I was looking at a map of our local stellar neighborhood, and it occured to me, the stars are really close, if one compares them to the size of some nebulae. So can it be, that the Sun, Alpha Centauri system, and all the other stars in the neighborhood actually formed from the same nebula? Have we discovered any information about this?

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    $\begingroup$ Many of the local stars are moving in different directions. I believe Alpha Centauri is moving towards us currently. So while it could be possible that some stars are the Sun's siblings, I think time has spread them very far apart to the point that they could be hundreds or even thousands of light years away now. $\endgroup$
    – Markitect
    Commented Feb 11, 2021 at 4:20
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    $\begingroup$ Even if you go just back millions of years our neighborhood would look a lot different, nevermind 4.5 billion. This is a good video that illustrates how that happens: youtu.be/XRlY6i42S9M $\endgroup$
    – eps
    Commented Feb 11, 2021 at 14:22

2 Answers 2


There are three main reasons why we can tell that local stars did not, for the most part, form from the same molecular cloud that the Sun formed from.

The first is that unless stars are born in a very tightly bound system such as a globular cluster (which the Sun is definitely not in), they will drift apart from their birth companions over time in slightly different orbits in the Milky Way galaxy. The Sun has gone around the center of the galaxy about twenty times since it was born; this is more than enough time for its initial cluster to become completely scrambled in azimuth (the angle in the plane of the galaxy). To see how this would work, imagine two stars with very similar orbits, one with a period of 200 million years and the other with a period of 210 million years. If they start off right next to each other, then after 2 billion years, the first star has made 10 complete orbits, while the second has made about 9.5 -- meaning it will now be on the other side of the galaxy from the first star.

You can also take the current velocities of nearby stars, and see that they can be rather different, which tells you they are not on very similar orbits. Alpha Centauri is moving about 20 km/s in our direction, while Barnard's star has a velocity in our direction of over 100 km/s (in both cases, they also have movement in perpendicular directions; they're not moving directly toward us). So these are stars which happen to be near each other now, but are on different orbits and therefore were almost certainly not near each other at any given time in the past.

The second reason is the one that David Hammen pointed out: stars born from the same clouds should have very similar chemical compositions ("metallicities" in astronomical parlance).

The third reason is that stars born from the same cloud should have the same age (to within a few million years, anyway), since molecular clouds tend to be disrupted by the star-formation process and don't keep forming stars for hundreds of millions or billions of years.

Looking at some of the very nearest stars: the Sun has what we define as "solar metallicity" and is about 4.6 billion years old. The stars of Alpha Centauri are slightly more metal-rich than the Sun and probably several hundred million years older. Barnard's Star is more metal-poor than the Sun (maybe only 1/3 the iron fraction), and about 5 billion years older. Sirius is significantly more metal-rich than the Sun (its iron fraction is about three times that of the Sun) and is only about 250 million years old.

So you can see that the Sun and the nearby stars are a heterogeneous group, and clearly not all born at the same time from the same patch of interstellar gas.

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    $\begingroup$ Why isn't Alpha Centauri younger if it is more metal-rich? $\endgroup$ Commented Feb 12, 2021 at 5:57
  • $\begingroup$ I have a question. If Barnard's Star has less heavy elements, doesn't that mean that it's from the previous generation of stars? $\endgroup$
    – Sam
    Commented Feb 12, 2021 at 13:25
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    $\begingroup$ @KeithMcClary The relationship between age and metallicity (i.e., younger stars tend to be more metal-rich) is only an approximate one, with lots of variation. There are plenty of stars approximately the same age as the Sun with higher metallicities and plenty with lower metallicities. $\endgroup$ Commented Feb 12, 2021 at 14:43
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    $\begingroup$ @Sam Well, Barnard's star is about 5 billion years older than the Sun, so it is indeed from an older generation ;-) $\endgroup$ Commented Feb 12, 2021 at 14:44

Is it possible that the Sun and all the nearby stars formed from the same nebula?

No, it is not.

Our Sun has marked differences in metals compared to the nearby stars. (In astronomy, every element higher than helium is a "metal".) The discovered and analyzed stars that appear to be most similar to the Sun are far, far away from the Sun. Our Sun apparently formed in what would soon become an open cluster. That means the Sun's sibling stars are spread far and wide. They are not nearby.

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    $\begingroup$ Can you please explain what you mean by "Our Sun apparently formed in what would soon become an open cluster." ? $\endgroup$
    – Sam
    Commented Feb 11, 2021 at 7:34
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    $\begingroup$ @Sam Open clusters tend to evaporate. $\endgroup$
    – PM 2Ring
    Commented Feb 11, 2021 at 9:41
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    $\begingroup$ @Sam Open clusters get disassociated rather quickly due to close encounters and ejections between neighboring stars. $\endgroup$
    – WarpPrime
    Commented Feb 11, 2021 at 13:22
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    $\begingroup$ @Sam - just to explain. What you say is certainly possible, but - amazingly - astronomers have looked in to this question. And they now know the answer. In fact our current neighbor stars WERE NOT made in the same origin as us. The answer here gives you the details. $\endgroup$
    – Fattie
    Commented Feb 11, 2021 at 15:12
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    $\begingroup$ In astronomy, every element higher than helium is a "metal" And yet you call them "rocky planets" with "metallic cores". Sloppy heathens... lol. I'm out to get some fresh air from our metallic atmosphere. $\endgroup$
    – J...
    Commented Feb 11, 2021 at 15:34

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