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# YY Sagittarii: an eclipsing binary star

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.7.

The variable star YY Sagittarii is an eclipsing binary system. The V (visual) light curve is shown in Carroll & Ostlie’s Fig. 7.2, and shows a maximum magnitude of 10.03, a primary minimum of 10.78 and a secondary minimum of 10.67 (values estimated from the graph). However, there is no flat bottom on the minima, which would indicate that the eclipse isn’t total, that is, neither star totally obscures the other during the eclipsing periods. The inclination is given in the figure as ${i=88.89^{\circ}}$which is around the value we might expect for a partial eclipse (that is, it’s near the minimum inclination at which a partial eclipse just occurs).

We can still estimate the temperature ratio of the two stars from the formula

$\displaystyle \frac{T_{B}}{T_{A}}=\left[\frac{1-B_{p}/B_{0}}{1-B_{s}/B_{0}}\right]^{1/4} \ \ \ \ \ (1)$

where ${B_{0}}$, ${B_{p}}$ and ${B_{s}}$ are the observed maximum, primary and secondary brightnesses. Since the eclipses aren’t total, the observed values of ${B_{p}}$ and ${B_{s}}$ are probably larger than what they would be if the eclipses were total, so the temperature ratio won’t be that accurate.

We first convert the magnitudes into brightness ratios:

 $\displaystyle \frac{B_{p}}{B_{0}}$ $\displaystyle =$ $\displaystyle 100^{\left(10.03-10.78\right)/5}=0.50\ \ \ \ \ (2)$ $\displaystyle \frac{B_{s}}{B_{0}}$ $\displaystyle =$ $\displaystyle 100^{\left(10.03-10.67\right)/5}=0.55 \ \ \ \ \ (3)$

This gives a temperature ratio of

$\displaystyle \frac{T_{B}}{T_{A}}=\left(\frac{0.5}{0.45}\right)^{1/4}=1.03 \ \ \ \ \ (4)$

As there isn’t much difference between the two minima, the temperatures of the two stars are almost the same.

# Spectroscopic, eclipsing binary stars: mass, radius and temperature

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.6.

Here’s another example of finding the parameters of the two stars in a spectroscopic, eclipsing binary system. The measured values are

 $\displaystyle P$ $\displaystyle =$ $\displaystyle 6.31\mbox{ yr}\ \ \ \ \ (1)$ $\displaystyle v_{A,r}$ $\displaystyle =$ $\displaystyle 5.4\mbox{ km s}^{-1}\ \ \ \ \ (2)$ $\displaystyle v_{B,r}$ $\displaystyle =$ $\displaystyle 22.4\mbox{ km s}^{-1} \ \ \ \ \ (3)$

The ratio of masses is

$\displaystyle \frac{m_{A}}{m_{B}}=\frac{v_{B,r}}{v_{A,r}}=4.15 \ \ \ \ \ (4)$

The sum of masses is

$\displaystyle m_{1}+m_{2}=\frac{P}{2\pi G}\left(\frac{v_{A,r}+v_{B,r}}{\sin i}\right)^{3} \ \ \ \ \ (5)$

As the binary is eclipsing, we’ll assume that ${i\approx90^{\circ}}$. We get

 $\displaystyle m_{A}+m_{B}$ $\displaystyle =$ $\displaystyle \frac{\left(6.31\right)\left(365.25\times24\times3600\right)}{2\pi\left(6.67\times10^{-11}\right)}\left(5.4\times10^{3}+22.4\times10^{3}\right)^{3}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.02\times10^{31}\mbox{ kg} \ \ \ \ \ (7)$

We can now get the individual masses:

 $\displaystyle m_{A}$ $\displaystyle =$ $\displaystyle 4.15m_{B}\ \ \ \ \ (8)$ $\displaystyle 5.15m_{B}$ $\displaystyle =$ $\displaystyle 1.02\times10^{31}\mbox{ kg}\ \ \ \ \ (9)$ $\displaystyle m_{B}$ $\displaystyle =$ $\displaystyle 1.98\times10^{30}\mbox{ kg}\approx M_{S}\ \ \ \ \ (10)$ $\displaystyle m_{A}$ $\displaystyle =$ $\displaystyle 8.23\times10^{30}\mbox{ kg}=4.14M_{S} \ \ \ \ \ (11)$

Thus the smaller star has a mass roughly that of the Sun.

The radii of the two stars in an eclipsing system can be found by measuring the times of the various stages of the eclipse. Suppose ${t_{a}}$ is the time when the eclipse starts, that is, it’s the time when one star just starts to go behind the other one. At this point, the light curve just begins to dip. When the first star is completely behind the other one (we’re assuming ${i=90^{\circ}}$ so all eclipses are total) at time ${t_{b}}$, the light curve reaches its minimum. If we know the velocity ${v}$ of one star relative to the other, this time interval tells us the radius of the eclipsed star:

$\displaystyle r_{B}=\frac{v}{2}\left(t_{b}-t_{a}\right) \ \ \ \ \ (12)$

The radius of the other star can be found from how long it takes the eclipsed star to re-emerge. The eclipsed star first touches the edge of the other star (as seen from Earth) at ${t_{a}}$ and begins to emerge at ${t_{c}}$ so the radius of the eclipsing star (the one in front) is

 $\displaystyle r_{A}$ $\displaystyle =$ $\displaystyle \frac{v}{2}\left(t_{c}-t_{a}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{v}{2}\left(t_{c}-t_{b}\right)+r_{B} \ \ \ \ \ (14)$

For our example, the measured times are ${t_{b}-t_{a}=0.58\mbox{ day}}$ and ${t_{c}-t_{b}=0.64\mbox{ day}}$. The relative velocity is ${v=v_{A,r}+v_{B,r}}$ (since the velocities are non-relativistic), so

 $\displaystyle r_{B}$ $\displaystyle =$ $\displaystyle \frac{\left(22.4+5.4\right)\times10^{3}}{2}\left(0.58\times24\times3600\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.97\times10^{8}\mbox{ m}=R_{S}\ \ \ \ \ (16)$ $\displaystyle r_{A}$ $\displaystyle =$ $\displaystyle \frac{0.64+0.58}{0.58}r_{B}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.47\times10^{9}\mbox{ m}=2.1R_{S} \ \ \ \ \ (18)$

The ratio of the temperatures of the two stars can also be estimated by measuring how much the flux received from the binary star dips in the two minima, if we assume the stars behave as blackbodies. If the smaller star B is hotter, then when it is eclipsed by star A, the light curve drops farther than when star B goes in front of star A, since B is brighter than A, and the same area of stellar surface is eclipsed in both cases. The amount of light received from the binary system when both stars are fully visible is

$\displaystyle B_{0}=k\left(L_{A}+L_{B}\right) \ \ \ \ \ (19)$

where ${L_{j}}$ is the luminosity of star ${j}$ and ${k}$ is a constant that depends on the distance and other factors such as the amount of dust between the binary and Earth, and so on.

The brightness ${B_{p}}$ of the primary minimum, when star B is fully eclipsed, is just star A on its own:

$\displaystyle B_{p}=kL_{A} \ \ \ \ \ (20)$

The brightness ${B_{s}}$ of the secondary minimum is the brightness of star B plus that of star A minus the bit that is obscured by star B:

 $\displaystyle B_{s}$ $\displaystyle =$ $\displaystyle kL_{B}+k\frac{\pi r_{A}^{2}-\pi r_{B}^{2}}{\pi r_{A}^{2}}L_{A}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle B_{0}-\frac{r_{B}^{2}}{r_{A}^{2}}B_{p} \ \ \ \ \ (22)$

Comparing the two light minima, we get

$\displaystyle \frac{B_{0}-B_{p}}{B_{0}-B_{s}}=\frac{L_{B}}{L_{A}}\frac{r_{A}^{2}}{r_{B}^{2}} \ \ \ \ \ (23)$

The luminosity is given in terms of the temperature by the Stefan-Boltzmann law:

$\displaystyle L=4\pi r^{2}\sigma T^{4} \ \ \ \ \ (24)$

where ${\sigma=5.670373\times10^{-8}\mbox{ W m}^{-2}\mbox{K}^{-4}}$ is the Stefan-Boltzmann constant and ${T}$ is the temperature. We thus have

$\displaystyle \frac{B_{0}-B_{p}}{B_{0}-B_{s}}=\left(\frac{T_{B}}{T_{A}}\right)^{4} \ \ \ \ \ (25)$

For our example, the brightnesses are measured in magnitudes, with the maximum brightness at magnitude 5.40, the primary minimum at 9.20 and the secondary minimum at 5.44. We can convert these into the flux of light:

 $\displaystyle \frac{B_{p}}{B_{0}}$ $\displaystyle =$ $\displaystyle 100^{\left(5.40-9.20\right)/5}=0.030\ \ \ \ \ (26)$ $\displaystyle \frac{B_{s}}{B_{0}}$ $\displaystyle =$ $\displaystyle 100^{\left(5.40-5.44\right)/5}=0.964 \ \ \ \ \ (27)$

The temperature ratio is

 $\displaystyle \frac{T_{B}}{T_{A}}$ $\displaystyle =$ $\displaystyle \left[\frac{1-B_{p}/B_{0}}{1-B_{s}/B_{0}}\right]^{1/4}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.28 \ \ \ \ \ (29)$

As we’d expect, the smaller, brighter star B is significantly hotter than star A.

# Spectroscopic binary stars: Zeta Phe

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.5.

As an example of estimating the masses of the components of a spectroscopic binary star, we’ll look at ${\zeta\mbox{ Phe}}$ in the constellation Phoenix. Its observed period is ${P=1.67\mbox{ days}}$ and the maximum radial velocities observed from Doppler shifts are (we’re assuming the orbits are circular, so the velocities of the stars are constant throughout their orbits):

 $\displaystyle v_{1,r}$ $\displaystyle =$ $\displaystyle 121.4\mbox{ km s}^{-1}\ \ \ \ \ (1)$ $\displaystyle v_{2,r}$ $\displaystyle =$ $\displaystyle 247\mbox{ km s}^{-1} \ \ \ \ \ (2)$

The formula for the sum of masses is

$\displaystyle m_{1}+m_{2}=\frac{P}{2\pi G}\left(\frac{v_{1,r}+v_{2,r}}{\sin i}\right)^{3} \ \ \ \ \ (3)$

As we don’t know the inclination angle ${i}$, the best we can do is find:

 $\displaystyle \left(m_{1}+m_{2}\right)\sin^{3}i$ $\displaystyle =$ $\displaystyle \frac{P}{2\pi G}\left(v_{1,r}+v_{2,r}\right)^{3}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(1.67\right)\left(24\times3600\right)}{2\pi\left(6.67\times10^{-11}\right)}\left(1.214\times10^{5}+2.47\times10^{5}\right)^{3}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.72\times10^{31}\mbox{ kg} \ \ \ \ \ (6)$

To find the mass ratio, we need the ratio of semimajor axes (or just radii, since we’re assuming the orbits are circular), which we can get from velocities and period.

 $\displaystyle r_{i}$ $\displaystyle =$ $\displaystyle \frac{v_{i,r}}{2\pi}P\ \ \ \ \ (7)$ $\displaystyle \frac{r_{1}}{r_{2}}$ $\displaystyle =$ $\displaystyle \frac{v_{1,r}}{v_{2,r}}=0.491\ \ \ \ \ (8)$ $\displaystyle m_{2}$ $\displaystyle =$ $\displaystyle \frac{r_{1}}{r_{2}}m_{1}=0.491m_{1}\ \ \ \ \ (9)$ $\displaystyle 1.491m_{1}\sin^{3}i$ $\displaystyle =$ $\displaystyle 1.72\times10^{31}\mbox{ kg}\ \ \ \ \ (10)$ $\displaystyle m_{1}\sin^{3}i$ $\displaystyle =$ $\displaystyle 1.153\times10^{31}\mbox{ kg}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 5.8M_{S}\ \ \ \ \ (12)$ $\displaystyle m_{2}\sin^{3}i$ $\displaystyle =$ $\displaystyle 5.66\times10^{30}\mbox{ kg}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.85M_{S} \ \ \ \ \ (14)$

Using the average value of ${\left\langle \sin^{3}i\right\rangle =\frac{3\pi}{16}}$ that takes into account the Doppler shift selection effect (the fact that the larger the inclination angle, the more likely it is that a spectroscopic binary will be observed), this gives us mass estimates for the components of ${\zeta\mbox{ Phe}}$:

 $\displaystyle m_{1}$ $\displaystyle =$ $\displaystyle 9.85M_{S}\ \ \ \ \ (15)$ $\displaystyle m_{2}$ $\displaystyle =$ $\displaystyle 4.84M_{S} \ \ \ \ \ (16)$

# Sirius binary star system

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.4.

As an example of using measurements of a binary star system’s properties to determine some of its physical characteristics, we’ll look at Sirius. Sirius is a visual binary (that is, both stars are visible directly) with a period of ${P=49.94}$ years. The parallax of Sirius is ${p=0.37921^{\prime\prime}\pm0.00158^{\prime\prime}}$. If we take the inclination angle to be ${i=0^{\circ}}$, that is, the orbital plane of the Sirius system is in the plane of the sky, then the angular semimajor axis of the reduced mass is ${\alpha=7.61^{\prime\prime}}$. The ratio of the distances of Sirius A (the bright star) and Sirius B (the white dwarf companion) from the centre of mass is ${a_{A}/a_{B}=0.466}$.

From the parallax, we can get the distance:

$\displaystyle d=\frac{1}{p}=2.637\mbox{ pc}=8.597\mbox{ ly} \ \ \ \ \ (1)$

We can then find the masses of the two stars. The mass ratio is

$\displaystyle \frac{m_{B}}{m_{A}}=\frac{a_{A}}{a_{B}}=0.466 \ \ \ \ \ (2)$

From Kepler’s third law we can get the sum of the masses:

$\displaystyle m_{A}+m_{B}=\frac{4\pi^{2}}{GP^{2}}a^{3} \ \ \ \ \ (3)$

where ${a}$ is the semimajor axis of the centre of mass. Since we know the angular semimajor axis and the distance, we have

 $\displaystyle a$ $\displaystyle =$ $\displaystyle \alpha d\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 7.61^{\prime\prime}\frac{\pi}{3600\times180}\times8.597\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.172\times10^{-4}\mbox{ ly}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.00\times10^{12}\mbox{ m} \ \ \ \ \ (7)$

The period is ${P=1.576\times10^{9}\mbox{ s}}$ so the sum of the masses is

 $\displaystyle m_{A}+m_{B}$ $\displaystyle =$ $\displaystyle \frac{4\pi^{2}\left(3.00\times10^{12}\right)^{3}}{\left(6.67\times10^{-11}\right)\left(1.576\times10^{9}\right)^{2}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.434\times10^{30}\mbox{ kg}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3.235M_{S} \ \ \ \ \ (10)$

where ${M_{S}}$ is the mass of the Sun. From this and 2, we get

 $\displaystyle m_{B}$ $\displaystyle =$ $\displaystyle 0.466m_{A}\ \ \ \ \ (11)$ $\displaystyle m_{A}$ $\displaystyle =$ $\displaystyle \frac{3.235}{1.466}M_{S}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2.21M_{S}\ \ \ \ \ (13)$ $\displaystyle m_{B}$ $\displaystyle =$ $\displaystyle 1.03M_{S} \ \ \ \ \ (14)$

[These values are somewhat higher than the currently accepted values of ${m_{A}=2.02M_{S}}$ and ${m_{B}=0.978M_{S}}$. The discrepancy arises from the currently measured angular semimajor axis being smaller at ${\alpha=7.50\pm0.04^{\prime\prime}}$, the period is slightly longer at ${P=50.09\pm0.055\mbox{ yr}}$ and the angle of inclination not being ${0}$, but rather ${i=136.53\pm0.43^{\circ}}$.]

We’ve already worked out the luminosity of Sirius A from its absolute bolometric magnitude of ${M_{A}=+1.36}$ and that of the Sun (${M_{S}=+4.74}$):

$\displaystyle \frac{L_{A}}{L_{S}}=100^{\left(M_{S}-M_{A}\right)/5}=22.49 \ \ \ \ \ (15)$

For the companion ${M_{B}=+8.79}$:

$\displaystyle \frac{L_{B}}{L_{S}}=100^{\left(M_{S}-M_{B}\right)/5}=0.024 \ \ \ \ \ (16)$

Given the surface temperature of Sirius B as ${T_{B}=24,790\mbox{ K}}$ and treating it as a blackbody, we can estimate its radius ${R_{B}}$ from the formula

$\displaystyle L_{B}=4\pi R_{B}^{2}\sigma T_{B}^{4} \ \ \ \ \ (17)$

where ${\sigma=5.670373\times10^{-8}\mbox{ W m}^{-2}\mbox{K}^{-4}}$ is the Stefan-Boltzmann constant. The luminosity of the Sun is

$\displaystyle L_{S}=3.839\times10^{26}\mbox{ watts} \ \ \ \ \ (18)$

so

 $\displaystyle R_{B}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{L_{B}}{4\pi\sigma T_{B}^{4}}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{0.024L_{S}}{4\pi\sigma T_{B}^{4}}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{0.024\times3.839\times10^{26}}{4\pi\left(5.670373\times10^{-8}\right)\left(24790\right)^{4}}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 5.85\times10^{6}\mbox{ m} \ \ \ \ \ (22)$

This is slightly less than the radius of the Earth (${6.371\times10^{6}\mbox{ m}}$) and only 0.0084 times the radius of the Sun (${6.963\times10^{8}\mbox{ m})}$.

# Eclipsing binary stars

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.3.

In attempting to determine the masses of the two stars in a binary system, one of the main problems is that we don’t know the angle of inclination ${i}$ between the orbital plane of the stars and the plane of the sky. A special case where a constraint can be put on this angle is when ${i}$ is close enough to ${90^{\circ}}$that the stars either partially or totally eclipse each other.

We can find the smallest angle at which the two stars just eclipse each other. The diagram shows the situation:

The stars have radii ${r_{1}}$ and ${r_{2}}$ and their centres are separated by a distance ${a}$, with the plane of the orbit at angle ${i}$. Suppose the distance from O to the centre of star 1 is ${\frac{a}{2}-\alpha}$, so that the distance from O to the centre of star 2 is ${\frac{a}{2}+\alpha}$. Then because the two triangles are similar

$\displaystyle \frac{r_{1}}{\frac{a}{2}-\alpha}=\frac{r_{2}}{\frac{a}{2}+\alpha}=\sin\left(\frac{\pi}{2}-i\right)=\cos i \ \ \ \ \ (1)$

Solving the first equation for ${\alpha}$ we get

$\displaystyle \alpha=\frac{a\left(r_{2}-r_{1}\right)}{2\left(r_{1}+r_{2}\right)} \ \ \ \ \ (2)$

Therefore the minimum inclination angle is given by

 $\displaystyle \cos i$ $\displaystyle =$ $\displaystyle \frac{r_{2}\left(r_{1}+r_{2}\right)}{\frac{a}{2}\left(r_{1}+r_{2}\right)+\frac{a}{2}\left(r_{2}-r_{1}\right)}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{r_{1}+r_{2}}{a} \ \ \ \ \ (4)$

[Actually, I suppose this result is a bit more obvious than this derivation shows, since if we move ${r_{1}}$ over so that it’s in line with ${r_{2}}$, we get a right triangle with the side opposite the angle ${\frac{\pi}{2}-i}$ of length ${r_{1}+r_{2}}$ and hypotenuse ${a}$.]

Example Suppose the two stars have ${r_{1}=10R_{S}}$, ${r_{2}=R_{S}}$ (${R_{S}}$ is the radius of the Sun) and ${a=2\mbox{ AU}}$. Then the minimum angle at which an eclipse occurs is

 $\displaystyle \cos i$ $\displaystyle =$ $\displaystyle \frac{11R_{S}}{2\mbox{ AU}}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{11\times696300\mbox{ km}}{2\times149597871\mbox{ km}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.0256\ \ \ \ \ (7)$ $\displaystyle i$ $\displaystyle =$ $\displaystyle 88.53^{\circ} \ \ \ \ \ (8)$

If this system is an eclipsing binary, we know ${i}$ to within ${1.5^{\circ}}$ so an accurate mass determination should be possible.

# Binary stars: determining masses and the mass function

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.2.

In a binary star system, we can get estimates of the masses of the two stars given enough measurements of their motions in their orbits.

Visual binaries

First, suppose the binary is a visual binary, that is, both stars are visible directly. This will occur if their separation is large enough and both components are bright enough relative to each other. Now suppose that the orbital plane is perpendicular to the line of sight, and that we observe the binary star through at least one of its periods of revolution. We can then measure the angle ${\alpha_{i},\; i=1,2}$ subtended by the semimajor axis ${a_{i}}$ of the orbit of each star. Then, from the centre of mass condition we can get the mass ratio:

 $\displaystyle m_{1}r_{1}$ $\displaystyle =$ $\displaystyle m_{2}r_{2}\ \ \ \ \ (1)$ $\displaystyle \frac{m_{1}}{m_{2}}$ $\displaystyle =$ $\displaystyle \frac{r_{2}}{r_{1}}=\frac{a_{2}}{a_{1}}=\frac{\alpha_{2}}{\alpha_{1}} \ \ \ \ \ (2)$

Note that we don’t need to know the distance to the stars to get their mass ratio.

To find the individual masses, we need another condition, which can be found in Kepler’s third law:

$\displaystyle P^{2}=\frac{4\pi^{2}}{G\left(m_{1}+m_{2}\right)}a^{3} \ \ \ \ \ (3)$

where ${P}$ is the period and ${a}$ is the semimajor axis of the orbit of the reduced mass: ${a=a_{1}+a_{2}}$. This time we need to know the actual length of ${a}$ rather than just its ratio to something else, so we do need the distance to the stars. The distance can be obtained by some other method, such as parallax.

However, the above analysis assumed that the orbital plane was perpendicular to the line of sight, that is, that its angle of inclination ${i=0}$. In practice, of course, this is rarely the case so we need to adjust the formulas to include this effect. For the special case (yes, another one) where the semiminor axis lies in the plane perpendicular to the line of sight, but the orbital plane is rotated about the semiminor axis by the angle ${i}$, the observed angles subtended by the semimajor axes of the two stars are both shortened by a factor of ${\cos i}$ (see Carroll & Ostlie’s Fig. 7.4), so from 2, the mass ratio can still be obtained by just taking the ratio of the observed angles, since the common factor of ${\cos i}$ cancels out. However, even if we know the distance ${d}$ to the stars, we must also know ${i}$ to use Kepler’s third law to get the sum of masses. Suppose ${\tilde{\alpha}=\tilde{\alpha}_{1}+\tilde{\alpha}_{2}}$ is the observed angle subtended by the reduced mass’s semimajor axis. Then the actual semimajor axis is

$\displaystyle a=\frac{d\tilde{\alpha}}{\cos i} \ \ \ \ \ (4)$

so

$\displaystyle m_{1}+m_{2}=\frac{4\pi^{2}}{GP^{2}}\left(\frac{d\tilde{\alpha}}{\cos i}\right)^{3} \ \ \ \ \ (5)$

We can estimate this by observing that if we project an ellipse tilted by an angle ${i}$ onto the observing plane, we get another ellipse with a different eccentricity (for example, if you tilt an ellipse at just the right angle, the projection will be a circle). More importantly, the projection of the foci will not coincide with the foci of the projected ellipse. In physical terms, this means that the projection of the true centre of mass won’t coincide with a focus of the projected ellipse, and this discrepancy can be detected by careful observation. The problem then is to figure out the actual ellipse and its inclination angle that gives the observed projection. Carroll & Ostlie leave the discussion at this point so we won’t pursue it here.

Spectroscopic binaries

If the two stars are too close together to be resolved visually, they can still be detected spectroscopically. If the inclination angle is greater than zero, some component of the stars’ orbital velocity is along the line of sight, so the spectral lines of the two stars are Doppler shifted relative to each other. If the actual velocity of star ${j}$ is ${v_{j}}$, the radial component is

$\displaystyle v_{j,r}=v_{j}\sin i \ \ \ \ \ (6)$

If the orbits are circular, then the velocity of each star is constant throughout its orbit and is

$\displaystyle v_{j}=\frac{2\pi r_{j}}{P} \ \ \ \ \ (7)$

so from 2 we can get the mass ratio from the radial velocity measurements:

$\displaystyle \frac{m_{1}}{m_{2}}=\frac{r_{2}}{r_{1}}=\frac{v_{2}}{v_{1}}=\frac{v_{2,r}}{v_{1,r}} \ \ \ \ \ (8)$

As before, we need to know ${i}$ to use Kepler’s third law to get the sum of the masses. In the circular orbit case, the semimajor axis ${a}$ is just the radius of the orbit, so

$\displaystyle a=a_{1}+a_{2}=\frac{P}{2\pi}\left(v_{1}+v_{2}\right) \ \ \ \ \ (9)$

From 3, we get

 $\displaystyle m_{1}+m_{2}$ $\displaystyle =$ $\displaystyle \frac{P}{2\pi G}\left(v_{1}+v_{2}\right)^{3}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{P}{2\pi G}\left(\frac{v_{1,r}+v_{2,r}}{\sin i}\right)^{3} \ \ \ \ \ (11)$

Note that we don’t need the distance ${d}$ to the star system, but we do need to be able to measure the radial velocities of both stars. This is possible unless one component is much brighter than the other, in which case only one set of spectral lines is visible. In this case, we have

 $\displaystyle v_{2,r}$ $\displaystyle =$ $\displaystyle v_{1,r}\frac{m_{1}}{m_{2}}\ \ \ \ \ (12)$ $\displaystyle \frac{m_{2}^{3}}{\left(m_{1}+m_{2}\right)^{2}}\sin^{3}i$ $\displaystyle =$ $\displaystyle \frac{P}{2\pi G}v_{1,r}^{3} \ \ \ \ \ (13)$

The last line places all the observable quantities on the RHS, with the unknowns on the left. The quantity of the RHS is called the mass function. Since ${\sin i\le1}$ and ${\frac{m_{2}^{2}}{\left(m_{1}+m_{2}\right)^{2}}=\frac{1}{\left(1+m_{1}/m_{2}\right)^{2}}<1}$, the LHS is always less than ${m_{2}}$ so the mass function gives a lower bound on the mass of the unseen star. If this lower bound is high enough, it can give strong evidence that the unseen companion is a black hole.

In the case where both stars are visible, so that both ${v_{1,r}}$ and ${v_{2,r}}$ can be measured, we can use 8 and 13 to find the actual masses if we know ${i}$. However, ${i}$ is usually unknown, so what is often done is to use an average value ${\left\langle \sin^{3}i\right\rangle }$ so that the two equations can be solved for the masses. As we’ll see, there is a strong correlation between the luminosities and masses of stars, so if we classify stars according to their luminosities (and surface temperatures), we can then use these two equations to get an estimate of the mass of stars in that class. When luminosity is plotted agains mass on a log-log graph, the result is a straight line.

In order to calculate ${\left\langle \sin^{3}i\right\rangle }$, we need to take into account that it is more likely that we will detect a spectroscopic binary if ${i}$ is larger, since for ${i}$ near zero, the orbital plane is essentially the observation plane and the radial velocities will be very small. To get an average value of ${\left\langle \sin^{3}i\right\rangle }$, we therefore should use a weighting function that emphasizes values of ${i}$ closer to ${90^{\circ}}$. In principle, we could use any monotonically increasing weighting function ${w\left(i\right)}$ in the range ${0\le i\le\frac{\pi}{2}}$, but for the sake of argument, we can use ${w\left(i\right)=\sin i}$. This is normalized since

$\displaystyle \int_{0}^{\pi/2}w\left(i\right)di=1 \ \ \ \ \ (14)$

The average then becomes

 $\displaystyle \left\langle \sin^{3}i\right\rangle$ $\displaystyle =$ $\displaystyle \int_{0}^{\pi/2}w\left(i\right)\sin^{3}i\; di\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi/2}\left(\sin i\right)\left(\sin^{3}i\right)di\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\frac{3i}{8}-\cos i\left(\frac{3}{8}\sin i+\frac{1}{4}\sin^{3}i\right)\right]_{0}^{\pi/2}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3\pi}{16}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 0.589 \ \ \ \ \ (19)$

Another possible weighting function is just the (normalized) straight line

 $\displaystyle w\left(i\right)$ $\displaystyle =$ $\displaystyle \frac{8}{\pi^{2}}i\ \ \ \ \ (20)$ $\displaystyle \left\langle \sin^{3}i\right\rangle$ $\displaystyle =$ $\displaystyle \frac{8}{\pi^{2}}\int_{0}^{\pi/2}i\left(\sin^{3}i\right)di\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{56}{9\pi^{2}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 0.63 \ \ \ \ \ (23)$

# Binary star orbits: relation of semimajor axes

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 7, Problem 7.1.

As a significant fraction of star systems contain double stars, a lot of the effort in studying stellar properties has been concentrated on binary star systems. In a binary system, it’s easiest to analyze the orbits of the two stars in the centre of mass frame. In this frame, the two-body problem can be reduced to a one-body problem in which the reduced mass ${\mu=m_{1}m_{2}/\left(m_{1}+m_{2}\right)}$ orbits a fixed mass ${M=m_{1}+m_{2}}$ located at the centre of mass. The reduced mass’s distance ${r}$ from ${M}$ is an ellipse (for bound orbits) with semimajor axis ${a}$ and eccentricity ${e}$.

The two stars are always on opposite sides of the centre of mass at distances

 $\displaystyle d_{1}$ $\displaystyle =$ $\displaystyle \frac{m_{2}}{M}r\ \ \ \ \ (1)$ $\displaystyle d_{2}$ $\displaystyle =$ $\displaystyle \frac{m_{1}}{M}r \ \ \ \ \ (2)$

[I’m using ${d_{1}}$ and ${d_{2}}$ instead of ${r_{1}}$ and ${r_{2}}$ to avoid confusion with the position vectors ${\mathbf{r}_{1}}$ and ${\mathbf{r}_{2}}$ which give the positions of ${m_{1}}$ and ${m_{2}}$ relative to some arbitrary origin. ${d_{1,2}}$ are always measured from the centre of mass.]

Since the masses of the stars are constants (in this simple analysis, anyway) the distances ${d_{1,2}}$ are scaled down copies of ${r}$ so they too must describe ellipses with the same eccentricity ${e}$ as the ellipse followed by the reduced mass.

For a point lying on the major axis of an ellipse, the distance from the farthest focus is ${2a-p}$, where ${a}$ is the semimajor axis and ${p}$ is the perihelion distance (distance from the focus to the opposite end of the ellipse). When the two stars are aligned at opposite ends of their respective major axes (that is, when they are farthest apart, or at aphelion), then the distance of star ${i}$ from the centre of mass is

$\displaystyle d_{i,a}=2a_{i}-p_{i} \ \ \ \ \ (3)$

where ${a_{i}}$ is the semimajor axis of star ${i}$ and ${p_{i}}$ is its perihelion (closest approach) distance. Since ${d_{1}+d_{2}=r}$ at every point in the orbit, then at aphelion

 $\displaystyle r_{a}$ $\displaystyle =$ $\displaystyle d_{1,a}+d_{2,a}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left(a_{1}+a_{2}\right)-\left(p_{1}+p_{2}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2a-p \ \ \ \ \ (6)$

where ${a}$ is the semimajor axis of the reduced mass and ${p}$ is its perihelion distance (the last line follows because all the ellipses are similar (in the geometric sense) to each other).

At perihelion

 $\displaystyle d_{i,p}$ $\displaystyle =$ $\displaystyle p_{i}\ \ \ \ \ (7)$ $\displaystyle r_{p}$ $\displaystyle =$ $\displaystyle d_{1,p}+d_{2,p}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle p_{1}+p_{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle p \ \ \ \ \ (10)$

Comparing the two results, we see that

$\displaystyle a=a_{1}+a_{2} \ \ \ \ \ (11)$

That is, the semimajor axis of ${\mu}$ is the sum of the semimajor axes of the two individual stars. Because the eccentricities are the same, a similar relation holds for the semiminor axes:

$\displaystyle b=b_{1}+b_{2} \ \ \ \ \ (12)$

# Resolving two single slit light sources: graphical example

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 6, Problem 6.17.

We can get a feel for the effects of single slit diffraction by drawing a few plots of an idealized situation. Suppose we have two parallel vertical slits and we place a separate light source (to avoid any double-slit interference patterns) behind each slit. The diffraction pattern produced by slit 1 (which we’ll assume is aligned so that its central peak appears at an angle ${\theta=0}$ relative to the normal to the slit) is given by

$\displaystyle \frac{I\left(\theta\right)}{I_{0}}=\frac{\sin^{2}\left(\beta/2\right)}{\left(\beta/2\right)^{2}} \ \ \ \ \ (1)$

where

$\displaystyle \beta\equiv\frac{2\pi D}{\lambda}\sin\theta \ \ \ \ \ (2)$

and ${D}$ is the width of the slit, with ${\lambda}$ being the wavelength of the light (which we’ll take to be the same for both light sources). A plot of the diffraction pattern for slit 1 is (with ${I_{0}=1}$):

Now suppose we position slit 2 so that its central maximum falls at the second minimum on the left of slit 1, that is, at ${\beta=-4\pi}$. Its diffraction pattern is given by

$\displaystyle \frac{I\left(\theta\right)}{I_{0}}=\frac{\sin^{2}\left(\left[\beta+4\pi\right]/2\right)}{\left(\left[\beta+4\pi\right]/2\right)^{2}} \ \ \ \ \ (3)$

and looks like this:

With both lights on at the same time, the combined diffraction pattern is the sum of the two:

There’s a clear separation of the two light sources with a minimum of ${I=0}$ at ${\beta=-2\pi}$.

For circular apertures, the Rayleigh criterion requires the second light source’s central maximum to be no closer than the first minimum of the first light source. In our single slit experiment, we would place the second slit so that the central maximum occurs at ${\beta=-2\pi}$:

$\displaystyle \frac{I\left(\theta\right)}{I_{0}}=\frac{\sin^{2}\left(\left[\beta+2\pi\right]/2\right)}{\left(\left[\beta+2\pi\right]/2\right)^{2}} \ \ \ \ \ (4)$

Adding this to the first light source gives:

In this case, the two lights would be separated by only a slight dimming in intensity.

Incidentally, the fact that the two maxima both have an intensity of 1.0 occurs because we’ve chosen the separation of the slits such that the maximum of one slit falls at a minimum of the other. If we choose some other separation such as setting the second slit’s maximum at ${\beta=-3\pi}$, we get a curve like this:

The two maxima are now slightly greater than 1.0, and there are no absolute minima, where the intensity drops to zero.

# Optical interferometry: the SIM PlanetQuest project

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 6, Problem 6.15.

Interferometry can be used with optical as well as radio telescopes. The angular resolution ${\Delta\theta}$ between two neighbouring sources is given in terms of the observation angle ${\theta}$, the wavelength of the light ${\lambda}$ and the baseline distance ${d}$ between the telescopes by

$\displaystyle \Delta\theta=\frac{\lambda}{d\cos\theta} \ \ \ \ \ (1)$

For observations near the zenith ${\theta\approx0}$ and ${\cos\theta\approx1}$ so the resolution is

$\displaystyle \Delta\theta=\frac{\lambda}{d} \ \ \ \ \ (2)$

For radio telescopes, ${d}$ needs to be quite large (kilometres, typically) since radio wavelengths are in the centimetre to metre range. Optical wavelength are much shorter, so we can get good resolution with the two telescopes much closer together.

This principle is being used in the SIM PlanetQuest space telescope which is, at the time of writing (August 2015) still under construction. It is planned to have a resolution of ${4\times10^{-6\;\prime\prime}}$.

Example 1 To get a feel for how good this resolution is, suppose we could use SIM to watch grass grow from a distance of 10 km. At that distance, SIM could resolve a distance of

 $\displaystyle D$ $\displaystyle =$ $\displaystyle \left(10^{4}\mbox{ m}\right)\times\frac{\pi\left(4\times10^{-6}\right)}{180\times3600}\mbox{ rad}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.94\times10^{-7}\mbox{ m} \ \ \ \ \ (4)$

If grass grows at 2 cm per week, this is ${3.31\times10^{-8}\mbox{ m s}^{-1}}$ so SIM could detect a change in the length of the grass after around 6 seconds.

Example 2 Now suppose we use SIM for parallax measurements, with a baseline of the diameter of Earth’s orbit. In this case, the distance to the object is given in parsecs if the parallax angle is measured in arc-seconds:

$\displaystyle d=\frac{1}{p}\mbox{ pc} \ \ \ \ \ (5)$

The parallax angle is half the measured change in angular position from one end of the orbit to the other, so with SIM’s resolution, the smallest parallax angle is ${2\times10^{-6}}$, giving a maximum distance of

$\displaystyle d_{max}=5\times10^{5}\mbox{ pc} \ \ \ \ \ (6)$

Given that the Sun is around 8000 pc from the centre of the Milky Way, SIM should be able to measure parallax angles for any object in the galaxy, provided it’s bright enough to be visible to SIM’s telescopes. However, it can’t quite detect parallax for the Andromeda galaxy, as it is around ${7.8\times10^{5}\mbox{ pc}}$ away.

Example 3 Given that SIM can detect stars down to apparent magnitude 20, could it actually see the Sun from a distance ${d_{max}}$? The Sun’s absolute magnitude is ${M=+4.74}$ so its apparent magnitude at ${d_{max}}$ is

 $\displaystyle m$ $\displaystyle =$ $\displaystyle M+5\left(\log d_{max}-1\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle +28.2 \ \ \ \ \ (8)$

Sadly, the Sun wouldn’t be visible to SIM from that distance.

How about an intrinsically bright star like Betelgeuse, with ${M=-5.14}$? The maximum distance at which it could be seen by SIM is

 $\displaystyle \log d$ $\displaystyle =$ $\displaystyle \frac{m-M}{5}+1\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{20-\left(-5.14\right)}{5}+1\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.028\ \ \ \ \ (11)$ $\displaystyle d$ $\displaystyle =$ $\displaystyle 1.07\times10^{6}\mbox{ pc} \ \ \ \ \ (12)$

Thus if Betelgeuse were at a distance ${d_{max}}$, SIM could see it and measure its parallax.

Reference: Carroll, Bradley W. & Ostlie, Dale A. (2007), An Introduction to Modern Astrophysics, 2nd Edition; Pearson Education – Chapter 6, Problems 6.12 – 6.14.

The angular resolution of a single radio telescope isn’t particularly good. Using the Rayleigh criterion for resolution, a telescope of diameter ${D}$ observing at wavelength ${\lambda}$ can resolve objects separated by an angle ${\theta}$ given by

$\displaystyle \theta=1.22\frac{\lambda}{D} \ \ \ \ \ (1)$

For example, if a telescope with ${D=25\mbox{ m}}$ observes at the 21 cm line of hydrogen, the resolution is

 $\displaystyle \theta$ $\displaystyle =$ $\displaystyle 1.22\frac{0.21}{25}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.010248\mbox{ rad}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.587^{\circ} \ \ \ \ \ (4)$

The resolution can be greatly improved by using interferometry. The idea behind interferometry is similar to that of the diffraction grating. Suppose we have two telescopes separated by a baseline distance ${d}$, and both telescopes are observing an object that makes an angle ${\theta}$ with the normal to the ground. The radio waves travelling to one telescope will go a distance ${L=d\sin\theta}$ further than those to the other telescope. If ${L=\left(n-\frac{1}{2}\right)\lambda}$ the signal at one telescope is exactly out of phase with the signal at the other. Since ${\lambda}$ is known (it’s the wavelength at which we’re doing the observing), we can adjust the angle ${\theta}$ of the telescopes until the combined signal vanishes.

The angle by which the telescopes must be changed to move from one minimum (or maximum) to the next is found from

 $\displaystyle \sin\theta$ $\displaystyle =$ $\displaystyle \frac{L}{d}\ \ \ \ \ (5)$ $\displaystyle \cos\theta\;\Delta\theta$ $\displaystyle =$ $\displaystyle \frac{\Delta L}{d}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{d}\left[\left(n+1-\frac{1}{2}\right)\lambda-\left(n-\frac{1}{2}\right)\lambda\right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\lambda}{d}\ \ \ \ \ (8)$ $\displaystyle \Delta\theta$ $\displaystyle =$ $\displaystyle \frac{\lambda}{d\cos\theta} \ \ \ \ \ (9)$

If the two telescopes are separated by the diameter of the Earth, then for ${\lambda=21\mbox{ cm}}$ and observations near ${\theta=0}$ we have

 $\displaystyle \Delta\theta$ $\displaystyle =$ $\displaystyle \frac{0.21\mbox{ m}}{1.2742\times10^{7}\mbox{ m}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.65\times10^{-8}\mbox{ rad}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.0034^{\prime\prime} \ \ \ \ \ (12)$

Thus the resolution is about a million times better than for a single telescope on its own.

Several arrays of radio telescopes have been built. The Very Large Array (VLA) in New Mexico consists of 27 dishes, each 25 m in diameter, spread over a circle 27 km in diameter. Along with the increased resolution, the VLA has a total collection area of ${27\times\pi\times\left(12.5\right)^{2}=13253\mbox{ m}^{2}}$ which is equivalent to a single dish of diameter 130 m.

Another large array is situated at the Atacama Large Millimeter Array (ALMA) in Chile. It consists of 50 12 m diameter antennas with an additional 16 smaller dishes, and was completed in 2013. The 50 main dishes provide ${\binom{50}{2}=1225}$ distinct baselines for pairs of telescopes.