Riemann and Ricci Tensors?

[Freeman]

I understand the Ricci tensor is the trace of the Riemann tensor, but I don't fully understand the Riemann tensor; it's the tangent tensor of the curved area?

 

What is the ricci tensor, then, exactly in layman's terms? The reason I ask is because I am trying to understand the lefthand side of Einstein's field equation; I understand the stress energy tensor (as explained in the Feynman Lectures in Physics).

 

What exactly does the Einstein tensor ([math]G_{ab} = R_{ab} - \frac{1}{2}g_{ab}R[/math] where [math]R = g^{ab}R_{ab}[/math]) signify in layman's terms?

[□h=-16πT]

The Ricci tensor is a contraction of the Riemann curvature tensor on its first and third indices, i.e. [math]R_{\alpha\beta}=\Sigma_{\nu}R^{\nu}_{\alpha\nu\beta}[/math] and because of the contraction it is symmetric. The Ricci scalar is the only contraction one can form from just the Riemann tensor, the contraction of any of the other indices is simply the Ricci tensor with various sign changes.

 

The Ricci scalar is the simplest scalar one can form from the Riemann tensor. It is a further contraction of the Riemann tensor

 

[math]R=\Sigma_{\alpha, \beta}g^{\alpha\beta}R_{\alpha\beta}=\Sigma_{\nu, \alpha, \mu, \beta}R^{\nu\alpha\mu\beta}R_{\nu\alpha\mu\beta}[/math]

 

The Einstein tensor is a symmetric divergence free tensor that essentially describes the curvature of the manifold under consideration. The divergence free characteristic follows from the Bianchi identies and demonstrates energy-momentum conservation since

 

[math]

G^{\alpha\beta}=kT^{\alpha\beta}[/math]

[math]\nabla _{\beta}G^{\alpha\beta}=k\nabla _{\beta}T^{\alpha\beta}=0[/math]

[math]k\nabla _{\beta}T^{\alpha\beta}=0

[/math]

 

Thus energy-momentum is conserved.

 

The derivation of the field equations originally given by Einstien appeals to such nature of the Einstein tensor as constructed from the Ricci tensor and scalar:

 

It being second order in the metric, as the Newtonian field equation is second order in the potential. The Einstein tensor is second order in the metric as it is constructed from Ricci tensors which in turn are constructed from the Riemann tensor, which consists of terms involving first derivatives and quadratic terms in the Christoffel connection (in GR any way) that is constructed from first order terms in the metric.

 

Symmetric, thus in the equality of G with T both sides are symmetric in their indices.

 

G embodies the curvature, thus formalising Einstein's idea of gravity being the result of inertial motion along a curved space-time.

 

Energy-momentum conservation by virture of the Bianchi identities.

 

The rigorous and formal alternative to this intuitive derivation implements the principle of least action on the Einstein-Hilbert Lagrangian density

 

[math]{\cal L}=R\surd(-det[g_{\alpha\beta}])[/math]

 

This is acreditable to Hilbert, who collaborated with Einstein during the development of GR.

[□h=-16πT]

There are a few ways of defining the Riemann tensor, all of which are equivalent. The first is the term arising in the commutator of covariant derivatives

 

[math]\lbrack \nabla_{\alpha}, \nabla_{\beta} \rbrack V^{\nu}

=R^{\nu}_{\mu\alpha\beta}V^{\mu}[/math] (Summation running over repeated indices)

 

This is for a torsion free Lorentzian connection, where the connection coefficient (Christoffel connection [of the second kind] in GR) is symmetric in its contravariant indices. For manifolds with torsion there is another term subtracted from the end.

 

Another is as follows: parrallel transport a vector about a closed loop and upon it returning to its initial position the difference of the two defines the loss of parrallelism, i.e. curvature.

 

The other can be seen from the equation of geodesic deviation.

 

The gist of it is that it is a quantitative measure of the curvature of a manifold. One can define the Riemann tensor for non-coordinate bases using the Weyl curvature tensor.

[□h=-16πT]

In relativity we use greek indices. You've used latin indices in your expression for the Einstien tensor etc. It's no biggy, I just thought I'd let you know in case you didn't already. Latin indices are usually used for spinor indices.

 

Oh and the Lagrangian I gave above is for the vacuum. To couple gravity to matter you just add a Lagrangian for matter and include a normalisation constant [math]\frac{1}{8\pi}[/math] (units in which G=c=1).