Hilbert Book Model Project/Quaternionic Field Equations/Nabla Operators/nl

<Hilbert Book Model Project/nl

Nabla speeltuin

De ruimtelijke nabla en quaternionische nabla zijn bijzondere operatoren die een belangrijke rol spelen bij de partiële differentiaalvergelijkingen die het gedrag bepalen van velden in het Hilbert Book Model.

Hier behandelen we drie soorten nabla operatoren.

1. ruimtelijke nabla ${\textstyle {\vec {\nabla }}={\vec {i}}{\frac {\partial {}}{\partial {x}}}+{\vec {j}}{\frac {\partial {}}{\partial {y}}}+{\vec {k}}{\frac {\partial {}}{\partial {z}}}}$
2. quaternionische nabla ${\textstyle \nabla ={\frac {\partial {}}{\partial {\tau }}}+{\vec {i}}{\frac {\partial {}}{\partial {x}}}+{\vec {j}}{\frac {\partial {}}{\partial {y}}}+{\vec {k}}{\frac {\partial {}}{\partial {z}}}=\nabla _{r}+{\vec {\nabla }}}$
3. Dirac nabla ${\textstyle \nabla ={\frac {\partial {}}{\partial {\tau }}}+\mathbb {I} \left\{{\vec {i}}{\frac {\partial {}}{\partial {x}}}+{\vec {j}}{\frac {\partial {}}{\partial {y}}}+{\vec {k}}{\frac {\partial {}}{\partial {z}}}\right\}=\nabla _{r}+\mathbb {I} \,{\vec {\nabla }}}$

De Dirac nabla speelt een rol bij de interpretatie van de  Dirac vergelijking.

Eigenschappen van de ruimtelijke nabla operator

Het nabla product is niet zonder meer associatief. Dus

$\nabla (\nabla ^{*}\psi )=\nabla ^{*}(\nabla \psi )=(\nabla _{r}\nabla _{r}+\langle {\vec {\nabla }},{\vec {\nabla }}\rangle )\psi \neq (\nabla \nabla ^{*})\psi =(\nabla ^{*}\nabla )\psi =(\nabla _{r}\nabla _{r}-{\vec {\nabla }}{\vec {\nabla }})\psi =(\nabla _{r}\nabla _{r}+\langle {\vec {\nabla }},{\vec {\nabla }}\rangle \mp {\vec {\nabla }}\times {\vec {\nabla }})\psi$

(1)

Nabla in verschillende coördinatenstelsels 

De ruimtelijke nabla bestaat in verschillende coördinatenstelsels. In deze sectie wordt de weergave van de quaternionische nabla voor cartesische coördinatensystemen en polaire coördinatensystemen.

${\vec {\nabla }}=\{{\frac {\partial {}}{\partial {x}}},\,{\frac {\partial {}}{\partial {y}}},\,{\frac {\partial {}}{\partial {z}}}\}={\frac {\partial {}}{\partial {x}}}{\vec {\hat {x}}}+{\frac {\partial {}}{\partial {y}}}{\vec {\hat {y}}}+{\frac {\partial {}}{\partial {z}}}{\vec {\hat {z}}}$

(2)

${\vec {\nabla }}a_{r}={\frac {\partial {}}{\partial {x}}}{\vec {\hat {x}}}+{\frac {\partial {}}{\partial {y}}}{\vec {\hat {y}}}+{\frac {\partial {}}{\partial {z}}}{\vec {\hat {z}}}={\frac {\partial {a_{r}}}{\partial {\rho }}}{\vec {\hat {\rho }}}+{\frac {1}{\rho }}{\frac {\partial {a_{r}}}{\partial {\theta }}}{\vec {\hat {\theta }}}+{\frac {1}{\rho \sin(\theta )}}{\frac {\partial {a_{r}}}{\partial {\phi }}}{\vec {\hat {\phi }}}$

(3)

Here $\{\rho ,\theta ,\phi \}$ zijn de coördinaten met  $\{{\vec {\hat {\rho }}},{\vec {\hat {\theta }}},{\vec {\hat {\phi }}}\}$ als coördinaat assen.

$\langle {\vec {\nabla }},{\vec {a}}\rangle ={\frac {\partial {a_{x}}}{\partial {x}}}{\vec {\hat {x}}}+{\frac {\partial {a_{y}}}{\partial {y}}}{\vec {\hat {y}}}+{\frac {\partial {a_{z}}}{\partial {z}}}{\vec {\hat {z}}}={\frac {1}{\rho ^{2}}}{\frac {\partial {(\rho ^{2}a_{\rho })}}{\partial {\rho }}}+{\frac {1}{\rho \sin(\theta )}}{\frac {\partial {(a_{\theta }\sin(\theta )}}{\partial {\theta )}}}+{\frac {1}{\rho \sin(\theta )}}{\frac {\partial {a_{\phi }}}{\partial {\phi }}}$

(4)

$\langle {\vec {\nabla }},{\vec {\nabla }}\rangle f_{r}={\frac {1}{\rho ^{2}}}{\frac {\partial {(\rho ^{2}{\frac {\partial {f_{r}}}{\partial \rho }})}}{\partial {\rho }}}+{\frac {1}{\rho ^{2}\sin(\theta )}}{\frac {\partial {(\sin(\theta ){\frac {\partial {f_{r}}}{\partial {\theta }}}}}{\partial {\theta )}}}+{\frac {1}{\rho ^{2}\sin ^{2}(\theta )}}{\frac {\partial ^{2}{f_{r}}}{\partial {\phi ^{2}}}}$

(5)

${\vec {\nabla }}\times {\vec {a}}=({\frac {\partial {a_{z}}}{\partial {y}}}-{\frac {\partial {a_{y}}}{\partial {z}}}){\vec {\hat {x}}}+({\frac {\partial {a_{x}}}{\partial {z}}}-{\frac {\partial {a_{z}}}{\partial {x}}}){\vec {\hat {y}}}+({\frac {\partial {a_{y}}}{\partial {x}}}-{\frac {\partial {a_{x}}}{\partial {y}}}){\vec {\hat {z}}}$

(6)

$\qquad ={\frac {1}{\rho \sin(\theta )}}({\frac {\partial {a_{\phi }}}{\partial {\phi }}}-{\frac {\partial {(a_{\theta }\sin(\theta )}}{\partial {\theta )}}}){\vec {\hat {\rho }}}+({\frac {1}{\rho ^{2}}}{\frac {\partial {(\rho ^{2}a_{\rho })}}{\partial {\rho }}}-{\frac {1}{\rho \sin(\theta )}}{\frac {\partial {a_{\phi }}}{\partial {\phi }}}){\vec {\hat {\theta }}}+({\frac {1}{\rho \sin(\theta )}}{\frac {\partial {(a_{\theta }\sin(\theta )}}{\partial {\theta )}}}-{\frac {1}{\rho ^{2}}}{\frac {\partial {(\rho ^{2}a_{\rho })}}{\partial {\rho }}}){\vec {\hat {\phi }}}$

(7)

${\vec {\nabla }}\times {\vec {a}}={\frac {1}{\rho \sin(\theta )}}({\frac {\partial {a_{\phi }}}{\partial {\phi }}}-{\frac {\partial {(a_{\theta }\sin(\theta )}}{\partial {\theta )}}}){\vec {\hat {\rho }}}+({\frac {1}{\rho ^{2}}}{\frac {\partial {(\rho ^{2}a_{\rho })}}{\partial {\rho }}}-{\frac {1}{\rho \sin(\theta )}}{\frac {\partial {a_{\phi }}}{\partial {\phi }}}){\vec {\hat {\theta }}}+({\frac {1}{\rho \sin(\theta )}}{\frac {\partial {(a_{\theta }\sin(\theta )}}{\partial {\theta )}}}-{\frac {1}{\rho ^{2}}}{\frac {\partial {(\rho ^{2}a_{\rho })}}{\partial {\rho }}}){\vec {\hat {\phi }}}$

(8)

Speciale formules

De ruimtelijke nabla operator blijkt gedrag geldt voor alle quaternionische functies waarvoor de eerste orde partiële differentiaalvergelijking bestaat.

Hier gehoorzaamt het quaternionische veld $a=a_{r}+{\vec {a}}$ aan de eis dat de eerste orde partiële differentiaal $\nabla a$ bestaat.

$\Delta \ {\overset {\underset {\mathrm {def} }{}}{=}}\ \nabla ^{2}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ \langle {\vec {\nabla }},{\vec {\nabla }}\rangle \ {\overset {\underset {\mathrm {def} }{}}{=}}\ {\frac {\partial {}^{2}}{\partial {x}^{2}}}+{\frac {\partial {}^{2}}{\partial {y}^{2}}}+{\frac {\partial {}^{2}}{\partial {z}^{2}}}$

(9)

$\langle {\vec {\nabla }},{\vec {\nabla }}a_{r}\rangle =\langle {\vec {\nabla }},{\vec {\nabla }}\rangle a_{r}=\nabla ^{2}a_{r}$

(10)

$\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {a}}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ \nabla ^{2}{\vec {a}}$

(11)

$\langle {\vec {\nabla }}\times {\vec {\nabla }},{\vec {a}}\rangle =0$

(12)

${\vec {\nabla }}\times {\vec {\nabla }}a_{r}={\vec {0}}$

(13)

$\langle {\vec {\nabla }},{\vec {\nabla }}\times {\vec {a}}\rangle =0$

(14)

${\vec {\nabla }}\times ({\vec {\nabla }}\times {\vec {a}})={\vec {\nabla }}\langle {\vec {\nabla }},{\vec {a}}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {a}}$

(15)

${\vec {\nabla }}({\vec {\nabla }}a)={\vec {\nabla }}({\vec {\nabla }}\times a-\langle {\vec {\nabla }},{\vec {a}}\rangle +{\vec {\nabla }}a_{r})={\vec {\nabla }}\times ({\vec {\nabla }}\times {\vec {a}})-{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {a}}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle a_{r}={\vec {\nabla }}\langle {\vec {\nabla }},{\vec {a}}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {a}}-{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {a}}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle a_{r}=-\langle {\vec {\nabla }},{\vec {\nabla }}\rangle a$

(16)

Voor constante ${\vec {k}}$ en parameter $q=q_{r}+{\vec {q}}=\{q_{r},q_{x},q_{y},q_{z}\}$ geldt

${\vec {\nabla }}\langle {\vec {k}},{\vec {q}}\rangle ={\vec {k}}$

(17)

$\langle {\vec {\nabla }},{\vec {q}}\rangle =3$

(18)

${\vec {\nabla }}\times {\vec {q}}={\vec {0}}$

(19)

${\vec {\nabla }}\,|{\vec {q}}|={\frac {\vec {q}}{|{\vec {q}}|}};\quad \nabla \,|q|={\frac {q}{|q|}}$

(20)

${\vec {\nabla }}{\frac {1}{|{\vec {q}}-{\vec {q}}^{'}|}}=-{\frac {{\vec {q}}-{\vec {q}}^{'}}{|{\vec {q}}-{\vec {q}}^{'}|^{3}}};\ \nabla {\frac {1}{|q-q^{'}|}}=-{\frac {q-q^{'}}{|q-q^{'}|^{3}}}$

(21)

$\langle {\vec {\nabla }},{\vec {\nabla }}{\frac {1}{|{\vec {q}}-{\vec {q}}^{'}|}}\rangle =\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\frac {1}{|{\vec {q}}-{\vec {q}}^{'}|}}=-\langle {\vec {\nabla }},{\frac {{\vec {q}}-{\vec {q}}^{'}}{|{\vec {q}}-{\vec {q}}^{'}|^{3}}}\rangle =4\,\phi \,\delta ({\vec {q}}-{\vec {q}}^{'})$

(22)

Dit geeft de overeenkomsten tussen de Poissonvergelijking en Green's functie..

$\nabla \ q=1-3;\quad \nabla ^{*}q=1+3;\quad \nabla q^{*}=1+3$

(23)

The term $({\vec {\nabla }}\times {\vec {\nabla }})f$ geeft the kromming van veld $f$.

The term $\langle {\vec {\nabla }},{\vec {\nabla }}\rangle f$ geeft de stress van het veld $f$

Eerste orde partiële differentiaalvergelijking

$\varphi =\varphi _{r}+{\vec {\varphi }}=\nabla \psi =(\nabla _{r}+{\vec {\nabla }})(\psi _{r}+{\vec {\psi }})=\nabla _{r}\psi _{r}-\langle {\vec {\nabla }},{\vec {\psi }}\rangle +\nabla _{r}{\vec {\psi }}+{\vec {\nabla }}\psi _{r}{\color {Red}\pm }{\vec {\nabla }}\times {\vec {\psi }}$

(24)

De vergelijking is een quaternionische eerste orde partiële differentiaalvergelijking.

De vijf termen aan de rechterzijde geven de componenten die de volledige eerste orde verandering vormen.

Zij vertegenwoordigen deelgebieden van het veld $\varphi$, en vaak krijgen ze speciale namen en symbolen.

Subvelden

${\vec {\nabla }}\psi _{r}$ is de gradient van $\psi _{r}$

$\langle {\vec {\nabla }},{\vec {\psi }}\rangle$ is de divergentie van ${\vec {\psi }}$.

${\vec {\nabla }}\times {\vec {\psi }}$ is de rotatie van ${\vec {\psi }}$

$\varphi _{r}=\nabla _{r}\psi _{r}-\langle {\vec {\nabla }},{\vec {\psi }}\rangle$

(25)

(Vergelijking (25) heeft geen equivalent in de vergelijkingen van Maxwell!)

${\vec {\varphi }}=\nabla _{r}{\vec {\psi }}+{\vec {\nabla }}\psi _{r}\pm {\vec {\nabla }}\times {\vec {\psi }}=\mp {\vec {B}}-{\vec {E}}$

(26)

${\vec {E}}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ -\nabla _{r}{\vec {\psi }}-{\vec {\nabla }}\psi _{r}$

(27)

${\vec {B}}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ {\vec {\nabla }}\times {\vec {\psi }}$

(28)

${\vec {J}}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ {\vec {\nabla }}\times {\vec {B}}-\nabla _{r}{\vec {E}}$

(29)

Afleiding van hogere orde vergelijkingen

Met behulp van de eigenschappen van de ruimtelijke nabla operator voert een interessante tweede orde partiële differentiaalvergelijking.

$\langle {\vec {\nabla }},{\vec {B}}\rangle =\langle {\vec {\nabla }},{\vec {\nabla }}\times {\vec {\psi }}\rangle =0$

(30)

$\langle {\vec {\nabla }},{\vec {E}}\rangle =-\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\psi _{r}\rangle =-\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle \psi _{r}$

(31)

$\langle {\vec {\nabla }},{\vec {B}}-{\vec {E}}\rangle =\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle +\langle {\vec {\nabla }},{\vec {\nabla }}\rangle \psi _{r}$

(32)

$\nabla _{r}{\vec {E}}=-\nabla _{r}\nabla _{r}{\vec {\psi }}-\nabla _{r}{\vec {\nabla }}\psi _{r}$

(33)

${\vec {\nabla }}\times {\vec {E}}=-{\vec {\nabla }}\times (\nabla _{r}{\vec {\psi }})-{\vec {\nabla }}\times ({\vec {\nabla }}\psi _{r})=-\nabla _{r}{\vec {\nabla }}\times {\vec {\psi }}=-\nabla _{r}{\vec {B}}$

(34)

${\vec {\nabla }}\times {\vec {B}}={\vec {\nabla }}\times ({\vec {\nabla }}\times {\vec {\psi }})={\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {\psi }}$

(35)

${\vec {\nabla }}\times {\vec {B}}-{\vec {\nabla }}\times {\vec {E}}={\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {\psi }}+\nabla _{r}{\vec {B}}$

(36)

${\vec {\nabla }}\times ({\vec {B}}-{\vec {E}})-\nabla _{r}({\vec {B}}-{\vec {E}})={\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {\psi }}-\nabla _{r}\nabla _{r}{\vec {\psi }}-\nabla _{r}{\vec {\nabla }}\psi _{r}$

(37)

${\vec {\nabla }}\times ({\vec {B}}-{\vec {E}})-\nabla _{r}({\vec {B}}-{\vec {E}})=-(\langle {\vec {\nabla }},{\vec {\nabla }}\rangle +\nabla _{r}\nabla _{r}){\vec {\psi }}-\nabla _{r}{\vec {\nabla }}\psi _{r}+{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle$

(38)

${\vec {\nabla }}\times ({\vec {B}}-{\vec {E}})-\langle {\vec {\nabla }},{\vec {B}}-{\vec {E}}\rangle -\nabla _{r}({\vec {B}}-{\vec {E}})=\nabla ^{*}{\vec {\varphi }}=-(\langle {\vec {\nabla }},{\vec {\nabla }}\rangle +\nabla _{r}\nabla _{r}){\vec {\psi }}-\nabla _{r}{\vec {\nabla }}\psi _{r}+{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle \psi _{r}$

(39)

Verder

${\vec {J}}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ {\vec {\nabla }}\times {\vec {B}}-\nabla _{r}{\vec {E}}={\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {\psi }}+\nabla _{r}\nabla _{r}{\vec {\psi }}+\nabla _{r}{\vec {\nabla }}\psi _{r}=(\nabla _{r}\nabla _{r}-\langle {\vec {\nabla }},{\vec {\nabla }}\rangle ){\vec {\psi }}+\nabla _{r}{\vec {\nabla }}\psi _{r}+{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle$

(40)

Uit bovenstaande formules volgt dat de Maxwell vergelijkingen geen volledige set vormen. Natuurkundigen gebruiken gauge vergelijkingen om Maxwell vergelijkingen completer te maken.

Afleiden van tweede orde partiële differentiaalvergelijking 1 

$\varphi =\nabla \psi =(\nabla _{r}+{\vec {\nabla }})\psi$

(41)

$\varphi _{r}=\nabla _{r}\psi _{r}-\langle {\vec {\nabla }},{\vec {\psi }}\rangle$

(42)

${\vec {\varphi }}=\nabla _{r}{\vec {\psi }}+{\vec {\nabla }}\psi _{r}\pm {\vec {\nabla }}\times {\vec {\psi }}$

(43)

${\vec {\nabla }}\varphi _{r}={\vec {\nabla }}(\nabla _{r}\psi _{r}-\langle {\vec {\nabla }},{\vec {\psi }}\rangle )=\nabla _{r}{\vec {\nabla }}\psi _{r}-{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle$

(44)

$\nabla _{r}\varphi _{r}=\nabla _{r}(\nabla _{r}\psi _{r}-\langle {\vec {\nabla }},{\vec {\psi }}\rangle )=\nabla _{r}\nabla _{r}\psi _{r}-\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle$

(45)

$\nabla ^{*}\varphi _{r}=-\nabla _{r}{\vec {\nabla }}\psi _{r}+{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle +\nabla _{r}\nabla _{r}\psi _{r}-\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle$

(46)

${\vec {\nabla }}{\vec {\varphi }}={\vec {\nabla }}(\nabla _{r}{\vec {\psi }}+{\vec {\nabla }}\psi _{r}\pm {\vec {\nabla }}\times {\vec {\psi }}\rangle )=-\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle \pm \nabla _{r}{\vec {\nabla }}\times {\vec {\psi }}-\langle {\vec {\nabla }},{\vec {\nabla }}\rangle \psi _{r}\mp \langle {\vec {\nabla }},{\vec {\nabla }}\times {\vec {\psi }}\rangle +{\vec {\nabla }}\times {\vec {\nabla }}\times {\vec {\psi }}$

(47)

${\vec {\nabla }}{\vec {\varphi }}=-\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle \pm \nabla _{r}{\vec {\nabla }}\times {\vec {\psi }}-\langle {\vec {\nabla }},{\vec {\nabla }}\rangle \psi _{r}+{\vec {\nabla }}\times {\vec {\nabla }}\times {\vec {\psi }}=-\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle \pm \nabla _{r}{\vec {\nabla }}\times {\vec {\psi }}-\langle {\vec {\nabla }},{\vec {\nabla }}\rangle \psi _{r}+{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {\psi }}$

(48)

$\nabla _{r}{\vec {\varphi }}=\nabla _{r}(\nabla _{r}{\vec {\psi }}+{\vec {\nabla }}\psi _{r}\pm {\vec {\nabla }}\times {\vec {\psi }})=\nabla _{r}\nabla _{r}{\vec {\psi }}+\nabla _{r}{\vec {\nabla }}\psi _{r}\pm \nabla _{r}{\vec {\nabla }}\times {\vec {\psi }}$

(49)

$\nabla ^{*}{\vec {\varphi }}=\nabla _{r}\nabla _{r}{\vec {\psi }}+\nabla _{r}{\vec {\nabla }}\psi _{r}\pm \nabla _{r}{\vec {\nabla }}\times {\vec {\psi }}+\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle \mp \nabla _{r}{\vec {\nabla }}\times {\vec {\psi }}+\langle {\vec {\nabla }},{\vec {\nabla }}\rangle \psi _{r}-{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle +\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {\psi }}$

(50)

$\nabla ^{*}{\vec {\varphi }}=\nabla _{r}\nabla _{r}{\vec {\psi }}+\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {\psi }}+\nabla _{r}{\vec {\nabla }}\psi _{r}+\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle +\langle {\vec {\nabla }},{\vec {\nabla }}\rangle \psi _{r}-{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle$

(51)

$\nabla ^{*}\varphi =-\nabla _{r}{\vec {\nabla }}\psi _{r}+{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle +\nabla _{r}\nabla _{r}\psi _{r}-\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle +\nabla _{r}\nabla _{r}{\vec {\psi }}+\langle {\vec {\nabla }},{\vec {\nabla }}\rangle {\vec {\psi }}+\nabla _{r}{\vec {\nabla }}\psi _{r}+\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle +\langle {\vec {\nabla }},{\vec {\nabla }}\rangle \psi _{r}-{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle$

(52)

$\chi =\nabla ^{*}\varphi =(\nabla _{r}-{\vec {\nabla }})\varphi =(\nabla _{r}-{\vec {\nabla }})(\nabla _{r}+{\vec {\nabla }})\psi =(\nabla _{r}\nabla _{r}+\langle {\vec {\nabla }},{\vec {\nabla }}\rangle )(\psi _{r}+{\vec {\psi }})+{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {\psi }}\rangle +\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\nabla _{r}\langle {\vec {\nabla }},{\vec {\psi }}\rangle -\nabla _{r}{\vec {\nabla }}\psi _{r}+\nabla _{r}{\vec {\nabla }}\psi _{r}$

(53)

Het merendeel van de termen verdwijnen.

$\chi =\nabla ^{*}(\nabla \psi )=(\nabla _{r}-{\vec {\nabla }})((\nabla _{r}+{\vec {\nabla }})(\psi _{r}+{\vec {\psi }}))=(\nabla _{r}\nabla _{r}+\langle {\vec {\nabla }},{\vec {\nabla }}\rangle )\psi$

(54)

Afleiden van tweede orde partiële differentiaalvergelijking 2

We voegen de complexe imaginaire basegetal  ${\textstyle \mathbb {I} ={\sqrt {-1}}}$ bij de ruimtelijke nabla operator ${\textstyle {\vec {\nabla }}}$.

$g=(\nabla _{r}+\mathbb {I} {\vec {\nabla }})f$

(55)

$g_{r}=\nabla _{r}f_{r}-\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle$

(56)

${\vec {g}}=\nabla _{r}{\vec {f}}+\mathbb {I} {\vec {\nabla }}f_{r}\pm \mathbb {I} {\vec {\nabla }}\times {\vec {f}}$

(57)

$\mathbb {I} {\vec {\nabla }}g_{r}=\mathbb {I} {\vec {\nabla }}(\nabla _{r}f_{r}-\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle )=\nabla _{r}\mathbb {I} {\vec {\nabla }}f_{r}-\mathbb {I} {\vec {\nabla }}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle$

(58)

$\nabla _{r}g_{r}=\nabla _{r}(\nabla _{r}f_{r}-\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle )=\nabla _{r}\nabla _{r}f_{r}-\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle$

(59)

$(\nabla _{r}-\mathbb {I} {\vec {\nabla }})g_{r}=-\nabla _{r}\mathbb {I} {\vec {\nabla }}f_{r}+\mathbb {I} {\vec {\nabla }}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle +\nabla _{r}\nabla _{r}f_{r}-\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle$

(60)

$\mathbb {I} {\vec {\nabla }}{\vec {g}}=\mathbb {I} {\vec {\nabla }}(\nabla _{r}{\vec {f}}+\mathbb {I} {\vec {\nabla }}f_{r}\pm \mathbb {I} {\vec {\nabla }}\times {\vec {f}}\rangle )=-\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle \pm \nabla _{r}\mathbb {I} {\vec {\nabla }}\times {\vec {f}}-\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle f_{r}\mp \langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\times {\vec {f}}\rangle +\mathbb {I} {\vec {\nabla }}\times \mathbb {I} {\vec {\nabla }}\times {\vec {f}}$

(61)

$\mathbb {I} {\vec {\nabla }}{\vec {g}}=-\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle \pm \nabla _{r}\mathbb {I} {\vec {\nabla }}\times {\vec {f}}-\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle f_{r}+\mathbb {I} {\vec {\nabla }}\times \mathbb {I} {\vec {\nabla }}\times {\vec {f}}=-\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle \pm \nabla _{r}\mathbb {I} {\vec {\nabla }}\times {\vec {f}}-\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle f_{r}+\mathbb {I} {\vec {\nabla }}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle -\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle {\vec {f}}$

(62)

$\nabla _{r}{\vec {g}}=\nabla _{r}(\nabla _{r}{\vec {f}}+\mathbb {I} {\vec {\nabla }}f_{r}\pm \mathbb {I} {\vec {\nabla }}\times {\vec {f}})=\nabla _{r}\nabla _{r}{\vec {f}}+\nabla _{r}\mathbb {I} {\vec {\nabla }}f_{r}\pm \nabla _{r}\mathbb {I} {\vec {\nabla }}\times {\vec {f}}$

(63)

$(\nabla _{r}-\mathbb {I} {\vec {\nabla }}){\vec {g}}=\nabla _{r}\nabla _{r}{\vec {f}}+\nabla _{r}\mathbb {I} {\vec {\nabla }}f_{r}\pm \nabla _{r}\mathbb {I} {\vec {\nabla }}\times {\vec {f}}+\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle \mp \nabla _{r}\mathbb {I} {\vec {\nabla }}\times {\vec {f}}+\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle f_{r}-\mathbb {I} {\vec {\nabla }}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle +\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle {\vec {f}}$

(64)

$(\nabla _{r}-\mathbb {I} {\vec {\nabla }}){\vec {g}}=\nabla _{r}\nabla _{r}{\vec {f}}+\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle {\vec {f}}+\nabla _{r}\mathbb {I} {\vec {\nabla }}f_{r}+\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle +\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle f_{r}-\mathbb {I} {\vec {\nabla }}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle$

(65)

$(\nabla _{r}-\mathbb {I} {\vec {\nabla }})g=-\nabla _{r}\mathbb {I} {\vec {\nabla }}f_{r}+\mathbb {I} {\vec {\nabla }}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle +\nabla _{r}\nabla _{r}f_{r}-\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle +\nabla _{r}\nabla _{r}{\vec {f}}+\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle {\vec {f}}+\nabla _{r}\mathbb {I} {\vec {\nabla }}f_{r}+\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle +\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle f_{r}-\mathbb {I} {\vec {\nabla }}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle$

(66)

$\zeta =(\nabla _{r}-\mathbb {I} {\vec {\nabla }})g=(\nabla _{r}-\mathbb {I} {\vec {\nabla }})(\nabla _{r}+\mathbb {I} {\vec {\nabla }})f=(\nabla _{r}\nabla _{r}+\langle \mathbb {I} {\vec {\nabla }},\mathbb {I} {\vec {\nabla }}\rangle )(f_{r}+{\vec {f}})+\mathbb {I} {\vec {\nabla }}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle -\mathbb {I} {\vec {\nabla }}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle +\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle -\nabla _{r}\langle \mathbb {I} {\vec {\nabla }},{\vec {f}}\rangle -\nabla _{r}\mathbb {I} {\vec {\nabla }}f_{r}+\nabla _{r}\mathbb {I} {\vec {\nabla }}f_{r}$

(67)

$\zeta =(\nabla _{r}-\mathbb {I} {\vec {\nabla }})g=(\nabla _{r}-\mathbb {I} {\vec {\nabla }})(\nabla _{r}+\mathbb {I} {\vec {\nabla }})f=(\nabla _{r}\nabla _{r}-\langle {\vec {\nabla }},{\vec {\nabla }}\rangle )(f_{r}+{\vec {f}})-{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {f}}\rangle +{\vec {\nabla }}\langle {\vec {\nabla }},{\vec {f}}\rangle +\mathbb {I} (\nabla _{r}\langle {\vec {\nabla }},{\vec {f}}\rangle -\nabla _{r}\langle {\vec {\nabla }},{\vec {f}}\rangle -\nabla _{r}{\vec {\nabla }}f_{r}+\nabla _{r}{\vec {\nabla }}f_{r})$

(68)

$\zeta =(\nabla _{r}\nabla _{r}-\langle {\vec {\nabla }},{\vec {\nabla }}\rangle )f$

(69)

Dus ook deze quaternionische tweede orde partiële differentiaalvergelijking splitst in twee eerste orde partiële differentiaalvergelijkingen. Maar dit zijn geen quaternionische partiële differentiaalvergelijkingen 

$g=(\nabla _{r}+\mathbb {I} {\vec {\nabla }})f;\qquad \zeta =(\nabla _{r}-\mathbb {I} {\vec {\nabla }})g$

(70)