If we happen to see a binary's orbit almost edge-on, the stars take turns passing in front of each other, and the total light dips twice each orbit. Algol in Perseus does this: every 2.87 days it fades by more than a magnitude over several hours, when its larger, cooler companion covers the hot main star. Combined with the speeds from spectroscopic binaries, eclipses measure the sizes of stars directly.
Why it works
Picture a small star of radius $R_s$ crossing behind a large star of radius $R_\ell$ at relative speed $v$. Label four moments: first contact $t_a$; the small star just fully hidden $t_b$; it starts to reappear $t_c$; last contact $t_d$. Between $t_a$ and $t_b$ the small star slides its own diameter, so $2R_s = v(t_b - t_a)$. Between $t_a$ and $t_c$ its leading edge crosses the whole large star, so $2R_\ell = v(t_c - t_a)$. For the depths, use the fact that each square meter of a star's surface shines in proportion to $T^4$. In both eclipses (when one disk lies completely across the other) the hidden area is the same, $\pi R_s^2$. Only the surface that gets hidden differs.
Key idea
$R_s = \tfrac12 v(t_b - t_a)$ and $R_\ell = \tfrac12 v(t_c - t_a)$. The deeper, primary eclipse hides the hotter star's surface, and $$\frac{\text{primary depth}}{\text{secondary depth}} = \left(\frac{T_{\text{hot}}}{T_{\text{cool}}}\right)^{4}.$$
Worked examples
Worked example: Sizes from timing
In an edge-on binary the relative speed is 300 km/s. The small star takes 1.5 h from first contact to full eclipse, and 9.0 h from first contact until it begins to reappear. Find both radii.
- $R_s = \tfrac12(300\text{ km/s})(5400\text{ s}) = 8.1 \times 10^{5}$ km $= 1.16\,R_\odot$.
- $R_\ell = \tfrac12(300\text{ km/s})(32{,}400\text{ s}) = 4.9 \times 10^{6}$ km $= 7.0\,R_\odot$.
Answer: About $1.2\,R_\odot$ and $7.0\,R_\odot$.
Worked example: Temperatures from depths
A binary's primary eclipse removes 50% of the light and its secondary removes 8%. Both are total or annular. What is $T_{\text{hot}}/T_{\text{cool}}$?
- Depth ratio $= 0.50/0.08 = 6.25$.
- $T_{\text{hot}}/T_{\text{cool}} = 6.25^{1/4} = 1.58$.
Answer: The hotter star is about 1.6 times hotter.
Worked example: Which eclipse is deeper?
A small B star (20,000 K) orbits a large K giant (4,500 K). Which eclipse is the primary one?
- Either way the hidden area is the B star's disk.
- When the B star goes behind the giant, we lose a disk of 20,000 K surface. When it passes in front, we lose a disk of 4,500 K surface: $(20{,}000/4500)^4 \approx 390$ times less light per square meter.
Answer: The primary (deep) eclipse is when the small hot star is hidden behind the giant.
Common mistake
Thinking the deeper eclipse happens when the bigger star is hidden. The bigger star is never fully hidden; in both eclipses the same area, the small star's disk, is covered. The depth depends on the surface brightness ($\propto T^4$) of whichever surface is covered, so the primary eclipse always hides the hotter star. Check: if both stars had the same temperature, the two eclipses would be equally deep.
Real stars are darker near their edges (limb darkening), which rounds the bottoms of the dips. Fitting that shape refines the radii.