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1

Depending on the shape of the universe the luminosity distance is given by : \begin{equation} d_L(z) = \left\{ \begin{array}{rl} \frac{(1 + z) c}{H_0 \sqrt{|\Omega_k|}} \sin \left[ \sqrt{|\Omega_k|} \int _0 ^z \frac{dz'}{H(z')/H_0} \right] & \mbox{for $k = 1$} \\ \frac{(1 + z) c}{H_0} \int _0 ^z \frac{dz'}{H(z')/H_0} & \mbox{for $k = 0$} \\ ...


2

The upper mass limit for a quark star depends on your assumptions and ranges between 1 and 2 solar masses (cf. this paper (arXiv link) from 2001). It seems to me that the reason for the similarity to neutron stars' mass range is that it both compact objects satisfy the TOV equation, $$ \frac{dp}{dr}=-\frac{G}{r^2}\left[\rho+\frac{p}{c^2}\right]\left[M+4\pi ...


2

There is some debate about whether it exists or not, but there has been some research into what is called a quark star. This article (should be a free link, but here is the arXiv version in case) suggests that A recent calculation for cold and dense QCD strange quark matter including corrections to order $O\left(\alpha_s^2\right)$ indicates that ...


5

[...] $\Delta^+ \rightarrow p + \pi^0$, [...] $\Delta^+ \rightarrow n + \pi^+$, which process is favored: the proton and neutral pion or neutron and charged pion [?] Since the kinematics (and corresponding "phase space" factors) for the two final states are presumably as good as equal, the evaluation of the branching ratio $$\text{BR} := ...


0

I'm not that firm with the concept of blazars, but as with all super-massive astrophysical objects like neutron stars, black holes and all their subclasses, their heaviness forces the particles to highly-relativistic states. There you just can't use the hypothesis of "charged billiard balls" for atoms any more - a neutron star for example is only stable ...


0

Most of it would've become radiation- mostly photons and some neutrinos. Some of that can be seen in the cosmic microwave background. The rest would've gone into the kinetic energy of remaining matter particles (i.e., heat). We don't actually know what dark matter is made of, but we do know what the reaction products of matter/anti-matter annihilation are ...


1

The word energy tends to be used in a rather vague way, and typically to mean something exotic. In the context of particle reactions energy either means photons or the kinetic energy of the particles leaving the reaction. For example an electron and anti-electron annihilate to produce two photons. By contrast the annihilation of a proton and anti-proton is ...


2

Your question can't be answered because the qualifier when it was only the size of our solar system is meaningless. The size of the universe is a rather vague concept. The universe may well be infinite (it's unlikely we'll ever know for sure) in which case it was always infinite and it doesn't have a size. You could ask about the size of the observable ...


0

One of the most troublesome of the anomalies with which the Big Bang is marred, is that it is not possible to physically describe the Big Bang process in an infinite universe. The reason for this impossibility is that it is equally impossible to speak of the concept of concurrency everywhere in such a universe. Yes, in such a universe even the concept of ...


5

SSC: synchrotron self-Compton BBC: Compton upscattered blackbody radiation; Compton upscattering of stellar blackbody photons XC: upscattering of photons emitted by the accretion flow; accretion flow photons From: http://arxiv.org/abs/1307.1309 and http://arxiv.org/abs/1403.4768


5

The obvious answer is hydrogen and helium plasma but the nuclear fusion can also create heavier elements. Are these heavier elements a significant portion of the core? As said in dmckee's answer, no, the core of the Sun is much too cool (about ~15 000 000 K) to burn any other than hydrogen into helium. The triple-alpha process, which converts helium ...


4

Herein a filler answer until one of our experts gets around to giving us a more detailed picture. The short answer remains "hydrogen and helium", plus what every metalicity the star started with. The reason is that at the temperatures of the sun's core production of the next stable step (carbon) is many orders of magnitude slower than helium production. ...


3

Well in principle, it's pretty easy. The example of the Sun is often taught in introductory astronomy laboratory courses. The Sun is a nice clean example because it is so bright compared to anything that might contaminate its spectrum. You just put some sunlight through a diffraction grating (or slit, or prism) that you've previously calibrated so that you ...


8

The lack of the electric field in modeling plasmas stems from the Lorentz force, $$ \mathbf F=q\mathbf E+q\boldsymbol\beta\times\mathbf B $$ where $\boldsymbol\beta=\mathbf v/c$. For most astrophysical plasmas, the force is zero, so we have that $$ \mathbf E=-\boldsymbol\beta\times\mathbf B $$ So any time we see an electric field, we can simply replace it ...


25

Many astrophysical plasmas are well modeled as perfect conductors. Ideal MHD assumes this limit. As a result, there is no electric field in the fluid's rest frame. In other frames, we generally have $\vec{E} = -\vec{v} \times \vec{B}$, so there is an electric field. However, the perfect conductivity constraint means we don't have to model the electric field ...


0

There exists a basic asymmetry: there are no magnetic monopoles of the number and dimensions that the electric monopoles exist ( there are theories with magnetic monopoles and people are looking for them but we are talking of masses much larger than electrons and quarks) . Hand waving, (because I have not checked the math just extending the symmetry that ...


1

Probably because matter on a large scale is electrically neutral and therefore the electric effects cancel out .This asymmetry arises from the fact that atom as a whole is electrically neutral.


1

The gray model results from starting with the assumption that we have a plane-parallel slab: The light ray from the source (i.e., the star's atmosphere) travels at some angle, $\theta$, from normal, $z=0$. Since the light is coming from an angle, we need to account for that by modifying the radiative transfer equation to have a vertical optical depth, ...


10

What you are looking for is called the stellar mass function by astronomers. It is the distribution of masses for stars. There is a nice review of the definitions, measurements, and basic theory in Galactic Stellar and Substellar Initial Mass Function, Chabrier 2003, PASP 115 763. It discusses both the initial mass function (IMF) and the present-day mass ...


1

I think you've got the whole reason. We have good evidence that meteor showers are caused by lots of tiny pieces of fluffy comet detritus, which has no chance of reaching the ground; a bright bolide, however, is much more likely to have a substantial rocky or metallic mass.


2

This would ultimately be more a problem of signal processing than physics. The situation is detecting a signal at a very low signal to noise ratio. At the broadband level, the noise (starlight) is several orders of magnitude more intense than the signal (the explosion). The only hope would be some sort of spectral technique , taking advantage of spectral ...


3

To calibrate our expectations, consider the largest nuclear weapon ever detonated, the Tsar Bomba. It's yield was at most about $58$ megatons TNT equivalent, or about $2.43\times10^{24}$ erg. Now, let's consider a smallish star, something like Gliese 581, which is reasonably nearby, small and faint, and has a planetary system (of some sort: the number of ...


1

The two books on my shelf that I regularly thumb through are: Galactic Dynamics by Binney and Tremaine Galaxy Formation and Evolution by Mo, van den Bosch and White You can probably tell from the titles that neither is a general astronomy text. I find both to be excellent graduate-level texts on their topics. Galactic Dynamics is a classic from the '70s, ...


4

I agree it looks like a transverse wave - like the ripples on a pond. But I believe you are fooled by a simple thing: the waves you are looking at look like "illuminated ripples" but are in reality just changes in temperature (changes in brightness of the sun's surface). If you have a shock wave traveling out across the surface of the sun, what happens? The ...


1

I only know one particular reason: The transition responsible for the H1 line is highly forbidden and shows an extreme lifetime (10^7 years), so the absorption rate in interstellar clouds, which can be very opaque for every other radiation, is very small. Looking at the H1 line allows you to see objects which are for example hidden behind dust clouds that ...


-4

Galaxies stay together because gravity is about A BILLION TIMES faster than the speed of light.* *This is pure heresy to Einsteinians of course (disciples of Einsteins general and special relativity theories), since they firmly believe that nothing travels faster than light. Not even gravity. If this were true however,it would make galaxies like Hercules A ...


1

First: Two different concepts of point and direction are used in the question. One concept is "point of big bang" and "direction of (wrt) bing bang", second is "point of observer" and "direction of (wrt) observer". The (assumed) initial nig bang point is (wrt to the observer) not a point now, it is a whole volume/surface, so a volume/surface has many ...


1

A dense object like a planet, star, or black hole would blur the image of the universe. what we see, instead, is a very smooth and homogenous image. I guess this would still leave open the prospect of some sort of diffuse object of sufficient mass to bend all the light coming into us, but there are reasons to believe that no object with a mass this large ...


2

Not all galaxies are disk-shaped, but some certainly are. (Some others are spiral, etc.) For one thing, we see a lot of galaxies, and several of them look exactly like they would if they were disk-shaped and we were just seeing them from different angles. Some seem circular because we are seeing them head-on, while others seem more linear or elliptical ...


0

Assuming that the radiation source is a blackbody then redshift stretches all the wavelengths by a factor $(1+z)$ as per your formula. You are then correct to say that an application of Wien's law will yield a temperature which is lower by a factor of $(1+z)$. Hence the microwave background appears now to have a temperature of 2.73K, but it was emitted ...


0

I guess what is meant is mass loss powered by non-radiative heating occurring above the photosphere. i.e The outer atmosphere is heated by magnetic Alfven waves and causes a wind to stream away from the star. The same magnetic field is responsible for heating a chromosphere - a hotter layer above the photosphere. Here is a paper by Schroder and Cuntz ...



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