11

Close, but not quite right - the blue light is indeed emission from CO$^+$, but it's from the CO$^+$ ions themselves, with no need for recombination to CO; that (ionized) molecule has a strong set of energy transitions around 425 nm (4250 Angstroms), which is in the blue part of the visible spectrum: Spectrum of Comet C/2016 R2 (Pan-STARRS), Figure 2 from ...


10

If you look on the main SIMBAD page, you will see the following in the box titled "Content": The SIMBAD astronomical database provides basic data, cross-identifications, bibliography and measurements for astronomical objects outside the solar system. (emphasis mine) The Moon and the Sun are inside the Solar System, thus they are out-of-scope for ...


9

As your question is based on the plot you posted, I suggest you to look for a lower wavelength range of the atmospheric electromagnetic absorption. A quick search in google gave me this paper, which says: The importance of molecular nitrogen as the most abundant species in the Earth's atmosphere is evident. The strong absorption bands in the range 80–100 nm ...


8

Essentially what they did was assume that normally when observing with their telescope the spectral absorptions they see are due to the Earth's atmosphere. Which is a pretty good assumption. They then normalize the data to those absorption, so if there was any phosphine gas within the package of atmosphere they are looking through, it will be taken into ...


7

You need to compare it with the spectrum of a similar galaxy at a known redshift, that would probably enable you to identify features with known rest wavelengths. If you can find such a template, then the best way of estimating a redshift for a galaxy spectrum like this, consisting of mostly weak and blended absorption features, is to cross-correlate your ...


7

You are correct that the characteristic emission and absorption lines we see in stars' spectra are from electrons that are bound to atoms making transitions between different energy levels. That is possible because the elements in a star's photosphere are not fully ionized. Hydrogen - the easiest element to fully ionize because its nucleus only has a ...


5

I think you can understand that factor as follows: When photons lose energy, they're spread out over a larger wavelength range. Since there is a fixed number of photons, that means that the number of photons per observed wavelength bin decreases by a factor of (1+z), and hence the flux density, which is really what $f_\lambda$ is, decreases by this factor.


5

They have both 21cm and mm-wave observations for those systems. Caption to Fig.1 The hollow squares correspond to two 21 cm and molecular absorption systems. By "system" they are referring to sets of absorption features caused by multiple clouds of material along the line of sight to a single quasar. Each cloud imposes its own set of absorption features....


5

As @ELNJ answer pointed out, fully ionized the atoms at the star surface are not. It is not hot enough. Star cores are another case, but we usually don't see them. There, both pressure and temperature make impossible the existence of the usual atoms. Atoms and molecules usually emit their characteristic wavelengths because of the electrons' energy levels... ...


5

Yes, radio spectra have been used extensively to find the distances and locations of HI regions and molecular clouds inside the Milky Way. Observations of the 21 cm hydrogen line and/or several carbon monoxide lines (in particular, $\text{CO}(1\to0)$) enable us to make radial velocity measurements of clouds within the galaxy. From there, some geometry (see ...


5

You cannot have free protons without electrons. Plasmas, in general, are electrically neutral. It is usually electrons that dominate the scattering (note that a point-like charge cannot absorb a photon and conserve energy and momentum) in a plasma at low photon energies. That is basically due to their much lower masses (classically you can think of the ...


4

It depends what lines you are measuring and in what kind of star. When you measure the RV from a spectral line, you are measuring an intensity-weighted average RV over the region where the line is formed. For a star like the Sun, the photospheric lines are all formed within a layer no thicker than about a 1000 km and the differences in RV with depth in the ...


4

Yes, nebulae can often have very distinct colours. What produces those colours can depend on what elements are in the nebula material and what the temperature and density are. Generally speaking, green colours in a nebula are due to forbidden transitions in ionised Oxygen, though can feature the hydrogen $\beta$ Balmer line. Red colours can be due to ...


4

A recombination line is a special case of an emission line. Emission lines An emission line is any spectral feature that rises above the continuum — i.e. the average amplitude of the spectrum (in some wavelength region) — and is due to atomic transitions (where "atomic" include atoms in molecules and dust grains, and "transitions" may be electronic, ...


3

This figure is from the paper "Phosphine as a bio signature gas in exoplanet atmospheres". It shows the absorption cross section of Phosphine compared to other molecules. We can see that Phosphine has a distinct enough profile from the others molecules in the 7.8-11.5 microns range, with the exception of NH3. Probing from 2-11.5 microns should ...


3

If you are doing chi-squared fitting, then a value for the uncertainty (or an estimate) should have been returned by the software. If not, you could perhaps manually estimate by fixing the line centre at various values, allowing the other parameters to be fitted, and seeing where the minimum chi-squared increases by 1 over the global minimum. Failing that, ...


3

If you know how to reduce raw spectrum data, you may look at the official archives of big observatories (ESO, MAST, CADC). But a quicker and much easier way could be to use the following portal: http://archive.eso.org/scienceportal/home Typing the name of some Wolf Rayet stars will lead you to reduced data and thus spectra.


2

To get the flux of an SED through a particular filter, you actually multiply the the SED by the filter's response. Talking about convolution in this context is a bit of a misnomer. Basically, for each wavelength, you look at what fraction of the light will go through the filter, and you sum up the values you get at those wavelengths. The division by $\int ...


2

The "radius" of the Sun is basically where the optical depth to radiation at any particular wavelength is about 1 (or some times 2/3 is used). This is known as the photosphere and is from where the light that we measure comes from. The "thickness" of the photosphere - i.e. the depth over which the optical depth changes from being negligible to very large is ...


2

In this case itseems to mean that the depth of the line is 7 times its error bar below the continuum level. Impossible to answer. You say it can't be done, but the authors say that they fitted a Gaussian. You either use a rough estimate (attributable to Cayrel de Strobel 1988) of $$\Delta {\rm EW} \sim 1.5\frac{\sqrt{RP}}{{\rm SNR}},$$ where $R$ is the ...


2

It is difficult to say, there isn't a description in the paper as to how they arrive at those numbers. Given that the equivalent widths of the lines have signal to noise levels that mean they are only greater than zero at significance levels of 2-3, then it isn't clear these are meaningful numbers at all. However, here is how it might have been done. You fit ...


2

On the sky at night programme on bbc 4 yesterday, the scientists explained how they were able to detect different levels of phosphine at the equator vs the poles of Venus. For me this indicates that the gas isn’t being detected locally


2

Sure, but it is not straightforward. There's a lot of degeneracies involved, such as many molecules sharing similar absorption bands, the presence of clouds, scattering, absorption happening at many different pressure ranges, solving the chemical equilibirum equations of the species and the temperature-profile modifying the overall spectrum as well. There's ...


2

You are doing it incorrectly if you are trying to cross-disperse your Fig.2. You should be cross-dispersing your Fig.1. A "cross-dispersed" spectrum requires requires the dispersive elements to be at right angles. Your Fig.1 shows the dispersed spectrum from the echelle with overlapping orders. To separate the orders you view Fig.1 with the cross-...


2

You're asking two related but distinct questions here: Well, what about nearly-massive-enough-to-be-stellar objects? Why would there not be more of these - and the inevitable satellites involved in something of that mass - than there are solar systems? and also I would expect them to be in abundance, so, could this explain the dark matter mystery? For ...


2

Brown dwarfs (ie sub-stellar objects that are too small to support hydrogen fusion) do exist and some brown dwarfs have been found to host planets. An example of this is the brown dwarf 2M1207. It is a very dim object, about 25 Jupiter masses. It is still quite hot, so not completely "dark", and it hosts a planet with a mass of 3-10 Jupiters. ...


2

This is not exclusive to spectroscopy applied to astronomy but general. Matter can interact with electromagnetic waves spanning a very wide range of frequency (energy). Also matter can emit electromagnetic radiation when in a kind of excited state. Due to the internal mechanism of absorption/emission it happens that the spectral characteristics can be ...


1

The SM developers seem to have anticipated such issues. In your nm output sample, T sm_ylabel is the native C function, and T sm_ylabel_ is its wrapper to be called from Fortran. gfortran usually appends single underscores to external symbol names. If the MOOG object files were in fact built with the gfortran options in the question, the symbol references ...


1

Visualizing a cross-dispersed echelle spectrum directly is a nice experiment to do (I do it with my upper-level astronomy students when I teach about this) but it's a little tricky because your intuition from other types of gratings can lead you astray in terms of how to arrange things for viewing. The key is to realize that the blaze angle of an echelle ...


1

You can do this kind of thing reasonably easily, even as amateur astronomer. See e.g. this link. There also have been programmes by the Americal Meteor society and there are a lot of published papers to that end, e.g. here Two reasons it has not been done more widlely: It's a difficult and very costly endevour to make high-resolution spectroscopy at very ...


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