# Tag Info

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This is just a comment, not an answer, and I'm not allowed to add comments in this question. But in the context I've seen them, we would normally expect breakings of diffeomorphism invariance in Modified Theories of Gravity (such as Massive Gravity), but Galileons are the nicest example of a transformation in which we can reinstate diffeomorphism invariance. ...

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I am not sure what you mean by if something has more energy it should "slow time" more than something with low energy But even if I just neglect that for a moment. I'll use a description of waves for visual aid and simplicity. Imagine in "not slowed down time" a wave has a period (and frequency) of 1 in this time. So there is one wave crest and one ...

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It's probably best to not think of single photons as sources of gravitational energy. For one thing, most bulk electromagnetic fields are not eigenstates of the photon number operator. For another thing, the thing that couples to the gravitational field is the energy density of the field. This density is proportional to the intensity of the field, ...

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What if you had a proton travelling at .99999c towards a heavy object? Would it have to keep accelerating or would the acceleration of the proton slow down to zero and only it's mass would increase? It would keep accelerating, but there's a problem. See what Einstein said in the second paragraph about the speed of light being spatially variable: The ...

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Good question. Although the speed of gravity is approx. 3x10^8 meters per second, the acceleration at the Earth's surface is as you describe ~9.8m/s^2. If you choose the Earth as the heavy mass your proton in question is approaching, and forgo magnetic influence, then you would realize that the increase in acceleration is negligible as compared to the ...

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In order for that proton to reach the speed of light, it would need an infinite amount of energy. Perhaps if the proton reached the center of a black hole, converging with the singularity, it would cease to accelerate, but it would not travel neither faster or at the speed of light.

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You should look at it like an asymptote. Yes the proton would accelerate but it would probably accelerate to .999991c or more likely less due to the massive energy required to accelerate something so fast already. Therefore you could always keep accelerating your particle but it would never cross the Light-Barrier.

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You asked: "Does one imply another?" No. Neither implies the other. However, I think there are benefits to first being clear what the ideas are, particularly since I think each idea actually already assumes an arrow of time. In the first case, you start with an arrow of time that only earlier times affect later times, and then end up strengthening that to ...

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Imagine a stone in your hand. Shake it. Its resistance to motion is inertial mass. He calls it answering mas because it is how much the stone answers to your force. Now drop your stone. It is pulled by gravity. The force of pull is due to gravitational mass, what he calls calling mass. He calls it calling mass because that is how much the gravity of a body ...

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Let's say you assume that the neutron star is spherically symmetric, e.g., ignore the effects of rotation. Then for a radial trajectory in the resulting Schwarzschild spacetime, the calculation is actually not quite wrong, although you must be careful in interpreting it. The reason is that orbits in a Schwarzschild spacetime have an effective potential that ...

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A black hole is a region of spacetime enclosed by an event horizon. Thus, the singularity, while a fact about black holes as far as we understand them, is not an defining feature of what black hole is. Therefore, it makes more sense to try to calculate volume (and hence indirectly, density) of a black hole according to the extent of the event horizon rather ...

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Using the Schwarzschild radius for this purpose makes sense, because this is the radius of a sphere which becomes a black hole, if it has the given density. For example a sphere made of air at Earth density does not become a black hole if its radius is 1 meter. But if the radius is big enough, it will actually become a black hole. Even though the density is ...

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Our model for spacetime is that of a manifold, which is the mathematical term for something that looks like $\mathbb{R}^n$ in any zoomed-in patch, and where all these patches are stitched together in a sensible way. On our manifold we have $n$ coordinates -- real numbers that describe each point and vary smoothly from point to point. We also add to our ...

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Having read John Rennie's answer above, I'm going to give an answer that's hopefully of the same sense, which hopefully makes sense, but which hopefully brings out an issue. 1. Is the free fall acceleration the same as the coordinate acceleration for a hypothetical observer at rest on the star surface? Yes and no. Yes because the falling body falls ...

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I'm guessing your questions all amount to whether general relativistic effects become important at the surface of a neutron star. To answer this we can compare the flat space metric (in polar coordinates): $$ds^2 = -c^2dt^2 + dr^2 + r^2 d\Omega^2 \tag{1}$$ with the Schwarzschild metric that describes the geometry outside a spherically symmetric mass: $$... 1 Quite apart from whether a point singularity can exist inside a ring singularity there's the little detail that a solid ring (and a ring singularity will behave as one) in orbit about a central body is unstable. If one point is infinitesimally closer to the center (and it will--quantum uncertainty) there will be more attraction there and the difference will ... 0 In General Relativity, Einstein established that gravity is due to the curvature produced by objects in space. That's not quite right I'm afraid. The YouTube video shows the rubber-sheet analogy, which isn't ideal, but it will do. Because the force of gravity depends on the slope, not the curvature. The steeper the slope the stronger the force of gravity. ... 1 Over the real numbers, any non-degenerate quadratic form is determined (up to a change of basis) by its signature, which consists entirely of 1s and -1s. 0 Your mistake is assuming that the curvature is due to size. The curvature is due to mass. So if an object were taken off of the earth, the earth's curvature would decrease and there would be some curvature away from the wart, due to the mass of the object. 0 Using a rubber sheet to visualize gravity may be confusing in some cases, as the deformation of the sheet is affected by the size of an object, while gravity is not. The rubber sheet analogy is only a visual representation of gravitation outside a massive body. As gravity is proportional to mass, not size, the analogy becomes awkward if applied to a large ... 3 You have linked a YouTube video that shows the rubber sheet analogy for the curvature of spacetime. The trouble is that while the rubber sheet analogy is not a bad way for beginners to get a rough idea what is going on, it can be very ,misleading if you push it too far. In this case I suspect it has lead you to imagine the sheet wrapping round the surface of ... 16 There are only four known stable black hole geometries: Schwarzschild, Reissner-Nordstrom, Kerr and Kerr-Newman. We expect that any random assemblage of matter dense enough to form a black hole will relax into one of these four geometries by emission of gravitational waves. None of these geometries has two distinct singularities, so (as far as we know) it is ... 14 I think the closest model to what you're talking about would be two colliding black holes, during the intermediate period where their horizons had merged, but the central objects had not yet collided. These systems are very different from isolated black holes, as they give off significant gravitational radiation, and have horizons that rapidly change in ... 1 There is a book "Exact Solutions of Einstein’s Field Equations" by Hans Stephani, Dietrich Kramer, Malcolm MacCallum, Cornelius Hoenselaers and Eduard Herlt where classified solutions are given. Table of contents: http://assets.cambridge.org/97805214/61368/toc/9780521461368_toc.pdf 4 Is time an illusion? No. I think it's best to think of it as something like heat. You know what heat is, especially if you put your hand on a stove: szzz aaargh! Heat is definitely not some illusion. However it is an "emergent property". Think about the kinetic theory of gases. The temperature of a hot gas is something like a measure of the average kinetic ... 1 Normally, when combining Lagrangians, we often leave the constant multiplying factor to be determined by experiment. For example, if \mathcal{L}_{k} is the kinetic term (for a system of charges and the electromagnetic field), and we choose to describe the electromagnetic coupling by \mathcal{L}_{int} = A_\mu J^\mu, then we combine them as ... 0 An answer you might find satisfactory is that our model of spacetime is "larger" than what we observe. Yes, there is a preferred direction which gives us a (1,3) signature, but actual, real objects in spacetime must travel on timelike curves (we normalize the length of their geodesics to -1). These timelike curves only transverse part of the entire ... 0 I used to wonder how spacetime curvature could cause you to start moving when you weren't moving already. I got the idea that the path you followed could be bent by the curvature of spacetime, but what if you start out standing still? (Standing still relative to a particular massive body, that is.) If you understand all the other answers that have been ... 0 That's just strong equivalence principle. Standing still on a planet is getting accelerated upward. Light is going at the speed c only for inertial observers, that is, for the ones free falling. It is a big misconception to claim that light doesn't "slow down" as it ascends, for you standing still on the planet. Where do you think that red shift comes from ... 1 IMHO it's important to look hard at the ontology of what's actually there and take care to distinguish between reality and abstraction. For example: I was reading about the light cone in relativity... Relativity is just about the best-tested theory we've got. I "root for relativity". But I will say this: a light cone is an abstract thing. You cannot point ... 0 IMHO, of course Time has singular direction for all real life events. E.g. living bodies, human bodies, etc. age and cannot be still or start moving the opposite direction. (The Curious Case of Benjamin Button is a fiction. Lets not mix facts with fiction here.) Causality principle is true and so is the second law of thermodynamics. But the former is more ... 0 Another way of looking at user40330's answer is to think of the inverse metric as the map from the space of one-forms (or differentials, if you prefer) and mapping them to the space of vectors (or directional derivatives, if you prefer that language), and then thinking of the metric as the inverse of this map. Namely$$g^{-1}({ d}v)=g^{ab}v_{b} = v^{a} = ...

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The tensor algebra is symmetric between one-forms and vectors. One could start with defining any of them first and then obtain the rest of the things. The inverse metric tensor is a linear map that takes two one forms on a manifold and maps into $\mathbb{R}.$ $g^{\mu \nu}: A_\mu,B_\nu \rightarrow \mathbb{R}$ It of course tranforms like a vector with ...

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No, you don't have to re-introduce the luminiferous aether (or aether of any sort) in order to make space/time and General Relativity sensible. You do not have to postulate any sort of spatial filling or philosophical substratum in order to keep General Relativity logical and experimentally intact. It would be helpful for you to conceptualize space/time as ...

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No, the argument is not correct. The spatial "conformal" coordinate $R$ in which, together with the conformal time $\tau$, the angle of the light rays is 45 degrees is not $\rho$ but nothing else than $r$: $$ds^2 = -dt^2 + a(t)^2 dr^2 = a(t)^2 (-d\tau^2 + dr^2)$$ If you want a diagram with $\tau$ on the vertical axis where the light rays are drawn at 45 ...

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I've been looking at this Java archive General Relativity (GR) Package written by Wolfgang Christian, Mario Belloni, and Anne Cox It includes a lot of simple programs about Newtonian mechanics, special relativity and general relativity, including the aforementioned GROrbits. It doesn't permit custom metrics - you are limited to Schwarzschild ...

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For particle/light motion in 2D space, my nomination would be GROrbits It's free and requires a JVM to run, there is also a web start version for the brave ;) Sorry but I've never found anything aimed at visualizing metrics or curvature (apart from plotting programs of course).

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To understand how light is affected by gravity, it helps to think of light as energy. So let's ask a basic question: when it come to light, what is energy? By the Planck-Einstein relation, we know $$E = h \nu$$ where $\nu$ is the frequency of the light and $h$ is Planck's constant. So when we talk about the energy of light, keep that in mind. Also note ...

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There are several possible approaches to this question, but I've always been a fan of the one taken by Edwin Jaynes in his 1965 paper Gibbs vs Boltzmann Entropies. (See sections V and VI for the discussion, which I think can be read in isolation from the rest of the paper.) Here he derives the second law from the empirical fact that we as scientists and ...

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Ref. 1 defines a local translation on spacetime $M$ as a diffeomorphism. The word local means here point-dependent ($x$-dependent). An infinitesimal diffeomorphisms $\xi^{\mu}(x)$ can be identified with a vector field. Note that in flat Minkowski spacetime we called affine transformations for global Poincare transformations. The word global means here ...

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Surely therefore our locally measured time is not the cosmological time t but rather the conformal time T? I don't think so. Our locally measured time is our locally measured time. If you had a clock that started ticking when the big bang occurred, the clock reading would be 13.8 billion years. If it displayed the conformal time, the clock reading would be ...

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$$\nabla \epsilon_{...}=\nabla \sqrt{g}[...]=gg^{\mu\nu}\nabla g_{\mu\nu}[...]=0$$ In the last step we use Jacobi's formula for differentiating a determinant and $\nabla g_{\mu\nu}=0$. Here $[...]$ describes permutations and consists of $-1,0,1$ as you see from this article.

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This is correct. Each point on the rod will follow a hyperbola, so the path of the rod through time will look like this. When b (time) is zero, the rod is at rest and evenly spaced, but at any other value the rod is contracted more in the direction it's accelerating away from.

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I don't think so. The whole point about the equivalence principle, is that gravity is indistinguishable from inertia. It is rooted in the fact that gravitational and inertial mass are the same. See this answer. This is not the case for the electromagnetic interaction. Two bodies with different charges but identical masses will not have the same acceleration ...

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Firstly, the gravitational field inside the Earth, decreases with depth. To a first approximation, you can use the shell theorem for spherically symmetric mass distributions to argue that the gravitational field at some depth is due only to the mass enclosed within a sphere interior to that depth. If we further make the crude assumption that the Earth's ...

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You are absolutely correct that three-dimensional sections of space-time do not satisfy Euclidean geometry -- they are not flat. However, they are almost flat. On room-sized scales the curvature is very small indeed. I don't know the exact context where you read that "tridimensional space sections of space time continuum (whatever its number of dimensions) ...

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Q1. In the case of a uniform spherical distribution you cannot sense anything further away from the center than you are. This is directly derived from Gauss' law for gravity. At the center you do not feel any net gravitational force from earth at all. Think about it this way: earth is pulling you up from all directions exactly the same way, so all the ...

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Just because you can "derive" the same physical phenomenon in two different settings does not necessarily make them related, or allow you to just set things equal. In particular, the time dilation formula familiar from special relativity (i.e. $T = \frac{T_0}{\sqrt{1-v^2/c^2}}$) is only valid for uniformly moving observers in flat space. The spacetime ...

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As noted, the sun isn't on this path. But, the gravity from a black hole is not radically increased. If the sun became a black hole, its mass wouldn't change, and we would experience the same gravitational force. (I am not sure how close you have to get before you start to see the GR-type effects. But it's pretty close.)

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At the distance of the Earth, the gravitational pull would be the same: its mass does not change just because it's concentrated in a small region of space, and the gravitational field of the black sun at that distance would therefore be unchanged. Our friends at NASA have answered the question essentially the same way: It would exert no more ...

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