New answers tagged

0

I will just sketch this out. The Killing equation $\xi_{a;b} + \xi_{b;a} = 0$ tells you that $\nabla^i\xi^k$ is antisymmetric in indices. I am going to write this as the commutator between $U$ the vector along which we differentiate $\xi$ with $\nabla_U$ so in compact notation we have $$ Q = -\frac{1}{\kappa}\int[U, \xi]d\sigma $$ I now use Stokes' rule to ...


0

Your first equation looks a bit like GR with a dilaton. IIRC, the analogous equations in supergravity will naturally have spinors coupling to the dilaton.


0

Detecting the emitted photons from a charge accelerating in Earth's gravity looks like a hopeless task, given that the radiated power is tiny (see formula 9 in Anna's answer). But we can try to exploit the fact that the number of emitted photons always diverges no matter how small the acceleration. While very low energy photons (so-called "soft photons") ...


2

Let me answer this question systematically. 1)The first is how to think of causality and locality. Locality (sometimes people use locality to talk about microcausality but that's not very important) is the statement that two events cannot communicate with each other if they are separated by spacelike distances. So what does this mean? Suppose you have a ...


3

The true meaning of all the trouble is that coordinates are just tools for calculations and don't have any intrinsic meaning in GR. You should never interpret the coordinates alone. These can be very misleading and in particular become singular in some regions of spacetime (because coordinates are local). And never trust the labels of coordinates: Fact that ...


0

In special relativity the transition from one frame to another is given by the Lorentz boosts. This is not quite the same as an acceleration, but a transformation that relates observations on one frame with another. We can think of an acceleration as being a succession of infinitesimal Lorentz boosts that map one frame to another. The infinitesimal distance ...


1

There is some misunderstanding here. whether a freely falling charge radiate photons, how strongly and relative to which frame of reference it does or does not radiate if you mean a charge in free fall. In this calculation:, from the conclusion It is found that the "naive" conclusion from the principle of equivalence - that a freely falling ...


3

The radiation emitted by an accelerated charge depends on the boundary conditions on the fields at infinity. When one takes this into account properly, then accelerated observers will agree with inertial observers about the emitted radiation (after trivial transforms are applied). Any treatment which purports to show that in the accelerated observer's frame ...


2

Assuming that Classical Electrodynamics (Maxwell's Equations) holds, the answer is that the inertial observer would see the radiation while the non-inertial observer would NOT. The question you are asking is basically the following paradox: https://en.wikipedia.org/wiki/Paradox_of_a_charge_in_a_gravitational_field This paradox has been analyzed and ...


0

All motion is relative because - motion is phenomena of change of position of an object. Now if you think about it - change of position has always to be relative to something, otherwise, how do you know that the position changed? Thus all detected motion is relative. There can be absolute motion, but that can not be detected. For example, suppose you are ...


2

The conformal transformation $g'_{\mu\nu} = e^{-2\sigma}g_{\mu\nu}$, $sigma = sigma(x)$ leads to the transformation of the Ricci scalar $$ R' = e^{2\sigma}R - 12e^{2\sigma}(2\sigma_{,\mu}^{,\nu} - 2\sigma_{,\mu}\sigma^{,\mu} $$ Since $\phi' = e^{\sigma}$ then $$ \frac{1}{12}\phi'^2R' = \frac{1}{12}\phi^2 R - \phi^2(2\square\sigma - ...


3

First of all we need the equation: \begin{equation}\require{Amsmath} 2\nabla_{[a} \nabla_{b]} K_{cd} = R_{abc}{}^e K_{ed} + R_{abd}{}^e K_{ce} = 2 R_{ab(c} K_{d)e}\tag{1}.\label{eq:KT} \end{equation} We have: $$R_{d(ba}{}^e K_{c)e} = \frac{1}{3}(R_{db(a}{}^e K_{c)e} + R_{da(b}{}^e K_{c)e} + R_{dc(b}{}^e K_{a)e}) = \frac{1}{3} (\nabla_{[d} \nabla_{b]} K_{ac} ...


7

Thanks to the hint given by knzhou I figured out that if one wants to give the particle a proper initial velocity of $v_0$, the initial velocity in terms of Schwarzschild coordinates $v_i$ would then be $$\dot{\theta}(0)\cdot r(0) = \frac{{v\perp}_0}{ \color{green}{\sqrt{ 1-v_0^2/c^2}}\cdot \color{blue}{\sqrt{1-r_s/r_0}}}$$ for the transversal component, ...


17

There seem to be several confusions here. Massive and massless particles behave qualitatively differently, even if the massive particle is traveling very fast. The minimum radius for a stable orbit for a massive particle is $3 r_s$. Circular orbits above this radius are all stable. Massless particles only have circular orbits at the photon sphere, $(3/2) ...


3

Indeed, $f$ is a symmetric form, since $\omega$ and $\omega '$ are Grassmann-even: $$(\text dx \wedge \text d y)\wedge (\text d z \wedge \text d t)=(\text d z \wedge \text d t)\wedge(\text dx \wedge \text d y)$$etc.. Now, to calculate the signature, you should find a basis which diagonalizes $\omega$, the dimension of the space is $6$. A basis is given ...


2

Yes the Schwarzschild metric describes the spacetime geometry around the Earth, and I describe how to use the geodesic equation to describe objects falling in Earth's gravity in How does "curved space" explain gravitational attraction?. An example of how the Schwarzschild metric describes the Earth's gravitational field is the time dilation of GPS ...


0

Easy way Let me first state the straight-forward way to do this computation. $$ \langle \nabla_a \nabla_b V, \partial_c\rangle = \partial_a \langle \nabla_b V, \partial_c \rangle - \langle \nabla_aV, \nabla_a \partial_c\rangle = \partial_a (\nabla_bV)_c - (\nabla_bV)_d \Gamma_{ac}^d $$ First equality follows from compatibility, second equality uses ...


2

I will answer this with a simple example. Let us consider the metric for weak gravity, $$ ds^2 = \left(1 - \frac{2GM}{rc^2}\right)c^2dt^2 - dr^2 -r^2d\Omega^2. $$ The $g_{tt}$ metric element is largest by a factor of $c^2$ and we have $$ \Gamma^r_{tt} = \frac{1}{2}g^{rr}\partial_r g_{tt} = \frac{GM}{r^2}. $$ Now let us work with the geodesic equation that is ...


4

As stated in the comment by Peter Diehr, the question is in principle no different whether you ask it for electromagnetic, gravitational or any other kind of wave. The wave's entropy is simply the conditional Shannon entropy of the specification needed to define the wave's full state given knowledge of its macroscopically measured variables. A theoretical ...


0

A spacetime is a manifold with two different, and technically independant, structures : a metric, which describes the distances between two points, and a connection, which describes how to transport a vector "straight". In general relativity, there is a formula to link the two, but at their core, the two notions are independant. A free particle is one that ...


0

Since I'm in a car right now I can't double check my answer, but I'll try my best not to lie to you. To begin with I am a little unclear as to what you are asking. In most circumstances (i.e if the region under consideration is geodesically complete, which in physics is the case if there is no singularity present) any two points can be connected by a ...


2

The answer is a definite yes and no. Gravitational waves have entropy in that we can think of them travelling from their source to our detector as a channel that sends units of information in the sense of the Shannon formula. The ringing of our detector is then the reception of that information. The Shannon formula $S=-k\sum_n p_n log(p_n)$ would give a ...


0

The quasinormal mode frequencies are complex-valued numbers. For a given type of field (for example, scalar, vector, gravitational, fermionic, etc) there are a discrete infinity of these frequencies which are of course independent of the metric used to describe the background. If the background possess some symmetry, for example spherical symmetry, then the ...


0

A slightly different and, perhaps, more simple proof follows. $\def\lie{\mathit{£}}$ For $K^a$ a Killing vector, we have (Kostant formula), $$\nabla_a \nabla_b K^c = -R_{bca}{}^d K^d.$$ Using this, we may prove that the covariant and the Lie derivatives commute: $$\lie_K \nabla_a X^b = \nabla_a \lie_K X^b,$$ for arbitrary $X^a$. We have: ...


0

The spatial part of a metric is just a 3-sphere: $$ d\Omega_3^2 = d\chi^2 + \sin^2\chi d\Omega_2^2 \, ,$$ and in fact the metric on a $n$-dimensional sphere can always be written in a resursive way using the metric on an $(n-1)$-dimensional sphere in the same manner: $$ d\Omega_n^2 = d\chi^2 + \sin^2\chi d\Omega_{n-1}^2 \, .$$ In these coordinates the ...


4

The Einstein field equations $$ R_{\mu\nu}~-~\frac{1}{2}Rg_{\mu\nu}~=~8\pi GT_{\mu\nu} $$ for zero stress energy means that the Ricci Curvature $R_{\mu\nu}$ is proportional to the metric with $R_{\mu\nu}~=~\frac{1}{2}Rg_{\mu\nu}$. This is called an Einstein spacetime, and for a constant Ricci scalar $R~=~R_{\mu\nu}g^{\mu\nu}$ this is a spacetime of constant ...


1

I do not buy into the view that spacelike geodesics are not physical entities or, at best, can only be understood in tachyon terms. Mathematically spacelike geodesic paths are well understood in 3+1 spacetime. They can be computed in principle given a metric. Given 2 events in a local (but not infinitesimal) part of manifold they can be joined by a unique ...


0

Let $v^\mu \equiv \frac{d}{dt} (x^\mu \circ \gamma)$, where $x^\mu$ is a chart, and a tangent vector $u^a$, such that $v^a u_a = 0$, $u^a u_a \neq 0$ for certain $t$. Such a vector always exists for $\dim X > 1$. For $\gamma$ a geodesic, we parallel-transport $u^a$, hence $v^a \nabla_a (v^b u_b) = 0$ for every $t$ in the definition of the curve; ...


4

You can easily see this isn't the case by considering the special case of the stress-energy tensor equal to zero i.e. the vacuum solutions. These include the Minkowski metric, which is flat, but also the Schwarzschild and Kerr metrics and of course gravitational waves.


1

Traversable - Overlapping (actually intersecting) region would not be Traversable even if the gravity at some parts of the region may be zero. For exampple, between earth and moon, gravity will be zero at some point. That does not mean something in that region can go out of earth/moon system. As soon as an observer leaves that region, it either falls towards ...


2

Torsion is not frame dragging. Torsion is having an anti-symmetric spacetime connection. As you do parallel transport in general relativity (GR) you drag frames the frames roll as they move. With torsion they would twist. The connection is GR is the Christopher symbols, symmetric in the two bottom indices. The torsion is an anti-symmetric tensor. It will ...


0

If the event horizons overlap you get one big horizon. EM forces can not counteract gravity if the curvature is too large since the force required to counteract gravity becomes infinite at the horizon. You can see this in the equation $$F=\frac{G\cdot M\cdot m}{r^2\cdot\sqrt{1-r_s/r}} $$ which becomes infinite at the horizon $r_s$. Since from the outside ...


1

The answer is that we don't know. Why? Because the theory of gravity which we have and use, GR, has a singularity. Things which should be finite in a physical theory, like the density, become infinite. And theories with a singularity are simply wrong, they need a modification, and this modification is necessary not only at the singularity itself, but already ...


4

The "theory" describing the black hole interior (in the classical approximation) is the same theory that implies the existence of the black holes, namely the general theory of relativity. As the OP correctly said, the singularity at the event horizon is a coordinate singularity – one that is an artifact of a bad choice of coordinates. When a coordinate ...


0

An orbiting observer is a bit problematic because there are no stable orbits for $r \le 3r_s$, so let's instead consider an observer hovering at some distance $r$ from the black hole. In that case as $r \rightarrow r_s$ the blue shift does indeed $\rightarrow\infty$ and the observer would indeed be roasted. But this shouldn't be surprising. The acceleration ...


1

The reason to consider such metrics was not a particular interpretation, but that for this ansatz one could hope to find some exact solutions. It is known as the Kerr-Schild metric, and its role in the process of finding exact solutions has been described by Kerr in Wiltshire, Visser, Scott (eds.), The Kerr Spacetime. One can, of course, try to develop ...


2

Negative energy or mass is not forbidden in Relativity, but gravity is not a force but geometry, so if you have a negative mass it would repell positive mass as well as negative mass, just like positive mass would attract negative and positive mass all together. If you place a positive and a negative mass near each other the positive mass would attract the ...


0

Not sure about the etiquette of this but I think I can now answer my own question. Please post if there is a better answer. The problem is that I misunderstood the meaning of the statement "the geodesic is the path the optimises the proper time". What this means is that given two endpoints, say $x_0^\alpha$ and $x_1^\alpha$ a geodesic is a path ...


1

our aging is directly proportional to metabolism of body and division,growth and death of cells of body and if gravity has some effect on the rate of above aspects astronauts will be definitely younger


0

Actually, only a gravitational wave and not a uniform gravitational field is composed of gravitons. I think that just like en electromagnetic wave, a gravitational wave is a sum of sinusodial waves with gravitons of different energy for different wavelengths in that sum. An isolated blackhole doesn't emit gravitons but a pair of orbiting blackholes does. I ...


4

We can always find such a vector field when $n=2m$ and $M$ is orientable. Proof. Let $\gamma:[0,a]\to M$ be the geodesic in question, with $\gamma(0)=\gamma(a)=p$. Let $P_t:T_{p}M\to T_{\gamma(t)}M$ be the parallel transport along this geodesic. From Picard-Lindelöf, it is clear that $P_a\gamma'(0)=\gamma'(a)=\gamma'(0)$. Thus $P_a$ is an isomorphism of ...


4

Actually, the metric variational definition for the stress-energy tensor (due to Hilbert, as remarked by Qmechanic) is an universal improvement procedure for the canonical stress-energy tensor (and hence not always concides with the latter), in a sense which will be made precise below. Such a procedure is necessary because the canonical stress-energy tensor, ...


2

Well, you cannot take any ol' matter theory in flat Minkowski space and stick in a curved metric tensor $g_{\mu\nu}$ in the matter action as you like, if that's what you're implying. The caveat is that the resulting matter action $S_{\rm m}[\Phi, g]$ should be a general relativistic diffeomorphism-invariant functional. Then the Hilbert stress-energy-momentum ...


2

The first answer has all the results, but I will try to show some calculations, cause I have been writing them since there was no answer. It is known from General Theory of Relativity (GTR) that the closer you are to a massive object - the slower the time goes. On the other hand Special Theory of Relativity (STR) gives us the next statement: the faster you ...


4

This is anwered in Gravity on the International Space Station - General Relativity perspective, where we learn that time dilation in the ISS with respect to Earth equator is 1.00000000028655. So after 17 years for us, the astronauts would come back younger by about 0.15 seconds than if they have stayed on the ground. Note that a full GR treatment is ...


-1

The most Natural way where to base units on is to base them on the Planck length and Planck time wich are universal lengths. So you can set the Planck length and Planck time equal to one. They are not based on human measures or alien measures. Every scientific thinking being in the universe would agree on how big these units are. And because all other units ...


2

The acceleration should be $$a = \frac{G\cdot M}{r^2 \cdot \sqrt{1-r_s/r}}$$ with $r$ as the height above the center of mass and the Schwarzschildradius $$r_s = \frac{2\cdot G\cdot M}{c^2}$$ The force to hold the ball at rest is $$F=m\cdot a$$ As one can see it now takes an infinite force and energy to keep a body at a fixed height when $r=r_s$.


0

A very peculiar fact is that in a compact space THERE IS a preferred inertial system. Indeed even if locally there is no way to single out a preferred inertial system, globally you can do it. Is the topology that tells you that an observer doing a loop around a torus is topologically different from an observer moving around simply connected loops. So for ...


2

The commutator of two vector fields $n^a$ and $X^b$ is $[n,X]^a = n^b \nabla_b X^a - X^b \nabla_b n^a$. Since this vanishes, it follows that $n^b \nabla_b X^a = X^b \nabla_b n^a$, and the second step follows from there. I don't have my copy of Wald in front of me, but I'm 99% sure that the commutator of two vector fields is defined in terms of the ...


6

The metric reads, restoring $c$, $$\mathrm{ds}^2 = -(1+gz/c^2)^2 c^2\mathrm{dt}^2 + \mathrm{dz}^2 + \mathrm{dx}^2\:.$$ The metric can be used to determine the geodesics by means of the associated quadratic Lagrangian $$L = -(1+gz/c^2)^2 c^2\dot{t}^2 + \dot{z}^2 +\dot{x}^2\tag{0}$$ where the dot denotes the derivative with respect to the affine parameter $s$. ...



Top 50 recent answers are included