 TraceFreeRicciTensor - Maple Help

Tensor[TraceFreeRicciTensor] - calculate the trace-free Ricci tensor of a metric tensor

Calling Sequences

TraceFreeRicciTensor(g)

TraceFreeRicciTensor(g, C)

TraceFreeRicciTensor(g, R)

Parameters

g    - the metric tensor on the tangent bundle of a manifold

C    - the curvature tensor of the metric $g$

R    - the Ricci tensor of the metric $g$ Description

 • Let be a metric tensor with associated Ricci tensor R and Ricci scalar S. The trace-free Ricci tensor P is the symmetric, rank 2 covariant tensor with components , where  is the dimension of the underlying manifold. It is trace-free with respect to the metric  in the sense that where are the components of the inverse metric.
 • This command is part of the DifferentialGeometry:-Tensor package, and so can be used in the form TraceFreeRicciTensor(...) only after executing the command with(DifferentialGeometry) and with(Tensor) in that order.  It can always be used in the long form DifferentialGeometry:-Tensor:-TraceFreeRicciTensor. Examples

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

Example 1.

In this example we calculate the trace-free Ricci tensor for a metric.

 M1 > $\mathrm{DGsetup}\left(\left[x,y,z\right],M\right)$
 ${\mathrm{frame name: M}}$ (2.1)
 M > $g≔\mathrm{evalDG}\left(y\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dx}+z\mathrm{dy}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}+\mathrm{dz}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dz}\right)$
 ${g}{:=}{y}{}{\mathrm{dx}}{}{\mathrm{dx}}{+}{z}{}{\mathrm{dy}}{}{\mathrm{dy}}{+}{\mathrm{dz}}{}{\mathrm{dz}}$ (2.2)

Calculate the trace-free Ricci tensor for the metric directly.

 M > $P≔\mathrm{TraceFreeRicciTensor}\left(g\right)$
 ${P}{:=}{-}\frac{{1}}{{12}}{}\frac{\left({2}{}{{y}}^{{2}}{-}{z}\right){}{\mathrm{dx}}{}{\mathrm{dx}}}{{{z}}^{{2}}{}{y}}{+}\frac{{1}}{{12}}{}\frac{\left({{y}}^{{2}}{+}{z}\right){}{\mathrm{dy}}{}{\mathrm{dy}}}{{z}{}{{y}}^{{2}}}{+}\frac{{1}}{{4}}{}\frac{{\mathrm{dy}}{}{\mathrm{dz}}}{{y}{}{z}}{+}\frac{{1}}{{4}}{}\frac{{\mathrm{dz}}{}{\mathrm{dy}}}{{y}{}{z}}{+}\frac{{1}}{{12}}{}\frac{\left({{y}}^{{2}}{-}{2}{}{z}\right){}{\mathrm{dz}}{}{\mathrm{dz}}}{{{z}}^{{2}}{}{{y}}^{{2}}}$ (2.3)

We check that is trace-free by computing the inverse metric and using the ContractIndices command.

 M > $h≔\mathrm{InverseMetric}\left(g\right)$
 ${h}{:=}\frac{{\mathrm{D_x}}{}{\mathrm{D_x}}}{{y}}{+}\frac{{\mathrm{D_y}}{}{\mathrm{D_y}}}{{z}}{+}{\mathrm{D_z}}{}{\mathrm{D_z}}$ (2.4)
 M > $\mathrm{ContractIndices}\left(h,P,\left[\left[1,1\right],\left[2,2\right]\right]\right)$
 ${0}$ (2.5)

The same calculation can be done with the TensorInnerProduct command.

 M > $\mathrm{TensorInnerProduct}\left(g,g,P\right)$
 ${0}$ (2.6)

Example 2.

The third calling sequence can be applied to any rank 2 symmetric tensor to construct a trace-free, rank 2 symmetric tensor.

 M > $\mathrm{A1}≔\mathrm{evalDG}\left(\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&s\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dy}\right)$
 ${\mathrm{A1}}{:=}{\mathrm{dx}}{}{\mathrm{dy}}{+}{\mathrm{dy}}{}{\mathrm{dx}}$ (2.7)
 M > $\mathrm{A2}≔g$
 ${\mathrm{A2}}{:=}{y}{}{\mathrm{dx}}{}{\mathrm{dx}}{+}{z}{}{\mathrm{dy}}{}{\mathrm{dy}}{+}{\mathrm{dz}}{}{\mathrm{dz}}$ (2.8)
 M > $\mathrm{A3}≔\mathrm{evalDG}\left(\mathrm{dx}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&t\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{dx}\right)$
 ${\mathrm{A3}}{:=}{\mathrm{dx}}{}{\mathrm{dx}}$ (2.9)

The tensor is already trace-free, so its trace-free part is itself.

 M > $\mathrm{TraceFreeRicciTensor}\left(g,\mathrm{A1}\right)$
 ${\mathrm{dx}}{}{\mathrm{dy}}{+}{\mathrm{dy}}{}{\mathrm{dx}}$ (2.10)



The trace-free part of the metric itself is always 0.

 M > $\mathrm{TraceFreeRicciTensor}\left(g,\mathrm{A2}\right)$
 ${0}{}{\mathrm{dx}}{}{\mathrm{dx}}$ (2.11)

The trace-free part of is

 M > $T≔\mathrm{TraceFreeRicciTensor}\left(g,\mathrm{A3}\right)$
 ${T}{:=}\frac{{2}}{{3}}{}{\mathrm{dx}}{}{\mathrm{dx}}{-}\frac{{1}}{{3}}{}\frac{{z}{}{\mathrm{dy}}{}{\mathrm{dy}}}{{y}}{-}\frac{{1}}{{3}}{}\frac{{\mathrm{dz}}{}{\mathrm{dz}}}{{y}}$ (2.12)
 M > $\mathrm{TensorInnerProduct}\left(g,g,T\right)$
 ${0}$ (2.13) See Also