<p>Let's say we have a spherical shell of some material emitting light, much bigger than the star it surrounds. If we look right down the center of the shell, our line of sight takes us through only a small amount of gas - not enough for there to be significant emission. On the other hand, if we look at the nebula near the outer edge, our line of sight takes us through substantially more material:</p>
<p><a href="https://i.sstatic.net/tBpLs.png" rel="noreferrer"><img src="https://i.sstatic.net/tBpLs.png" alt="Model of nebula shell" /></a>
<br>
<sup>Inspired in part by an image in <a href="https://slate.com/technology/2011/07/a-nearly-perfect-circle-in-space.html" rel="noreferrer">this piece by Phil Plait</a>.</sup></p>
<p>If you look through more light-emitting material, you'll see stronger emission than if you look through less material, so in this toy model, we would perceive the edges to have strong emission. So a spherical nebula wouldn't lead to uniform emission throughout our cross-sectional view of it.</p>
<p>This simple spherical model holds for some cases, but certainly not all, particularly because many nebulae aren't spherical (<a href="https://en.wikipedia.org/wiki/Planetary_nebula" rel="noreferrer">Wikipedia</a> makes the claim that only ~20% are truly spherical, which I haven't been able to verify.) For example, we know that <a href="https://www.scifac.hku.hk/kwok/publication_reprint/ring%20nebula.pdf" rel="noreferrer">the Ring Nebula isn't actually a sphere</a> because the central surface brightness is much lower than the projection effect I described above would predict. Something else clearly must be going on. In this case, the discrepancy is solved by the fact that the Ring Nebula is a bipolar nebula, with its axis along our line of sight, <a href="https://www.forbes.com/sites/startswithabang/2018/05/14/the-ring-is-a-lie-ring-nebula-not-a-ring-after-all/" rel="noreferrer">with a dense ring of gas around its equator</a>. Therefore, there's more material near its edges than its center from our perspective, and so we see more emission.</p>

CC BY-SA 4.0
2021-05-12T01:25:37.210Z
@HDE_226868 I knew there was something on the line like that, too, you mentioned there! Definitely helped me to visualize it in my head how the nebula looks. Thanks for your input!
kiaat
CC BY-SA 4.0
2021-05-13T17:58:55.347Z
I did not know this!  Thanks a lot, HDE
Fattie
CC BY-SA 4.0
2021-05-13T19:21:50.667Z
@kiaat I think this is the same geometry that gives soap bubbles a similar appearance.
user3067860
CC BY-SA 4.0
2021-05-15T12:31:25.380Z
@user3067860 Well, that is highly probable, but the appearance of a soap buble isn't that prominent in that regard. What I exactly mean is: The Fresnel equations are contributing mostly to the look of soap bubbles, while the nebula emit light through excitiation. So, soap bubbles reflect light, nebula emit light. But yeah, thinking of the Fresnel-Reflection, the pattern looks the same as the nebula!
kiaat
CC BY-SA 4.0
2021-05-11T22:02:33.903Z
2021-05-11T22:02:33.903Z
HDE 226868
<p>The answer is quite simple: you can also see the other side of the room (or maybe through the fume or haze above your oven) when you create a tasty meal) - even when the space between you and the wall is not empty: it is filled with air. But air is thin enough that you can see through even when it might be thick enough to both, absorbe a bit light and emmit a bit light. A maybe more astronomical analogy: from space you see that Earth has a thin layer through which you can also look :)</p>
<p>In space the densities we talk about are MUCH less then here on Earth in air. Even a planetary nebula is by terrestrial standards still an extremely good vacuum (100.. 100000 particles / cubic centimetre compared to 10 billion billion particles per cubic centimetre of air).</p>

CC BY-SA 4.0
2021-05-12T09:54:43.313Z
Or to put it in relation: Even if our whole solar system up to the Kuiper belt would be filled like a nebula - the absorption would still be a lot less than Earths own atmosphere on a very clear day on the highest mountain.
asdfex
CC BY-SA 4.0
2021-05-11T20:42:35.003Z
2021-05-11T20:42:35.003Z
planetmaker
<h2>Short Answer</h2>
<p>Thin gas in a nebula only absorbs some portion of star light corresponding to a small subset of the overall visible spectrum, according to the corresponding molecular composition of the nebula. For even these limited spectral ranges, the star is still often visible in a nebula since the gas can be too thin to absorb all the energy emitted by the star.</p>
<h2>Long Answer</h2>
<p>The opacity of gas is frequency dependent according to its composition and temperature.  For example, <a href="https://arxiv.org/abs/1311.3201" rel="noreferrer">Monteiro et al.</a> [2013] in <em>Spatially resolved physical and chemical properties of the planetary nebula NGC 3242</em> show a typical spectrum figure from a particular pixel of the <a href="https://en.wikipedia.org/wiki/NGC_3242" rel="noreferrer">NGC 3242</a> (Ghost of Jupiter) nebula (away from the central star):</p>
<p><a href="https://i.sstatic.net/k04fc.png" rel="noreferrer"><img src="https://i.sstatic.net/k04fc.png" alt="enter image description here" /></a></p>
<p>The peaks and troughs in the above figure correspond to particular molecules in the gas nebula emitting and absorbing energy in the visible wavelength ranges of light (from 4000-7000 angstroms).  The biggest peaks around 5000A wavelength are probably <a href="https://en.wikipedia.org/wiki/Doubly_ionized_oxygen" rel="noreferrer">doubly ionized  oxygen</a>.  See: <a href="https://astronomy.stackexchange.com/questions/19656/why-are-o-iii-lines-so-prominent-in-the-spectra-of-emission-nebulae-when-the-amo">Why are O III lines so prominent in the spectra of emission nebulae when the amount of oxygen relative to hydrogen is a million times smaller?</a> for further detail.</p>
<p>Stars, on the other hand, tend to have a more broad emission spectrum consistent with a thermal blackbody.  Here is an example from the <a href="https://cas.sdss.org/dr6/en/proj/basic/spectraltypes/stellarspectra.asp" rel="noreferrer">Sloan Digital Sky Survey</a>:</p>
<p><a href="https://i.sstatic.net/vRK58.png" rel="noreferrer"><img src="https://i.sstatic.net/vRK58.png" alt="enter image description here" /></a></p>
<p><strong>So, along the light frequencies for which the nebular gas is mostly transparent, we can certainly expect to see the central star.</strong></p>
<p><a href="https://arxiv.org/abs/1807.01787" rel="noreferrer">Barria &amp; Kimeswenger</a> [2018], in <em>HST/WFPC2 imaging analysis and Cloudy modelling of the multiple shell planetary nebulae NGC 3242, NGC 6826 and NGC 7662</em> show greyscale figures derived from filtered Hubble Telescope Images at light wavelengths corresponding to various molecular emissions.  Here is an example for NGC 3242 (Ghost of Jupiter):</p>
<p><a href="https://i.sstatic.net/1x8jS.png" rel="noreferrer"><img src="https://i.sstatic.net/1x8jS.png" alt="enter image description here" /></a></p>
<p>The darker pixels in the image correspond to higher energy emissions.  The star varies in energy intensity compared to the gas by frequency, but is still visible above in all four wavelengths.  This is not the case for <a href="https://en.wikipedia.org/wiki/NGC_7662" rel="noreferrer">NGC 7662</a> (Snowball Nebula), which is fully obscured for some narrow chosen spectrum of Oxygen and Nitrogen emissions.</p>
<p><a href="https://i.sstatic.net/Olv6e.png" rel="noreferrer"><img src="https://i.sstatic.net/Olv6e.png" alt="enter image description here" /></a></p>
<p>However, again in the full optical spectrum, the central star for NGC 7662 is clearly visible:</p>
<p><a href="https://i.sstatic.net/EDSnM.jpg" rel="noreferrer"><img src="https://i.sstatic.net/EDSnM.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-05-11T21:35:29.327Z
As suggestion, an educational answer would first define the optical depth and the transmission function (OP's question was "Why can you see through the nebula?"!) before delving into details of the opacity...
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-05-12T01:24:19.013Z
@AtmosphericPrisonEscape It's alright, I still got it. Luckily, I'm a computational visualistics student. I work with computer graphics and ray-tracing and stuff. But I see that others might get a problem with the intuitive approach of Connor_Garcia when looking into this question. Thanks for your input!
kiaat
CC BY-SA 4.0
2021-05-11T21:31:30.487Z
2021-05-11T21:52:16.543Z
2021-05-11T21:52:16.543Z
Connor Garcia
<p>I recently read the book &quot;An Introduction to Planetary Nebulae&quot; by Jason J. Nishiyama. Although I'm not an astronomy student, I could at least understand the written texts and less the physical formulas. So, I got a nice basis knowledge now -- or that's what I like to think.</p>
<p>Now, I browsed through some pictures of planetary nebulae. Mostly from Hubble and such. I know that they synthesize their images by combining infrared + visible light + uv-light, but I still can't comprehend why these telescopes are able to see INSIDE a nebulae when the shape of mentioned nebulae are spherical? For example, the planetary nebula: NGC 3242 (Ghost of Jupiter); NGC 7293 (Helix Nebula); NGC 6543 (Cat's Eye).</p>
<p><a href="https://i.sstatic.net/t6TqW.png" rel="noreferrer"><img src="https://i.sstatic.net/t6TqW.png" alt="NGC 3242 (Ghost of Jupiter)" /></a> <a href="https://i.sstatic.net/lYNe3.jpg" rel="noreferrer"><img src="https://i.sstatic.net/lYNe3.jpg" alt="NGC 7293 (Helix Nebula)" /></a> <a href="https://i.sstatic.net/3gmxC.jpg" rel="noreferrer"><img src="https://i.sstatic.net/3gmxC.jpg" alt="NGC 6543 (Cat's Eye)" /></a></p>
<p>You can see the exposed white dwarf. And it looks like a thin shell envelopes that star, while the space between the star and the shell is scarce, like see-through. But wouldn't that image imply that the planetary nebula is disk-shaped? If the nebulae is spherical, then shouldn't there be &quot;more nebulae&quot; between the star and the shell? Or shouldn't the inner space be tinted in the color of the nebulae, because we look at a white dwarf that's behind a shell? I've drawn my problem with Paint too, sry for my artistic skills: Pattern A is what I thought it should be; Pattern B is what I think it looks like.</p>
<p><a href="https://i.sstatic.net/WB4OB.png" rel="noreferrer"><img src="https://i.sstatic.net/WB4OB.png" alt="enter image description here" /></a> <a href="https://i.sstatic.net/enFYL.png" rel="noreferrer"><img src="https://i.sstatic.net/enFYL.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-05-13T17:30:39.843Z
TLDR; Nebulas don't look like what you see in movies and on TV
Jason Goemaat
CC BY-SA 4.0
2021-05-13T17:58:09.640Z
the question in your title is quite different from the question "why is it ring-looking?"
Fattie
CC BY-SA 4.0
2021-05-13T18:06:16.943Z
Ground-based telescopes are not really different in this regard.
fraxinus
CC BY-SA 4.0
2021-05-15T12:33:40.653Z
@JasonGoemaat I clearly stated that the images are synthesized. But that wasn't my question.
kiaat
CC BY-SA 4.0
2021-05-15T12:34:12.613Z
@Fattie "Why do nebula look elliptical?" wasn't my question.
kiaat
CC BY-SA 4.0
2021-05-15T12:36:37.880Z
@fraxinus The images I took were from Space Telescopes. I would state two different things, if I took Space Telescope images and then talk about Ground-based one, in the context of my question.
kiaat
<observational-astronomy><nebula>
Why can space telescopes see through a planetary nebula?
kiaat
# Why can space telescopes see through a planetary nebula?

(Asked by: username_0 on 2021-05-11)

I recently read the book "An Introduction to Planetary Nebulae" by username_1 J. Nishiyama. Although I'm not an astronomy student, I could at least understand the written texts and less the physical formulas. So, I got a nice basis knowledge now -- or that's what I like to think.

Now, I browsed through some pictures of planetary nebulae. Mostly from Hubble and such. I know that they synthesize their images by combining infrared + visible light + uv-light, but I still can't comprehend why these telescopes are able to see INSIDE a nebulae when the shape of mentioned nebulae are spherical? For example, the planetary nebula: NGC 3242 (Ghost of Jupiter); NGC 7293 (Helix Nebula); NGC 6543 (Cat's Eye).

[![NGC 3242 (Ghost of Jupiter)](https://i.sstatic.net/t6TqW.png)](https://i.sstatic.net/t6TqW.png) [![NGC 7293 (Helix Nebula)](https://i.sstatic.net/lYNe3.jpg)](https://i.sstatic.net/lYNe3.jpg) [![NGC 6543 (Cat's Eye)](https://i.sstatic.net/3gmxC.jpg)](https://i.sstatic.net/3gmxC.jpg)

You can see the exposed white dwarf. And it looks like a thin shell envelopes that star, while the space between the star and the shell is scarce, like see-through. But wouldn't that image imply that the planetary nebula is disk-shaped? If the nebulae is spherical, then shouldn't there be "more nebulae" between the star and the shell? Or shouldn't the inner space be tinted in the color of the nebulae, because we look at a white dwarf that's behind a shell? I've drawn my problem with Paint too, sry for my artistic skills: Pattern A is what I thought it should be; Pattern B is what I think it looks like.

[![enter image description here](https://i.sstatic.net/WB4OB.png)](https://i.sstatic.net/WB4OB.png) [![enter image description here](https://i.sstatic.net/enFYL.png)](https://i.sstatic.net/enFYL.png)

**Tags:** observational-astronomy, nebula

> **Comments:**
> **username_1**: TLDR; Nebulas don't look like what you see in movies and on TV  
> **username_2**: the question in your title is quite different from the question "why is it ring-looking?"  
> **username_3**: Ground-based telescopes are not really different in this regard.  
> **username_0**: @username_1 I clearly stated that the images are synthesized. But that wasn't my question.  
> **username_0**: @username_2 "Why do nebula look elliptical?" wasn't my question.  
> **username_0**: @username_3 The images I took were from Space Telescopes. I would state two different things, if I took Space Telescope images and then talk about Ground-based one, in the context of my question.  

---

## 3 answers

### Answer by username_8 on 2021-05-11. Score: 20 | Accepted answer

## Short Answer

Thin gas in a nebula only absorbs some portion of star light corresponding to a small subset of the overall visible spectrum, according to the corresponding molecular composition of the nebula. For even these limited spectral ranges, the star is still often visible in a nebula since the gas can be too thin to absorb all the energy emitted by the star.

## Long Answer

The opacity of gas is frequency dependent according to its composition and temperature. For example, [Monteiro et al.](https://arxiv.org/abs/1311.3201) [2013] in *Spatially resolved physical and chemical properties of the planetary nebula NGC 3242* show a typical spectrum figure from a particular pixel of the [NGC 3242](https://en.wikipedia.org/wiki/NGC_3242) (Ghost of Jupiter) nebula (away from the central star):

[![enter image description here](https://i.sstatic.net/k04fc.png)](https://i.sstatic.net/k04fc.png)

The peaks and troughs in the above figure correspond to particular molecules in the gas nebula emitting and absorbing energy in the visible wavelength ranges of light (from 4000-7000 angstroms). The biggest peaks around 5000A wavelength are probably [doubly ionized oxygen](https://en.wikipedia.org/wiki/Doubly_ionized_oxygen). See: [Why are O III lines so prominent in the spectra of emission nebulae when the amount of oxygen relative to hydrogen is a million times smaller?](https://astronomy.stackexchange.com/questions/19656/why-are-o-iii-lines-so-prominent-in-the-spectra-of-emission-nebulae-when-the-amo) for further detail.

Stars, on the other hand, tend to have a more broad emission spectrum consistent with a thermal blackbody. Here is an example from the [Sloan Digital Sky Survey](https://cas.sdss.org/dr6/en/proj/basic/spectraltypes/stellarspectra.asp):

[![enter image description here](https://i.sstatic.net/vRK58.png)](https://i.sstatic.net/vRK58.png)

**So, along the light frequencies for which the nebular gas is mostly transparent, we can certainly expect to see the central star.**

[Barria & Kimeswenger](https://arxiv.org/abs/1807.01787) [2018], in *HST/WFPC2 imaging analysis and Cloudy modelling of the multiple shell planetary nebulae NGC 3242, NGC 6826 and NGC 7662* show greyscale figures derived from filtered Hubble Telescope Images at light wavelengths corresponding to various molecular emissions. Here is an example for NGC 3242 (Ghost of Jupiter):

[![enter image description here](https://i.sstatic.net/1x8jS.png)](https://i.sstatic.net/1x8jS.png)

The darker pixels in the image correspond to higher energy emissions. The star varies in energy intensity compared to the gas by frequency, but is still visible above in all four wavelengths. This is not the case for [NGC 7662](https://en.wikipedia.org/wiki/NGC_7662) (Snowball Nebula), which is fully obscured for some narrow chosen spectrum of Oxygen and Nitrogen emissions.

[![enter image description here](https://i.sstatic.net/Olv6e.png)](https://i.sstatic.net/Olv6e.png)

However, again in the full optical spectrum, the central star for NGC 7662 is clearly visible:

[![enter image description here](https://i.sstatic.net/EDSnM.jpg)](https://i.sstatic.net/EDSnM.jpg)

> **Comments:**
> **username_9**: As suggestion, an educational answer would first define the optical depth and the transmission function (OP's question was "Why can you see through the nebula?"!) before delving into details of the opacity...  
> **username_0**: @username_9 It's alright, I still got it. Luckily, I'm a computational visualistics student. I work with computer graphics and ray-tracing and stuff. But I see that others might get a problem with the intuitive approach of username_8 when looking into this question. Thanks for your input!  

---

### Answer by username_4 on 2021-05-11. Score: 32

Let's say we have a spherical shell of some material emitting light, much bigger than the star it surrounds. If we look right down the center of the shell, our line of sight takes us through only a small amount of gas - not enough for there to be significant emission. On the other hand, if we look at the nebula near the outer edge, our line of sight takes us through substantially more material:

[![Model of nebula shell](https://i.sstatic.net/tBpLs.png)](https://i.sstatic.net/tBpLs.png)
  
Inspired in part by an image in [this piece by Phil Plait](https://slate.com/technology/2011/07/a-nearly-perfect-circle-in-space.html).

If you look through more light-emitting material, you'll see stronger emission than if you look through less material, so in this toy model, we would perceive the edges to have strong emission. So a spherical nebula wouldn't lead to uniform emission throughout our cross-sectional view of it.

This simple spherical model holds for some cases, but certainly not all, particularly because many nebulae aren't spherical ([Wikipedia](https://en.wikipedia.org/wiki/Planetary_nebula) makes the claim that only ~20% are truly spherical, which I haven't been able to verify.) For example, we know that [the Ring Nebula isn't actually a sphere](https://www.scifac.hku.hk/kwok/publication_reprint/ring%20nebula.pdf) because the central surface brightness is much lower than the projection effect I described above would predict. Something else clearly must be going on. In this case, the discrepancy is solved by the fact that the Ring Nebula is a bipolar nebula, with its axis along our line of sight, [with a dense ring of gas around its equator](https://www.forbes.com/sites/startswithabang/2018/05/14/the-ring-is-a-lie-ring-nebula-not-a-ring-after-all/). Therefore, there's more material near its edges than its center from our perspective, and so we see more emission.

> **Comments:**
> **username_0**: @username_4 I knew there was something on the line like that, too, you mentioned there! Definitely helped me to visualize it in my head how the nebula looks. Thanks for your input!  
> **username_2**: I did not know this!  Thanks a lot, HDE  
> **username_5**: @username_0 I think this is the same geometry that gives soap bubbles a similar appearance.  
> **username_0**: @username_5 Well, that is highly probable, but the appearance of a soap buble isn't that prominent in that regard. What I exactly mean is: The Fresnel equations are contributing mostly to the look of soap bubbles, while the nebula emit light through excitiation. So, soap bubbles reflect light, nebula emit light. But yeah, thinking of the Fresnel-Reflection, the pattern looks the same as the nebula!  

---

### Answer by username_6 on 2021-05-11. Score: 8

The answer is quite simple: you can also see the other side of the room (or maybe through the fume or haze above your oven) when you create a tasty meal) - even when the space between you and the wall is not empty: it is filled with air. But air is thin enough that you can see through even when it might be thick enough to both, absorbe a bit light and emmit a bit light. A maybe more astronomical analogy: from space you see that Earth has a thin layer through which you can also look :)

In space the densities we talk about are MUCH less then here on Earth in air. Even a planetary nebula is by terrestrial standards still an extremely good vacuum (100.. 100000 particles / cubic centimetre compared to 10 billion billion particles per cubic centimetre of air).

> **Comments:**
> **username_7**: Or to put it in relation: Even if our whole solar system up to the Kuiper belt would be filled like a nebula - the absorption would still be a lot less than Earths own atmosphere on a very clear day on the highest mountain.  

---

