Converting the Coordinates of Proper Motion

In addition to converting the equatorial coordinates of a star into Ecliptic coordinates or Galactic coordinates, we also want to make the same conversions of the star’s proper motion. We usually get proper motion described in standard equatorial coordinates, the rates at which the star’s Right Ascension and Declination are changing. Converting those rates into the corresponding rates in Ecliptic coordinates and Galactic coordinates necessitates a procedure different from the one by which we converted the location of the star on the sky. Again, though, we have recourse to the techniques of spherical trigonometry. As an example I will use 61 Cygni, a binary star whose impressive proper motion hints that it may properly be a member of the Milky Way’s halo.

Ecliptic Coordinates

We begin this calculation by determining the angle at which the ecliptic meridian crosses the equatorial meridian at the star’s location. We use essentially the same calculation that we would use to describe the geodesic extending between two points on a spherical surface, one of the points in our case being the ecliptic north pole. The calculation involves a spherical triangle whose sides are arcs measured from the center of the sphere; one side extends from the target star to the equatorial north pole (the co-declination, 90 -δ), one side extends from the equatorial north pole to the ecliptic north pole (the obliquity of the ecliptic, 23.4393 degrees), and one side extending from the ecliptic north pole to the target star (the ecliptic co-latitude, 90 -λ).

To calculate the angle between the meridians where they cross each other at the target star we use the sine rule. We have

(Eq’n 1)

in which

(Eq’ns 2)

With sinα
and cosα
we can treat the conversion of the projection of the proper motion vector on the
sky as an exercise in plane geometry, the size of the proper motion (very much
less than one arc-minute) making it a good approximation. The ecliptic north
pole has a right ascension of six hours (90 ), so we note that the angle alpha
measures the degree to which the ecliptic coordinate grid is rotated
counterclockwise relative to the equatorial coordinate grid. Values of alpha
greater than 180 can also be represented by a rotation of the ecliptic grid
clockwise relative to the equatorial grid. For the counterclockwise rotation we
can convert the proper motion components
∂_{t}ρ
(the change in right ascension) and
∂_{t}δ
(the change in declination) into ∂_{t}θ
(the change in ecliptic longitude) and
∂_{t}λ
(the change in ecliptic latitude). The relevant transformation equations are

(Eq’ns 3)

In those equations ∂_{t
}represents the partial time derivative, expressing the rate at which the
suffixed variable changes. Note that if alpha goes to zero, the declination and
the ecliptic latitude coincide and the right ascension and the ecliptic
longitude coincide. If alpha goes to plus 90˚,
the right ascension turns into positive ecliptic latitude and the declination
turns into negative ecliptic longitude.

In the case of 61 Cygni we calculate
α=-27.9883
, which means that the ecliptic grid appears rotated about 28 degrees clockwise
from the equatorial grid. We also have
∂_{t}ρ=4.13
arc-sec/yr and ∂_{t}δ=3.2
arc-sec/yr, so Equations 3 give us
∂_{t}θ=5.149
arc-sec/yr and ∂_{t}λ=0.8875
arc-sec/yr.

If we want to calculate the angle between the proper motion vector and a vector pointing due ecliptic north, we take the change in ecliptic latitude as our cosine, take the change in ecliptic longitude as our sine, and calculate the associated tangent. The arc-tangent gives us the angle by which we must turn a line point due ecliptic north counterclockwise to make it coincide with the proper motion vector. For 61 Cygni we calculate that angle as 80.22˚.

Galactic Coordinates

This calculation proceeds just like the one above. Again I have to resort to spherical trigonometry to plot the arcs connecting three points – the galactic north pole, the equatorial north pole, and the location of the target star – on the celestial sphere. To calculate the angle between the equatorial and galactic meridians where they cross each other at the target star’s position, we again use the sine rule;

(Eq’n 4)

in which Lp= the co-declination of the galactic north pole (62.87˚), Δρ= the difference between the right ascensions of the target star and the galactic north pole, and 90˚-L= the galactic co-latitude of the target star.

As we did with the ecliptic coordinates, with sinβ
and cosβ,
we can treat the conversion of the projection of the proper motion vector on the
sky as an exercise in plane geometry, the size of the proper motion (very much
less than one arc-minute) making it a good approximation. The galactic north
pole has a right ascension of 12 hours 51.4 minutes (192.85˚),
so we note that the angle beta measures the degree to which the galactic
coordinate grid is rotated counterclockwise relative to the equatorial
coordinate grid. Values of beta greater than 180˚
can also be represented by a rotation of the galactic grid clockwise relative to
the equatorial grid. For the counterclockwise rotation we can convert the proper
motion components ∂_{t}ρ
(the change in right ascension) and
∂_{t}δ
(the change in declination) into ∂_{t}σ
(the change in galactic longitude) and
∂_{t}ϕ
(the change in galactic latitude). The relevant transformation equations are

(Eq’ns 5)

In the case of 61 Cygni we calculate
β=+22.3678
, which means that the galactic grid appears rotated about 22-1/3 degrees
counterclockwise from the equatorial grid. We also have
∂_{t}ρ=4.13
arc-sec/yr and ∂_{t}δ=3.2
arc-sec/yr, so Equations 5 give us
∂_{t}σ=
2.60 arc-sec/yr and ∂_{t}ϕ=4.53
arc-sec/yr.

If we want to calculate the angle between the proper motion vector and a vector pointing due galactic north, we take the change in galactic latitude as our cosine, take the change in galactic longitude as our sine, and calculate the associated tangent. The arc-tangent gives us the angle by which we must turn a line point due galactic north counterclockwise to make it coincide with the proper motion vector. For 61 Cygni we calculate that angle as 29.85˚.

Calculating a Star’s Perihelion

We begin this calculation by noticing that the components
of the star’s velocity and the distances involved in the calculation form two
similar right triangles lying on the same plane. The total speed V_{T}
through space relative to Sol and the distance from Sol to the star D_{T},
the longest lines in the triangles, correspond to the triangles’ hypotenuses. On
the space triangle we have an angle gamma between the perihelion distance D_{ph}
and D_{T}. When the star goes through its perihelion, its total velocity
makes a right angle with the line D_{ph}. Relative to gamma, then, D_{ph}
represents the cosine of the right triangle, whose sine is represented by the
side D_{V} traced by the star going between its present location and its
perihelion. At present the star’s total velocity has two components, the radial
motion and the proper motion, that form the sides of the velocity triangle. In
that triangle the proper motion represents the cosine of gamma and the radial
motion represents the sine of gamma. Thus, the proper motion component V_{pm}
of the star’s speed and the perihelion distance D_{ph} between the star
and Sol correspond to each other. We can thus calculate the length of the
perihelion distance by multiplying the distance between Sol and the target star
by the ratio between the proper motion component of the star’s speed and its
total speed. We have that calculation as

(Eq’n 6)

The time elapsed between the present and the star’s perihelion passage comes from dividing the remaining side of the space triangle (obtained through the Pythagorean theorem) by the star’s total speed. The perihelion passage lies in the future if the star has a negative radial velocity and in the past if the radial velocity is positive.

For 61 Cygni the relevant figures are the radial speed (V_{r}=-63.9
km/sec), the speed of proper motion (V_{pm}=86.307 km/sec), and the
total speed (V_{T}107.42 km/sec), and the distance between 61 Cygni and
Sol (D_{T}11.36 lightyears). We then calculate D_{ph}=9.127
lightyears. The distance traveled then comes out as 6.7581 lightyears, which the
star will cross in 18,859 years.

Calculating a Binary Star’s Eclipses

Imagine representing a binary star’s orbital plane with an
elliptical disc that has a straight line lying in it (which we call the stars’
inclination axis), that line passing through the system’s barycenter. We hold
the disc before us in such a way that the inclination axis lies on the plane of
the sky. At that line the orbital plane (the disc) and the plane of the sky make
an angle equal to the stars’ orbits’ inclination. Now imagine drawing another
straight line, also lying on the plane of the sky, perpendicular to the
inclination axis. That line, of length D_{i}, extends in the direction
of the stars’ proper motion to a line parallel to the inclination axis, a line
on which the stars will have an inclination of 90˚.
We want to calculate how long it will take the stars to reach that latter line,
on which line they will exist as an eclipsing binary.

On first impression we believe that we can calculate the
time between now and the stars’ existence as an eclipsing binary by taking the
difference between 90˚
and the current inclination of the stars’ orbits and divide it by the proper
motion expressed in degrees per year. Simple geometry tells us that when the
stars move one degree across the sky, the inclination of their orbits will
change by one degree. Further, if we have an angle mu between the proper motion
vector and the line D_{i}, then we must divide that calculation of the
time by the cosine of mu. We get mu as the difference between the position angle
and the proper motion, both referred to equatorial north. That’s too simplistic,
but it gives us a good starting point for our analysis.

The analysis is complicated by the fact that the proper motion does not remain unchanged over thousands of years. For our purposes we can assume that the space velocity of the star system does not change over the time intervals that we consider (less than one million years), but our changing perspective on that velocity will change the radial component and the proper motion.

To gain a correct result easily, we must use the kilometers-per-second velocity and not the angular velocity that we normally use in representing proper motion. The star’s motion traces a straight line through space, so if we can determine the length of that line, a simple division by the star’s total speed will give us the time we want.

Calculation of the star’s total space velocity requires
that we express the proper motion in kilometers per second. We convert
arc-seconds per year into kilometers per second by converting the proper motion
into radians per second and then multiplying the result by the distance (in
kilometers) between Sol and the star. We get radians per second when we multiply
the proper motion by pi and then divide by 180, then 3600, and then the number
of seconds in a year. We get the distance in kilometers by multiplying the
distance in lightyears, D_{T}, by 299,792.458 and then by the number of
seconds in a year. Seconds per year cancel each other, so we get

(Eq’n 7)

The Pythagorean theorem then gives us the total space velocity.

Now we want to calculate the length of the line that the
star traverses between the present and the time it becomes an eclipsing binary.
Start by assuming that mu equals zero, that the proper motion lies parallel to
the projection of the stars’ orbital axis on the sky. To calculate the distance
D_{i}’=D_{V} that the star moves we use the sine rule. Between
the line D_{T} extending from Sol to the stars now and the line D_{B}
extending from Sol to the stars when they are eclipsing each other we have the
angle gamma, which equals 90˚
minus the inclination of the stars’ orbital plane. We have an angle eta between
D_{T} and the line D_{V} traced by the stars as they move from
the end of D_{T} to the end of D_{B}. We calculate eta as 90˚
plus the arc-tangent of the radial velocity divided by the proper-motion
velocity. And between the lines D_{B} and D_{V} we have an angle
omega equal to 180˚
minus the sum of eta and gamma. The sine rule tells us that D_{V}
divided by the sine of gamma equals D_{T} divided by the sine of omega.
So we calculate D_{V} as

(Eq’n 8)

Finally we divide that distance by the cosine of mu to obtain the actual distance moved and divide by the total space velocity.

Referring back to our elliptical disc, hold it so that one side tilts more or less away from us and imagine the proper motion vector emerging from the system’s barycenter. If the proper motion vector emerges on that side facing away from us, then the binary phase lies in the past. If the proper motion vector emerges from the side facing more or less toward us, the binary phase lies in the future.

The stars exist as an eclipsing binary for only a few centuries as a rule, so we can use a very simplistic calculation based on the assumption that the stars’ proper motion does not change during that time. We simply need to calculate how long it takes for the plane of the stars’ orbits to tilt by an amount that brings the stars into and then out of the ability to eclipse each other. We calculate the angle of that tilt by taking the sum of the stars’ diameters and dividing it by the sum of the mean radii of the stars’ orbits. We then get the time by dividing that angle by the system’s proper motion and the cosine of mu. It’s a crude calculation, but it gets us close enough to the fully precise answer that we can take it as a good first estimate.

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