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Calculating Stellar Positions (from other points in space)

Coordinate translation

Now you're ready to calculate, for example, what the night sky from Sirius or Epsilon Eridani looks like. To do this, you'll need to know two things about the stars you'll be considering:

  1. Their positions
  2. Their brightnesses

Let's tackle the positions first.

Positions

Step 1: Get Cartesian coordinates for the stars

First, I'll introduce a little terminology that should help keep some ideas straight:

  1. Reference star: The star that you'll be "traveling to" and calculating the positions of other stars from. Its Cartesian coordinates are xR, yR and zR.
  2. Target Star: Any star you're looking at from the reference star. Its Cartesian coordinates are xT, yT, and zT.
  3. Transformed Target Coordinates: The coordinates of the target star as seen from the reference rather than from Earth. They are x'T, y'T, and z'T.

Suppose you're interested in figuring out what the night sky looks like from Barnard's Star, the second closest star to the Sun. The reference coordinates are therefore those of Barnard's Star, which has \alpha = 17.9636 hours, \delta = 4.668 degrees, and d = 1.82 pc:

  1. xR = -0.017 pc
  2. yR = -1.815 pc
  3. zR = +0.148 pc

Let's figure out where Proxima Centauri would be. Its coordinates (from the Calculating Stellar Positions page) are:

  1. xT = -0.472 pc
  2. yT = -0.361 pc
  3. zt = -1.151 pc

Step 2: Figure out the transformed target coordinates

This is easier than it might seem! To find out the transformed target coordinates, as seen from the reference star, just subtract the reference's coordinates from the target's coordinates.

For Proxima Centauri:

  1. x'T = -0.472 - (-0.017) = -0.455 pc
  2. y'T = -0.361 - (-1.815) = +1.454 pc
  3. z't = -1.151 - (0.148) = -1.299 pc

Step 3: If desired, convert the transformed Cartesian
coordinates to spherical coordinates

The transformed target coordinates may be all you need if you have a computer program that'll plot the stars. If you want to draw star charts by hand, especially if you want to use common map projections, it will be useful to convert the Cartesian coordinates to something else.

The most useful transformation is to something called general spherical coordinates, which are very similar to equatorial coordinates. In fact, they're essentially identical -- the main reason for using them instead of equatorial coordinates is to avoid confusion between Earth-based measurements (which use the equatorial system) and calculations like these which apply far from
Earth. The spherical coordinates are:

  1. Azimuth \theta
  2. Altitude \phi
  3. Distance r

\theta is like right ascension: it measures the angle around the Cartesian coordinate system, starting at the x-axis, as seen from the reference star. \phi is like declination: it measures the angle "up" or "down". r is the target star's distance from the reference.

To transform Cartesian coordinates to spherical coordinates:

  1. Calculate two distances:
    1. r = sqrt(x'T2 + y'T2 + z'T2)
    2. rxy = sqrt ( x'T2 +
      y'T2)
  2. Apply two formulas:
    1. \phi = tan-1 (z'T / rxy);
    2. \theta = tan-1 (y'T / x'T)

This gives two angles similar to right ascension and declination (or, for that matter, longitude and latitude), which is what many map projections will call for. If you're using a computer rather than a hand calculator, the inverse tangent function will probably return a value in radians rather than degrees -- you may want to convert. You will use the distance r to get the star's brightness later.

Make sure the angle \theta is in the correct quadrant. If you are using the standard ATAN function on most calculators and computers, \theta will be calculated correctly only if x'T is positive. If it's negative, you'll generally have to add 180 degrees to the result to get the correct angle. Also, if x'T is positive and y'T is negative, ATAN will calculate a negative value for the angle; in that case, add 360 to bring it into the range 0 - 360 degrees. If you are using a calculator or a programming language with an ATAN2 function, use it instead:

\theta = atan2 (y'T, x'T)

ATAN2 automatically puts the angle into the correct quadrant, in the range 0 - 360 degrees.

As an example: let's find the general spherical coordinates for Proxima Centauri as seen from Barnard's Star. Recall its transformed target coordinates:

  1. x'T = -0.472 - (-0.017) = -0.455 pc
  2. y'T = -0.361 - (-1.815) = +1.454 pc
  3. z't = -1.151 - (0.148) = -1.299 pc
  • r = sqrt( (-0.455)2 + (1.454)2 + (-1.299)2)
  • r = 2.002 pc.
  • rxy = sqrt( (-0.455)2 + (1.454)2)
  • rxy = 1.524 pc.
  • \phi = tan-1 (-1.299 pc / 1.524 pc);
  • \phi = -40.44 degrees.
  • \theta = tan-1 (1.454 pc / -0.455 pc);
  • \theta = -72.62 degrees.

Ugh. x'T is negative and the angle is outside the
usual range 0 - 360. Add 180 to bring \theta to the correct quadrant:

  • \theta = 107.37 degrees.

Brightnesses

You now have the position of a star, in two different types of coordinates. The next step is the brightness.

Step 1: Get the distance to the target star from the
reference

If you calculated spherical coordinates, you have the distance, r, already. If not, calculate it:

  • r = sqrt(x'T2 + y'T2 + z'T2)

Step 2: Find the star's apparent magnitude

As seen from Earth, a target star has an apparent magnitude, V. From a different reference star, the target star will have a different apparent magnitude, V'. Once you know the star's distance from the reference, r, as well as its distance from the Sun, d, you can use either of two formulas, depending on whether you know the target's apparent magnitude V (as seen from Earth), or its absolute magnitude MV:

  1. Earth-based apparent magnitude: V' = V - 5 log10 (d/r)
  2. Absolute magnitude: V' = MV - 5 log10 (10/r)

As an example, what is Proxima Centauri's apparent magnitude from Barnard's Star? From Earth, V = +11.01 and d = 1.29 pc, and the transformed coordinates of Proxima Centauri are:

  1. x'T = -0.472 - (-0.017) = -0.455 pc
  2. y'T = -0.361 - (-1.815) = +1.454 pc
  3. z't = -1.151 - (0.148) = -1.299 pc
  • r = sqrt( (-0.455)2 + (1.454)2 +
    (-1.299)2)
  • r = 2.002 pc.
  • V' = +11.01 - 5 log10 (1.29/2.002)
  • V' = +11.96

Proxima Centauri is dimmer than it is from Earth.

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