OWAO 2024, Теория — Задача 1. Crossroad of Worlds

• 59000 yr

  The official solution gives two moments: after about 59 thousand years and after about 148 thousand years (Method 2: 0.055 and 0.141 million years); either one earns full marks, acceptable range ±5 % each.

The stars are in the plane of the celestial equator. Since there are no components of proper motion in declination, the movement of stars also occurs in the plane of the celestial equator, $\mu = |\mu_\alpha|$.

Let us determine the distances to each of the stars and their tangential velocities:

We also calculate the overall velocity of each star relative to the Sun:

Currently, the angular distance between the stars is

We need to determine when this distance will decrease to $0.5\theta_0 = 12^{\circ} \times 0.5 = 6.0^{\circ}$.

Since the radial velocities of the stars are negative, the stars are currently moving towards the Sun. The signs of the components of their proper motion in right ascension differ, so the stars really will come “closer” in the Earth’s sky.

Method 1. Let us introduce a coordinate system centered on the Sun as shown in Figure 1. After a time $t$ from the current moment, the coordinates of the star $A$ will be

The angle between the full velocity vector of the star $B$ and its tangential velocity is $\arctan \frac{|V_{\tau,B}|}{|V_{r,B}|}$, therefore the angle of the velocity vector of the star $B$ to the $X$-axis is

After a time $t$ from the current moment, the coordinates of the star $B$ will be

The angle $\theta(t)$ between the directions to the stars is calculated as the angle between the vectors $\mathbf{SA}(t)$ and $\mathbf{SB}(t)$:

We will measure distances in parsecs, time in thousands of years and velocities in parsec per thousand years. Next, we substitute the numerical data into the expressions for the coordinates:

So our goal is to find the roots of the function

The equation is similar to the 4th degree equation. We will solve it numerically.

For geometric reasons, the equation must have exactly 2 roots. At some point, the stars will be on the same line of sight for the observer. Therefore, an angular distance of 6 degrees will occur both before and after such a moment.

First, let us define the ranges of the argument $t$ at which the function $f(t)$ changes its sign. To do this, we calculate the values of the function on some grid:

We can find roots on the intervals $[40 : 60]$ and $[140 : 160]$ using the bisection method:

As a result, the angular distance of $6^{\circ}$ is reached after 59 thousand years and 148 thousand years from the current moment. Due to emerging rounding errors, the answers may differ from those indicated above by several thousand years.

Method 2. The geometrical framework here is the same as used in Method 1, but the calculations are much simpler.

Firstly, we adopt the local coordinate system for each star $i \in A, B$ at the current moment. We denote the first coordinate $x_i$ being along the line of sight at the initial moment of time $t = 0$. Second coordinate in the equatorial plane $y_i$ is measured along the perpendicular to the $x_A$ direction and is positive in the direction counterclockwise from polar axis.

Then the initial coordinates for each star $i$ are:

Velocities measured along those axes are the radial and the tangential one respectively.

After a time $t$ passes the coordinates should be

Calculating all the velocities and distances we get the coordinates (in parsec) expressed as function of time $t$ (in million years):

Coordinate frame of star $A$ is rotated relatively the coordinate frame of star $B$ through the angle $\theta_0$ counterclockwise. To transform the coordinates into the same coordinate system we perform the rotation of coordinates $(x_B, y_B) \to (x'_B, y'_B)$:

After that we should express the angle $\theta$ between $A$ and $B$ as a function of time. Using the given cartesian coordinates we can write a dot product of the position vectors:

The answers can be found using bisection method. Note that we have 2 answers, any of them will give participant a full score, since the angle of $6^{\circ}$ between star can occur before and after the alignment of such stars on the sky.

Such answers are $t_1 = 0.055$, $t_2 = 0.141$ in millions of years.

The results are slightly different because the calculations in this problem are sensitive to rounding.

Method 3. Using the same systems of coordinates as were in Method 2, one can say that the angle star travelled on the sky up to time $t$ from initial moment corresponding to parameters given in the table is

From the signs of proper motions and radial velocities it is obvious that the stars are moving towards each other. Then the equation to solve for the first moment of time is

For the second moment of time

Marking Scheme:

1. Proper geometric model, including positions and velocities of the stars — 4 pt.
2. Determining the tangential velocity of each star — 2 pt. + 2 pt.
3. Formulation of the equation(s) for the geometric model — 4 pt. Points may be awarded even if the model is incorrect.
4. Solution of the equation(s) — 6 pt. The numerical method used is not described — $-2$ pt.
   0 points if the model is incomplete (it is impossible to obtain a meaningful answer).
5. Numerical answers — 1 pt. + 1 pt. The acceptable range is $\pm 5$ % each.

Note. A participant could also say that radial velocities are too small to make a difference in answers and simply transform the last equation for the first moment of time to $\frac{1}{2}\theta_0 = (|\mu_{\alpha,A}| + |\mu_{\alpha,B}|)\,t$. The answer in this approximation is $t \approx 6.3 \times 10^4$ years. It is out of the 5% boundary specified in the marking scheme. If the participant proves that the radial velocities are small, then points are awarded for the understanding of limitations of the model. Otherwise, no more than 6 points are awarded for simply using the proper motions.