New answers tagged

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Contrary to what has just been claimed, there is a general definition of a center of mass in General Relativity, which was proposed by Dixon (W.G. Dixon, Il Nuovo Cimento 34, 317 (1964)) and whose existence and uniqueness were proved by Beiglböck (W. Beiglbock, Commun. Math. Phys. 5, 106 (1967)). These are very old papers, so that people have been working ...


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I will use the notation $\theta^i$ for the dual basis since $\omega$ is reserved for another important form. Since $\mathrm{d}\theta^i$ is a 2-form, we may expand it in the basis $\theta^i$ itself: $$\mathrm{d}\theta^i=-\frac{1}{2}C^i{}_{jk}\theta^j\wedge\theta^k$$ But the first structure equation gives $$\mathrm{d}\theta^i=-\omega^i{}_j\wedge\theta^j$$ ...


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But, how does one solve for a metric for that given situation? There are many possible metrics. For vacuum solutions you have Minkowski space as a solution. Another is a solution is a spacetime with a gravitational wave going in the $+\hat x$ direction as a plane wave filling all of spacetime. Another is like Minkowski space locally, but which is ...


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In practice, given a stress-energy tensor $T_{\mu\nu}$, we may attempt to find solutions to the Einstein field equations using perturbation theory. The basic idea is to expand around a known solution $g_{\mu\nu}$ by a perturbation $h_{\mu\nu}$. In the case of a flat background, $$\delta G_{\mu\nu} = 8\pi G \delta T_{\mu\nu} = \partial_\mu \partial_\nu h - ...


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He meant that You may think an orange is present in space when it actually doesn't exist. You imagine coordinates of space which don't exist. Can you see a dream? Similarly, here he meant space and time don't exist if you don't imagine them. But once you think about it, it exist. Have you seen space and time? You just imagine it out of nothingness and ...


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This an attempt to give a more detailed explanation since the question really is quite fundamental and has mostly been explained by referring to the impossibility of a co-moving observer detecting any effects of the non accelerated linear motion whatever the speed might be. Its the same as saying you must just trust Einstein without explaining the mechanism ...


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If you are in a flat space(time), i.e. without any source of gravitation, in a spaceship, and you emit a ray of light across the spaceship, both the spaceship and the light will be in the same frame of reference. The frame will be inertial - not accelerated - and therefore the light will follow a straight path. Yet if a boost is applied to the spaceship, it ...


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John G. Cramer discussed G4V in a recent Analog Alternate View Column (Mar. 2016), and how Advanced LIGO data could possibly falsify G4V, General Relativity or even both of them (Their predicted gravity wave signatures signatures differ.) Cramer also stated that there would be no dark energy since G4v explains distant receding Type IIa supernova dimming as ...


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You are thinking of the entangled state as two objects, one of which is stationary With respect to us and one of which is moving at near $c$ With respect to us. Viewed this way it is natural to think that one object must be time dilated With respect to the other. However the entangled state is not two objects - it is a single extended object described by a ...


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When he fires he will see the white spot where it was the light time away so it will have moved into the right position when it is hit at instant later. By the way I dont understand fedino's remark about light speed having a vertical component. Are we talking about the photons momentum vector, the Poyinting vector or what?


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OP's proposal (v2) is a special case of Finsler geometry with $n=3$. The main idea is to replace the quadratic metric tensor $g^{(2)}_{\mu_1\mu_2}$ for pseudo-Riemannian manifolds, which defines (infinitesimal, possibly imaginary) distance on the manifold via $$ds ~=~ \sqrt[2]{g^{(2)}_{\mu_1\mu_2}dx^{\mu_1}dx^{\mu_2}},$$ with (possibly a sequence of) ...


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The amount of deflection of light (the bending of the null geodesic) passing a star is a deflection angle of 4m/R where m is the mass of the star and R is the radius of the star. The mass of the sun is 2 x 1030 kgs, where as the mass of the Earth is 6 x 1024 kilograms. The radius of the earth is 6 x 106 meters and the radius of the sun is 7 x 108 meters. ...


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This is how you do the calculation. The elapsed time on an observer's clock is called the proper time, $\tau$, and it is calculated by integrating the metric: $$ c^2d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)c^2dt^2 - \frac{dr^2}{1-\frac{2GM}{c^2r}} - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 $$ In this case we'll assume all motion is radial so $d\theta = ...


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The fact that the clock near the earth would run continually slower - i.e. the difference between the two would grow the more time they are seperated - is enough to be equivalent to different rates of acceleration. It is not like it runs more slowly for at bit and then runs at the same rate as the other one, but slightly behind.


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I think you have all the right pieces to answer the question, here are a few hints that should be of some use. You say that you picked coordinates $ \{v^{\mu} \}$. It seems to me that they should instead be called $ \{ x^{\mu} \}$, as that is what you're taking partial derivatives with respect to. As you correctly pointed out, you are working with ...


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I'll reduce your question to its simplest expression: "What is mass?" And give you my best, simplest answer:"It is a measurement of how much an entity opposes acceleration or deceleration". I believe that in the end it all comes to that...


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We know that the Levi-Civita connection satisfies $\nabla_a g_{bc} = 0$ and the product rule. The definition of the inverse metric $g^{ab}$ is $g^{ab}g_{bc} = \delta^a_c$. Therefore, we have: $$\begin{align} 0 &= \nabla_a \delta^b_c \\ &= \nabla_a (g^{bd}g_{dc}) \\ &= (\nabla_a g^{bd}) g_{dc} + g^{bd} \nabla_a g_{dc} \\ &= (\nabla_a g^{bd}) ...


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Stars orbiting black holes (I assume that's what you mean) and observed from afar will have their light doppler shifted due to (i) gravitational redshift; (ii) the relativistic doppler effect due to their orbital motion. Effect (i) becomes more important the closer a star gets to the event horizon of the black hole. The redshifted frequency is given by ...


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Yes, they absolutely would. In general, a light-ray which passes at a minimum distance $x$ to the BH will have all of the same effects as a light-ray emitted at the same distance $x$.


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It does not matter whether you are moving relative to anything else that you can observe or just in a fixed reference frame in a completely empty space. Special relativity treats both scenarios as the same because you experience no acceleration/gravity. No matter what state of motion you are in your time will pass with the same speed as always so in a ...


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Both vertical lines in this diagram, i.e. one "after" the evaporation as well as the vertical line in the left lower corner (and all vertical lines without "teeth" in all Penrose diagrams in the world) describe the vicinity of the point $r=0$ in polar coordinates. One can't "cross" these lines because there are no points with the radial coordinate $r\lt 0$. ...


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The most common definition of a black home is the portion of the spacetime manifold $\mathcal{M}$ that is $\mathcal{M} - J^-(ℐ^+)$, the manifold minus the causal past of null future infinity. That is, it's the region of spacetime where no signal can escape to infinity at some point in the future. This requires you to know the global structure of spacetime ...


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My somewhat naive (but understandable) contribution. Begin with flat Euclidean space ( Earth with zero gravity ) Switch on the g force. Spacetime curves and the [euclidean] radius of earth diminish 9 mm. ( earth is sucked in down the gravitational pit ) This diminish earths area on the order of A=2Rr (2 * 10^7 * 10^-2 = 10^5 sqm) The [euclidean] volume of ...


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We do not have an absolute rate for time flow. So, slow or fast depend in what reference frame you are fixing to define what is fast and what is slow. What make sense is asking if we can make the rate of time flow slowly as possible when compared with the proper time for example. And the answer to this question is yes. Any referential frame with constant ...


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Your experiment looks ok to me. There is no resolution of the paradox, so it's one way to look at the root of the 'quantize gravity problem'. Another experiment with the same detector: If the detector is in on a shelf, when it drops off the shelf, it is supposed to mysteriously stop seeing gravitons, even though the 'flux' has not changed. Note that the ...


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The ray will hit the right hand wall at its midpoint, and both the passenger and the outside observer will agree on this. The equivalence principle states that the laws of Physics in a free fall and in an inertial moving elevator are the same, meaning that a passenger cannot expect an experiment to distinguish between the two (but deduce its motion ...


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For clarity I think is best to start with Minkowski spacetime. The equation we are trying to solve to understand the radiation of a point particle is: $$\square A^{b}=j^{b}$$ with the gauge $\nabla_{a}A^{a}=0$ and $j^{b}$ is the current density. The potential \begin{eqnarray} A^{b}(t,x)&=&\int G^{b}_{a}(t,x,t'x')j^{a}(t',x')dx'^{3}dt\\ ...


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An observer outside the elevator would not see the laser beam as being parallel to the floor of the elevator. Imagine you are playing ball with a friend. You stand to the west of him and he runs from south to north as you throw him the ball. You throw the ball as he is directly east of you; in order for him to catch the ball, you must throw it northeast, ...


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Both schools, as described by you, are wrong. There seem to be lack of agreement regarding conditions (like gravitational acceleration and time dilation) at the event horizon. Gravitational acceleration isn't a thing. The metric evolves according to the Einstein equation, test particles move along geodesics determined by their tangents, and the ...


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The answer to your first question can be answered regardless of your misunderstanding of Big Bang. The time passing before atoms begin to form is 379,000 years, which is the time it took for the temperature to drop sufficiently for atoms not to be constantly ionized. – pela, in a comment.


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https://en.wikipedia.org/wiki/Geodesic_deviation This is due to tidal forces. The equations of geodesic deviation which are valid for objects that are small compared to the (radius) of curvature are the easiest way to qualitatively see that. If you plug in the coefficients for the Schwarzschild metric in the conventional distant observer form you will get ...


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In principle yes, though the situation isn't as clear cut as you describe. If you could confine a volume of matter within some volume then gradually heat it by adding energy to it then at some point the total energy density would exceed the density required to form a black hole and at that point the matter would start to collapse into a black hole. However ...


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Observe that $$ \frac{\mathrm{d}\tilde{U}^\mu}{\mathrm{d}s} = \frac{\mathrm{d}}{\mathrm{d}s}\frac{\mathrm{d}x^\mu}{\mathrm{d}s} = \frac{\mathrm{d}}{\mathrm{d}s}\left(U^\mu\frac{\mathrm{d}\tau}{\mathrm{d}s}\right) = \frac{\mathrm{d}U^\mu}{\mathrm{d}\tau}\left(\frac{\mathrm{d}\tau}{\mathrm{d}s}\right)^2 + U^\mu\frac{\mathrm{d}^2\tau}{\mathrm{d}s^2}$$ and $$ ...


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Pulling together what's been said in various comments: 1) General relativity admits models where spacetime is foliated by spacelike leaves, all of which are indexed by a global time coordinate. The simplest of these models is Minkowski space. All of your observations about models with comoving observers apply equally well to Minkowski space, so if you ...


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Suppose two observers, Alice and Bob, are moving relative to each other since the beginning of the universe. While they do it, they construct the chronologies of all the events of the universe, as they record them in their frame of reference. They will construct different chronologies. However, and this is key, each can reconstruct the other's chronology. ...


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A comoving observer and an observer that has been moving at $0.866c$ since Big Bang will disagree on their measured age of the Universe by a factor of 2. While both measurements are correct, we can say that the comoving observer measures a more "natural" age of the Universe. For instance, the comoving observer is the only observer who will measure the ...


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To continue Matas' comment: the definition of temperature is a quantity dependent on mean energy. It's sort of ugly when the particles all go near-relativistic, but there is no upper bound. Interestingly, you can extend the definition to things like magnetic particle orientation states. In this case, beyond a certain magnetic field strength, all the ...


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The space between galaxies isn't that much more empty than what we have here. There is some curvature, enough so that it causes gravitational lensing, even enough to make us suspect there's some extra matter (dark matter) we can't detect by other means. What you're asking is directly measurable: it's the gravitational blueshift from whatever light sources ...


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1) Going to Rindler coordinates, in the near horizon limit of the schwarzschild black hole, you get that the surface gravity is precisely the constant acceleration of the Rindler observer. Indeed this is the equivalence principle at work: gravitation "=" acceleration. For a static observer you can do an explicit calculation by defining the ...


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See the question Laws and theories and the answers to it. The terms law and theory are somewhat vaguely defined so your question doesn't make sense. Relativity, both special and general, is well enough tested that physicists regard it as an excellent working description of the universe. However it can only be an effective theory since it does not take into ...


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The derivation of the gravitational lensing equation requires more maths than you'll learn at school, and indeed more than you'll learn in your first year or two as an undergraduate. General relativity isn't taught in any detail until your postgraduate studies, or possibly in the final year of your degree. The full equation is: $$ \theta = \frac{4GM}{c^2b} ...


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Gravity is acceleration. Einstein's equivalence principle says that gravity (with the vector pointing toward the center of the mass) is equivalent to actual movement with acceleration pointed "outward". That's why we observe gravitational blueshift. Now, blueshift means that the frequency of the photon received is increased as compared to its frequency at ...


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There are actually some nifty simulations that show what you would see: http://jila.colorado.edu/~ajsh/insidebh/intro.html (Had to post as 'answer' because I don't have enough reputation to comment)


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The answer to this question is covered in the book "Exploring black holes: Introduction to General Relativity" by Taylor & Wheeler (2000), within the framework of classical General Relativity. If we are talking about a supermassive black hole, such that a free-falling observer can survive tidal forces as they approach the event horizon and the ...


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Suppose we start with an orthonormal cartesian coordinate basis with unit vectors $\left( \hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}} \right)$. Suppose we wish to rotate to a new coordinate system (may or may not be a coordinate basis, see the following answer which has a good discussion on the difference) with unit vectors $\left( ...


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The convention we pick here will interact with the convention we have for matrix multiplication in the following way: If we have matrices $A$ and $B$ and we use the usual convention that the matrix multiplication $AB$ multiplies the rows of $A$ with the columns of $B$ then we have either $$(AB)_{ij}=\sum_kA_{ik}B_{kj}\tag{#}$$ or ...


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Technically: It will be full dark. you won't able to see light in any direction because all light will be absorb after event horizon. Update: The light you will see inside the black hole it will be the light before you hit the event horizon. Does someone falling into a black hole see the end of the universe? front view(full black) side view(half black) ...


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A test particle in general relativity moves along a worldline. If the world line is differentiable, then it has a tangent. If the test particle is similar to a massive particle, that worldline has tangents that are always timelike. And you could make a unit tangent. That unit tangent lives in the 4d tangent space. As does the energy-momentum vector. In ...


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Since most the universe is empty space The short answer to this surprise is to see that the density $\rho = \frac M V = \frac M {\frac 4 3 {{\frac{2 G M}{c^2}}^{3}} \pi} = \frac{positive-constant} {M^2}$ decreases quickly as the mass increases. Now let's play with the values of size and mass that one can find on Wikipedia and other publications. ...


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The proportionality between angular momentum and mass of the Kerr black hole can be shown directly by performing the Komar integral for angular momentum. In fact, you find that $J=Ma$. The parameter $a$ in the Kerr metric is thus the angular momentum per unit mass. I am not familiar with the relation between $M$ and $R_S$



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