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) ...


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

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 ...


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

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. ...


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 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

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

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 ...


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

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 ...


2

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 ...


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 ...


1

It's a wave, and like any other wave, it loses flux (power over area) with the square of the distance. That is, if you double the distance it travels, your wave will have a quarter of the flux it had. If you triple the distance, your wave will have a ninth of the flux it had. This is a property of waves and can be observed in light intensity, sound, and any ...


1

While it's not recommended, your comments cover enough ground that rather than make a long answer, I'll address them here. If my answer is unsatisfactory, you might want to create a new question, clarifying the specifics. A star 3 or 5 times the size of our sun will gradually burn up its hydrogen and collapse to form a black hole big enough to swallow ...


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