# Tag Info

## New answers tagged general-relativity

2

Special thanks to Qmechanic and David Bar Moshe for providing their answers for this post. Combining Qmechanic's and David Bar Moshe's insights, I finally figured out the confusing element in the problem. The covariant derivative can be written in terms of an eigendecomposition of the partial derivative, the Christoffel symbols can be written in terms of ...

3

Yes, your second guess is more or less correct. In GR, perturbing the metric is the usual way of doing perturbation theory. One writes for the true metric $g_{\mu\nu}$ an expansion of the form $$g_{\mu\nu} = g^{(0)}_{\mu\nu}+h_{{\mu\nu}}+O(h^2),$$ where $g^{(0)}_{\mu\nu}$ is known and usually called background metric. One then substitutes this into the ...

0

'Spooky' action at a distance bothered even Newton, and his verbal explanation of gravity was just as poor or even worse than anyone else's. But his math worked well enough to predict the elliptical orbits of the outer planets. After Einstein thought he had worked out general relativity, Eddington led an expedition to the Crimea to take data on the ...

10

When the apple was detatched from the branch of the tree, it was stationary, so it did not have to follow any geodesic curve. Even when at rest in space, the apple still advances in space-time. Here is a visualization of the falling apple in distorted space-time: http://www.youtube.com/watch?v=DdC0QN6f3G4

0

The Einstein-Hilbert action of general relativity, to make the variational principle fully rigorous, must be supplemented by a boundary term, $$S = \frac{1}{8\pi G} \int_{\partial M} d^3 x \sqrt{-h} \, K$$ where $h_{\mu \nu}$ is the first fundamental form of a submanifold which we take to be $\partial M$, the boundary of the spacetime manifold. The cuvature ...

1

Not everything needs to follow geodesic Spacetime curvature available to it. With external force, you can prevent a particle from following Spacetime curvature. Only "freely" falling particles follow Spacetime curvature available to them. So, when you see a stationary object not following Spacetime curvature, it's because an external force is preventing it ...

3

As to the first paragraph, gravity shows up as geodesic deviation; initially parallel geodesics do not remain parallel. Since, for a freely falling particle, the proper acceleration (the reading of an accelerometer attached to the particle) is zero, it is correct to say that a particle whose worldline is a geodesic has no proper acceleration. But it is not ...

40

To really understand this you should study the differential geometry of geodesics in curved spacetimes. I'll try to provide a simplified explanation. Even objects "at rest" (in a given reference frame) are actually moving through spacetime, because spacetime is not just space, but also time: apple is "getting older" - moving through time. The "velocity" ...

1

You have the right basic idea. But it gets simpler to visualize if you just drop the ball, or throw it vertically. Then there is just one spatial dimension to consider, and you can directly compare the paths in space and in space-time, like shown here: http://www.youtube.com/watch?v=DdC0QN6f3G4

1

Both neutrinos and anti-neutrinos are affected by gravity (same magnitudes and direction). The 1987 Supernova event was the first instance of neutrino and anti-neutrino detection of a source outside of our solar system. Not only were they detected, but the neutrino event recorded occurred a short interval of time AFTER the visible light and gamma ray burst ...

3

I) The vielbein $e^a{}_{\mu}$ in the Cartan formalism is an intertwiner $$\tag{1} g_{\mu\nu}~=~e^a{}_{\mu} ~\eta_{ab} ~e^b{}_{\nu}$$ between the curved (pseudo) Riemannian metric $g_{\mu\nu}$ and the corresponding flat metric $\eta_{ab}$. Here $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while $a,b,c, \ldots,$ are so-called flat indices. ...

2

There are already many good answers. Tensors and their decomposition in terms of simple tensors are important to virtually any topic of physics. I) For starters, functional analysis and operator theory is used in almost all branches of physics. And rank-one operators $$\tag{1} T~\in~{\cal L}(V;W)~\cong~ V^{*}\otimes W$$ are important objects here. E.g. ...

0

This thread came up some years ago, I saw it referenced and would like to posit an idea. My belief, backed up with math here, is that this cannot be lens. Lensing always results in rings, sometimes faint, sometimes not. There would be some evidence of a ring in the Hubble image, which is quite deep and quite fully resolved. There is none. So what could this ...

9

I can easily construct an example of smooth tensor field over a manifold whose "rank" changes depending on the point. My idea relies upon the following elementary proposition. I stress that the notion of "rank" used here is that introduced within the original question and not the standard one. Proposition. Consider a $n$-dimensional real vector space $V$ ...

0

Inertial Frames of reference are fractals. You can imagine each Frame of reference as a box within a box within a box etc You can zoom in or out. The observer in each inertial "box" see the behavior of matter according to the laws of "classical"Mechanics == That is clocks run normally, mass is constant as is length. Example a car traveling at constant ...

0

Is this an accurate description, or is it complete nonsense? "If you throw a ball, it will move along a parabola. Initially its vertical speed will be high, then it will slow down, and then speed up again as it approaches the ground. More accurately: The ground (as well as anything rigidly "connected to the ground") would move along a parabola wrt. ...

0

Think of it this way: Imagine you are in a rocket ship that has transparent portholes (very small ones) diametrically opposed on the body of the ship, and the ship is accelerating at g. At some instant during flight, a laser beam pulse enters one of the portholes, and as soon as that event occurs, it is detected and another laser apparatus emits a laser ...

2

I'll raise some issues. Firstly you say In reality... Do you contrast something against something else here? It implies you say there the preceding sentence If you throw a ball, it will move along a parabola. Initially its vertical speed will be high, then it will slow down, and then speed up again... wasn't right, but that sentence seems pretty ...

5

To me, the best way of describing it is as follows (semi-stolen from Carroll's Spacetime & Geometry): we all know from high school physics that, when no force acts on an object, it should not change its velocity $v$. In other words, the velocity vector tangent to the objects trajectory through space and time (i.e. world line) undergoes parallel ...

6

Yes, that's a fair description of what happens though of course from the ball's perspective it isn't moving - the rest of the universe is moving around it. However statements like this, while true, give little feel for what's going on. Actually it's extraordinarily difficult to get an intuitive feel for the way spacetime curvature works (or at least I find ...

2

Similar questions have cropped up on this site many times, and the debate surrounding them is usually fractious because people misunderstand each other's use of words like exist. One of the lessons of General Relativity is that any observer has to choose a locally convenient coordinate system that may not be globally convenient. We on Earth (quite sensibly) ...

0

As I understand Hawking, the point that he is trying to make is that Newton's theory of gravity is actually incomplete and needs to be improved. So Hawking finishes on page 36, saying that Einstein tried to find that better theory of gravity for 6 years and finally in 1915 published his general theory of relativity. In this theory it takes now also 8 minuets ...

3

I don't see your problem. We are dealing in hypothetical situations that lead to paradoxes and inconsistencies, so there is no problem with postulating what would happen if...?, even if the "if..." is impossible. He could have as easily said what would happen if the sun moved suddenly, we would see it move after 8 minutes, but gravitationally feel it move ...

0

As stated before our planets magnetic field is too weak to play a significant role. Centrifugal force just describes conservation of momentum, that means that a body (e.g. earth) wants to go in a straight line forever, unless a force (e.g. gravity from the sun) acts on it. The force that forces a body on a curved path is usually called centripetal force ...

1

To observe "nothing special" near a black hole, you would have to be staring very intently away from it. If you could see it as more than a "missing" point source, you could see the entire universe wrapped around it in perfect Einstein rings. The closer you approached the black hole, the more it would intrude into your remaining field of vision. Around ...

1

It's a matter of what you mean by "see". Even for a distant observer, it will take a small amount of time for the gravitational redshift effect to become essentially infinite. If your collapsing gas star redshifts to the point where it won't emit a single photon in the age of the universe, it may not have yet technically "redshifted to zero", but it has ...

4

Equation (13) expresses the metric on an embedded hypersurface given by the relations $y^k = y^k(x^a)$. However, the equation for the inverse metric (4-th equation) is in general not correct. Take for example a hypersurface defined by: $y^1 = x^1$, $y^2 = x^2$, $y^3 = x^2$. In our case, the partial derivative of $x^2$ with respect to $y^2$ or $y^3$ ...

5

Let there be given a manifold $(M,\nabla)$ equipped with a (not necessarily torsionfree) tangent bundle connection $\nabla$. I got the (possibly faulty) impression from reading the first lines in OP's question formulation (v18) that OP is asking: Is it possible that the local coordinate expression for the covariant derivative of a co-vector/one-form ...

3

The idea that the pendulum would drop instantly isn't even true of short, Earth-bound pendula: c.f. various Internet videos about dropping slinkies (toy springs). The reason why slinkies drop in this way is essentially the same reason why an idealised pendulum (strong enough to hold itself together, albeit maybe not as stretchy as a slinky) would not ...

1

In most cases, it doesn't really make sense to talk about a lowered effective mass caused by sitting in a gravitational potential well, since the equivalence principle says that locally the spacetime looks flat, and hence it looks like the gravitational field vanishes. However, in certain special cases, there is a sensible notion of energy that is ...

1

According to the following paper and commentaries, general relativity can be derived from a standard model matter field equation combined with some other consistency criteria. If new matter such as dark matter is found then the given procedure could give a new theory of gravity or it might just lead back to general relativity. How quantizable matter ...

5

I don't think I can rigorously prove that simulation engines don't need to worry about the (possibly? I don't know if there's a reliable measurement) finite speed of gravity, but I can offer some lines of thought that point in that direction. I'll start with your question 3. Suppose that gravity does have finite speed equal to $c$. Your question seems to be ...

7

It's tempting to think of gravity as some kind of interaction between the two bodies involved - maybe some form of signal (gravity wave?) sent between the two bodies. If this were the case then you would indeed have to allow for a propagation delay as the signals were sent between the two bodies. However this is not how gravity works. A massive object ...

2

Magnetic field doesn't play a part. In the case of the earth, the magnetic field is incredibly weak and can't attract much. Gravity and centrifugal forces have to do with the mass of a body, not the magnetic pole strength. Magnetic fields are different beasts entirely which have to do with the velocity of charged particles or their magnetic pole strength.

7

Special relativity is used in the SM formulation. It is kinematics, so somehow more basic than interactions between bodies. A QFT derivation of General Relativity has been the Holy Grail of the field for many years. In the early times, Feynman, Dirac, and the others tackled this problem, but after decades of failures it was more or less considered ...

0

There is not a universal rest frame. There is, however, a galactic rest frame. Because you can look up at the stars, falsely assume that they do not change, and count your rotations that way. However, that method is only as reliable as the premise that the stars don't move, which they do slightly. The Hafele-Keating experiment used a variant of this, ...

2

There is a really nice derivation of this identity using differential forms, and it completely avoids all the messiness of the Christoffel symbols. The nice thing about differential forms is that the exterior derivative can be computed using any derivative operator, so it allows us to compare the expressions we get using the covariant derivative to the ...

0

In Special Relativity CTCs can't exist (or at least I don't think so) but General Relativity has solutions that include CTCs. The best known is probably Gödel's solution for a rotating universe. The Alcubierre drive could also be used to construct CTCs, as could any FTL mechanism. Also see the Tipler cylinder, and probably many other examples I can't ...

1

It would be more correct to say that distant galaxies appear than to say they disappear. Based upon the accepted big bang theory, there are galaxies that formed early in the universe from which light has not yet reached us, but that will reach us in the future. On the other hand, accelerating expansion of the universe could cause light emitted after a ...

2

This is based on the observation that, given some vector $V^\mu$, $$\nabla_\mu V^\mu=\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}V^\mu)$$ We can show explicitly that this is true: $$\nabla_\mu V^\mu=\partial_\mu V^\mu +\Gamma^\mu_{\mu\lambda}V^\lambda$$ Let's examine the last term: $$\Gamma^\mu_{\mu\lambda}=\frac{1}{2}g^{\mu\rho}(\partial_\mu ... 2 Given a pseudo-Riemannian manifold (M,g), the Laplace-Beltrami operator acts on scalar functions. The formula for the Laplace-Beltrami operator follows from the formula$$\Gamma^{\nu}_{\mu\nu}=\partial_{\mu}\ln\sqrt{|g|}$$for the Levi-Civita connection. 1 As I am not allowed to comment for lack of reputation and cannot find a way to message I would like to point you to a reasonably recent source on the Kerr metric. I am by no means an expert but from what I have read and from what I understand the "Lines" of space time do twist and become unstable at the Cauchy Horizon. From what I gather the rotation isn't ... 4 What you said is only true if the hypersurface is space-like or time-like. If a non-null hypersurface is defined by f(x) =  constant, then the normal to the hypersurface is given by$$ n_\alpha \propto \partial_\alpha f $$The fact that the hypersurface is non-null implies$$ g^{\alpha\beta} \partial_\alpha f \partial_\beta f = \varepsilon\neq 0 $$... 2 Finding an action that gives you dust when varied is actually kind of tricky. Part of the reason it is hard is that for scalar matter, the action is usually proportional to the pressure,and the pressure vanishes for dust. I'm not sure where you found your action for dust, but I think you didn't make any calculational errors, it just isn't the right action. ... 1 Usually the Newtonian limit is described as taking v << c but a much better way to express it is saying that the kinetic energy is much less than the rest energy$$ \frac{1}{2}m v^2 << m c^2 $$of course this runs into trouble when we talk about photons since we don't have a well defined concept of velocity, in the Newtonian sense. This is ... 1 For a particle of fixed mass m moving in a fixed gravitational potential \phi(\vec{r}) the motion is independent of the mass of the particle. The equations are$$ \vec{F}=-m\nabla\phi $$and$$ \vec{F} = \frac{d\vec{p}}{dt} = m \frac{d\vec{v}}{dt} $$It's clear that the m's cancel when combining these equations. So from this point of view it doesn't ... 1 It's just a rescaling of the time coordinate. Define t = f(\eta) and {\bar a}(\eta) = a(f(\eta)). Then,$$ds^{2} = -{\dot f}^{2}d\eta^{2} + {\bar a}^{2}d^{3}{x}$$Thus, if f(s) satisfies \frac{df}{d\eta} = a(f(\eta))\rightarrow \eta = \int \frac{dt}{a}, then you have transformed into conformal coordinates, and there is no special meaning for the ... 2 When you say "without altering the actual momentum of it" is that really true?$$ E^2 = p^2c^2 + m^2c^4  so for a photon $E = pc$, since rest mass is zero. Now according to your first "traditional" calculation of m, we would have $E = pc = m_1c^2$, and therefore $p=m_1c$, where $m_1$ is mass according to the first "traditional" calculation. For your ...

0

Because you have $D$ coordinates, so you get $D$ choices for choosing coordinates -- i.e., I choose eastern standard time, and cylindrical coordinates centered on the sun, measured in kilometers and radians. This would make the solar system metric go from having ten independent components to six.

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