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1

You seem to be asking for a comparison of the angular momentum carried by gravitational waves in classical general relativity with the the angular momentum carried by quantum gravitons. For the analogous problem in electromagnetism, you should read Beth's Mechanical Detection and Measurement of the Angular Momentum of Light, from early in the days of ...


-2

GPS satellite navigation system doesn't use, doesn't need and doesn't prove Einstein's General Relativity. The GPS satellites use classical (Newtonian) relativistic principles to work. These are the same relativistic principles that make sense in the everyday world, that most people equate with 'common sense'. GPS calculates positions based on geometric ...


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The speed of light can be slower than c. wikipedia: Propagation_of_light_in_non-inertial_reference_frames So there can be things going faster in one frame than the slow light of another frame.


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There is a subtle difference between "local" and "global" (or apparent) superluminal travel. Kaku is only correct in the global sense. Local superluminal travel, in the sense of increasing one's speed to exceed the speed of light, is strictly prohibited as special relativity holds in any sufficiently local frame. However, general relativity (with the most ...


5

The cosmological constant has an interesting history behind it. Originally, when Einstein introduced his theory of general relativity in the early 20th century, the Einstein Field Equation, which was the equation for the gravitational field, described gravity as the effect of the curvature of space-time due to the presence of matter and energy. Perhaps you ...


3

The cosmological constant is important for at least two reasons. Our universe is currently asymptotically evolving towards a universe where a constant energy density dominates the total energy density. The cosmological constant can be interpreted as exactly this. Therefore, analysis of the current state of our universe relies heavily on the concept of a ...


1

This is the basic idea of brane-world approaches to string theory. I don't know their current state.


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The answer is "maybe". Anomalous gravitational effects in black holes just exist beyond a region called "event horizon". After the collapse, only bodies moving very closely the "event horizont" can feel these bizzare gravitational effects. That's a good example about it: if the sun collapses to a black hole right now, gravitationaly, nothing would change for ...


1

Here are several thought experiments (and what happens in each). I'll ignore relativistic effects like time distortion - not for your sake, but mine :) The earth collapses to a black hole beneath our feet. We fall with it, and end up inside a black hole, presumably dead. The earth collapses to a black hole beneath our feet, but we stay in the same place. ...


0

As a geometric quantity, the value of an "angle" can be determined and expressed in a coordinate-free way: Given three pairwise space-like events, "$A$", "$B$" and "$C$", and given the positive real numbers $\frac{s^2[ A C ]}{s^2[ A B ]}$, $\frac{s^2[ A C ]}{s^2[ B C ]}$ and $\frac{s^2[ B C ]}{s^2[ A B ]} = \frac{s^2[ A C ]}{s^2[ A B ]} / \frac{s^2[ A C ...


3

There are both physical and formal reasons to introduce the spin connection. Physically, we know that there are spin 1/2 particles. A spin 1/2 field cannot be described by anything built from 4-vector fields. You can realize this for example by that 4-vector fields (and so anything built from them) returns to their original value after a $2\pi$ rotation ...


0

In situations like this we generally assume the mass of the moving body is negligable compared to the mass of the black hole. In that case the spacetime curvature can be assumed to be just due to the black hole, and it's described by the Schwarzschild metric. Any moving body moves according to the spacetime curvature at its location. This is important to ...


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I don't know exactly where to start either but one could think of the black body radiation coming from a source far away that was emitted by a source at temperature $T_1$ where $g_{00}$ had some value say $g_{00}^1$. Now,if we make use of the fact that $g_{00}$ influences the photon frequency via gravitational Doppler effect, then imagining balance of ...


1

We will work with the metric supplied, $$\mathrm{d}s^2 = A(r)^2\mathrm{d}t^2 -B(r)^2 \mathrm{d}r^2 -r^2 \mathrm{d}\theta^2 - r^2 \sin^2\theta \, \mathrm{d}\phi^2$$ I will assume that omitting the fourth spatial coordinate (which I have added) is simply a typo in the original post. I have also redefined the arbitrary functions for convenience. We choose a ...


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In answer to the edit, any transitions due to single-graviton exchange will involve energies that are just impossibly small. To convince yourself of this, remember that the energy levels of the hydrogen atom are given by: $$E = \frac{\mu k^{2}e^{4}}{2\hbar^{2}n^{2}} = \frac{13.6\,\,{\rm eV}}{n^{2}}$$ If you do the same for two solar mass neutron stars ...


2

If it is a photon, then you know that $x^{a}x_{a}$ is always zero. You have measured the spatial direction of the photon with your mirror apparatus, so you know the values of the $x^{i}(0)$ at some time${}^{1}$, which we will call zero. Furthermore, we can infer the value of $x^{0}$ at this time from the fact that $x^{a}x_{a} = 0$. Then, all we need is to ...


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Yes indeed to all your questions: mutually orbitting binaries do spin down, the system's orbital angular momentum thus decreases with time and the loss of energy and angular momentum is almost certainly owing to the emission of gravitational waves. Look up the Hulse-Taylor binary system: its spin-down has been carefully observed and measured since its ...


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A common procedure to determine the spin of the excitations of a quantum field is to first determine the conserved currents arising from quasi-symmetries via Noether's theorem. For example, in the case of the Dirac field, described by the Lagrangian, $$\mathcal{L}=\bar{\psi}(i\gamma^\mu \partial_\mu -m)\psi $$ the associated conserved currents under a ...


2

If you linearise the theory such that $$ g^{\mu \nu}(x) = \eta^{\mu \nu} + h^{\mu \nu}(x) $$ say, you will find that your quantum of gravitation is this tensor $h^{\mu \nu}(x)$. Then clearly it has two free indices, and is what we call a 'spin-2 particle'. The maths to do the linearisation and prove that it transforms as a spin-2 particle would under ...


0

There exists semi-classical computations of the orbit of an electron in Schwarzschild spacetime, such as "The Gravitational Analogue to the Hydrogen Atom" by Koch, Kober and Bleicher. The electron has a non-vanishing wavefunction at the singularity, but on the other hand, the Hamiltonian isn't hermitian. This corresponds to the case of the electron hitting ...


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Your suspicions are correct: It is wrong! At least as it is written presently. Let us start form the embedding of manifold $$\imath_t : F_t \ni p \mapsto p \in M\:.$$ It induces and embedding of corresponding tangent bundles: $$T\imath_t : TF_t \ni (p,v) \mapsto (p, d\imath_t (v)) \in TM$$ The latter can only preserve the vectors tangent to $F_t$ seen ...


-3

Black holes have light not escape them Beacuse. A black hole has so much mass and gravity can't escape it, not even light! The fastest thing in the universe!


2

I'm guessing that when you say: What is now the spatial distance between the two particles? You mean the proper distance. The coordinate distance is of course just $L$. The proper distance is the distance you would measure if you sat at radius $R+L$ and let out a tape measure until it reached radius $R$. To calculate the proper distance start with the ...


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We can evaluate it from the observer's rest frame. Then $v^\mu = (1, 0, 0, 0)$ and $u_\mu = (\gamma , -\gamma\mathbf v)$ where $\gamma = \frac{1}{\sqrt{1-v^2}}$ and $\mathbf v$ is the relative velocity. (We put $c = 1$.) Then $$T_{\mu\nu}v^\mu v^\nu = \rho_0 \gamma^2.$$ However, $\rho_0$ is the density in the rest frame of the fluid. In the observer's rest ...


4

I henceforth assume $c=1$. Consider a 3-volume $\Delta \Sigma_0$ at rest with the dust. For the rest observer (this is its definition) $u^\mu$ has only (unit) temporal component. The energy (i.e the mass) associated with that portion of system is $\Delta \Sigma_0 \rho_0$. The 4-momentum of that portion is therefore $\Delta p^\mu := \Delta \Sigma_0 \rho_0 ...


0

Let's say you're the one accelerating towards the other person who is at rest, so after meeting him you're clocks wouldn't agree because of time dilation, right? Let's make the thought experiment more precise. In some inertial frame of reference, a moving clock is located at $x = 0$ when both the coordinate time $t$ and the moving clock time $\tau$ ...


0

All inertial reference frames are equivalent. This is the most basic assumption of Special Relativity as well as Newtonian Mechanics. This means that if you are in an inertial reference frame, say, a car moving with constant velocity, you can never tell if the car is moving or not (unless you look out of the window of course). This is not true for a ...


0

Time dilation is linked to motion. Be it from acceleration, or velocity, or both. It is because the speed of light is invariant for all observers. If this speed is the same, then what changes is 'time.' As for the who experiences time dilation, the answer is both of you. You both feel time dilation with respect to each other. Time dilation is not 'whoah ...


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I want to add my personal understanding of the concept of reference frame. In the articles: Marmo, G., Preziosi, B. (2006). The Structure Of Space-Time: Relativity Groups. International Journal of Geometric Methods in Modern Physics, 03(03), 591-603. Marmo, G., Preziosi, B. (2005) Objective existence and relativity groups. Symmetries in Science XI, ...


0

The difference between unaccelerated (which includes stationary) and accelerated state is that in case of the latter you can actually feel (and measure) the force causing acceleration. Example: Without looking out the window, you cannot tell whether the train you are in is currently moving or not (relative to Earth) unless there is some acceleration ...


0

The question is somewhat ill-defined. This answer is based on what I gather the OP is asking. No. Generically, the covariant derivative of a tensor is zero only if the entire tensor is zero. Note how the covariant derivative is defined (taking example of a 1-form) $$ (\nabla_\mu A)_\nu = \partial_\mu A_\nu - \Gamma^\lambda_{\mu\nu} A_\lambda $$ Now, ...


1

You might find the following paper useful: http://arxiv.org/abs/astro-ph/0310808v2 In the figure 2 on page 7 you see the plots of different $v(z)$ relations, among them the classical and the special relativistic ones that you have used in your calculations. You can see that the classical $v(z)$ relation intersects the $v=1c$ line for some redshift $z$, this ...


1

Re your edited question, this is just simple spherical geometry. If the initial separation is $d$ then the separation at time $t$ is $d \cos(vt/r)$, where $r$ is the radius of the sphere, $v$ is the vehicle speed and $t$ is time. The diagram shows a cross section through the poles. The vehicle is driving north at a velocity v, so the distance it drives in ...


0

The situation you are describing is very similar to (but not exactly like) a Schwarzschild-de Sitter universe. That is a spacetime that is flat, infinite in size, expanding with a cosmological constant, and contains a massive body (such as a black hole). The metric for such a spacetime is: ...


0

Anywhere there is energy there is time dilation. But you have used a linear approximation - which may hide the super - tiny effect of time change as a wave runs though a region of space. In other words, if there was a beam of gravity waves, and one person was in the waves, the other not, the person who experienced the waves would have a small difference ...


0

The paper suggested by Trimok seems to answer your question. The paper gives an entropy for the observable universe of: \begin{equation} S_{obs U}= 3.1×10^{104} k \approx 10^{104}bits \end{equation} where $k$ is the Boltzmann constant and $S_{obs}$ is the entropy. However I would like to answer your two questions with a back of the envelope calculation. ...


0

A model i used to explain it is this. It is particularly true of non-euclidean geometry. All space is curved. Curvature is a measure of circumference per angle. Straight lines divide the circumference. A large mass would cause space near it to become more negatively curved, which would create more circumference in the direction towards it. Gravity can ...


1

May I suggest that your premise that "Space isn't moving so as to push or rotate matter" is the source of your problem in appreciating the General Relativity explanation of gravity. The GR perspective is that the very fabric of space is accelerating from outers space towards the centre of the earth. A freely falling object feels no force. It has not moved ...


1

They are not the same thing. Suppose you are standing still and your friend is moving at 5 meters per second on a train. Let's say your frame is the $K'$ frame and your friend's frame is the $K$ frame. Now you throw a ball in the direction of the trains motion at eight meters per second. Then the speed you see is $\mathbf{v}'=8\mathrm{m/s} \hat{x}$, where ...


0

This is more a comment than an answer. The condition for the conjecture to hold also require that the connection is torsion free. If you have a metric connection, but is not torsion free, you can build in an holonomic frame the torsion tensor: \begin{equation} T^{k}_{ij}=\Gamma^{k}_{ij}-\Gamma^{k}_{ji} \end{equation}


0

Einstein's equivalence principle would suggest that in this case not the space, but rather Earth itself is expanding. That would be consistent with Einstein's words about the reversal of the vector of the gravitational acceleration (the famous lift thought experiment). Obviously, in order to keep such theory coherent and things in proportion, one would need ...


0

As a guideline we may look to Einstein's maxim that All our well-substantiated space-time propositions amount to the determination of space-time coincidences [such as] encounters between two or more recognizable material points. (Instead of "recognizable material points" other authors also use equivalent terminology such as "substantial points", ...


1

Note that adding boundary terms to the action does not change the equations of motion! It merely changes the boundary conditions which you have to impose in order for the variational principle to yield the equations of motion. That said, the role of the Gibbons-Hawking-York boundary term is not to cancel the all the surface integrals you obtain from varying ...


1

Your question cannot be answered because ultimately it depends on experiment. In general relativity the gravitational constant, $G$, is assumed constant and the geometry of spacetime is derived on this basis. At the moment observation suggests the metric we obtain using GR is a good description of the universe (provided you believe in dark energy), so the ...


2

The answer is yes, and in fact I've described how this works in my answer to another question of yours: If you shoot a light beam behind the event horizon of a black hole, what happens to the light?. I won't repeat the working from that question here, but it might be worth a comment on exactly how the idea works. When you solve the equations of GR you get ...


0

The time dilation is not caused by moving with respect to the space. In special relativity the time dilation is measured only by observers that move with respect to each other and it is symmetrical (both of them measure the clock of the other one slow down). This symmetry shows that they will never agree on who is actually "at rest" in space and who is ...


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This is more of an extended comment than an answer, since I haven't had time to look up all the papers referenced in Felber's article. Hopefully this discussion will be a useful clarification if nothing else. It should be well known to any GR enthusiast (whether a formal student or not) that as viewed by a distant observer it takes an infinite time for any ...


2

One of the tricky things with general relativity is that different observers may use different coordinate systems and measure very different things. The exterior geometry of any static spherically symmetric object is described by the Schwarzschild metric: $$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{r_s}{r}\right)} + r^2 ...


0

Without getting complicated, Einstein predicted and proved that electro-magnetic wave/photons describe a "curvature" of space-time known as a "geodesic relative to a body of mass. Nothing escapes geodesics that we know so far.


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All mass and light waves(photons) experience an accelerating field and describe what Einstein called "geodesics". Note that the term "force" is not used. See "The principal of equivalence"--Einstein's great thought game break through. No force necessary. Absolutely no "pulling."



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