# Tag Info

## New answers tagged general-relativity

0

I'm no physicist, but I think that event horizon will form, but nothing will ever fall there. So no information is destroyed, just very inaccessible. See How can anything ever fall into a black hole as seen from an outside observer? "All the oblects close enough to the horizon remain "sticked" to it and follow all the changes of the BH form" Imagine two ...

3

When we say that they are unit vectors, we mean that the proper length is equal to one. The proper lengths of the two vectors are $$\gamma_{ab} t^a t^b=1,\quad \gamma_{ab}n^a n^b=1$$ and should be equal to one, i.e. $1\to 1$, at all times. (In the Minkowski signature, one of these squared lengths is minus one, but that won't change anything about the text ...

2

I know that on the smallest scales, general relativity predicts that space-time is flat It's part of the Einstein Equivalence Principle that on a small scale spacetime is approximately described by the Minkowski metric, but you need to be clear that this is an assumption used to construct general relativity and it is not a prediction of general ...

1

I know that on the smallest scales, general relativity predicts that space-time is flat. This is not the case. First of all, General Relativity assumes that the spacetime on the smallest scales is flat. This is a pre-condition for GR mathematical apparatus to start to work. So, this cannot be the prediction of the theory. Second, if we try to predict ...

-1

Cool question. Gravitational waves can only be emitted from objects in very specific configurations - say, for example, binary neutron stars. Other sources include asymmetric systems - Wikipedia lists some - but binary systems such as the Hulse-Taylor binary are some of the most well-studied. So let's imagine that such a system is in place by a black hole. ...

1

"Quantum foam" is a physical phantasy, for which no evidence exists, so far. Having said that, contradictions between actual theories are normal and nothing to be upset about. Every theory has a range of applications, which is partly defined inside the theory and partly by its experimental limits. It is well understood, that Newtonian mechanics has nothing ...

2

I think it's fair to say that there is no consensus on the status of firewalls (which one might define as an answer to the question, "what does an observer falling through the horizon of an old black hole experience?"). I haven't read the paper you cite in detail, but a quick look through suggests that it doesn't address the AMPS argument. A resolution ...

0

I am posting this answer - based purely upon old comments from other users - to get this question out of the 'Unanswered' section of our site, following David Z's suggestion from a meta post. Note that this answer is a CW, and I am therefore not trying to gain reputation with this answer. As has been shown in this paper, pointed out by Matthew in the ...

13

First the somewhat misleading rubber sheet analogy: You've probably seen the bending of spacetime described as the deformation of a rubber sheet. Be careful taking this too literally as the sheet bending doesn't illustrate the bending of time, only space, and in any case it's not that good a mathematical model. Anyhow, it should be obvious that the ...

-2

No, every object's trajectory doesn't curve equally...for instance, light travels a null geodesic, which is very different from an orbiting planet. Your first example is quite correct, and the the path does depend on the momentum.But this applies to the second example as well...if the train is moving very fast, it won't bed at all. Since compared to gravity, ...

1

As in the case of Electromagnetism, in General Relativity you can perform two gauge transformations (diffeomorphisms). The first one being $x'^\mu(x)=\Lambda^\mu(x)$ you have four gauge functions which fix the gauge, for example the harmonic one $$\Gamma^\mu=\partial_\alpha(\sqrt{g}g^{\alpha\mu})=0.$$ Now it can be shown that exist a residual gauge ...

1

What you read is correct. I am not sure if those were the exact words of your teacher but according to the general theory of relativity, sun doesn't "attract" the photon (or any other body for that matter). In fact gravity is not even a real force. Let me briefly state what the theory of relativity has to say about gravity without going into the complicated ...

5

To properly understand what is going on you need to understand general relativity. Massless particles, like photons, travel on null geodesics and mass bends spacetime so the null geodesics are not straight lines. The problem is that neither you nor your teacher understand general relativity so this isn't a very convincing argument. But here is an argument to ...

2

Yes, gravitational waves travel along light-like geodesics, and short-wavelength gravitational radiation will be bent around gravitating bodies just like light. For details, see chapter 35, "Propagation of Gravitational Waves" in Misner, Thorne and Wheeler, especially section 35.14, "Effect of Background Curvature on Wave Propagation", exercise 35.15, ...

6

Despite my comment, on second look your second equation, attributed to Lambourne, is always identically zero. This is because you multiply the symmetric tensor $$\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}$$ against $R^{\mu}{}_{\nu\beta\gamma}$, and the riemann tensor is antisymmetric on those last two indices, and tracing a symmetric tensor ...

3

EDIT2: I've just seen Jerry Schirmer's answer, which confirms to me that Lambourne's equation is incorrect. I would mark his as the accepted answer. I'll leave the below for reference, though it is entirely false! Indeed, as pointed out by Peter in the comments, Lambourne uses the same convention as MTW for the Riemann tensor (see equation 3.35). Funnily, ...

0

I believed I found the answer to my own question. My reasoning seems to be right, the junction conditions obtained by a small integration over the shell are not altered by adding electromagnetism. However, to find the dynamics, these equations must be appended by at least one more equation, for example the Hamiltonian constraint, and these are indeed changed ...

3

For the benefit of brevity I'll give a straight answer: the second Weyl scalar is not related to electromagnetic radiation in any way. As per my previous comment, a simple proof is the Schwarzschild (or Kerr) spacetime. It describes a stationary black hole, without electromagnetic (or even gravitational) radiation, yet there is a basis in which all Weyl ...

0

I don't know if I'm right, but here is an attempt to estimate one effect that might be relevant. If a 133Cs atom of mass $m$ is in thermal equilibrium with blackbody radiation at temperature $T$, then it has an average kinetic energy $(1/2)mv^2=(3/2)kT$. This will cause Doppler shifts. The longitudinal Doppler shift cancels out on the average, but the ...

0

Some bits and pieces on angular momentum: Angular momentum is that which is conserved in rotationally invariant systems, just like energy is that which is conserved in time translation invariant systems and momentum is that which is conserved in space translation invariant system. This is the essence of Noether's theorem. The analogue in QFTs are ...

1

In general relativity the fundamental equation is roughly $$\textrm{Curvature} = \textrm{Matter content}$$ .In the context of general relativity it does not make sense to ask "why" matter curves spacetime, because this is the most basic assumption in the theory. It makes more sense to ask why one would try to construct a theory where gravitation is a ...

2

As photons have energy, gravity affects light rays, turning their path from straight to curved, and changing their energy (frequency/color). In classical relativity light always travels in a straight zero-lenght line, with phase speed = $c$. If you include gravity in your relativistic model you this is not true anymore. As this is an important property of ...

2

This is a statement about a congruence of null geodesics. We are looking for a conjugate point, which is just a place where the null geodesics cross each other. The theorem is putting a bound on how far you can advance the affine parameter $\nu$ along the geodesics before the conjugate point occurs (this is what is meant by affine parameter distance). ...

0

$dx^\sigma$ in the formula means that the infinitely small transport is considered. For two infinitely close points the path dependence vanishes (unless you add some finitely sized loop to the infinitely small displacement). When you want to make the transort along some finite path, you should integrate that formula and get $$\Delta A^\mu=\int_L ... 2 I learned my GR from Landau and Lifshitz Classical Theory of Fields, 2nd edition. Even at 402 (4th Edition) pages it is kind of breathless. The interesting thing about it is the first half is special relativity and electrodynamics which dovetails into the 2nd half which is GR. One has to persivere because it's terse but not too terse. Like Weinberg it has a ... 1 In general the base vectors (\vec{e}_{i}) are not constant, e.g., in polar coordinates the radial vector do not point in the same direction (unlike the Cartesian base vectors \hat{i},\hat{j},\hat{k}. If one takes the j-th base vector \vec{e}_{j} and consider its change if one moves on the direction defined by the i-th vector, mathematically this is ... 0 My apologies that I can't follow your question enough to answer everything, in particular I can't see how you project your images. If you extended them in what look like straight lines to you, you'll get radial lines expanding from your position, but it seemed like you wanted a Cartesian grid. However I do want to answer the part about strain and its ... 0 The positive mass theorem is a theorem, so you make some assumptions and fix a deductive methodology and then get a conclusion. Common assumptions include an energy condition and asymptotic flatness. The conclusion is usually something like that the ADM mass is positive. I think the idea is intuitively explained by an example. If you look at the ... 1 General Relativity allows black holes of any size (though making a small one might be hard or worse than hard). So as a thought experiment that means that you can consider a small black hole that curves spacetime exactly as much as the sun does. This black hole would be much smaller than the sun, but to us out here spacetime would look the same (except we ... 0 First of all, dcgeorge’s contrast of the rest mass vs the field mass is erroneous. The concept of rest mass requires the center-of-momentum frame of reference. Rest mass is indifferent about whether the massive object is a “particle”, a “field”, or both things combined. The problem is that curved spacetime doesn’t admit inertial frames of reference. The only ... 0 The answer is more simple than you think. Time is that, which is measured by (technologically suitable) clocks. Physical theories will simply tell you how clocks behave under certain conditions. This is purely descriptive. There is not a single physical theory out there, that gives a microscopic description of time, although the similarity of time with ... 1 Okay, I am going to try and give this a shot, but this is most probably not going to be a decisive answer. Let us operate with the term event time and duration and consider only special relativity (SR). The conclusions of general relativity should be the same for reasonable space-times. (e.g. without closed time-like curves etc.) We expect event time to ... 1 Duration is certainly a more physical concept than time. Duration is something you may measure between timelike separated events while time is always something you compute by adding up duration measurements + an arbitrary constant to fix the origin. Duration is experimental and relational while time (e.g. GPS time) is an abstract a posteriori ... 1 For a stationary massive object like Earth or the Sun, the gravity well does not change from the expansion of space. The gravity well at one time is the same as the gravity well at future times for the same mass. Any minute force imposed by the expansion of space does not constitute a change in the gravity well itself. The well is not stretched or ... 0 If we assume that general relativity is the correct theory for this case, and there are currently no indications, that I am aware of, that it isn't, then the expansion of the universe adds a small modifying term to gravity wells. I doubt that it is measurable at the scale of the solar system. The current best estimate for the Hubble constant is ... 2 If you are asking about the mechanism that causes gravitation, it's mass. And energy. Why? General relativity doesn't say why. It says what happens. If you are asking why Einstein chose to use the equivalence principle as a guiding concept in his development of general relativity, in a very real sense he had no other choice. There's a general concept that ... 0 Planes of simultaneity in special relativity don't really mean much of anything. The real physical structure of spacetime is in the light cones. The takeaway from "relativity of simultaneity" is not that there are "different time orderings for different observers", but rather that there is no meaningful time ordering for spacelike separated events. They ... 0 It is easiest to directly derive the form of the vector fields from the boundary conditions for asymptotically flat spacetimes. See for example this paper http://arxiv.org/abs/1001.1541 or this one http://arxiv.org/abs/1106.0213 Infinitesimally diffeomorphisms act on the metric via lie derivative \mathcal{L}_{\xi}g_{\mu \nu}. In the case of BMS, you ... 11 What you're asking about is the existence of surfaces of simultaneity. In SR, surfaces of simultaneity can be defined by measurement procedures such as Einstein synchronization, and they turn out to depend on one's frame of reference. In GR it gets a lot tougher to do this. We don't even have global frames of reference. It turns out that what you need in ... 10 It sounds like you're interested in when a spacetime admits a Cauchy surface. The answer is that every spacetime that is globally hyperbolic has this property. This was proved by Geroch in 1970 (article here, see Section 5). This includes most of the textbook relativistic spacetimes --- Schwarzschild, Kerr, FLRW, and many others. But there are some ... 2 The question is what do we need the matter content of the universe for. As I understand it, in the usual case we want to find the conserved quantity associated with a certain conserved current gained by the projection of the energy-momentum tensor into a Killing vector, as for example in the paper by Abott and Deser. The requirement of asymptotical ... 3 Matt Visser's How to Wick rotate generic curved spacetime is a great reference on this subject, which basically summarizes a lot of folklore on the subject. Addendum (Summary of Paper). This turns out to be an important problem in quantum gravity and QFT in curved spacetime for the obvious reason ("How do we know the usual tricks still work in curved ... -1 The expansion of the universe is not a force. Forces don't pull things apart at a particular speed; they change the speed by a particular acceleration. The speed itself is just inertia. It's no different in cosmology: there is nothing actively pulling things apart at the speed given by Hubble's law; that speed is just leftover momentum from the big bang, as ... 1 Based on Baez's simplified presentation of the field equations, all that would matter would be \rho+3p, and mutual attraction is described by \ddot{V}<0, which only requires \rho+3p>0. So if we observe mutual attraction, there is clearly no reason to think it implies p<0; making p more positive would increase the attraction. In general, ... 0 The expansion does lead to a kinetic energy term that can be (at least partially) extracted. For objects that are bound to each other, it would lead to a classic acceleration term, which, of course, is equivalent to a classic pseudo-force. In an expanding universe, any two objects that are bound by a potential, are therefor experiencing an additional (albeit ... 1 The notion of kinetic energy is ill-defined in the spacetimes where you have a time-dependent cosmological expansion. If you somehow attached two galaxies to each other with a spring, however, the expansion of the universe would do "work" against that spring, as there would be a force requied to keep the proper distance of the two galaxies fixed. At ... 2 For 1., you have, by applying the Leibnitz rule for covariant derivatives : \nabla_{d}(\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}) \\= \nabla_{d}(\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} \delta_{c'}^{c}) \\= \nabla_{d}(\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} g^{ca}g_{ac'}) \\=(\nabla_{d}\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l}) ... 1 Since an (k,\ell) tensor can be viewed as linear combinations of pertinent tensor products of (1,0) tensors (=vectors) and (0,1) tensors (=covectors), and by linearity and Leibniz rule for the covariant derivative, it is enough to consider the latter. The condition (1) in manifestly covariant notation then reduces to$$ ...

2

A more physical attempt: In general relativity, the metric tensor represents local clock and ruler measurments. If I multiply the metric tensor by a scalar constant, it should be obvious that this is inequivalent (in general) to a set of coordinate transformations, but, at the same time, I'm affecting local clock and ruler measurments (the ratio of the ...

4

General relativity is only conformally invariant in two dimensions. This can be proven by making the transformation $g_{ab} \rightarrow \phi g_{ab}$, and seeing what transformation Einstein's equation${}^{1}$ makes. What you will find is that Einstein's equation will MOSTLY transform, but you will get terms proportional to $(d-2)(d-1)$ and derivatives of ...

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