APOD: Zeta Oph: Runaway Star (2011 Feb 04)

[Devil Particle]

"Why should science encourage the mistaken idea that blue is a cooler color than red?"

It's not just the scientific community that uses this convention. Fluorescent, Metal Halide and LED light sources (and sometimes even incandescent) are given color temperature ratings. A "cool white" light source is 4100K while a "warm" light source is 3000K or lower. A "daylight" fluorescent bulb has a 6500K rating.

(K = Kelvin)

[neufer]

when it comes to the Sun's "true color", the orange version of the Sun is only marginally better than the blue one.
The main difference would seem to be many people's lower tolerance for "false blue color" than for "false red or orange color".

One shouldn't look at the Sun.
[youtube]http://www.youtube.com/watch?v=G8JymP6h ... r_embedded[/youtube]

[Ann]

Noel,

And no, you didn't make me think about it, other than, "Oh, carp; not again." (The whole color thing...) I know some people don't like colors being used to demonstrate science rather than something else, because they keep saying so, but to what end, I really don't know. Saying so isn't going to stop the science from being done (thank goodness).

There are so many other things in the world to be offended about; I just can't see being offended because scientific images are used for science and thus are colored accordingly. I don't get that at all; obviously, this is my failing.

Since the "some people" that Owlice mentioned probably refers to me, let me point out that I haven't protested against Friday's APOD. I, too, agree that it shows something that can't be seen in visible light, and I, too, agree that it must be shown in false color (or mapped color, or assigned color, or whatever) to be visible as a picture to us at all. I haven't protested against the picture of Zeta Ophiuchi. I don't exactly love the image of Zeta Ophiuchi, because it is hard for me to absolutely love a false-color picture, but yes, I do like it. Quite a lot. Moreover, if I had belonged to the team who produced this picture, I wouldn't have asked them to change the coloring of it.

But there are times when I think that false color is being used in a way that I really disapprove of. Look at this famous picture of the "Pillars of Creation" in the Eagle Nebula, Messier 16:

Click to view full size image

Yes, this is an amazing picture. I, too, was incredibly impressed when I saw it, and when I saw all the fantastic structural detail in the pillars. But there is one thing which I strongly disapprove of, and that is the pink color of all the stars. I find that choice of color for the stars extremely jarring, in view of the fact that most of the stars here are very hot and blue, although of course their apparent color has been reddened by dust. But even their dust-reddened color is much bluer than the Sun.

I think the reason for the pink color of the stars in the picture is that Hubble imaged the pillars in the light of oxygen, which was colored blue, hydrogen, which was colored green, and sulphur, which was colored red. But the sensitivity of the detection of sulphur was apparently much greater than the sensitivity of the detection of oxygen and hydrogen. So when Hubble looked at the stars, it simply saw a lot more sulphur than hydrogen and oxygen in them, and the stars came out pink.

I'm happy to say that the pink color of hot stars didn't really catch on. Several people regularly post images of nebulae in the Hubble palette here at Starship Asterisk, but many people make an effort not to make their stars come out pink. I'm glad. Please note, too, that all the stars are colored blue in Friday's APOD. So are all the stars in Zeta Ophiuchi's vicinity hot and blue, then? No, certainly not, since hot Zeta Ophiuchi itself is a runaway star and not at all surrounded by its hot siblings. Neverteless, compared with the temperature of the surrounding dust, all the stars in the picture are hot, and it is right and fitting that they should be colored blue in this image.

Ann

[owlice]

But there are times when I think that false color is being used in a way that I really disapprove of.

As I said,

they keep saying so, but to what end, I really don't know.... I don't get that at all; obviously, this is my failing.

[Chris Peterson]

But there are times when I think that false color is being used in a way that I really disapprove of. Look at this famous picture of the "Pillars of Creation" in the Eagle Nebula, Messier 16:

Yes, this is an amazing picture. I, too, was incredibly impressed when I saw it, and when I saw all the fantastic structural detail in the pillars. But there is one thing which I strongly disapprove of, and that is the pink color of all the stars. I find that choice of color for the stars extremely jarring, in view of the fact that most of the stars here are very hot and blue, although of course their apparent color has been reddened by dust. But even their dust-reddened color is much bluer than the Sun.

It wasn't a "choice". The stars are not what is interesting in the image. The image intent is to show the nebula; the star colors are simply a consequence of the color mapping used in the nebula.

I think the reason for the pink color of the stars in the picture is that Hubble imaged the pillars in the light of oxygen, which was colored blue, hydrogen, which was colored green, and sulphur, which was colored red. But the sensitivity of the detection of sulphur was apparently much greater than the sensitivity of the detection of oxygen and hydrogen. So when Hubble looked at the stars, it simply saw a lot more sulphur than hydrogen and oxygen in them, and the stars came out pink.

No, that's not quite right. Stars are basically white, continuum sources. Yes, we see a bit of color because they aren't perfectly white, but nevertheless, compared to anything like an emission line, we can consider them white. So when you do narrow band imaging, all stars will look practically the same color, because most continuum thermal sources will have a very similar intensity balance between the different filters.

[NoelC]

So when you do narrow band imaging, all stars will look practically the same color, because most continuum thermal sources will have a very similar intensity balance between the different filters.

That seems to fly in the face of the observation of many "Hubble" palette images. We simply see many of them with pink/magenta looking stars.

Why?

I've always assumed that the sulfur emissions are generally much LESS intense than hydrogen and oxygen, causing the astrophotographers to take longer exposures and/or stretch the data further in order to not have a substantially blue/green looking nebula and present a more "balanced" looking image.

I'm with Ann, though, I've never liked a "Hubble" palette image as much as visual spectrum work. Artistically they may be very pretty, but they just feel WRONG.

-Noel

[neufer]

So when you do narrow band imaging, all stars will look practically the same color, because most continuum thermal sources will have a very similar intensity balance between the different filters.

That seems to fly in the face of the observation of many "Hubble" palette images. We simply see many of them with pink/magenta looking stars.

Why?

Bright stars tend to have Balmer absorption lines.

Other things (e.g., H II regions, planetary nebulae, spiral and irregular galaxies, and AGNs) have Balmer emission lines.

Since the red H-alpha Balmer line contributes the green color to the "Hubble" palette bright stars tend to end up being green deficient or purple.

<<Balmer lines can appear as absorption or emission lines in a spectrum, depending on the nature of the object observed. In stars, the Balmer lines are usually seen in absorption, and they are "strongest" in stars with a surface temperature of about 10,000 kelvin (spectral type A). In the spectra of most spiral and irregular galaxies, AGNs, H II regions and planetary nebulae, the Balmer lines are emission lines.>>

[Chris Peterson]

So when you do narrow band imaging, all stars will look practically the same color, because most continuum thermal sources will have a very similar intensity balance between the different filters.

That seems to fly in the face of the observation of many "Hubble" palette images. We simply see many of them with pink/magenta looking stars.

Why?

I'm a little confused here. I said that in Hubble palette images, stars look pink. You said that flies in the face of observation, they appear pink?

I think the reason that most stars have a similar color is largely unrelated to any emission or absorption at the narrow bands chosen. I think it is mainly just a consequence of where the three wavelength samples are coming from on the similar blackbody spectrum shared by most stars.

[dougettinger]

Zeta Oph is traveling at 24Kilometers per SECOND.

Where is it going and why does it not hit any thing?

Its "solar wind" is hitting something..the gas & dust shown in the APOD.

You are traveling at 30Kilometers per SECOND in orbit around the sun.

Where are you going and why don't you hit any thing

Zeta Oph's proper motion according to this APOD is 24 km/s. I am assuming its combined vector velocity is closer to the velocity of the Sun which is over 200 km/s. Can the bow or shock wave be part of the supernova remnant that propelled this star into some anomalous direction and not the solar wind from the star itself ? Of course, the bow wave may be a combination of both.

Doug Ettinger, Pittsburgh, PA 02/06/11

[zloq]

I think the reason that most stars have a similar color is largely unrelated to any emission or absorption at the narrow bands chosen. I think it is mainly just a consequence of where the three wavelength samples are coming from on the similar blackbody spectrum shared by most stars.

The magenta is just due to scaling the three channels differently so the nebulosity is rich in color. Since Ha dominates the emission, and since it is shown green, it is suppressed relative to the red (Sii) and blue (Oiii). The magenta is just a consequence of aesthetically scaling the channels so the image looks most impressive.

The choice of colors in the Hubble palette was somewhat arbitrarily set so that RGB are in the same order as the three wavelengths, Sii (Red), Ha (shorter wavelength red) and Oiii (green). I think the first time this convention was used was in the "Pillars" image around 1995.

Professional images in the Hubble palette usually allow arbitrary linear scaling of the channels to make the picture look good, and sometimes standard transforms such as log() are applied to enhance the dynamic range - but there is rarely any arbitrary manipulation of the channels. Amateurs, on the other hand, tend to go nuts manipulating the channels to bring out colors and detail with nonlinear "curves."

Either way - these are just aesthetic renditions of the data and the scientific value isn't in the depicted "color" but in the quantitative photometric values for each emission line.

zloq

[NoelC]

I'm a little confused here. I said that in Hubble palette images, stars look pink. You said that flies in the face of observation, they appear pink?

I'm sorry; I read you as saying they should be white:

No, that's not quite right. Stars are basically white, continuum sources. Yes, we see a bit of color because they aren't perfectly white, but nevertheless, compared to anything like an emission line, we can consider them white.

-Noel

[NoelC]

The magenta is just due to scaling the three channels differently so the nebulosity is rich in color. Since Ha dominates the emission, and since it is shown green, it is suppressed relative to the red (Sii) and blue (Oiii). The magenta is just a consequence of aesthetically scaling the channels so the image looks most impressive.

Thank you. That's what I have always thought.

-Noel

[neufer]

<<824 Anastasia is a main belt asteroid orbiting the Sun. It is approximately 34.14 km in diameter. It was discovered on March 25, 1916 by Grigory Neujmin at Simeiz Observatory in Ukraine. It is named in memory of Anastasia Semenoff, an acquaintance of the discoverer.

On April 6, 2010, 824 Anastasia had the distinction of causing the brightest asteroid occultation ever predicted for North America for an asteroid of its size. The asteroid occulted the naked-eye star ζ Ophiuchi over a path stretching from the Los Angeles area to Edmonton, Alberta.>>

[Chris Peterson]

I'm sorry; I read you as saying they should be white:

White to the eyes, not to an imager. That's because they are thermal sources, with blackbody curves that in almost all cases are close to white. We only see a slight deviation from white with our eyes- a hint toward blue or a hint toward red. "White" is a color, which is something that is really only meaningful to our eyes. "White" doesn't mean that if you make a trichromic image that the three selected wavelengths will be the same intensity. That's why the stars come out with a magenta bias- the natural balance of the stellar continuum convolved by whatever weighting is used to balance the three colors in the emission zones just happens to come out that way.

My real point was that I don't think it has anything to do with absorption or emission lines, but is related to the sampling of the thermal continuum.

[neufer]

My real point was that I don't think it has anything to do with absorption or emission lines, but is related to the sampling of the thermal continuum.

And my real point was that I think that the stars not being white has everything to do with absorption & emission lines.

Balmer absorption lines

---

Balmer Jump in the spectra of the hot star alpha Lyra (Vega), observed by Barbier and Chalonge (1939).

---

[Chris Peterson]

And my real point was that I think that the stars not being white has everything to do with absorption & emission lines.

I disagree. Both absorption and emission lines are only a few hundredths of a nanometer wide. Narrow band filters have notches that are 5-10nm wide. With nebular emission sources, the very narrow emission line dominates, because it is usually much stronger than reflected or scattered continuum light. But for stellar sources, the continuum energy around the emission/absorption line dominates, by several orders of magnitude (that's why you can't see any Ha structure at all looking at the Sun with one of these filters). And what does the continuum source look like? For all but the coolest stars, the three filtered wavelengths lie on the long side of the blackbody peak, along a section with a steep negative slope. The OIII filter will pass a much higher intensity than the Ha filter or SII filter. So for all stars, when mapped to the Hubble palette, you'll get strong blue, with an approximately equal (but much smaller) mix of green and red. This is most of the way towards the star color seen in most Hubble palette images. You get the rest of the way when you normalize the three color channels to optimize the contrast.

[zloq]

Narrow band filters have notches that are 5-10nm wide. With nebular emission sources, the very narrow emission line dominates, because it is usually much stronger than reflected or scattered continuum light. But for stellar sources, the continuum energy around the emission/absorption line dominates, by several orders of magnitude (that's why you can't see any Ha structure at all looking at the Sun with one of these filters). And what does the continuum source look like? For all but the coolest stars, the three filtered wavelengths lie on the long side of the blackbody peak, along a section with a steep negative slope. The OIII filter will pass a much higher intensity than the Ha filter or SII filter. So for all stars, when mapped to the Hubble palette, you'll get strong blue, with an approximately equal (but much smaller) mix of green and red. This is most of the way towards the star color seen in most Hubble palette images. You get the rest of the way when you normalize the three color channels to optimize the contrast.

I don't know why you two are so insistent that a false color rendition of narrow band images captures something fundamental about star color - either as blackbodies or as objects with absorption bands. You seem to think the magenta of stars is largely inherent in the imaging process - and is only helped by "normalizing" the three color channels. In fact, based on the filters used, the "intensity" in the images is very different from what you predict.

J. Hester's article on the image can be found in AJ 111 #6, 1996. p. 2349. Equal exposures were made with three filters, and the filters have different bandpass. The Ha filter is 2.2nm bandpass, the Oiii is 2.7, and the Sii is 4.7. For a solar blackbody spectrum at 5800K, if you use a wavelength dependent (not frequency) form of the blackbody curve you can then multiply by the bandpass to get an estimate of power delivered through each filter. Sii (red) has greatest power, while Oiii (blue) is next at 68%, and Ha is least at 47%. This is very different from your prediction, which ignored bandpass of the filters. In fact, the change in blackbody power from 500nm to 660 is only about 14% (power per wavelength, not frequency) and would be a small factor compared to bandpass and camera sensitivity.

Hester does not provide details on the color composite image, but the gray scale images he shows have sqrt() applied to increase the dynamic range. My guess is that each exposure had sqrt() applied, and then the three channels were combined with scaling to make it look colorful. Although the power from stars would be weakest in Ha (green channel), as I said, the power from the nebulosity would be dominated by Ha, so that it would be combined with less weighting compared to Sii and Oiii - resulting in magenta for objects like stars that are relatively "white." When you combine this scaling with the original square roots - and the different filter bandpasses - the result is very far from a pure RGB image that captures inherent weighting of the objects - whether stars or nebulae.

As I said - the star color is what you get when you combine the channels, which are dominated by Ha colored green, with weighting that makes the nebulosity look most colorful to fill the gamut.

zloq

[Chris Peterson]

I don't know why you two are so insistent that a false color rendition of narrow band images captures something fundamental about star color - either as blackbodies or as objects with absorption bands. You seem to think the magenta of stars is largely inherent in the imaging process - and is only helped by "normalizing" the three color channels. In fact, based on the filters used, the "intensity" in the images is very different from what you predict...

Of course, the color of the stars in a false color image does tell us something about the stars. But otherwise, your analysis is very similar to mine, and comes to the same conclusion, so I guess we are largely in agreement.

[zloq]

But otherwise, your analysis is very similar to mine, and comes to the same conclusion, so I guess we are largely in agreement.

Umm - no. You said the filters are 5-10nm wide... and you didn't consider the different bandpass as being very important in how much continuum gets through. I pointed out the filters are much narrower and very different widths for each channel - and that has a huge effect on the received power. You imply the "steep slope" of the blackbody curve "gets most of the way" to the magenta color. I point out that the difference in blackbody power is only 14% - and the different bandpass of the filters completely dominates since it is a factor of >2. You said that Oiii would dominate the channels - I point out that Sii is by far the strongest for stars due to bandwidth, while Ha would be strongest for typical Nebulae due to emission strength.

So I think the only thing we agree on is that stars in the final image are magenta - and my analysis is based on actual values relevant to the situation. It's important to maintain a sense of proportion when several factors are playing a role.

In simple terms - I would just say the stars are effectively white and the nebula is dominated by Ha. Therefore the Ha is turned down in the nebula during the Photoshop color composition stage, which leaves the stars mostly red and blue - i.e. magenta.

zloq

[Chris Peterson]

Umm - no. You said the filters are 5-10nm wide... and you didn't consider the different bandpass as being very important in how much continuum gets through. I pointed out the filters are much narrower and very different widths for each channel - and that has a huge effect on the received power.

The numbers I gave are for the filters typically used by imagers, and which often show up in APODs. You looked up the actual bandpasses of the HST filters used. They are a little bit narrower, but the difference is insignificant in considering Art's suggestion that the reduced red signal is caused by Ha absorption. The important point here is that the width of the spectral lines is very, very small compared with the width of the filters- whether 10nm, 5nm, or 3nm wide. What the filters are sampling is continuum signal. And I agree with you completely that if the filters have different bandpasses, that is another factor in determining the relative intensities of the three sampled blackbody segments.

You imply the "steep slope" of the blackbody curve "gets most of the way" to the magenta color. I point out that the difference in blackbody power is only 14% - and the different bandpass of the filters completely dominates since it is a factor of >2. You said that Oiii would dominate the channels - I point out that Sii is by far the strongest for stars due to bandwidth, while Ha would be strongest for typical Nebulae due to emission strength.

Our arguments are similar, just the conclusions vary. One major reason for that is that in a star forming region like this, most stars are closer to 10,000K than 5,000K. That has the effect of greatly increasing the OIII intensity with respect to the other two. Take the 10,000K blackbody and look at the filter intensities, convolved by the filter widths; take the square root of the values to allow for the transfer curve used, and you get almost exactly the R, G, and B values that can be measured off the published image.

The Hubble palette is mostly used in active, star-forming regions. Because most of those stars are hot, the shape of the blackbody, combined with the filter characteristics, produces magenta stars. In general, when we see magenta stars in an image using the Hubble palette, we know we're looking at hot stars.

[zloq]

Take the 10,000K blackbody and look at the filter intensities, convolved by the filter widths; take the square root of the values to allow for the transfer curve used, and you get almost exactly the R, G, and B values that can be measured off the published image.

The Hubble palette is mostly used in active, star-forming regions. Because most of those stars are hot, the shape of the blackbody, combined with the filter characteristics, produces magenta stars. In general, when we see magenta stars in an image using the Hubble palette, we know we're looking at hot stars.

You are missing many key issues - and I give up.

For others reading - once again - the Ha in nebulosity is often very strong when doing narrow band imaging, so the green channel is suppressed by the "artist" to make the nebula colors look nice in Photoshop - and the white stars end up magenta. Those stars could be unrelated foreground stars at whatever temperature - they all tend to look magenta because the suppression of the Ha channel is so severe as a result of this aesthetic choice. The stars are effectively white, and the nebula is effectively "green" (Ha-strong) - so the magenta of the stars is more a statement that the nebula is a strong emitter of Ha than anything about the stars.

zloq

[Chris Peterson]

For others reading - once again - the Ha in nebulosity is often very strong when doing narrow band imaging, so the green channel is suppressed by the "artist" to make the nebula colors look nice in Photoshop - and the white stars end up magenta.

That may be one factor (which I identified in my first post on the subject). But the more important point is that you don't need to do this to get magenta stars. Any hot star imaged with the HST using the Hubble palette will be magenta, just combining the raw data into a color image, with no additional processing at all.

[Ann]

zloq wrote:
J. Hester's article on the image can be found in AJ 111 #6, 1996. p. 2349. Equal exposures were made with three filters, and the filters have different bandpass. The Ha filter is 2.2nm bandpass, the Oiii is 2.7, and the Sii is 4.7. For a solar blackbody spectrum at 5800K, if you use a wavelength dependent (not frequency) form of the blackbody curve you can then multiply by the bandpass to get an estimate of power delivered through each filter. Sii (red) has greatest power, while Oiii (blue) is next at 68%, and Ha is least at 47%. This is very different from your prediction, which ignored bandpass of the filters. In fact, the change in blackbody power from 500nm to 660 is only about 14% (power per wavelength, not frequency) and would be a small factor compared to bandpass and camera sensitivity.

zloq wrote:
For others reading - once again - the Ha in nebulosity is often very strong when doing narrow band imaging, so the green channel is suppressed by the "artist" to make the nebula colors look nice in Photoshop - and the white stars end up magenta. Those stars could be unrelated foreground stars at whatever temperature - they all tend to look magenta because the suppression of the Ha channel is so severe as a result of this aesthetic choice.

zloq, thank you very much. This is exactly the point I was trying to make, although you provided the detailed information that I lacked.

This is my point. The Hubble palette produces magenta stars. The main reason for this is that the SII filter samples a wider range of wavelengths than either the OIII or Ha filters. The way the filters work is the reason for the magenta color of the stars in the Hubble palette, and moreover, the stars will look magenta whatever their intrinsic color may be. Therefore the Hubble palette provides no information about the stars themselves, other than that they are, in fact, stars. Also, to me, the magenta color is jarring.

Ann

[Ann]

To show an example of what I mean with star colors as shown in RGB and star colors in the Hubble palette, let's compare two images of the Rosette Nebula:

Click to view full size image

This is an RGB image of the Rosette Nebula. There is a "double row of stars" seen in front of the center of the Rosette. The one on the lower left in this "double row of stars" is noticably more orange than the others. And indeed, this star, HD46241, is a K-type foreground star. The blue stars all belong to the cluster of hot stars ionizing the nebula.

In this Hubble palette image, the orange foreground star and the blue stars inside the nebula all seem to be more or less the same color:



In all fairness I must point out, however, that not all RGB images of the Rosette nebula manage to capture the color difference between the orange and the blue stars.

Ann

[Chris Peterson]

This is my point. The Hubble palette produces magenta stars.

Yes, but "magenta" describes a fairly wide color range. With the Hubble palette, there is a significant variation in star color, despite the initial impression that they are all the same. This is readily obvious if you study the images closely, or look at how the RGB ratios vary from star to star (using a graphics tool).

The main reason for this is that the SII filter samples a wider range of wavelengths than either the OIII or Ha filters. The way the filters work is the reason for the magenta color of the stars in the Hubble palette, and moreover, the stars will look magenta whatever their intrinsic color may be.

This is where the analysis goes a bit wrong. The reason the stars look magenta is because red and blue are the dominant contributors. That happens for two different reasons: blue maps to OIII, which for a hot star is intrinsically about twice as intense as either Ha or SiII. Even though SiII is half as intense, its filter (with the HST- this will not be the case for all imagers) is twice as wide. So the end result is that a hot star will have about the same intensity for OIII (blue) and SiII (red). That would be pure magenta, of course, but the narrow Ha filter also produces some signal, so the magenta is contaminated by a green signal which is about half as strong as the red or blue.

Because of all this, there is a strong dependence between stellar temperature and the color in the image. A cooler star will have less OIII compared with the two red wavelengths, and the signal from SiII will increasingly dominate, pushing the color towards red. And in fact, that's just what you see with Hubble palette images made from HST data- stars range from magenta to red. That happens to be a color axis that our eyes don't discriminate very well, so it isn't all that apparent without some close examination. But once you realize this is going on, you can see that there is a good deal of information about the intrinsic color of the star that can be extracted visually, and even more if you measure the RGB ratios.

It is worth noting that most amateurs who work with narrow band imaging use matched filter sets that provide nearly the same bandwidth for all three emission lines. So in those images, the SiII contribution isn't doubled, but is as low as the Ha contribution. The result is that stars are much less magenta, with their intrinsic color mapping to a blue-magenta axis instead. And there is a final complication, as zloq pointed out, that the imager may artificially change the color balance between channels for aesthetic reasons.