16

To first order, the relative abundances of the heavier elements to iron (for instance) are constant. So the metal content of a star is shorthand for the content of any element heavier than He. (NB: we now know this is not true in many circumstances and elements can be grouped by synthesis process - for example we can talk about "alpha elements" - O,...


11

Radiative energy transport continues. The point is that the radiative flux, which is proportional to $dT/dr$ can be overtaken when the temperature gradient achieves the adiabatic value and convection starts. Once convection is started, it is very efficient and the majority of energy flux will be transported by convection. Details Broadly speaking, radiative ...


6

In 1812 Fraunhofer measured a set of absorption lines in the sun. It wasn't until later that Kirchoff and Bunsen figured out that these absorption lines matched emission features from metals they were burning in the lab. (http://www.chemteam.info/Chem-History/Kirchhoff-Bunsen-1860.html) In chemistry, most things in the periodic table are known as "metals"( ...


6

There is no 'history' behind it. Stellar physics is less than a 100 years old. Thus, it is not a terminology that came to be due to some anecdotal reason. This is the way it has been since the start. Why? Because we don't care. Why, really? Hydrogen and Helium are more important and abundant, so we need something to measure the others, which can be grouped ...


6

You seem to be confusing the simple mathematical relationship between radius and surface area, and the more complex relationship between mass and size. If you double the radius of a sphere the surface area quadruples. This is pure maths, and is not particular to stars. The volume is multiplied by 8. But in a star, increasing the amount of matter by a ...


6

I think that there isn't a strict answer to this question. However, I believe the answer is that there's a difference between the core of a hydrogen-burning star and the core of a protostar or star-forming, gas cloud. For a hydrogen-burning star, the core, as you say, is the region of the star where fusion is taking place. This is surrounded by the ...


4

Magnesium does not have its own fusion shell inside stars. If you have a look at the Nuclear Binding Energies per nucleon(NBE) of Elements, you will notice one trend: Most of the elements that form a shell have a higher value of NBE locally. Nuclear binding energy is the minimum energy that is required to disassemble the nucleus of an atom into its ...


4

In terms of mean angular velocity, the distribution of rotation rates among main sequence stars is well known. Allen (1963) compiled data on mass, radius, and equatorial velocity, which was then expanded upon by McNally (1965), who focused on angular velocity and angular momentum. It became clear that angular velocity increases from low rates for spectral ...


4

I have not the qualification to answer the question in its whole but the question is interesting (I worked on Be Stars which are episodically surrounded by an decretion disk and which rotates at nearly critical velocities. The phenomenon in Be stars is different from accreting stars. The only consequences of subcritical velocity is a flattened envelope and ...


3

Even though it arised for historical reasons outlined in the other answers, the distinction between metals and non-metals as defined by astronomers does continue to make sense today. Metals are formed in stars and supernovas, whereas non-metals preexist stars. Therefore, the distinction is relevant when considering nucleosynthesis.


3

The spectrum of a star is almost certainly a unique fingerprint. Even though stars are born in clusters, formed from the reasonably chemically homogeneous environment of a giant molecular cloud, there are likely to be small differences in their local environment. Furthermore their formation environments, and the stars themselves at later times, can be ...


3

The argument goes something like this. Hydrostatic equilibrium means that the local pressure gradient is proportional to the local density multiplied by a latitude-dependent local gravity. If the pressure just depends on density and temperature, this means that those quantities will also just depend on latitude and therefore will be constant along an ...


2

From the same Wikipedia page: This means that equatorial regions of a star will have a greater centrifugal force when compared to the pole. The centrifugal force pushes mass away from the axis of rotation, and results in less overall pressure on the gas in the equatorial regions of the star. This will cause the gas in this region to become less dense, and ...


2

$\nabla B^2 \sim B^2/l,$ where $l$ is a length scale on which $B^2$ varies. In SI units, $B^2 \sim 4\times 10^{-8}$ T$^2$ and $\mu_0 = 4\pi \times 10^{-7}$, so an order of magnitude for $\nabla \cdot P_B$ is $4\times 10^{-8}/(8\pi \times 10^{-7}\times 10^{6}) \sim 10^{-8}$ Pa/m. The density of the solar photosphere (at optical depth unity in the visible ...


2

Even the optical spectrum alone has lots of goodies. In addition to chemical abundance, size, and other outer properties from emission and absorption lines you can get the rotational rate from a narrow line's doppler profile, and very careful spectroscopy information on vibrational modes of the star through spectral asteroseismology. Also most stars are ...


2

The question is compromised by saying that you allow arbitrarily perfect measurements. If we have a bolometer that can measure the amount of flux from a star, at a distance that is known to arbitrary accuracy, with arbitrarily good spatial resolution, then what we do is measure the bolometric luminosity from a 1 m$^2$ area at the centre of the stellar disk. ...


1

There is no hard and fast answer. To be treated as an idea gas (your title question), the particles in your gas should be point-like and they should be non-interacting if you are to use the ideal gas approximation. This means you can take the mean separation ($\sim n^{-1/3}$) and compare that with the size of the particles - it should be much bigger. You ...


1

It is quite easy. In fact you do not need a bolometer. You just need to perform Intensity measurements in several parts of the spectrum, and then fit these to a theoretical black body spectrum. Three uses to be enough if it does not happen that you are measuring on a spike or valley in the spectrum caused by an emission or absorption line. The black body ...


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