Data Analysis & Uncertainty

An answer without an uncertainty is not a measurement

You time twenty swings of a pendulum, divide by twenty, and get a period. You put it into $g = 4\pi^2 L/T^2$ and out comes 9.81 m/s². It looks like an answer. It is not one yet, because you have no idea whether the next student down the bench would have got 9.80 or 9.4.

Everything below is about that gap. A measured number is really a short interval, and once you do algebra to it, the interval has to come along for the ride. There is a second half to the job too: real experiments almost never hand you one number, they hand you a table. Getting a physical constant out of a table means drawing a straight line through it, and if the data are not straight, changing what you plot until they are.

The physics in the last unit was about what exists. This is about what you can honestly claim to know.

Where quadrature comes from

Lay two rods end to end. The first is $L_1 = 2.50 \pm 0.03$ m and the second is $L_2 = 1.20 \pm 0.04$ m. The total is 3.70 m. What is its uncertainty?

The lazy answer is 0.07 m, the two errors added. But that is the worst case: it assumes that whenever rod one reads long, rod two also reads long. The two measurements were made separately with no connection between them. Half the time the errors point the same way and add, half the time they point opposite ways and cancel. Averaging over all the ways they can land, the typical size of the combined error is not $\delta_1 + \delta_2$ but the hypotenuse of a right triangle with those legs. That is quadrature, and for a sum or difference it says

Here that is $\sqrt{0.03^2 + 0.04^2} = \sqrt{0.0009 + 0.0016} = \sqrt{0.0025} = 0.05$ m. So $3.70 \pm 0.05$ m. Notice the total error is smaller than the sum and larger than either piece alone.

The same formula holds for a difference, because an error of unknown sign does not care whether you add or subtract the quantity. That fact has teeth. If $t_1 = 12.48 \pm 0.05$ s and $t_2 = 12.31 \pm 0.05$ s, then $\Delta t = 0.17$ s with uncertainty $\sqrt{0.05^2 + 0.05^2} = 0.071$ s. Each time was good to 0.4 percent; their difference is good to only $0.071/0.17 = 42$ percent.

Products, quotients and powers

Sums combine absolute errors. Products combine fractional errors, which are the error divided by the quantity, usually quoted as a percent.

Here is why. Write the area of a plate as $A = ab$. Nudge $a$ by $\delta a$ and $b$ by $\delta b$:

The last term is tiny, a small number times a small number, so drop it. The change in the area is $a\,\delta b + b\,\delta a$. Divide through by $A = ab$ and the clutter cancels:

Then apply the independence argument from before and combine those two fractional errors in quadrature rather than straight. For a general product of powers $Q = a^{m} b^{n}$, the same expansion gives each exponent as a multiplier:

The exponent multiplies the fractional error before it is squared. A quantity that enters squared is twice as damaging as one that enters linearly; a cube root is only a third as damaging. This single fact decides where you should spend your effort in a lab.

For the plate, $a = 12.0 \pm 0.2$ cm and $b = 8.0 \pm 0.3$ cm. The fractional errors are $0.2/12.0 = 1.667\%$ and $0.3/8.0 = 3.75\%$. Both exponents are 1, so $\delta A/A = \sqrt{1.667^2 + 3.75^2} = \sqrt{2.78 + 14.06} = \sqrt{16.84} = 4.1\%$.

That figure is worth sitting with. When you square small numbers, the small ones vanish. A term that is a third the size of the biggest one contributes a ninth as much to the variance. In practice only the largest one or two terms matter, and if you want a better result you must attack the largest.

Worked example 1: g from a pendulum

A pendulum of length $L = 1.000 \pm 0.005$ m is timed over 20 full swings, giving $20T = 40.12 \pm 0.20$ s. Find $g$ and its uncertainty from $g = 4\pi^2 L/T^2$.

First the period. Dividing by 20 divides both the value and its uncertainty: $T = 2.006$ s and $\delta T = 0.010$ s. This is why you time twenty swings instead of one: your reaction time error is fixed at around 0.2 s no matter how many swings you count, so spreading it over twenty cuts the fractional error by twenty.

The central value: $g = 4\pi^2(1.000)/(2.006)^2 = 39.478/4.024 = 9.811$ m/s².

Now the fractional errors. In $L$: $0.005/1.000 = 0.500\%$. In $T$: $0.010/2.006 = 0.4985\%$. But $T$ appears squared and in the denominator, so its exponent is $-2$ and it enters with a multiplier of 2:

So $\delta g = 0.01115 \times 9.811 = 0.109$ m/s², and you report $g = 9.81 \pm 0.11$ m/s². Round the uncertainty to one or two significant figures and round the value to match its last decimal place. Never write $9.8106 \pm 0.1094$; you are claiming a precision the third digit already rules out.

Averaging, and how much averaging buys you

Repeat a measurement $N$ times and you get a spread. The standard deviation $\sigma$ of that spread describes a single reading, and it does not shrink as you take more data. What shrinks is the uncertainty of the mean, called the standard error:

The $\sqrt{N}$ is the same quadrature idea: you are adding $N$ independent errors and dividing by $N$, which gives $\sqrt{N}\sigma/N$. Five stopwatch readings of 10.1, 10.3, 9.8, 10.2, 10.0 s have mean 10.08 s and standard deviation 0.192 s, so the mean is good to $0.192/\sqrt{5} = 0.086$ s.

Because it is $\sqrt{N}$ and not $N$, averaging is an expensive way to buy precision. To take a spread of $\sigma = 0.24$ s down to 0.050 s in the mean you need $N = (0.24/0.050)^2 = 23.0$, so 24 trials, and the next factor of two would cost you 96.

Counting experiments give you $\sigma$ for free. Radioactive decays arrive independently, so a count of $N$ has uncertainty $\sqrt{N}$. Record 2500 decays and the fractional uncertainty is $\sqrt{2500}/2500 = 50/2500 = 2.0\%$. To halve that you need four times the counts.

When you have two results of different quality, do not average them plainly. Weight each by $w_i = 1/\sigma_i^2$, so a measurement twice as precise counts four times as much:

For $g = 9.79 \pm 0.03$ and $g = 9.83 \pm 0.02$: $w_1 = 1/0.0009 = 1111$ and $w_2 = 1/0.0004 = 2500$. The average is $(1111 \times 9.79 + 2500 \times 9.83)/3611 = 35453/3611 = 9.818$ m/s², pulled toward the better measurement. Its uncertainty is $1/\sqrt{3611} = 0.0166$ m/s², smaller than either input.

To ask whether two results agree, take their difference over the quadrature sum of their errors. For $9.81 \pm 0.03$ and $9.74 \pm 0.04$, that is $0.07/\sqrt{0.03^2+0.04^2} = 0.07/0.05 = 1.4$. A disagreement of 1.4 combined standard deviations is unremarkable; around 3 you should start looking for a systematic error.

Make the data straight before you fit it

Your eye is good at straight lines and terrible at curves. So is least squares. The move is always the same: do algebra to the model until it reads $Y = mX + c$ for some $X$ and $Y$ you can compute from the data.

For a power law $y = Ax^{n}$, take logs of both sides: $\log y = n \log x + \log A$. Plot $\log y$ against $\log x$ and the slope is the exponent, the intercept is $\log A$. Drag on a sphere at $F = 0.80$ N when $v = 2.0$ m/s and $F = 6.5$ N when $v = 6.0$ m/s gives $n = \ln(6.5/0.80)/\ln(6.0/2.0) = 2.0956/1.0986 = 1.91$.

For an exponential $A = A_0 e^{-t/\tau}$, take logs of the vertical axis only: $\ln A = \ln A_0 - t/\tau$. Plot $\ln A$ against $t$, slope $-1/\tau$.

Sometimes no logs are needed, just better axes. For a mass on a spring, $T = 2\pi\sqrt{m/k}$ gives $T^2 = (4\pi^2/k)m$, so $T^2$ against $m$ is a line through the origin with slope $4\pi^2/k$. A measured slope of 3.95 s²/kg gives $k = 4\pi^2/3.95 = 9.99$ N/m. For uniform acceleration, $v^2 = v_0^2 + 2ax$ means $v^2$ against $x$ has slope $2a$, so a slope of 7.30 gives $a = 3.65$ m/s².

And for anything with a sum in the denominator, try reciprocals.

Worked example 2: a saturating response

A response obeys $y = ax/(1+bx)$. It rises and then flattens, so neither a log-log nor a semilog plot will straighten it. Take reciprocals instead:

So $1/y$ against $1/x$ is straight, with slope $1/a$ and intercept $b/a$. A student's line passes through $(1/x, 1/y) = (1.00, 2.00)$ and $(0.250, 1.00)$. The slope is $(2.00 - 1.00)/(1.00 - 0.250) = 1.00/0.750 = 1.333$. Since the slope is $1/a$, you get $a = 1/1.333 = 0.75$.

Two habits make this kind of problem quick. Always ask what the slope means in terms of the constants before you compute it, and always check whether the constant you want is the slope, its reciprocal, or the intercept divided by the slope.

Least squares, and the uncertainty of a slope

A best-fit line is the one that makes the sum of the squared vertical residuals as small as possible. A residual $r_i = y_i - (mx_i + b)$ is how far a point sits above the line. Squaring it means an outlier hurts a lot, and means the answer has a clean closed form:

The intercept formula says the fit always passes through the centre of mass of the data, the point $(\bar{x}, \bar{y})$. If theory forces the line through the origin instead, drop the intercept and the minimisation gives $m = \sum x_i y_i / \sum x_i^2$.

Worked example 3: slope, intercept, and the error in the slope

Take the five points $(1.0, 2.1)$, $(2.0, 3.9)$, $(3.0, 6.2)$, $(4.0, 7.9)$, $(5.0, 10.1)$.

The means are $\bar{x} = 15/5 = 3.00$ and $\bar{y} = 30.2/5 = 6.04$. Build the two sums. For the bottom, the deviations in $x$ are $-2, -1, 0, 1, 2$, so $\sum (x_i - \bar{x})^2 = 4 + 1 + 0 + 1 + 4 = 10$. For the top, the deviations in $y$ are $-3.94, -2.14, 0.16, 1.86, 4.06$, so the products are $7.88 + 2.14 + 0 + 1.86 + 8.12 = 20.00$.

So $m = 20.00/10 = 2.00$ and $b = 6.04 - 2.00 \times 3.00 = 0.04$.

Now the uncertainty. The fitted values are 2.04, 4.04, 6.04, 8.04, 10.04, so the residuals are $+0.06, -0.14, +0.16, -0.14, +0.06$ and $\sum r_i^2 = 0.0036 + 0.0196 + 0.0256 + 0.0196 + 0.0036 = 0.0720$. The standard formula is

giving $\sigma_m = 0.049$. The $N - 2$ is there because two parameters were already spent fitting the line, so five points leave only three independent checks on it. Two points would give $N - 2 = 0$ and an infinite uncertainty, which is right: a line through two points has no evidence at all about its own reliability.

Worked example 4: a time constant from a semilog fit

An amplitude decays as $A = A_0 e^{-t/\tau}$, measured as 100 mm at 0 s, 74 mm at 10 s, 55 mm at 20 s and 41 mm at 30 s. Fit $\ln A$ against $t$.

The logs are $4.6052, 4.3041, 4.0073, 3.7136$. Then $\bar{t} = 15.0$ and $\overline{\ln A} = 16.6301/4 = 4.1575$.

Bottom sum: deviations in $t$ are $-15, -5, 5, 15$, squares $225 + 25 + 25 + 225 = 500$. Top sum: deviations in $\ln A$ are $0.4477, 0.1466, -0.1502, -0.4440$, so the products are $-6.715 - 0.733 - 0.751 - 6.659 = -14.858$.

Slope $= -14.858/500 = -0.02972$ per second. Since the slope is $-1/\tau$, $\tau = 1/0.02972 = 33.7$ s. The half-life follows from the same fit: $A$ halves when $t/\tau = \ln 2$, so $t_{1/2} = 0.693 \times 33.7 = 23.3$ s.

One warning about log fits. Taking logs also transforms the error bars: a constant percentage error in $y$ becomes a constant absolute error in $\ln y$, which is exactly what plain least squares assumes. But a constant absolute error in $y$ becomes a huge error bar on the smallest points, and an unweighted log fit will then take those noisy tail points far too seriously.

If you need the uncertainty in a fitted exponent, work backwards through the same relation. Probe $y = Ax^n$ at two $x$ values a factor of ten apart with 2.0 percent uncertainty on each $y$. Then $n = \Delta \ln y/\Delta \ln x$, and $\delta(\ln y) = \delta y/y = 0.020$ at each end, so $\delta(\Delta \ln y) = \sqrt{0.020^2 + 0.020^2} = 0.0283$. Dividing by $\Delta \ln x = \ln 10 = 2.303$ gives $\delta n = 0.0123$.

Reading a line off a plotted graph

Exam questions often give you two points on a plotted line rather than raw data. Take the slope from them directly. A velocity-time line through $(2.0, 5.0)$ and $(8.0, 17.0)$ has slope $(17.0 - 5.0)/(8.0 - 2.0) = 2.0$ m/s², which is the acceleration.

On a log-log plot, read the slope the same way but remember what the axes hold. A line through $(\log_{10} x, \log_{10} y) = (0.50, 1.20)$ and $(2.00, 3.00)$ has slope $(3.00 - 1.20)/(2.00 - 0.50) = 1.80/1.50 = 1.20$, so $n = 1.20$. For the prefactor, use the intercept at $\log_{10} x = 0$: $\log_{10} A = 1.20 - 1.20 \times 0.50 = 0.60$, so $A = 10^{0.60} = 3.98$. The prefactor is never the number you read straight off the axis; you have to undo the log.

Finally, significant figures. A quotient carries the figures of its worst input. With $I = 0.035$ A (two figures) and $V = 12.0$ V (three), $R = 12.0/0.035 = 342.857$, which you report as 340 ohm.

How to recognise these problems

The tells are blunt once you know them.

A value written with a $\pm$, or a "percent uncertainty", or the word "tolerance", means propagation. Decide first whether the combination is a sum or a product. Sums and differences take absolute errors in quadrature. Products, quotients and powers take fractional errors in quadrature, each multiplied by its exponent. Nothing else is needed for most of them.

A table of $x$ and $y$ with more than three rows and a request for a slope means least squares. Compute $\bar{x}$ and $\bar{y}$ first; everything else is deviations from those.

A model of the form $y = Ax^n$, $y = A e^{kx}$ or anything with $x$ in a denominator, plus the words "linearize", "what should the student plot" or "straight line", means transform the axes. Logs of both sides for a power law, logs of one side for an exponential, reciprocals for a saturating curve.

Repeated readings of the same quantity mean standard error, and the answer will contain a $\sqrt{N}$. Several independent published results mean a weighted average with $w = 1/\sigma^2$. Counts from a detector mean $\sqrt{N}$ again, this time as the uncertainty itself.

And whenever a problem asks you to design rather than compute (how precisely must I time this, how many runs do I need) write the propagation formula and solve it for the term you control. The algebra is the same, run backwards.

Next comes a different kind of power tool: Lagrangian mechanics, which replaces forces and free body diagrams with a single scalar function and a recipe for differentiating it.