<blockquote>
  <p>Light that is not light</p>
</blockquote>

<p>That's meaningless.  All light is electromagnetic radiation.  A finite part of the infinitely large range of the electromagnetic spectrum is visible light.  So you should talk about EM radiation and to discuss the visible spectrum just say visible spectrum.</p>

<p>Stars emit lots of energy at frequencies that are outside the visible range.</p>

<blockquote>
  <p>Dark Matter</p>
</blockquote>

<p>Dark matter does <em>not</em> mean that it's just not radiating in the visible part of the spectrum.  It means it's not radiating in any part of the EM spectrum much.  If dark matter exists (and that's not certain) it won't interact with most matter much at all.</p>

<p>If dark matter were in star-like objects, it sounds very unlikely that there would not be any EM radiation as a result of the interactions that would result in a star-like object.  We'd detect it if it were there.  So I don't think that's likely.</p>

CC BY-SA 3.0
2017-02-15T20:41:43.357Z
Is the EM spectrum "infinitely large"? I'd have thought it was limited in both directions.
Chappo Hasn't Forgotten
CC BY-SA 3.0
2017-02-15T21:19:26.090Z
Limited in one direction, but infinitely large, at least in principle.  Given an assumption of a finite universe that "infinite" lacks meaning.  And as wavelength grows larger making a measurement of it becomes less practical, even perhaps impossible.  As as frequency increases so does energy, and that presumably has a practical limit as well.  But in the abstract theoretical sense, an infinite range of values.
StephenG - Help Ukraine
CC BY-SA 3.0
2017-02-15T22:46:14.070Z
In theory, extremely long wavelengths are limited only by the size of the universe, but this radiation would have an extremely low energy, reducing asymptotically with wavelength - so would there be a quantum limit? In the other direction, are extremely short wavelengths limited by the Planck length, or do we just say "yes it could be smaller but we don't know how to describe this"?
Chappo Hasn't Forgotten
CC BY-SA 3.0
2017-02-15T23:10:53.183Z
The [plank length](https://en.wikipedia.org/wiki/Planck_length) isn't a hard limit in physics.  It's constant you can define which seem representative of the order of magnitude at which our existing theories stop working well.  In principle we hit measurement problems as a limit at this range, but then a theory "beyond" current QFT might change that idea.  Nobel prize for the first completely correct answer to what that theory is.
StephenG - Help Ukraine
CC BY-SA 3.0
2017-02-15T02:05:29.620Z
2017-02-15T02:05:29.620Z
StephenG - Help Ukraine
<p>You mix up a lot of things in your assumptions.</p>
<blockquote>
<p>Stars convert mass into energy.</p>
<p><span class="math-container">$E = mc^2$</span></p>
</blockquote>
<p>The way stars produce energy is nuclear fusion. One main process there is the <a href="https://en.wikipedia.org/wiki/Triple-alpha_process" rel="nofollow noreferrer">tripple-alpha process</a>, which releases a net energy of about <span class="math-container">$E \sim 7$</span> MeV (that is not <em>that</em> much)</p>
<blockquote>
<p>So can they emit so much energy in a form of light that is not light?</p>
</blockquote>
<p>By <em>light</em> you are speaking of <strong>visible light</strong>. But radiation goes over all frequencies and wavelength, not just the one the human eye can see (which is roughly <span class="math-container">$\lambda = 400 - 700$</span> nm).
Here you want to check out the theory behind <a href="https://en.wikipedia.org/wiki/Black-body_radiation" rel="nofollow noreferrer">black body radiation</a>, which describes - more or less - the intensity of the light at different wavelengths/frequencies at a given temperature.</p>
<p>For an overwiev, look at this figure. The sun's surface temperature ist about 6000 Kelvin, so you can infer, that it's radiation maximum is in the range, which the human eye would perceive as green.
<a href="https://i.sstatic.net/bWXBI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bWXBI.png" alt="Planck's law at different temperatures [from wikipedia]" /></a></p>

CC BY-SA 3.0
2017-02-15T10:25:59.440Z
2017-02-15T10:25:59.440Z
2020-06-17T09:47:40.653Z
rtime
<blockquote>
  <p>So can they emit so much energy in a form of light that is not light? Resulting in a star that cannot be detected using our senses(+ a telescope).</p>
</blockquote>

<p>Presumably by "light that is not light" you mean light outside of the visible spectrum. This is why the fields of gamma ray astronomy, x-ray astronomy, ultraviolet astronomy, infrared astronomy, microwave astronomy, and radio astronomy exist. Astronomy is not limited to the visible spectrum, and there's lots of interesting stuff to "see" from using a telescope that senses light outside of the visible spectrum.</p>

<p>The hottest of stars emit most of their radiation as x-rays and ultraviolet light, but they also emit a lot of visible light. These very hot stars are quite large, short-lived, and very rare. The coolest of stars emit most of their radiation in the infrared, but they also emit some light in the visible range. These very cool stars are quite small, very long-lived, and extremely numerous. Their cool temperatures makes them rather hard to see, even in the infrared.</p>

<blockquote>
  <p>If so what problems could this star pose? Could this be dark matter?</p>
</blockquote>

<p>The very smallest of those cool stars, along with black holes, cool neutron stars, brown dwarfs, and rogue planets collectively form a set of candidates for dark matter called "massive astrophysical compact halo objects", or MACHOs for short. However, theoretical calculations along with observations says that MACHOs can account for at most about 20% of dark matter.</p>

CC BY-SA 3.0
2017-02-16T02:43:39.157Z
2017-02-16T02:43:39.157Z
David Hammen
<p>I will just point out two things that haven't really been discussed above. </p>

<p>Firstly, stars generally fall into two categories. 'Hot stars' are those that convert mass into energy by nuclear fusion, and 'cold stars' are those that have ended the nuclear burning phase, e.g. white dwarfs and neutron stars. These still have a finite temperature and therefore will emit radiation. As you can imagine, however, since they no longer produce energy, eventually they will become actually cold, hence not radiating (or radiating very little). </p>

<p>The second point to note is that radiations from stars far away are redshifted as they reach us, and they are likely to be attenuated by dust. Therefore, even if the radiation itself when emitted is visible light, by the time it gets to us it may no longer be. However, there are still many other different spectroscopic techniques that will allow us to 'see' it.</p>

CC BY-SA 3.0
2017-02-21T20:22:15.203Z
WDs and NSs are far from cold, having temperatures of tens, and millions, of Kelvins, respectively. Also, redshift and reddening are sort of beside the point here.
pela
CC BY-SA 3.0
2017-02-21T21:03:11.440Z
@pela This is why I said "eventually they will become actually cold". The names "hot stars" and "cold stars" have been defined clearly in my answer; they do not refer to the temperatures, though I admit the terminology may somewhat be misleading. As for redshift - the question asks about detection, so I would not say it is irrelevant. I would admit that reddening is probably slightly beside the point.
John
CC BY-SA 3.0
2017-02-21T21:23:06.383Z
Okay, sorry, I misunderstood.
pela
CC BY-SA 3.0
2017-02-21T19:58:31.567Z
2017-02-21T19:58:31.567Z
John
<p>Stars convert mass into energy. Even converting the slightest mass into energy is immense because the speed of light is so great and $E = mc^2$. This means that stars have to emit large amounts of energy. So can they emit so much energy in a form of electromagnetic radiation that is not visible light? Resulting in a star that cannot be detected using our senses(+ a telescope). If so what problems could this star pose?</p>

CC BY-SA 3.0
2017-02-16T14:07:10.313Z
Have you looked at a star recently?
Carl Witthoft
<light><fundamental-astronomy><space><spectroscopy>
Can some stars not emit any energy in the visible spectrum?
user15836
# Can some stars not emit any energy in the visible spectrum?

(Asked by: username_0 on 2017-02-15)

Stars convert mass into energy. Even converting the slightest mass into energy is immense because the speed of light is so great and $E = mc^2$. This means that stars have to emit large amounts of energy. So can they emit so much energy in a form of electromagnetic radiation that is not visible light? Resulting in a star that cannot be detected using our senses(+ a telescope). If so what problems could this star pose?

**Tags:** light, fundamental-astronomy, space, spectroscopy

> **Comments:**
> **username_1**: Have you looked at a star recently?  

---

## 4 answers

### Answer by username_2 on 2017-02-15. Score: 7

> Light that is not light

That's meaningless. All light is electromagnetic radiation. A finite part of the infinitely large range of the electromagnetic spectrum is visible light. So you should talk about EM radiation and to discuss the visible spectrum just say visible spectrum.

Stars emit lots of energy at frequencies that are outside the visible range.

> Dark Matter

Dark matter does *not* mean that it's just not radiating in the visible part of the spectrum. It means it's not radiating in any part of the EM spectrum much. If dark matter exists (and that's not certain) it won't interact with most matter much at all.

If dark matter were in star-like objects, it sounds very unlikely that there would not be any EM radiation as a result of the interactions that would result in a star-like object. We'd detect it if it were there. So I don't think that's likely.

> **Comments:**
> **username_3**: Is the EM spectrum "infinitely large"? I'd have thought it was limited in both directions.  
> **username_2**: Limited in one direction, but infinitely large, at least in principle.  Given an assumption of a finite universe that "infinite" lacks meaning.  And as wavelength grows larger making a measurement of it becomes less practical, even perhaps impossible.  As as frequency increases so does energy, and that presumably has a practical limit as well.  But in the abstract theoretical sense, an infinite range of values.  
> **username_3**: In theory, extremely long wavelengths are limited only by the size of the universe, but this radiation would have an extremely low energy, reducing asymptotically with wavelength - so would there be a quantum limit? In the other direction, are extremely short wavelengths limited by the Planck length, or do we just say "yes it could be smaller but we don't know how to describe this"?  
> **username_2**: The [plank length](https://en.wikipedia.org/wiki/Planck_length) isn't a hard limit in physics.  It's constant you can define which seem representative of the order of magnitude at which our existing theories stop working well.  In principle we hit measurement problems as a limit at this range, but then a theory "beyond" current QFT might change that idea.  Nobel prize for the first completely correct answer to what that theory is.  

---

### Answer by username_4 on 2017-02-15. Score: 4

You mix up a lot of things in your assumptions.

> Stars convert mass into energy.
>
> $E = mc^2$

The way stars produce energy is nuclear fusion. One main process there is the [tripple-alpha process](https://en.wikipedia.org/wiki/Triple-alpha_process), which releases a net energy of about $E \sim 7$ MeV (that is not *that* much)

> So can they emit so much energy in a form of light that is not light?

By *light* you are speaking of **visible light**. But radiation goes over all frequencies and wavelength, not just the one the human eye can see (which is roughly $\lambda = 400 - 700$ nm).
Here you want to check out the theory behind [black body radiation](https://en.wikipedia.org/wiki/Black-body_radiation), which describes - more or less - the intensity of the light at different wavelengths/frequencies at a given temperature.

For an overwiev, look at this figure. The sun's surface temperature ist about 6000 Kelvin, so you can infer, that it's radiation maximum is in the range, which the human eye would perceive as green.
[![Planck's law at different temperatures [from wikipedia]](https://i.sstatic.net/bWXBI.png)](https://i.sstatic.net/bWXBI.png)


---

### Answer by username_5 on 2017-02-16. Score: 4

> So can they emit so much energy in a form of light that is not light? Resulting in a star that cannot be detected using our senses(+ a telescope).

Presumably by "light that is not light" you mean light outside of the visible spectrum. This is why the fields of gamma ray astronomy, x-ray astronomy, ultraviolet astronomy, infrared astronomy, microwave astronomy, and radio astronomy exist. Astronomy is not limited to the visible spectrum, and there's lots of interesting stuff to "see" from using a telescope that senses light outside of the visible spectrum.

The hottest of stars emit most of their radiation as x-rays and ultraviolet light, but they also emit a lot of visible light. These very hot stars are quite large, short-lived, and very rare. The coolest of stars emit most of their radiation in the infrared, but they also emit some light in the visible range. These very cool stars are quite small, very long-lived, and extremely numerous. Their cool temperatures makes them rather hard to see, even in the infrared.

> If so what problems could this star pose? Could this be dark matter?

The very smallest of those cool stars, along with black holes, cool neutron stars, brown dwarfs, and rogue planets collectively form a set of candidates for dark matter called "massive astrophysical compact halo objects", or MACHOs for short. However, theoretical calculations along with observations says that MACHOs can account for at most about 20% of dark matter.


---

### Answer by username_6 on 2017-02-21. Score: 3

I will just point out two things that haven't really been discussed above.

Firstly, stars generally fall into two categories. 'Hot stars' are those that convert mass into energy by nuclear fusion, and 'cold stars' are those that have ended the nuclear burning phase, e.g. white dwarfs and neutron stars. These still have a finite temperature and therefore will emit radiation. As you can imagine, however, since they no longer produce energy, eventually they will become actually cold, hence not radiating (or radiating very little).

The second point to note is that radiations from stars far away are redshifted as they reach us, and they are likely to be attenuated by dust. Therefore, even if the radiation itself when emitted is visible light, by the time it gets to us it may no longer be. However, there are still many other different spectroscopic techniques that will allow us to 'see' it.

> **Comments:**
> **username_7**: WDs and NSs are far from cold, having temperatures of tens, and millions, of Kelvins, respectively. Also, redshift and reddening are sort of beside the point here.  
> **username_6**: @username_7 This is why I said "eventually they will become actually cold". The names "hot stars" and "cold stars" have been defined clearly in my answer; they do not refer to the temperatures, though I admit the terminology may somewhat be misleading. As for redshift - the question asks about detection, so I would not say it is irrelevant. I would admit that reddening is probably slightly beside the point.  
> **username_7**: Okay, sorry, I misunderstood.  

---

