Orbits and Masses of Stars Types of Binary Stars Visual Binaries Spectroscopic Binaries/Doppler Effect Eclipsing Binary Stars Main Sequence/Mass-Luminosity Relation Homework 2: Due Today Homework 3: Due Classtime, Thursday, Feb. 20 Reading for Today: Chapters Reading for Next Lecture: Chapter 7

If we measure periods in years, semi-major axis in AU, and masses in solar masses, we can simplify this to: Measurements of binary stars can be used to derive the sum of the masses of stars. If you measure the relative sizes of their orbits ( a 1 and a 2 ) or measure their velocities (v 1 and v 2 ), then remember: v 1 /v 2 = a 1 /a 2 = m 2 /m 1 Last lecture we found for binary stars a general relationship between a (a = a 1 +a 2 ), P and the star's masses:

\ Binary Stars: About half of the stars we see in the sky are binary stars or multiple star systems (our Sun is a single star). Variety of ways we can detect binary stars: Visual Binary Stars Spectroscopic Binary Stars Eclipsing Binary Stars We will start with visual binary stars, you can directly detect and follow each star of the system. An example is shown to the right.

Visual Binary Stars: Two examples: Alpha Centauri A&B and Sirius A &B. Can see both stars and can follow orbit over time. Sirius A is a main sequence star and its binary companion, Sirius B, is a white dwarf star. At visible wavelengths, Sirius B is very faint relative to Sirius A, however since the white dwarf star is much hotter, it is brighter than Sirius A at x-ray wavelengths.

We can follow the positions of Sirius A & B over time. Remember that a line between the two stars always passes through the center of mass. For Sirius A&B, there is an overall motion of the center of mass relative to the background stars (proper motion). The star closest to the center of mass is the most massive star, in this case Sirius A. Note that the orbital periods for visual binaries can be long.

To the right shows the orbits of Sirius A & B with the center of mass fixed. The center of mass should be at the focus of each star's elliptical orbits. However, the orbits look a little peculiar, why is that ?? What we observe is the apparent orbit. Note that the projection of a tilted circle is an ellipse and the projection of a tilted ellipse is still an ellipse with just a different eccentricity. The problem is that we are not seeing the orbit face on, we view it from some arbitrary angle (the orbit can be tilted in two different planes).

How do we solve for inclination to derive e, a 1, and a 2 ? A tilted circular orbits would appear elliptical, however the motions would be quite different from the case of a true elliptical orbit. With sufficient observations can find P, inclination, e, and the semi-major axes of both stars. For an elliptical orbit: r(periastron)/r(apastron) = v(apastron)/v(periastron)

The apparent orbit of the less massive star relative to the more massive star is often plotted (see Sirius A and B). The offset of the massive star from the ellipse axis allows one to compute the inclination of the orbit. The true orbit of Sirius B relative to Sirius A is shown with an arbitrary orientation after correction for tilts.

The apparent orbit for Alpha Centauri B is shown below. The more massive stars is a G2 V star (like the Sun) and its companion is a K1 V star. Proxima Centauri (M5 V) may be a third member of this system. The true orbit, after correction for tilts is also shown in the figure to the right.

After solving for the inclination of the orbit, one can fit for the eccentricity (e), the semi-major axes of both stars (a 1 and a 2 ), and the period (P). Of course we need to know the distance to the star to convert angular scales to sizes. With a = a 1 + a 2 the sum of the masses is just given by: m 1 + m 2 = 4π 2 a 3 /(GP 2 ) In addition, since we know a 1 and a 2 we also know the ratio: m 1 /m 2 = a 2 /a 1 With both the sum of the masses and ratio of the masses, we can determine the mass of each stars.

Example: Sirius A & B Brightest star in the sky (d = 2.64 pc) Sirius A: classification A1V, m v = -1.4, L = 26 L ⊙, T = 9900 K, R = 1.7 R ⊙ Sirius B: white dwarf star, m v = 8.4, L = L ⊙, T = 24,800 K, R = R ⊙ Orbital Parameters: inclination = 43.4°, e = 0.59, P = yrs, a A = 6.18 AU, a B = 13.6 AU, a = AU m A + m B = a 3 (AU)/P 2 (yrs) = (19.78) 3 /(50.05) 2 = 3.09 M ⊙ However: m A /m B = a B /a A = 2.20, therefore: m A = 2.12 M ⊙ and m B = 0.97 M ⊙

Spectroscopic Binary Stars: For most binary stars, it is often impossible to distinguish the two stars as as they are too close. However, we can still tell if they are a binary star system from their spectrum. We see that the spectrum is a blend of the spectral lines from two different spectral type stars. From this we infer the presence of two stars and can even determine the spectral type of each. More interesting, we can detect the shifts of their spectral line due to the Doppler Effect as the stars orbit their center of mass.

Doppler Effect For all types of waves, if the source of the waves is moving, then a Doppler effect is produced. In the case of sound waves, the change in wavelength (or frequency) is detected as a change in the pitch of the sound. Probably all experienced this effect if you have listened to a train whistle as the train passes by. Doppler Effect Animation

The Doppler Effect also occurs for moving light sources. Blue Shift (to higher frequencies) Red Shift (to lower frequencies) Moving toward the observer Moving away from the observer The light of a moving source is blue or red shifted by  / 0 = v/c 0 = actual wavelength emitted by the source  Wavelength change due to Doppler effect v = velocity of light source relative to observer Thus, measurements of the shift in wavelength allows us to measure the radial velocity of a light source away or toward.

The varying radial component of the velocities of the two stars produce changes in the wavelength of the absorption lines with time. Double-Line Spectroscopic Binary Stars:

Below is a movie of the changing spectra of a double-line spectroscopic binary star 57 Cygni. The feature at a wavelength of about 6565 angstroms is the Balmer α absorption line of hydrogen (n = 2 to n = 3). The splitting occurs when the star are moving in opposite radial directions. Note the other line shows similar behavior.

Below is another example of a double-line spectroscopic binary star.

We can measure the radial velocity of each star as a function of time.

The above velocity curve is for circular orbits and the velocity variations are sinusoidal. The offset velocity is the overall radial motion of the pair. The velocity amplitude for the low mass star 1 is larger than star 2. Remember that the ratio of amplitudes is the inverse ratio of masses.

Note for elliptical orbits the velocity curve is not sinusoidal. The deviations from sinusoidal allows us to determine the eccentricity of the orbit. Again the offset velocity is the overall radial motion of the pair. However, there is one big uncertainity in these orbits – what do you think this is ??

Problem: with spectroscopic binary stars cannot solve for the inclination of the orbit. Only measure the radial component of the velocity: v 1r = v 1 sin i and v 2r = v 2 sin i where i is the inclination of the orbit. If i = 0° (face-on) we would not be able to measure any radial velocity change, and if i = 90° (edge-on) we would measure the correct orbital velocity. If we assume it is edge-on, we can measure a lower limit to the velocity and thus a lower limit to the orbit size (a) and a lower limit to the sum of the masses. Although we only measure the radial component of the velocity, the following holds: v 1r /v 2r = m 2 / m 1 Thus the ratio of radial velocities gives the mass ratio.

Eclipsing Binary Stars If the binary orbit is in the plane of the sky (inclination = 90°), then the stars could eclipse each other. Distinctive and periodic variations in the apparent brightness of binary system. Each orbit will have two minima and two maxima. If the stars are not identical, then the minima are of different depth. We can directly measure the period of the orbit.

Eclipsing Binary Stars Example of an eclipsing binary star. The primary eclipse is deeper than the secondary eclipse because star 2 is cooler than star 1. Per unit area, star 2 only produces 25% of the light of star 1.

Spectroscopic Eclipsing Binary Stars If a star is both a double-line spectroscopic binary and the stars undergo eclipses we can solve for all of the stellar parameters. If eclipsing, then the inclination must be nearly 90°. We can use the previous relations for double-line spectroscopic binaries to find the masses of the stars. Since we have eclipse timing and we know the velocities, we can solve for the sizes of each of the stars. Such systems provide the most fundamental data on the properties of stars.

Return to the Main Sequence Once we solve for the masses of stars on the main sequence, we find that the stars are ordered by mass. The lowest mass stars are the M-type stars with masses of 0.1 M ⊙ and highest mass stars are O-type stars with masses of about 60 M ⊙. For stars in this stage of their evolution, if you specify the mass, there is only one configuration (surface temp., luminosity and size) that makes them stable.

Return to the Main Sequence Not surprising there is a relation between mass and luminosity for stars on the main sequence. This is power-law relation – the luminosity is proportional to mass to some power. Over all mass, the power is approximately 3.5. Therefore: L/L ⊙ ~ (M/M ⊙ ) 3.5

Number of Stars by Mass Once we have calibrated the mass of stars, we can then try to ask the question how are stars distributed by mass (stellar mass function) ? Above is the initial stellar mass function - the number of stars formed as a function of mass in a given volume of space. Find many more low mass. For masses > 0.5 M ⊙, the number (N) of stars as a function of mass (M): N(M) ∝ M Thus, there are 200 times more 1 M ⊙ stars than 10 M ⊙.