Hot answers tagged

28

The Sun is currently turning hydrogen into helium. There are no other nuclear reactions taking place at any significant rate in the Sun. The Sun will not start to make heavier elements until it reaches the tip of the red giant branch in about 7 billion years time. The elements heavier than helium that are present in the Sun were almost all made inside other ...


12

The photosphere of the sun does produce an emission spectrum (a Planck spectrum according to its temperature of about 6000K). It is only that the atmosphere above the photosphere (the chromosphere) scatters light out of the line of sight at those frequencies where the scattering cross section is very high (i.e. at the atomic resonance frequencies of a given ...


7

The hotter layers above the solar photosphere do have an emission spectrum. The emission spectrum is much fainter than the visible photosphere and so is not easily seen through broadband filters in the optical spectrum, though it can be observed through very narrow filters centred on the emission lines (e.g. H$\alpha$ from the chromosphere) in question. The ...


7

You are essentially correct, but here's a more in-depth explanation: Both units are a measure of the total flux $F$ of a light source in some interval, e.g. a spectral line or a broader region given by a filter. The flux is obtained by integrating a spectrum’s flux density, i.e. the flux per "bin". A spectrum is a distribution of photons as a ...


6

Collisional broadening - which includes van der Waals and Stark broadening - is more important in the higher gravity, higher pressure/density atmospheres of dwarf stars (a factor of 100-1000 higher for dwarfs vs giants of the same photospheric temperature). These collisional effects effectively "truncate" radiative emission and absorption processes,...


6

There is the NIST Atomic Spectra Database where you could browse by elements. This the reverse approach, meaning that you have to first query element by element and then see which of the lines you find in the spectrum - that's how we used to do it in our labs during studies: Then, there is another NIST site where you can enter up to 4 spectral lines in ...


5

I have read that the reason why the sun produces an absorption spectrum is because the temperature drops as you go away from the center, such that as the various layers of the atmosphere of the sun absorb certain wavelengths, the re-emitted light will have a smaller intensity than the absorbed one, causing a dip in the spectrum (ie an absorption spectrum). ...


5

Okay, well, calcium is just an easily recognizable spectral feature, so I guess that's why they used it. The y-axis shows the relative intensity. I am assuming this is a spectrum that is normalized or scaled down so that the highest value of the intensity is 1, but this is not so important. The laboratory wavelengths of the absorption lines are shown in red ...


5

I was wondering why the absorbtion lines of the template are broader then those of the galaxy, since it actually should be the other way arround. You are correct that it should be the other way around. The reason the plot looks confusing is that you are not actually plotting the galaxy spectrum in the top panel; you are plotting some combination of noise ...


4

Surely the sun possesses calcium in its atmosphere, as well as in its bulk volume. This plot, based on the data published in Asplund et al.,(2009), shows what elements can be found in the solar atmosphere: And we can read off that the abundance [Ca]/[Si] = 0.1 for example. Elements in stellar atmospheres can occur both in absorption and emission in stellar ...


4

Here is another alternative/supplement to NIST. The VALD atomic line database, which is specifically for astrophysical applications. I believe this database does contain some molecular data too.


4

Yes, the spectra from SuperCam along with that from many other instruments is available from the NASA Planetary Data System (PDS) Geosciences Node. The overall landing page for all of the Mars2020 instruments is here and the one specifically for SuperCam is here. Reading the documentation is highly recommended as PDS archive bundles such as this can be quite ...


3

Initially there is dense gas surrounded by an "atmosphere" of less dense gas. This atmosphere is being rapidly ejected by the nova, so you see the absorption spectrum of the thin outer parts of the nova, lit by the continuous spectrum of the dense central part. As the nova proceeds, the gas thins, and now the central part no longer emits a ...


3

I think I can sort of answer your first/second question. It's a bit hard to guess what your background is, but I hope you have seen or derived somewhere that the $a_{lm}$ coefficients can be written as $$\oint \Theta(\hat{x}) Y_{lm}^*(\hat{x}) d\hat{x}$$ where $\Theta$ is the temperature fluctuation (as seen in the CMB) and $Y_{lm}^*$ is (the complex ...


3

You seem to have all the ingredients apart from the variables of what size your detector pixels are (either physically or binned in software/hardware) and the angular extent of the object you are taking a spectrum of. The basic trade-off, as you say, is between flux and spectral resolution, but there are limits to that trade off. You should not reduce your ...


3

'Absorption' lines are caused by resonance scattering (scattering the radiation out of the line of sight, see illustration below), and resonance scattering has a very large cross section of roughly $10^{-12} cm^2$. This means that even for a thin layer of 10km ($10^6 cm$) you need only a density of >$10^6 /cm^3$ of an element for the layer to become ...


3

The strength of an absorption feature in the stellar spectrum is dependent on the amount of that element that is in the photosphere but it also depends on the atomic structure of the element and the conditions of temperature and density in the photosphere. For example the CaII lines need there to be singly ionised calcium ions in the photosphere. This ...


3

This is just a Fourier transform: (let $\boldsymbol{x}=\boldsymbol{r}_2-\boldsymbol{r}_1$) $$ \begin{aligned} \langle \delta(\boldsymbol{k}_1)\delta(\boldsymbol{k}_2) \rangle&=\int\int d^3r_1d^3r_2\langle \delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2) \rangle e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}e^{-i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\\ &...


3

What I am actually after is to answer more scientifically the children's question: "What happens if I throw something in the Sun?" and my answer was always "It burns." but I am now wondering about the "How does it burn?" It doesn't burn. What happens when an object such as an asteroid or comet impacts the Sun is in a sense the ...


2

The three common techniques used for aquiring the spectra of exoplanets and their atmosphereseres are: Transmission: The brightness of a star decreases as the subject planet moves in front of it. If the planet has an atmosphere, it absorbs the suns emitted light. By measuring the brightness decrease at different wavelengths, the wavelength dependent ...


2

If you look for a simple answer by easy principles and not technicalities this would be the picture... When the exoplanet transit between our point of sight and its star, its disk block part of the light. This results in a dimming of the star light we receive. The latter can be analyse as we do for light of any source. It is well approximated by a black body ...


2

Maybe somebody can help me understanding the following quote intuitively: However, by looking at the ratio of two different but related lines - those of iron - we found the ratio itself related to temperature. And it did so in a consistent and predictable way. A particular atom can only be at integer quantum states (Hydrogen is depicted here for simplicity)...


2

Stars behave like blackbodys. Not perfect idealized blackbodies, however, the spectrum of a star is close enough to the standard blackbody spectrum. Reason why you can use the Wien's Law to calculate an estimate of its surface temperature: $\lambda_{\rm max} = (0.29 {\rm\, cm\, K}) / T$ Where $\lambda_{\rm max}$ is the frequency of maximum measured emission ...


2

Pretty sure from my stellar nucleosynthesis days that the p-p I chain is the dominant form of nucleosynthesis in the sun, but the p-p II, p-p III, and p-p IV chains also occur, just to a much lesser extent. Those will make Be, B, Li. But, Na and Fe - and other heavier elements - mostly come from the sun not being a first-generation star: Previous supernovae ...


2

Imagine your line as a rectangle of width $w$ and depth $d$ relative to a normalised continuum. Without scattered light, the area blocked off by the line is $wd$ and if the continuum level is normalised to 1, then the equivalent width is $wd$. Now add 5% scattered light. The height of the continuum is 1.05 (but we're going to renormalise it) and the depth ...


2

One thing you might try doing is to use the SkyServer Navigate interface to see if the object was imaged by SDSS. Enter the name in the "name" box in the upper left and then click on the "Resolve" button. If an image with the galaxy shows up, click on the "Object with spectra" checkbox in the "Drawing options" panel on ...


2

I think the velocity scale is calculated something like this: $$v_r = c\left( \frac{\lambda - \lambda_0}{\lambda_0}\right),$$ where $\lambda$ is the observed wavelength and $\lambda_0$ is wavelength of the Mg II line, corrected for the redshift of the quasar ($z=0.868)$ it is observed towards. The confusing thing is, for Fig.~3 in the cited paper, that the ...


2

The reason that you know it is an emission spectrum, rather than an absorption spectrum, is that there are peaks rather than dips. This is by definition: absorption lines are where light was absorbed by the atom, so you see a dip in the spectrum; emission lines are where the atom released a photon (whose energy corresponds to the energy difference of its ...


2

The continuum spectrum could be described as "blue" as it is skewed towards shorter wavelengths. There is an excess of blue light compared to red. You can see this in the way that the continuum part of this spectrum curves up towards the left and down towards the right.


Only top voted, non community-wiki answers of a minimum length are eligible