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Sandbox Physics
A dark planet bends starlight

AC23 · L2

What happens when a stellar disk crosses a caustic?

Sandbox Physics ·

A caustic crossing brightens a star by creating and stretching image area. A finite stellar disk spreads that change over time. To calculate its light, we first map the sky through both masses, then add the brightness over every resulting image.

Put both masses on one angular map

Use the Einstein radius of the total lens mass M=M1+M2M=M_1+M_2 as the angular unit. Define q=M2/M1q=M_2/M_1, the normalized masses m1=1/(1+q)m_1=1/(1+q) and m2=q/(1+q)m_2=q/(1+q), and the separation ss. Put their center of mass at zero and the binary axis along the horizontal coordinate.

a1=−m2s,a2=m1s,m1+m2=1.a_1=-m_2s,\qquad a_2=m_1s,\qquad m_1+m_2=1.
The lighter companion lies farther from the center of mass. Moving it also moves the primary enough to keep that center fixed.

An image has two coordinates. Write them as the complex number z=x+iyz=x+iy, where the imaginary unit satisfies i2=−1i^2=-1. Its conjugate z‾=x−iy\overline z=x-iy reverses the vertical sign. This notation packages an ordinary pair of real coordinates; it does not introduce another physical dimension.

Write the source direction as ζ=ξ+iη\zeta=\xi+i\eta. The two gravitational deflections add, giving

ζ=z−m1z‾−a1−m2z‾−a2.\zeta=z-\frac{m_1}{\overline z-a_1}-\frac{m_2}{\overline z-a_2}.
Each term points toward a mass and falls as the inverse angular distance.

For example, 1/(z‾−a)=[(x−a)+iy]/[(x−a)2+y2]1/(\overline z-a)=[(x-a)+iy]/[(x-a)^2+y^2]. The real and imaginary parts are just the horizontal and vertical deflection components. A program can evaluate the equation using two real numbers throughout.

Reading: Bozza et al., VBMicroLensing, arXiv:2410.13660v1; A&A 694 A219 (2025)

An image pair appears at a critical direction

Now move a small image-plane square through the same equation. Usually it maps to a small source-plane patch. At special image directions the map crushes one local dimension: the signed area determinant vanishes. Those directions form critical curves in the image plane.

J(z)=1−∣m1(z−a1)2+m2(z−a2)2∣2.J(z)=1-\left|\frac{m_1}{(z-a_1)^2}+\frac{m_2}{(z-a_2)^2}\right|^2.
The complex derivative combines the deflection gradients. Its modulus determines the local area change.

Map the critical directions back through the lens equation. Their source directions form the caustics. A point source can have three or five physical images in a generic binary configuration. Crossing a fold of a caustic creates or destroys two images, one of each orientation. The total light adds ∣1/J∣|1/J| from every physical image.

The point-source value becomes singular on a caustic. That mathematical singularity tells us that a source with no size is no longer the right object. A stellar disk supplies light from a range of source directions. Its integrated image area remains finite.

Reading: Bozza et al., VBMicroLensing, arXiv:2410.13660v1; A&A 694 A219 (2025) · Bozza et al. (2018), VBBinaryLensing finite-source methods

Add the surface brightness over the images

Let the source disk have radius ρ\rho in Einstein-radius units and center ζc\zeta_c. Surface brightness at radius rr from its center is I(r)I(r) in an arbitrary but fixed radiance unit. An integral adds contributions from tiny pieces of area; d2zd^2z denotes one image-plane piece.

AFS(ζc)=∫all imagesI ⁣(∣ζ(z)−ζc∣) d2z∫source diskI(r) d2ζ.A_{\mathrm{FS}}(\zeta_c)=\frac{\displaystyle\int_{\text{all images}} I\!\left(|\zeta(z)-\zeta_c|\right)\,d^2z}{\displaystyle\int_{\text{source disk}} I(r)\,d^2\zeta}.
Trace each image-plane element back to the source. It contributes only when it lands inside the stellar disk.

For a uniform star the intensity cancels. Magnification is the sum of all image areas divided by the original disk area. This gives a direct calculation that never has to connect a newly born image branch to the wrong old branch.

A numerical image can use the same area contributions. If a calculated square overlaps several display pixels, distribute its integrated light by overlap area. Adding the linear radiance values stored in that grid recovers the photometric total, apart from the separately measured storage rounding. A contrast transformation can help the eye see faint images; its colors must be decoded before doing photometry. Refining a picture and refining a flux are parts of the same image calculation.

Reading: Bozza et al. (2018), VBBinaryLensing finite-source methods · Bozza et al., VBMicroLensing, arXiv:2410.13660v1; A&A 694 A219 (2025)

Choose a stellar edge profile with a known mean

A useful linear limb law is expressed in terms of μ=1−r2/ρ2\mu=\sqrt{1-r^2/\rho^2}. The center has μ=1\mu=1 and the edge has μ=0\mu=0. Set the disk-averaged intensity to I‾\overline I and choose 0≤Γ≤10\le\Gamma\le1.

I(r)=I‾[1−Γ+32Γ1−r2ρ2].I(r)=\overline I\left[1-\Gamma+\frac32\Gamma\sqrt{1-\frac{r^2}{\rho^2}}\right].
Gamma zero is a uniform disk. Gamma one gives a dark edge and a center 1.5 times the mean brightness.

The normalization follows by adding circular annuli. An annulus of radius rr and width drdr has area 2πr dr2\pi r\,dr. The disk average of the square-root factor is two thirds, so the average of the bracket is one.

∫0ρI(r) 2πr dr=πρ2I‾.\int_0^\rho I(r)\,2\pi r\,dr=\pi\rho^2\overline I.
Changing limb darkening redistributes a fixed unlensed total. It does not secretly brighten the entire source.

Some software uses the ordinary linear coefficient aa. Matching the same normalized profile gives a=3Γ/(2+Γ)a=3\Gamma/(2+\Gamma). Entering Gamma directly into a field expecting the other coefficient would change the star being modeled.

Reading: Bozza et al. (2018), VBBinaryLensing finite-source methods

Calculate an aligned disk before a binary crossing

A uniform star centered behind one point mass gives an independent check. Map its circular outer boundary through the two scalar roots from the preceding chapter. The bright image is an annulus with inner and outer radii

rout=ρ2+4+ρ2,rin=ρ2+4−ρ2.r_{\mathrm{out}}=\frac{\sqrt{\rho^2+4}+\rho}{2},\qquad r_{\mathrm{in}}=\frac{\sqrt{\rho^2+4}-\rho}{2}.
The two traced boundaries enclose a single ring-shaped image.
Aaligned=rout2−rin2ρ2=ρ2+4ρ.A_{\mathrm{aligned}}=\frac{r_{\mathrm{out}}^2-r_{\mathrm{in}}^2}{\rho^2}=\frac{\sqrt{\rho^2+4}}{\rho}.
Subtract the inner empty area once, then divide by the original stellar disk area.

For ρ=0.100\rho=0.100 the result is 20.025020.0250. The point-source formula at perfect alignment would be infinite. The finite ring is an area we can calculate.

For a binary worked example, fix s=1.20s=1.20, q=0.00300q=0.00300, Γ=0.538\Gamma=0.538, a vertical source track at horizontal direction 0.3666670.366667, and time (t−t0)/tE=0.0200(t-t_0)/t_E=0.0200. The independent image-area calculation gives A=3.90, 3.56, 3.17A=3.90,\ 3.56,\ 3.17 for source radii 0.0400, 0.0800, 0.1500.0400,\ 0.0800,\ 0.150. These are synthetic calculations at the same source center.

A larger star mixes a broader range of directions around this particular high-magnification feature. The full curves show how that averaging also broadens the feature. The effect depends on position: near a feature, a larger disk can also include newly bright directions. At another position its magnification may rise.

135-0.300.3
  • ρ=0.04\rho=0.04
  • ρ=0.08\rho=0.08
  • ρ=0.15\rho=0.15
Own synthetic calculation, with identical lens and trajectory. Horizontal coordinate: (t−t0)/tE(t-t_0)/t_E; vertical: source magnification. Each of the 183 states is integrated over its finite disk. Lines connect sampled states; they do not certify every intermediate time.

Reading: Bozza et al. (2018), VBBinaryLensing finite-source methods

Average over the exposure after following the motion

Let the source move along ζc(t)=eiα[(t−t0)/tE+iu0]\zeta_c(t)=e^{i\alpha}[(t-t_0)/t_E+iu_0]. The angle α\alpha rotates its track relative to the binary axis. The signed offset u0u_0 gives its closest distance to the center of mass. The crossing time tEt_E sets the clock.

⟨A⟩(t)=1Δt∫t−Δt/2t+Δt/2AFS ⁣(ζc(t′)) dt′.\langle A\rangle(t)=\frac1{\Delta t}\int_{t-\Delta t/2}^{t+\Delta t/2} A_{\mathrm{FS}}\!\left(\zeta_c(t^\prime)\right)\,dt^\prime.
A rectangular exposure averages the moving images over its actual time window. Zero exposure recovers the instantaneous value.

The image can be averaged with those same positive time weights. The result is a blurred set of image areas whose total still matches the measured-light prediction. An unrelated smoothing filter on a finished curve would lose that connection.

The interactive calculation uses a static dark binary, straight source motion and geometric optics. The source profile and exposure are explicit. The offline event exercise later adds the observer’s changing position and compares it with this simpler trajectory.

Reading: Bozza et al. (2018), VBBinaryLensing finite-source methods · Bozza et al., VBMicroLensing, arXiv:2410.13660v1; A&A 694 A219 (2025)

Exercises and explanations

  1. For a uniform aligned source with ρ=0.100\rho=0.100, why does the calculation subtract the inner boundary area instead of adding two filled disks?

    Read the explanation

    The source encloses the point lens. Its two mapped boundaries enclose one annulus. Filling both outlined disks would count the central empty region as light. Subtracting the inner area gives the actual image.

  2. A program draws a wider arc when the source grows but keeps a point-source magnification. What comparison would reveal the mistake?

    Read the explanation

    Integrate the rendered image radiance and divide by the unlensed source total. It should agree with the reported light. The finite ring example gives an analytic test; the three-radius binary example tests a changing finite source.

  3. At Γ=1\Gamma=1, what are the normalized central and edge intensities, and does their arithmetic mean equal the disk mean?

    Read the explanation

    The center is 1.5 times the disk mean and the edge is zero. Their arithmetic mean is not the area-weighted disk mean. Annuli grow in area with radius; the stated polar integral gives a mean of one.

Try the prediction

Predict how an extended star changes a crossing

  1. Pause the planetary event at a fixed time. Increase only the angular source radius and compare the image area and the width of the light-curve feature.
  2. In the advanced controls, add a short exposure. Explain which physical motion makes the image blur and why the curve averages the same light.

Sources and calculation scope

  1. Bozza et al. (2018), VBBinaryLensing finite-source methods

    Sections 2–4 and equations 14–20, 25–26 explain contour area, error refinement and radial source profiles. The pinned LGPL code remains a separately downloaded offline reference.

  2. Bozza et al., VBMicroLensing, arXiv:2410.13660v1; A&A 694 A219 (2025)

    Read equations 1–16 and section 6 before interpreting binary roots, parity, critical curves or a missing image pair. This is a methods reading, not a replication of its benchmark suite.

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