. A square doubles the percentage uncertainty, a cube triples it and a square root halves it. For small uncertainties (1+p)² ≈1 + 2p, which is where the factor comes from.

Exact numbers (2, π, ½, a count of 20 oscillations) have no uncertainty and contribute nothing.

A constant multiplier changes the value but not the percentage uncertainty: T = (time for 20 swings) ÷ 20 has the same percentage uncertainty as the time.

Mixed calculations such as c = E ÷mΔθ: absolute uncertainty of the difference first, then its percentage, then add percentages.

Rounding. Give the final absolute uncertainty to 1 (or 2) significant figures and the value to the same decimal place.

The quantity with the highest power usually dominates the total, so it deserves the most precise instrument: for a wire, measure the diameter with a micrometer at several places and average.

WORKED EXAMPLE · QUESTION

For a wire, R = 6.4 Ω ± 0.1 Ω, diameter d = 0.28 mm ± 0.01 mm and length l = 0.850 m ± 0.002 m. Calculate the resistivity ρ = RA/l, where A = πd²/4, and its absolute uncertainty.

SOLUTION

Step 1 The value: A = π(0.28 × 10⁻³)² ÷ 4 = 6.16 × 10⁻⁸ m², so ρ = 6.4 × 6.16 × 10⁻⁸ ÷ 0.850 = 4.64 × 10⁻⁷ Ω m.

Step 2 Percentage uncertainty of each measurement: %R = 0.1 ÷ 6.4 × 100 = 1.6%; %d = 0.01 ÷ 0.28 × 100 = 3.6%; %l = 0.002 ÷ 0.850 × 100 = 0.24%.

Step 3 d is squared, so its percentage counts twice; add everything: %ρ = 1.6 + 2 × 3.6 + 0.24 = 8.9%.

Step 4 Back to an absolute uncertainty: Δρ = 8.9% × 4.64 × 10⁻⁷ = 0.41 × 10⁻⁷ Ω m.

ρ = (4.6 ± 0.4) × 10⁻⁷ Ω m. The uncertainty is quoted to 1 s.f., so the value is rounded to the same decimal place (4.6, not 4.64).

Once a result has an uncertainty, you can ask whether it is any good. Two different percentages answer two different questions.

Percentage difference < percentage uncertainty: the accepted value lies inside the range of your result, so the result agrees with (is consistent with) it within the experimental uncertainty.

Percentage difference > percentage uncertainty: the accepted value lies outside your range. The uncertainties you estimated cannot explain the gap, so there is probably a systematic error, or an uncertainty you have not allowed for. Say what it might be and which way it pushes the result.

Equivalent check: does the accepted value fall inside value ± uncertainty?

With no accepted value (two methods, or a tolerance such as "470 μF ± 10%"), compare the two ranges: if they overlap, the values are consistent.

WORKED EXAMPLE · QUESTION

A student finds the p.d. at which a red LED just starts to conduct, V = 1.74 V ± 0.05 V. The LED's wavelength is given as 630 nm ± 10 nm. She uses h = eVλ/c. (a) Calculate h and its percentage uncertainty. (b) Decide whether her value agrees with the accepted value, 6.63 × 10⁻³⁴ J s, and comment.

SOLUTION

(a) e and c are constants with no uncertainty: h = 1.60 × 10⁻¹⁹ × 1.74 × 630 × 10⁻⁹ ÷ 3.00 × 10⁸ = 5.85 × 10⁻³⁴ J s. Add the percentage uncertainties: 0.05 ÷ 1.74 × 100 + 10 ÷ 630 × 100 = 2.9% + 1.6% = 4.5%.

(b) Percentage difference: (6.63 − 5.85) ÷ 6.63 × 100 = 12%, more than twice the 4.5% uncertainty. The value does not agree: (5.85 ± 0.26) × 10⁻³⁴ J s reaches only 6.11. A systematic error is likely. Judging the "switch-on" p.d. by eye gives a value that is too low, which makes h too low.

Most Unit 6 results come from the gradient or intercept of a straight-line graph, so the uncertainty is found from the graph itself.

Error bars. A vertical bar through each point, ± the uncertainty in y (horizontal too if x is uncertain), from the resolution or half-range of repeats.

Best-fit line. As at AS, with points scattered evenly about it; it should pass through (almost) all the error bars.

Steepest and shallowest lines. The steepest and least steep straight lines that still pass through all the error bars. In practice they usually join the bottom of the first bar to the top of the last bar, and the top of the first to the bottom of the last.

Uncertainty in the gradient ≈ (steepest − shallowest) ÷ 2. Another common method uses |best − worst|, with whichever extreme line differs more from the best fit; either way, show the gradients you used.

Uncertainty in the intercept: half the difference between the intercepts of the steepest and shallowest lines. A long extrapolation makes it far less certain than the gradient.

Percentage uncertainty in the gradient = uncertainty ÷ best gradient × 100%; it feeds into anything calculated from the gradient.

WORKED EXAMPLE · QUESTION

Air is sealed in a flask of fixed volume and heated in a water bath. Pressure p is read from a gauge with an uncertainty of ± 1.0 kPa.

The best-fit line passes through (20 °C, 107.3 kPa) and (90 °C, 132.8 kPa). (a) Calculate its gradient and the temperature at which it predicts p = 0. (b) Use the steepest and shallowest lines to find the uncertainties. (c) Compare the result with −273 °C.

SOLUTION

(a) gradient = (132.8 − 107.3) ÷ 70 = 0.364 kPa °C⁻¹. p = 0 when θ = 20 − 107.3 ÷ 0.364 = −275 °C.

(b) Steepest: (134.0 − 106.4) ÷ 70 = 0.394; shallowest: (132.0 − 108.4) ÷ 70 = 0.337. Uncertainty = (0.394 − 0.337) ÷ 2 = ± 0.029 kPa °C⁻¹ (7.8%). Extrapolate each line to p = 0: steepest 20 − 106.4 ÷ 0.394 = −250 °C; shallowest 20 − 108.4 ÷ 0.337 = −302 °C, so the intercept is uncertain by ± 26 °C.

(c) Result (−275 ± 26) °C.

Percentage difference from −273 °C = 0.7%, far inside the ± 9% range: the data agree with −273 °C.

Exponential and power-law relationships give curves, which are hard to judge by eye. Taking logs turns both into straight lines whose gradient and intercept give the constants. First, the log skills themselves.

Two bases. ln uses base e (≈ 2.718): ln eˣ = x. log (or lg) uses base 10: log 10ˣ = x. Use the one the question uses, and stay with it.

The three rules (true for either base): log(ab) = log a + log b, log(a/b) = log a - log b, log(aⁿ) = nlog a.

Units. Only a pure number has a log, so divide by the unit first: ln (V / V), log (T / s). Logs have no unit.

Significant figures. Give a log to the same number of decimal places as the number has significant figures. 5.83 V (3 s.f.) → ln 5.83 = 1.763; 0.64 V (2 s.f.) → ln 0.64 = −0.45. The digits before the decimal point only say where the decimal point was.

Error bars on a log graph. Δ(ln y) ≈Δy ÷y, the fractional uncertainty. For 2.38 V ± 0.05 V: ln 2.43 = 0.888 and ln 2.33 = 0.846 either side of ln 2.38 = 0.867, about ± 0.021.

The bars grow as y gets smaller.

Changing base. log y = 0.434ln y. So if you plot log y for an exponential, the gradient is 0.434k, not k: another reason to use ln for exponentials.

An exponential change has a constant ratio in equal intervals of x: the same fraction is lost every 15 s, every half-life, every millimetre of absorber. A falling curve could be many things, so the curve alone is never convincing.

Take natural logs of y = Aeᵏˣ:

This matches y = mx + c with ln y against x.

A straight line means the relationship is exponential; the gradient is k and the intercept ln A, so A = eⁱⁿᵗᵉʳᶜᵉᵖᵗ, with the unit of y.

For a decay k is negative: the gradient gives the decay constant (−λ) or −1/RC.

Half-life (time to halve) = ln 2 ÷ |gradient|.

Correct the data first if the physics needs it: subtract the background count from a count rate, or the room temperature from a cooling temperature, before taking logs. Uncorrected data give a curve on the log graph.

WORKED EXAMPLE · QUESTION

A capacitor discharges through a 100 kΩ resistor; a data logger records the p.d. V across it. The capacitor is labelled 470 μF ± 10%.

(a) Show that a graph of ln V against t should be a straight line. (b) Use the graph to determine the capacitance. (c) Decide whether the result is consistent with the label.

SOLUTION

(a) Take natural logs of V = V₀ e^(-t/RC): ln V = ln V₀ - 1/RCt, which matches y = c + mx. A straight line of gradient −1/RC and intercept ln V₀.

(b) Gradient from the triangle, values read from the line: (−0.45 − 2.19) ÷ (120 − 0) = −0.0220 s⁻¹. RC = 1 ÷ 0.0220 = 45.5 s, so C = 45.5 ÷ (100 × 10³) = 455 μF (4.55 × 10⁻⁴ F).

(c) Percentage difference from the label: (470 − 455) ÷ 470 × 100 = 3.2% < 10%: consistent with the label.

The intercept, 2.19, gives V₀ = e^(2.19) = 8.94 V, matching the first reading. The time for the p.d. to halve is ln 2 × RC = 31.5 s.

For y = kxⁿ with n unknown, y against x is a curve you cannot read n from. Take logs of both sides:

Plot log y against log x; a straight line confirms a power law.

The gradient is the power n and has no unit. Either base works and gives the same n; just use the same base on both axes.

The intercept is log k, but only where log x = 0. The log x axis rarely starts at 0, so calculate it: log k = log y − n log x using a point on the line, then k = 10^(log k).

If theory predicts the power (0.5 for a pendulum, −2 for an inverse-square law), compare your n with it using its uncertainty from the extreme lines.

If you already know n, a direct linear graph (T² against l) is simpler; use log–log when n itself is the question.

WORKED EXAMPLE · QUESTION

Astronomers measure the orbital radius r and period T of five moons of a distant planet. They expect T = krⁿ.

(a) Use the graph to determine n. (b) For circular orbits, T = (2π/√(GM)) r^(1.5). Determine the mass M of the planet.

SOLUTION

(a) Gradient from the triangle: n = (5.705 − 4.579) ÷ (8.85 − 8.10) = 1.126 ÷ 0.75 = 1.50.

(b) log k from a point on the line, with n = 1.50: log k = 4.579 − 1.50 × 8.10 = −7.571, so k = 10^(−7.571) = 2.69 × 10⁻⁸ s m^(−1.5). Match k to 2π/√(GM) and rearrange: M = 4π² ÷ (Gk²) = 4π² ÷ (6.67 × 10⁻¹¹ × (2.69 × 10⁻⁸)²) = 8.2 × 10²⁶ kg.

M ∝ 1/k² = 10⁻²ˡᵒᵍᵏ, so an error of 0.01 in log k changes M by about 5%: keep three decimal places in logs.

1 Measurements, uncertainties and graphs (AS, Unit 3)

Half the resolution, half the range, single percentage uncertainties and linearising to y = mx + c are the starting point for everything on this page.

Capacitors (Unit 4) and nuclear decay, oscillations and gravitation (Unit 5)

Capacitor discharge, Q = Q₀e^(−t/RC), is the classic ln graph. ln (count rate) against t gives the decay constant; log–log graphs test T ∝ l^(0.5) and T ∝ r^(1.5); error bars and extreme lines judge absolute zero.

3 The A2 core practicals

Capacitor discharge, gas pressure, gamma absorption by lead and resonance all finish with the analysis practised here. Planning, precision and accuracy are on 6.1.

1. a = 2.0 ± 0.1 and b = 5.0 ± 0.2. What is the percentage uncertainty in ab, and in a/b?

2. A length has a percentage uncertainty of 2%. What is the percentage uncertainty in its cube, and in its square root?

3. Two lengths are 25.3 cm ± 0.1 cm and 24.1 cm ± 0.1 cm. Find their difference and its percentage uncertainty. What does this suggest?

4. You suspect y = kxⁿ with unknown n. What do you plot, and how do you find n and k?

5. A reading is 3.40 V. Write ln (V / V) to the correct precision.

6. A result is (−275 ± 26) °C and the accepted value is −273 °C. Does it agree? Would it still agree if the uncertainty were ± 1 °C?

1. %a = 0.1 ÷ 2.0 × 100 = 5% and %b = 0.2 ÷ 5.0 × 100 = 4%. So 5% + 4% = 9% for both: percentage uncertainties add for products and for quotients.

2. 3 × 2% = 6% for the cube; ½ × 2% = 1% for the square root.

3. 1.2 cm ± 0.2 cm, which is 0.2 ÷ 1.2 × 100 = 17%. Small differences between large readings are very uncertain; measure the difference directly or make it larger.

4. log y against log x. A straight line confirms the power law; its gradient is n; log k = log y − n log x for any point on the line, so k = 10ˡᵒᵍᵏ.

5. 3.40 has 3 significant figures, so the log gets 3 decimal places: ln 3.40 = 1.224.

6. Percentage difference = 2 ÷ 273 × 100 = 0.7% < 26 ÷ 275 × 100 = 9.5%, so it agrees. With ± 1 °C the range −276 °C to −274 °C excludes −273 °C, so it would not agree.