The TransformTetrad commands applies transformations of coordinates of a specific type to a tetrad. These transformations, related to the optional arguments, are all Lorentz transformations of the tetrad (local) system of references.

The transformed tetrad returned by TransformTetrad is not automatically set as the current tetrad. That allows for experimenting with different transformations. To set the output of TransformTetrad as the current tetrad, either pass to TransformTetrad an additional argument settetrad, or pass its output to Setup as in $\mathrm{Setup}\left(\mathrm{e\_}=\mathrm{..\; output\; of\; TransformTetrad\; ..}\right)$.

The tetrad being transformed does not need to be indicated, in which case it is, by default, a null tetrad related to the current value of the spacetime metric g_, that is, one that satisfies the tetrad definition ${𝔢}_{a,\mathrm{\mu }}{𝔢}_b^{\mathrm{\mu }}={\mathrm{\eta }}_{a,b}$, ${𝔢}_{a,\mathrm{\mu }}{𝔢}_{\mathrm{\nu }}^a=g_{\mathrm{\mu },\mathrm{\nu }}$. and where the tetrad metric is formed with the null vectors of the Newman-Penrose formalism. To see what tetrad (among the infinitely many possible ones) is actually used as a starting point by default, see Tetrads[e_].

Instead of the default action of transforming e_, you can also indicate any other tetrad to be transformed. For that purpose, indicate the tetrad as first argument T, either as a matrix (whose components are assumed to correspond to the all-covariant tetrad e_[a, mu]), or as an equation where the left-hand side is the tensor e_[a, mu] itself.

When a tetrad is indicated, before proceeding, a test verifying whether it satisfies the tetrad definition equations is performed using the command IsTetrad, and if the tetrad definition is not satisfied an error interruption happens. Sometimes, the indicated T is in fact a tetrad but the computer fails in verifying that, perhaps missing a simplification; in these cases you can indicate to not perform this verification by passing the optional argument verifygiventetrad = false.

When the tetrad indicated to be transformed is an orthonormal tetrad (for the corresponding definitions, see IsTetrad), first a null tetrad is derived from it using the formulas shown in NullTetrad, then transformed.

Two different formats for the output of TransformTetrad are possible: as a matrix that can be set as the value of the tetrad e_ using Setup, or as a list of null vectors (option output = nullvectors).

Covering the effect of the 6-parameter Lorentz transformations on the null vectors of the Newman-Penrose formalism, the predefined transformations implemented are:

I) Rotations of Class I leave the null vector l_ unchanged (option nullrotationwithfixedl_, introduces a global transformation complex parameter E)

II) Rotations of Class II leave the null vector n_ unchanged (option nullrotationwithfixedn_, introduces a global transformation complex parameter B)

III) Rotations of Class III leave the directions of l_ and n_ unchanged and rotate m_ and mb_ on the (m_, mb_)-plane. To these transformations correspond the options boostsn_l_plane, that introduces a global transformation real parameter A,

and also spatialrotationsm_mb_plane, that introduces a global transformation real parameter Omega).

To transform a null tetrad using any of these standard transformations, pass any (one or more) of the corresponding optional arguments, for example as in TransformTetrad(spatialrotationsm_mb_plane). When indicating many transformations by passing many of these optional arguments together, the transformations are performed in the order these optional arguments are passed.

NOTE (I): You can always pass these optional arguments by passing only a few identifying letters of these long concatenations of words, and when there is only one match your input will be interpreted correctly. For example: Both TransformTetrad(spatial) and TransformTetrad(m_mb) are interpreted as TransformTetrad(spatialrotationsm_mb_plane).

NOTE (II): Transformations of Class I and II introduce complex parameters E and B. When these parameters are going to be adjusted to obtain a certain form of the tetrad, or the transformed tetrad is going to be simplified, it is convenient to represent them by their real and imaginary parts, for example using $E_x=\mathrm{\Re }\left(E\right)$ and $E_y=\mathrm{\Im }\left(E\right)$ and setting assumptions $\mathrm{Assume}\left(E_x::\mathrm{real}\right)$ and $\mathrm{Assume}\left(E_y::\mathrm{real}\right)$ using Assume and the same for B and the real parameters A and omega.

Other frequent transformations used that have a keyword associated that you can use to indicate them are those that preserve the l_ direction plane (option preservel_direction, introduces the global transformation parameters A, E, Omega) and those that preserve the n_ direction plane (option preserven_direction, introduces the global transformation parameters A, B, Omega)

Finally, you can also indicate any particular transformation by passing it as a set or list of transformation equations (see TransformCoordinates), in which case the null tetrad is viewed as the corresponding list of four spacetime 4-vectors [l_, n_, m_, mb_] and transformed accordingly using the TransformCoordinates command.

Provided that the spacetime is not conformally flat (for which all the components of the Weyl tensor are equal to zero), TransformTetrad can also transform a tetrad into a canonical form in one go. For that purpose, use the canonicalform optional argument. The resulting tetrad is such that the Weyl scalars associated to it are of one and only one of the following five forms related to the PetrovType of the spacetime (see reference [3] page 388)

The process of rewriting a tetrad into canonical form involves a combination of rotations of Classes I, II and III, automatically deduced and applied by TransformTetrad.

The canonical form in the table above is not the only possible one and, depending on the spacetime, a different canonical form would be more convenient. For example, a preferred form for Petrov type "I" would be to have ${\mathrm{\Psi }}_1=1,{\mathrm{\Psi }}_2\ne 0,{\mathrm{\Psi }}_4\ne 0$ instead of ${\mathrm{\Psi }}_3=1,{\mathrm{\Psi }}_1\ne 0,{\mathrm{\Psi }}_2\ne 0$. You can still use TransformTetrad to produce any desired canonical form different than the default one presented in the table above by taking into account the way the Weyl scalars are changed by these three classes of rotation, that is:

I) Rotations of Class I leave ${\mathrm{\Psi }}_0$ unchanged

I) Rotations of Class II leave ${\mathrm{\Psi }}_4$ unchanged

I) Rotations of Class III leave ${\mathrm{\Psi }}_2$ unchanged

$\mathrm{with}\left(\mathrm{Physics}\right)\:$$\mathrm{with}\left(\mathrm{Tetrads}\right)$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\mathrm{Setting}\mathrm{lowercaselatin\_ah}\mathrm{letters\; to\; represent}\mathrm{tetrad}\mathrm{indices}$

$\mathrm{Defined\; as\; tetrad\; tensors}\left(\mathrm{see\; ?Physics,tetrads}\right),{𝔢}_{a,\mathrm{\mu }},{\mathrm{\eta }}_{a,b},{\mathrm{\gamma }}_{a,b,c},{\mathrm{\lambda }}_{a,b,c}$

$\mathrm{Defined\; as\; spacetime\; tensors\; representing\; the\; NP\; null\; vectors\; of\; the\; tetrad\; formalism}\left(\mathrm{see\; ?Physics,tetrads}\right),l_{\mathrm{\mu }},n_{\mathrm{\mu }},m_{\mathrm{\mu }},{\stackrel{\&conjugate0;}{m}}_{\mathrm{\mu }}$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\left[\mathrm{IsTetrad}\,\mathrm{NullTetrad}\,\mathrm{OrthonormalTetrad}\,\mathrm{PetrovType}\,\mathrm{SegreType}\,\mathrm{TransformTetrad}\,\mathrm{WeylScalars}\,\mathrm{e\_}\,\mathrm{eta\_}\,\mathrm{gamma\_}\,\mathrm{l\_}\,\mathrm{lambda\_}\,\mathrm{m\_}\,\mathrm{mb\_}\,\mathrm{n\_}\right]$

$\mathrm{Setup}\left(\mathrm{mathematicalnotation}=\mathrm{true}\right)$

$\left[\mathrm{mathematicalnotation}=\mathrm{true}\right]$

$\mathrm{g\_}\left[\mathrm{sc}\right]$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\mathrm{Systems\; of\; spacetime\; coordinates\; are:}\left\{X=\left(r\,\mathrm{\theta }\,\mathrm{\phi }\,t\right)\right\}$

$\mathrm{Default\; differentiation\; variables\; for\; d\_,\; D\_\; and\; dAlembertian\; are:}\left\{X=\left(r\,\mathrm{\theta }\,\mathrm{\phi }\,t\right)\right\}$

$\mathrm{Setting}\mathrm{lowercaselatin\_is}\mathrm{letters\; to\; represent}\mathrm{space}\mathrm{indices}$

$\mathrm{The\; Schwarzschild\; metric\; in\; coordinates}\left[r\,\mathrm{\theta }\,\mathrm{\phi }\,t\right]$

$\mathrm{Parameters:}\left[m\right]$

$\mathrm{Signature:}\left(\mathrm{-\; -\; -\; +}\right)$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$g_{\mathrm{\mu },\mathrm{\nu }}=\left[\begin{array}{cccc}\frac{r}{2m-r}& 0& 0& 0\\ 0& -r^2& 0& 0\\ 0& 0& -r^2{\sin \left(\mathrm{\theta }\right)}^2& 0\\ 0& 0& 0& \frac{r-2m}{r}\end{array}\right]$

$\mathrm{e\_}\left[\right]$

${𝔢}_{a,\mathrm{\mu }}=\left[\begin{array}{cccc}\frac{\mathrm{-I}\sqrt{r}}{\sqrt{2m-r}}& 0& 0& 0\\ 0& r& 0& 0\\ 0& 0& r\sin \left(\mathrm{\theta }\right)& 0\\ 0& 0& 0& \frac{\mathrm{-I}\sqrt{2m-r}}{\sqrt{r}}\end{array}\right]$

$\mathrm{l\_}\left[\mathrm{definition}\right]$

$l_{\mathrm{\mu }}{l}_{\phantom{}\phantom{\mathrm{\mu }}}^{\phantom{}\mathrm{\mu }}=0,l_{\mathrm{\mu }}{n}_{\phantom{}\phantom{\mathrm{\mu }}}^{\phantom{}\mathrm{\mu }}=1,l_{\mathrm{\mu }}{m}_{\phantom{}\phantom{\mathrm{\mu }}}^{\phantom{}\mathrm{\mu }}=0,l_{\mathrm{\mu }}{\stackrel{\&conjugate0;}{m}}_{\phantom{}\phantom{\mathrm{\mu }}}^{\phantom{}\mathrm{\mu }}=0,g_{\mathrm{\mu },\mathrm{\nu }}=l_{\mathrm{\mu }}{n}_{\mathrm{\nu }}+l_{\mathrm{\nu }}{n}_{\mathrm{\mu }}-m_{\mathrm{\mu }}{\stackrel{\&conjugate0;}{m}}_{\mathrm{\nu }}-m_{\mathrm{\nu }}{\stackrel{\&conjugate0;}{m}}_{\mathrm{\mu }}$

$\mathrm{TransformTetrad}\left(\mathrm{nullrotationwithfixedl\_}\right)$

$\left[\begin{array}{cccc}\frac{\stackrel{\&conjugate0;}{E}\sqrt{2}\sqrt{r}}{2\sqrt{2m-r}}& \frac{\mathrm{I}}{2}\sqrt{2}r& \frac{\sqrt{2}r\sin \left(\mathrm{\theta }\right)}{2}& \frac{\stackrel{\&conjugate0;}{E}\sqrt{2}\sqrt{2m-r}}{2\sqrt{r}}\\ \frac{\sqrt{2}\sqrt{r}}{2\sqrt{2m-r}}& 0& 0& \frac{\sqrt{2}\sqrt{2m-r}}{2\sqrt{r}}\\ \frac{\sqrt{2}\sqrt{r}\left(E\stackrel{\&conjugate0;}{E}-1\right)}{2\sqrt{2m-r}}& \frac{\mathrm{I}}{2}\sqrt{2}r\left(E+\stackrel{\&conjugate0;}{E}\right)& \frac{\sqrt{2}\sin \left(\mathrm{\theta }\right)r\left(E-\stackrel{\&conjugate0;}{E}\right)}{2}& \frac{\sqrt{2}\sqrt{2m-r}\left(E\stackrel{\&conjugate0;}{E}+1\right)}{2\sqrt{r}}\\ \frac{E\sqrt{2}\sqrt{r}}{2\sqrt{2m-r}}& \frac{\mathrm{I}}{2}\sqrt{2}r& -\frac{\sqrt{2}r\sin \left(\mathrm{\theta }\right)}{2}& \frac{E\sqrt{2}\sqrt{2m-r}}{2\sqrt{r}}\end{array}\right]$

$\mathrm{Assume}\left(E::\mathrm{real}\right)$

$\left\{E::\mathrm{real}\right\}$

$\mathrm{map}\left(\mathrm{eval}\,\right)$

$\left[\begin{array}{cccc}\frac{E\sqrt{2}\sqrt{r}}{2\sqrt{2m-r}}& \frac{\mathrm{I}}{2}\sqrt{2}r& \frac{\sqrt{2}r\sin \left(\mathrm{\theta }\right)}{2}& \frac{E\sqrt{2}\sqrt{2m-r}}{2\sqrt{r}}\\ \frac{\sqrt{2}\sqrt{r}}{2\sqrt{2m-r}}& 0& 0& \frac{\sqrt{2}\sqrt{2m-r}}{2\sqrt{r}}\\ \frac{\sqrt{2}\sqrt{r}\left(E^2-1\right)}{2\sqrt{2m-r}}& \mathrm{I}E\sqrt{2}r& 0& \frac{\sqrt{2}\sqrt{2m-r}\left(E^2+1\right)}{2\sqrt{r}}\\ \frac{E\sqrt{2}\sqrt{r}}{2\sqrt{2m-r}}& \frac{\mathrm{I}}{2}\sqrt{2}r& -\frac{\sqrt{2}r\sin \left(\mathrm{\theta }\right)}{2}& \frac{E\sqrt{2}\sqrt{2m-r}}{2\sqrt{r}}\end{array}\right]$

$\mathrm{IsTetrad}\left(\right)$

$\mathrm{Type\; of\; tetrad:}\mathrm{null}$

$\mathrm{true}$

$\mathrm{Setup}\left(\mathrm{e\_}=\right)$

$\left[\mathrm{tetrad}=\left\{\left(1\,1\right)=\frac{E\sqrt{2}\sqrt{r}}{2\sqrt{2m-r}}\,\left(1\,2\right)=\frac{\mathrm{I}}{2}\sqrt{2}r\,\left(1\,3\right)=\frac{\sqrt{2}r\sin \left(\mathrm{\theta }\right)}{2}\,\left(1\,4\right)=\frac{E\sqrt{2}\sqrt{2m-r}}{2\sqrt{r}}\,\left(2\,1\right)=\frac{\sqrt{2}\sqrt{r}}{2\sqrt{2m-r}}\,\left(2\,4\right)=\frac{\sqrt{2}\sqrt{2m-r}}{2\sqrt{r}}\,\left(3\,1\right)=\frac{\sqrt{2}\sqrt{r}\left(E^2-1\right)}{2\sqrt{2m-r}}\,\left(3\,2\right)=\mathrm{I}E\sqrt{2}r\,\left(3\,4\right)=\frac{\sqrt{2}\sqrt{2m-r}\left(E^2+1\right)}{2\sqrt{r}}\,\left(4\,1\right)=\frac{E\sqrt{2}\sqrt{r}}{2\sqrt{2m-r}}\,\left(4\,2\right)=\frac{\mathrm{I}}{2}\sqrt{2}r\,\left(4\,3\right)=-\frac{\sqrt{2}r\sin \left(\mathrm{\theta }\right)}{2}\,\left(4\,4\right)=\frac{E\sqrt{2}\sqrt{2m-r}}{2\sqrt{r}}\right\}\right]$

$\mathrm{gamma\_}\left[\mathrm{definition}\right]$

${\mathrm{\gamma }}_{a,b,c}={▿}_{\mathrm{\nu }}\left({𝔢}_{a,\mathrm{\mu }}\right){𝔢}_{b\phantom{\mathrm{\mu }}}^{\phantom{b}\mathrm{\mu }}{𝔢}_{c\phantom{\mathrm{\nu }}}^{\phantom{c}\mathrm{\nu }}$

$\mathrm{g\_}\left[\left[12\,8\,1\right]\right]$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\mathrm{Systems\; of\; spacetime\; coordinates\; are:}\left\{X=\left(x\,y\,z\,t\right)\right\}$

$\mathrm{Default\; differentiation\; variables\; for\; d\_,\; D\_\; and\; dAlembertian\; are:}\left\{X=\left(x\,y\,z\,t\right)\right\}$

$\mathrm{The}\left[x\,y\,z\,t\right]\mathrm{metric\; in\; coordinates\; \%2}$

$\mathrm{Parameters:}\left[A\,B\right]$

$\mathrm{Comments:}\mathrm{cosmological\; constant\; =\; 1/B^2\; =\; 1/A^2,\; k\; =\; 1,\; kprim\ⅇ\; =\; 1}$

$\mathrm{Domain:}\left[0<x\&comma;x<\mathrm{\pi }\&comma;0<z\&comma;z<\mathrm{\pi }\right]$

$\mathrm{Assumptions:}\left[B=A\&comma;0<A\&comma;0<B\right]$

$\mathrm{Resetting\; the\; signature\; of\; spacetime\; from}\left(\mathrm{-\; -\; -\; +}\right)\mathrm{to}\left(\mathrm{+\; +\; +\; -}\right)\mathrm{in\; order\; to\; match\; the\; signature\; in\; the\; database\; of\; metrics}$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$g_{\mathrm{\mu },\mathrm{\nu }}=\left[\begin{array}{cccc}{A}^2& 0& 0& 0\\ 0& A^2{\sin \left(x\right)}^2& 0& 0\\ 0& 0& B^2& 0\\ 0& 0& 0& -B^2{\sin \left(z\right)}^2\end{array}\right]$

$\mathrm{Assume}\left(0<z\mathbf{and}z\le \frac{1}{2}\mathrm{\pi }\&comma;0<x\mathbf{and}x\le \frac{1}{2}\mathrm{\pi }\right)$

$\left\{z::\left(0\&comma;\frac{\mathrm{\pi }}{2}\right]\right\},\left\{x::\left(0\&comma;\frac{\mathrm{\pi }}{2}\right]\right\}$

$\mathrm{PetrovType}\left(\right)$

$''D''$

$\mathrm{SegreType}\left(\right)$

$''D'',''[(1,1)(11)]''$

a) the other metric has the same classification (the Segre classification is necessary only when the Petrov type is "O")

b) a transformation of the spacetime coordinates (that you can perform with TransformCoordinates) exists and transforms one set of scalars into the other one

$\mathrm{e\_}\left[\right]$

${𝔢}_{a,\mathrm{\mu }}=\left[\begin{array}{cccc}-\frac{\sqrt{2}A}{2}& 0& 0& \frac{\sqrt{2}B\sin \left(z\right)}{2}\\ 0& \frac{\sqrt{2}A\sin \left(x\right)}{2}& \frac{\mathrm{I}}{2}\sqrt{2}B& 0\\ 0& \frac{\sqrt{2}A\sin \left(x\right)}{2}& -\frac{\mathrm{I}}{2}\sqrt{2}B& 0\\ \frac{\sqrt{2}A}{2}& 0& 0& \frac{\sqrt{2}B\sin \left(z\right)}{2}\end{array}\right]$

$\mathrm{Weyl}\left[\mathrm{scalarsdefinition}\right]$

$\mathrm{\&psi;\_\_0}=C_{\phantom{}\phantom{\mathrm{\mu },\mathrm{\nu },\mathrm{\alpha },\mathrm{\beta }}}^{\phantom{}\mathrm{\mu },\mathrm{\nu },\mathrm{\alpha },\mathrm{\beta }}{l}_{\mathrm{\mu }}{m}_{\mathrm{\nu }}{l}_{\mathrm{\alpha }}{m}_{\mathrm{\beta }},\mathrm{\&psi;\_\_1}=C_{\phantom{}\phantom{\mathrm{\mu },\mathrm{\nu },\mathrm{\alpha },\mathrm{\beta }}}^{\phantom{}\mathrm{\mu },\mathrm{\nu },\mathrm{\alpha },\mathrm{\beta }}{l}_{\mathrm{\mu }}{n}_{\mathrm{\nu }}{l}_{\mathrm{\alpha }}{m}_{\mathrm{\beta }},\mathrm{\&psi;\_\_2}=C_{\phantom{}\phantom{\mathrm{\mu },\mathrm{\nu },\mathrm{\alpha },\mathrm{\beta }}}^{\phantom{}\mathrm{\mu },\mathrm{\nu },\mathrm{\alpha },\mathrm{\beta }}{l}_{\mathrm{\mu }}{m}_{\mathrm{\nu }}{\stackrel{\&conjugate0;}{m}}_{\mathrm{\alpha }}{n}_{\mathrm{\beta }},\mathrm{\&psi;\_\_3}=C_{\phantom{}\phantom{\mathrm{\mu },\mathrm{\nu },\mathrm{\alpha },\mathrm{\beta }}}^{\phantom{}\mathrm{\mu },\mathrm{\nu },\mathrm{\alpha },\mathrm{\beta }}{l}_{\mathrm{\mu }}{n}_{\mathrm{\nu }}{\stackrel{\&conjugate0;}{m}}_{\mathrm{\alpha }}{n}_{\mathrm{\beta }},\mathrm{\&psi;\_\_4}=C_{\phantom{}\phantom{\mathrm{\mu },\mathrm{\nu },\mathrm{\alpha },\mathrm{\beta }}}^{\phantom{}\mathrm{\mu },\mathrm{\nu },\mathrm{\alpha },\mathrm{\beta }}{n}_{\mathrm{\mu }}{\stackrel{\&conjugate0;}{m}}_{\mathrm{\nu }}{n}_{\mathrm{\alpha }}{\stackrel{\&conjugate0;}{m}}_{\mathrm{\beta }}$

$\mathrm{l\_}\left[\right]$

$l_{\mathrm{\mu }}=\left[\begin{array}{cccc}\frac{\sqrt{2}A}{2}& 0& 0& \frac{\sqrt{2}B\sin \left(z\right)}{2}\end{array}\right]$

$\mathrm{n\_}\left[\right]$

$n_{\mathrm{\mu }}=\left[\begin{array}{cccc}-\frac{\sqrt{2}A}{2}& 0& 0& \frac{\sqrt{2}B\sin \left(z\right)}{2}\end{array}\right]$

$\mathrm{m\_}\left[\right]$

$m_{\mathrm{\mu }}=\left[\begin{array}{cccc}0& \frac{\sqrt{2}A\sin \left(x\right)}{2}& \frac{\mathrm{I}}{2}\sqrt{2}B& 0\end{array}\right]$

$\mathrm{mb\_}\left[\right]$

${\stackrel{\&conjugate0;}{m}}_{\mathrm{\mu }}=\left[\begin{array}{cccc}0& \frac{\sqrt{2}A\sin \left(x\right)}{2}& -\frac{\mathrm{I}}{2}\sqrt{2}B& 0\end{array}\right]$

$\mathrm{Weyl}\left[\mathrm{scalars}\right]$

$\mathrm{\&psi;\_\_0}=\frac{A^2+B^2}{4A^2{B}^2},\mathrm{\&psi;\_\_1}=0,\mathrm{\&psi;\_\_2}=\frac{A^2+B^2}{12A^2{B}^2},\mathrm{\&psi;\_\_3}=0,\mathrm{\&psi;\_\_4}=\frac{A^2+B^2}{4A^2{B}^2}$

$\mathrm{TransformTetrad}\left(\mathrm{canonicalform}\right)$

$\left[\begin{array}{cccc}\frac{\sqrt{2}A}{4}& -\frac{\mathrm{I}}{4}\sqrt{2}A\sin \left(x\right)& \frac{\sqrt{2}B}{4}& -\frac{\sqrt{2}B\sin \left(z\right)}{4}\\ \frac{\mathrm{I}}{2}\sqrt{2}A& -\frac{\sqrt{2}A\sin \left(x\right)}{2}& -\frac{\mathrm{I}}{2}\sqrt{2}B& -\frac{\mathrm{I}}{2}\sqrt{2}B\sin \left(z\right)\\ -\frac{3\mathrm{I}}{8}A\sqrt{2}& -\frac{3\sqrt{2}A\sin \left(x\right)}{8}& \frac{\mathrm{I}}{8}B\sqrt{2}& -\frac{\mathrm{I}}{8}B\sin \left(z\right)\sqrt{2}\\ -\frac{\sqrt{2}A}{4}& -\frac{\mathrm{I}}{4}\sqrt{2}A\sin \left(x\right)& -\frac{3\sqrt{2}B}{4}& -\frac{3\sqrt{2}B\sin \left(z\right)}{4}\end{array}\right]$

$\mathrm{Setup}\left(\mathrm{e\_}=\right)$

$\left[\mathrm{tetrad}=\left\{\left(1\&comma;1\right)=\frac{\sqrt{2}A}{4}\&comma;\left(1\&comma;2\right)=-\frac{\mathrm{I}}{4}\sqrt{2}A\sin \left(x\right)\&comma;\left(1\&comma;3\right)=\frac{\sqrt{2}B}{4}\&comma;\left(1\&comma;4\right)=-\frac{\sqrt{2}B\sin \left(z\right)}{4}\&comma;\left(2\&comma;1\right)=\frac{\mathrm{I}}{2}\sqrt{2}A\&comma;\left(2\&comma;2\right)=-\frac{\sqrt{2}A\sin \left(x\right)}{2}\&comma;\left(2\&comma;3\right)=-\frac{\mathrm{I}}{2}\sqrt{2}B\&comma;\left(2\&comma;4\right)=-\frac{\mathrm{I}}{2}\sqrt{2}B\sin \left(z\right)\&comma;\left(3\&comma;1\right)=-\frac{3\mathrm{I}}{8}A\sqrt{2}\&comma;\left(3\&comma;2\right)=-\frac{3\sqrt{2}A\sin \left(x\right)}{8}\&comma;\left(3\&comma;3\right)=\frac{\mathrm{I}}{8}B\sqrt{2}\&comma;\left(3\&comma;4\right)=-\frac{\mathrm{I}}{8}B\sin \left(z\right)\sqrt{2}\&comma;\left(4\&comma;1\right)=-\frac{\sqrt{2}A}{4}\&comma;\left(4\&comma;2\right)=-\frac{\mathrm{I}}{4}\sqrt{2}A\sin \left(x\right)\&comma;\left(4\&comma;3\right)=-\frac{3\sqrt{2}B}{4}\&comma;\left(4\&comma;4\right)=-\frac{3\sqrt{2}B\sin \left(z\right)}{4}\right\}\right]$

$\mathrm{Weyl}\left[\mathrm{scalars}\right]$

$\mathrm{\&psi;\_\_0}=0,\mathrm{\&psi;\_\_1}=0,\mathrm{\&psi;\_\_2}=\frac{-A^2-B^2}{6A^2{B}^2},\mathrm{\&psi;\_\_3}=0,\mathrm{\&psi;\_\_4}=0$

perform a Class III transformation using TransformTetrad(boostsn_l_plane, spatialrotationsm_mb_plane)

set the resulting tetrad using Setup as done above

recompute  the Weyl scalars: you obtain again scalars with only ${\mathrm{\Psi }}_2\ne 0$:

$\mathrm{TransformTetrad}\left(\mathrm{boostsn\_l\_plane}\&comma;\mathrm{spatialrotationsm\_mb\_plane}\right)$

$\left[\begin{array}{cccc}\frac{3{\&ExponentialE;}^{\mathrm{-I}\mathrm{\Omega }}\sqrt{2}A}{8}& -\frac{3\mathrm{I}}{8}{\&ExponentialE;}^{\mathrm{-I}\mathrm{\Omega }}\sqrt{2}A\sin \left(x\right)& -\frac{{\&ExponentialE;}^{\mathrm{-I}\mathrm{\Omega }}\sqrt{2}B}{8}& \frac{{\&ExponentialE;}^{\mathrm{-I}\mathrm{\Omega }}\sqrt{2}B\sin \left(z\right)}{8}\\ \frac{-\frac{\mathrm{I}}{4}\sqrt{2}A}{\mathrm{A\_\_1}}& \frac{\sqrt{2}A\sin \left(x\right)}{4\mathrm{A\_\_1}}& \frac{-\frac{3\mathrm{I}}{4}\sqrt{2}B}{\mathrm{A\_\_1}}& \frac{-\frac{3\mathrm{I}}{4}\sqrt{2}B\sin \left(z\right)}{\mathrm{A\_\_1}}\\ \frac{\mathrm{I}}{4}\mathrm{A\_\_1}\sqrt{2}A& \frac{\mathrm{A\_\_1}\sqrt{2}A\sin \left(x\right)}{4}& \frac{\mathrm{I}}{4}\mathrm{A\_\_1}\sqrt{2}B& -\frac{\mathrm{I}}{4}\mathrm{A\_\_1}\sqrt{2}B\sin \left(z\right)\\ -\frac{{\&ExponentialE;}^{\mathrm{I}\mathrm{\Omega }}\sqrt{2}A}{2}& -\frac{\mathrm{I}}{2}{\&ExponentialE;}^{\mathrm{I}\mathrm{\Omega }}\sqrt{2}A\sin \left(x\right)& \frac{{\&ExponentialE;}^{\mathrm{I}\mathrm{\Omega }}\sqrt{2}B}{2}& \frac{{\&ExponentialE;}^{\mathrm{I}\mathrm{\Omega }}\sqrt{2}B\sin \left(z\right)}{2}\end{array}\right]$

$\mathrm{Setup}\left(\mathrm{e\_}=\right)$

$\left[\mathrm{tetrad}=\left\{\left(1\&comma;1\right)=\frac{3{\&ExponentialE;}^{\mathrm{-I}\mathrm{\Omega }}\sqrt{2}A}{8}\&comma;\left(1\&comma;2\right)=-\frac{3\mathrm{I}}{8}{\&ExponentialE;}^{\mathrm{-I}\mathrm{\Omega }}\sqrt{2}A\sin \left(x\right)\&comma;\left(1\&comma;3\right)=-\frac{{\&ExponentialE;}^{\mathrm{-I}\mathrm{\Omega }}\sqrt{2}B}{8}\&comma;\left(1\&comma;4\right)=\frac{{\&ExponentialE;}^{\mathrm{-I}\mathrm{\Omega }}\sqrt{2}B\sin \left(z\right)}{8}\&comma;\left(2\&comma;1\right)=\frac{-\frac{\mathrm{I}}{4}\sqrt{2}A}{\mathrm{A\_\_1}}\&comma;\left(2\&comma;2\right)=\frac{\sqrt{2}A\sin \left(x\right)}{4\mathrm{A\_\_1}}\&comma;\left(2\&comma;3\right)=\frac{-\frac{3\mathrm{I}}{4}\sqrt{2}B}{\mathrm{A\_\_1}}\&comma;\left(2\&comma;4\right)=\frac{-\frac{3\mathrm{I}}{4}\sqrt{2}B\sin \left(z\right)}{\mathrm{A\_\_1}}\&comma;\left(3\&comma;1\right)=\frac{\mathrm{I}}{4}\mathrm{A\_\_1}\sqrt{2}A\&comma;\left(3\&comma;2\right)=\frac{\mathrm{A\_\_1}\sqrt{2}A\sin \left(x\right)}{4}\&comma;\left(3\&comma;3\right)=\frac{\mathrm{I}}{4}\mathrm{A\_\_1}\sqrt{2}B\&comma;\left(3\&comma;4\right)=-\frac{\mathrm{I}}{4}\mathrm{A\_\_1}\sqrt{2}B\sin \left(z\right)\&comma;\left(4\&comma;1\right)=-\frac{{\&ExponentialE;}^{\mathrm{I}\mathrm{\Omega }}\sqrt{2}A}{2}\&comma;\left(4\&comma;2\right)=-\frac{\mathrm{I}}{2}{\&ExponentialE;}^{\mathrm{I}\mathrm{\Omega }}\sqrt{2}A\sin \left(x\right)\&comma;\left(4\&comma;3\right)=\frac{{\&ExponentialE;}^{\mathrm{I}\mathrm{\Omega }}\sqrt{2}B}{2}\&comma;\left(4\&comma;4\right)=\frac{{\&ExponentialE;}^{\mathrm{I}\mathrm{\Omega }}\sqrt{2}B\sin \left(z\right)}{2}\right\}\right]$

$\mathrm{Weyl}\left[\mathrm{scalars}\right]$

$\mathrm{\&psi;\_\_0}=0,\mathrm{\&psi;\_\_1}=0,\mathrm{\&psi;\_\_2}=\frac{-A^2-B^2}{6A^2{B}^2},\mathrm{\&psi;\_\_3}=0,\mathrm{\&psi;\_\_4}=0$

$\mathrm{g\_}\left[\left[12\&comma;35\&comma;1\right]\right]$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\mathrm{Systems\; of\; spacetime\; coordinates\; are:}\left\{X=\left(x\&comma;y\&comma;z\&comma;u\right)\right\}$

$\mathrm{Default\; differentiation\; variables\; for\; d\_,\; D\_\; and\; dAlembertian\; are:}\left\{X=\left(x\&comma;y\&comma;z\&comma;u\right)\right\}$

$\mathrm{The}\mathrm{Kaigorodov\; (1962)}\mathrm{metric\; in\; coordinates}\mathrm{Cahen\; (1964)}$

$\mathrm{Parameters:}\left[\mathrm{\Lambda }\right]$

$\mathrm{Assumptions:}\left[\mathrm{\Lambda }<0\right]$

$\mathrm{Signature:}\left(\mathrm{+\; +\; +\; -}\right)$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$g_{\mathrm{\mu },\mathrm{\nu }}=\left[\begin{array}{cccc}{\&ExponentialE;}^{4z}& 2{\&ExponentialE;}^z& 0& {\&ExponentialE;}^{-2z}\\ 2{\&ExponentialE;}^z& 2{\&ExponentialE;}^{-2z}& 0& 0\\ 0& 0& -\frac{3}{\mathrm{\Lambda }}& 0\\ {\&ExponentialE;}^{-2z}& 0& 0& 0\end{array}\right]$

$\mathrm{Assume}\left(0<z\&comma;0<\mathrm{\Lambda }\right)$

$\left\{z::\left(0\&comma;\mathrm{\infty }\right)\right\},\left\{\mathrm{\Lambda }::\left(0\&comma;\mathrm{\infty }\right)\right\}$

$\mathrm{PetrovType}\left(\right)$

$''III''$

$\mathrm{e\_}\left[\right]$

${𝔢}_{a,\mathrm{\mu }}=\left[\begin{array}{cccc}-\frac{\mathrm{I}}{2}\sqrt{2}{\&ExponentialE;}^{2z}& \mathrm{-I}{\&ExponentialE;}^{-z}\left(1+\sqrt{2}\right)& 0& -\frac{\mathrm{I}}{2}{\&ExponentialE;}^{-4z}\left(2+\sqrt{2}\right)\\ 0& 0& \frac{\frac{\mathrm{I}}{2}\sqrt{6}}{\sqrt{\mathrm{\Lambda }}}& \frac{\mathrm{I}}{2}\sqrt{2}{\&ExponentialE;}^{-4z}\\ 0& 0& \frac{\frac{\mathrm{I}}{2}\sqrt{6}}{\sqrt{\mathrm{\Lambda }}}& -\frac{\mathrm{I}}{2}\sqrt{2}{\&ExponentialE;}^{-4z}\\ -\frac{\mathrm{I}}{2}\sqrt{2}{\&ExponentialE;}^{2z}& \mathrm{-I}{\&ExponentialE;}^{-z}\left(\sqrt{2}-1\right)& 0& -\frac{\mathrm{I}}{2}{\&ExponentialE;}^{-4z}\left(-2+\sqrt{2}\right)\end{array}\right]$

$\mathrm{Weyl}\left[\mathrm{scalars}\right]$

$\mathrm{\&psi;\_\_0}=\mathrm{\Lambda }\left(-\frac{11}{4}+2\sqrt{2}\right),\mathrm{\&psi;\_\_1}=-\frac{3\mathrm{\Lambda }}{4}+\frac{\mathrm{\Lambda }\sqrt{2}}{2},\mathrm{\&psi;\_\_2}=-\frac{\mathrm{\Lambda }}{4},\mathrm{\&psi;\_\_3}=\frac{3\mathrm{\Lambda }}{4}+\frac{\mathrm{\Lambda }\sqrt{2}}{2},\mathrm{\&psi;\_\_4}=\mathrm{\Lambda }\left(-\frac{11}{4}-2\sqrt{2}\right)$

$\mathrm{Setup}\left(\mathrm{e\_}=\mathrm{TransformTetrad}\left(\mathrm{canonicalform}\right)\right)\&colon;$

$\mathrm{Weyl}\left[\mathrm{scalars}\right]$

$\mathrm{\&psi;\_\_0}=0,\mathrm{\&psi;\_\_1}=0,\mathrm{\&psi;\_\_2}=0,\mathrm{\&psi;\_\_3}=1,\mathrm{\&psi;\_\_4}=0$

$\mathrm{e\_}\left[\right]$

${𝔢}_{a,\mathrm{\mu }}=\left[\begin{array}{cccc}\frac{\left(55\sqrt{2}+14\right){\&ExponentialE;}^{2z}}{64}& \frac{\left(22\sqrt{2}+21\right){\&ExponentialE;}^{-z}}{32}& \frac{\sqrt{3}\left(43\sqrt{2}+44\right)}{64\sqrt{\mathrm{\Lambda }}}& \frac{11{\&ExponentialE;}^{-4z}}{8}\\ \frac{\left(11\sqrt{2}-6\right)\mathrm{\Lambda }{\&ExponentialE;}^{2z}}{32}& -\frac{11{\&ExponentialE;}^{-z}\mathrm{\Lambda }}{16}& \frac{11\sqrt{\mathrm{\Lambda }}\sqrt{6}}{32}& 0\\ \frac{\left(-6+5\sqrt{2}\right){\&ExponentialE;}^{2z}}{4\mathrm{\Lambda }}& \frac{\left(-1+2\sqrt{2}\right){\&ExponentialE;}^{-z}}{2\mathrm{\Lambda }}& \frac{\sqrt{3}\left(\sqrt{2}+4\right)}{4{\mathrm{\Lambda }}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}& \frac{2{\&ExponentialE;}^{-4z}}{\mathrm{\Lambda }}\\ \frac{\left(-2+\sqrt{2}\right){\&ExponentialE;}^{2z}}{2}& -{\&ExponentialE;}^{-z}& \frac{\sqrt{6}}{2\sqrt{\mathrm{\Lambda }}}& 0\end{array}\right]$