# Tag Info

## Hot answers tagged general-relativity

56

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" ...

21

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

13

They rotate because they are produced by matter that has net angular momentum, and angular momentum is conserved in axially symmetric space-time. So, there's nothing unusual making them rotate that's different from any other physics. However, you are absolutely right to object that rotation of an infinitesimally small point wouldn't make much sense. In ...

10

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 $V$ ...

6

There is a conserved quantity for geodesics which comes from the fact that the metric $g_{ab}$ is (trivially) a Killing tensor, i.e. $$\nabla_{(c}g_{ab)} = 0.$$ Any tensor $\xi_{ab}$ that satisfies $\nabla_{(c}\xi_{ab)}=0$ gives rise to the conserved quantity $\epsilon = u^a u^b\xi_{ab}$, which is preserved along geodesics for which $u^a$ is the tangent ...

6

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 ...

4

Well, relativistic effects are accounted for in all space travel. These effects were apparently minor enough during the Apollo missions to rely on standard Newtonian equations and minor course corrections enroute, but are pronounced enough on longer missions within the solar system to account for several kilometer errors. General relativity is routinely ...

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 ...

3

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

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

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 But note that this doesn't involve any intrinsic space-time ...

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. ...

1

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

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 ...

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