# Tag Info

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Einstein's equations can be loosely summarized as the main relation between matter and the geometry of spacetime. I will try to give a qualitative description what every term in the equation signifies. I will, however, have to warn potential readers that this will not be a short answer. Furthermore, I will refrain from trying to derive the equations in ...

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In relativity there is no standard-clock that tells you which time is "right". That's the point about relativity. There is no need for a absolute reference to compare with. Everything is just the way you observe it (that is, relative to you). Things may slightly differ from observer to observer but the qualitative behaviour stays the same just as classical ...

6

In general relativity, rather than a two objects exerting a gravitational force on each other, the two objects are both part of the stress-energy tensor. This tensor determines the shape of spacetime (via the spacetime metric), and the spacetime metric determines what the geodesics are (roughly speaking, the metric determines how an object will move when no ...

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N-body simulations in full general relativity are difficult because gravity is a field theory and because it is non-linear. Let's deal with the field theory part first. In Newtonian mechanics gravity is static. The field itself has no energy or momentum, no degrees of freedom at all. It is simply an instantaneous force law between all matter. Remove the ...

3

There is a relatively fast approach to computing the Riemann tensor, Ricci tensor and Ricci scalar given a metric tensor known as the Cartan method or method of moving frames. Given a line element, $$ds^2 = g_{\mu\nu}dx^\mu dx^\nu$$ you pick an orthonormal basis $e^a = e^a_\mu dx^\mu$ such that $ds^2 = \eta_{ab}e^a e^b$. The first Cartan structure ...

3

Einstein's equation relates the matter content (right side of the equation) to the geometry (the left side) of the system. It can be summed up with "mass creates geometry, and geometry acts like mass". For more detail, let's consider what a tensor is. A two-index tensor (which is what we have in Einstein's equation), can be thought of as a map which takes ...

3

The singularities are in a sense mere artefacts of your coordinate system, but do describe important surfaces. The reason your metric was ill defined up on those surfaces is that you picked a metric that was nice when you are far far from the black hole (all those $1/r$ terms get small and it looks like the SR metric for a spherical coordinate system). ...

2

You are correct that any theory can be written in diffeomorphism invariant language. Diffeomorphism invariance does not have anything to do with GR, except that it is hard (impossible?) to write down the theory in a gauge-fixed way. What sets GR apart from other theories, aside from the fact that the metric is dynamical, is that it does not allow for ...

2

$g^{\mu\nu}$ is the inverse of the metric $g_{\mu\nu}$; and the expression is for a general metric, not for the Minkowski metric. In 4 dimensions, if $g = \det(g_{\mu\nu})$ $$\det(\sqrt{-g} g^{\mu\nu}) = (\sqrt{-g})^4\det(g^{\mu\nu}) = \frac{g^2}{\det(g_{\mu\nu})} = g$$

2

Let me clear up a few misconceptions. The edge of our observable universe would contain information from the beginning of the universe, since it is a particle horizon. However, the edge of the observable universe is not currently visible to us. What we can currently see only goes as far back as the recombination era, when electrons first joined with nuclei ...

2

The basic point is that you can't shine light away from the centre of a black hole once you are inside the event horizon. Far away from the black hole, light cones are oriented so that propogating light from an event can travel anywhere inside a cone bounded by two lines at 45 degrees in a standard (Schwarzschild coordinates) space-time diagram. Nearing ...

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1) The spacelike hypersurface has three spacelike directions tangent to it. Any vector that is normal to all three spacelike directions in the eneveloping space is necessarily timelike. Equivalently, the spacelike surfaces can be thought to be labeled by a function $\tau$ which gives the "time coordinate"'s value on those surfaces. the normal to the ...

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The short answer is that calculating the Riemann Tensor is a grind. It will take a while, no matter what way you do it. Presumably you're doing the Schwarzschild metric in the standard (Schwarzschild) coordinates, so you're aided by the fact that the metric tensor is diagonal. This means that $R^\alpha_{\beta \gamma \delta} = g^{\alpha \alpha}R_{\alpha ... 1 I think Wald gets it right and Gaul and Rovelli get it wrong. Active and passive have very different feels to them, since active is moving points and passive is just re-choosing coordinates. But the whole point of coordinates is that they describe the points very well, and so mapping the points is completely equivalent to unmapping the coordinates. There ... 1 What local forces cause the damage? Tidal forces. Tidal forces can only be neglected in extremely small regions, and you have an extended body. More details follow. You suppose an extended body, bound together, interacting gravitationally with a black hole. With the center of mass staying outside the photon sphere, moving on a more than barely unbounded ... 1 The picture you provide is a difficult problem. However there is a simple example consistent with the words you wrote. You can make a thin spherical shell and place an electric charge Q on it. Outside the shell the metric will be the Reissner–Nordström metric:$$... 1 There is no reference object that transcends all inertial frames of reference. Everything in this universe has an inertial frame of reference, and none of them are privileged. If there were any object that existed independently of the relativistic effects of acceleration/gravity or of observer movement, then theoretically it could provide a reference to ... 1 Are there another views on what dark matter/energy is? Yes, but you tend not to hear about them because there's a presumption that dark matter consists of particles. This is despite Einstein saying "the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy" in The Foundation of the General Theory of ... 1 No, or at least not under normal circumstances. Calculating the gravitational force between light beams turns out to be rather complicated, so let's use an analogous but simpler system. Instead of a spherically symmetric pulse of light consider an explosion that throws out a spherically symmetric cloud of light particles. At any time$t\$ the cloud of ...

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