Alpha Centauri AB is the closest star system to Earth (4.366 ly), followed closely by Barnard's star (5.988 ly). The closest star system to Alpha Centauri is Luhman 16 (3.8 ly from α Cen). So I am wondering, what is the distance from α Centauri AB to Barnard's star, and more generally, from one star to a different one (all below 10 pc)?

  • 1
    $\begingroup$ See this physics.stackexchange.com/questions/224950/… $\endgroup$
    – seVenVo1d
    Commented Oct 29, 2020 at 22:11
  • $\begingroup$ I am not an expert in this, but think about the spherical coordinates. You have two position vectors to the stars $\vec{r_1}$ and $\vec{r_1}$. So the distance between them is simply. $||\vec r_1 - \vec r_2||$. We know their radial distance $r$. Here the problem is that, you need to convert Right ascension and Declination to $\theta$ and $\phi$. But maybe above post helps as well. $\endgroup$
    – seVenVo1d
    Commented Oct 29, 2020 at 22:14
  • $\begingroup$ As Layla said, you can use the spherical trig formula in Rob's answer to find the angular distance between the 2 stars. And then you can plug that angle and the distances from Earth to the 2 stars into the cos rule of plane trig to find the distance between them. $\endgroup$
    – PM 2Ring
    Commented Oct 30, 2020 at 6:00

2 Answers 2


To find the distance from one star to another, we need three things for both of the stars: their right ascensions, declinations, and the distance from Earth to those stars.

So, let's get those things:

From the Wikipedia page on Alpha Centauri:

$RA = 14^h\:39^m\:36.49400^s$

$DEC = -60^{\circ}\:50'\:0.23737''$

$R = 4.37\:\rm{ly}$ (you gave 4.366, some other sources give 4.367... I'm going to stick with 4.37)

and for Barnard's Star:

$RA = 17^h\: 57^m\: 48.49303^s$

$DEC = +04^{\circ}\: 41'\: 36.2072''$

$R = 5.958 \: \rm{ly}$ (again, you gave a slightly different value, I'm sticking with Wikipedia for now)

where RA is right ascension, DEC is declination, and R is radial distance from Earth to the target star.

Now, by themselves, it is relatively difficult for us to obtain an actual distance. What I would do is convert these to rectangular coordinates, and then it's a matter of using the 3-d distance formula.

First, however, we need to convert RA and DEC into units like radians or degrees.

For right ascension, we can use the general formula:

$degrees = 15 (h + \dfrac{m}{60} + \dfrac{s}{3600})$

and for declination:

$degrees = deg + \dfrac{m}{60} + \dfrac{s}{3600}$

(when the declination is negative, however, multiply all terms in the formula by -1)

So, for Alpha Centauri AB, we have:

$RA = 15 (14 + \dfrac{39}{60} + \dfrac{36.49400}{3600}) \approx 219.902^{\circ}$

$DEC = -1 (60 + \dfrac{50}{60} + \dfrac{0.23737}{3600}) \approx -60.833^{\circ}$

and for Barnard's Star, we have:

$RA = 15 (17 + \dfrac{57}{60} + \dfrac{48.49303}{3600}) \approx 269.452^{\circ}$

$DEC = 04 + \dfrac{41}{60} + \dfrac{36.2072}{3600} \approx 4.693^{\circ}$

Now, to convert from spherical to rectangular coordinates, we have to define which of RA, DEC, and R can be assigned to $r$, $\theta$, and $\phi$. R should be $r$ - that's pretty straightforward. Since RA can be thought of as "celestial longitude", we'll assign it to $\theta$, and thus declination will be $\phi$.

To clarify, I'm defining $\phi$ as the angle from the xy-plane - so a $\phi$ of $\dfrac{\pi}{2}$ would mean pointing straight upwards. I know some sources define $\phi$ as the angle complementary to that angle (so, $\dfrac{\pi}{2}$ - the angle from the xy-plane), but for astronomical purposes, I think the definition I'm using is more intuitive and easier to work with.

We can then use the conversions:

$x = r\cos{\theta}\cos{\phi}$

$y = r\sin{\theta}\cos{\phi}$

$z = r\sin{\phi}$

So, for Alpha Centauri AB:

$x = 4.37 \cos{219.902^{\circ}} \cos{−60.833^{\circ}} \approx -1.634\: \rm{ly}$

$y = 4.37 \sin{219.902^{\circ}} \cos{−60.833^{\circ}} \approx -1.366\: \rm{ly}$

$z = 4.37 \sin{−60.833^{\circ}} \approx -3.816\: \rm{ly}$

and for Barnard's Star:

$x = 5.958 \cos{269.452^{\circ}} \cos{4.693^{\circ}} \approx -0.057\: \rm{ly}$

$y = 5.958 \sin{269.452^{\circ}} \cos{4.693^{\circ}} \approx -5.938\: \rm{ly}$

$z = 5.958 \sin{4.693^{\circ}} \approx 0.487\: \rm{ly}$

And now, finally, we can use the distance formula for 3-d:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}$

So, the distance between Alpha Centauri AB and Barnard's Star is:

$d = \sqrt{(-1.643 + 0.057)^2 + (-1.366 + 5.938)^2 + (-3.816 - 0.487)^2} \approx\mathbf{6.476\,ly}$

Well, that was certainly tedious - but it's a process that you can standardize to pretty much any star, or really, any two astronomical objects:

First, convert RA and DEC to degrees.

Second, assign R, RA, and DEC to the spherical coordinates $r$, $\theta$, and $\phi$.

Third, convert spherical coordinates to rectangular coordinates.

Lastly, use the distance formula with the two sets of $x$, $y$, and $z$ coordinates.

Hope this helps. :)

  • 2
    $\begingroup$ Nice explanation. you can even write a code about it, so that you just put the inputs and it tells you the distance. $\endgroup$
    – seVenVo1d
    Commented Oct 30, 2020 at 6:49
  • 4
    $\begingroup$ It's a bit less tedious if you don't convert to Cartesian coordinates, and just use the cos rule on the angular distance and the radial distances to the 2 stars. $\endgroup$
    – PM 2Ring
    Commented Oct 30, 2020 at 7:18
  • $\begingroup$ As d_e noticed, the z component of Alpha Centauri should be negative. $\endgroup$
    – PM 2Ring
    Commented Oct 29, 2021 at 5:48
  • $\begingroup$ @PM2Ring ahh you're right. Thank you d_e for noticing that as well - let me correct that. $\endgroup$
    – sforsingh
    Commented Oct 29, 2021 at 19:21

Just to add another method. As hinted in a comment one can find the angular distance between the starts. How to find the angular distance?

Imagine the spherical triangle formed by the locations of the two stars and the North or South pole. Let's use the South pole in this case. We have: 1) the angle at the pole which is the difference in RA's: 3h 18m 12s (=49.55°) - let's mark this angle with $p$; 2) we have also the arcs next to this angle - their lengths are the declination differences from the declination of -90. That is 29.16° (for AlphaC) - Let's mark it with $A$ and 94.69° (for Barnard's) - Let's mark it with $B$. Now, we are looking for the third arc - $X$. the length of the third arc is basically the angular distance from the center of the Earth. According to Spherical law of cosines:

$$\cos (X) = \cos(A)\cos(B) + \sin(A)\sin(B)\cos(p) \Rightarrow$$

$$\cos(X) = \cos(29.16°)\cos(94.69°) + \sin(29.16°)\sin(94.69°)\cos(49.55°) \Rightarrow$$

$$X = 75.89°$$.

Now, we can use the cosine theorem for the Euclidian Triangle (Earth-center, Star A, Star B). As we have 2 sides length and the angle between them which results in $6.49ly$ (using the distances from the question))


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