Times of minima and changing periods

Builds on: Sections 8.4 and D8
Reading time: 30 min
Exercises: 3
Olympiad problems: 1

Near Vega, in Lyra, the star β Lyrae rises and falls between magnitude 3.3 and 4.4 every 12.94 days, as two stars eclipse each other. Its period is not quite steady. It grows by about 19 seconds every year, so that eclipses predicted with the period of two centuries ago now arrive about four months late. The cause is gas pouring from one star onto the other, which changes the orbit. Small changes in a period build up over many cycles, and the times of eclipses are among the most precise clocks in astronomy. In this section we see what they reveal.

8.5.1When eclipses come early or late

Before you readMake a guess first

A binary's period grows by 1 second each century. Over a century of watching, how late does the last eclipse arrive compared with the prediction from the period at the start?

Times of minimum are compared with a prediction, the ephemeris, made with a fixed period. The observed minus the calculated time, O − C, plotted against the cycle number $E$ , is the O − C diagram of Section D8. If the period used is slightly wrong, the points lie on a straight line whose slope is the error in the period ((D.11)). If the period itself is changing steadily, each cycle differs from the last by the same small amount, and the points lie on a parabola.

ResultA steadily changing period(8.7)

\begin{aligned} T_{E} & = T_{0} + P_{0} E + \frac{1}{2} P_{0} \dot{P} E^{2} \\ O - C & = \frac{1}{2} P_{0} \dot{P} E^{2} \end{aligned}

$T_{E}$	the time of the minimum in cycle $E$
$T_{0}, P_{0}$	the time of the minimum in cycle 0 and the period then
$\dot{P}$	the rate at which the period changes, in seconds per second, or in seconds of period per year when $P_{0}$ is in years
$O - C$	the observed time minus the time calculated with the fixed period $P_{0}$

The period grows by

P_{0} \dot{P}

in each cycle. A parabola that opens upwards means a lengthening period and one that opens downwards a shortening period. With both a wrong period and a changing one, O − C is a straight line plus a parabola.

8.5.2A third star

If a third star orbits the eclipsing pair, the pair itself moves round the centre of mass of all three. When the pair is on the far side, its light has further to travel and every eclipse arrives late, and on the near side early. The O − C diagram then shows a wave with the period of the third star's orbit, while the true period of the pair does not change at all. This is the light-travel-time effect. The largest delay is the time light takes to cross $a_{12} \sin i$ , where $a_{12}$ is the semi-major axis of the pair's orbit round the centre of mass of the three stars. The same effect, in the timing of Jupiter's moons, gave the first measurement of the speed of light, in 1676.

Figure 8.5Reading an O − C diagram

Error in the period usedChange of period per centuryLight-time wave

Shape
Last eclipse

The figure shows the O − C diagram of an eclipsing binary with a period of 2.000 days, followed for 10 000 cycles, about 55 years. The dots are measured times of minimum with a scatter of a minute or two, and the line is the model. An error in the period used for the predictions tilts the points into a straight line, a changing period bends them into a parabola, and a third star adds a wave with a period of 15 years. The figure can be set to , , , or .

Quick checkThe size of a light-time wave

An eclipsing pair moves round the centre of mass of a triple system on an orbit with $a_{12} \sin i = 1.00$ au. How many minutes early or late can its eclipses be?

minutes

Example 8.9A period from a parabola

A binary with a period of 2.000 days is timed for 8000 cycles. With the starting period, the O − C diagram is a parabola through the origin, reaching 30.0 minutes at the last cycle. At what rate is the period changing?

Solution

By (8.7), $\frac{1}{2} P_{0} \dot{P} E^{2} = 1800$ s at $E = 8000$ , so the period grows in each cycle by $P_{0} \dot{P} = 2 \times 1800 / 8000^{2} = 5.63 \times 10^{- 5}$ s. There are $365.25 / 2.000 = 182.6$ cycles in a year.

AnswerThe period grows by

5.63 \times 10^{- 5} \times 182.6 = 1.03 \times 10^{- 2}

s a year, about 1 second a century.

8.5.3Mass flowing between the stars

In a close binary, one star may swell until gas spills from it onto the other, as Section 8.6 describes. If no mass or angular momentum leaves the system, the total mass and the angular momentum stay the same while the masses change, and the orbit must adjust. Moving mass from the heavier star to the lighter one shrinks the orbit and shortens the period, and moving it from the lighter star to the heavier one does the opposite.

ResultThe period of a binary exchanging mass(8.8)

\frac{Δ P}{P} = \frac{3 Δ M_{1} (M_{1} - M_{2})}{M_{1} M_{2}}, \frac{Δ a}{a} = \frac{2}{3} \frac{Δ P}{P}

$Δ P, Δ a$	the changes of the period and of the separation
$Δ M_{1}$	the mass gained by star 1, which star 2 loses
$M_{1}, M_{2}$	the masses of the two stars

This holds for circular orbits when the mass moved is small and none escapes. If the heavier star gains mass,

Δ P

is positive and the period lengthens. The second relation comes from Kepler's third law with the total mass fixed.

Show the derivation

The angular momentum of two point masses on circular orbits a distance $a$ apart, turning at $ω$ , is $L = M_{1} M_{2} a^{2} ω / (M_{1} + M_{2})$ . By Kepler's third law $ω^{2} a^{3} = G (M_{1} + M_{2})$ , which with the total mass fixed makes $a \propto ω^{- 2 / 3}$ , so $L \propto M_{1} M_{2} ω^{- 1 / 3}$ . Keeping $L$ fixed,

\frac{Δ M_{1}}{M_{1}} + \frac{Δ M_{2}}{M_{2}} - \frac{1}{3} \frac{Δ ω}{ω} = 0.

With $Δ M_{2} = - Δ M_{1}$ and $Δ P / P = - Δ ω / ω$ , this gives $Δ P / P = 3 Δ M_{1} (1 / M_{1} - 1 / M_{2})$ , which is the result. Finally $a \propto ω^{- 2 / 3} \propto P^{2 / 3}$ gives the second relation.

Example 8.10How fast β Lyrae exchanges massfinish it

β Lyrae has a period of 12.94 days that grows by 19 s a year. The star gaining mass has about 13 solar masses and the one losing it about 3. At what rate is mass flowing between them?

The start

In one year the period grows by the fraction $Δ P / P = 19 / (12.94 \times 86 400) = 1.70 \times 10^{- 5}$ . By (8.8), $Δ M_{1} = (Δ P / P) M_{1} M_{2} / (3 (M_{1} - M_{2}))$ .

Your stepNow work out

Δ M_{1}

for one year.

× 10⁻⁵ solar masses a year

Common errorsSection 8.5

A straight O − C diagram does not mean that the period is changing. It means that the period used for the predictions is slightly wrong.
A parabola that opens upwards means a period that grows, since the eclipses fall further and further behind the prediction.
Mass moving onto the heavier star lengthens the period. Mass moving onto the lighter star shortens it.
A wave in the O − C diagram from a third star does not mean that the eclipsing pair's period is changing. Only the travel time of the light changes.

ExercisesSection 8.53 questions

8.5.1
In an O − C diagram, the points lie on a straight line that rises by 1.0 minute every 100 cycles. By how much is the true period longer than the one used?
s
8.5.2
The O − C diagram of an eclipsing binary shows a wave of semi-amplitude 5.0 minutes. What is $a_{12} \sin i$ , in astronomical units?
au
8.5.3
Mass flows from the heavier star of a binary to the lighter one, with no mass lost from the system. What happens to the period?

Olympiad problemsPast IOAA problems, restated1 problem

2007A binary exchanging mass10 points

Adapted from IOAA 2007 (Chiang Mai), theory problem 3. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

Two stars of masses $M_{1}$ and $M_{2}$ , a distance $D$ apart, move on circular orbits round their centre of mass with angular speed $ω$ . Mass flows steadily from one star to the other, which slowly changes the period and the separation. Treat the stars as points and ignore their spins. For small $x$ and $y$ , $(1 + x)^{n} \approx 1 + n x$ and $(1 + x) (1 + y) \approx 1 + x + y$ .

(a) Find the total angular momentum and the kinetic energy of the system. [2 points]

(b) Find the relation between $ω$ and $D$ . [2 points]

(c) In a time $Δ t$ , star 1 gains a mass $Δ M_{1}$ . Find $Δ ω$ in terms of $ω$ , $M_{1}$ , $M_{2}$ and $Δ M_{1}$ . [3 points]

(d) In one binary $M_{1} = 2.9$ solar masses, $M_{2} = 1.4$ solar masses and the period is 2.49 days. After 100 years the period has grown by 20 s. Find $Δ M_{1} / (M_{1} Δ t)$ , per year. [1.5 points]

(e) Which way is mass flowing? [0.5 points]

(f) Find $Δ D / (D Δ t)$ , per year. [1 point]

I solved it without the solution

Answer(a) $L = M_{1} M_{2} D^{2} ω / (M_{1} + M_{2})$ and the kinetic energy is $M_{1} M_{2} D^{2} ω^{2} / (2 (M_{1} + M_{2}))$ . (b) $ω^{2} = G (M_{1} + M_{2}) / D^{3}$ . (c) $Δ ω = 3 ω Δ M_{1} (M_{2} - M_{1}) / (M_{1} M_{2})$ . (d) $2.9 \times 10^{- 7}$ per year. (e) From $M_{2}$ to $M_{1}$ , the lighter star to the heavier. (f) $6.2 \times 10^{- 7}$ per year.