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In general relativity, instead of momentum, you have local stress-energy tensor, which contains energy density, momentum and stress components. But, these are local - there are in general no conserved integral quantities (such as total energy, momentum,... across a large chunk of space-time) because (loosely speaking), conservation of a quantity from one ...


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No, the age of the universe doesn't depend on the observer. What depends on the observer is the "perceived" time that has passed since the Big Bang. What you are asking is if the conformal time and the age of the universe are the same and the answer is negative as you can see in that link.


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Certain condensed matter systems show emergent behavior that is similar to general relativity: see this for example. Also, in fluid mechanics, sound waves can become trapped behind an "event horizon" called an acoustic black hole. Finally, the Einstein field equations are essentially the only possible classical equations of motion for a massless spin-two ...


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General relativity is a theory of gravity; as such, it makes predictions about gravity. However, general relativity does make predictions about time and physical entities such as black holes. Some of the predictions general relativity did make: Gravitational waves exist (proved by LIGO last year) Black holes exist Light bends (proved in 1919 by an ...


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In a quantum world, all oscillations are quantized (if they are truly periodic, not just approximately periodic). For example, oscillations of a solid are quantized and the quanta are called phonons. That doesn't mean that the model of the solid as a lattice of (quantum) atoms is wrong, or even that it's an approximation with a limited domain of validity. On ...


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Does the discovery of a photon and development of quantum electrodynamics make Maxwellian electrodynamics redundant? Not a bit. Each physical theory has its domain of applicability. Electrodynamics successfully describes the macro phenomena of electricity and magnetism which are very much obscured when you look at them from the point of view of QED. ...


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If the graviton is detected and its cross section and other properties and interactions are measured well find out something about quantum gravity and maybe even how to unify gravity and the other forces. As @CuriousOne says we are pretty far from discovering it, nobody is looking for it, and we don't have any way of even coming within 10 orders of magnitude ...


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A 2013 paper by Shtanov and Sahni (already mentioned by Ben Crowell in the comments) says that the modes grow exponentially in conformal coordinates, and Barrow et al overlooked the fact that the conformal time changes very little during and after inflation. A 2014 preprint by Tsagas, one of the authors of the original paper, cites Shtanov and Sahni and ...


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To add to Ocelo7's answer, the transverse part is, as I have seen it used by e.g. Ellis, used to refer to components that are orthogonal to a future-pointing and geodesic null vector field spanning the past light cones of a particular world line -- the central observer (e.g. our world line). They are defined on parts of spacetime that is observeable by the ...


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Take the covariant derivative of the equation $X^aX_a=-1$. The RHS becomes zero so we have $$2X_a\nabla_b X^a=0\implies X^aB_{ab}=0.$$ The other equation, $X^bB_{ab}=0$, is the geodesic equation, so it doesn't hold for just any $X^a$. Let's consider the situation at some point $p\in M$. Then $X^a$ is a prime candidate for the timelike basis vector of $T_pM$....


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I can answer to the first part of your question. A metric with harmonic coefficients is for example the FLRW metric for an universe with positive curvature. In this case the metric takes the form ($c=1$): $$ds^2 = dt^2 - a^2(t) \left(dr^2 + \frac{1}{\sqrt{k}} \sin(r\sqrt{k}) d\Omega^2 \right)$$


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Since you mention the following in one of your comments I'm less interested in Einsteins historical struggles and would love a more modern perspective on how to get to this insight. I hereby unashamedly ignore history, and offer instead a quick plausibility argument. Let's start with the equivalence principle which, loosely speaking, says that a (...


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According to the fuzzball proposal in string theory, black hole are actually horizonless and regular solutions. For some systems in five dimensions made with bound states of intersecting branes this has been already proved directly in supergravity: there are solutions without horizons and singularities, with the same asymptotic charges (Mass, Angular ...


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This is not what I, and I would posit most physicists, understand as a physical treatment of what general covariance is in physics. General covariance is that the equations look the same in any coordinate frame - any meaning that the transformations can be any function. The only limitation is that the functions be differentiable, maybe n or infinite times (...


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At best things are pretty speculative. Cumrun Vafa has proposed that black holes have condensates of tachyons. In some sense you can understand this without much complexity. The Schwarzschild metric has a physical singularity that is a spatial surface. The Penrose conformal diagram for the Schwarzschild metric illustrates this The bosonic string has two ...


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The development of general relativity has led to a lot of misconceptions about the significance of general covariance. It turns out that general covariance is a manifestation of a choice to represent a theory in terms of an underlying differentiable manifold. Basically, if you define a theory in terms of the geometric structures native to a differentiable ...


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I have communicated with both of these fellows. The mathematics is based on Donaldson's theorem that in four dimensions there exists an infinite number of atlases of charts on a manifold that are homeomorphic but not diffeomorphic. I am not able to go into the mathematics, for it is pretty deep. It centers around the moduli space of self-dual connections $SD/...


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The holes you are used to seeing most likely occur on Earth. Since Earth's surface is a 2-dimensional object, holes in it are also 2-dimensional. Given that we live in a 3-dimensional world (spatially/geometrically), these holes would then be traversable; leading to a "pit" below. Since it isn't realistic to have a pit that goes completely through Earth, you ...


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for getting Christofel symbols we should notice that if two vectors are parallel transported along any curve then the inner product between them remains constant under parallel transport. so you should write your sentence for 3 parameter a, lambda and nu cyclical,and then you obtain the correct Christofel.


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An object swallowed billions of years ago thus will retain some incredibly tiny solid angle over which a sufficiently vertical photon will eventually reemerge into the outside world... For a galactic center black hole, how exactly does "pump" mass-energy from the singularity back out to the event horizon... ? ... the apparent horizon at which the ...


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This question has already been answered by a Randall Monroe of XKCD and Dr. Cindy Keeler of the Niels Bohr Institute. It forms a Naked Singularity. Which is an infinitely dense object, from which light can escape. Source: https://what-if.xkcd.com/140/ You Reissner–Nordström metric for this question, as opposed to the more well known Schwarzschild metric. ...


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I think this is one of those situations where it is cleanest to think relativistically and concentrate on gravity as spacetime geometry, not gravity as a force. If the stress energy tensor's components reach a such a magnitude that an horizon forms, then the mass/energy inside the horizon is doomed to stay inside evermore. This is a question of spacetime ...


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The strong force is responsible for neutron stars not becoming black holes. Although attractive at nuclear densities, it becomes repulsive at somewhat higher densities. This repulsion is many times greater than the electromagnetic repulsion between protons at similar densities. Yet we know that if a star is too large, it's supernova event will lead to a ...


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I wonder of you are overthinking this. Wald says: If the universe had always expanded at its present rate that is, $\dot{a}$ is a constant and independent of time. In that case the value of $a$ at time $t$ after the Big Bang is simply: $$ a = \dot{a} t $$ So if you define $T$ by $T = a/\dot{a}$ then $T$ is necessarily the age of the universe.


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For the case $\Omega_M + \Omega_\Lambda = 1 = \Omega_\text{total}$, which is a good approximation for $t \gtrsim 10^8 \; \text{yr}$, an explicit formula for a(t) is $$a(t) = \left[ \frac{\Omega_{M,0}}{\Omega_{\Lambda,0}} \sinh^2 \left( \frac32 \sqrt{\Omega_{\Lambda,0}} \, H_0 \, t \right) \right]^{\frac13}$$ or, plugging in numbers from Pulsar's answer, $$...


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A spin-2 field in 4D is not five-dimensional. The standard spin-2 object transforms in the $(1,1)$-representations of the Lorentz group where the $(m,n)$-labels are half-integers labelling an equivalent representation of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, see the Wikipedia article on representations of the Lorentz group for more information. The spin ...


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According to the abstract of a paper at https://arxiv.org/abs/hep-th/0604072, Breakdown of local physics in string theory at distances longer than the string scale is investigated. Such nonlocality would be expected to be visible in ultrahigh-energy scattering. The results of various approaches to such scattering are collected and examined. No evidence ...


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You statement $R = a^2 \gamma$ is wrong at least because of the dimensions. The correct one would be the following: if the 2 observers had initial coordinate distance $r$, then the physical distance between them is $R = a r$. Then, follows: $$ R(\tau) = a(\tau) r $$ $$ \frac{d R}{d \tau} = \frac{d a}{d \tau} r = \frac{d a}{d \tau} \frac{R}{a} $$ which ...


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I've answered the questions I can below. @John Rennie already gave a link to Lemaitre's paper. Here's a summary from John Gribbin's The Scientists, on page 596-597: ...the Belgian astronomer Georges Lemaitre (1894-1966), who was also an ordained priest, independently published similar solutions [to those of Aleksandr Friedmann's] to Einstein's ...


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The time dilation factor with respect to an observer at infinity is $$\sqrt{1-\frac{\text{2 G M}}{\text{c}^2\text{ r}}}$$ so if we plug in G=1, c=1, r=10 and M=+1 we get the clocks running slower by a factor of 0.8944 if they are in a distance of 10GM/c² from the center of the positive mass. If we change the sign of M to M=-1 we get a time dilation factor ...


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The arrow of time is believed to be related to the fact that the universe started in a state of low entropy and is evolving towards a state of larger entropy. The effect of negative mass will not change this. The reason is that any model of negative mass will leave the initial state of the universe as as state of low entropy. A rather uniform distribution of ...


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The geodesic equation is given by, \begin{equation} \frac{\mathrm d^2}{\mathrm d\tau^2}x^{\mu}+\Gamma^{\mu}_{\lambda\sigma}\frac{\mathrm dx^{\lambda}}{\mathrm d\tau}\frac{\mathrm dx^{\sigma}}{\mathrm d\tau}=0 \end{equation} which is a set of 4 equations for $x^{i}$. $\Gamma^{i}_{jk}$ tells us about the curvature of the space time which can be written in ...


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The energy $h\nu$ of a photon is simply the contraction $$ h\nu = g_{\alpha\beta} k^\alpha u^\beta $$ of its momentum $k^\mu$ with the frame's velocity $u^\mu$. The frequency shift is then given by $$ 1 + z = \frac{\nu_S}{\nu_O} = \frac{(g_{\alpha\beta} k^\alpha u^\beta)_S}{(g_{\sigma\rho} k^\sigma u^\rho)_O} $$ where we denote the velocities of both source ...


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Given a metric $$ \mathrm{d}s^2 = g_{MN}\mathrm{d}x^M\mathrm{d}x^N$$ in $n$ dimensions you find the decomposition down to $d$ dimensions by rewriting the $n$-dimensional metric in terms of objects with no, one, and two $d$-dimensional indices (in the following indicated by Greek letters): $$ \mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu + 2 A_{\...


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To understand this you need to understand what we mean by speed. If I want to measure positions and times I need to set up a coordinate system. For example I can take my stopwatch and my metre rule and construct some Cartesian axes $t$, $x$, $y$ and $z$, then I can describe every point in spacetime by its position in my coordinates $(t,x,y,z)$. Once I have ...


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Is this simply the ratio of the angular momentum that the blackhole is observed to have as a ratio of the maximal angular momentum as limited by some Physics (Kerr Metric?)? Yes, exactly. For a spinning black hole there are two event horizons, an inner and an outer horizon. The positions of the horizons are given by: $$ r = \tfrac{1}{2}\left(r_s \pm \sqrt{...


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Yes, the dimensionless spin such as $0.2$ in this case is simply the ratio $$ a= 0.2 = \frac{|\vec J_{\rm measured}|}{|\vec J_{\rm max}(M)|}$$ where the denominator is the maximum angular momentum allowed for the same value of the mass (as the measured mass). For the $d=4$ Kerr black hole, the maximum (the angular momentum of the extremal Kerr black hole) is ...


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A metric defines a flat space within an open neighborhood $U$ if and only if the Riemann tensor $R$ vanishes over that neighborhood. So you simply have to calculate $R$ and check that it vanishes in the neighborhood. The only if part of the assertion is clear. The if part is probably more interesting to you and indeed a simple proof of the if shows you (in ...


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After a binary merger the resulting black hole will experience a ringdown. The ringdown is characterized by a spectrum of quasi-normal modes (QNM) which, according to GR (specifically black hole perturbation theory), depend only on the final black hole mass, angular momentum, and charge. This is effectively the no-hair theorem. QNMs are oscillations that ...


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Check out this figure from the most recent LIGO paper. This is the reconstructed gravitational wave signal determined by the analysis. Focus on the zoom of the end of the signal. This is the merger. The merger part of the GW151226 looks a lot like the merger of GW150914. The differences tell us about the sources of the gravitational waves in each case....


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I think you are supposed to simply argue that since the metric is "constant" in the sense of Eq. (C.2.3) on $\Sigma_t$, the curvature should be a constant as well. However, here is a more sophisticated point of view. Our definition of homogeneity is that the isometry group $\mathrm{Iso}(\Sigma_t,h)$ is transitive, i.e. given $p,q\in\Sigma_t$ there exists $\...


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Ever after Maxwell unified electricity and magnetism, the goal of unification is a holy grail for theoretical physics, for all forces. The macroscopic forces of electricity and magnetism emerge from the underlying quantized level of nature, photons, electrons and all the elementary particles in the standard model table enter with fields in quantum field ...


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The action for classical gravitation is the Hilbert-Palatini action $$ S = \int d^4x\sqrt{-g}R $$ for $R$ the Ricci scalar. The Goto or Polyakov action is of the form $$ S = -\frac{1}{4\pi\alpha'}\int d\tau d\sigma g_{\mu\nu}\left(\frac{\partial X^\mu}{\partial\tau}\frac{\partial X^\nu}{\partial\tau}~-~\frac{\partial X^\mu}{\partial\sigma}\frac{\partial X^\...


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The curvature in general relativity is determined completely by the metric (since the metric determines the Levi-Civita connection). Since the metric is constant on the homogeneous space so too must the curvature be.


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The question is a bit weird because from that distance, the photons would take so long to come back, I am not even sure a whole species could survive long enough to see one photon come back. You have to realize that even for the Sun, as relatively close as it is, photons take on average 8min 20sec to arrive. So you're always seeing the Sun as it was over ...


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A good resource is M.D. Scadron, "Advanced Quantum Theory" from 1979. Gravitons in Minkowski-space have energy and momentum and couple to gravitons. Iteration (at tree level) reproduces the Einstein theory. This was first shown by Gupta 1952. Of course, there are problems in higher loop order, the gravition theory is not renormalizable.


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On top of the other excellent answers I'd like to point out that the accretion rate of dark matter particles is believed to be much smaller. The reason matter in accretion disk is being accreted rapidly is because they lose energy from electromagnetic radiation. For dark matter particles, in practice the only way it can be accreted is if the particle ...


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I suppose you are asking about locally inertial frames? Chapter 2.4: Postulate (2) of general relativity implies that at each point of spacetime it is possible to choose locally inertial coordinates: $\xi^m$ Say you have coordinates $x^\mu$ and you want to transform to inertial coordinates $\xi^m$ which are in locally inertial frame $ds^2=\eta_{m ...


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The local escape velocity is $$v_{esc} = \sqrt{\text{2 G M}/\text{r}}$$ At infinty you observe that velocity slower by a factor of $$1-\text{r}_s/\text{r}$$ so at infinity you observe $$\text{v}_{esc} = \sqrt{\text{2 G M}/\text{r}} \cdot (1-\text{r}_s/\text{r})$$ because of gravitational length contraction radial to the mass and gravitational time ...


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If "open" means nonpositive spatial curvature (which is the usual meaning), then no. The experimentally measured spatial curvature is consistent with zero, but the error margin includes values on both sides of zero. There's no conflict in the Friedmann equations between accelerating expansion and positive spatial curvature. If "open" means "never ...



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