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If a star appears to be very faint, does that imply that it's really far away, or that its intrinsic luminosity is small?

More precisely: If I look up into the cosmos and see a star that is very faint, which is more likely to be true: (1) its intrinsic luminosity is low, or (2) the star is far away?

Emilio Pisanty
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Kalis
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  • Uh, you can't decide between these two if you don't know how far away the star is. – ACuriousMind Feb 12 '15 at 16:00
  • What I am trying to ask is; If I look up into the cosmos and see a star that is very faint, which is more likely to be true?; (1) it's luminosity is low. (2) it's far away. – Kalis Feb 12 '15 at 16:02
  • Without any quantitative values this is unanswerable. And even with them, it essentially comes down to looking up the distribution of luminosity among stars and the distribution of their distance, and calculating some probabilities. There's little physics here. – ACuriousMind Feb 12 '15 at 16:08
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    Linked - http://physics.stackexchange.com/questions/163999/of-the-9096-visible-stars-90-are-how-close/164021#164021 – ProfRob Feb 13 '15 at 00:10
  • The closest star to us (other than our own Sun) is Proxima Centauri. It's a red dwarf; it's luminosity is low. You might be able to see it on a very clear, moonless night with a very good set of binoculars. UY Scuti, a red supergiant, is about the same color and same apparent magnitude as Proxima Centauri, but it's over 2000 times as far away from us as Proxima Centauri. Without extra information such as parallax, there's know way to know whether a very faint red star is a nearby red dwarf or a distant red giant. – David Hammen Feb 13 '15 at 20:03
  • No way to know for sure. But you can say which is more likely. – ProfRob Feb 13 '15 at 21:44
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    This is the key historical problem in observational astronomy. It took a lot of effort and insight to figure out the answer, and whole books have been written on how to devise tricks in order to get independent measurements of either the distance to a star or its absolute luminosity. – Zo the Relativist Feb 13 '15 at 23:28

4 Answers4

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There are several ways of answering this question.

  1. We may have additional information (not including a parallax!). For instance, if we know the surface temperature of the star and its gravity, both of which can be estimated directly from spectroscopy, then the type of star leads to a direct prediction of its absolute luminosity. This in turn leads to a distance estimate.

  2. Stars have a distribution of tangential proper motions on the sky. In general, the closer the star, the more likely it is to have a large proper motion. Often we can calculate the "reduced proper motion plot", usually defined as the apparent magnitude plus 5 times the (base 10) log of the proper motion in arcseconds per year. A plot of reduced proper motion versus the colour of a star gives you a strong indication of whether it is a giant or a dwarf and this classification can then be used to estimate the absolute luminosity and hence distance.

  3. If you know nothing else, then you can construct a plot of luminosity versus distance for star of known distance and brightness. A star of a given brightness will have a distribution of possible distances that can be used as a probabilistic estimator of the true distance. I attach a diagram showing how well (or not) this is likely to work. This uses the Hipparcos revised parallax catalogue. I plot distance (in parsecs) versus apparent visual magnitude. The spread in distance for a given magnitude is very large. There is only a very weak correlation between apparent brightness and actual distance. I also colour code the points according to their B-V colour. There are distinct bands. i.e. If we know the colour of a star we can do much better, but there are also ambiguities introduced if you do not know whether a star is a giant or dwarf. Hence, there are two bands for stars with B-V of about 1.2-1.3 - a (closer) dwarf sequence and a more distant giant sequence.

Hipparcos distance vs magnitude

If we ignore extinction and assume sources are distributed uniformly through space then the number $N$ of sources with a given intrinsic luminosity that are found with flux $S$ varies as $N(S) \propto S^{-5/2}$. Because this is true at any luminosity it is also true for any spatially uniform distribution of intrinsic luminosities. The relevance to this question is that for stars of any type you find many more apparently faint examples than bright ones and the reason for this is that they are included from a much greater volume of space at larger distances. Under these restrictive assumptions it is always more likely that a faint object is found near to the limiting distance, $d_{max}$, of its detectability - to be exact, the average distance will be $d_{max}/2^{1/3}$.

The question however is more subtle than this and is affected by extinction, spatial anisotropy and the relative densities of different kinds of star. The probability that a faint star is close or far away very much depends on which direction you are looking. For example, if we are looking out of the Galactic plane, then a faint star is more likely to be nearby, because there are few stars at great distances in that direction. On the other hand, if you look towards the Galactic plane, you might also be able to say that a faint star is unlikely to be very distant, but this time, it is because extinction by dust limits the horizons of your observation. On the other hand, if you look in the infrared, then the star is much more likely to be distant, because you can see (through dust) much further and sample a larger volume at greater distance.

EDIT: 13/02/15

Just to give chapter and verse on this. Since the question appears to be asking about the faintest visible (naked eye) stars, I selected all objects with $5.5<V<6.5$ in the revised Hipparcos catalogue. The first plot below shows their (normalised) distance distribution. It is reasonably symmetric in log distance. The median distance of a faint naked eye star is 440 light years (i.e half are further way than this). Thus your best estimate of distance in the absence of any other information would be about 440 light years, but with a factor of $\sim 2.5$ as a 1-sigma uncertainty. Thus a "faint star" by this definition is most likely to have an absolute magnitude of about zero, and therefore be $\sim 100$ times brighter than the Sun. However, plotting the stars at around the most probable distance on a HR diagram (see next plot) we see they are not the most intrinsically luminous stars. They are mostly red giants and also some main sequence stars a few times more massive than the Sun. This is because the brightest stars are rare, so even though they can be seen in a larger volume of space, there are still not many of these rare objects in that space.

Finally, I reinforce the point I made about Galactic latitude. The bottom plot compares the (normalised) distance distributions for stars with Galactic latitude less than $\pm 15$ degrees (low Galactic latitude), with those that are more than 45 degrees out of the Galactic plane (high Galactic latitude). This plot perfectly shows the point I made about direction-dependence. Looking out of the Galactic plane we see the normalised probability distribution peaks at a smaller distance and cuts off sharply at 2000 light years as the Galactic disk "peters out".

Distance probability distribution

Hipparcos HR diagram

Distance probability for low and high Glactic latitude

ProfRob
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  • Presumably bright sources are more likely to be visible? Otherwise, I do not understand your second-to-last paragraph. – Emilio Pisanty Feb 12 '15 at 18:06
  • Sorry, I'm not sure we're talking about the same thing. When you say "Faint sources are more common", you mean fainter observed sources? Presumably for fixed observed flux $S$, intrinsically brighter stars are more common. – Emilio Pisanty Feb 12 '15 at 18:41
  • @EmilioPisanty S is observed flux. Fainter sources are more common. This is true whatever the distribution of intrinsic luminosities. – ProfRob Feb 12 '15 at 18:58
  • I don't doubt that, but I don't see how that is useful in addressing the question. We don't care whether we can see more faint stars or not, we are given a star and want to estimate how far or faint it is. $S$ is (the only thing that's) given, and we don't need to worry about its distribution over the whole star population. – Emilio Pisanty Feb 12 '15 at 19:54
  • @EmilioPisanty I have made a small edit, but I think the original argument is quite clear. In a uniform spatial distribution we see more faint stars because they are scattered over a larger volume and are therefore more likely to be at a larger distance. – ProfRob Feb 12 '15 at 21:52
  • Apologies, but I am still confused by that paragraph. Are fainter sources more common because there are in fact more faint stars than bright ones? Or is this a volume effect? If the latter, are you saying that more of the stars we observe are observed to be faint, or that more of the stars we observe are intrinsically faint? I don't want to poke holes in your content as I'm sure that you know the physics better than I do, but if I'm still confused then I don't think the OP will do much better (judging from the question). – Emilio Pisanty Feb 12 '15 at 22:23
  • In that case, I still don't see how this relates to the question. We know the star, so we are interested in the population with that observed brightness, and not in all the stars visible to the naked eye (or any given instrument). – Emilio Pisanty Feb 12 '15 at 22:47
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    That's a lot clearer. (I'm not a fan of the false colour in your diagram, though.) – Emilio Pisanty Feb 13 '15 at 19:08
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A faint star can either be far away or be faint to begin with, and without more data to go on there is simply no way to tell.

This is a big problem in astronomy, and measuring distances is one of the main challenges in understanding any given system; for more details look up the cosmic distance ladder. If it's a star you can see with the naked eye, chances are that its distance from Earth can be measured relatively simply using the parallax method, which means that we know how far away it is, and you can simply look up this distance and from it work out how intrinsically bright it is. Without this external data, though, it's not an easy thing.

For some stars, though, you can get some idea by looking at the colour. This is because the colour of the star is in general related to its instrinsic brightness. This was discovered by Hertzsprung and Russell, who measured the intrinsic brightness of stars and plotted it against their colour (or more specifically, their temperature). The result is a big diagonal streak with a large population:

Image source

This diagonal streak is known as the Main Sequence, and if you know that a star is in it then you know that the bluer it is, the higher its internal brightness. Thus if you see two Main Sequence stars of the same brightness then the bluer one will be the more distant one.

The problem is, however, that you can't know for sure that a star is in the main sequence. A reddish star, for example, could be a small, cool Main Sequence star, or it could be a bloated red giant with a much bigger surface area and therefore a much higher total luminosity. (Because of the Stefan-Boltzmann law, the brightness of a patch of star surface of a constant area is only a function of its temperature). Similarly, a bluish star can be a massive, young main-sequencer, or it can be a smallish star that has run out of fuel and shrunk down into a white dwarf. Without having more information, it's simply not possible to tell for sure.


If you're OK with a probabilistic sort of statement, though, then there are indeed more things you can say. If you look at the HR diagram above, you can see that the different populations can have radically different numbers of stars in them; for example, there's generally rather few supergiant stars. There are a number of things which affect the numbers of different stars that we can see:

  • volume effects (where a brighter star is visible over longer distances, which means that more of the stars we see tend to be intrinsically brighter),

  • intrinsic density effects (i.e. the overall probability of a star to be of a given type, so for instance in the main sequence brighter stars are less common),

  • local density effects (as the local environment can differ from galaxy-wide properties), and

  • absorption effects (where stars in the galactic plane are partially obscured by dust, appearing fainter than they otherwise would be)

among others. However, if you're looking at a given star and you want to be able to say things about it, then we can simply compile the statistics for the stars we can actually see, at the brightness we see them at, and then use those to try and see how much you can say about your star.

The diagram below, adapted from Rob Jeffries' answer, shows the distance (in light years, on a logarithmic scale) and the visual magnitude of the stars that are visible to the naked eye (so bright stars are to the left and faint ones to the right). The colour of each star indicates its B-V index, and it is a rough indication of the visual colour of the star. (See the code used to produce it in the revision history.)

Distance vs visual magnitude for naked-eye stars

There's a few things to notice here. The first is that there is a pretty sizeable spread in the distances, with most visible stars between 20 and 2,000 light years. Any further than that, and they'll generally be too faint for us to see; any closer, and there simply isn't much volume between here and 20 ly away to fit many stars.

Additionally, there is definitely structure in the way the different colours occupy the diagram. Unfortunately, there's relatively little you can actually say with it, because the different star colours are still pretty mixed. Over a larger population, as depicted by Rob and with the kind of stars you might see in a telescope, there are much clearer bands, but if you restrict yourself to naked-eye objects there's rather less you can say. Nevertheless, a faint star is somewhat more likely to be intrinsically bright and far away if it is very red or very blue, whereas a whiter star is somewhat more likely to be not-so-bright intrinsically, and somewhat closer by. To the small extent to which you can say anything, though!

Emilio Pisanty
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    Meanwhile, your argument in the last paragraph does not acknowledge the critical point that intrinsically luminous stars have a much lower density than intrinsically faint stars. Thus although they can be seen to much greater distances, they are intrinsically much rarer. Massive main sequence stars have $r=10^{4}$ compared to a solar-type stars, but they are not $10^{6}$ times more common in magnitude limited surveys! – ProfRob Feb 12 '15 at 22:07
  • I had sort of dealt with that earlier in the text, but I have clarified that section. If it is still inaccurate I'll be happy to know why. – Emilio Pisanty Feb 12 '15 at 22:30
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    The thing that is incorrect is to use the $r^{3/2}$ factor to argue that a faint star is more likely to be an intrinsically luminous, but distant star. The volume of space goes up by a factor $r^{3/2}$, but the density of (main sequence) stars goes something like $N(r) \propto r^{-1.4}$. So these almost cancel out. That's why the distribution of main sequence stars in a magnitude limited HR diagram is reasonably uniform. Your HR diagram conflates a magnitude limited and a volume limited survey so can't be used to judge the relative proportions of stars in either. – ProfRob Feb 12 '15 at 23:15
  • The situation is much more nuanced than that. The argument works better for red giants which are comparatively more common than equally luminous main sequence stars. But the $N(r)$ factor must still be calculated. – ProfRob Feb 12 '15 at 23:18
  • +1 from me. This is a question that requires a lot of pieces to answer in detail. This answer covers most of them, certainly well beyond what I imagine the OP had in mind... – Kyle Oman Feb 12 '15 at 23:29
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You need some standard candles and spectral analysis or you have to measure the parallax, otherwise you can't tell for sure. If the object is far enough away that its velocity due to the expansion of the universe is bigger than the peculiar velocity you get a good estimate for the distance by measuring its redshift.

Yukterez
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Chances are it's really far away. Whilst stars are fairly evenly spread over the faint-to-bright scale, simple geometry dictates that the vast majority of stars are very, very far away, so for any random faint star (with no other information available) the chances are it's down to distance rather than being at the particularly faint end of the range.

This is all assuming we're using a big telescope. If your question specifically refers to naked eye observations, then most of the stars you can see in the sky are very close, within a few thousand lightyears, and most of those are very luminous, even the "faint" ones, so for naked eye observations distance is also the primary factor.

The Geoff
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    It is much more complicated than this. – ProfRob Feb 12 '15 at 17:55
  • Yeah, but rather get into how we get around the problem of why a star appears faint I went for a literal answer to the question as clarified in God's follow up comment. – The Geoff Feb 12 '15 at 18:09
  • No, I mean that if you are looking out of the Galactic plane, what you have said is not (necessarily) true. – ProfRob Feb 12 '15 at 18:28
  • "fairly evenly distributed over the faint to bright scale" by luminosity perhaps, but certainly not by number!!! – Kyle Oman Feb 12 '15 at 23:24