# Tag Info

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All models of gamma-ray bursts involve extremely energetic phenomena: particular types of supernovae, the coalescence of binary compact objects, strong magnetar flares, or tidal disruption events. It turns out that these events are quite rare - so rare, in fact, that GRBs would be expected to occur in a low-redshift Milky Way-like galaxy at a rate of only ...

23

You correctly state that neutrinos do not interact too often. The physical parameter describing that is the effective cross-section. So what you observe in a detector is not the neutrino itself, but secondary particles, e.g. muons. Colloquially put, you may regard anything with high mass (density) in between the neutrino source and your instrument (to detect ...

9

High energy muon neutrinos occasionally interact and produce a muon. Energy and momentum must be conserved in the process and the muon heads off in the same direction as the neutrino. The relativistic muon can then be tracked by a network of detectors which are sensitive to the Cerenkov radiation produced when muons travel faster than the speed of light in ...

8

Long and short GRBs are thought to arise from different types of event, involving different types of star. Therefore the question you should be asking is why are their event rates so similar?! Long duration bursts are thought to be produced in the death throes of rapidly, rotating massive stars - the hypernova model. Short GRBs are thought to be produced ...

7

The current supernova is a supernova of type Ia. Supernovae of type Ia are used as standard candles for distance estimates, especially used to determine the Hubble constant. Hence by a better calibration of this kind of supernovae, more about the reliability and accuracy of distance estimates can be learned. The expansion rate (in relation to the distance) ...

7

Let's start by making some points clear: 1. We don't know what the Big Bang was. Rather, we know that the Universe is expanding. If you extrapolate backwards, you'd expect the Universe to be denser and denser. More specifically, we talk about this as a change in the scale factor $a$, and this gets smaller and smaller as we look further back in time. ...

6

tl;dr: Yes. Feynman's beads on a string argument The other answers skirt what I think is the issue that the OP is asking about. In a lossless medium a spherical wave packet itself, caused by a disturbance, will not "loose energy" itself. If you integrate over a large volume you get a constant energy versus time. Of course the flux per unit area will ...

6

There are multiple solutions to general relativity which allow for multiple different types of black holes. The "normal" black hole you see most people talk about, with a zero-volume, point singularity, is known as the Schwarzschild black hole. If the black hole is spinning, the Schwarzschild solution no longer applies and you're talking about a different ...

6

The article you've read is not quite accurate/correct. A more correct pictue is as follows: A star may approach a super-massive black hole (SMBH) so closely that the tidal forces of the SMBH tear it appart. The distance to the SMBH at which this happens is often referred to as the tidal radius. For a (non-rotating) SMBH with a mass in excess of about $10^8$...

5

Your question touches on a few points. First, yes, he was flippant, but the risk was super-low. The simplest way to explain this is that nothing happens in CERN that doesn't happen all over the universe and in the upper atmosphere of the Earth or on the surface of all the moons, planets and stars every day. For example, the Oh My God particle was ...

4

I think it is referring to the speed and Lorentz factor $(\beta = v/c$ and $\gamma = [1-\beta^2]^{-1/2})$ of the gas as a whole. Within the gas, there could be particles moving with a variety of velocities. So if you pick up a ball of gas at 10,000 K (ouch) and throw it at 100 m/s then the bulk speed is 100 m/s, but obviously the particles in the gas have ...

4

There are several types of supernova and ways that the core can collapse. Lets take an extreme case in which gamma-ray photodisintegration destroys all of the heavy elements (Si, Fe and Ni, etc) and breaks them all up into protons, neutrons and electrons. Each nucleus releases all of its binding energy, about 9 MeV per nucleon mass or 0.9% of the rest mass....

4

The most favoured WIMPS at the moment are probably neutralinos, see http://en.wikipedia.org/wiki/Neutralino These particles are purely hypothetical at the moment. The mass estimates in the above Wikipedia article for the lightest neutralino range between 10 and 10,000 GeV, meaning that the production rates in SNs will be much lower than with an assumed 1 ...

4

Don't think of the singularity as being an object made of matter. A black hole is a vacuum solution to the relativity equations. That means there is nothing inside the black hole. A black hole doesn't contain matter, but it still has mass. The mass of the black hole can be observed in the curvature of space-time around the black hole. It also has angular ...

4

The non-thermal S-Z effect is caused by inverse Compton scattering of the CMB photons from a non-thermal population of electrons - i.e. electrons that have high energies not because they are hot, but because they have been accelerated non-thermally. The usual mechanisms are accelerating by electromagnetic fields and the Lorentz force. The rest-mass energy ...

3

In a word, no. The universe was certainly quite hot for the first few hundred thousand years, but most of the important events of the Big Bang happened in the first few seconds, when the universe was extremely hot. By the time the universe was about 10 seconds old it was cool enough for protons and neutrons to combine into deuterium and thence helium, as ...

3

The magnetic field of the rapidly rotating neutron star interacts with the material coming from the other star in the binary. This results in a transfer of angular momentum, spinning the neutron star down but accelerating the material out of the system. The effect is somewhat like a garden sprinkler seen from above, with material flung out ("propelled") of ...

3

One way to think of a black hole is that it is what is left behind when some matter (or energy) collapses so far that an event horizon forms. After that, no information of any kind can get out past the event horizon, so what happens inside has no effect on the rest of the universe. The externally visible properties of the black hole (basically the ...

3

The interpretation you suggest in the second paragraph is incorrect. It is understandable, since there is a debate in the literature - different papers come to potentially contradicting conclusions. "Excluding a possibility that the event is associated with substantial gamma-ray radiation, directed towards the observer" simply means that no observable GRB ...

3

The answer is pair production. Once photon energies exceed 1.02 MeV it is possible to spontaneously create an electron-positron pair in the presence of an atomic nucleus (to conserve momentum). In general for high energy photon interactions with matter you need to consider the photoelectric effect, Compton scattering and pair production. The former is more ...

3

You're correct that "$p$-wave" in this context means that the charmonium has orbital angular momentum $L=1$. The principal observable effect is that two-particle states with even $L$ have even parity, while those with odd $L$ change sign under parity. If you were to skim the Particle Data Group's list of charmonia (or the $c\bar c$ section of the short list)...

2

According to our current understanding, there is no lower bound for the energy needed to make a black hole. Any object, no matter how small, if compressed enough, could in theory form a black hole. Make it small enough, and it would not require much energy at all. But here's the main issue. Black holes have this thing called Hawking radiation: they ...

2

I'm not entirely sure what you mean, but the (Planck's) formula for blackbody radiation is given by $$S_{\lambda} = \frac{8 \pi h c}{\lambda^5} \frac{1}{e^{hc/\lambda kT} - 1}$$ where $h$ is in $\mathrm{J \cdot s}$, $c$ in $\mathrm{m/s}$, $\lambda$ in $\mathrm{m}$, $k$ in $\mathrm{J/K}$ and $T$ in $\mathrm{K}$. So, the temperature is just in kelvin, not in ...

2

In empty space, just like a light wave, they spread out, becoming less intense as they get further from their source, but never vanishing completely. At some stage the waves from a distant event might become undetectable in local noise, or so weak that quantum effects might become relevant, but essentially they never die out. As @uhoh points out, they do ...

2

This is a difficult question to answer. Physics really starts after the big bang. Scientist don't know about the laws, if any, before the first instance after the big bang. The time before and during the big bang is really the province of philosophy.

1

Not being an expert in star formation, I found a well-written paper summary from which I conclude that typical star formation rates range between $6 \ldots 24 M_\odot / yr$. The blog quotes the following graph by M. Boquien, V. Buat, and V. Perret, see https://arxiv.org/abs/1409.5792 In this paper we investigate in isolation the impact of a variable star ...

1

I found a database that contains all GRBs from April 1991-August 1991. However, it has 1637 registered GRBs so it would probably be enough for your purposes. Here is the link-Goddard Space Center GRB Archive. Also, just a fair warning, that website is incredibly old and very difficult to use, it took me nearly fifteen minutes to navigate to the actual ...

1

There are theories which suggest that the universe is like a giant spring, constantly expanding and collapsing. Its possible that pre-big-bang, there was another universe which collapsed, only to re-expand into the universe we know now. But in reality we have no idea

1

In short, there are no nice standard formulas for this. One can make some order-of-magnitude calculations, though. The key formula you need is the inverse-square law: the intensity of a spherical energy source falls off with the inverse square of distance. $$I(r)=\frac{I}{r^2}.$$ The useful thing is that if you know that some source with intensity $I_1$ ...

1

I think you mean that kappa = sigma / m. The sigma is the "cross section," like the area you expose when you slice a watermelon in half. The m is the mass of the particle that has the cross section in question (it's not literally the cross section of the physical particle, but an effective cross section for interaction with light), or more often, the mass ...

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