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Well i'm not into dark matter, but i am into entropy and stuff, so i will post an answer. How one is supposed to measure the entropy before and after, by counting micro-configurations, by counting volume/size, all these together? i suggest one or both of the above may give you an answer as to how the 2nd law may still be valid. No need for radiation (and ...


0

Dark matter does not radiate photons by definition, but as I said in the comment to CuriousOne, dark matter may not have electromagnetic radiations to first order, but it does have gravitational radiation. The current Big Bang model accepts an effective gravitational interaction and thus the existence of gravitons, i.e. elementary particles of mass zero and ...


1

A free dark matter cloud (without the presence of ordinary matter) will simply not "collapse" the same way a radiating gas cloud does. In both cases total momentum, angular momentum and energy are conserved, but in the case of a gas cloud the photons can carry away some of the angular momentum and most of the energy, in case of a dark matter cloud they ...


1

I think the assumption that radiation is required for a collapse in general is mistaken. Think about a cloud of gas. If it is going to gravitationally collapse it must have a negative total energy; if it doesn't parts of the gas will fly off. If it has a negative total energy then there is some finite maximum size for the gas cloud, where it only has ...


4

A brief overview of stellar evolution can be depicted in the following image: (From here which says it is originally from an encyclopedia; click here for larger image). The heavier stars (top track) have very short life times (a few million years) because they run through hydrogen, helium, carbon+oxygen, ..., iron fusion in the core. Once a particular ...


4

You are neglecting two important facts. The first one is that stars, toward the end of their lives, return to the interstellar medium (ISM) a lot of their initial mass, but now enriched with heavy elements produced by nuclear reactions inside the stars themselves. In this way, younger stars which form from the ISM begin their life with a larger fraction ...


2

It is convention and laziness (and I'm as guilty as anyone). In fact cgs units should not be used (according to the International Astronomical Union), in the same sense as Pluto should not be called a planet. Both were decisions made by the IAU. In the case of units, the IAU unambiguously endorsed the use of SI units, except for a short list of defined units ...


0

Light with wavelength $\lambda\gg10\,\mathrm{km}$ can pass around a neutron star thanks to diffraction, even if the star is made of a perfectly absorbing material. That's exceptionally low-energy radio waves, though, and pretty different from what you probably had in mind.


1

OK, let's try a back of the envelope... http://www.slac.stanford.edu/econf/C0805263/Slides/Budge.pdf states that the total anti-neutrino count from a supernova is on the order of 1e58 and the neutrino energy is on the order of 40MeV. Let's imagine all of these neutrinos would go trough earth, which is the worst case scenario. The nuclear cross section of ...


1

It is difficult to keep track on these things, specially since quasars are also highly variable. Trying to answer the question of the post title, I found this example that sound pretty impressive: $7\times 10^{14}\,{\rm L_\odot}$, or $1.4\times 10^{41}$ W. The ESO press release refers to the kinetic luminosity, or the kinetic energy of the outflow per unit ...


1

The Jeans equations can be a bit tricky. Their simplest form (in cartesian coordinates, with no particular assumptions) is: $$\frac{\partial\nu}{\partial t}+\frac{\partial(\nu\bar{v_i})}{\partial x_i} = 0$$ $$\nu\frac{\partial\bar{v_j}}{\partial t}+\nu\bar{v_i}\frac{\partial\bar{v_j}}{\partial x_i} = -\nu\frac{\partial\Phi}{\partial ...


1

Let's suppose the broadening mechanism is van der Waals or Stark broadening - something where the energy levels of individual atoms are perturbed. In this case you could use the following argument. Divide the line profile up into groups of atoms which share the same perturbation and treat each of these as a subpopulation with a different energy gap and ...


0

The flux (defined as power per unit area) coming from the surface of the star is given by $\pi I_{\nu}$, where in the case of a blackbody, the specific intensity $I_{\nu}$ is given by the Planck function. The total luminosity of the star is therefore $\pi I_{\nu} 4\pi R^2$. At the earth, the received flux is therefore $\pi I_{\nu} 4 \pi R^2/(4\pi r^2)$, ...


1

The answer given by Kyle refers of course only to the surface or photospheric temperature of the neutron star - the temperature of the layer from which photons can escape to reach an observer. In these outer layers the relationship between temperatures and particle motions is more-or-less consistent with the "everyday" Maxwell-Boltzmann picture referred to ...


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ever since the big bang,there universe has been expanding.other stars continues to burst and expand when they get hotter.so yes other stars wil disappear in the future and no longee be seen with naked eyes.


-1

The plasma around Sgr A$^*$ is emitting in X-rays which indicates 10$^7$ K temperatures. Densities vary between 10$^6$ atoms/cm$^3$ and 10$^8$ atoms/cm$^3$, depending on which point along the accretion disk (pdf link)--possibly also depends on who you ask. I don't know that we can answer this question for many wavelengths as there can be a lot of stuff in ...


20

First, strictly speaking a neutron star is not a nucleus since it is bound together by gravity rather than the strong force. Measuring a surface temperature for any star is deceptively simple. All that is needed is a spectrum, which gives the luminous flux (or similar quantity) as a function of photon wavelength. There will be a broad thermal peak somewhere ...


1

Let me add something to the second part of the question. The evidence for the existence of a black hole starts with the observation that there is a very compact, massive object that is not emitting as much light as a "normal" star of that mass. This in itself does not rule out a neutron star, because it may well be as the OP proposes that the neutron star ...


6

Dark matter as far as gravitational forces go , has the same behavior as normal matter. That is how it was discovered and defined. By balancing gravitational forces in the motion of galaxies etc, it was found that more matter was needed than the matter estimated from the luminosity of the bodies. It was observed that the trajectories would not fit the ...


0

The majority of the sources that LIGO (and other gravitational wave detectors) are aiming for are astrophysical (e.g. neutron stars, black holes, supernovae, pulsars). The expected cosmological gravitational radiation from standard inflationary models (see this recent article from P. Steinhardt) would be very weak in the LIGO band (10-1000 Hz). There are a ...


2

The answer is yes! And it has been used in this way many times. To pick a random, recent example by Vogt et al. (2014) - a 4-planet system orbiting a K0V star. http://arxiv.org/abs/1404.7462 The problem is akin to separating out multiple frequencies in a blended chord and is accomplished with fourier analysis techniques. A useful public resource ...


21

A typical giant galaxy, such as the one you've provided a picture of, has a radius of something like $10\;\rm kpc$ (kiloparsec - $1\;\rm pc \approx 3.2\;ly$). A supermassive black hole hosted in such a galaxy has a mass of something like $10^6-10^9\;\rm M_\odot$ (solar mass, $1\;\rm M_\odot \approx 2\times10^{30}\; kg$). The monstrous billion solar mass ...



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