Vector triple product in 4 dimensions












0












$begingroup$


I have rewritten an equation in 3 dimensions
$$ vec{f} + nabla a + b nabla c = vec{0} $$
where $a,b,c$ are unknown functions to be determined and $vec{f}$ is a known vector by taking the curl
$$nabla times vec{f} + nabla b times nabla c = vec{0}$$
and using the the property of the triple product $vec{u} cdot (vec{u} times vec{v}) = 0$ to give 3 first-order linear partial differential equations, 1 for each of the unknown functions
$$ nabla b cdot (nabla times vec{f}) = 0 \ nabla c cdot (nabla times vec{f}) = 0 \ nabla a cdot (nabla times vec{f}) + vec{f} cdot (nabla times vec{f}) = 0 ; .$$
The 1st-order PDE's allowed me to reduce the number of independent variables to 2.





I am now faced with the same problem in 4 dimensions.



Question:



Is there a way of applying a similar method to derive 1st-order PDE's for each of the unknown functions?



Addendum:



I am thinking that this can be solved more easily using index notation. If the original equation is
$$ f_{k} + partial_{k} a + b partial_{k} c = 0_k $$
then one can perform the following steps (to obtain an equation for $b$)
$$ partial_j b , partial_l left( f_{k} + partial_{k} a + b partial_{k} c right) varepsilon^{i j k l} textbf{1}_i = 0 $$
where $textbf{1}_i$ is supposed to mean a vector of ones and $varepsilon^{i j k l}$ is the Levi-Civita symbol in 4 dimensions. Some terms drop out and and we are left with
$$ partial_j b , partial_l f_{k} varepsilon^{i j k l} textbf{1}_i = 0 ; .$$



Do you think this is correct?










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$endgroup$

















    0












    $begingroup$


    I have rewritten an equation in 3 dimensions
    $$ vec{f} + nabla a + b nabla c = vec{0} $$
    where $a,b,c$ are unknown functions to be determined and $vec{f}$ is a known vector by taking the curl
    $$nabla times vec{f} + nabla b times nabla c = vec{0}$$
    and using the the property of the triple product $vec{u} cdot (vec{u} times vec{v}) = 0$ to give 3 first-order linear partial differential equations, 1 for each of the unknown functions
    $$ nabla b cdot (nabla times vec{f}) = 0 \ nabla c cdot (nabla times vec{f}) = 0 \ nabla a cdot (nabla times vec{f}) + vec{f} cdot (nabla times vec{f}) = 0 ; .$$
    The 1st-order PDE's allowed me to reduce the number of independent variables to 2.





    I am now faced with the same problem in 4 dimensions.



    Question:



    Is there a way of applying a similar method to derive 1st-order PDE's for each of the unknown functions?



    Addendum:



    I am thinking that this can be solved more easily using index notation. If the original equation is
    $$ f_{k} + partial_{k} a + b partial_{k} c = 0_k $$
    then one can perform the following steps (to obtain an equation for $b$)
    $$ partial_j b , partial_l left( f_{k} + partial_{k} a + b partial_{k} c right) varepsilon^{i j k l} textbf{1}_i = 0 $$
    where $textbf{1}_i$ is supposed to mean a vector of ones and $varepsilon^{i j k l}$ is the Levi-Civita symbol in 4 dimensions. Some terms drop out and and we are left with
    $$ partial_j b , partial_l f_{k} varepsilon^{i j k l} textbf{1}_i = 0 ; .$$



    Do you think this is correct?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have rewritten an equation in 3 dimensions
      $$ vec{f} + nabla a + b nabla c = vec{0} $$
      where $a,b,c$ are unknown functions to be determined and $vec{f}$ is a known vector by taking the curl
      $$nabla times vec{f} + nabla b times nabla c = vec{0}$$
      and using the the property of the triple product $vec{u} cdot (vec{u} times vec{v}) = 0$ to give 3 first-order linear partial differential equations, 1 for each of the unknown functions
      $$ nabla b cdot (nabla times vec{f}) = 0 \ nabla c cdot (nabla times vec{f}) = 0 \ nabla a cdot (nabla times vec{f}) + vec{f} cdot (nabla times vec{f}) = 0 ; .$$
      The 1st-order PDE's allowed me to reduce the number of independent variables to 2.





      I am now faced with the same problem in 4 dimensions.



      Question:



      Is there a way of applying a similar method to derive 1st-order PDE's for each of the unknown functions?



      Addendum:



      I am thinking that this can be solved more easily using index notation. If the original equation is
      $$ f_{k} + partial_{k} a + b partial_{k} c = 0_k $$
      then one can perform the following steps (to obtain an equation for $b$)
      $$ partial_j b , partial_l left( f_{k} + partial_{k} a + b partial_{k} c right) varepsilon^{i j k l} textbf{1}_i = 0 $$
      where $textbf{1}_i$ is supposed to mean a vector of ones and $varepsilon^{i j k l}$ is the Levi-Civita symbol in 4 dimensions. Some terms drop out and and we are left with
      $$ partial_j b , partial_l f_{k} varepsilon^{i j k l} textbf{1}_i = 0 ; .$$



      Do you think this is correct?










      share|cite|improve this question











      $endgroup$




      I have rewritten an equation in 3 dimensions
      $$ vec{f} + nabla a + b nabla c = vec{0} $$
      where $a,b,c$ are unknown functions to be determined and $vec{f}$ is a known vector by taking the curl
      $$nabla times vec{f} + nabla b times nabla c = vec{0}$$
      and using the the property of the triple product $vec{u} cdot (vec{u} times vec{v}) = 0$ to give 3 first-order linear partial differential equations, 1 for each of the unknown functions
      $$ nabla b cdot (nabla times vec{f}) = 0 \ nabla c cdot (nabla times vec{f}) = 0 \ nabla a cdot (nabla times vec{f}) + vec{f} cdot (nabla times vec{f}) = 0 ; .$$
      The 1st-order PDE's allowed me to reduce the number of independent variables to 2.





      I am now faced with the same problem in 4 dimensions.



      Question:



      Is there a way of applying a similar method to derive 1st-order PDE's for each of the unknown functions?



      Addendum:



      I am thinking that this can be solved more easily using index notation. If the original equation is
      $$ f_{k} + partial_{k} a + b partial_{k} c = 0_k $$
      then one can perform the following steps (to obtain an equation for $b$)
      $$ partial_j b , partial_l left( f_{k} + partial_{k} a + b partial_{k} c right) varepsilon^{i j k l} textbf{1}_i = 0 $$
      where $textbf{1}_i$ is supposed to mean a vector of ones and $varepsilon^{i j k l}$ is the Levi-Civita symbol in 4 dimensions. Some terms drop out and and we are left with
      $$ partial_j b , partial_l f_{k} varepsilon^{i j k l} textbf{1}_i = 0 ; .$$



      Do you think this is correct?







      pde partial-derivative vector-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 22:37







      Crenguta

















      asked Dec 19 '18 at 18:57









      CrengutaCrenguta

      456




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