New answers tagged

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Just to offer a counterpoint: That the law of physics are the same for all observers is what is meant by "relativity". Now for any given observer, the distinction between past, present and future is not an illusion. And, there is no observer seeing spacetime from the outside, so no observer can say anything about the whole of spacetime. So the two ...


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You don't. Two given spacetimes can have their metrics written in the same way but may have different coordinate ranges. A simple example is just a spacetime with spatial coordinate identified , such as the cylinder spacetime : $$ds^2 = -dt^2 + d\theta$$ Identical to Minkowski space, which is its universal cover. Of course, two things to watch out for : ...


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It is well understood that if gravity is a 'long range force' its quanta's mass is zero. That is the graviton. Excuse the term in quotation marks, its the easiest way to say what is known. It really means the gravitational field, at infinity in aan asymptotic flat spacetime goes like 1/r But the fact is that the graviton mass could indeed be greater than 0, ...


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The concept of 'straight' is a bit ill defined in GR and has no real definition. In fact in a sense the geodesics themselves be seen as 'straight' lines; they are the shortest paths connecting 2 points (this is what in normal Euclidean space would be a 'straight line') In the LC connection they are the integral curves of some vector field $V$ with $ ...


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The conservation of $\vec{k}\cdot\vec{u}$ only holds in the test particle limit. That is, it considers the metric to be unaffected by the motion of the particle. In this limit, there are no gravitational waves, since the metric has no time-varying quadrupole. If you want to see gravitational waves, you need to allow the metric to evolve dynamically, ...


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There is a treatment of lowering a string through a Rindler horizon here, (which contains a brief discussion on the extent to which the approximation is representative).


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Prahar is correct, but here are two more things to note. If spacetime is an $n$-dimensional Lorentzian manifold $(M,g)$, let $\{E_1,\dotsc, E_n\}$ be an orthonormal frame, i.e. $E_i\in\Gamma(TM), T_pM=\mathrm{span}\,\{E_i\lvert_p\}$, and $g(E_i,E_j)=\eta_{ij}$, where $\eta=\mathrm{diag}\,(-1,1,\dotsc, 1)$ is the Minkowski matrix. Then, if $M$ is orientable ...


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What is more fundamental: Geometry and Topology or physical matter? None of the above. Like CuriousOne said, geometry and topology are mathematical disciplines rather than real objective things. And whilst people talk about electrons and quarks as "fundamental" particles, they aren't really fundamental because after annihilation they aren't there any ...


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Take a future-directed timelike curve $\gamma= \gamma(\tau)$, $\tau$ being the proper time along $\gamma$ in the spacetime $M$. Assume that $p = \gamma(0)$ is the initial point of $\gamma$. Fermi coordinates adapted to $\gamma$ are constructed this way. Consider an orthonormal basis of $T_pM$ with $e_0$ parallel to $\dot{\gamma}$. Transport the basis ...


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Well when there is no answer available, I do not think it hurts to try to come up with one. As far as existence is concerned, Geometry and Topology can exist without physical matter, but not vice versa. In fact, there is so much empty space without matter, but no matter without being in space. Empty space has some Geometry and Topology. However, to ...


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The question is given as a practice problem in Poisson's book "Relativists Toolkit" (p. 159, problem #7) along with some helpful hints. The tensor $\gamma_{ab}$ (which is not a tensor at all) must be defined in the following way: Let $h_{ab}$ be the induced 3-metric on the spacelike hypersurface $\Sigma$, and let on its boundary, $\partial \Sigma$, be ...


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The problem you describe is explored in the wikipedia page on spaghettification. https://en.wikipedia.org/wiki/Spaghettification


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Yes, you are taking the reversal thing too seriously. If you used Kruskal-Szekeres coordinates then those coordinates don't flip from timelike to spacelike. The flip is just because you chose bad coordinates. If you had flat boring Minkowski spacetime you could pick a coordinate system where a coordinate flips from spacelike to timelike across some surface. ...


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The coordinate invariant volume element/measure on a manifold with metric $g$ is $$ d^d x \sqrt{|g|} $$ By coordinate invariance, I mean that if I choose to work in a different coordinate system $x'$, then both the metric determinant changes as does the measure $d^dx$. But they change in a way so as to cancel each other out. In other words $$ d^d x' ...


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Let us for simplicity work in units where the speed of light $c=1$ is equal to one, and assume that there is no cosmological constant $\Lambda=0$. A spherically symmetric vacuum solution to the EFE of the form $$\tag{1} ds^2~=~g_{tt}(r)dt^2 + g_{rr}(r)dr^2 +r^2 d\Omega^2,$$ such that it asymtotically becomes Minkowski space $$\tag{2} ...


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"Presumably, the usual method for speeds v≪c is to consider a Newtonian n-body approach and simply integrate the equations of motion with the gravitational inverse square law, using a numerical scheme such as Runge-Kutta or, something more sophisticated, like a symplectic integrator?" It might work for non-relativistic but for more than two bodies it is ...


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Conserved quantities in GR In GR, energy (or mass) is typically an ill-defined concept. In flat spacetime, we define energy as the conserved quantity corresponding to time translational symmetry. Extending this to GR is quite tricky mainly because, what one is calling time is already observer dependent (this is of course also true in flat spacetime, but at ...


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Actually, it's NOT true that in SR the speed of light in vacuum is the same for all observers, regardless of the motion of the light source. This is true only for inertial observers. The same applies for GR, in which the generalization is a "freely falling frame" (a local inertial frame without effects of gravity). A good reference: Speed of Light


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Correct; in general the speed of light is constant only as measured by local inertial observers. As an extreme example, consider a photon emitted from a galaxy far, far away, in our direction. Although it moves away from the galaxy in the direction of the Milky Way, the expansion of space makes it increase its distance from us. Eventually, however, it will ...


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The explosion that you have seen is actually 4 dimensional representation of the universe. if we are representing universe in 4D then big bang had happened at a point and is expanding as a hollow sphere.But in 3D the big bang should have happened in every point to the universe and is expanding to every direction.This interpretation is using Friedman model of ...


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Of course cellular phone systems could work without GPS. In fact, cell phones DID work before they carried GPS chips! Bottom Line (moved to the top): if there was never GPS or an understanding of General Relativity, the system would have been developed just fine without it. Spend time in a city now with narrow roads and tall buildings (or inside those ...


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According to general relativity, Time is not absolute . But in most of the electronic devices which we use, Time is taken as absolute (For easy calculations) in clock speeds.And when these devices interacts with satellites which are away from earth the difference in time would be a problem in accuracy. Hence the general relativity is used to correct the time ...


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Is there another way to conclude the Schwarzschild solution has a mass M It's not so much a conclusion as a definition. From Schutz in "A first course in general relativity", section 8.4 "Newtonian gravitational fields", pages 207 - 208: Any small body, for example a planet, that falls freely in the relativistic source's gravitational field ...


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Peter is right. The tensor nature requires it to be a spin 2 field, and the graviton is its presumed quanta. But there have been and are theories of gravity that include a spin 0 field. Brans-Dicke theory was one (I think mostly or fully disporved), and some theories for dark energy are spin 0 - quintessence is one, it assumes the cosmological constant is ...


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Your intuition that the Einstein equations are equations for the metric tensor, not for the manifold is mostly on the right track, but the details are wrong. That core bit of intuition is best phrased, I think, as saying that the Einstein equations are local equations for the geometry of the manifold. That is, they tell you that, whatever manifold your ...


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Your mistake is that you left out the summation over double indices (which would give another delta). Since you have to sum over all possiblities of e.g. $c$, one of these possibilities is when c=d and only this term remains. I'll give a simplified example: $$\frac{\partial}{\partial{x^d}}({Q_{bc}}x^c) = \frac{\partial}{\partial{x^d}}({Q_{b0}}x^0) + ...


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Is this analogy of Hawking Radiation correct? No. I know you read stuff like this on the internet, in pop-science articles, and even in textbooks. But it's not correct at all. Sorry. I'll go through it step by step: Within the ergosphere of the black hole, virtual pairs of particles and anti-particles are constantly appearing No they aren't. ...


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The ECE theory is a breakthrough that can be improved in certain ways. The criticisms in wikipedia and in the papers published are just B.S. Carrying out certain experiments, we reach the conclusion that the extra term is correct and experiments match the results of ECE better than the classical one. We will work on that, but convince others is not our ...


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For force carriers the interacting field theory determines the spin. A scalar field yields spin 0; the Higgs is the only example; a vector field yields spin 1, the photon, W, and Z are examples; a tensor field yields spin 2. Since gravitational field theory requires a tensor field for General Relativity, quantized gravity, in the weak-field, linearized ...


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$du,dv$ are light-like, i.e. they could, in principle, be viewed as some affine parameters of some light-rays. However, we will focus (in the spirit of the usual coordinate-nature analysis) on what is the nature of either $u,v$ constant. I.e., we want to know what is the nature of $u,v$ constant hypersurfaces and derive the nomenclature from this. We will ...


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The square root of the determinant of the metric can be understood as a particular function of the components of the metric $g_{ab}$ $$\sqrt{-g} =f(g_{ab})$$ By the chain rule we of course have $$\nabla_a \sqrt{-g} = \nabla_a f(g_{bc}) = \frac{df}{d g_{bc}} \nabla_a g_{bc}$$ But we know that $\nabla_a g_{bc}=0$ so that of course $\nabla_a \sqrt{-g} =0$. This ...


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John above spoke about the law of equivalence to refute your idea, however, a new theory by a British university teacher claims that there is a difference between "an elephant and a gnat" falling down (John's example). MIT wrote about it the other day too: ...


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As the comments have suggested, the problem is that your description of virtual particles appearing to vacuum fluctuations is wrong. Have a look at my answer to Black holes and positive/negative-energy particles for more on this. There isn't an explanation of what is really going on that is accessible to the non-quantum field theory nerd (though I have ...


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The vacuum Einstein equation is just $$ G^{\mu\nu} = 0 \,.$$ Of course, that does not help much, if one does not specify this tensor. That is given by $$ G^{\mu\nu} = R^{\mu\nu} - \frac 12 \mathcal Rg^{\mu\nu} \,.$$ Then we need to specify the Ricci tensor and the Ricci scalar. Those are $$ \mathcal R = g^{\nu\beta} R_{\nu\beta} \qquad\text{and}\qquad ...


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Once you have the metric in u,v coordinates, you notice that $g_{uv}=0$ and this shows that the basis vectors $e_u$ and $e_v$, respectively tangent to the v=const and u=const lines (no typo, you see why?) are orthogonal.


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That quote requires some modification for it to make sense: "General Relativity basically says that the reason I am sticking to the floor is that the path of maximal aging between 'here now' and 'here tomorrow' is through Earth's center."


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Which comes first for gravity: mass or space-time? Neither I'm afraid. As Einstein said, the mass of a body is a measure of its energy-content, and it's actually a concentration of energy that causes gravity. Hence a massless photon has a (very weak) gravitational field. On top of that there's a bit of an issue with spacetime as per CuriousOne's ...


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That is awesome! And it makes complete sense too! (other than a possible misusage of the word "distance"). Let's have a look at the equations of motion of you in Earth's curved spacetime, assuming that your feet are not touching the ground: $$ \frac{\mathrm d^{2}x^{\mu}}{\mathrm ds^{2}}+\Gamma^{\mu}_{\nu\sigma}(x(s))\ \frac{\mathrm dx^{\nu}}{\mathrm ...


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What GR says is correct: the straight line between, say, London today and London tomorrow is not the curve that spends all the time between in London: whether it actually passes through the centre of the Earth I'm not sure, and it depends on how fast you are moving as well as where you are. The caveat is that the straight line (geodesic) not the shortest ...


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It makes sense as a "visual" description. In GR, free particles with mass move on time-like geodesics. A common description of geodesics are such curves that locally minimalize path length, but this desciption comes from Riemannian geometry, not Lorentzian geometry, which GR is. In Lorentzian geometry, timelike geodesics are those that locally maximalize ...


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No it cannot be explained why or how matter can curve space. Unfortunately like Dominect suggest many have the attitude that we should not even be concerned with the why.


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A photon that escapes from black hole's neighborhood does work on the black hole. The photon causes the black hole not to be in the photon's gravity well after the photon has escaped. In other words the photon increases the potential energy of the black hole. The following paragraph may not be science, I just want to say something sane, as opposed to ...


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The curvature of space-time is provided by the solutions of Einstein equation \begin{equation} R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R - \lambda g_{\mu\nu} = 8\pi G T_{\mu\nu}, \end{equation} where $R_{\mu\nu},R$ denotes respectively the Ricci tensor and Riemann scalar. It is important to note that these quantities are provided as a function of the metric ...


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Should physics make an ultimate answer to the why questions? In my opinion, it is not the physicist's aim, and moreover it is beyond the scope of physics. Physics mostly builds theories as our tools to understand and predict some aspect of the surrounding infinitely complicated world. Sometimes we come to a theory that is of such a generic applicability ...


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This is just ordinary potential energy from first semester physics -- when the photon is close to the black hole, it's deep in the potential well. As it goes away from the black hole, it picks up gravitational potential energy, so therefore, it must lose kinetic energy. For a photon, the kinetic energy is given by the Planck formula $E = hf$, so the photon ...


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Since due gravitational time dilation something takes forever to fall inside a black hole from the perspective of an outside observer, there is nothing in it yet that could come out. The light which is emitted from outside the horizon does reach an outside observer, but since it hasn't yet fallen in, it technically doesn't escape from inside the black ...


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Here's a heuristic calculation: Let $\{E_i\}$ be an orthonormal frame ($g(E_i,E_j)=\epsilon_i\delta_{ij}, \epsilon_i=\pm 1$). Then $\mu$ is the canonical volume form $\sqrt{g}\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$ iff $\mu(E_1,\dotsc, E_n)=1$. Then $$(\nabla_X\mu)(E_1,\dotsc,E_n)=\nabla_X(\mu(E_1,\dotsc,E_n))-\sum \mu(E_1,\dotsc,\nabla_X ...


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The time dilation relative to a stationary observer is given by: $$ \frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}} $$ where $d\tau/dt$ is the ratio of the elapsed time on the moving object to the elapsed time for the stationary observer. For any value of $v$ greater than zero the ratio $d\tau/dt \lt 1$ meaning there is some time dilation though in practice ...


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Stephen Hawking says that black holes emit radiations known as Hawking Radiations. This is not yet proven but tachyons might be Hawking Radiation particles since they travel faster than light.


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$f(a)$ is not an independent degree of freedom from $R$ or from $a$. You have a system of equations here, and if you solve for the exact form of $a(t)$, you are not likely to admit solutions that have $f(R(a(t))) = 0$ for finite $t$. If you do, however, it's likely an indication of geodesic incompleteness of the solution.



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