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 ...


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 ...


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 ...


6

You can probably get most if not all of your questions answered by perusing the main DESI web site, which I encourage you to check out. There is, for example, a nice video describing the assembly of the main focal plane elements (the fibers and the associated robot positioners) here. But in simple terms: the circular focal plane is divided into ten wedges (...


6

In general, you can't. If obtaining spectra in regions where there is expected to be a spatially varying background then you either need to do long-slit spectroscopy so that you have a good measurement of the ISM contribution either side of your source, or you do integral field spectroscopy with the same idea. The problem is that the line strengths for the ...


5

Supplemental to @PeterErwin's answer, some more details on the five thousand "robots". Each fiber has a circular "patrol area" with a diameter of 12 millimeters, and these are located on a hexagonal array with a pitch (nearest neighbor distance) of 10.3 millimeters. Motion is implemented with eccentric axis (Θ–Φ) kinematics. Instead of x-y or r-Θ which use ...


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....


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

Depending on the type of camera you use, you may be able to photograph stars without a telescope. I know that there are apps for cell phones that allow you to take long exposures, and there are some point-and-shoot cameras that have a "bulb" setting or allow multi-second exposures, and certainly an SLR-type will allow it. You'll need to stablize ...


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, ...


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 ...


3

The role of the microshutters is NOT to act as pinholes or coded aperture and to produce a focused image, similar to a mirror or lens. Its job is just to separate the the light from different sources. This is achieved simply by opening only the shutters corresponding to the sources of interest. See the following image: This is important for spectroscopy as ...


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.


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

$\nu_0$ is the frequency you would expect for the absorption/emission line in the absence of any broadening. i.e. It is the centre of the line profile. $\delta \nu$ is just $\nu - \nu_0$. (i.e. it is the separation in frequency from the centre of the line profile and is equivalent to $f -f_0$ in the second expression). $\Delta \nu_{th}$ is a measure of the ...


2

Slipher 1912 and Lowell 1912 used a one-prism spectrograph on the 24-inch refractor at Lowell Observatory in this way. Their 1911 observations of Uranus yielded a period of 10.8 hours, which made sense given the polar flattening previously observed by others. They knew the obliquity of Uranus's satellites' orbits and assumed that its equator was in a ...


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

I know what an optically think/thick medium is... Okay so this isn't much more complicated. A medium or material can be optically dense or opaque at one wavelength, but fairly transparent at a different wavelength. If you look at the dark plastic window on a remote control for a TV or other appliance, you can't see through it. It's optically dense at ...


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

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 ...


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 ...


1

There are many ways to measure a star's motion, radial velocity (wobble) can be measured using doppler spectroscopy. The first exoplanet discovered by this method was 51 Pegasi b by Michel Mayor and Didier Queloz, who discovered the planet on December 1995. A drawback of this method is that it can only detect the movement of a star towards or away from the ...


1

Since this seems to be a kind of niche question I will answer and leave some comments for future souls with the same problem. To whom it may interest that may be searching for the same problem. The solution is the basic $M = m + 5( \log (parallax) + 1)$. Parallax in arcsec. The reason I'm going with K instead of V is due to a more independent ...


1

To use the Doppler shift, you need to know by how much the light is blue/red-shifted. Suppose you have a light curve where the wavelength varies between $\lambda_1$ and $\lambda_2$. Using these wavelengths, the Doppler shift will give you an interval of radial velocities. This gives you a value for K that you can plug into your last equation to derive $M_p$...


1

This is exactly what is done for template fitting. The shifted SED is one of the intermediate data products of template fitting codes such as Le Phare or Phosphoros. For instance, with Phosphoros, you can shift your SEDs to any number of redshifts, then find the shifted SEDs (also called model grids) in the IntermediateProducts folder in the Phosphoros ...


1

After a lot of research, I found the answer to the second question, since the FITS files in sdss Images have more than 1 primary hdus, you have to pass the following fits images like an array *.fits[0] to the swarp program, where the first hdu in the fits file contains the image pixel. solution here For the first question, yes the method to do that is ...


1

This is a supplemental answer to @MikeG's excellent answer. I've just plotted Uranus' axis direction versus the direction we view it at, to put those dates into context. The "solstice of 1901-02" is clearly seen in the plot. Plotted using Python script based on Skyfield: import numpy as np import matplotlib.pyplot as plt from skyfield.api import Loader, ...


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