Tag Info

Hot answers tagged

26

At first many people didn't care much for black holes. But later people showed that they were pretty unavoidable features of the theory of general relativity and that theory made other quite precise predictions that were tested and found good. So when you are told that black holes are required if you have GR and GR looks like the best game in town then it ...


15

...why do we trust black hole physics? ... (physics which is derived by combining quantum mechanics and GR such as Hawking Radiation, things relating to the Information Paradox, etc. ) Formally, there isn't quite a reason to because we've not observed these things yet. But that's also perfectly okay as well because that is how science sometimes works: ...


5

It was the first to fit the observations (e.g. the anomalous precession of Mercury), while having no free parameters to "adjust". The only parameter is the gravitational constant, which was already known with high precision in 1916. Its every prediction since either has been confirmed or is consistent with observations without any massaging. It has ...


4

Let me try to break down the question into several parts, in the context of seeking a gravity theory that satisfies an action principle. That is, we are looking for a Lagrangian density that describes the theory. First, the equivalence principle tells us that the gravitational field must couple universally to matter. Second, the theory has to be (at least) ...


4

About evidence supporting the existence of Event Horizons in these very compact objects, here are some news from the well known Cygnus X-1, one of the most studied compact objects and the most promising candidate for a stellar collapse black hole: ... evidence of just such an event horizon may have been detected in 1992 using ultraviolet (UV) ...


3

This paper is interesting. It uses the method of calculating the number of nucleons in the neutron star, $N$, based on the radius, $r$, the number density as a function of radius, $n(r)$, and the metric function $\lambda$, which comes from the equations of general relativity: $$N=\int_0^R 4\pi r^2e^{\lambda/2}n(r)dr=\int_o^R4\pi r^2 ...


3

General relativity (GR) turned out to be a great mathematically beautiful theory with amazingly accurate experimental predictions/observations (e.g, bending of light, precession of Mercury, etc). This theory naturally provides some simple solutions which are called black holes. In that sense one should take them seriously as they come from a firmly ...


3

In principle, no, you cannot make a Dyson sphere which is indistinguishable from the CMB. The reason is fairly simple. Let's start with a blackbody DS which encloses nothing at all, and is so far from any nearby stars that no noticeable radiation reaches it. Since it is surrounded by CMB with an effective temperature of 2.75 K, it will reach an equilibrium ...


2

Cinsider an observer who is stationary at point $A$. Because they are stationary $dr = d\theta = d\phi = 0$ and the metric becomes: $$ ds = \sqrt{g_{00}(r_A, \theta_A, \phi_A)} dt $$ And likewise for an observer at point $B$: $$ ds = \sqrt{g_{00}(r_B, \theta_B, \phi_B)} dt $$ The relative change in the frequency of the light is simply the relative time ...


2

The gravitational mass of a neutron star is quite a lot less than its baryonic rest mass (plus the mass associated with the kinetic energy of its contents), because a bound neutron star, by definition, must have a total energy (the sum of its internal energy and gravitational potential energy) that is less than zero. In a “normal star” this is also true, ...


1

Time might be relative, aging (time passing) is absolute. Run around, jump into a rocket, speed up and circle a few times around a black hole, and do whatever else you fancy, all observers will agree how much you have aged in the process. Here, for 'aging' you can read 'proper time': the time that has passed according to your wristwatch.


1

Classical Lagrangian field theory deals with fields $\phi: M \to N$, where $M$ is spacetime and $N$ is the target-space of the fields. We shall for convenience call $M$ and $N$ the horizontal and the vertical space, respectively. The metric $g$ can be viewed as a classical field of this kind. OP is asking about finding the Euler-Lagrange equations. In that ...


1

If you built the sphere, at the optimum radial distance, then insulated the exterior as much as possible, would the gravitational redshift provided by the black hole not act to sort out your problem for you, if you want to dissipate it, as far as co-ordinate observers at any reasonable distance were concerned? Would accretion discs and the massive gravity ...


1

The general theory of relativity predicts that kinetic energy will contribute to gravitational mass. Here is a paper that explores the gravitational effect of kinetically energetic particles within a system: http://arxiv.org/PS_cache/gr-qc/pdf/9909/9909014v1.pdf. Here is an interesting article by Frank Helle on the production of gravity by relativistic ...


1

The relevant part of the book is the section titled Motion through Spacetime in chapter 2. I'll copy the paragraph, but it's a bit long so feel free to skip over it: Einstein proclaimed that all objects in the universe are always traveling through spacetime at one fixed speed—that of light. This is a strange idea; we are used to the notion that objects ...


1

One of the basic principles in relativity is that spacetime always looks locally flat. By this I mean that if you restrict your observations to a small region surrounding you the curvature becomes negligable. You can make the effects of curvature arbitrarily small by making the region you consider arbitrarily small. The point of this is that for your clock ...


1

Mass is a Lorentz invariant quantity! The relativistic mass is not the real mass, it is is just called relativistic "mass" for obvious reasons. This term is abandoned by most textbooks, as it often causes this confusion.


1

Let there be given a general covariant matter action $$S~=~ \int \! d^4x~ {\cal L}, \qquad {\cal L}~=~e L, \qquad L~=~L(\Phi,\nabla_a\Phi). \tag{1}$$ The main strategy will be to demand that the matter fields $\Phi^A$ carry flat rather than curved indices$^1$. This is achieved with the help of a vielbein $e^a{}_{\mu}$, where $$g_{\mu\nu}~=~e^a{}_{\mu} ...



Only top voted, non community-wiki answers of a minimum length are eligible