The design of products of tensorial expressions that have contracted indices got enhanced. The idea: repeated indices in certain subexpressions are actually dummies. So suppose $T_{a,b}$ and $B_b$ are tensors, then in $T_{\mathrm{trace}}=T_{a\phantom{a}}^{\phantom{a}a}$, $a$ is just dummy, therefore $T_{a\phantom{a}}^{\phantom{a}a}{B}_a=T_{b\phantom{b}}^{\phantom{b}b}{B}_a$ is a well defined object. The new design automatically maps input like $T_{a\phantom{a}}^{\phantom{a}a}{B}_a$ into $T_{b\phantom{b}}^{\phantom{b}b}{B}_a$.

This shows the automatic handling of collision of indices

Consider now the case of three tensors

The product above has indeed the index $a$ repeated more than once, therefore none of its occurrences got automatically transformed into contravariant in the output, and Check detects the problem interrupting with an error message

However, it is now also possible to indicate, using parenthesis, that the product of two of these tensors actually form a subexpression, so that the following two tensorial expressions are well defined, where the dummy is automatically replaced making that explicit

This change in design makes concretely simpler the use of indices in that it eliminates the need for manually replacing dummies. For example, consider the tensorial expression for the angular momentum in terms of the coordinates and momentum vectors, in 3 dimensions

Define $L_j,p_k$ respectively representing angular and linear momentum

Introduce the tensorial expression for $L_a$

The left-hand side has one free index, $a$, while the right-hand side has two dummy indices $b$ and $c$

If we want to compute${‖\overrightarrow{L}‖}^2\={L_a}^2$we can now take the square of (11) directly, and the dummy indices on the right-hand side are automatically handled, there is now no need to manually substitute the repeated indices to avoid their collision

The repeated indices on the right-hand side are now $a,b,c,d,e$

A new keyword in Setup: differentialoperators, allows for defining differential operators (not necessarily linear) with respect to indicated differentiation variables, so that they are treated as noncommutative operands in products, as we do with paper and pencil. These user-defined differential operators can also be vectorial and/or tensorial or inert. When desired, one can use Library:-ApplyProductOfDifferentialOperators to transform the products in the function application of these operators. This new functionality is a generalization of the differential operators ${\partial }_{\mathrm{\mu }}$ and ${▿}_{\mathrm{\mu }}$, and can used beyond Physics.

A new routine Library:-GetDifferentiationVariables also acts on a differential operator and tells who are the corresponding differentiation variables

Example:

In Quantum Mechanics, in the coordinates representation, the component of the momentum operator along the x axis is given by the differential operator

 $\mathrm{p\_\_x}\=-i \mathrm{\ℏ}\frac{\partial }{\partial x}$

The purpose of the exercises below is thus to derive the commutation rules, in the coordinates representation, between an arbitrary function of the coordinates and the related momentum, departing from the differential representation

$p_n=-i\mathrm{\hslash }{\partial }_n$

${\left[\mathit{F}\left(\mathit{X}\right)\mathbf{,}\overrightarrow{\mathit{p}}\right]}_{\mathbf{-}}\mathbf{=}\mathrm{i} \mathrm{\ℏ} \mathbf{\nabla }\mathit{F}\left(\mathit{X}\right)$

>  $\mathrm{restart}\:\mathrm{with}\left(\mathrm{Physics}\right)\:\mathrm{with}\left(\mathrm{Physics}\left[\mathrm{Vectors}\right]\right)\:\mathrm{interface}\left(\mathrm{imaginaryunit} \= i\right)\:$   Start setting the problem: –   all of$x,y,z,\mathrm{p\_\_x},\mathrm{p\_\_y},\mathrm{p\_\_z}$ are Hermitian operators –   all of $x,y,z$ commute between each other –   tell the system only that the operators $x\, y\, z$ are the differentiation variables of the corresponding (differential) operators $\mathrm{p\_\_x},\mathrm{p\_\_y},\mathrm{p\_\_z}$ but do not tell what is the form of the operators   >  $$\begin{gathered} \mathrm{Setup}\left(\mathrm{mathematicalnotation} \= \mathrm{true}\, \\ \mathrm{differentialoperators}\=\left\{\left[\mathrm{p\_}\,\left[x\,y\,z\right]\right]\right\}\, \\ \mathrm{hermitianoperators}\=\left\{p\,x\,y\,z\right\}\, \\ \mathrm{algebrarules}\=\left\{\mathrm{\%Commutator}\left(x\,y\right)\=0\,\mathrm{\%Commutator}\left(x\,z\right)\=0\,\mathrm{\%Commutator}\left(y\,z\right)\=0\right\}\, \\ \mathrm{quiet}\right) \end{gathered}$$ $\left[\mathrm{algebrarules}=\left\{{\left[x,y\right]}_{-}=0\,{\left[x,z\right]}_{-}=0\,{\left[y,z\right]}_{-}=0\right\}\,\mathrm{differentialoperators}=\left\{\left[\overrightarrow{p}\,\left[x\,y\,z\right]\right]\right\}\,\mathrm{hermitianoperators}=\left\{p\,x\,y\,z\right\}\,\mathrm{mathematicalnotation}=\mathrm{true}\right]$ (15) Assuming $F\left(X\right)$ is a smooth function, the idea is to apply the commutator ${\left[F\left(X\right),\overrightarrow{p}\right]}_{-}$ to an arbitrary ket of the Hilbert space $\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$, perform the operation explicitly after setting a differential operator representation for $\overrightarrow{p}$, and from there get the commutation rule between $F\left(X\right)$ and $\overrightarrow{p}$.   Start introducing the commutator, to proceed with full control of the operations we use the inert form %Commutator >  $\mathrm{alias}\left(X \= \left(x\,y\,z\right)\right)\:$ >  $\mathrm{CompactDisplay}\left(F\left(X\right)\right)$ $F\left(X\right)\mathrm{will\; now\; be\; displayed\; as}F$ (16) >  $\mathrm{\%Commutator}\left(F\left(X\right)\,\mathrm{p\_}\right)\cdot \mathrm{Ket}\left(\mathrm{\ψ}\,X\right)$ ${\left[F,\overrightarrow{p}\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ (17) For illustration purposes only (not necessary), expand this commutator >  $\=\mathrm{expand}\left(\right)$ ${\left[F,\overrightarrow{p}\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩=F\overrightarrow{p}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩-\overrightarrow{p}F\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ (18) Note that  $\overrightarrow{p}$, $F\left(X\right)$ and the ket $\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ are operands in the products above and that they do not commute: we indicated that the coordinates $x\, y\, z$ are the differentiation variables of $\overrightarrow{p}$. This emulates what we do when computing with these operators with paper and pencil, where we represent the application of a differential operator as a product operation.   This representation can be transformed into the (traditional in computer algebra) application of the differential operator when desired, as follows: >  $\=\mathrm{Library}:-\mathrm{ApplyProductsOfDifferentialOperators}\left(\right)$ ${\left[F,\overrightarrow{p}\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩=F\overrightarrow{p}\left(\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩\right)-\overrightarrow{p}\left(F\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩\right)$ (19) Note that, in $\overrightarrow{p}\left(F\left(X\right)\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩\right)$, the application of $\overrightarrow{p}$ is not expanded: at this point nothing is known about  $\overrightarrow{p}$ , it is not necessarily a linear operator. In the Quantum Mechanics problem at hands, however, it is. So give now the operator  $\overrightarrow{p}$ an explicit representation as a linear vectorial differential operator (we use the inert form %Nabla, $\nabla$, to be able to proceed with full control one step at a time) >  $\mathrm{p\_}≔f\to -\mathrm{i}\cdot \mathrm{\hslash }\cdot \mathrm{\%Nabla}\left(f\right)$ $\overrightarrow{p}≔f\mapsto -\mathrm{i}\mathrm{\hslash }\left(\nabla f\right)$ (20) The expression (19) becomes >  $$ ${\left[F,\overrightarrow{p}\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩=-\mathrm{i}\mathrm{\hslash }F\nabla \left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩+\mathrm{i}\mathrm{\hslash }\nabla \left(F\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩\right)$ (21) Activate now the inert operator $\nabla$ and simplify taking into account the algebra rules for the coordinate operators $\left\{{\left[x,y\right]}_{-}=0\,{\left[x,z\right]}_{-}=0\,{\left[y,z\right]}_{-}=0\right\}$ >  $\mathrm{Simplify}\left(\mathrm{value}\left(\right)\right)$ ${\left[F,\overrightarrow{p}\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩=\mathrm{i}\mathrm{\hslash }\stackrel{\wedge }{i}{F}_x\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩+\mathrm{i}\mathrm{\hslash }\stackrel{\wedge }{j}{F}_y\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩+\mathrm{i}\mathrm{\hslash }\stackrel{\wedge }{k}{F}_z\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ (22) To make explicit the gradient in disguise on the right-hand side, factor out the arbitrary ket $\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ >  $\mathrm{Factor}\left(\right)$ ${\left[F,\overrightarrow{p}\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩=\mathrm{i}\mathrm{\hslash }\left(F_x\stackrel{\wedge }{i}+F_y\stackrel{\wedge }{j}+F_z\stackrel{\wedge }{k}\right)\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ (23) Combine now the expanded gradient into its inert (not-expanded) form >  $\mathrm{subs}\left(\left(\mathrm{Gradient}\=\mathrm{\%Gradient}\right)\left(F\left(X\right)\right)\,\right)$ ${\left[F,\overrightarrow{p}\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩=\mathrm{i}\mathrm{\hslash }\nabla F\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ (24) Since (24) is true for all$\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$, this ket can be removed from both sides of the equation. One can do that either taking coefficients (see Coefficients) or multiplying by the "formal inverse" of this ket, arriving at the (expected) form of the commutation rule between $F\left(X\right)$ and $\overrightarrow{p}$ >  $\cdot \mathrm{Inverse}\left(\mathrm{Ket}\left(\mathrm{\psi }\,x\,y\,z\right)\right)$ ${\left[F,\overrightarrow{p}\right]}_{-}=\mathrm{i}\mathrm{\hslash }\nabla F$ (25)

Tensor notation, ${\left[\mathrm{X\_\_m}\mathbf{\,}{\mathit{P}}_{\mathit{n}}\right]}_{\mathbf{-}}\mathbf{=}\mathbf{i}\mathbf{ }\mathbf{\hslash }{\mathit{g}}_{\mathit{m}\mathbf{,}\mathit{n}}$

The computation rule for position and momentum, this time in tensor notation, is performed in the same way, just that, additionally, specify that the space indices to be used are lowercase Latin letters, and set the relationship between the differential operators and the coordinates directly using tensor notation. You can also specify that the metric is Euclidean, but that is not necessary: the default metric of the Physics package, a Minkowski spacetime, includes a 3D subspace that is Euclidean, and the default signature, (- - - +), is not a problem regarding this computation.   >  $\mathrm{restart}\; \mathrm{with}\left(\mathrm{Physics}\right)\:\mathrm{interface}\left(\mathrm{imaginaryunit} \= i\right)\:$ >  $$\begin{gathered} \mathrm{Setup}\left(\mathrm{mathematicalnotation}\=\mathrm{true}\, \\ \mathrm{coordinates} \= \mathrm{cartesian}\, \\ \mathrm{spaceindices} \= \mathrm{lowercaselatin}\, \\ \mathrm{algebrarules}\=\left\{\mathrm{\%Commutator}\left(x\,y\right)\=0\,\mathrm{\%Commutator}\left(x\,z\right)\=0\,\mathrm{\%Commutator}\left(y\,z\right)\=0\right\}\, \\ \mathrm{hermitianoperators} \= \left\{X\,P\,p\right\}\, \\ \mathrm{differentialoperators}\=\left\{\left[P\left[m\right]\,\left[x\,y\,z\right]\right]\right\}\, \\ \mathrm{quiet}\right) \end{gathered}$$ $\left[\mathrm{algebrarules}=\left\{{\left[x,y\right]}_{-}=0\,{\left[x,z\right]}_{-}=0\,{\left[y,z\right]}_{-}=0\right\}\,\mathrm{coordinatesystems}=\left\{X\right\}\,\mathrm{differentialoperators}=\left\{\left[P_m\,\left[x\,y\,z\right]\right]\right\}\,\mathrm{hermitianoperators}=\left\{P\,p\,t\,x\,y\,z\right\}\,\mathrm{mathematicalnotation}=\mathrm{true}\,\mathrm{spaceindices}=\mathrm{lowercaselatin}\right]$ (26) Define now the tensor $P_m$ >  $\mathrm{Define}\left(P\left[m\right]\,\mathrm{quiet}\right)$ $\left\{{\mathrm{\gamma }}_{\mathrm{\mu }}\,P_m\,{\mathrm{\sigma }}_{\mathrm{\mu }}\,X_{\mathrm{\mu }}\,{\partial }_{\mathrm{\mu }}\,g_{\mathrm{\mu },\mathrm{\nu }}\,{\mathrm{\gamma }}_{a,b}\,{\mathrm{\delta }}_{\mathrm{\mu },\mathrm{\nu }}\,{\mathrm{\epsilon }}_{\mathrm{\alpha },\mathrm{\beta },\mathrm{\mu },\mathrm{\nu }}\right\}$ (27) Introduce now the Commutator, this time in active form, to show how to reobtain the non-expanded form at the end by resorting the operands in products >  $\mathrm{Commutator}\left(X\left[m\right]\,P\left[n\right]\right)\cdot \mathrm{Ket}\left(\mathrm{\ψ}\,x\,y\,z\right)$ ${\left[X_m,P_n\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ (28) Expand first (not necessary) to see how the operator $P_n$ is going to be applied >  $\=\mathrm{expand}\left(\right)$ ${\left[X_m,P_n\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩=X_m{P}_n\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩-P_n{X}_m\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ (29) Now expand and directly apply in one ago the differential operator $P_n$ >  $\=\mathrm{Library}:-\mathrm{ApplyProductsOfDifferentialOperators}\left(\right)$ ${\left[X_m,P_n\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩=X_m{P}_n\left(\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩\right)-P_n\left(X_m\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩\right)$ (30) Introducing the explicit differential operator representation for $P_n$, here again using the inert ${\partial }_n$ to keep control of the computations step by step >  $P\left[n\right]≔f\to -\mathrm{i}\cdot \mathrm{\ℏ}\cdot \mathrm{\%d\_}\left[n\right]\left(f\right)$ $P_n≔f\mapsto -\mathrm{i}\mathrm{\hslash }{\partial }_n\left(f\right)$ (31) The expanded and applied commutator (30) becomes >  $$ ${\left[X_m,P_n\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩=-\mathrm{i}\mathrm{\hslash }{X}_m\left({\partial }_n\left(\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩\right)\right)+\mathrm{i}\mathrm{\hslash }\left({\partial }_n\left(X_m\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩\right)\right)$ (32) Activate now the inert operators ${\partial }_n$ and simplify taking into account Einstein's rule for repeated indices >  $\mathrm{Simplify}\left(\mathrm{value}\left(\right)\right)$ ${\left[X_m,P_n\right]}_{-}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩=\mathrm{i}\mathrm{\hslash }{g}_{m,n}\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ (33) Since the ket $\left|{\mathrm{\psi }}_{x\,y\,z}\right.⟩$ is arbitrary, we can take coefficients (or multiply by the formal Inverse of this ket as done in the previous section). For illustration purposes, we use  Coefficients and note how it automatically expands the commutator >  $\mathrm{Coefficients}\left(\,\mathrm{Ket}\left(\mathrm{\psi }\,x\,y\,z\right)\right)$ $X_m{P}_n-P_n{X}_m=\mathrm{i}\mathrm{\hslash }{g}_{m,n}$ (34) One can undo this (frequently undesired) expansion of the commutator by sorting the products on the left-hand side using the commutator between $X_m$ and $P_n$ >  $\mathrm{Library}:-\mathrm{SortProducts}\left(\,\left[P\left[n\right]\,X\left[m\right]\right]\,\mathrm{usecommutator}\right)$ ${\left[X_m,P_n\right]}_{-}=\mathrm{i}\mathrm{\hslash }{g}_{m,n}$ (35) And that is the result we wanted to compute.   Additionally, to see this rule in matrix form, >  $\mathrm{TensorArray}\left(-\right)$ $\left[\begin{array}{ccc}{\left[x,P_1\right]}_{-}=\mathrm{i}\mathrm{\hslash }& {\left[x,P_2\right]}_{-}=0& {\left[x,P_3\right]}_{-}=0\\ {\left[y,P_1\right]}_{-}=0& {\left[y,P_2\right]}_{-}=\mathrm{i}\mathrm{\hslash }& {\left[y,P_3\right]}_{-}=0\\ {\left[z,P_1\right]}_{-}=0& {\left[z,P_2\right]}_{-}=0& {\left[z,P_3\right]}_{-}=\mathrm{i}\mathrm{\hslash }\end{array}\right]$ (36) In the above, we use equation (35) multiplied by -1 to avoid a minus sign in all the elements of (36), due to having worked with the default signature (- - - +); this minus sign is not necessary if in the Setup at the beginning one also sets  $\mathrm{signature}=\mathrm{`+\; +\; +\; -`}$   For display purposes, to see this matrix expressed in terms of the geometrical components of the momentum $\overrightarrow{p}$ , redefine the tensor $P_n$ explicitly indicating its Cartesian components >  $\mathrm{Define}\left(P\left[m\right]\=\left[\mathrm{p\_\_x}\,\mathrm{p\_\_y}\,\mathrm{p\_\_z}\right]\,\mathrm{quiet}\right)$ $\left\{{\mathrm{\gamma }}_{\mathrm{\mu }}\,P_m\,{\mathrm{\sigma }}_{\mathrm{\mu }}\,X_{\mathrm{\mu }}\,{\partial }_{\mathrm{\mu }}\,g_{\mathrm{\mu },\mathrm{\nu }}\,{\mathrm{\gamma }}_{a,b}\,{\mathrm{\delta }}_{\mathrm{\mu },\mathrm{\nu }}\,{\mathrm{\epsilon }}_{\mathrm{\alpha },\mathrm{\beta },\mathrm{\mu },\mathrm{\nu }}\right\}$ (37) >  $\mathrm{TensorArray}\left(-\right)$ $\left[\begin{array}{ccc}{\left[x,\mathrm{p\_\_x}\right]}_{-}=\mathrm{i}\mathrm{\hslash }& {\left[x,\mathrm{p\_\_y}\right]}_{-}=0& {\left[x,\mathrm{p\_\_z}\right]}_{-}=0\\ {\left[y,\mathrm{p\_\_x}\right]}_{-}=0& {\left[y,\mathrm{p\_\_y}\right]}_{-}=\mathrm{i}\mathrm{\hslash }& {\left[y,\mathrm{p\_\_z}\right]}_{-}=0\\ {\left[z,\mathrm{p\_\_x}\right]}_{-}=0& {\left[z,\mathrm{p\_\_y}\right]}_{-}=0& {\left[z,\mathrm{p\_\_z}\right]}_{-}=\mathrm{i}\mathrm{\hslash }\end{array}\right]$ (38) Finally, in a typical situation, these commutation rules are to be taken into account in further computations, and for that purpose they can be added to the setup via >  $\mathrm{Setup}\left(\right)$ $\left[\mathrm{algebrarules}=\left\{{\left[x,\mathrm{p\_\_x}\right]}_{-}=\mathrm{i}\mathrm{\hslash }\,{\left[x,\mathrm{p\_\_y}\right]}_{-}=0\,{\left[x,\mathrm{p\_\_z}\right]}_{-}=0\,{\left[x,y\right]}_{-}=0\,{\left[x,z\right]}_{-}=0\,{\left[y,\mathrm{p\_\_x}\right]}_{-}=0\,{\left[y,\mathrm{p\_\_y}\right]}_{-}=\mathrm{i}\mathrm{\hslash }\,{\left[y,\mathrm{p\_\_z}\right]}_{-}=0\,{\left[y,z\right]}_{-}=0\,{\left[z,\mathrm{p\_\_x}\right]}_{-}=0\,{\left[z,\mathrm{p\_\_y}\right]}_{-}=0\,{\left[z,\mathrm{p\_\_z}\right]}_{-}=\mathrm{i}\mathrm{\hslash }\right\}\right]$ (39) For example, from herein computations are performed taking into account that >  $\left(\mathrm{\%Commutator} \= \mathrm{Commutator}\right)\left(x\, \mathrm{p\_\_x}\right)$ ${\left[x,\mathrm{p\_\_x}\right]}_{-}=\mathrm{i}\mathrm{\hslash }$ (40) >  $$

There are 991 metrics in the database of solutions to Einstein's equations, based on the book "Exact solutions to Einstein's equations". One can check this number via

For the majority of these solutions, the book also presents, explicit or implicitly, the form of the Energy-Momentum tensor. New in Maple 2018, we added to the database one more entry indicating the components of the corresponding EnergyMomentum tensor, covering, in Maple 2018.0, 686 out of these 991 solutions.

The design of the EnergyMomentum tensor got slightly adjusted to take these new database entries into account, so that when you load one of these solutions, if the corresponding entry for the EnergyMomentumTensor is already in the database, it is automatically loaded together with the solution.

In addition, it is now possible to define the tensor components using the Define command, or redefine any of its components using the new Library:-RedefineTensorComponent routine (see Physics,Library)

Examples

Consider the metric of Chapter 12, equation number 16.1

New, the covariant components of the EnergyMomentum tensor got automatically loaded, given by

One can verify this checking for the tensor's definition

Take now the tensor components of the first defining equation of (44)

where $\mathrm{\κ0}\=8 \mathrm{\pi }$ is related to Newton's constant.

To see the continuity equations for the components of ${\mathrm{{\rm T}}}_{\mathrm{\mu },\mathrm{\nu }}$, use for instance the inert version of the covariant derivative operator D_ and the TensorArray command

The EnergyMomentum tensor can also be (re)defined in any particular way (a correct definition must satisfy ${▿}_{\mathrm{\mu }}\left({\mathrm{{\rm T}}}_{\phantom{}\phantom{\mathrm{\mu }}\mathrm{\nu }}^{\phantom{}\mathrm{\mu }\phantom{\mathrm{\nu }}}\right)\=0$).

Example:

Define the EnergyMomentum tensor indicating the functionality in the definition in terms of $W$ to be constant energy (i.e. no functionality) and the flux density $S_{\mathrm{\mu }}$ and stress ${\mathrm{\sigma }}_{\mathrm{\mu },\mathrm{\nu }}$ tensors depending on $X\.$ For this purpose, use the new option minimizetensorcomponents to make explicit the symmetry of the stress tensor ${\mathrm{\sigma }}_{\mathrm{\mu },\mathrm{\nu }}$

The symmetry of ${\mathrm{\sigma }}_{\mathrm{\mu },\mathrm{\nu }}$ is now explicit in that its matrix form is symmetric

The new routines for testing tensor symmetries

The symmetry is regarding interchanging positions of the 1st and 2nd indices

So this is the form of the EnergyMomentum with all its components - but for the total energy - depending on the coordinates

Define now ${\mathrm{{\rm T}}}_{\mathrm{\mu },\mathrm{\nu }}$ with these components

To see the continuity equations for the components of ${\mathrm{{\rm T}}}_{\mathrm{\mu },\mathrm{\nu }}$, use again the inert version of the covariant derivative operator D_ and the TensorArray command

For a more convenient reading, present the result as a vector column