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0

Let us do the RHS first. This just gives us a derivative on the metric: $$\frac{\partial L}{\partial x^\lambda}=\frac{1}{2}\partial_\lambda g_{\mu\nu}\dot x^\mu\dot x^\nu$$ The first derivative on the LHS is essentially a derivative of a square, thus $$\frac{\partial L}{\partial \dot x^\lambda}=g_{\mu\lambda}(x(\lambda))\dot x^\mu$$ where we have made the ...


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Let $\Omega^2=\exp\phi/2$ so that $\gamma=\Omega^2 g$. Then we have the standard formula $$R_\gamma=\Omega^{-2}\big(R_g-2(n-1)\Delta\ln\Omega-(n-2)(n-1)[\nabla\ln\Omega]^2\big)$$ This is proven in any number of General Relativity texts, but the one in R.M. Wald, General Relativity (1984), is particularly easy to follow and is proved in the form shown here. ...


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I can only answer with opinion rather than facts, but for what it's worth: Is travelling through a worm hole really possible (if we can find one stable worm hole)? No. Wormholes are science fiction. So is time travel. See A World without Time: The Forgotten Legacy of Godel and Einstein. A clock doesn't literally clock up the flow of time like some cosmic ...


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We know that light is massless so why does a black hole's gravity attract light? Because gravity doesn't just attract objects with mass. It alters the path of light too. Because gravity is caused by a concentration of energy which "conditions" the surrounding space, altering its metrical properties, whereupon we talk about spacetime curvature. But note that ...


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A photon has a rest mass of nought (where the rest mass $m$ is the Lorentz-invariant quantity in the four-momentum's Minkowski norm squared $E^2/c^2 - p^2 = m^2 c^2$). However, a lightfield of energy $E$ gravitates and itself has a gravitational source equivalent to a mass $E/c^2$. Also, a system of photons has a nonzero rest mass (see reference), as does ...


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light is supposed to possess relativistic moving mass even though it does not possess any rest mass. m2c2=M2c2-M2v2 where m is rest mass and M is relativistic mass and v=c. this gives m = 0 , but M is not zero the value of M can be calculated from the experimental data on radiation pressure.


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Mm.... At first I repeated your calculation and got the same answer as yours, then I checked the paper you gave and found it consist with (32 a) and it seems not a typo, so I read it from begining - oh brother it's not 2+1 gravity - -b it's 3+1 gravity and you should treat $d\Omega^2$ more carefully: $r^2 d\Omega^2=r^2d\theta^2+r^2\sin^2\theta d\phi^2$ so ...


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You have the order wrong in which you "get" things. You get the energy-momentum tensor from your specific matter theory. You do not know what $g_{ab}$ is. Then, given some general assumptions about your spacetime, you write down an ansatz for the metric. For Schwarzschild, we have time independence and $\mathrm{SO}(3)$ isometry. Then you calculate the Ricci ...


2

anna v's nice answer didn't go where I expected: the big acceleration at the LHC isn't in the accelerator, it's in the collisions. Let's suppose we have a proton in the LHC that undergoes an elastic, billiard-ball type collision and ends up with its original momentum in the opposite direction: \begin{align} \vec p_\text{initial} &= +7\,\mathrm{TeV}/c ...


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I'll prove a formula that is probably easier to use for this. \begin{equation} \begin{split} \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} g^{\mu\nu} \partial_\nu \phi \right) &= \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} \right) g^{\mu\nu} \partial_\nu \phi + \partial_\mu \left( g^{\mu\nu} \partial_\nu \phi \right) \\ &= ...


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All timelike geodesics in Minkowski spacetime start at past timelike infinity and end at future timelike infinity. The worldlines of Rindler observers are not geodesics, whereas the worldlines of Minkowski observers are. Heuristically think of a flat Euclidean plane. There are plenty of inextendible curves that don't go to infinity, but all geodesics start ...


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If the observer is not in free-fall, the metric-tensor $g_{\mu,\nu}(s)$ at the observer's position, expressed in local coordinates around the observer, will not be $\eta_{\mu,\nu}$. Your first assumption about the path $(\gamma)$ is wrong. I guess what you are aiming at is the notion of the space of coordinates around a point, which is indeed a flat space ...


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So how would the gravity engendered by the electron affect the photon if it can’t slow down? Somewhat counter-intuitively, the ascending photon speeds up in line with the increasing "coordinate speed of light". See this PhysicsFAQ article where Don Koks writes this: "Given this situation, in the presence of more complicated frames and/or gravity, ...


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The energy of a photon is given by the equation E = hf where h is Planck's constant and f is frequency. The energy would decrease, making the frequency decrease (since h is constant). So, if the photon was blue light, then it would get redder and redder as time when on. There is a point, however, when your system eventually stops working. This is because the ...


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The energy (i.e. frequency) of the photon will change as it travels thru a gravitational gradient, by Einstein and so: as the photon goes away its color will be redshifted. What experiment proved that the electron is the source of a gravitational field? none afaik. edit post: the total energy budget ( electron + grav field + photon) is compromised if ...


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Two thoughts: You have to be careful about what you mean with the "same physical laws as our own" in one spatial dimension. If you write down Einstein gravity in 1+1 dimension you realize that it is completely topological, i.e. there are no local excitations, no gravitons, no equations of motion etc. It is not surprising then that this theory looks ...


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After further research and thought, I think I can now offer an answer to my own question. I first restate the key question, then state the conclusions and finally comment on them in the context of clarifications of the original “Background and Commentary” arising from the responses of Timaeus and JerrySchirmer, who I thank for their input. The answer I am ...


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First possible point of view: In the Pound–Rebka experiment the redshift / blueshift of photons is measured in small distances. This experiment one explain by the influence of gravitational field on the photon: "When the photon travels through a gravitational field, its frequency and therefore its energy will change due to the gravitational ...


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This is yet another instance of taking the ubiquitous balloon analogy too far. See, while it's a wonderful way to express the expansion of the universe, there are some misconceptions that arise from it: We live in a universe of finite size (we don't know, but we think not) and non-zero curvature (according to WMAP, we don't, or at least we think we don't) ...


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From scholarpedia: The Unruh effect is a surprising prediction of quantum field theory: From the point of view of an accelerating observer or detector, empty space contains a gas of particles at a temperature proportional to the acceleration. Direct experimental confirmation is difficult because the linear acceleration needed to reach a temperature 1 K ...


1

No definite answer to this question, the effect in some sources is accepted and other sources dispute it. From Wikipedia: The hypothetical Unruh effect (or sometimes Fulling–Davies–Unruh effect) is the prediction that an accelerating observer will observe black-body radiation where an inertial observer would observe none. In other words, the ...


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With current technologies the Kerr parameter could not be precisely estimated yet. The best result today is a Kerr factor $a=0.52$. Genzel, R., Schoedel, R., Ott, T., Eckart, A., Alexander, T., Lacombe, F., Rouan & Aschenbach, B., Near-infrared flares from accreting gas around the supermassive black hole at the Galactic Centre (2003), ...


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If you mean by "our universe" the matter in spacetime we are able to reach and observe then you are right. The universe will become more and more finite for us unless someone will invent a "warp drive" or "wormhole" (currently the probability for it is very low). According to research you have ca. 100bn years time before all others galaxies will be gone ...


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You raise an interesting point about the role of experiment and falsifiability in science. Despite a long-standing anomaly in Mercury's perihelion, Newton's theory of gravity itself wasn't heavily questioned, let alone rejected or falsified. Rather, auxiliary assumptions were concocted that saved Newton's theory, such as an erroneous mass of Venus, a planet ...


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As far as cosmology is concerned, the book which I consider to be THE best for a mathematical treatment of cosmology, is AK Raychaudhuri's "General relativity, astrophysics, and cosmology". It is excellently presented, Raychudhuri doesn't shy away from the math, and the old-school style makes it all the more elegant. So, I would STRONGLY recommend it. I ...


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Galaxies are not moving away from us, it is the space between us and the galaxies (and everything, in general) that is continually expanding. This is allowed to happen faster than the speed of light, because no object actually crosses the light speed barrier in the process. So consequentially, the universe has no size constraint like the one you've stated.


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Just a quick preliminary answer, I will fix it later. The connection to general relativity is a change of variables in which the metric is replaced by a "spin connection" and a "frame field". These quantities can then be arranged in a new matrix, so the metric field has been rewritten as a different matrix-valued field, and the transformations ...


1

One thing you can use the curvature tensor for is to detect singularities in the spacetime. For the Schwarzschild solution, the simpler curvature scalars formed from the Ricci tensor, $R$ and $R_{ab} R^{ab}$ vanish everywhere due to the fact that this is a vacuum solution to the Einstein equations. But since you have the full Riemann tensor at your ...


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Update: According to this paper, "On the Interpretation of the Redshift in a Static Gravitational Field", the answer I give below is a common but misleading interpretation. The classical phenomenon of the redshift of light in a static gravitational potential, usually called the gravitational redshift, is described in the literature essentially in ...


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I now know what to do for the case where the four roots $b_1, b_2, b_3, b_4$ are distinct. Write $\rho_1^A := o^A + b_1 i^A \\ \rho_2^A := o^A + b_2 i^A \\ \rho_3^A := a^A + b_3 i^A \\ \rho_4^A := a^A + b_4 i^A$ Then $\Psi_0 = 0$ implies the 4 equations for $\alpha , \beta , \gamma, \delta$: $\alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B ...


3

Some redshifts are an example of the Doppler effect, familiar in the change in the apparent pitches of sirens and frequency of the sound waves emitted by speeding vehicles. A redshift occurs whenever a light source moves away from an observer. The energy balance is with the source of the photons. If the source is moving away the photons have less energy ...


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The short answer is: in ultra-deep gravity wells one has to bring relativistic considerations into the picture. The result of this is that an 'event horizon' forms in ultra-deep gravity wells such that an observer looking from a distance into the gravity well can only look till a finite depth. In loose terms, that specific finite depth corresponds to the ...


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Without invoking relativity, let's look at energy. The force of gravity between two objects goes as $$F= \frac{GMm}{r^2}$$ Which implies that the force is weaker when you are far away and stronger when you get closer. potential energy is the integral of force, and we know the sum of potential and kinetic energy is constant. Putting potential energy =0 at ...


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You make two wrong assumptions in your question, namely that if an object is accelerating the velocity would keep increasing ad infinitum without limit, and that the acceleration due to gravity on earth is always $9.8 m/s^2$ these are both not the case. First of. The theory of relativity doesn't allow for objects that have mass to go faster than the speed ...


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If you are working with the Newman Penrose null tetrad, i.e. with respect to which the metric looks like $g=\left(\begin{array}{cccc}0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 1 & 0\end{array}\right)$ then the standard notation is $\lbrace l,n,m,\overline{m}\rbrace$, where $l,n$ distinguish the ...


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The problem with you argument is that you are standing on the planet! When you make it denser and denser and eventually turn it into a black hole, you would still be standing on it, ie be outside the event horizon, so as you are saying correctly the light will escape. If you would like to think about being inside the event horizon then it's wrong to imagine ...


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The issue here is that the Schwarzschild coordinates are divided into two disconnected patches by the coordinate singularity at $r=2m$. There is no physical connection and the two pieces can be viewed as separate solutions. The Eddington-Finkelstein coordinates take the outer solution and extend it beyond the horizon, but inside the horizon it is different ...


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if by "gauge transformation in the vielbein" you mean the local lorentz transformation that acts on one of the indices of the vielbein, then the answer is no. Because this local symmetry is in addition to the general coordinate transformation, and not part of it. In other words all veilbeins that are related by a gauged lorentz transformation correspond to ...


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The question of AdS (in)stability is indeed a hot topic in current research of the AdS/CFT correspondence. It is a field that ties together many interesting subjects: Gravity in AdS (i.e a confining box), thermalization in QFTs, the theory of non-linear differential equations and their perturbative treatment, turbulence etc. This explains the explosion of ...


1

Choosing $\dot{v}>0$ is equivalent to choosing an arrow of time in your spacetime. Purely from GR, there is now way to determine the arrow of time and therefore it is not possible to prove $\dot{v}>0$. The logic is the following: You observe that there are two equivalence classes of timelike vector fields (which you call future- and past-directed). You ...


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As you mentioned, $\gamma$ is a causal curve. The space-time causal structure is given by the semi-definite (pseudo-Riemannian) metric. But locally we can always choose a Galilean basis in which $g_{\mu \nu}$ is just the Minkowski tensor (this is called the equivalence principle, as you probably know). The structure of the Minkowski space arises naturally ...


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Re a recent variant of the question: Physicist using mc² to calculate positive energy and −MmG/R to calculate negative energy to calculate total energy of the universe but I heard that one of them is relatively invariant and other one is not. Is that true? Physicists don't use mc² to calculate positive energy, that's merely a mass-energy equivalence ...


0

I believe gravity's rainbow relates to gravitational lensing, which can be likened to refraction. In normal refraction we see a rainbow spectrum, because the degree of refraction depends on the photon E=hf energy. However in gravitational lensing, we do not. Gravity's rainbow is a speculation that we might. I'm not fond of it because it's akin to saying ...


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I'm afraid I can't answer either question, but can I point out that a torus with a major and minor rotation is arguably emulating a Dirac bispinor, and could be useful for something other than gravitomagnetism. Note that Thomson and Tait contrived smoke rings in an early theory, and demonstrated attraction and repulsion and annihilation, see ...


0

To 1.: You don't need a plasma for this effect. As the article states this 'pulling through the hole effect' is a frame-dragging effect. The apparent force (seen in the Lab-frame) results from the object falling along the spacetime-geometry, however in its own frame there are no g-forces as a free fall is always force-free. So the statement you put in bold ...


0

Doesn't the Schwarzschild metric combined with Hawking radiation imply that nothing ever gets past the event horizon of a black hole? No. First of all, let's set Hawking radiation aside, we don't need it. Let's look at this: An observer looking from the outside will never see the particle cross the event horizon, even if he/she looked for an arbitrarily ...


0

If, as is canonically accepted, the black hole evaporates within a finite time, then the freely falling particle will be released from its gravitational attraction. By the premise of the question, in the observer frame the particle can not have passed the event horizon during the intervening period. This would also solve the quantum information paradox, as ...


0

The galaxy rotation problem is the most obvious, but the advent of computers has allowed this to be fudged (in my view) with complex distributions of dark matter. The challenge is to find a 'crucial' anomaly that can't be fudged in this way. Possibilities include: globular clusters which show a similar rotation anomaly (Scarpa et al., 2006) but are too small ...


0

In this paper they speak of separating the negative energy density out of a vacuum. This is something we barely understand as it is, and to say that it could lead to anything like a warp-drive is a tremendous leap of faith.


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The definition of conformal time is actually $$ a\,d\eta = dt\Leftrightarrow d\eta=\frac{dt}{a} $$ which gives you the correct result.



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