Turning curves into straight lines

Builds on: Sections S3, S4 and D3
Reading time: 50 min
Exercises: 6
Olympiad problems: 1

Through a pair of binoculars held steady, Jupiter shows up to four points of light strung out on either side of it, the moons that Galileo found in 1610. Io goes round Jupiter in less than two days, while Callisto, more than four times farther out, takes nearly seventeen. Kepler's third law says that the square of a moon's period is proportional to the cube of its distance, but a graph of period against distance is a curve, and a curve can neither be checked by eye nor measured with a ruler. In this section we learn to choose what to plot so that a curved relation becomes a straight line, to which the methods of Sections D2 to D4 apply, and to pick the best of the several ways that are usually possible.

D5.1Why we want a straight line

Before you readMake a guess first

Callisto is 4.46 times as far from Jupiter as Io. How many times as long does it take to go round?

A straight line is easier to work with than any curve. We can see at once whether points lie along one, we can draw it with a ruler, and its slope and intercept, with their uncertainties from the steepest and shallowest lines of Section D3, are two numbers that can carry the unknown constants of a model. So when a model is curved, we look for a quantity $Y$ worked out from each measurement, and another quantity $X$ , such that the model becomes $Y = m X + c$ with the unknown constants in $m$ and $c$ . Finding such quantities is called linearising the model. A graph of $Y$ against $X$ then tests the model, because the points lie along a straight line only if the model is right, and its slope and intercept give the constants.

ResultThree common ways to linearise(D.5)

\begin{aligned} y = A x^{n} & \Rightarrow \log y = n \log x + \log A \\ y = A e^{k x} & \Rightarrow \ln y = k x + \ln A \\ y = \frac{A}{x} + B & \Rightarrow y = A \frac{1}{x} + B \end{aligned}

$x, y$	the measured quantities
$A, B, k, n$	constants of the model, to be found from the line

The first line is straight in a graph of

\log y

against

\log x

, with slope

n

. The second is straight in a graph of

\ln y

against

x

, with slope

k

, and with ordinary logarithms it becomes

\log y = 0.434 k x + \log A

, so the slope is then

0.434 k

. The third is straight in a graph of

y

against

1 / x

, with slope

A

and intercept

B

D5.2Choosing between straight lines

Before you readMake a guess first

Kepler's third law can be made straight in two ways, as $T^{2}$ against $a^{3}$ or as $\log T$ against $\log a$ . Which is better for a graph drawn by hand of moons whose distances range over a factor of 60?

The table gives the semi-major axes $a$ and the periods $T$ of six moons of Jupiter, from the four that Galileo saw to the small inner moon Amalthea and the distant Himalia.

Moon	$a$ (thousand km)	$T$ (day)
Amalthea	181.4	0.498
Io	421.8	1.769
Europa	671.1	3.551
Ganymede	1070.4	7.155
Callisto	1882.7	16.689
Himalia	11 461	250.56

Section S3 showed that a power law is a straight line on logarithmic axes, and Kepler's third law can be made straight in a second way as well. For a small body on a nearly circular orbit of radius $a$ round a mass $M$ , (S.19) gives $T^{2} = (4 π^{2} / G M) a^{3}$ , so a graph of $T^{2}$ against $a^{3}$ is a straight line through the origin with slope $4 π^{2} / G M$ . Figure D.4 shows the moons both ways.

Figure D.4Six moons of Jupiter on three kinds of axesinteractive

Plot

Slope of the best straight line
Largest distance of a point from that line
Points in the left tenth of the plot

The figure shows the periods of the six moons of the table against the semi-major axes of their orbits, from JPL's table of the mean orbits of the planetary satellites. The dashed line is the best straight line through the points, found by the method of Section D6, and the table below the graph says how far the points stray from it and how many of them are crowded into the left tenth of the plot. On the five inner moons, crowded near the origin, lie below the best straight line, because the points follow a curve that bends upwards. On they lie on a straight line, but five of the six moons sit within the first half per cent of each axis. On they lie on a straight line and spread across the whole plot.

Both straight lines test Kepler's law, but they are not equally good for a graph drawn by hand. On the graph of $T^{2}$ against $a^{3}$ , Callisto's point lies only 0.44% of the way along each axis, so the five inner moons cannot be told apart, and the line is set almost entirely by Himalia. On the logarithmic graph every moon counts equally, and the slope comes out as 1.500, which tests the power of the law instead of assuming it. In general a good choice of axes spreads the points evenly, which logarithms do when the values range over several factors of ten, and puts the quantity we want into the slope or the intercept, or into something simply worked out from them.

Example D.10The mass of Jupiter from its moons

The best line on the graph of $\log (T / day)$ against $\log (a / km)$ has slope 1.500 and passes through the centre of the points C, at $\log a = 6.012$ and $\log T = 0.829$ . Find the mass of Jupiter from (S.19), with $G = 6.674 \times 10^{- 11}$ N m² kg⁻².

Solution

Every point on the line describes a possible orbit, so a moon at $a = 10^{6.012}$ km $= 1.028 \times 10^{6}$ km would take $T = 10^{0.829}$ days $= 6.745$ days. In SI units $a = 1.028 \times 10^{9}$ m and $T = 6.745 \times 86 400$ s $= 5.828 \times 10^{5}$ s, so

\begin{aligned} G M & = \frac{4 π^{2} a^{3}}{T^{2}} = \frac{4 π^{2} \times (1.028 \times 10^{9})^{3}}{(5.828 \times 10^{5})^{2}} \\ = 1.26 \times 10^{17} m^{3} s^{- 2}, \end{aligned}

and $M = 1.26 \times 10^{17} / 6.674 \times 10^{- 11} = 1.89 \times 10^{27}$ kg.

AnswerThe mass of Jupiter is

1.89 \times 10^{27}

kg, close to the accepted

1.90 \times 10^{27}

kg. We read the line at C, the best-determined point of any line through the points, as in Section D3. The intercept, at

\log a = 0

, lies more than five units of

\log a

outside the data.

D5.3Exponential laws

Before you readMake a guess first

The pressure of the air falls by roughly the same factor for every 6 km we climb. Which graph of pressure $p$ against height $z$ is close to a straight line?

When a quantity changes by the same factor in every equal step of another, it follows an exponential law, $y = A e^{k x}$ . The pressure of an atmosphere at a single temperature is one example, $p = p_{0} e^{- z / H}$ by (S.27), where $H$ is the scale height. Taking natural logarithms gives $\ln p = \ln p_{0} - z / H$ , a straight line in a graph of $\ln p$ against $z$ with slope $- 1 / H$ . A graph with a logarithm on one axis only is called a semi-logarithmic graph.

Figure D.5The pressure of the air on two kinds of axesinteractive

Plot

Slope of the best straight line
Largest distance of a point from that line
Points in the left tenth of the plot

The figure shows the pressure of the International Standard Atmosphere, the average atmosphere that aircraft altimeters are calibrated against, at every 2 km of height up to 20 km. On the points follow a curve that flattens with height. On they lie close to a straight line. Below 11 km, where the air grows colder with height, they bend slightly, and above 11 km, where its temperature stays at 216.65 K, they lie exactly on a straight line, as the formula for an atmosphere at a single temperature says.

Two points on the straight part, at 12 km and 18 km, where the pressure is 193.3 hPa and 75.05 hPa, give the scale height. The difference of their logarithms is $\ln (193.3 / 75.05) = 0.946$ , so

H = \frac{18 km - 12 km}{0.946} = 6.34 km,

which is the value (S.27) gives for air at 216.65 K. The best line through all eleven points gives 6.8 km instead, an average over air of different temperatures.

Magnitudes are logarithms already, so a light source whose flux falls exponentially gives a straight line when its magnitude is plotted against time, with no further work. A flux $F = F_{0} e^{- t / τ}$ gives $m = m_{0} + 1.086 t / τ$ , since $2.5 \log e = 1.086$ . A few months after a supernova explodes, most of its light comes from the decay of radioactive cobalt-56, and it fades in just this way.

Quick checkA fading supernova

A supernova fades from magnitude 15.20 to 17.15 over 200 days. If its flux follows $F = F_{0} e^{- t / τ}$ , what is $τ$ ?

days

D5.4Rearranging a model

Before you readMake a guess first

A thin lens obeys $1 / d_{o} + 1 / d_{i} = 1 / f$ , where $d_{o}$ and $d_{i}$ are the distances of an object and its image. Which graph of measured distances is a straight line?

Many models fit none of the three forms of (D.5), but they can still be rearranged. The aim is to gather the measured quantities into $X$ and $Y$ and leave each unknown as a slope or an intercept. The lens formula of (S.28) is one example, and the guess above shows how it becomes straight. An unknown starting point is another. When a model contains a time measured from a moment we do not know, such as the instant two stars merge, the unknown moment goes into the intercept, and the problem at the end of this section works through such a case.

Example D.11The focal length of a lensfinish it

A student places an object at distances $d_{o}$ of 30, 40, 60 and 100 cm from a lens and finds sharp images at distances $d_{i}$ of 59.5, 40.3, 29.8 and 25.1 cm behind it. Find the focal length of the lens.

The start

The reciprocals $1 / d_{o}$ are 0.0333, 0.0250, 0.0167 and 0.0100 cm⁻¹, and the reciprocals $1 / d_{i}$ are 0.0168, 0.0248, 0.0336 and 0.0398 cm⁻¹. The four points of $1 / d_{i}$ against $1 / d_{o}$ lie close to a straight line of slope $- 1.0$ , as the formula says, and the best line through them crosses the vertical axis at 0.0499 cm⁻¹.

Your stepNow find

f

D5.5Error bars on the new axes

Before you readMake a guess first

A flux is measured as 50 ± 5 units. When $\log F$ is plotted instead, how far does its error bar reach on either side?

When we plot $Y$ instead of $y$ , each error bar has to be carried to the new axis by the rule of small changes, $σ_{Y} = | d Y / d y | σ_{y}$ , as in Section D4. For a logarithm this gives $σ_{\log y} = 0.434 σ_{y} / y$ , so error bars that are all the same length on $y$ come out long for small values and short for large ones. For a reciprocal it gives $σ_{1 / y} = σ_{y} / y^{2}$ . A hand-drawn line should then follow the points with short error bars more closely than the others. When an error bar is a large fraction of its value, its two ends are best carried to the new axis separately, as Section D4 explains, which gives a bar that is longer on one side than on the other.

D5.6Logarithmic graph paper

The IOAA syllabus mentions graph paper with logarithmic scales. On such paper the lines are spaced by their logarithms, every factor of ten takes the same length, called a cycle or a decade, and values are plotted as they are, without taking their logarithms first. Since the paper has already taken the logarithms, the slope of a line on log–log paper is worked out from values read off its scales,

n = \frac{\log y_{2} - \log y_{1}}{\log x_{2} - \log x_{1}} .

Measuring the slope in centimetres gives the right answer only when a decade has the same length on both axes. Semi-logarithmic paper has one logarithmic scale and one ordinary scale, for exponential laws. For two points read off it, $(\log y_{2} - \log y_{1}) / (x_{2} - x_{1})$ is the slope $0.434 k$ of (D.5).

Common errorsSection D5

A slope measured in centimetres on logarithmic paper is wrong whenever a decade has a different length on the two axes. It has to be worked out from the logarithms of values read off the scales.
With ordinary logarithms an exponential law gives a slope of $0.434 k$ , and taking this slope for $k$ makes $k$ 2.303 times too small.
Error bars cannot be copied from $y$ to $\log y$ unchanged. Each one is carried to the new axis as in Section D4, and on a logarithmic axis they come out of different lengths.
A choice of axes that crowds most of the points into one corner hides how well they fit, and the line is then set by the one or two points far from the rest.
The intercept of a logarithmic graph lies where $\log x = 0$ , which can be far outside the data. The line is better read at the centre of the points.

ExercisesSection D56 questions

D5.1
The luminosities $L$ of main-sequence stars are thought to grow as a power of their masses $M$ . Which graph tests this and gives the power as its slope?
D5.2
A quantity follows $y = A e^{- t / τ}$ . On a graph of $\log y$ against $t$ , the line falls by 0.30 over 10 days. What is $τ$ ?
days
D5.3
A value is measured as $y = 200 \pm 10$ . How long is its error bar on either side when $1 / y$ is plotted?
D5.4
On log–log graph paper a straight line passes through the points (2, 30) and (50, 3000), read off the scales. What is its slope?
D5.5
The speed $v$ of a comet at a distance $r$ from the Sun obeys $v^{2} = G M (2 / r - 1 / a)$ , where $a$ is the semi-major axis of its orbit. Which graph is a straight line whose intercept gives $a$ ?
D5.6
In the standard atmosphere the pressure is 1013.25 hPa at sea level and 794.96 hPa at a height of 2 km. What scale height do these two values give?
km

Olympiad problemsPast IOAA problems, restated1 problem

2022The chirp of two merging neutron stars27 points

Adapted from IOAA 2022 (Kutaisi), data-analysis problem 1, parts 1.2.1 to 1.2.3. We have restated it in our own words and kept the original data. The original problem first asked for the values to be read from an image of the signal, and we give the values that the official solution read from it. The solution was last checked on 20 October 2026.

On 17 August 2017 the LIGO detectors recorded the gravitational waves of GW170817, from two neutron stars spiralling into each other. The table gives the frequency $f$ of the waves at times $t$ before the merger.

$t$ (s)	$f$ (Hz)	$t$ (s)	$f$ (Hz)
−28	39.1	−12	53.8
−26	39.1	−10	56.7
−24	41.3	−8	63.1
−22	41.3	−6	70.1
−20	43.5	−4	78.0
−18	45.9	−2	101.7
−16	48.4	0	182.4
−14	51.0

While the frequency is low, the way it rises depends on the masses $m_{1}$ and $m_{2}$ of the two stars only through their chirp mass,

\begin{aligned} M & = \frac{(m_{1} m_{2})^{3 / 5}}{(m_{1} + m_{2})^{1 / 5}} \\ = \frac{c^{3}}{G} {(\frac{5}{96} π^{- 8 / 3} f^{- 11 / 3} \dot{f})}^{3 / 5}, \end{aligned}

where $\dot{f}$ is the rate at which the frequency rises. (a) Find how $f$ depends on $t$ , using the fact that if $x^{n} \dot{x} = k$ then $x^{n + 1} / (n + 1) = k t + C$ , where $k$ and $C$ are constants. This was worth 3 marks. (b) By plotting a suitable graph, find the chirp mass and the uncertainty that comes from the uncertainty of the slope. This was worth 22 marks. (c) The detectors measure masses multiplied by $1 + z$ , where $z$ is the redshift of the source, and the host galaxy NGC 4993 has $z = 0.009783$ . Find the chirp mass in the frame of the source. This was worth 2 marks. Use $G = 6.674 \times 10^{- 11}$ N m² kg⁻², $c = 2.998 \times 10^{8}$ m/s and a solar mass of $1.989 \times 10^{30}$ kg.

I solved it without the solution

AnswerThe chirp mass is 1.201 ± 0.015 solar masses as measured, and 1.189 ± 0.015 solar masses in the frame of the source. The official solution finds 1.20 solar masses with a 95% range of ±0.03, which is twice our uncertainty of one standard deviation.