Interacting binaries and novae

Builds on: Sections 3.9, 8.4 and 8.5
Reading time: 30 min
Exercises: 3
Olympiad problems: 1

On 14 August 2013 the Japanese amateur astronomer Koichi Itagaki found a new star of magnitude 6.8 in the small constellation Delphinus, close to the Summer Triangle, where only a star of about magnitude 17 had been seen before. Two days later it reached magnitude 4.5, visible to the eye from a dark site, and over the following months it faded away. Nova Delphini 2013 was not a new star. It was a white dwarf in a close binary, which had gathered hydrogen from its companion until the hydrogen exploded on its surface. In this section we see how gas passes from one star to the other, and what it can do when it arrives.

8.6.1Roche lobes

Before you readMake a guess first

In a close binary, one star swells up as it ages. What happens when its outer layers reach the point between the two stars where their pulls balance?

In the frame that turns with a binary, each star sits in a hollow of the effective potential, which adds the energy of the turning to the gravity of both stars (Section 3.9). The surface of equal effective potential through the inner Lagrange point L1 makes a figure eight round the two stars, and each half is that star's Roche lobe. Gas inside a star's lobe belongs to that star. A star that grows to fill its lobe spills gas through L1 onto its companion.

Figure 8.6The Roche lobes of a binary

Mass ratio

m / (M + m)

L1 and L2 from m
Stability

The figure shows lines of equal effective potential round two stars, a heavier M and a lighter m, in the frame turning with them. The line through L1, between the stars, outlines the two Roche lobes. The lighter star has the smaller lobe, so in a pair of similar stars it fills its lobe first if it is the larger. The figure can be set to , , or .

Binaries are named by how full their lobes are. In a detached binary both stars lie well inside their lobes and live much as single stars do. In a semi-detached binary one star fills its lobe and passes gas to the other, as in Algol and β Lyrae. In a contact binary both stars overfill their lobes and share one envelope, as in the W Ursae Majoris stars, whose light curves change smoothly all the time.

ResultThe size of a Roche lobe(8.9)

\frac{r_{L}}{a} = \frac{0.49 q^{2 / 3}}{0.6 q^{2 / 3} + \ln (1 + q^{1 / 3})}, q = \frac{M_{1}}{M_{2}}

$r_{L}$	the radius of a sphere with the same volume as the Roche lobe of star 1
$a$	the separation of the two stars
$M_{1}, M_{2}$	the masses of star 1 and of its companion

This is Eggleton's approximation, good to about 1% for every mass ratio. For equal masses each lobe has

r_{L} = 0.38 a

. The heavier star always has the larger lobe.

Quick checkA lobe for equal masses

Two stars have equal masses. What fraction of their separation is the radius of each Roche lobe?

8.6.2The Algol paradox

The heavier a star, the faster it uses its fuel and the sooner it swells into a giant (Chapter 7). In Algol, though, the cooler star is a swollen subgiant of only about 0.7 solar masses, while its companion of about 3.2 solar masses is still on the main sequence. Two stars born together should not have aged the wrong way round. The answer is mass transfer. The star that is now lighter was once the heavier one. It swelled first, filled its Roche lobe and gave most of its mass to its companion. As (8.8) shows, the orbit first shrank and then widened again as the masses passed through equality.

8.6.3Novae

When the star receiving gas is a white dwarf, the stream through L1 cannot fall straight onto it, because the gas carries the angular momentum of the orbit. It spirals in through a flat, hot accretion disc instead. Such pairs are called cataclysmic variables. Hydrogen builds up on the white dwarf's surface, squeezed and heated by its strong gravity, until after thousands of years the layer reaches tens of millions of kelvin and fuses in a runaway. The star brightens by 8 to 15 magnitudes in a day or two. This is a nova. The ejected shell, a hundred-thousandth of a solar mass or so, flies out at hundreds to thousands of kilometres per second, and the white dwarf survives to do it again. A nova is far less powerful than a supernova, which destroys its star.

Novae that brighten more fade faster. The time a nova takes to fade by 2 magnitudes from its peak is called $t_{2}$ , and by 3 magnitudes $t_{3}$ . Their peak absolute magnitudes follow from these times, so a nova is a rough standard candle.

ResultPeak brightness and rate of decline of a nova(8.10)

\begin{aligned} M_{0} & = - 11.32 + 2.55 \log t_{2} \\ M_{0} & = - 11.99 + 2.54 \log t_{3} \end{aligned}

$M_{0}$	the absolute visual magnitude of the nova at its peak
$t_{2}, t_{3}$	the times in days for the nova to fade by 2 and by 3 magnitudes from its peak

These are fits to novae of known distance by Downes and Duerbeck (2000), with a scatter of about half a magnitude. Other fits of a similar kind exist, and a problem will give the one to use. The distance follows from the distance modulus with extinction ((4.14)).

Example 8.11The distance of a fast nova

A nova peaks at visual magnitude 6.0 and fades by 2 magnitudes in 4.0 days. The extinction towards it is 1.0 magnitude. How far away is it?

Solution

By (8.10), $M_{0} = - 11.32 + 2.55 \log 4.0 = - 9.785$ . The distance modulus corrected for extinction is $m - M - A_{V} = 6.0 + 9.785 - 1.0 = 14.785$ , so $d = 10^{(14.785 + 5) / 5} = 10^{3.957}$ pc, or 9060 pc.

AnswerThe nova is about 9.1 kpc away, on the far side of the Milky Way's central regions if it lies in that direction.

Example 8.12How large a nova shell looksfinish it

A nova 3.0 kpc away throws off a shell at 1500 km/s. How large does the shell look on the sky, as a diameter, 30 days after the outburst?

The start

The shell's radius after 30 days is $1500 \times 30 \times 86 400 = 3.89 \times 10^{9}$ km, which is 26.0 au.

Your stepNow turn the diameter into an angle at 3.0 kpc.

mas

Common errorsSection 8.6

A nova is not the birth of a new star, and it does not destroy its star. It is an explosion on the surface of a white dwarf, which survives and can erupt again.
A Roche lobe is not a sphere. The radius of (8.9) is that of a sphere with the same volume.
The lighter star has the smaller Roche lobe, so in Algol it is the lighter star that fills its lobe and gives gas to the heavier one.
A nova that fades fast was brighter at its peak, not fainter, so a small $t_{2}$ means a more negative $M_{0}$ .

ExercisesSection 8.63 questions

8.6.1
Star 1 of a binary has half the mass of star 2. What fraction of the separation is the radius of star 1's Roche lobe?
8.6.2
A nova rises from magnitude 16 to magnitude 6. By what factor has its brightness grown?
times
8.6.3
In Algol the less massive star is the more evolved one. Why?

Olympiad problemsPast IOAA problems, restated1 problem

2019Nova Delphini 201345 points

Adapted from IOAA 2019 (Keszthely), data-analysis problem 1. We have restated it in our own words and kept the original data. The original asked for readings from a light curve and a set of spectra, and for a graph. We give the readings of the official solution instead and leave out those parts, so the points add up to 45 of the original 60. The solution was last checked on 21 October 2026.

The nova V339 Delphini was discovered on MJD 56518.584, where MJD is the Julian date minus 2 400 000.5. Its light curve peaked at visual magnitude $4.50 \pm 0.05$ and it took $t_{2} = 11 \pm 1$ days to fade by 2 magnitudes and $t_{3} = 23 \pm 1$ days to fade by 3. Three fits give its peak absolute magnitude.

$M_{0} = - 7.92 - 0.81 \arctan [(1.32 - \log t_{2}) / 0.23]$ , by Della Valle and Livio (1995).
$M_{0} = - 11.32 + 2.55 \log t_{2}$ , by Downes and Duerbeck (2000).
$M_{0} = - 11.99 + 2.54 \log t_{3}$ , by the same authors.

The colour excess towards the nova is $E (B - V) = 0.184 \pm 0.035$ . Six spectra near the peak show the Hα line, of rest wavelength 6562.82 Å, with a P Cygni profile, an emission peak and a blueshifted absorption made in the outer part of a spherical shell moving towards us. The absorption lies at these wavelengths.

MJD	56518.986	56519.813	56520.843	56521.835	56522.829	56523.827
Wavelength (Å)	6527	6531	6534	6537	6542	6544

(a) Find $M_{0}$ from each fit, their mean, and their standard deviation as its uncertainty. [5 points]

(b) Find the extinction $A_{V}$ and its uncertainty, with $R_{V} = 3.1$ . [4 points]

(d) Find the radial velocity of the shell from each spectrum. [14 points]

(e) Assuming that the shell was negligibly small at the first spectrum, find its radius at the last, in astronomical units. [7 points]

(f) Find the angular diameter of the shell at the last spectrum, about 5 days after discovery. [4 points]

I solved it without the solution

Answer(a) $M_{0} = - 8.61 \pm 0.07$ . (b) $A_{V} = 0.57 \pm 0.11$ . (c) $d = 3.2 \pm 0.3$ kpc. (d) From $- 1636$ km/s on the first night to $- 860$ km/s on the last. (e) About 3.4 au. (f) About 2.1 milliarcseconds. The official solution takes the mean of only the first and last speeds, 1248 km/s, and gets 3.5 au and 2.2 milliarcseconds, which are as good within the accuracy of the readings.