Astronomy · Stars with companions and shared origins
Eclipsing systems and combined light
Interpret two luminous stars’ eclipses by adding both components’ light and subtracting only the hidden contribution, then test dilution and geometry assumptions.
Continue in the approved chapter order. Necessary vocabulary, quantities and assumptions are explained in the lesson.
Back to the chapterAfter reading, you can
- Explain why an eclipse changes a system’s combined received flux.
- Calculate two conditional eclipse depths when both components emit light.
- Identify information needed to infer radii and inclination from a light curve.
Both stars can contribute to the measured flux
An eclipsing binary is oriented so that one component passes in front of the other along the observer’s line of sight during part of its orbit. The measured light curve records the system’s received flux as a function of time. It need not resolve the two stars as separate images.
The flux outside eclipse includes both stars. During an eclipse, the foreground component can hide part or all of the background component’s projected luminous area while still contributing its own light. The fractional loss therefore depends on surface brightness as well as geometry.
Surface brightness in this lesson means the received flux contribution per unit projected disk area in one specified observing band at a common observer distance. An actual stellar disk need not have uniform brightness: limb darkening, spots and other effects can change the distribution.
A light-curve dip is evidence to be interpreted, not a radius measurement by itself. The orbit, the contributions of both stars, instrumental calibration and possible additional light enter a model that connects the observed flux to physical sizes.
Specify radii, surface brightness and the reference flux
Consider an original model of two opaque circular luminous disks with radii R_A = R₀ and R_B = 0.5R₀. Let their uniform received surface-brightness contributions be S_A = 4S₀ and S_B = S₀ in the selected band. R₀ and S₀ are arbitrary positive reference quantities.
Define U₀ = S₀πR₀² as a received-flux unit. Then F_A = 4U₀ and F_B = 0.25U₀. With no additional light, the out-of-eclipse reference is F_out = 4.25U₀ = 17U₀/4.
Assume the distance difference between the components is negligible for their received-flux scale. Ignore limb darkening, distortion, spots, reflection, beaming, lensing and time-dependent intrinsic emission. This is a supplied projected-geometry comparison, not a fitted binary or a complete orbit.
Define eclipse depth as δ = (F_out − F_eclipse)/F_out in this band with this out-of-eclipse reference. Both the numerator and denominator must use the same observing configuration. A different band can change the two stars’ brightness ratio and hence the depth.
| Component | Radius | Uniform surface brightness | Out-of-eclipse flux |
|---|---|---|---|
| A | R₀ | 4S₀ | 4U₀ |
| B | 0.5R₀ | S₀ | 0.25U₀ |
Subtract the background component’s hidden contribution
At exact central alignment, the smaller disk’s area π(0.5R₀)² is one quarter of A’s disk area. When B is in front of A, the removed light is A’s surface brightness times that overlap area: ΔF = 4S₀ × πR₀²/4 = U₀.
B’s own light remains included. The resulting system flux is 3.25U₀ and the depth is 1/4.25 = 4/17, about 0.235294. The smaller disk’s area fraction 1/4 is not itself the normalized system depth because the reference also contains B’s light.
At the opposite central alignment, A fully hides B. The removed contribution is all of B’s 0.25U₀, giving remaining flux 4U₀ and depth 0.25/4.25 = 1/17, about 0.058824.
The same two radii therefore give different eclipse depths depending on which luminous surface is hidden. Neither eclipse requires a change in either star’s intrinsic luminosity in this teaching model. The difference follows from the received brightness contributions and overlap geometry.
| Foreground component | Hidden background flux | Remaining system flux | Depth relative to 17U₀/4 |
|---|---|---|---|
| B in front of A | 1U₀ | 13U₀/4 | 4/17 |
| A in front of B | U₀/4 | 4U₀ | 1/17 |
The denominator includes unresolved light too
Now add a constant uneclipsed flux F₃ = U₀ in the observing aperture, keeping every disk property and eclipse geometry fixed. This could represent a separately modelled third light contribution; the example does not identify a real companion or contaminating source.
The out-of-eclipse total becomes 5.25U₀ = 21U₀/4. The hidden contributions remain U₀ and U₀/4 in the two central alignments because the additional light is stipulated to remain visible.
The corresponding depths are 4/21 and 1/21, both smaller than before. This is dilution: the same absolute flux loss is divided by a larger total reference. A shallower dip therefore does not necessarily mean a smaller occulting radius.
A real analysis must constrain the extra contribution and its variability in the same band and aperture. It should also distinguish reduced depth from a change in the eclipse geometry, surface-brightness ratio or intrinsic stellar behaviour. Several explanations can otherwise fit a similar flux curve.
| Central alignment | Depth without extra light | Depth with constant U₀ |
|---|---|---|
| B in front of A | 4/17 ≈ 0.235294 | 4/21 ≈ 0.190476 |
| A in front of B | 1/17 ≈ 0.058824 | 1/21 ≈ 0.047619 |
Both stars shine during an eclipse
Scroll sideways to see the full diagram. Its explanation is below.
Durations and shapes provide additional constraints
Eclipse duration depends on how quickly the projected separation changes and how the two disk boundaries cross. The shape of ingress and egress carries geometric information, but it is also affected by surface-brightness structure and the sampling of the observations.
For a known orbital model, eclipse fitting can constrain radii relative to an orbital scale and the inclination. Spectroscopic component velocities can provide complementary dynamical information. A physically dimensioned radius requires the appropriate length or velocity–time scale, not just a fractional flux dip.
Eccentricity changes the separation and projected speed at eclipse, so equal time intervals or identical durations should not be assumed universally. Finite stellar sizes also permit eclipses over a range of near-edge-on orientations; the observation does not force an exactly edge-on orbit.
A convincing system model explains both the out-of-eclipse light and the eclipse details, with compatible spectra, geometry and uncertainties. When light or mass moves between companions, their histories require the additional interacting-binary physics developed in the next lesson.
Worked examples
Keep foreground light in the reference
Why is the depth with B centrally in front of A equal to 4/17 rather than 1/4?
- The obscured fraction of A’s area is 1/4, so the hidden A flux is 1U₀.
- The outside-eclipse total includes A and B: 4U₀ +0.25U₀ =4.25U₀.
- Divide the hidden flux by the total: 1/4.25 =4/17.
The depth is 4/17 because the normalized system flux includes the smaller star’s own emission.
Change only the supplied extra flux
What is the central depth when A hides B and constant extra light U₀ is present?
- The hidden B contribution remains 0.25U₀.
- The reference total becomes 4 +0.25 +1 =5.25U₀.
- Calculate 0.25/5.25 =1/21.
The depth is 1/21, about 0.047619, under the stipulated constant-extra-light model.
Try the reasoning
When both components emit light, is normalized eclipse depth always just the blocked area fraction of the larger disk?
Show the answer and reasoning
No.
The background surface brightness and the light of all visible components enter the system flux ratio.
If a fixed hidden flux is divided by a larger total reference flux, does the fractional depth become larger or smaller?
Show the answer and reasoning
Smaller.
The added uneclipsed light enlarges the denominator while the stipulated loss stays fixed.
Must the reference and eclipse flux use a consistent band and observing configuration?
Show the answer and reasoning
Yes.
Band-dependent brightness ratios and aperture contributions affect the normalized depth.
Does a flux dip alone provide a unique physical radius in metres?
Show the answer and reasoning
No.
A radius inference also needs a geometry and brightness model plus a compatible physical scale.
What to carry forward
- An eclipse subtracts hidden background light from the combined system flux.
- Luminous companions and additional light change the interpretation of a normalized depth.
- Eclipse shapes and timing become physical constraints through a consistent orbital and surface-brightness model.
References
Teaching resources and research supporting this lesson’s explanations and examples.