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Sandbox Physics
Measuring the depth of the sky

L2 · Stellar astrometry

Derive a parallax ellipse from Earth’s orbit

Sandbox Physics ·

A parallax ellipse is the observer’s orbit projected onto the sky’s tangent plane. Define the directions first, then project Earth’s position onto two sky axes. The ellipse follows directly.

Define the geometry before measuring an angle

Place the Sun at the origin and Earth’s orbit in the horizontal ecliptic plane. Give the star ecliptic longitude zero and latitude β\beta. Its unit direction vector is n=(cos⁡β,0,sin⁡β)\boldsymbol n=(\cos\beta,0,\sin\beta). A unit vector records direction with length one.

Let rr be distance in parsecs. Earth follows a circle of radius one astronomical unit, completing it in one year. At time tt years, its position in astronomical units is R=(cos⁡2πt,sin⁡2πt,0)\boldsymbol R=(\cos 2\pi t,\sin 2\pi t,0). Time zero places Earth on the positive first axis.

To measure a tiny change in sky direction, imagine a small sheet of paper perpendicular to the original sightline. This is the tangent plane. Its east axis points towards increasing ecliptic longitude; its north axis points towards increasing ecliptic latitude.

ex=(0,1,0),ey=(−sin⁡β,0,cos⁡β)\boldsymbol e_x=(0,1,0),\qquad\boldsymbol e_y=(-\sin\beta,0,\cos\beta)
These two unit axes are perpendicular to each other and to the star’s direction. Within a sufficiently small patch of sky, angular displacements add along these axes.

Move the observer and project the new sightline

Write the star’s position as DnD\boldsymbol n, now using astronomical units for DD. From the displaced observer, the sightline before normalization is Dn−RD\boldsymbol n-\boldsymbol R. Since the distance is vastly larger than the orbit, retain only first order in their ratio.

Motion along the sightline mainly changes distance. The perpendicular part changes direction. The direction change is therefore −[R−(R⋅n)n]/D-[\boldsymbol R-(\boldsymbol R\cdot\boldsymbol n)\boldsymbol n]/D. A dot product multiplies matching components and adds them; with a unit direction, it gives the component along that direction.

Project this change onto east and north. Both axes are perpendicular to n\boldsymbol n, so the along-sightline term drops out. What remains is the negative projection of the observer’s position. The angle subtended by one astronomical unit is the parallax.

Δx=−ϖsin⁡(2πt),Δy=ϖsin⁡βcos⁡(2πt)\Delta x=-\varpi\sin(2\pi t),\qquad\Delta y=\varpi\sin\beta\cos(2\pi t)
Use the same angular unit for both coordinates and parallax. The minus sign expresses the apparent shift opposite to the observer’s sideways motion.

Reading: Lindegren et al.

A line and a circle are limiting ellipses

The eastward semiamplitude is ϖ\varpi; the northward semiamplitude is ϖ∣sin⁡β∣\varpi|\sin\beta|. At zero latitude, northward motion vanishes. At an ecliptic pole, both amplitudes agree. At intermediate latitudes, the path is an ellipse. Switching from positive to negative latitude reverses the sense of motion.

Take a star twenty parsecs away at latitude thirty degrees. Its parallax is fifty milliarcseconds and its northward semiamplitude is twenty-five. Select observations six months apart at the extremes of the eastward motion: their separation is one hundred milliarcseconds. The star’s actual distance has stayed fixed; the viewing direction has changed.

ϖ [mas]=1000r [pc],a=50 mas,b=50sin⁡30∘=25 mas\varpi\,[\mathrm{mas}]=\frac{1000}{r\,[\mathrm{pc}]},\qquad a=50\,\mathrm{mas},\quad b=50\sin30^\circ=25\,\mathrm{mas}
The parsec uses the one-astronomical-unit parallax geometry. Stellar angles are so small that replacing the tangent of the angle by the angle in radians is extremely accurate.

Add the star’s continuing drift

Let x0x_0 and y0y_0 be tangent positions at the reference epoch. Proper motions μx\mu_x and μy\mu_y are angular displacements accumulated per year. Treating them as constant over the observation span gives a starting position, a linear drift and the annual parallax.

For example, eastward motion of eighteen milliarcseconds per year and northward motion of 7.2 move the ellipse’s centre by thirty-six and 14.4 milliarcseconds after two years. The annual oscillation persists, but the full track no longer closes.

Once distance is known, angular proper motion also gives transverse speed. Convert angular speed to radians per year and multiply by distance. One astronomical unit per year is about 4.74047 kilometres per second, giving the familiar conversion below.

x(t)=x0+μxt+ϖPx(t),y(t)=y0+μyt+ϖPy(t)x(t)=x_0+\mu_xt+\varpi P_x(t),\quad y(t)=y_0+\mu_yt+\varpi P_y(t)
The two sinusoidal projections are abbreviated as parallax factors. The model has two starting positions, two proper motions and one parallax.
v⊥[km s−1]=4.74047 r[pc] μ[arcsec yr−1]v_\perp[\mathrm{km\,s^{-1}}]=4.74047\,r[\mathrm{pc}]\,\mu[\mathrm{arcsec\,yr^{-1}}]
Total proper motion is the square root of the sum of the squared tangent components. Twenty parsecs and one arcsecond per year give about 94.8 kilometres per second.

When the geometric model needs more physics

A circular orbit makes the projection easy to see. Real astrometry uses precise ephemerides, and a spacecraft is not exactly at Earth’s centre. Precision modelling also includes aberration, gravitational light deflection, reference frames and instrument calibration.

Constant proper motion is a finite-span approximation. A nearby star with substantial radial speed can show perspective acceleration. An unresolved binary can add motion of its light-weighted centre. The next chapter fits the single-star model; the final paper guide returns to these more demanding cases.

Reading: Lindegren et al. · JPL Horizons system manual

Try the geometry

Predict the projection before changing the direction

  1. Set distance to twenty parsecs and proper motion to zero. Predict how the two semiaxes change between zero and ninety degrees latitude.
  2. Compare the ecliptic and polar presets, then choose negative latitude and inspect the sense of motion.
  3. Keep the distance fixed and set eastward motion to eighteen. Explain the centre’s displacement after two years.

Exercises and explained answers

A star’s maximum annual excursion is forty milliarcseconds. What is its distance?

The full excursion is twice the parallax. The parallax is twenty milliarcseconds, giving one thousand divided by twenty: fifty parsecs.

Find the semiaxes at latitude minus thirty degrees and parallax forty milliarcseconds.

The eastward semiaxis is forty milliarcseconds and the northward length is twenty. Negative latitude changes the sign of the northward phase, while semiaxis lengths remain positive.

Double distance while keeping angular proper motion fixed. What happens to transverse speed?

It doubles, because transverse speed is distance times angular speed. This differs from moving the same constant-speed star farther away, which halves its angular proper motion.

Sources and calculation scope

  1. Lindegren et al., Gaia EDR3 — The astrometric solution, arXiv:2012.03380v1

    Sections 3.1, 5.3–5.7; equation 23 and figures 10–12. The fixed single-source model and covariance conventions apply to the basic DR3 astrometry inherited from EDR3.

  2. JPL Horizons system manual

    Reference-frame and observer-vector conventions for a future ephemeris-based extension. The browser uses a declared circular orbit, not a Horizons ephemeris.

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