Spectroscopic binaries and the mass function

Builds on: Sections 8.1, 8.2 and S4
Reading time: 45 min
Exercises: 3
Olympiad problems: 2

In the constellation Cygnus a blue supergiant star of ninth magnitude shows nothing unusual through a telescope. Its spectral lines, though, swing back and forth every 5.6 days, by about 75 km/s each way, so it is orbiting something. The companion gives out X-rays but no visible light, and in 1972 this system, Cygnus X-1, became the first strong case for a black hole. The argument rested on the star's velocity curve alone, which sets a smallest possible mass for the companion. In this section we see how much a velocity curve can tell us.

8.3.1Velocity curves

Before you readMake a guess first

A star moves on a circular orbit seen edge on. What shape does a graph of its velocity towards or away from us against time have?

The radial velocity of a star, its speed towards or away from us, comes from the Doppler shift of its spectral lines ((S.30)). In a binary, each star's radial velocity has a steady part $γ$ , the motion of the whole system through space, and a part that repeats with the orbit. A graph of it against time is the star's velocity curve. Half the difference between its highest and lowest values is the semi-amplitude $K$ . Only the motion along our line of sight shows, so a tilted orbit has its velocities reduced by $\sin i$ , where $i$ is the inclination, 90° for an orbit seen edge on.

ResultThe velocity curve of a binary star(8.4)

\begin{aligned} v_{r} & = γ + K [\cos (ν + ω) + e \cos ω] \\ K & = \frac{2 π a_{1} \sin i}{P \sqrt{1 - e^{2}}}, \frac{K_{1}}{K_{2}} = \frac{M_{2}}{M_{1}} \end{aligned}

$v_{r}$	the radial velocity of the star, positive when it moves away from us
$γ$	the radial velocity of the centre of mass
$K$	the semi-amplitude of the velocity curve
$ν$	the true anomaly, the angle of the star round its orbit from periastron
$ω$	the argument of periastron, the angle from the line of nodes to periastron in the orbit's plane
$e, P$	the eccentricity and the period
$a_{1}$	the semi-major axis of the star's orbit round the centre of mass
$i$	the inclination of the orbit

For a circular orbit

e = 0

, and the curve is a sine of semi-amplitude

K = 2 π a_{1} \sin i / P

. In a double-lined binary the two curves are mirror images about

γ

, with semi-amplitudes in the inverse ratio of the masses.

Figure 8.3Two stars and their velocity curves

Mass of star 2Eccentricity

e

Periastron angle

ω

Inclination

i

Watch

Semi-amplitudes
Ratio
Mass function of star 2

The figure shows a binary of a star 1 of one solar mass and a star 2 of chosen mass, with a period of 10 days. Above, the orbits are seen from above their plane round the centre of mass B, with Earth far below, so a star moving up the page moves away from us. Below are their velocity curves, star 1 heavy and star 2 thin, with

γ = 0

. Running the orbit moves both stars and the dashed line of the present moment. An eccentric orbit gives a skewed curve, with a sharp swing at periastron, and its shape changes with

ω

. The inclination shrinks both curves by

\sin i

. The figure can be set to , , , , or .

8.3.2The mass function

Kepler's third law ((8.1)) needs the size of the relative orbit, which is $a = a_{1} (M_{1} + M_{2}) / M_{2}$ . A velocity curve gives $a_{1} \sin i$ but not $\sin i$ itself. Putting $a_{1} \sin i$ from (8.4) into Kepler's law gathers everything we cannot measure on one side.

ResultThe mass function(8.5)

f (M) = \frac{M_{2}^{3} \sin^{3} i}{(M_{1} + M_{2})^{2}} = \frac{P K_{1}^{3} (1 - e^{2})^{3 / 2}}{2 π G}

$f (M)$	the mass function, found from the velocity curve of star 1 alone
$M_{1}, M_{2}$	the masses of the seen star and of its companion
$K_{1}$	the semi-amplitude of the seen star's velocity curve
$P, e, i$	the period, the eccentricity and the inclination

Since

\sin i

is at most 1 and

M_{1}

is positive, the companion's mass is always more than

f (M)

, and usually much more. If both curves are seen,

M_{1} \sin^{3} i = P (1 - e^{2})^{3 / 2} K_{2} (K_{1} + K_{2})^{2} / (2 π G)

, and the same with

K_{1}

and

K_{2}

swapped for

M_{2} \sin^{3} i

. An eclipse shows that

i

is close to 90°, and then the two masses follow.

Quick checkThe mass function of Cygnus X-1

The supergiant in Cygnus X-1 has a circular orbit with period 5.60 days and semi-amplitude 75.0 km/s. What is the mass function, in solar masses?

solar masses

Example 8.5Why the companion is a black hole

The supergiant in Cygnus X-1, from its spectrum and luminosity, has at least 20 solar masses. Using the mass function of 0.245 solar masses, find the smallest mass its companion can have.

Solution

The companion's mass is smallest when $\sin i = 1$ , and the mass function then says $M_{2}^{3} / (20 + M_{2})^{2} = 0.245$ . Trying values, $M_{2} = 5$ gives $125 / 625 = 0.200$ and $M_{2} = 6$ gives $216 / 676 = 0.320$ , and the equation is met at $M_{2} = 5.4$ .

AnswerThe companion has at least 5.4 solar masses, more than the largest possible mass of a neutron star, about 2 to 3 solar masses, and it gives out no light of its own, so it is a black hole. Later studies found the supergiant to have about 40 solar masses and the orbit to be inclined by about 27°, which make the black hole about 21 solar masses.

Example 8.6A double-lined binaryfinish it

A double-lined binary has a circular orbit with a period of 10.0 days, and its two velocity curves have semi-amplitudes $K_{1} = 50.0$ km/s and $K_{2} = 100.0$ km/s. Find $M_{1} \sin^{3} i$ .

The start

By (8.5) with $e = 0$ , $M_{1} \sin^{3} i = P K_{2} (K_{1} + K_{2})^{2} / (2 π G)$ , with $P = 8.64 \times 10^{5}$ s, $K_{2} = 1.000 \times 10^{5}$ m/s and $K_{1} + K_{2} = 1.500 \times 10^{5}$ m/s.

Your stepNow work it out in kilograms and in solar masses.

solar masses

Common errorsSection 8.3

The mass function is not the mass of either star. It is a smallest possible value for the companion's mass, reached only for $\sin i = 1$ and a seen star of no mass.
The semi-amplitude $K$ is half the difference between the largest and smallest velocities, measured from $γ$ , not the full range of the curve.
The steady velocity $γ$ of the centre of mass has to be taken off before reading the orbital motion.
A velocity curve alone never gives $\sin i$ . The masses come out multiplied by $\sin^{3} i$ unless an eclipse or a visual orbit gives the inclination.

ExercisesSection 8.33 questions

8.3.1
A star in a circular orbit with a period of 20.0 days has a velocity curve of semi-amplitude 30.0 km/s. What is $a_{1} \sin i$ , in astronomical units?
au
8.3.2
The two velocity curves of a double-lined binary have semi-amplitudes of 40 km/s and 60 km/s. What is the ratio of the mass of the star with the smaller curve to the mass of the other?
8.3.3
A velocity curve spends most of the period slowly falling and then swings up steeply over a short time. What does this tell us?

Olympiad problemsPast IOAA problems, restated2 problems

2014The mass function from a Doppler shift

Adapted from IOAA 2014 (Suceava), theory problem P4. We have restated it in our own words and kept the original data. The original calls the system a visual binary, though it is found from its spectrum, and the book gives no official solution, so the solution here is our own. The solution was last checked on 21 October 2026.

A star in a binary is observed for several months. Its spectral lines shift back and forth in a sine curve with a period of $T = 7$ days, and the largest fractional shift is $z = Δ λ / λ = 0.001$ . The orbits are circular, and the companion has mass $M_{2}$ .

(a) Show that the mass function, $f = M_{2}^{3} \sin^{3} i / (M_{1} + M_{2})^{2}$ , equals $T (v_{1} \sin i)^{3} / (2 π G)$ , where $v_{1} \sin i$ is the largest speed of the star along the line of sight.

(b) Find the mass function, using $c = 3 \times 10^{8}$ m/s and $G = 6.67 \times 10^{- 11}$ N m² kg⁻².

I solved it without the solution

Answer(a) As shown. (b) The mass function is $3.9 \times 10^{31}$ kg, about 20 solar masses, so the companion has at least 20 solar masses.

2015Neutron star or black hole?

Adapted from IOAA 2015 (Magelang), theory problem L2. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

Two stars A and B, of masses $m_{A}$ and $m_{B}$ , are a distance $d$ apart and move on circular orbits round their centre of mass, in the $x y$ plane with the centre of mass at the origin.

(a) Find the speed and the angular speed of star A. [30%]

(b) An observer far away in the $y z$ plane looks at the stars along a direction at an angle $θ$ to the $z$ axis. The velocity of A along the line of sight is $K \cos (ω t + ε)$ , with $K$ and $ε$ positive. Express $K^{3} / (ω G)$ in terms of $m_{A}$ , $m_{B}$ and $θ$ . [30%]

(c) Star A has a mass of 30 solar masses. Star B gives out X-rays and is a neutron star if its mass is less than 2 solar masses, or a black hole if it is more. The observer measures $K^{3} / (ω G) = M_{⊙} / 250$ . The angle $θ$ is not known. For which values of $θ$ is B a black hole? [40%]

I solved it without the solution

Answer(a) $v_{A} = m_{B} \sqrt{G / ((m_{A} + m_{B}) d)}$ and $ω = \sqrt{G (m_{A} + m_{B}) / d^{3}}$ . (b) $K^{3} / (ω G) = m_{B}^{3} \sin^{3} θ / (m_{A} + m_{B})^{2}$ . (c) B is a black hole if $\sin θ$ is less than 0.8, that is for $θ$ less than 53.1° or more than 126.9°.