I heard that we must know the Weyl tensor for fully describing the curvature of the 4-dimensional space-time (in space-time with less dimensions it vanishes, so I don't interesting in cases of less dimensions). So I have the question: what is physical (or geometrical) sense of the Weyl tensor and why don't we need only Riemann tensor for describing the curvature? Does it connected with gravitational waves directly?
[Physics] Why do we must know the Weyl tensor for 4-dimensional space-time
differential-geometrygeneral-relativitygravitational-wavesmetric-tensor
Related Solutions
Dear John, let me post the same thing that Marek has said as a standard answer.
Einstein's equations are not equations for the Riemann tensor because the Riemann tensor's components are not independent fields. Instead, Einstein's equations are differential equations for the metric tensor.
In 4 dimensions, the metric tensor has 10 components - a symmetric tensor - and Einstein's equations have 10 components - a symmetric tensor - too. It doesn't matter that the Riemann tensor has 20 components because these 20 functions of space and time are calculated from the 10 component functions of the metric tensor and its (first and second) derivatives.
In fact, the 10-10 counting is oversimplified. Four "differential combinations" of Einstein's equations vanish identically because $\nabla_\mu R^{\mu\nu}=0$ is an identity (that always holds, even if the equations of motion are not satisfied). The same identity holds for the corresponding other tensors that are added to the Ricci tensor in Einstein's equations.
So instead of 10 equations, the Riemann equations are, in some sense, just 6 independent equations. That means that they don't determine the metric completely: they leave 4 functions undetermined and these are exactly the 4 functions that you may choose arbitrarily to specify a diffeomorphism, mapping one solution into another (equivalent) solution.
Up to the coordinate transformations which are always allowed to be made, initial conditions for the metric and its first derivative determine the metric tensor - and therefore the whole Riemann tensor - everywhere in the future. One doesn't need any Newtonian equations as a "mandatory supplement" in general relativity. That doesn't mean that the Newtonian limit is unimportant: of course, it is one of the most important approximate consequences of general relativity.
The direct calculation of the derivatives isn't that hard. But one can also quickly see the values of the Riemann tensor for a sphere – and similar simple shapes – by using the definition of the Riemann tensor via the parallel transport of vectors. $$\delta V^\alpha = R_{\alpha\beta\gamma\delta}V^\beta d\Sigma^{\gamma\delta}$$ Around a point of the sphere $S^d$, the transport around an area given by $d\Sigma^{\gamma\delta}$ for fixed values of the indices (locally orthonormal basis) allows you to see that all of this is happening on an $S^2$ only. The other dimensions are unaffected. That's why you get $R_{\alpha\beta\gamma\delta}$ equal to $(1/a^2)$ times $g_{\alpha\gamma}g_{\beta\delta}-g_{\alpha\delta}g_{\beta\gamma}$. Effectively, the antisymmetrized pair of indices $\alpha\beta$ has to be the same as the pair $\gamma\delta$. I didn't assume anything special about the point; all points on a sphere are equally good by a symmetry. So the Ansatz for the Riemann tensor has to hold everywhere.
Note that it's multiplied by $1/a^2$, the inverse squared radius of the sphere. Many of your formulae omit it; moreover, you are using a confusing symbol $R$ for the radius which looks like the Ricci scalar – a different thing.
In $d=2$, the Riemann tensor only has one independent component and the formula for the Riemann tensor in terms of the metric tensor above actually holds for any surface if $1/a^2$ is replaced by $R/2$. Note that the two-sphere has (Ricci scalar) $R=2/a^2$. Also, the Ricci tensor is $R_{ij}=Rg_{ij}/2$ in $d=2$ so that the vacuum Einstein equations are obeyed identically.
For $d=3$, the Riemann tensor has 3 independent components, just like the Ricci tensor, so the Riemann tensor may be written in terms of the Ricci tensor. That's not true for higher dimensions, either.
Best Answer
The Riemann tensor encapsulates all information about the 4-dimensional space-time. This information can generally divided into two sectors:
Information about the curvature of space-time due to the existence of matter. This is given by the Ricci tensor according to the Einstein equation $$ R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = 8 \pi G T_{\mu\nu} $$
Information about the structure of gravitational waves in the space-time. This is given by the trace-free part of the Riemann tensor, namely the Weyl tensor. Often, we are not quite interested in the exact structure of the space-time, but only if gravitational waves can exist or their structure. In these cases, one studies the Weyl tensor rather than the Ricci tensor. For example, in the setup of quantum gravity, one requires to study the asymptotic structure of spacetime. In these theories, a good understanding of the Weyl tensor is more important.