DifferentialGeometry/Tensor/BelRobinson - Maple Help

Tensor[BelRobinson] - calculate the Bel-Robinson tensor

Calling Sequences

BelRobinson(g, W, indexlist)

Parameters

g         - a metric tensor on a 4-dimensional manifold

W         - (optional) the Weyl tensor of the metric $g$

indexlist - (optional) the keyword argument indexlist = ind, where ind is a list of 4 index types "con" or "cov"

Description

 • The Bel-Robinson tensor ${B}_{\mathrm{ijhk}}$ is a covariant rank 4 tensor defined in terms of the Weyl tensor ${W}_{\mathrm{ijhk}}$ on a 4-dimensional manifold by (see, for example, Penrose and Rindler Vol. 1)

The Bel-Robinson tensor is totally symmetric: ${B}_{\mathrm{ijhk}}={B}_{\mathrm{jihk}}={B}_{\mathrm{hjik}}={B}_{\mathrm{kjhi}}$ . The Bel-Robinson tensor is trace-free: ${g}^{\mathrm{ij}}{B}_{\mathrm{ijhk}}=0$. If ${g}_{\mathrm{ij}}$ is an Einstein metric, that is, ${R}_{\mathrm{ij}}={\mathrm{Λg}}_{\mathrm{ij}}$ (where ${R}_{\mathrm{ij}}$ is the Ricci tensor for the metric ${g}_{\mathrm{ij}}$ and $\mathrm{Λ}$ is a constant), then the covariant divergence of Bel-Robinson vanishes:   Here ${\nabla }_{l}$ denotes the covariant derivative with respect to the Christoffel connection for ${g}_{\mathrm{ij}}$.

 • The keyword argument indexlist = ind allows the user to specify the index structure for the Bel-Robinson tensor. For example, with indexlist = ["con", "con", "con", "con"], the contravariant form ${B}^{\mathrm{ijhk}}$ is returned. The default output is the purely covariant form (as above).
 • This command is part of the DifferentialGeometry:-Tensor package, and so can be used in the form BelRobinson(...) only after executing the commands with(DifferentialGeometry); with(Tensor); in that order. It can always be used in the long form DifferentialGeometry:-Tensor:-BelRobinson.

Examples

 > $\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):$$\mathrm{with}\left(\mathrm{Tensor}\right):$

Example 1.

First create a 4-dimensional manifold M and define a metric $g$on $M$. The metric shown below is a homogenous Einstein metric (see (12.34) in Stephani, Kramer et al).

 > $\mathrm{DGsetup}\left(\left[x,y,z,u\right],M\right)$
 ${\mathrm{frame name: M}}$ (2.1)
 M > $g≔\mathrm{evalDG}\left(\mathrm{exp}\left(z\right)\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dx}+\mathrm{exp}\left(-2z\right)\left(\mathrm{dy}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}+\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&s\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{du}\right)-\frac{3}{\mathrm{\Lambda }}\mathrm{dz}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dz}\right)$
 ${g}{:=}{{ⅇ}}^{{z}}{}{\mathrm{dx}}{}{\mathrm{dx}}{+}\frac{{{ⅇ}}^{{-}{2}{}{z}}}{{2}}{}{\mathrm{dx}}{}{\mathrm{du}}{+}{{ⅇ}}^{{-}{2}{}{z}}{}{\mathrm{dy}}{}{\mathrm{dy}}{-}\frac{{3}}{{\mathrm{\Lambda }}}{}{\mathrm{dz}}{}{\mathrm{dz}}{+}\frac{{{ⅇ}}^{{-}{2}{}{z}}}{{2}}{}{\mathrm{du}}{}{\mathrm{dx}}$ (2.2)

Calculate the Bel-Robinson tensor for the metric $g$.  The result is clearly a symmetric tensor.

 M > $B≔\mathrm{BelRobinson}\left(g\right)$
 ${B}{:=}\frac{{{\mathrm{\Lambda }}}^{{2}}{}{{ⅇ}}^{{2}{}{z}}}{{4}}{}{\mathrm{dx}}{}{\mathrm{dx}}{}{\mathrm{dx}}{}{\mathrm{dx}}$ (2.3)

Use the optional keyword argument indexlist to calculate the contravariant form of the Bel-Robinson tensor.

 M > $\mathrm{B1}≔\mathrm{BelRobinson}\left(g,\mathrm{indexlist}=\left["con","con","con","con"\right]\right)$
 ${\mathrm{B1}}{:=}{4}{}{{ⅇ}}^{{10}{}{z}}{}{{\mathrm{\Lambda }}}^{{2}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}$ (2.4)

The tensor B is trace-free.

 > $h≔\mathrm{InverseMetric}\left(g\right)$
 ${h}{:=}{2}{}{{ⅇ}}^{{2}{}{z}}{}{\mathrm{D_x}}{}{\mathrm{D_u}}{+}{{ⅇ}}^{{2}{}{z}}{}{\mathrm{D_y}}{}{\mathrm{D_y}}{-}\frac{{\mathrm{\Lambda }}}{{3}}{}{\mathrm{D_z}}{}{\mathrm{D_z}}{+}{2}{}{{ⅇ}}^{{2}{}{z}}{}{\mathrm{D_u}}{}{\mathrm{D_x}}{-}{4}{}{{ⅇ}}^{{5}{}{z}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}$ (2.5)
 > $\mathrm{ContractIndices}\left(h,B,\left[\left[1,1\right],\left[2,2\right]\right]\right)$
 ${0}{}{\mathrm{dx}}{}{\mathrm{dx}}$ (2.6)

The covariant divergence of the tensor B1 vanishes.  To check this, first calculate the Christoffel connection C for the metric g and then calculate the covariant derivative of B1.

 > $C≔\mathrm{Christoffel}\left(g\right)$
 ${C}{:=}{-}{\mathrm{D_x}}{}{\mathrm{dx}}{}{\mathrm{dz}}{-}{\mathrm{D_x}}{}{\mathrm{dz}}{}{\mathrm{dx}}{-}{\mathrm{D_y}}{}{\mathrm{dy}}{}{\mathrm{dz}}{-}{\mathrm{D_y}}{}{\mathrm{dz}}{}{\mathrm{dy}}{+}\frac{{\mathrm{\Lambda }}{}{{ⅇ}}^{{z}}}{{6}}{}{\mathrm{D_z}}{}{\mathrm{dx}}{}{\mathrm{dx}}{-}\frac{{\mathrm{\Lambda }}{}{{ⅇ}}^{{-}{2}{}{z}}}{{6}}{}{\mathrm{D_z}}{}{\mathrm{dx}}{}{\mathrm{du}}{-}\frac{{\mathrm{\Lambda }}{}{{ⅇ}}^{{-}{2}{}{z}}}{{3}}{}{\mathrm{D_z}}{}{\mathrm{dy}}{}{\mathrm{dy}}{-}\frac{{\mathrm{\Lambda }}{}{{ⅇ}}^{{-}{2}{}{z}}}{{6}}{}{\mathrm{D_z}}{}{\mathrm{du}}{}{\mathrm{dx}}{+}{3}{}{{ⅇ}}^{{3}{}{z}}{}{\mathrm{D_u}}{}{\mathrm{dx}}{}{\mathrm{dz}}{+}{3}{}{{ⅇ}}^{{3}{}{z}}{}{\mathrm{D_u}}{}{\mathrm{dz}}{}{\mathrm{dx}}{-}{\mathrm{D_u}}{}{\mathrm{dz}}{}{\mathrm{du}}{-}{\mathrm{D_u}}{}{\mathrm{du}}{}{\mathrm{dz}}$ (2.7)
 > $\mathrm{nablaB1}≔\mathrm{CovariantDerivative}\left(\mathrm{B1},C\right)$
 ${\mathrm{nablaB1}}{:=}{-}\frac{{2}{}{{\mathrm{\Lambda }}}^{{3}}{}{{ⅇ}}^{{8}{}{z}}}{{3}}{}{\mathrm{D_z}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{dx}}{-}\frac{{2}{}{{\mathrm{\Lambda }}}^{{3}}{}{{ⅇ}}^{{8}{}{z}}}{{3}}{}{\mathrm{D_u}}{}{\mathrm{D_z}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{dx}}{-}\frac{{2}{}{{\mathrm{\Lambda }}}^{{3}}{}{{ⅇ}}^{{8}{}{z}}}{{3}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_z}}{}{\mathrm{D_u}}{}{\mathrm{dx}}{-}\frac{{2}{}{{\mathrm{\Lambda }}}^{{3}}{}{{ⅇ}}^{{8}{}{z}}}{{3}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_z}}{}{\mathrm{dx}}{+}{24}{}{{ⅇ}}^{{10}{}{z}}{}{{\mathrm{\Lambda }}}^{{2}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{D_u}}{}{\mathrm{dz}}$ (2.8)
 > $\mathrm{Divergence}≔\mathrm{ContractIndices}\left(\mathrm{nablaB1},\left[\left[1,5\right]\right]\right)$
 ${\mathrm{Divergence}}{:=}{0}{}{\mathrm{D_x}}{}{\mathrm{D_x}}{}{\mathrm{D_x}}$ (2.9)

The divergence of the Bel-Robinson tensor is not automatically zero; the divergence vanishes when the metric g is an Einstein metric.  To check this, compute the Ricci tensor of g.

 > $R≔\mathrm{RicciTensor}\left(g\right)$
 ${R}{:=}{\mathrm{\Lambda }}{}{{ⅇ}}^{{z}}{}{\mathrm{dx}}{}{\mathrm{dx}}{+}\frac{{\mathrm{\Lambda }}{}{{ⅇ}}^{{-}{2}{}{z}}}{{2}}{}{\mathrm{dx}}{}{\mathrm{du}}{+}{\mathrm{\Lambda }}{}{{ⅇ}}^{{-}{2}{}{z}}{}{\mathrm{dy}}{}{\mathrm{dy}}{-}{3}{}{\mathrm{dz}}{}{\mathrm{dz}}{+}\frac{{\mathrm{\Lambda }}{}{{ⅇ}}^{{-}{2}{}{z}}}{{2}}{}{\mathrm{du}}{}{\mathrm{dx}}$ (2.10)
 M > $\mathrm{evalDG}\left(R-\mathrm{\Lambda }g\right)$
 ${0}{}{\mathrm{dx}}{}{\mathrm{dx}}$ (2.11)

The Weyl tensor, if already calculated, can be used to quickly compute the Bel-Robinson tensor.

 > $W≔\mathrm{WeylTensor}\left(g\right)$
 ${W}{:=}{-}\frac{{\mathrm{\Lambda }}{}{{ⅇ}}^{{-}{z}}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dy}}{}{\mathrm{dx}}{}{\mathrm{dy}}{+}\frac{{\mathrm{\Lambda }}{}{{ⅇ}}^{{-}{z}}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dy}}{}{\mathrm{dy}}{}{\mathrm{dx}}{-}\frac{{3}{}{{ⅇ}}^{{z}}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dz}}{}{\mathrm{dx}}{}{\mathrm{dz}}{+}\frac{{3}{}{{ⅇ}}^{{z}}}{{2}}{}{\mathrm{dx}}{}{\mathrm{dz}}{}{\mathrm{dz}}{}{\mathrm{dx}}{+}\frac{{\mathrm{\Lambda }}{}{{ⅇ}}^{{-}{z}}}{{2}}{}{\mathrm{dy}}{}{\mathrm{dx}}{}{\mathrm{dx}}{}{\mathrm{dy}}{-}\frac{{\mathrm{\Lambda }}{}{{ⅇ}}^{{-}{z}}}{{2}}{}{\mathrm{dy}}{}{\mathrm{dx}}{}{\mathrm{dy}}{}{\mathrm{dx}}{+}\frac{{3}{}{{ⅇ}}^{{z}}}{{2}}{}{\mathrm{dz}}{}{\mathrm{dx}}{}{\mathrm{dx}}{}{\mathrm{dz}}{-}\frac{{3}{}{{ⅇ}}^{{z}}}{{2}}{}{\mathrm{dz}}{}{\mathrm{dx}}{}{\mathrm{dz}}{}{\mathrm{dx}}$ (2.12)
 > $\mathrm{BelRobinson}\left(g,W\right)$
 $\frac{{{\mathrm{\Lambda }}}^{{2}}{}{{ⅇ}}^{{2}{}{z}}}{{4}}{}{\mathrm{dx}}{}{\mathrm{dx}}{}{\mathrm{dx}}{}{\mathrm{dx}}$ (2.13)