{Note: If your browser does not distinguish between "a,b" and "α, β" (the Greek letters "alpha, beta") then I am afraid you will not be able to make much sense of the equations on this page.}
Geocentric or diurnal parallax varies with the
daily spinning of the Earth around its axis.
Annual parallax is caused by the Earth's yearly orbit around the Sun.
The Earth shifts by 2a from side to side,
where a is the radius of the Earth's orbit (assumed circular) = 1 Astronomical
Unit.
For the star S1, the maximum shift occurs
as the Earth moves from position E1 to E2.
If the distance between the Sun and the star S1 is r,
then we define the annual parallax as Π, where tan(Π) = a/r.
And since a/r is always extremely small,
we may write Π = a/r (in radians).
If the star is not at S1, but at some other
arbitrary position S2,
then the shift in position as Earth moves from E1 to E2
will appear less.
Let the direction from the Sun to the star make an angle θ
with the line E1E2 .
But the star appears at angle θ' from Earth at E1.
By plane trigonometry,
sin(θ-θ')/a = sin(θ')/r
so
sin(θ-θ') = sin(θ') a/r
= sin(θ') sin(Π)
Since (θ-θ') is a very small angle,
we can replace θ' by θ, and write
θ-θ' = Π sin(θ)
The apparent shift is towards the Sun,
and it alters the star's ecliptic longitude λ
(so this is another occasion for using ecliptic coordinates).
If the star is not in the plane of the ecliptic,
there is a shift in ecliptic latitude β too.
The star X at (λ, β) is shifted to X' at (λ+Δλ, β+Δβ)
along a great circle arc towards the position of the Sun S.
XX' is the parallactic shift Π sin(θ).
We need to find the shifts Δλ and Δβ.
UX is the arc of a small circle centred on the ecliptic pole
K,
passing through the star X.
The length of arc UX is Δλ cos(β) – the shift in longitude.
The length of arc UX' is -Δβ – the shift in latitude.
Consider the tiny triangle UXX' as a plane right-angled
triangle,
and denote the angle at X by the arbitrary symbol ψ:
UX = XX' cos(ψ) = Π sin(θ) cos(ψ)
UX' = XX' sin(ψ) = Π sin(θ) sin(ψ)
In other words,
Δλ cos(β) = Π sin(θ) cos(ψ)
Δβ = -Π sin(θ) sin(ψ) (equations 1)
To eliminate θ and ψ from these two equations,
we use the spherical triangle KXS.
First, by the sine rule:
sin(90°+ψ/sin(90°) = sin(λS-λ)/sin(θ)
i.e. sin(θ) cos(ψ) = sin(λS-λ)
(expression 2)
where λS is the ecliptic longitude of the Sun.
Then, by the cosine rule:
cos(90°) = cos(θ) cos(90°-β) + sin(θ) sin(90°-β) cos(90°+ψ)
i.e. 0 = cos(θ) sin(β) - sin(θ) cos(β) sin(ψ)
So: sin(θ) sin(ψ) = cos(θ) sin(β)/cos(β)
To get rid of the cos(θ) on the right-hand side of this
expression,
we use the cosine rule again:
cos(θ) = cos(90°-β) cos(90°) + sin(90°-β) sin(90°) cos(λS-λ)
i.e. cos(θ) = cos(β) cos(λS-λ)
Substituting this expression for cos(θ) in the previous
equation:
sin(θ) sin(ψ) = cos(β) cos(λS-λ) sin(β) /cos(β)
i.e. sin(θ) sin(ψ) = cos(λS-λ) sin(β)
(expression 3)
Now we can substitute these expressions (2) and (3) in equations (1), to get:
Δλ cos(β) = Π sin(λS-λ)
Δβ = Π cos(λS-λ) sin(β)
This is actually the formula for an ellipse, of the
form:
x = a cos(θ), y = b sin(θ)
where x is the shift parallel
to the ecliptic [ Δλ cos(β) ]
y is the shift perpendicular to the ecliptic [ Δβ ]
and θ is temporary shorthand for [ 90° - (λS-λ)
].
So we have a = Π and b
= Π sin(β)
In other words, this parallactic ellipse has
semi-major axis Π, parallel to the ecliptic,
and semi-minor axis Π sin(β), perpendicular to the ecliptic.
So, during the year,
the star appears to trace out a parallactic ellipse,
which is a reflection of the Earth's orbit.
For a star on the ecliptic (β = 0°) it reduces to a straight line;
for a star at the pole of the ecliptic (β = 90°) it becomes a circle.
The size of a star's parallactic ellipse yields its
distance,
in units of parsecs (parallax-seconds):
r (in parsecs) = 1 / Π (in arc-seconds),
so a star at 1 parsec would have parallax Π = 1 arc-second.
(In fact, no star is this close.)
Exercise:
A star’s true position is
Right Ascension 6h 0m 0s, declination 0° 0' 0",
and it lies at a distance of 25 parsecs.
On the date of the Spring Equinox,
how far will it appear to be shifted by annual parallax,
and in what direction?
Click here for the answer.