Most binaries are too far away, or too close together, for any telescope to split them. Their spectra still give them away: the lines of each star shift back and forth as it moves toward and away from us. This builds on the Doppler shift of light and on visual binaries.
Why it works
Both stars circle their center of mass with the same period. If the center of mass is at rest, the total momentum is zero: $M_1v_1 = M_2v_2$. So the star with the larger speed is the lighter one, and the ratio of speeds gives the mass ratio. For circular orbits seen edge-on, each star covers a circle of radius $a_i = v_iP/(2\pi)$, so the separation is $a = (v_1 + v_2)P/(2\pi)$. Kepler's third law, $M_1 + M_2 = 4\pi^2a^3/(GP^2)$, then gives $$M_1 + M_2 = \frac{(v_1 + v_2)^3P}{2\pi G}.$$
Key idea
Speeds come from Doppler shifts, $v_r = c\,\Delta\lambda/\lambda_0$. The mass ratio is $M_1/M_2 = v_2/v_1$, and for an edge-on orbit $M_1 + M_2 = (v_1 + v_2)^3P/(2\pi G)$. If the tilt $i$ is unknown, the measured speeds are $v\sin i$ and you get $(M_1 + M_2)\sin^3 i$, a lower limit.
Worked examples
Worked example: A speed from a shift
The H$\alpha$ line (656.28 nm) of one star in a binary swings back and forth by up to $0.088$ nm. What is the star's speed along our line of sight?
- $v_r = c\,\Delta\lambda/\lambda_0 = 3.00 \times 10^{5}\text{ km/s} \times 0.088/656.28$.
- $= 40$ km/s.
Answer: The radial-velocity amplitude is about 40 km/s.
Worked example: Weighing an eclipsing pair
An eclipsing binary (orbit edge-on, circular) has radial-velocity amplitudes $v_1 = 40$ km/s and $v_2 = 80$ km/s and period $P = 10.0$ days. Find the separation and both masses.
- Mass ratio: $M_1/M_2 = v_2/v_1 = 2.0$; star 1 is the heavier one.
- $P = 8.64 \times 10^{5}$ s and $v_1 + v_2 = 1.20 \times 10^{5}$ m/s, so $a = (v_1 + v_2)P/(2\pi) = 1.65 \times 10^{10}$ m $= 0.110$ AU.
- $$M_1 + M_2 = \frac{(1.20 \times 10^{5})^3(8.64 \times 10^{5})}{2\pi(6.67 \times 10^{-11})} = 3.56 \times 10^{30}\text{ kg}.$$
- That is $1.79\,M_\odot$; split 2 : 1, $M_1 = 1.19\,M_\odot$ and $M_2 = 0.60\,M_\odot$.
Answer: $a = 0.110$ AU; $M_1 \approx 1.2\,M_\odot$, $M_2 \approx 0.60\,M_\odot$.
Worked example: When the tilt is unknown
Suppose the same speeds were measured for a binary with no eclipses, and later the tilt turned out to be $i = 60^\circ$. What are the true total mass and the masses?
- The measured speeds were $v\sin i$, so the formula gave $(M_1 + M_2)\sin^3 i = 1.79\,M_\odot$.
- $\sin^3 60^\circ = 0.650$, so $M_1 + M_2 = 1.79/0.650 = 2.75\,M_\odot$.
Answer: About $2.8\,M_\odot$ in all: $1.8$ and $0.92\,M_\odot$. The ratio is unchanged.
Common mistake
Thinking the faster-moving star is the heavier one. It is the other way around: the heavier star sits closer to the center of mass and moves on a smaller, slower circle. Check with $M_1v_1 = M_2v_2$: a large speed must go with a small mass. (A planet and its star are the extreme case: the star barely moves.)