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0

Exactly what here is referred to be "faster than light?"  By fast than light it means that if you put a light beam to your left and to your right and you ball take off like racers running towards a take across a finish line that you get to the finish line first. And how does this work? Are you clumping together a punch of exotic matter at the ...


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I think the comment was talking about a race type situation. For instance if you had a 50m dash with a bunch of runner lined up horizontally and running towards a horizontal finish line with some tape across the finish line they could race and see who hits the tape first. It is exactly (and only that sense) that the warp drive arrives first. Let's say ...


1

If you and your friends took some helicopters to the north pole and went up and then took off in different directions and flew at the same altitude you would feel like you were being bent towards each other, but yet as you all started to approach the south pole you would notice that you were all moving a way from other at first, you were all moving parallel ...


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If you would still like to obey some matter fields, there is a possibility still, via topological shenanigans *. If the metric is totally flat until the hypersurface, you may still get changes in the matter field if this hypersurface develops a Cauchy horizon at some point in the future, either by removing pieces of the manifold (ie, naked singularities) ...


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It is an extremely complicated problem. Imposing restrictions on the stress energy tensor will give you different possible spacetimes (which spacetimes is also quite a complex problem), but if those restriction apply in all cases is quite hard to prove. The stress energy tensor still has to obey the field equations of the various matter fields that exist, ...


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According to Newtonian gravity if you made 100 million cannon balls (10kg each) and shipped them to deep space assembled them with some explosives and exploded it with enough force so none get hurt but they all went out spherically to a maximum of 2 miles (if you don't know what a mile is, just read it as a km and none of the physics will be different but ...


1

It is helpful to look at another simpler classical field theory, electromagnetism. Electromagnetism has equations for how electromagnetic fields change in time, they are fairly simple compared to general relativity, namely $$\frac{\partial \vec E}{\partial t}=\frac{1}{\mu_0\epsilon_0}\vec \nabla \times \vec B, \text{ and }\frac{\partial \vec B}{\partial ...


1

No, the curvature outside does not increase, it can decrease though. First lets consider a simpler case. You have a small spherical ball of neutral hydrogen at rest in an otherwise empty and infinite (asymptotically flat) spacetime that is not in equilibrium (and larger than equilibrium size). It contracts, it gets smaller, it develops some pressure, but ...


4

There are two issues. One issue is that some people try to oversell the "no hair" theorem to areas beyond where it applies, they apply a long term analysis of final states as if it applies in the short term. The second issue is that in the shirt term we have two objects, the black hole and the incoming object. We have to think about both. I'll go into ...


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The curvature of spacetime is determined by the stress-energy tensor, and in the stress-energy tensor we do not distinguish between matter and energy. The two are treated as equivalent and interconverted using Einstein's famous equation $E = mc^2$. The Schwarzschild radius of a mass is conventionally written as: $$ r_s = \frac{2Gm}{c^2} \tag{1} $$ where ...


-4

Is a black hole's mass uniformly distributed? One hears conflicting answers to this. One article I rather like is the mathspages Formation and Growth of Black Holes. See this bit: "Historically the two most common conceptual models for general relativity have been the 'geometric interpretation' (as originally conceived by Einstein) and the 'field ...


-1

In the mathematical models of black holes that we use, there is a parameter M which is related to the mass-energy of the black hole. There is NO hint of distribution of mass-energy inside the black hole. In particular: a non-rotanting BH is spherically symmetric (these BH probably do not exist in nature, since all astronomical objects rotate and pick up ...


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A non-rotating black hole can be treated as spherically symmetric using the Schwarzschild metric. A rotating black hole has an axis of symmetry and can be represented with the Kerr metric. Treatments of black holes using either of these would make the assumption that the "test particle" you are considering does not influence the metric (is much less ...


0

Trivially no. Consider two such sets in a flat space-time with the same number of events. Each event in the second set is associated with an event in the first event such that taking the earliest event in each set to define the origin of a frame of reference the 4-vectors of position of events in the second set are twice the four vectors of position of the ...


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You can only recover conformally related spacetimes from its null geodesics, that is, the class of spacetimes related by the transformation $g_{\mu\nu} \rightarrow \Omega^2(x) g_{\mu\nu}$ which possess a different matter content


4

The short answer is yes. An object moving at non-relativistic velocity $\vec{v}$ in a weak gravitational field will have a proper time $\Delta \tau$ elapse that is related to the time $\Delta t$ on distant clocks (far from the Earth) by the equation $$ \frac{\Delta \tau}{\Delta t} \approx 1 - \left( \frac{1}{2} \frac{v^2}{c^2} + \frac{G M_E}{r c^2} \right) ...


0

The analogy between how a massive object bends a rubber sheet and how sources of mass-energy-momentum bend spacetime is a poor one. It's essentially comparing the mechanical deformation of a two dimensional sheet in space to the curvature of four dimensional (3+1) spacetime, which have a mathematical similarity, but not a physical one. In Newton's theory of ...


1

“Proper mass” and “gravitational binding energy” in general relativity There are some issues with proper mass. I think the best way to remind yourself of this is to say to yourself proper mass isn't proper at all. I'm reading Robert Wald's "General Relativity" and after the discussion of the Schwarzschild Solution it goes on to talk about ...


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I think in general there are a wide class of models that should pass cosmological and astrophysical constraints. The Living Review on this is a good place to look. In section 14 they explain the basic constraints on allowed types of $f(R)$ models. It seems like they amount to $f_{,R} > 0$ and $f_{,RR}>0$ when $R>R_0$, where $R_0$ is the scalar ...


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 ...


0

To add to Johannes's succinct answer: there are two other things about time that we believe to be absolute: The topology of any web of causal links between events in space time and The direction in time of causal links. If we postulate that a cause-effect relationship can only propagate at a maximum speed of $c$, then if a cause comes before an effect ...


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.


0

What is the binding energy of a neutron star? What Rob said is about right. It's about a fifth of the original mass-energy. See Wikipedia: "Its mass fraction gravitational binding energy would then be 0.187". Neutrons which constitute a neutron star have a rest mass that is greater when separated from the star because they are bound with a certain ...


0

If time is relative, how could time pass? It doesn't really pass. That's just a figure of speech. Footballers pass. Buses pass. But there is no physical thing called time that actually passes. Instead things move. Things like light and planets and planes and people and hearts and blood and electrochemical signals in your brain. And pendulums and cogs ...


3

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, ...


4

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 ...


5

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) ...


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...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: ...


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 ...


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 ...


27

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 ...


0

I don't think so Dirk. GRBs don't happen for nothing. See this 2001 paper by Friedwardt Winterberg. I think it's essentially correct. Why? Take a look at Einstein saying light curves because the speed of light is spatially variable. Also see Shapiro, and this Baez article: Einstein talked about the speed of light changing in his new theory. In the ...


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

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 ...


0

The volume form is $dt f(\theta,\phi)d\theta d\phi=(dt)(Rd\theta )(R\sin\theta d\phi).$ However, you already have that factor from $\sqrt{|h|}$ so your original equation might have meant to have the coordinate differential, not the 3-volume differential, in which case you should just use $d^3x=dtd\theta d\phi$ instead of the volume element $R^2\sin\theta ...


0

If you want to make a flat space theory for a topologically trivial manifold you can do it in the standard ways if your metric was very very close to the Minkowski metric. Yours is not. So for instance you aren't going to be able to ignore higher order terms in $h$ since in one of the $y$ directions your $h$ blows up. You can still compute an $h$ field by ...


0

Usually, the variation of a filed $\delta\phi$ is defined to be \begin{equation} \delta\phi\left(x\right)=\phi^{'}\left(x\right)-\phi\left(x\right) \end{equation} where the new field $\phi^{'}$ and old field $\phi$ are evaluated at the same point, if we take a active transformation point of view. So I think $\delta g_{\mu\nu}$ should be ...


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 ...


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) ...


0

Like Jimself, I know I've seen this somewhere (and will try to dig it up), but in the meantime I'll give you the answer off the top of my head. Can't guarantee this is entirely correct until I do dig some things up, but some parts are true (confident in the flat Universe part!). As long as you're interested in our Universe, your idea will actually work, ...


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

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 ...


0

Everywhere in the observerable universe is affected by gravity and since you can't escape it, that means it is always pulling on you. So every objecct in every place will be moving relative to everything else, so an absolute state of rest is impossible. If you could edit your post to quote the part of the book where he says "stationary", you will probably ...


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 ...


0

Actually the clocks tick at the same rate. You probably forgot that distances do not stay the same when you rotate the clock. (In non-Euclidean space turning a rod changes its length)


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} ...


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

I got the answer: if an index appears twice, it implies summation.


3

The Alcubierre drive is like an escalator. Space expands behind the ship, and this is like the moving stairs coming out of the floor behind you, while space contracts in front of the ship, like the moving stairs disappearing into the floor in front of you. This is not, however, a description of the actual Alcubierre process. We live in de Sitter space, ...


0

No, not in general. Things are only negligible for certain purposes. For example, a field that obeys the Klein-Gordon equation has the stress-energy tensor:$$T^{\mu\nu} = \frac{\hbar^2}{m} \left (g^{\mu \alpha} g^{\nu \beta} + g^{\mu \beta} g^{\nu \alpha} - g^{\mu\nu} g^{\alpha \beta} \right ) \partial_{\alpha}\bar\psi \partial_{\beta}\psi - g^{\mu\nu} m c^2 ...



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