Difference between revisions of "Another CMD explanation attempt"

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(Created page with "''This comes from an email conversation; I am copying it here because I liked the logic, and want to capture it, but I don't have time at just this instant to integrate it or...")
 
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Remember this video:
 
Remember this video:
http://www.youtube.com/watch?v=mjfSy2kHWqQ&list=UUQfN0BTwaSZ6ABsQcieCpdA&index=8&feature=plcp
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https://www.youtube.com/watch?v=VHMTB95b5us&list=PLjCjDYabTFm9b9jQd4hcZPAnFWYsIjs2D&index=7&t=0s
 
this is a series of blackbodies (stand-ins for stars) with the locations of the bands marked. B-V can be mapped to temperature because those bands are near the peak of the blackbody energy distribution, so B-V changes rapidly as temperature changes. K should mostly still see photosphere in many cases, though it too can be contaminated by circumstellar emission. Longer wavelengths will sample more of the disk.  
 
this is a series of blackbodies (stand-ins for stars) with the locations of the bands marked. B-V can be mapped to temperature because those bands are near the peak of the blackbody energy distribution, so B-V changes rapidly as temperature changes. K should mostly still see photosphere in many cases, though it too can be contaminated by circumstellar emission. Longer wavelengths will sample more of the disk.  
  

Latest revision as of 19:46, 11 August 2020

This comes from an email conversation; I am copying it here because I liked the logic, and want to capture it, but I don't have time at just this instant to integrate it or find it a better home.

Think of a red star. What we mean by "red" is that the star is brighter in red than other colors of light. (Usually we restrict this to visible colors, but you now can extend this to other wavelengths of light.) It is still red if it close or far away. The brightness of red light divided by the brightness of blue light (the ratio of red to blue light) is something that won't change with distance. If the star is farther away it is both fainter in blue and fainter in red. The ratio of intensity doesn't change with distance.

What does change the ratio of the brightness in red to the brightness in blue is the temperature. If you remember the black body curves, as things get hotter they get bluer. These colors give us a measure of temperature. There are also some spectral line things that factor into this. The spectral line effects don't change with distance either. Between all of the effects we can isolate areas of color that indicate different types of stars.

Now we take the difference in magnitude. I have been talking about the ratio of brightness. Magnitudes are a logarithmic measure of brightness. This means that when we take the difference of magnitudes (the difference of logs) we are calculating the log of the ratio. (Log{a}-Log{b}=Log{a/b}) So we are basically taking the log of the ratio. The important part here is that we are calculating the ratio.

A measurement in a single band is a brightness. So when you say it's so many magnitudes at J band, that's a brightness (which can be converted to flux density units). (Technically, the definition of magnitude has in it an implicit comparison to Vega - a magnitude is an implicit flux ratio with Vega.)

It is often useful to take the same object and compare its brightness at two different bands. It's easiest to make such a comparison as a ratio of flux densities, e.g., a difference of magnitudes. Conventionally, you take the bluer band minus the redder band, e.g., B-V, J-K. That way, larger numbers are always redder, and smaller numbers are always bluer. Depending on which bands you pick, exactly, it tells you different things. B-V, for example, can be mapped to temperature. K-[8] is a good way to look for IR excesses, e.g., disks.

Note too that when you compare the brightness of the same object at two different bands, the distance cancels out. Consider a pair of identical objects. The one that is further away will be fainter. But they will both have the same color. Interpreting color-magntiude diagrams requires also worrying about the distances to all the objects in the plot. Color-color diagrams don't involve distances -- distances cancel out.

Remember this video: https://www.youtube.com/watch?v=VHMTB95b5us&list=PLjCjDYabTFm9b9jQd4hcZPAnFWYsIjs2D&index=7&t=0s this is a series of blackbodies (stand-ins for stars) with the locations of the bands marked. B-V can be mapped to temperature because those bands are near the peak of the blackbody energy distribution, so B-V changes rapidly as temperature changes. K should mostly still see photosphere in many cases, though it too can be contaminated by circumstellar emission. Longer wavelengths will sample more of the disk.

So - why do we make a plot like this? To compare the ratio of brightnesses at two different broadband filters for a set of many objects at once.

How do we pick the bands to use? By thinking about stuff like the movie, and making models of the sorts of sources you expect to care about.

How do you know what to expect? By making models, but ultimately by observing things you understand before applying it to things you don't understand. For example, this catalog consists of a pile of colors of known young objects. You can also find catalogs of "standard main sequence colors", establishing colors for a bunch of stars on the main sequence. You can use catalogs like this to establish where you expect young stars (or main sequence stars) to fall in any given combination, and then, with that knowledge, interpret the diagram of a new field.

In the infrared, it's easier to remember off the top of your head "where the main sequence should be", because all magnitudes are defined to be 0 when they are the same brightness as Vega (or, rather, what Vega would be if it didn't have a disk). Which means that if any *difference* between magnitudes is 0, the SED is the same shape as Vega (or, rather, what Vega would be if it didn't have a disk). So a star with K-x=0 where x is any IR band has no IR excess. K-x > 0 is redder than Vega, e.g., possibly a disk. K-x < 0 means either photometry problems, or it's not a star.