22

Hot dark matter would be made from very light, fast moving particles. Such particles could not possibly be gravitationally bound to any structure, but rather would be dispersed all across the universe. But dark matter is always "found" (or "inferred") either gravitationally bound to some visible structure (e.g. weak lensing detection of dark matter ...


20

Actually, the stars and nebulae that make up the spiral arm are only temporarily part of that spiral arm. Spiral arms are more like sound waves where individual particles move around a more or less stationary position. (Look for instance at the animation of longitudinal waves from Dan Russel, the red dots move a bit to the left and to the right around a ...


19

TL; DR Somewhere between now and a few hundred billion years time. (For a co-moving volume) Now read on. If stellar remnants are included, then the answer is very far in the future indeed, if and when the constituents of baryons begin to decay. So let's assume that "stars" means those things that are undergoing nuclear fusion reactions to power their ...


13

Not really, for the same reason that you cannot travel west by jumping up in the air and let Earth rotate underneath you, such that you land a little farther to the west. The reason is that standing on Earth's surface, you already have a velocity toward the east which matches exactly the speed of the surface. Thus, in the reference frame of Earth, you ...


13

Galaxies grow through cosmic time by accretion of the surrounding matter. Some of its mass increase happens through smooth accretion of gas, but much also happens through merging with small clumps of dark matter, gas, and stars, called satellite galaxies. This is called "minor merging". If merging galaxies are similar in size, it's called major merging. ...


12

Your calculation sounds correct. It is however based on assumptions that are non-trivial. An analogy with the Earth would give that the Earth rotated around the Sun 13.8 billion times since the Big Bang. Which is meaningless since the Earth was created only a few billion years ago. Our galaxy, the Milky-Way, may have had a long and quiet history since 10 ...


12

To add to Dieudonné's excellent answer, I'd like to say that spiral arms are only really prominent in the blue part of the spectrum (massive stars tend to be blue and short-lived), while in infrared wavebands, for example, spiral arms only appear as mild over-densities of 10-20%. Some galaxies have clear arms winding for almost 360$^\circ$ or even more (...


8

Well, it would be useful to define what a 'dead' galaxy is. Probably the most simple method would be a galaxy that is no longer producing new stars. We might also consider a galaxy that no longer produces significant light in the visual spectrum, or perhaps EMR across the entire spectrum. Generally, there's unlikely to be a firm line between living and ...


8

The half light radius is the radius from within which half the luminosity emerges. "Deprojected" means that the authors must have fitted some model to the 2D distribution of light, which can then be mathematically deprojected to give them a 3D model for luminosity as a function of radius, that they can then integrate to give a number for the half light ...


8

Since the second data release (DR2) of the European Space Agency's Gaia mission there has been a revolution in astrometry, including measuring the motion of the Andromeda Galaxy. On February this year van der Marel et al (also in ArXiv) published interesting results on that matter by using Gaia's DR2 measurements. The results reveal that the collision is ...


6

The cause for the oscillations perpendicular to the galactic plane is the gravity of the non-spherical mass distribution (needed for a plane Kepler ellipse) in the Milky Way. Simplified, there is a dense galactic plane. The density is not exactly known; therefore there is some uncertainty (a few million years) about the precise oscillation period. Details ...


6

Could dark energy (the mysterious accelerating expansion of the universe) be explained by "negative gravity"? But it already is "negative gravity". In general relativity, the stress-energy tensor $T_{\mu\nu}$ describes the energy, momentum, and stress of matter in spacetime. Through the Einstein field equation, it is connected with Ricci curvature $R_{\mu\...


6

The evidence for expansion is that the redshift is proportional to distance. The redshift of a galaxy can be divided into two components: that due to the cosmological expansion, which stretches the wavelength of light whilst it travels towards us; and a peculiar motion with respect to the cosmological expansion, which causes a straightforward doppler shift. ...


6

The Milky Way has arms that form due to density waves. Like the majority of spiral galaxies, the arms are trailing. Individual stars orbit in circles (roughly), neither towards or away from the centre. If you consider a common map of the Milky way (imagined from a point North of the Earth, Celestial North is not the same as Ecliptic North, which are both ...


5

No, for several reasons. there are no radial velocities in the required range observed, for example in the Milky Way galaxy. the reservoir of stars moving out from the centre would quickly be drained, so you would need to magically generate them at the centre -- this is much worse than postulating dark matter. you'd expect huge amounts of stars and gas that ...


5

Feng & Gallo have published a series of extremely similar papers, all of which essentially claim that they have "discovered" a major flaw in the way (some) astrophysicists think about rotation curves. Instead of assuming spherical symmetry, they try to solve for the mass distribution, using a rotation curve, without assuming spherical symmetry, instead ...


5

Globular clusters formed whilst the gas of the proto Milky Way was still approximately spherically distributed. The gas forms a dissipative system that loses energy and collapses (within the first billion years) to a disk whilst conserving angular momentum. Formed stars and clusters are essentially collisionless so the halo stars continue to have a ...


5

The reason is detailed in depth in this pdf, which contains the following diagram: Some key quantities: $R_0$: Distance from the observer to the center of the Milky Way $R$: Distance from target gas to the center of the Milky Way $V_0$: Velocity of the observer with respect to a certain reference frame $V$: Velocity of target gas with respect to the same ...


5

The collision timescale for a star in the solar neighborhood is1 $$t_c\simeq5\times10^{10}\text{ Gyr}\left(\frac{R}{R_{\odot}}\right)^{-2}\left(\frac{v}{30\text{ km s}^{-1}}\right)^{-1}\left(\frac{n}{0.1\text{ pc}^{-3}}\right)^{-1}$$ where $R$ is the radius of a star, $v$ is its speed relative to the stars around it, and $n$ is the stellar number density. ...


5

Equatorial coordinates have their equator and poles at the equator of the earth and the poles at the earth each projected onto the sky. Supergalactic coordinates on the other hand have their equator in the same plane as the "Supergalactic plane", which comes from the distribution of nearby galaxy clusters. And the poles just perpendicular to the the plane....


4

The answer is yes. Stellar collisions are events that occur mainly in Globular clusters due to their high stellar density. The likelihood of these events is highly dependent on the stellar density. As we know the galactic centers concentrate large amounts of matter and, of course, thousads of stars which are mainly in the old red main sequence. However, ...


4

The dipole in the microwave background indicates motion of the Milky Way and thus of the whole Local Group, at least, at about 600 km/s in a certain direction. The straightforward explanation is that the density irregularities nearby from superclusters and voids result in a net gravitational acceleration that, over the age of the universe, resulted in this ...


4

Particles in a gas approximate to point-like objects that interact roughly elastically through short-range forces when they collide, but are otherwise non-interacting. Stars interact gravitationally over long ranges, occasionally with each other, but always with the overall gravitational potential of the system. Sometimes people do talk thermodynamically ...


4

In simple terms: yes, the Milky Way's supermassive black hole (SMBH) is at the center of the galaxy, we know approximately where the center is (but not terribly precisely), and we should expect the SMBH to be there. You can define the Galactic Center using the orbits of stars and gas: what is the (average) center of their orbits? You can also use the ...


4

Not clear what the initial part of your question means. Objects can be in virial equilibrium without being in thermal equilibrium. A clear exception to the virial theorem would be any system that is gravitationally unbound. So you couldn't apply it to a supernova explosion or a dissolving cluster of stars. The examples you quote are not self-gravitating, ...


4

Galaxies are not so far from each other compared to their sizes as you might think. The typical distance between galaxies is a few Mpc (1 Mpc, or megaparsec, is roughly 3 million light-years). While the stellar disk of a galaxy like the Milky is only $\sim30\,\mathrm{kpc}$ across, its virial radius — determined primarily by its dark matter halo — is larger ...


3

On any practical timescale (ie say 100 billion years) a galaxy could not be said to have died. The majority of stars that ever formed in that galaxy will still be "alive", but they are dim, low-mass stars. However, some of those low-mass stars will have reached the ends of their main sequence lives and, if more massive than $\sim 0.4M_{\odot}$ will go ...


3

As I have discovered afterwards, the described bubble exists and is in fact a known phenomenon. For those, who want to read more, here is a wiki link http://en.wikipedia.org/wiki/Local_Bubble and the links therein. Otherwise, it is an underdensity in ISM (not only dust) in the region of Orion Arm (which is a minor spiral arm). The underdensity is up to ten ...


3

The $v_{GSR}$ in the linked paper seems to be the radial velocity (RV, in $km.s^{-1}$) with respect to the GSR. This information is mentioned in the Abstract section. I don't know exactly what kind of coordinates you have and how exactly are your velocities expressed (Cartesian vectors?), but what you probably need to do is to convert the velocity vector of ...


3

I posted an answer for this on Physics SE recently, but have also just had a query on this from another Astronomy SE answer, so I am adding this here for completness. You can approximate the plane of the galaxy as a disk made up of stars and gas, with a density $\rho(|z|)$, that decreases with absolute distance $|z|$ from the plane. If then assume that the ...


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