# Tag Info

10

Overall, a star stays more or less neutral. This is true for all stellar objects beside black holes. I am using a simple calculation that can be found in a footnote of https://arxiv.org/abs/1001.3294 on p. 11 chap. 2. Suppose the star has an overall charge of Z times the elementary charge, $Ze$, and we consider the Coulomb repulsion of a test particle, say ...

8

Background Stars are composed of plasmas, which are an ionized gas that exhibit a collective behavior much like a fluid. There are two important aspects of plasmas to keep in mind. The first is that they act like very highly conductive metals in that the electrons can move very freely in order to cancel out any charge imbalance. The consequence is that ...

7

We want the Newtonian limit of the Einstein Field equations for nonzero vacuum energy(=cosmological constant). As $\rho_\mathrm{vac}=\Lambda/4\pi G$ is a mass(=energy) density, Poisson equation is $$\Delta\Phi=4\pi G\rho(\boldsymbol r)-\Lambda \tag{1}$$ If we assume spherical symmetry, and point-like source $\rho\sim\delta(\boldsymbol r)$, the ...

5

Cosmic Rays are most often high-energy particles, mostly protons and alpha particles accelerated to high velocities by cosmic magnetic fields. They do not show up in the microwave wavelength range that comprise the CMB. As @ACuriousMind says in the comment, there is contamination in the CMB, but this is mainly due to Galactic dust and Bremsstrahlung from ...

5

It depends on what you mean by "real". Let's start with the Andromeda galaxy, which is easily visible with the unaided eye from the Northern hemisphere (and probably from a large part of the Southern hemisphere, too), but it does not look anything like what you are used seeing in photographs. It looks like a small, pale cloud in a dark patch of the sky and ...

4

Data is taken at different frequencies because different frequencies contain different information. The data here covers a factor of 9 in frequency (equivalently wavelength). Compare this to the factor of 2 accessible by human vision. Given two objects that have the same power at one end of the frequency range, they could very well differ at the other end. ...

3

The "strain" (usually $h$) is a measure of the 'distortion' (in the metric) from a gravitational wave (GW). The energy carried by GW (its 'luminosity' of sorts) scales like, $L_{gw} \propto h^2$, and for energy to be conserved as it radiates away, we need $L_{gw} \propto r^{-2}$, so we find that, $$h \propto r^{-1}$$ So your latter estimate is more ...

3

The surface temperature of the star has no direct bearing on its metal content. Most stars in the immediate vicinity of the Sun have a very similar metal content. What you are talking about is how this metal content affects the observed spectrum of the star. If the star's photosphere is very hot then the metals become ionised and you don't see the (for ...

3

When black holes form from the collapse of stars, we think they usually produce bright supernovae explosions which release tremendous amounts of energy. Some models suggest that a non-negligible fraction of stars which produce black holes may not produce normal supernovae, these are often called 'failed supernovae', see for example astrobites: Gone Without ...

3

Many spectral lines are very sensitive to the surface gravity of the star - which enables a distinction between dwarfs and giants, because a giant's surface gravity is factors of $\sim 100$ lower than that of a dwarf of the same temperature. The reason that surface gravity plays a role is via hydrostatic equilibrium; the densities and pressures in a gas ...

3

All the neutrino detectors we have or might build will have course angular resolution because they detect the direction of scattered products of neutrino interactions rather than the direction of the neutrinos themselves. Worse, the solar neutrinos are relatively low energy (a few MeV), which means the scattering angles are large. Yes, we could accumulate ...

3

You get an extra term that increases with r: $$a = -\frac{G\cdot M}{r^2} + j\cdot r$$ with j as the repulsive component.

2

First, let me explain the question a little further. The Hulse-Taylor binary is a binary system composed of two neutron stars orbiting each other. Each star is an extended body, and is in the gravitational field of the other, so should experience tidal forces, because one part of the star is closer than another to the opposite star, so the gravitational ...

2

Hints: Note that the derivative of the sign function $${\rm sgn}^{\prime}(z)~=~2\delta(z) \tag{A}$$ is twice the Dirac delta distribution. This fact seems to be at the heart of OP's question. Repeated differentiations of the Mestel disk potential $$\Phi~:=~ v_0^2 \ln(r+|z|), \qquad r~:=~\sqrt{R^2+z^2}, \tag{B}$$ leads to \frac{\partial \Phi}{\partial ...

2

Apart from the heating due to sound absorption, as per the comment by HolgerFiedler, I don't think you will find a mechanism that can radiate due to polarization effects in the medium. Any EM radiation would be at the frequency of your acoustic waves. With the difference between the speed of sound and the speed of light, that would be very-long-wave ...

1

You need to look in to the GZK cutoff. In 1966, Greisen, Kuzmin and Zatsepin calculated that above a threshold of $5\times10^{19} eV$ cosmic ray protons would lose energy to photo-pion production on the cosmic microwave background fairly rapidly. The consequence of this is that cosmic rays above that energy can't travel more than about $50 Mpc$ without ...

1

There are many mechanisms which can contribute to broadening spectral lines. Usually, one way or another, the atoms have a broad range of random velocities which cause doppler shifts of varying amounts, broadening the line. One the more fundamental cases is simple 'thermal broadening', where the velocity is from thermal motion. The hotter the gas is, the ...

1

A black hole $m\approx300,000M_{sun}$ at the center of galaxy $NGC 4845$ is thought to have a giant gas planet with the mass no larger than that of a brown star in orbit, though it is currently being drained into the black hole. (And by currently I mean what we are seeing right now, which happened a long time ago). The gas giant is thought to have been ...

1

The key word, to answer your question, is wavelength, as in: which wavelengths are you interested in? So far, GRBs have been seen in all wavebands, from TeV down to radio frequencies (of course, not all GRBs have been observed at all frequencies). The waveband in which GRBs release most of their energy is, unsurprisingly, that of $\gamma$ rays, with hard ...

1

In a realistic scenario, not very far, maybe just a few miles if you're discussing fragments large enough to leave impact craters. A fair bit further than that if you're talking smaller debris. The distance depending on how high the meteor broke apart and how fast and at what angle it was approaching. If you consider that the dense part of the atmosphere ...

1

Yes. Black Holes (BH) can grow from accreting anything with energy --- including dark matter (DM). I'm not entirely clear on the second part of your question, but probably the most important thing to keep in mind is that the black hole information paradox is still unresolved. Answering how information is not lost for any type of particle, including DM, ...

1

They both refer to the same thing. The habitable zone is also called the Goldilocks zone, a metaphor of the children's fairy tale of Goldilocks and the Three Bears, in which a little girl chooses from sets of three items, ignoring the ones that are too extreme (large or small, hot or cold, etc.), and settling on the one in the middle, which is ...

1

The question operates under a false premise. It is not the case that fitting a blackbody spectrum to the Sun gives you its "surface temperature". The Sun does not have a black body spectrum, although sometimes that approximation is made. A star also does not have a single "surface temperature". What you can do is divide the luminosity of the Sun by \$4\pi ...

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