Tag Info

New answers tagged

5

Enzo is fundamentally a grid-based finite-volume hydrodynamics code. That is, the domain is divided into cells, each is assigned various fluid properties (density, velocity, etc.), and at each timestep fluxes of those quantities across the interfaces between cells are used to update the quantities in the cells. It has a choice of particular methods for ...


5

Checking for electron degeneracy is a matter of comparing the Fermi kinetic energy with $kT$. If $E_F/kT \gg 1$, then you may assume the electrons are degenerate. The central density of the Sun is around $\rho=1.6\times 10^5$ kg/m$^3$ and the number of atomic mass units per electron is around $\mu_e =1.5$. The number density of electrons is therefore ...


3

The first part is handled well by Chris Drost - the kinetic particle energies are a lot larger than their interaction energies, so the gas can be considered (approximately) ideal. The last part - yes, as long as the Coulomb energy is a lot lower than the thermal energy then the protons or He ions can be considered an ideal gas with the appropriate average ...


7

According to a NASA page, the density in the middle of the Sun is about 150 g/cm3. That's about 9 × 1025 protons in a 1cm3 box, or 450 million to a side, and using that spacing for a voltage calculation reveals a typical interaction energy of 65 eV or so. (If you've never seen this unit before, that is the energy used by a 1V battery to move an electron's ...


3

The Lagrange points are only well defined for a two body system. The Earth-Sun Lagrange points exist only because the perturbations caused by the Moon and the other planets are relatively small. By contrast there are lots and lots of stars and dark matter between the Earth and Sgr A*, so this isn't even remotely a two body system. The mass of Sgr A* is ...


1

The above is the Boltzmann equation for annihilation/collisions in an expanding universe. Short answer The superscript in this case stands for equilibrium number density, as you pointed out. This means the $n_i^{(0)}$ stands for particle species $i$ number density in number density equilibrium, meaning equilibrium in which each particle species $i$ is ...


1

There are a couple of related questions: What elements can be created in the fusion process of different types of stars? What is the heaviest element possible produced in a supernova? though surprisingly I can't find an exact duplicate (which probably just means I didn't look hard enough). Iron is the most stable nucleus so in principle all other nuclei ...


1

The best evidence for the black hole at the center of the Milky Way comes from the simple Keplerian motion of nearby stars. Using orbital data deduced by scientists, I made a simulation of their motion. Please note that this simulation needs a browser that supports WebGL.


1

You are correct about the bias factor - because the dark matter distribution is not measured directly, but via tracers (galaxies), there may be some bias in the tracers. This is put into the analysis as an unknown bias parameter $b$ that needs to be fit. Now the second part of your question - why do we care about the dark matter? Briefly, it is because the ...


1

You should not confuse the terms light years $[ly]$ to be a unit of time, it is actually a unit of distance, the distance light will travel in vacuum in one year. When we speak of the age of the universe we use years, in particular about $13.7$ billion years. Now do to the expansion of the universe the distance form which the light comes to our eyes or ...


0

Due to inflation distances does not correspond directly with time ago. 13.7 billion years ago is actually what you see at the distance of about 48 billion light years away. What causes this discrepancy is also the same thing that causes redshifts, so things that are more redshifted also have more of a discrepancy between the naive conversion 2bn ly = 2bn ...


3

In the "microscopic" sense, the formation of the Sun and solar system do not depend strongly on dark matter. Looking at things on the scale of GMCs, you can get a Jeans-unstable situation that will lead to star formation without invoking dark matter. There's a step much earlier in the history of the Universe that needs dark matter, though. In the early ...


3

The sun's core has a density of of 150 g/cm³ (150 times the density of liquid water) at the center, and a temperature of close to 15,700,000 kelvin, or about 15,700,000 degrees Celsius; by contrast, the surface of the Sun is close to 6,000 kelvin. The core is made of hot, dense gas in the plasmic state, at a pressure estimated at 265 billion bar (26.5 ...


5

Good answer from Kyle. I will just add that there is a great deal of effort going into trying to discover "solar twins". These are stars with such similar parameters (including age inferred from the HR diagram or asteroseismology, which can be good to about 10% in the best cases) and photospheric compositions to the Sun, that it is thought likely they must ...


16

You're right that the Sun being 4.5 billion years old makes observations difficult. The Sun goes around the Galaxy about once every 225 million years, so since the Sun formed it has gone around the Galaxy perhaps 20 times. The trouble is that the Galaxy is not like the Solar System: stars don't go around on nice nearly circular orbits, everything is a bit ...


8

To ask what do physicists expect to accomplish with gravitational lensing is nowadays somewhat like asking what do biologists expect to accomplish with looking at things with microscopes. Gravitational lensing is a well established method used across astronomy and the main challenges the field itself has to tackle are mainly technicalities. But I will try ...


2

If you graph the temperature of your copper strip as a function of time you're going to get something like: This is because you have two effects. The light from the Sun heats the copper strip, but at the same time the strip cools. The equilibrium temperature (the dashed line) is the temperature at which the cooling balances the heating. If the intensity ...


2

I'll give this a try, though I don't have the time or inclination to work through the gory details myself. I'll start by defining the notation I'll use: Poisson's equation: $$\nabla^2\Phi({\bf x}) = 4\pi G \rho({\bf x})$$ Define a new potential by a constant offset, where $\Phi_0$ is the potential at infinity: $$\Psi=-\Phi+\Phi_0$$ And define the velocity ...


3

Not a definitive answer, but I note the following. The solar photospheric abundance of Thorium is indistinguishable from the abundances in the protosolar nebula deduced from meteorites (to within 0.1 dex, Asplund et al. 2009). Of course it might be thought possible that Th-depleted material in the centre of the Sun never makes it to the photosphere. ...


0

Most of the non-thermal electromagnetic energy is emitted by the neutron star as magnetic dipole radiation at the spin frequency of the pulsar (Crab: λ=107 m). No, this is not correct. The vast majority of electromagnetic radiation is produced only after the Poyting dominated cold relativistic wind passes through the termination shock. Here is ...



Top 50 recent answers are included