I know that the mass of a binary star system is given by Kepler's Law: $$\mathrm{m_1 + m_2 = \frac{4 \pi^2 r^3}{GT^2}}$$ Further we know that: $$\frac{r_2}{r_1} = \frac{v_2}{v_1} = \frac{m_1}{m_2}$$ Therefore if we are able to determine the period and velocity of the stars, we can then determine their mass. The period of the stars can be easily determined by the period of splitting of the spectroscopic binary's spectral lines. Also, it is possible to determine the velocity of the stars by the extent of red-shift/blue-shift of the spectral lines.

However, what if the binary stars weren't orbiting in a plane parallel to the observer, but rather on an angle? Can the velocity of the binary stars still be determined, and hence can its mass still be determined?

If this is not possible, is there any other means in which their mass can be determined?


2 Answers 2


In general, yes you need to know the orbital inclination angle $i$ in order to fully solve the orbit. The radial velocity amplitude $K$ is just modified to $K \sin i$ (where $i=0$ is a face-on orbit). Combining this with the orbital period and Keplerian orbits gives you the "mass function" $$ \frac{M_1^3 \sin^3 i}{\left(M_1 + M_2\right)^2} = \frac{K_{2}^3 \sin^3 i\ P_{orb}}{2\pi G},$$ where the right hand side can be measured from radial velocity data in a spectroscopic binary. If you have a velocity amplitude for both stars, then there is a similar expression with the labels reversed. Without $i$ this can then only tell you the mass ratio $M_1/M_2$.

There are several ways to break this degeneracy depending on what kind of binary system it is.

  1. In a visual binary system where you can observe the orbits, then the orbital path of both objects can be observed and the inclination of the orbit is directly measured. However, radial velocity amplitudes are not usually measurable (too small) and one relies on the absolute size of the orbit, which in turn requires a distance (parallax) estimate.

  2. In an eclipsing binary, then the shape and depth of the eclipses can be uniquely solved to give the inclination and hence the masses of the individual stars.

  3. In non-eclipsing close binary systems, or when one component is not seen, then ellipsoidal modulation of the seen component depends on the mass ratio and the inclination. Together with the radial velocity curve, this can then give unique masses for the components.

In general it is not possible to get any more than a mass ratio for the components of a double lined spectroscopic binary system (SB2), or the "mass function" (see above) of a single lined spectroscopic binary system (SB1).

To make further progress in these general cases you need an estimate of the primary mass. This can be done with reference to stellar evolutionary models. In principle, for an SB2, the mass ratio and the combined appearance of an object in the Hertzsprung-Russell diagram contain enough information to determine the masses of the individual components and the age of the system. In practice this is hard and there are degeneracies. A better way is to fit a combination of spectral type templates to the measured spectrum and hence estimate the spectral types and hence masses.

In an SB1 you really are stuck. The spectral type and position in the HR diagram give you $M_1$, but you will only have a lower limit to the unseen secondary mass. This is why it is difficult to estimate the masses of black holes in binaries - you need to know the inclination.

  • $\begingroup$ Thanks for your answer! So would you say that calculating mass of spectroscopic binaries would be more difficult than calculating the mass of visual binaries? $\endgroup$
    – Nanoputian
    Jun 26, 2016 at 6:27
  • $\begingroup$ @Nanoputian It is not so clear cut. A visual binary requires you to be able to clearly resolve both components; follow the orbit (perhaps over decades?) and have a precise distance to the system. This is one of the reasons that "Gaia" results will be important - distances for visual binaries. $\endgroup$
    – ProfRob
    Jun 26, 2016 at 8:45
  • $\begingroup$ So, would it be correct to say that spectroscopic binaries are more convenient to use to determine the mass of a star given that it is possible to measure all the require variables? $\endgroup$
    – Nanoputian
    Jun 26, 2016 at 11:19
  • $\begingroup$ @Nanoputian Yes. Most eclipsing binaries have short orbital periods (for obvious reasons) and the requisite measurements can be obtained quickly. It is now possible to determine masses (and radii) from such systems to of order 1%. Visual binaries have to be far enough apart (in angular terms) to resolve them, but not so far apart that they take centuries to orbit. On the other hand there are far more resolvable visual binaries than there are eclipsing short period binaries. $\endgroup$
    – ProfRob
    Jun 26, 2016 at 11:32

The apparent line-of-sight velocity (red shift / blue shift) is $v\cos\theta$ where $\theta$ is the angle between the plane of the stars' orbits and the line-of-sight line from the Earth.

  1. If the stars eclipse one another at a certain point in their orbit (eclipsing binaries) then we know that the Earth is in their orbital plane, so $\theta=0$ and the measured velocity is $v$.

  2. If the stars are visual binaries so that we can separate them telescopically, then we can measure the shape of the ellipse made by their orbit against the sky, and thus deduce $\theta$.

  3. If the stars are visual binaries and we are looking down directly onto the plane of their orbit, and we can know or guess their distance from us, then we may be able to measure $r$ directly. But there is a lot of estimation involved since distances are often a guess in themselves. Nevertheless, given that a range of masses make sense, and a range of distances make sense, sometimes "range and distance both have to make sense" can narrow down the possibilities quite nicely.

  4. Otherwise all we can measure is $v\cos\theta$. In some cases this is useful. For instance, suppose that we identify a particular class of binary stars and want to test the hypothesis "$v$ is the same for all these binaries". Then we can construct a distribution of $v\cos\theta$ for randomly chosen $\theta$, and compare that to the distribution of our measured values of $v\cos\theta$. If the distributions match then we have indeed confirmed the hypothesis and measured $v$.

Or to put it another way: if you make an assumption that implies a high $v$, higher than is ever observed, then the only way your assumption can match observation is if most binaries are face-on to us, and since there is no reason why binaries should face one way than another, that means that your assumption must be wrong.

Making deductions on a statistical basis when you can't rely on a single conclusive observation has a respectable place in astronomy. For instance, at one point the question "are quasars grouped in space?" was addressed by (a) measuring the number of quasar pairs close in the sky and (b) comparing this to the number that would be expected if quasars were randomly located. This provoked a lively discussion in the correspondence pages of Nature, because various groups of astronomers had a conflicting understanding of the relevant statistics and how they should work.

  • $\begingroup$ Line of sight velocity is v sin i Not v cos i. $\endgroup$ Oct 5, 2021 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.