Theory, worked examples and 20 previous-year questions with full solutions — IIT-JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards.
Theory
Every number obtained in a laboratory is incomplete until we know how far it can be trusted. This chapter develops the tools that an experimental physicist uses to make that judgement and to improve it. We begin with the nature of measurement and of experimental error, and with the rules for quoting a result. We then show how errors propagate from measured quantities into derived ones, and how the statistics of repeated measurements, in particular the Gaussian, binomial and Poisson distributions, set the uncertainty of a mean and of a count. The method of least squares allows a straight line to be fitted to data, and suitable transformations allow many non-linear laws to be tested in the same way. Counting experiments with nuclear detectors receive separate treatment, because their statistics, their dead-time losses and their background are of special importance. The second half of the chapter turns to the physics of the apparatus: the fundamental sources of electrical noise, the recovery of small signals by modulation and phase-sensitive (lock-in) detection, the common transducers and radiation detectors, the correct use of meters, oscilloscopes and four-probe methods, and finally charged-particle beams and vacuum techniques.
8.1 Measurement and experimental error
(a) Accuracy and precision
The error of a measurement is the difference between the measured value and the true value of the quantity. Since the true value is not known, the error itself cannot be known; what an experimenter can estimate is the uncertainty, a range within which the true value is likely to lie. In everyday usage the words error and uncertainty are used interchangeably, and we shall follow that usage where no confusion can arise.
Two distinct qualities of a set of measurements must be kept apart. The accuracy of a measurement describes how close it lies to the true value. The precision describes how closely repeated measurements agree with one another, that is, how small their scatter is, irrespective of where they lie. A set of measurements may be precise but inaccurate, when the readings cluster tightly about a wrong value, or accurate but imprecise, when the readings scatter widely about the right value. The familiar picture is that of shots at a target. A tight group of hits far from the centre shows high precision and poor accuracy; a loose group centred on the bull’s eye shows good accuracy and modest precision; hits scattered all over the board show neither.
(b) Systematic and random errors
The two qualities correspond to two kinds of error. Systematic errors shift every reading in the same direction by the same or a similar amount. They arise from a miscalibrated or wrongly zeroed instrument, from an incorrect theory of the apparatus (a neglected lead resistance, a neglected heat loss, an unknown loading by a meter), from the environment (temperature drift, stray magnetic fields), or from a consistent bias of the observer. Repeating the measurement does not reveal a systematic error, and averaging does not reduce it. It must be found by thought, by calibration against a standard, or by measuring the same quantity by an entirely different method. A systematic error limits the accuracy.
Random errors make successive readings scatter unpredictably on both sides of their mean. They arise from fluctuations that cannot be controlled: the reaction time of an observer starting and stopping a clock, electrical noise, mechanical vibration, the estimation of a reading between scale marks, and, in counting experiments, the intrinsically random arrival of the events themselves. Random errors limit the precision. Because they are equally likely to be positive or negative, they are revealed by repetition and are reduced by averaging, as we show in Section 8.3.
When many repeated measurements affected only by random errors are sorted into a histogram, the histogram is narrow if the precision is good and is symmetric and bell-shaped about its mean, approaching the Gaussian distribution of Section 8.3. A lopsided histogram, one that is cut off abruptly on one side, or one with no clear peak at all, is a sign that something other than random scatter is at work, for example a drift during the series or an instrument that saturates.
Exam tip. Precision is judged by the spread of the readings alone; accuracy by the distance of their mean from the true value. A small spread says nothing about the absence of systematic error.
(c) Least count of an instrument
Every instrument has a smallest division that it can resolve, its least count, and a single reading carries an uncertainty of the order of the least count. For a vernier calliper, suppose that \(n\) divisions of the vernier scale (VSD) coincide with \(n'\) divisions of the main scale (MSD). Then one vernier division is \(n'/n\) of a main-scale division, and the vernier constant is the difference
In the commonest design \(n'=n-1\), and the vernier constant reduces to \(\text{VC}=\text{MSD}/n\). A reading is the main-scale reading just before the vernier zero plus the number of the vernier division that coincides with a main-scale mark multiplied by the vernier constant. For a micrometer screw gauge the least count is the pitch of the screw (the distance advanced in one turn) divided by the number of divisions on the circular scale; a pitch of \(0.5\) mm and 50 divisions give \(0.01\) mm. Any reading obtained when the jaws are closed is a zero error, a systematic error that must be subtracted from every reading.
The least count decides which instrument is suitable for a given measurement. A vernier calliper resolves about \(0.1\) to \(0.02\) mm and a screw gauge about \(0.01\) mm, so neither can measure the diameter of a wire only a few micrometres thick. Such a diameter is measured optically instead: a thin wire illuminated by a laser of wavelength \(\lambda\) produces the same Fraunhofer pattern as a single slit of the same width (Babinet’s principle), and the spacing \(\Delta y\) of the dark fringes on a screen at distance \(D\) gives the diameter as \(d=\lambda D/\Delta y\).
(d) Significant figures and the quoting of results
A result is quoted as a best value together with its uncertainty, \(x=\bar x\pm\delta x\), in the same units. Because the uncertainty is itself only an estimate, it is rounded to one significant figure, or to two when its first digit is 1 (so that \(0.0236\) becomes \(0.02\), but \(0.14\) remains \(0.14\)). The best value is then rounded to the same decimal place as the uncertainty. Thus a calculator output of \(9.82147\pm0.02357\ \text{m s}^{-2}\) is reported as \((9.82\pm0.02)\ \text{m s}^{-2}\); digits beyond this carry no information, and quoting them misrepresents the quality of the measurement. Intermediate calculations should keep one or two extra digits, and rounding is done only at the end.
The fractional (relative) uncertainty \(\delta x/\lvert\bar x\rvert\), often given as a percentage, is the natural measure of the quality of a result, because it is independent of the units and, as we shall see, it is the quantity that propagates simply through products and powers.
Example 8.1. A vernier calliper has a main scale graduated in half-millimetres, and 25 divisions of its vernier coincide with 24 main-scale divisions. With the jaws closed, the vernier zero lies to the right of the main-scale zero and the 3rd vernier division coincides with a main-scale mark. When a rod is held between the jaws, the vernier zero lies just beyond the \(12.5\) mm mark and the 13th vernier division coincides with a main-scale mark. Find the vernier constant and the corrected diameter of the rod.
Solution. One main-scale division is \(0.5\) mm, and 25 vernier divisions cover 24 of them, so
The zero error is positive, because the vernier zero lies to the right of the main-scale zero when the jaws are closed, and its value is \(3\times0.02=0.06\) mm. The observed reading of the rod is
and subtracting the zero error gives the corrected diameter
\[d=12.76-0.06=12.70\ \text{mm}.\]
A single reading is uncertain by about one least count, so the result is \(d=(12.70\pm0.02)\) mm.
8.2 Propagation of errors
A quantity of interest is usually calculated from several measured quantities, and we need to know how their uncertainties combine. We consider a function \(q=f(x,y,\dots)\) of measured quantities \(x,y,\dots\) with small uncertainties \(\delta x,\delta y,\dots\).
(a) The general formula
For small errors a first-order Taylor expansion gives the change in \(q\) produced by changes \(\delta x,\delta y,\dots\):
Two different rules follow, according to what we assume about the errors. If we ask for the largest possible error, all the contributions must be taken with the same sign, and the maximum error is
If the errors are random and independent, they are as likely to partially cancel as to reinforce. Squaring the expansion and averaging over many repetitions, the cross terms \(\langle\delta x\,\delta y\rangle\) vanish for independent errors, while \(\langle\delta x^2\rangle=\sigma_x^2\) and so on. This gives the standard (probable) error, in which the contributions add in quadrature:
The quadrature sum is always smaller than or equal to the linear sum, and it is the correct estimate for independent random errors. The linear sum is a safe upper bound, and it is the appropriate rule when a question explicitly asks for the maximum error, or when the errors are known limits (such as a tolerance or a least count) rather than standard deviations. In an examination we must read the question carefully to see which is intended; often only one of the two results appears among the options.
(b) Sums, differences, products and powers
For a sum or difference \(q=x\pm y\), the partial derivatives are \(\pm1\), and the absolute errors combine:
The sign with which a quantity enters does not matter: the error of a difference is the same as that of a sum, never the difference of the errors. The difference of two nearly equal quantities therefore has a large fractional error, a fact of great practical importance in background subtraction (Section 8.5). For a linear combination \(q=ax+by\) with exact constants \(a,b\), we have \(\sigma_q^2=a^2\sigma_x^2+b^2\sigma_y^2\).
For a product of powers \(q=Cx^ay^b\cdots\) with an exact constant \(C\), it is quickest to take logarithms first, \(\ln q=\ln C+a\ln x+b\ln y+\cdots\), and then differentiate:
A quotient is a product with a negative power, so its fractional errors also add. A quantity raised to a high power dominates the error budget: in Poiseuille’s law the flow rate varies as \(r^4\), so a 1% error in the radius contributes 4% to the flow rate; and a square root halves the fractional error of its argument. Constants such as \(\pi\) or \(g\), which are known to far better accuracy than the measured quantities, contribute nothing. When one term is much larger than the others, the quadrature sum is essentially equal to it, and effort spent improving the smaller terms is wasted.
(c) Other functions
The same procedure applies to any function. For an exponential \(q=q_0e^{u}\) we have \(\delta q/q=\delta u\), so the fractional error of \(q\) is the absolute error of the exponent. For example, if \(\rho=\rho_0e^{\Delta/T}\) with independent errors in \(\Delta\) and \(T\), then \(\delta\rho/\rho=\delta(\Delta/T)=(\Delta/T)\sqrt{(\delta\Delta/\Delta)^2+(\delta T/T)^2}\), and the fractional error is multiplied by the size of the exponent. For a logarithm \(q=\ln x\), \(\delta q=\delta x/x\). For a trigonometric function \(q=\sin\theta\), \(\delta q=\lvert\cos\theta\rvert\,\delta\theta\), with \(\delta\theta\) in radians.
An important application is the growth of an error with time in an exponential law. A population that decays as \(N=N_0e^{-t\ln 2/T_{1/2}}\) has \(\ln N=\ln N_0-t\ln2/T_{1/2}\), so an error in the half-life gives
which grows in proportion to the number of elapsed half-lives. For large errors the linear formula is only approximate, and the exact ratio \(N_{\text{est}}/N_{\text{true}}=2^{-t/T_{\text{est}}+t/T_{\text{true}}}\) should be computed directly.
(d) Correlated errors
The quadrature rule assumes that the errors of the different variables are independent. If the same measured quantity appears more than once in a formula, the separate appearances are perfectly correlated, and the formula must first be written so that each independent variable appears explicitly; the general formula with partial derivatives then takes care of the correlation automatically. In general, for correlated variables with covariance \(\sigma_{xy}=\langle\delta x\,\delta y\rangle\),
The slope and intercept of a fitted straight line (Section 8.4) are an example of correlated quantities.
Example 8.2. The density of a metal cylinder is found from \(\rho=4m/(\pi d^2L)\), with \(m=(25.0\pm0.1)\) g, \(d=(1.00\pm0.01)\) cm and \(L=(4.00\pm0.02)\) cm. Find \(\rho\), its maximum error, and its standard error assuming the errors to be independent and random.
The fractional errors of the measured quantities are \(\delta m/m=0.1/25.0=0.004\), \(\delta d/d=0.01\) and \(\delta L/L=0.02/4.00=0.005\). Since \(\rho\propto m\,d^{-2}L^{-1}\), the diameter enters with power 2. The maximum fractional error is
so \(\sigma_\rho=0.17\ \text{g cm}^{-3}\) and \(\rho=(7.96\pm0.17)\ \text{g cm}^{-3}\). The diameter, measured to only 1% but entering squared, accounts for almost the entire error; to improve the result one should measure \(d\) more precisely, for example with a screw gauge.
Example 8.3. Two independently measured quantities are \(x=10.0\pm0.3\) and \(y=30.0\pm0.4\). Find the value and the standard error of \(f=x/(x+y)\).
Solution. The central value is \(f=10/40=0.250\). It is tempting to treat \(f\) as a quotient of \(x\) and \(s=x+y\), with \(\sigma_s=\sqrt{0.3^2+0.4^2}=0.5\), and to combine the fractional errors \(0.3/10=3\%\) and \(0.5/40=1.25\%\) in quadrature, giving \(3.25\%\), or \(\sigma_f=0.0081\). This is wrong, because \(x\) appears in both numerator and denominator and the two errors are correlated: an increase in \(x\) raises both, and partly cancels in the ratio. The correct procedure uses the partial derivatives with respect to the independent variables \(x\) and \(y\):
and \(f=0.250\pm0.006\), a fractional error of \(2.5\%\), smaller than the naive estimate.
8.3 Statistics of repeated measurements
(a) Mean, standard deviation and standard error
Suppose that a quantity is measured \(N\) times, with results \(x_1,x_2,\dots,x_N\) affected by random errors only. The best estimate of the true value is the mean
\[\bar x=\frac{1}{N}\sum_{i=1}^{N}x_i.\]
The spread of the individual readings about the mean is described by the standard deviation
The divisor \(N-1\) rather than \(N\) appears because the deviations are measured from the sample mean, which has itself been fitted to the data; one degree of freedom has been used up, and dividing by \(N\) would systematically underestimate the spread. For large \(N\) the difference is unimportant. The square of the standard deviation is the variance. The standard deviation characterises the precision of a single measurement: it does not decrease as more readings are taken, but settles to a value fixed by the method.
The mean is much better determined than any single reading. Since \(\bar x=(x_1+\dots+x_N)/N\) is a sum of \(N\) independent quantities, each with standard deviation \(s\), divided by \(N\), the propagation rule for sums gives
\[\boxed{\sigma_{\bar x}=\frac{s}{\sqrt N}}.\]
This is the standard error of the mean, the uncertainty that is quoted with \(\bar x\). It falls as \(1/\sqrt N\): to halve the uncertainty we need four times as many readings. The slow \(\sqrt N\) improvement means that averaging is an expensive way of reducing errors, and that a better method is often preferable.
A classic illustration is the timing of a pendulum. Every start–stop timing carries a random error \(\sigma\), from the reaction of the observer, which does not depend on the length of the interval. If one period is timed \(N\) times and the results are averaged, the period is uncertain by \(\sigma/\sqrt N\). If instead \(N\) periods are timed once and the total is divided by \(N\), the uncertainty is \(\sigma/N\), which is much smaller. The two methods give equal uncertainty only if the number \(N_A\) of single timings equals the square of the number \(N_B\) of periods timed at once, \(N_A=N_B^2\). Timing many cycles together is the better method whenever the timing error is independent of the interval.
(b) The Gaussian distribution
When a measurement is affected by many small, independent random disturbances, the central limit theorem ensures that the distribution of its results approaches the Gaussian or normal distribution
where \(\mu\) is the mean (the true value, in the absence of systematic errors) and \(\sigma\) the standard deviation, of which \(\bar x\) and \(s\) are the estimates from a sample. \(P(x)\,dx\) is the probability that a single measurement lies between \(x\) and \(x+dx\), and the distribution is normalised, \(\int_{-\infty}^{\infty}P(x)\,dx=1\). The curve, shown in Figure 8.1, is symmetric about \(\mu\), has its points of inflexion at \(\mu\pm\sigma\), and falls to half its maximum at \(\mu\pm1.177\sigma\), so that its full width at half maximum is \(\text{FWHM}=2\sqrt{2\ln2}\,\sigma=2.355\sigma\).
Figure 8.1 The Gaussian distribution. The dark band, \(\bar x\pm\sigma\), contains 68.3% of the area and the light band, \(\bar x\pm2\sigma\), contains 95.4%.
The probability that a measurement lies within \(\pm z\sigma\) of the mean is obtained from the error function, \(P(\lvert x-\mu\rvert<z\sigma)=\operatorname{erf}(z/\sqrt2)\). The values used most often are collected in the table.
Interval
\(\pm\sigma\)
\(\pm1.645\sigma\)
\(\pm1.96\sigma\)
\(\pm2\sigma\)
\(\pm2.576\sigma\)
\(\pm3\sigma\)
Probability inside
68.3%
90%
95%
95.4%
99%
99.7%
A one-sided tail is half of the probability outside the interval: for example, the probability that a measurement exceeds \(\mu+\sigma\) is \(15.9\%\), that it exceeds \(\mu+1.5\sigma\) is \(6.7\%\), and that it exceeds \(\mu+2\sigma\) is \(2.3\%\). Hence a result quoted as \(\bar x\pm\sigma_{\bar x}\) is a 68% confidence interval: there is roughly a two-in-three chance that the true value lies within it. Two measurements of the same quantity are regarded as consistent if they differ by less than about twice the standard error of their difference, and a discrepancy of more than \(3\sigma\) is strong evidence of a real difference or of a systematic error.
Example 8.4. Eight independent measurements of the period of an oscillator give \(2.04\), \(2.01\), \(1.98\), \(2.03\), \(2.00\), \(1.97\), \(2.02\) and \(2.03\) s. Find the mean, the standard deviation of a single reading and the standard error of the mean. How many readings would be needed to halve the standard error?
Solution. The sum of the readings is \(16.08\) s, so \(\bar T=2.010\) s. The deviations from the mean, in units of \(0.01\) s, are \(3,0,-3,2,-1,-4,1,2\), and the sum of their squares is \(44\times10^{-4}\ \text{s}^2\). Hence
The result is \(T=(2.010\pm0.009)\) s. Since \(\sigma_{\bar T}\propto1/\sqrt N\), halving it requires four times as many readings, that is 32.
Example 8.5. A batch of nominally \(1000\,\Omega\) resistors has values distributed normally with mean \(1000\,\Omega\) and standard deviation \(20\,\Omega\). What fraction of the resistors lie within \(\pm50\,\Omega\) of the nominal value, and what fraction exceed \(1040\,\Omega\)?
Solution. The interval \(\pm50\,\Omega\) corresponds to \(z=50/20=2.5\) standard deviations, so the fraction inside it is
or \(98.8\%\). A value of \(1040\,\Omega\) lies \(z=40/20=2\) standard deviations above the mean, and the one-sided tail beyond \(2\sigma\) is \(\tfrac12(1-0.954)=0.023\), so about \(2.3\%\) of the resistors exceed \(1040\,\Omega\).
(c) The weighted mean and the \(\chi^2\) test
Measurements of the same quantity with different uncertainties should not be averaged with equal weight. If the measurements \(x_i\) have standard errors \(\sigma_i\), maximising the joint Gaussian probability of the observed values, which amounts to minimising \(\sum_i(x_i-\mu)^2/\sigma_i^2\) with respect to \(\mu\), gives the weighted mean
A measurement twice as precise as another carries four times its weight. When all the \(\sigma_i\) are equal, the weighted mean reduces to the ordinary mean with error \(\sigma/\sqrt N\).
Whether a set of data is consistent with a model, here with a single common value, is judged by the chi-squared statistic
If the deviations are purely random and the \(\sigma_i\) correctly estimated, each term is on average about 1, and \(\chi^2\) is close to the number of degrees of freedom \(\nu\), the number of data points less the number of parameters fitted. The reduced chi-squared \(\chi^2/\nu\) should therefore be close to 1. A value much larger than 1 means that the model is wrong or the errors have been underestimated; a value much smaller than 1 means that the errors have been overestimated, or that the data are suspiciously regular.
Example 8.6. Two laboratories measure the acceleration due to gravity as \((9.79\pm0.02)\) and \((9.83\pm0.04)\ \text{m s}^{-2}\). Find the weighted mean and its standard error, and test whether the two results are consistent.
Solution. The weights are \(w_1=1/0.02^2=2500\) and \(w_2=1/0.04^2=625\ (\text{m s}^{-2})^{-2}\). Hence
for \(\nu=2-1=1\) degree of freedom. This is close to 1, so the two results are consistent. Equivalently, their difference \(0.04\) is only \(0.9\) times its standard error \(\sqrt{0.02^2+0.04^2}=0.045\).
(d) The binomial distribution
Many experiments consist of \(n\) independent trials, each of which has only two outcomes, “success” with probability \(p\) and “failure” with probability \(1-p\): a particle is detected or missed, an object is of type A or of type B, a candidate passes or fails. The probability of exactly \(k\) successes is given by the binomial distribution
This is how the efficiency of a detector, or the fraction of objects of a given kind in a sample, is quoted. The binomial spread assumes that \(p\) is the same for every trial. If the underlying probability itself varies from one occasion to another, for example the pass rate of an examination from year to year, the observed scatter is larger than the binomial value, and it is the observed (empirical) standard deviation that must be used to predict future outcomes.
(e) The Poisson distribution and counting statistics
When the number of trials is very large and the probability of success in each is very small, with the mean \(\lambda=np\) finite, the binomial distribution goes over into the Poisson distribution
This is the distribution of the number of independent random events in a fixed interval: radioactive decays (each of an enormous number of nuclei has a tiny chance of decaying in the interval), photons arriving at a detector, cosmic rays passing through a counter. Its essential property is that the variance equals the mean. Consequently, if \(N\) counts are recorded, the uncertainty in \(N\) is
A single count thus carries its own error estimate, without repetition. To measure a rate to 1% we need \(10^4\) counts, and to 5% we need 400 counts, whatever the counting time required to collect them. For large \(\lambda\) (more than about 20) the Poisson distribution is very nearly Gaussian with \(\sigma=\sqrt\lambda\).
The equality of mean and variance is a powerful check of data. Repeated counts from a steady source must scatter with a standard deviation close to \(\sqrt{\bar n}\). A set of counts whose variance is much smaller than their mean is “too good to be true” and suggests that the numbers were not obtained by genuine counting; a reported uncertainty much smaller than \(\sqrt{\bar n}\) for a single interval is equally unbelievable. A variance much larger than the mean indicates an extra source of fluctuation, such as an unstable detector or an intermittent noise source. The test is made quantitatively with the dispersion statistic \(\chi^2=\sum_i(n_i-\bar n)^2/\bar n\), which should be close to \(N-1\) for \(N\) genuine Poisson counts.
Example 8.7. (a) The background in a counter averages 3 counts in a 10 s interval. Find the probability that a given 10 s interval contains no counts, exactly 3 counts, and at least 2 counts. (b) In a test of a detector, 1000 particles known to pass through it produce 900 counts. Estimate its efficiency and the uncertainty.
Solution. (a) With \(\lambda=3\) the Poisson probabilities are
The efficiency is \(\varepsilon=0.900\pm0.009\). It would be wrong to use \(\sqrt{900}=30\) here, because the number of trials is fixed and the counts cannot exceed it; the binomial spread \(\sqrt{np(1-p)}=9.5\) is much smaller than the Poisson value whenever \(p\) is not small.
8.4 Fitting data: the method of least squares
(a) The best straight line
Suppose that we have measured \(N\) pairs of values \((x_i,y_i)\) which are expected to obey a linear law \(y=mx+c\), that the \(x_i\) are known accurately, and that each \(y_i\) has the same random uncertainty. The best line is the one that makes the sum of the squares of the vertical deviations (the residuals) \(d_i=y_i-(mx_i+c)\) as small as possible, as in Figure 8.2. For Gaussian errors this least-squares line is also the line of maximum probability.
Figure 8.2 A least-squares straight line through five data points. The fit minimises the sum of the squares of the vertical residuals \(d_i\) (dotted).
We minimise \(S=\sum_i(y_i-mx_i-c)^2\) by setting its partial derivatives to zero:
The second normal equation shows that the best line passes through the centroid \((\bar x,\bar y)\) of the data, so \(c=\bar y-m\bar x\). Measuring the variables from their means gives the slope in the form most convenient for hand calculation:
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