About the star in question...
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Gust644
i know what a star don´t hot enough to produce emission lines, but because in the spectro for that follow a pattern after the right swings?
http://skyserver.sdss.org/dr9/en/tools/chart/navi.asp?ra=185.806685026565&dec=0.4294607333333333&opt=Posted
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Budgieye
moderator
The big peak to the right is probably due to water is the Earth's atmosphere. Look at this spectrum. The green line shows interference in the Earth's atmosphere, especially water on the right side. The green line affects the black line and makes peaks appear.
http://cas.sdss.org/dr7/en/tools/explore/obj.asp?sid=723060327599570944
Yes, there are no strong emission lines in this spectrum. It is a hot AO blue star, about 30,000 degrees K, but that is still not hot enought for big emission lines. This is a weak spectrum from a dim star, with values barely above zero. The peaks are random noise. There are some absorption lines, where some gas has absorbed some light.
You are also correct, there is more light in 4000 and 5000, so it is definitely a blue star.
http://skyserver.sdss.org/dr14/en/tools/explore/Summary.aspx?id=1237648705114997100
Only very hot objects about a million degrees, produce emission peaks. eg
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supernovae,
active black holes,
galaxy collision.
More info here
Spectra guide for SDSS images in Galaxy Zoo Talk https://talk.galaxyzoo.org/#/boards/BGZ0000001/discussions/DGZ0000ulp?page=3&comment_id=53fef2ee3d5a77490c0001b6
It confused me, I thought you were talking about the galaxy at top left. Please start unrelated topics directly from the discussion page..
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Ghost_Sheep_SWR
in response to Budgieye's comment.
A bit of nitpicking to avoid confusion on my side; if I'm correct the green line ('empty sky measurement') is actually substracted from the original measurement to produce the black line in order to nullify sky noise. If it weren't the black line peaks would be much higher.
So the residual peaks between 8000Å-9000Å can either mean - imperfect substraction (common) or part of the peaks are real (unlikely).
I hope I got this right, right? 😄
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Budgieye
moderator
Duh,
"the lower green curve represents the error in the number of counts"
http://classic.sdss.org/gallery/gal_spectra.html
Re: Tutorial bits on galaxy spectra http://www.galaxyzooforum.org/index.php?topic=1923.msg35244#msg35244 by NGC3314 (Dr Keel)
In these plots, the lower green line represents the software's
estimate of the point-by-point error level. This tells you how
accurate the measurements of the object itself (the black line) are
likely to be. The error varies a lot across the spectrum for several
reasons. The detector and spectrograph have nonuniform senstivity with
wavelength, so the error from random counting statistics goes up where
it is less efficient. Also, the object spectrum is the difference of
what's seen through an optical fiber pointed right at the target and
an average of several optical fibers pointed at "blank sky", to
subtract the airglow (sort of a faint permanent aurora in the upper
atmosphere, which varies with time and location). Here again, the
error goes up where the night sky is brighter, and especially for
fainter objects. Throughout, it's important that the radiation is
detected as individual photons whose arrival is a stochastic process,
so the errors are often dominated by this process. Executive summary -
pay no attention to features in the object spectrum whose amplitude is
not a good deal larger than the error spectrum at that wavelength,
especially if the error spectrum shows a spike at the wavelength of
interest.
😃
Posted
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Ah, it rang a bell but couldn't find the thread again, now, here they are;
http://www.galaxyzooforum.org/index.php?topic=1923.msg156352#msg156352
So it seems now there are actually two explanations for the green line. Either substraction of sky noise from primary measurement or indication of error level of the primary measurement.
Posted