In the calorimetry practical of the Thermal Physics unit, a student finds the specific heat of iron. She counts the heat taken by the water but forgets the copper vessel. Her answer is \(440\,\text{J/(kg K)}\) instead of \(461\,\text{J/(kg K)}\). Is an error of \(21\) units large or small? The number alone cannot say.
Error measured in the unit of the quantity
Suppose the true value of a quantity is \(a\) and a measurement gives \(a_m\). The absolute error is the size of the difference, \(\Delta a = |a_m - a|\). It carries the unit of the quantity. The student's error is \(21\,\text{J/(kg K)}\). An error of \(1\,\text{cm}\) in the length of a room and in the thickness of a coin has the same absolute value.
The two errors of \(1\,\text{cm}\) are not equally serious. In a \(5\,\text{m}\) room it is 1 part in 500. In a coin \(2\,\text{mm}\) thick it is five times the whole quantity. So we divide the error by the value. The relative error is \(\Delta a/a\), a pure number. The percentage error is the relative error multiplied by \(100\).
The absolute error of a measurement is the magnitude of its difference from the true value. The relative error is the absolute error divided by the true value. The percentage error is the relative error expressed as a percentage.
For the forgotten vessel, the relative error is \(21/461 = 0.045\), and the percentage error is \(4.5\,\%\). That is serious for a careful practical, where the other errors are about one per cent. Now the question of the opening has an answer.
A length is measured as \(2.50\,\text{m}\) with an absolute error of \(0.01\,\text{m}\). What is the percentage error?
One student measures a \(1\,\text{m}\) rod with an error of \(1\,\text{cm}\). Another measures a \(10\,\text{m}\) corridor with an error of \(5\,\text{cm}\). Which measurement is better?
When nobody knows the true value
In a real experiment the true value is not given. The student measures the diameter of a wire at several places, as Searle's experiment asks, and gets six readings. The best value is their mean. The random errors are as often positive as negative, so they cancel in the sum.
Each reading differs from the mean by its deviation \(a_i - \bar a\). The deviations always add to zero, so their plain average says nothing. We average their sizes instead. This is the mean absolute error, \(\Delta a\). For the wire of Fig 2.1 it is \(0.06/6 = 0.01\,\text{mm}\).
The result is written \(d = 0.40 \pm 0.01\,\text{mm}\), a percentage error of \(0.01/0.40 = 2.5\,\%\). One more rule applies. If the readings agree so well that \(\Delta a\) is smaller than the least count, the least count is quoted instead. A digital balance that shows \(25.30\,\text{g}\) three times gives \(25.30 \pm 0.01\,\text{g}\), not \(\pm 0\).
Five readings of the period of a pendulum are \(2.04\), \(1.99\), \(2.01\), \(1.98\) and \(2.03\,\text{s}\). The mean is \(2.01\,\text{s}\). Find the mean absolute error, in seconds.
How wrong is an approximation?
The same tool judges a formula. Gravitational Energy and Orbits shows that a lift from the Earth's surface through a height \(h\) stores exactly \(\Delta U = mgh \cdot R/(R + h)\). The simple formula \(mgh\) is an approximation. Its absolute error is the gap between the two, and its relative error, measured against the exact value, is exactly \(h/R\).
In the figure above, the absolute error grows faster and faster with height. The relative error simply grows in proportion to \(h\). So the rule for using \(mgh\) is a rule about a percentage. For a climb of \(10\,\text{km}\), \(h/R = 10/6400\), about \(0.16\,\%\), smaller than the error of most measurements. For a satellite it is no longer small.
This is the worked example of Gravitational Energy and Orbits, lesson 1, done with the tool. A PSLV places a \(1000\,\text{kg}\) satellite at a height of \(1600\,\text{km}\). Take \(g = 9.8\,\text{m/s}^2\) and \(R = 6.4 \times 10^6\,\text{m}\). Find the absolute and the percentage error of \(mgh\).
\(mgh = 1000 \times 9.8 \times 1.6 \times 10^6 = 1.57 \times 10^{10}\,\text{J}\).
The exact rise is \(\Delta U = mgh \cdot R/(R + h) = 1.57 \times 10^{10} \times 0.8 = 1.25 \times 10^{10}\,\text{J}\).
The absolute error is \(1.568 \times 10^{10} - 1.254 \times 10^{10} = 3.1 \times 10^{9}\,\text{J}\). The relative error is \(3.14/12.54 = 0.25\).
The percentage error is \(25\,\%\), the physics chapter's "mgh claims 25 % more". It equals \(h/R = 1600/6400\).
A mountaineer climbs \(8.0\,\text{km}\) above the Earth's surface. Taking \(R = 6400\,\text{km}\), what percentage error does the formula \(mgh\) make in the rise of his potential energy?
In the calorimetry example of Calorimetry and Change of State, the heat gained is \(14\,608\,\text{J}\) with the vessel and \(13\,944\,\text{J}\) without it, and \(c = \text{heat}/31.7\). What is the percentage error of the answer that forgets the vessel?
A percentage error means something only on a scale whose zero is a true zero. A thermometer read to \(1\,\text{°C}\) at \(20\,\text{°C}\) is not "5 % wrong": the same reading is \(293\,\text{K}\), and \(1/293\) is \(0.34\,\%\). For a temperature, use the error of the rise or of the kelvin value. Near a true zero the relative error has no meaning at all: \(0.1\,\text{mm}\) in a length of \(0.1\,\text{mm}\) is \(100\,\%\).
Every practical of the book ends with a percentage error. In Calorimetry and Change of State it shows that a forgotten vessel costs about five per cent. In Searle's experiment each reading's uncertainty is turned into a fraction of the reading to find which one limits Young's modulus. In Gravitational Energy and Orbits it tells us where \(mgh\) may replace the exact rise in potential energy. The next lesson adds these fractions together when several readings meet in one formula.
Back to the forgotten vessel
The student's \(21\,\text{J/(kg K)}\) sounded small, but as a fraction it is \(4.5\,\%\). The careful answer, \(461\,\text{J/(kg K)}\), would carry an error of about one per cent from the thermometer. So the forgotten vessel is the largest error in her practical. Only the relative error could show this.
- Comparing two measurements by their absolute errors.Divide each error by its value. Compare relative or percentage errors.
- Averaging the signed deviations and finding a mean error of zero.The signed deviations always add to zero. Average their sizes.
- Quoting \(\pm 0\) when every reading agrees.The uncertainty is never below the least count. Quote the least count instead.
- Finding a percentage error of a temperature in °C.Use the kelvin value, or the error of a temperature difference.