# Tag Info

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I think many people here just too intelligent to see this point: because you do not have a better choice. It is simply not practical and not feasible to teach kids in high school about General Relativity. (um... expecting they know some tensor already? and understand space-time?) Besides, as mentioned by many others, the Newton approach is not so bad. In ...

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As I know, Field Theory, that to what appeals the topic creator cannot explain the very powerful gravitation fields . So trying to understand what happens with a photon there are inside the Black Hole in meaning of Field Theory, or Special Relativity, isn't a good idea. The Nature has no the alone space , and the alone time , you can abstractly image ...

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The laws of special relativity, of which the constant speed of zero mass particles in vacuum is a basic tenet , have been tested innumerable times with many experiments, particularly in particle physics. The Michelson Morley experiment has shown that there exists no luminiferous ether, i.e. there is no medium on which light propagates with this velocity c. ...

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is there anything that says that speed has to be a max limit of any kind? Yes, the so-called chronometric distance definition; whereby the (mutually equal) value of distance between two participants, $A$ and $B$, at rest to each other is evaluated as $$\ell[~A, B~] = \ell[~B, A~] := \frac{c_0}{2} ~ \tau A[~\text{signal}, \circledR B \circledR ... 1 Anything with mass and/or energy warps space-time. When Light with different energy level passes through gravitational field of a massive object, higher energy will be attracted under the influence more than lower energy light. But, Higher energy of light particle will have more inertia than low energy particle. And since light of different energy level ... 0 The principles of space time homogeneity and isotropy are dependent of one another. The reason of dependency of one another refers to the region of space time of being homogenous. Homogeneity in space time results from being symmetric, and what causes space time to be symmetrical is simply the Laws of Nature. Hence our space time, or the shell we are living ... -1 The question is why you would want to do this in the first place. The equations of motion that you obtain one a compact region still aren't the real equations of motion, since the compact region is a mathematical choice to simplify the formulation of the problem. It is implicitly understood that the true equations of motion are only obtained on the limit of ... 1 I know of two reasons for why we should consider gravity to be a force. The first is purely classical and Newtonian: tidal forces. Gravity is solely responsible for producing tidal forces, and they cannot be considered a fictitious force, whereas the usual acceleration due to gravity in some sense can always be thought of as fictitious. The way you know ... 1 The boundary of a subset of a topological space is abstractly defined as the set-theoretic difference between its closure and its interior. Since topological spaces in general have neither coordinates nor metrics, this notion is independent of the metric. Since the spacetime manifold is a manifold, it is a topological space (locally homeomorphic to ... 0 There is no problem with saying that we have a region with boundary as far as the underlying manifold goes. The problem is that you don't know the metric on the boundary or how to integrate. You should rephrase your question accordingly and in that case yes you need to be careful what boundary conditions you take as already said. 4 Let us fix a reference frame S, where a particle of charge q and velocity v lies. It can be experimentally proven that, if another such particle q' is present elsewhere in the universe, the initial one is subject to a force \textbf{F}=q\textbf{E}, where \textbf{E} can be measured and addressed to the other body q'. Likewise, if a current i ... 1 p=\frac{1}{3}\rho is the well-known equation of state of a photon gas. It may be derived by looking at the ultra-relativistic limit of the energy momentum tensor for a bunch of particles.^1 p=-\rho follows from the fact that the energy momentum tensor of \Lambda-style dark energy is proportional to the metric. Thus, at a point and in the proper ... 5 The geometry of spacetime is described by a function called the metric tensor. If you're starting to learn GR then any moment you'll encounter the Schwarzschild metric that describes the geometry outside a sphrically symmetric body. When you go inside the body the geometry is described by the (less well known) Schwarzschild interior metric. The exact form ... 0 It does not make sense to have \delta g_{\mu \nu}\neq 0. Actually, for finite region, even if the variation of metric is zero at boundary, the derivative of metric is not zero, it will contribute a surface term to the variation of actio, thus we have to add another surface term to cancel this contribution, which consists of exterior curvature of the ... -5 If gravity isn't a force, then why do we learn in school that it is? Because it is a force. It's just not a force in the Newtonian sense, wherein work = force x distance. When you drop a brick the "force" of gravity doesn't add any energy to the brick. Instead it converts potential energy into kinetic energy. This is different to what you do if you ... 3 Take a trace of Einstein equations (trace of g_{\mu \nu} is D), you obtain$$R - \frac{D}{2} R + D \Lambda = 0$$Or$$R=\frac{D \Lambda}{D/2-1}$$Then substitute this expression for R into full Einstein equations and you obtain trivially$$R_{\mu \nu } = \frac{\Lambda}{D/2 - 1} g_{\mu \nu}$$1 Following David Z's answer, the proof for the last paragraph is: since t is an affine parameter it satisfies: $$\frac{d^2x^a}{dt^2}-\Gamma^a_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}=0 \tag1$$ the parameter t' must be related in some way to t, that is:$$t'=t'(t) \tag2$$use the chain rule to get: ... 1 For better clarity, let's define the following: Axial direction = the direction the person & light beam are drawn into the BH. Radial direction = the direction perpendicular to the axial direction. If we, looking in the same direction as the person & light are being drawn into the BH, watch the light beam as it is drawn into the BH, we will see the ... 1 Here's a few methods to find the Einstein Field Equations : 1) The classical route The classical method is to note the similarity between the geodesic equation $$\ddot x^\sigma + {\Gamma^\sigma }_{\mu\nu} \dot x^\mu \dot x^\nu = 0$$ And the classical equation of motion for particles in a gravitational field : ... 2 Now there is a light ray moving outward at the speed of light. I'm afraid that isn't the case; within the event horizon of a Schwarzschild black hole, the radial coordinate is timelike and so, moving 'outward' toward the horizon is as impossible as moving 'backward' in time. This plain to see in the Kruskal–Szekeres coordinates: Image credit See ... 2 I think a possible analogy would be to imagine that the singularity is a waterfall. By emitting light, you are trying to send a signal upstream using a tame fish. Outside the event horizon the fish is able to make headway against the current. But the river flows so fast within the event horizon as it approaches the waterfall, that your fish ends up going ... -10 How does light behave within a black hole's event horizon? It doesn't behave at all. If the event horizon of a black hole is the distance from the center from within which light cannot escape, imagine a person with a flashlight falls into the black hole. I've explored this with a variety of relativists, and posed this question. The answer comes ... 3 Calculating the sum of the interior angles precisely woud be a big task as we'd need to compute the trajectory of the light ray and there isn't a convenient analytic expression for this. However we can easily calculate an upper limit for the interior angles. The key fact we need to know is that the deflection angle \theta of a light ray in the ... 4 Well, actually you are looking for a one-parameter group of diffeomorphisms (or isometries if referring to the boost vector field). This group is obtained by solving the differential equation$$\frac{dx}{ds}= X(x(s))\tag{1}$$with a generic initial condition z at s=0 in the manifold M (Minkowski spacetime in your example). X is your vector field on ... 2 In Feynman's Lectures on Physics (volume 2 chapter 42) he states that the field equation is equivalent to the following statement: For all local inertial observers, the scalar curvature of space at a point is proportional to the energy density at that point. Simple, right? By requiring the correct Newtonian limit the constant of proportionality can be ... 0 What is the general relativity explanation for why objects at the center of the earth are weightless? At that location spacetime is locally flat. See the Wikipedia Riemann curvature tensor article and look at the schematic on the right: CCASA image by Johnstone, see Wikipedia Let's imagine we could take away the Earth and look more closely. ... 7 The explanation is Birkhoff's theorem, which states that the Schwarzschild solution is the unique spherically symmetric vacuum solution in general relativity. An immediate result of this is that, just as in Newtonian gravity, a spherical shell does not contribute to the gravity experienced by an object within it. If this were otherwise it would suggest the ... 0 Comoving observers move along with the Hubble flow, and perceive the universe as having no Hubble expansion, due to an increasing scale factor. So, when you invoke a comoving observer, it's not surprising that the observer sees no redshifting. 0 Manifolds are defined such that locally they look like Euclidean space; this is why we call them smooth manifolds. A riemannian manifold is a manifold that locally has some inner product structure, ie a way of measuring length and angles. Lengths and angles are invariants, hence will have an invariant expression in terms of a local coordinate basis; and ... 0 In Riemannian geometry there is a beautiful theorem which states that a manifold with a symmetric connection is locally flat everywhere if and only if the curvature tensor vanishes. Therefore, in a locally flat coordinates such that \Gamma_{jk}^i=0, g_{ij} is constant throughout the chart and a linear transformation can be used to diagonalize the metric ... 5 I would like to take a slightly different angle on this question and point out that most physicists believe that gravity is in fact a force. The great triumph of modern particle physics, the standard model, contains the strong, weak, and electromagnetic forces. These forces are represented in the standard model by the presence of force carriers (spin 1 gauge ... 0 If ds^2=\eta_{\alpha \beta}d \xi^{\alpha} d \xi^{\beta} were true for all points of space, we would have no curvature, hence no gravity! Take for example a sphere (the Earth), locally we can measure distances by ds^2=dx^2+dy^2, but this can't hold for two arbitrary points on the sphere. In fact, this coordinate system changes from point to point ... 0 Let \mathcal{M} be the space time manifold, whose local charts (open sets) are described by U_i. A local coordinate frame S_i is a map \xi\colon U_i\mapsto \mathbb{R}^N such that \xi(m) = (x_1,\ldots,x_N) \in \mathbb{R}^N, m\,\in U_i. Let, moreover, g be a (0,2) rank tensor (the metric). A change of coordinates is any smooth invertible map ... 2 What you are confusing here is speed and velocity. Light speed is constant, but the velocity, which takes into account the direction as well as the speed is not. As an example of how something can accelerate without changing speed, consider the case of circular motion, where the acceleration of an object moving at a speed v in a circle of radius r is ... 0 The disconnect is between the first and second clauses of your first sentence: Light speed is constant, therefore experiences no acceleration Yes, the speed of light is a constant, but it experiences no acceleration in its direction of travel. Light definitely accelerates laterally when gravity pulls on it, which is why it curves when passing near ... 12 First, we notice that the paths traced by particles through spacetime under the influence of a gravitational field seem to depend only on their positions and velocities, i.e. they are independent of any identifiable charge or composition of the particles. It is almost as if the particles were moving along tracks carved into some curved surface. Although ... 0 So in a four-dimensional orientable space we have a [0\;4] orientation tensor$$\epsilon_{\alpha\beta\gamma\delta} = - \epsilon_{\beta\alpha\gamma\delta} = - \epsilon_{\gamma\beta\alpha\delta} = - \epsilon_{\delta\beta\gamma\alpha}$$Usually with respect to some basis we choose \epsilon_{0123} = 1 or so to finish off the specification of the whole ... 1 No, the straight beam will not magically turn into a curved beam. I suspect you have a slightly confused idea of what the curvature of spacetime means physically. Basically it means that a freely moving body will appear to accelerate relative to some distant observer. Conversely if we want stop the body from accelerating then we have to apply a force (i.e. ... 0 g_{\mu \nu}(x) means that g is a function of location (x) --- so it varies across the manifold, which is the problem. I think that if g \ne g(x), then necessarily g = \eta ... Hopefully someone else can chime in on that. -1 Don't forget that angular momentum is a psuedo vector. Recall what that means about properties under reflection, for example. 1 The equation you quote:$$ t' = t\sqrt{1-\frac{3GM}{rc^2}} \tag{1} $$gives the time relative to an observer at infinity. You want the time relative to an observer on the Earth's surface. You need to calculate:$$ t_\text{satellite} = t\sqrt{1-\frac{3GM}{r_\text{satellite}c^2}} $$and:$$ t_\text{Earth} = t\sqrt{1-\frac{2GM}{r_\text{Earth}c^2}}  ...

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Tim B. takes the position that this an example of lying to children. I completely disagree; in my view, what this is an example of is idealization, which is something that every model must do, in every branch of science. As George E.P.Box once wrote: Essentially, all models are wrong, but some are useful. It isn't lying, it's called doing science. ...

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I don't agree with the previous answer. Firstly, the OP's question isn't about the lagrangian formulation, it's about the Einstein equation : $$\tag{1} G_{\mu \nu} + \Lambda \, g_{\mu \nu} = -\; \kappa \, T_{\mu \nu}.$$ Secondly, there are stress-tensors that can't be derived from an action : fluids tensors (especially with ...

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You can also read about the test of 'frame dragging', which is predicted by General Relativity and not Newtonian gravity, confirmed by Gravity Probe B. Also, Newton's law of gravity is not so much 'wrong' as it is 'incomplete' (General Relativity reduces to Newtons law in the case of low energy/speeds).

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In addition, one more test will be the gravitational wave from binary, supernovae as well as supermass black hole.

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There are three classic tests of general relativity: the anomalous precession of Mercury's orbit the deflection of light by the Sun the gravitational redshift of light Newton's theory predicts zero precession in test (1) and zero redshift in test (3). For test (2) Newton's theory predicts a deflection half the size of the prediction in general ...

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Even if we restrict ourselves to a Newtonian conception of the world, forces do not exist. An essential thing that is not emphasized enough when teaching physics, is that physics (in all its wonder) is nothing but a mathematical model of the reality we perceive. Whether you are considering Newtonian mechanics, relativity, or quantum mechanics. There are ...

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There's a very real phenomenon called 'Gravitational Lensing', in which light is bent from its original trajectory by a massive enough cluster of matter (which curves the space-time around it). Moreover, it's bent by a different amount than predicted by a simply application of Newtonian ideas, as kindly pointed out by Rob Jeffries. Is this evidence enough? ...

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If we observe our school syllabus, almost all the physics that we learn is Newtonian physics. Everything from force to the laws of motion are all based on Newtonian ideas. And the general theory of relativity is a modern concept which in fact is more true. But you know the GTR is a difficult concept to understand for a child. So to make the course simple ...

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It's an example of "lie to children". https://en.wikipedia.org/wiki/Lie-to-children Because some topics can be extremely difficult to understand without experience, introducing a full level of complexity to a student or child all at once can be overwhelming. Hence elementary explanations are simplified in a way that makes the lesson more understandable, ...

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