Light curves and other time series

Builds on: Sections D3, D6 and D7
Reading time: 55 min
Exercises: 6
Olympiad problems: 1

Every 2 days and 21 hours, Algol in Perseus fades from magnitude 2.1 to 3.4 and back within about ten hours, as a fainter companion passes in front of it, and anyone who knows where to look can watch it happen with the naked eye. Its brightness measured again and again makes a light curve, one kind of time series, a list of measurements in the order they were made. In the data-analysis round time series come as light curves of variable stars and eclipsing binaries, radial-velocity curves of binaries and planets, and the flickering light of active galaxies, and the questions ask for a period, an amplitude, a time lag or a change in a period. In this section we learn how to answer each.

D8.1Periods from many cycles

Before you readMake a guess first

Two minima of an eclipsing binary are timed to within 0.005 days each, and they are exactly 100 cycles apart. How precisely does this give the period?

A period is best measured over many cycles. If two moments of the same phase, two minima of an eclipsing binary for example, lie $N$ cycles apart, the period is their difference divided by $N$ , and the uncertainty of the timing is divided by $N$ too. The catch is that $N$ has to be known exactly. A rough period, found from minima close together, predicts how many cycles fit between minima far apart, and the count is safe as long as the rough period is good to much better than half a period divided by the number of cycles. A count that is wrong by even one cycle gives a period that is wrong by far more than its apparent uncertainty.

Example D.13Counting the cyclesfinish it

Two minima of an eclipsing binary like Algol are observed 286.579 days apart, and minima a few nights apart show that the period is close to 2.87 days. Find the period.

The start

At 2.87 days a cycle, $286.579 / 2.87 = 99.85$ cycles fit between the two minima, so the number of cycles is 100. For the count to be safe the rough period has to be good to much better than $2.87 / 200 = 0.014$ days, which three minima a few nights apart easily give.

Your stepNow find the period.

days

D8.2Phase and folding

Before you readMake a guess first

Observations of a Cepheid are scattered over 80 days, one or two on each clear night. How can we see the shape of its light curve?

The phase of an observation made at time $t$ is how far through its cycle the star was, from 0 at a chosen reference moment $t_{0}$ , such as a maximum, to just below 1 a whole cycle later. It is the fractional part of the number of cycles since $t_{0}$ . Plotting a light curve against phase instead of time is called folding it. When the period is right every cycle lands on the same curve, and when it is wrong the cycles slide past each other and the points scatter. Trying many periods and keeping the one with the least scatter is a way of finding a period, and Figure D.8 lets us do it by hand.

ResultThe phase of an observation(D.10)

φ = \frac{t - t_{0}}{P} - ⌊ \frac{t - t_{0}}{P} ⌋

$t$	the time of the observation
$t_{0}$	a reference time, such as a maximum or a minimum
$P$	the period
$⌊ ⌋$	the whole number just below, so that $φ$ is the fractional part

Folded light curves are usually drawn over two cycles, from phase 0 to 2, so that the shape near phase 0 and 1 is seen whole.

Figure D.8Folding a light curve on a trial periodinteractive

Plot

Trial period

Scatter of the folded curve
Cycles in the 80 days of observations

The figure shows forty observations of a star like δ Cephei, made from a model of its light curve with its real period of 5.366 days and its real range of 3.48 to 4.37 magnitudes, at random times over 80 days and with random errors of 0.02 magnitudes. Brighter is up, as on every light curve. Plotted they look like a scatter. Folded on a trial period, each observation is placed at its phase and drawn twice, once in each of two cycles, and the table gives the typical difference in magnitude between neighbouring points. At the folded points scatter widely, at they almost line up, and at they trace one thin curve, with a quick rise to maximum and a slow fall.

The folded curves at 5.350 and 5.366 days look almost the same. Over 80 days, about 15 cycles, a trial period 0.016 days too short slides the last cycle against the first by only $15 \times 0.016 / 5.366 = 0.04$ of a cycle, so the curve hardly blurs. A run of observations twice as long would show the same error twice as clearly, which is why long runs give precise periods.

Quick checkThe phase of an observation

A Cepheid with a period of 5.366 days was at maximum at $t_{0}$ . What is its phase 20.0 days later, counting maximum as phase 0?

D8.3The O − C diagram

Before you readMake a guess first

The predicted minima of an eclipsing binary come later and later than the observed ones, by the same extra amount in each cycle. What does this mean?

A formula that predicts when minima happen is called an ephemeris. For a constant period it is $M_{calc} = M_{0} + P E$ , where $M_{0}$ is the time of one chosen minimum and $E$ , the epoch, counts the cycles since then, in whole numbers for the primary minima and halves for the secondary ones. The O − C diagram plots the observed time of each minimum minus the calculated one against $E$ . If the ephemeris is right, the points scatter about zero. If the period used is slightly wrong, they lie along a sloping line whose slope is the error in the period, and a line that crosses $E = 0$ above or below zero shows an error in $M_{0}$ . A straight-line fit, by eye or by the least squares of Section D6, therefore corrects both. A parabola shows a period that changes steadily, as it does when one star of a binary passes gas to the other, and a wave shows the light-travel time across the orbit of a third, unseen star.

ResultCorrecting an ephemeris(D.11)

\begin{aligned} M_{calc} & = M_{0} + P E \\ O - C & = Δ M_{0} + Δ P E \end{aligned}

$M_{0}, P$	the time of the chosen first minimum and the period used
$E$	the epoch, the number of cycles since $M_{0}$
$O - C$	the observed time of a minimum minus the calculated time
$Δ M_{0}, Δ P$	the corrections to $M_{0}$ and $P$ , the intercept and the slope of the straight line through the O − C diagram

The second line holds when the true period is constant. The improved ephemeris is

M_{0} + Δ M_{0} + (P + Δ P) E

Quick checkA period from an O − C diagram

On an O − C diagram the points lie along a straight line that rises by 0.020 days over 400 cycles. By how much should the period be increased?

days

D8.4Light curves and radial-velocity curves

Before you readMake a guess first

The radial velocity of a star in a binary swings between +35.2 km/s and −12.4 km/s. At what speed is the binary as a whole moving away from us?

From a light curve we read the period, from one feature to the next of the same kind, and the amplitude, the difference in magnitude between maximum and minimum. Its shape tells us what kind of star it is. A Cepheid rises quickly to maximum and fades slowly, an eclipsing binary like Algol stays steady between short dips, and a W Ursae Majoris binary, whose two stars touch, changes all the time with two nearly equal minima in each orbit. A shift between the curves of the same star in two colours, or between its light curve and its radial-velocity curve, is quoted as a fraction of the period.

A radial-velocity curve plots the speed of a star along our line of sight against time. For a circular orbit it is a sine curve, and two numbers describe it. Its semi-amplitude $K$ is half the full swing, the speed of the star round the centre of mass along the line of sight, and its middle, the systemic velocity $γ$ , is the speed of the whole binary. For the guess above, $K = (35.2 + 12.4) / 2 = 23.8$ km/s and $γ = (35.2 - 12.4) / 2 = 11.4$ km/s. Chapter 8 turns $K$ and the period into the masses of the stars.

D8.5Time lags and areas

Before you readMake a guess first

The light of an active galaxy's broad emission lines follows changes in its continuum 20 days later. Roughly how far is the gas that emits the lines from the source of the continuum?

When two curves rise and fall together, one later than the other, the delay is found by sliding one along the time axis until its peaks and dips line up with those of the other. IOAA 2020 gave the light curves of an active galaxy's continuum and of its broad Hβ emission line, and the line followed the continuum by about 20 days. The gas that emits the line is therefore about $c \times 20$ days $= 5.2 \times 10^{14}$ m $= 0.017$ pc from the centre, which at a distance of 100 Mpc is an angle of only $3.5 \times 10^{- 5}$ arcseconds, far too small to see directly.

Some problems ask for the area under a curve, such as the total energy radiated by an outburst, which is the integral of its luminosity over time. When the curve is known only at measured points, the trapezium rule of (S.13) joins the points with straight lines and adds up the areas of the strips beneath them. It is exact where the curve is straight and slightly off where it bends between points, so points closer together make it better.

Example D.14The energy of a supernova

The luminosity of a supernova, measured every 10 days from the explosion, is 0, 8, 6, 3 and 1.5 in units of $10^{35}$ W. Estimate the energy it radiates in these 40 days.

Solution

The four strips are each 10 days wide, so by (S.13)

\begin{aligned} E & \approx 10 \times (\frac{0 + 8}{2} + \frac{8 + 6}{2} + \frac{6 + 3}{2} + \frac{3 + 1.5}{2}) \\ = 177.5 in units of 10^{35} W day . \end{aligned}

A day is 86 400 s, so $E \approx 177.5 \times 10^{35} \times 86 400$ J $= 1.53 \times 10^{42}$ J.

AnswerThe supernova radiates about

1.53 \times 10^{42}

J in 40 days, as much as the Sun gives out in about 130 million years.

Common errorsSection D8

A period found from two minima far apart is badly wrong if the number of cycles between them is miscounted by even one. The count should be checked with a rough period first.
A light curve folded on half or twice the true period can also look tidy, with each cycle drawn twice or with two equal humps. The shape of the curve, and the depths of an eclipsing binary's two minima, decide between them.
O − C is the observed time minus the calculated one, and taking them the other way round reverses the sign of the correction to the period.
The semi-amplitude $K$ of a radial-velocity curve is half of the full swing from the largest velocity to the smallest.
Times of minima are usually given as heliocentric Julian dates, corrected for the time light takes to cross Earth's orbit, which can shift a time by up to about 8 minutes. A time from an ordinary clock needs the same correction before the two are compared.

ExercisesSection D86 questions

D8.1
A variable star with a period of 0.60 days was at maximum at day 2.40. What is its phase at day 5.95?
D8.2
The radial velocity of a star in a circular binary orbit swings between +35.2 km/s and −12.4 km/s. What is the systemic velocity of the binary?
km/s
D8.3
The broad emission lines of an active galaxy follow its continuum 30 days later. How far, in parsecs, is the gas that emits the lines from the centre?
pc
D8.4
A curve is measured at $x$ = 0, 1, 2, 3 and 4, where it has the values 0, 4, 6, 4 and 0. What area under it does the trapezium rule give?
D8.5
An eclipsing binary has a period close to 0.2734 days, and two of its primary minima are 275.05 days apart. What is its period?
days
D8.6
The points of an O − C diagram follow a parabola that opens upwards. What does this show?

Olympiad problemsPast IOAA problems, restated1 problem

2011The times of minima of V1107 Cassiopeiae25 points

Adapted from IOAA 2011 (Katowice), data-analysis problem 1. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

V1107 Cas is an eclipsing binary of the W Ursae Majoris type, whose two stars touch, so that its light changes all the time and it shows two nearly equal minima in each orbit. The table lists twelve observed times of minimum as heliocentric Julian dates, with their uncertainties.

No.	Date	Time of minimum (HJD)	Uncertainty (day)
1	22 December 2006	2 454 092.4111	0.0004
2	23 December 2006	2 454 092.5478	0.0002
3	23 September 2007	2 454 367.3284	0.0005
4	23 September 2007	2 454 367.4656	0.0005
5	15 October 2007	2 454 388.5175	0.0009
6	15 October 2007	2 454 388.6539	0.0011
7	26 August 2008	2 454 704.8561	0.0002
8	5 November 2008	2 454 776.4901	0.0007
9	3 January 2009	2 454 835.2734	0.0007
10	15 January 2009	2 454 847.3039	0.0004
11	15 January 2009	2 454 847.4412	0.0001
12	16 January 2009	2 454 847.5771	0.0004

(a) Find a first value of the period, assuming that it is constant and that the observations on any one night were continuous. (b) Using $M_{calc} = M_{0} + P E$ , where $E$ is a whole or half number of cycles counted from a chosen first minimum $M_{0}$ , draw an O − C diagram, with $E$ across and $O - C = M_{obs} - M_{calc}$ up. (c) From the diagram, improve $M_{0}$ and $P$ and estimate their uncertainties. (d) Predict the times of the minima between 19 h UT on 1 September 2011 and 2 h UT on 2 September 2011. The whole problem was worth 25 marks.

I solved it without the solution

AnswerThe improved ephemeris is $M_{calc} = 2 454 092.4115 + 0.2734127 E$ , with the period uncertain by 0.0000001 days and $M_{0}$ by 0.0002 days. The minima in the window fall at about 19 h 11 m and 22 h 28 m UT on 1 September 2011 and 1 h 45 m UT on 2 September. The official solution reads the line by eye and finds a period of 0.2734129 days and times one minute later, within the uncertainty of the prediction.