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A star is probably visible/detected by it's radiation. But that star may or may not belong to our own galaxy ... yet news reports speak of detecting a star/nova in a distant galaxy.

How does one determine whether the star she/he views belongs to Milky Way, or some other galaxy ... or is galactic orphan? Is it merely a matter of the distance to that star?

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Every ordinary star we are able to individually observe is a part of the Milky Way. Well, except for stars in a small number of very nearby galaxies but even galaxies such as Andromeda look like a "continuum" so we're not observing the stars individually although we see that the galaxy isn't just a point.

Only if a star goes nova (a lethal nuclear explosion of a white dwarf star) of supernova (a similar explosion but stronger), it may be observed outside the Milky Way. In all such cases we've experienced, one may always identify a galaxy at the same location that was known before the nova/supernova explosion. So the star going nova/supernova clearly belongs to that galaxy. Please note that distant galaxies look like dots – pretty much visually indistinguishable from stars in the Milky Way.

A star going nova has 50,000-100,000 times higher luminosity than the Sun; the number is even higher for a supernova. That's a sufficient increase of the luminosity for an exploding star in a distant galaxy to become "almost as bright" as the whole galaxy, well, not quite.

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Just want to clarify: The absolute furthest galaxies may look like dots, but the galaxies we can see supernovae in are generally extended objects in even moderate telescopes. –  Chris White Nov 1 '12 at 20:31
    
Dear Chris White, it's surely not true for most of the supernovae we are observing with a good purpose. Those cosmological type Ia supernovae that are used as standard candles to determine the expansion (and acceleration) of the Universe are usually billions of light years away, since they erupted up to 9 billion years ago: msnbc.msn.com/id/45975259/ns/technology_and_science-space/t/… - You don't see more than a point. –  Luboš Motl Nov 1 '12 at 20:54
    
Astrophysics is more than just cosmology, and indeed we study nearby SNe Ia just to figure out what they are, since no one knows for sure. Also most SNe are observed with $z < 1$ (including in fact all the SNe used by Riess and Perlmutter). Moreover, due to the turnover in the angular diameter distance as a function of redshift, galaxies will only ever appear so small. Though perhaps I have wandered a bit off topic from the OP. –  Chris White Nov 1 '12 at 22:39
    
Chris, the couterintuitive behaviour of the angular diameter distance above z~1.5 is known to me, but I have a question: why doesn't that translates into an increase in aparent brightness? Surface flux is supposed to be conserved by means of Liouville's theorem, no matter how counterintuitive it may seem. –  Eduardo Guerras Valera Nov 2 '12 at 3:14
    
Ouch, here is the answer to my own question: ned.ipac.caltech.edu/level5/March02/Ratra/Ratra4_2_4.html –  Eduardo Guerras Valera Nov 2 '12 at 12:39

The key is, that the intrinsic brightness of all supernovae (at least for the most important type 1a) is roughly the same: it peaks around magnitude -19. From the difference with the apparent magnitude (called distance modulus) the distance to the supernova can be derived, and then compared to the distance to the suspected host galaxy (derived from its redshift).

The problem is that for both distances, a model of universe must be assumed. But that is another question.

For nearby supernovae and novae, the size of the expanding photosphere can be used as an alternate method for distance estimates too. This case is less usual.

With some quasars the same question arises, because there is a galaxy in the line of sight. In this case, the difference in redshifts, and thus in distance, leaves no doubts, although both objects appear at the same position in the sky.

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Both galaxies and supernovae are redshifted. You just need to compare the redshift of the host galaxy to the supernova in question, and you can discern whether or not the supernova - or quasar - belongs to that galaxy or not. –  Ernie Dec 13 '12 at 17:05
    
@Ernie: no, you hardly ever measure supernovae redshifts. Distances are to supernovae are mostly derived from the extrapolated maximum brightness. It is uncommon to have a supernova so bright and nearby as to make detailed spectroscopic measures. Besides that, supernova lines are very broad, due to the high expansion velocity of the shell. And the more interesting ones for cosmology are the farthest ones, where integrated photometric measures are possible, but spectroscopy is harder to do. –  Eduardo Guerras Valera Dec 13 '12 at 18:15

The stars of our own galaxy are always much brighter than the stars of other galaxies. Just as a point of reference, the Milky way is about 100 thousand light years across. The nearest large galaxy, the Andromeda galaxy, is 2.2 million light years away. All of its stars would therefore be 20 times as far away than any star in our galaxy.

Here's a picture of the Andromeda galaxy, for example:

The stars that belong to the Andromeda galaxy are mostly not even recognizable as stars at all. They make up the fuzz of the disk. There are also two satellite galaxies that appear fuzzy in the image. Most other points of light are stars from our own galaxy, with a few faint fuzzies in the very distant background that are other, far more distant galaxies.

Every other galaxy (and there are billions) is many times farther away, with the exception of the large and small magellanic clouds, which are satellites of the Milky Way.

This coincidentally, was the reason why up until Edwin Hubble in the early 1900s, noone was able to determine whether galaxies were nebulae or separate, distinct objects from our own galaxy. It wasn't until the construction of the 100 inch Hooker telescope and the later 200 inch Mount Palomar telescope, that anyone was able to resolve any stars at all from these galaxies.

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