Deriving a unique derivative operator with torsion such that $triangledown g = 0$












1












$begingroup$


From Wald's General Relativity, Chapter 3 problem 1c), in the solution from http://www.physics.drexel.edu/~dcross/papers/wald.pdf:



$triangledown g = 0$ becomes again



$$triangledown_a g_{bc} = bar{triangledown}_ag_{bc} - C^d_{ab}g_{dc}- C^d_{ac}g_{bd} = bar{triangledown}_ag_{bc} - C_{cab}-C_{bac}$$



If we add the permutation (ab) and subtract the permutation (cab) as before we get



$$bar{triangledown}_a g_{bc} + bar{triangledown}_ag_{bc} - bar{triangledown}_ag_{bc} = T_{bac}-T_{abc}+2C_{c(ab)}$$



How have we gone form C's to T's?






derivatives differential-geometry general-relativity special-relativity







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    From Wald's General Relativity, Chapter 3 problem 1c), in the solution from http://www.physics.drexel.edu/~dcross/papers/wald.pdf:



    $triangledown g = 0$ becomes again



    $$triangledown_a g_{bc} = bar{triangledown}_ag_{bc} - C^d_{ab}g_{dc}- C^d_{ac}g_{bd} = bar{triangledown}_ag_{bc} - C_{cab}-C_{bac}$$



    If we add the permutation (ab) and subtract the permutation (cab) as before we get



    $$bar{triangledown}_a g_{bc} + bar{triangledown}_ag_{bc} - bar{triangledown}_ag_{bc} = T_{bac}-T_{abc}+2C_{c(ab)}$$



    How have we gone form C's to T's?






    derivatives differential-geometry general-relativity special-relativity







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      From Wald's General Relativity, Chapter 3 problem 1c), in the solution from http://www.physics.drexel.edu/~dcross/papers/wald.pdf:



      $triangledown g = 0$ becomes again



      $$triangledown_a g_{bc} = bar{triangledown}_ag_{bc} - C^d_{ab}g_{dc}- C^d_{ac}g_{bd} = bar{triangledown}_ag_{bc} - C_{cab}-C_{bac}$$



      If we add the permutation (ab) and subtract the permutation (cab) as before we get



      $$bar{triangledown}_a g_{bc} + bar{triangledown}_ag_{bc} - bar{triangledown}_ag_{bc} = T_{bac}-T_{abc}+2C_{c(ab)}$$



      How have we gone form C's to T's?






      derivatives differential-geometry general-relativity special-relativity







      share|cite|improve this question









      $endgroup$




      From Wald's General Relativity, Chapter 3 problem 1c), in the solution from http://www.physics.drexel.edu/~dcross/papers/wald.pdf:



      $triangledown g = 0$ becomes again



      $$triangledown_a g_{bc} = bar{triangledown}_ag_{bc} - C^d_{ab}g_{dc}- C^d_{ac}g_{bd} = bar{triangledown}_ag_{bc} - C_{cab}-C_{bac}$$



      If we add the permutation (ab) and subtract the permutation (cab) as before we get



      $$bar{triangledown}_a g_{bc} + bar{triangledown}_ag_{bc} - bar{triangledown}_ag_{bc} = T_{bac}-T_{abc}+2C_{c(ab)}$$



      How have we gone form C's to T's?






      derivatives differential-geometry general-relativity special-relativity




      derivatives differential-geometry general-relativity special-relativity






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 5 '18 at 12:32









      PermianPermian

      2,1961135




      2,1961135






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          Remember that the torsion in basically the antisymmetric part of $C$



          $$
          T_{abc} = C_{a[bc]} tag{1}
          $$



          So that



          begin{eqnarray}
          color{blue}{C_{bac} - C_{bca}} &=& color{blue}{C_{b[ac]} = T_{bac}} \
          color{red}{C_{abc} - C_{acb}} &=& color{red}{C_{a[bc]} = T_{abc}} \
          color{magenta}{C_{cab} + C_{cba}} &=& color{magenta}{2C_{c(ba)}} tag{2}
          end{eqnarray}



          With this in mind ($nabla g = 0$)



          begin{eqnarray}
          tilde{nabla}_ag_{bc} + tilde{nabla}_bg_{ac} - tilde{nabla}_cg_{ab} &=& [nabla_a g_{bc} + C_{cab} + C_{bac}] + [nabla_{b}g_{ac} + C_{cba} + C_{abc}] \
          && -[nabla_cg_{ab} + C_{bca} + C_{abc}] \
          &=& (color{magenta}{C_{cab} + C_{cba}}) + color{blue}{C_{bac} - C_{bca}} + (color{red}{C_{abc} - C_{acb}}) \
          &stackrel{(2)}{=}& color{magenta}{2C_{c(ba)}} + color{blue}{C_{b[ac]}} + color{red}{C_{a[bc]}} \
          &stackrel{(1)}{=}& color{magenta}{2C_{c(ab)}} + color{blue}{T_{bac}} + color{red}{T_{abc}}
          end{eqnarray}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ill have another look but I think statement is the bit im not seeing
            $endgroup$
            – Permian
            Dec 5 '18 at 14:07










          • $begingroup$
            Yeah I cant see how (1) is so obvious to you
            $endgroup$
            – Permian
            Dec 5 '18 at 20:43










          • $begingroup$
            @Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
            $endgroup$
            – caverac
            Dec 5 '18 at 20:45












          • $begingroup$
            where are the C's in 1a)?
            $endgroup$
            – Permian
            Dec 5 '18 at 20:47






          • 1




            $begingroup$
            It is a very fascinating subject, congratulations!
            $endgroup$
            – caverac
            Dec 7 '18 at 20:02











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Remember that the torsion in basically the antisymmetric part of $C$



          $$
          T_{abc} = C_{a[bc]} tag{1}
          $$



          So that



          begin{eqnarray}
          color{blue}{C_{bac} - C_{bca}} &=& color{blue}{C_{b[ac]} = T_{bac}} \
          color{red}{C_{abc} - C_{acb}} &=& color{red}{C_{a[bc]} = T_{abc}} \
          color{magenta}{C_{cab} + C_{cba}} &=& color{magenta}{2C_{c(ba)}} tag{2}
          end{eqnarray}



          With this in mind ($nabla g = 0$)



          begin{eqnarray}
          tilde{nabla}_ag_{bc} + tilde{nabla}_bg_{ac} - tilde{nabla}_cg_{ab} &=& [nabla_a g_{bc} + C_{cab} + C_{bac}] + [nabla_{b}g_{ac} + C_{cba} + C_{abc}] \
          && -[nabla_cg_{ab} + C_{bca} + C_{abc}] \
          &=& (color{magenta}{C_{cab} + C_{cba}}) + color{blue}{C_{bac} - C_{bca}} + (color{red}{C_{abc} - C_{acb}}) \
          &stackrel{(2)}{=}& color{magenta}{2C_{c(ba)}} + color{blue}{C_{b[ac]}} + color{red}{C_{a[bc]}} \
          &stackrel{(1)}{=}& color{magenta}{2C_{c(ab)}} + color{blue}{T_{bac}} + color{red}{T_{abc}}
          end{eqnarray}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ill have another look but I think statement is the bit im not seeing
            $endgroup$
            – Permian
            Dec 5 '18 at 14:07










          • $begingroup$
            Yeah I cant see how (1) is so obvious to you
            $endgroup$
            – Permian
            Dec 5 '18 at 20:43










          • $begingroup$
            @Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
            $endgroup$
            – caverac
            Dec 5 '18 at 20:45












          • $begingroup$
            where are the C's in 1a)?
            $endgroup$
            – Permian
            Dec 5 '18 at 20:47






          • 1




            $begingroup$
            It is a very fascinating subject, congratulations!
            $endgroup$
            – caverac
            Dec 7 '18 at 20:02
















          1












          $begingroup$

          Remember that the torsion in basically the antisymmetric part of $C$



          $$
          T_{abc} = C_{a[bc]} tag{1}
          $$



          So that



          begin{eqnarray}
          color{blue}{C_{bac} - C_{bca}} &=& color{blue}{C_{b[ac]} = T_{bac}} \
          color{red}{C_{abc} - C_{acb}} &=& color{red}{C_{a[bc]} = T_{abc}} \
          color{magenta}{C_{cab} + C_{cba}} &=& color{magenta}{2C_{c(ba)}} tag{2}
          end{eqnarray}



          With this in mind ($nabla g = 0$)



          begin{eqnarray}
          tilde{nabla}_ag_{bc} + tilde{nabla}_bg_{ac} - tilde{nabla}_cg_{ab} &=& [nabla_a g_{bc} + C_{cab} + C_{bac}] + [nabla_{b}g_{ac} + C_{cba} + C_{abc}] \
          && -[nabla_cg_{ab} + C_{bca} + C_{abc}] \
          &=& (color{magenta}{C_{cab} + C_{cba}}) + color{blue}{C_{bac} - C_{bca}} + (color{red}{C_{abc} - C_{acb}}) \
          &stackrel{(2)}{=}& color{magenta}{2C_{c(ba)}} + color{blue}{C_{b[ac]}} + color{red}{C_{a[bc]}} \
          &stackrel{(1)}{=}& color{magenta}{2C_{c(ab)}} + color{blue}{T_{bac}} + color{red}{T_{abc}}
          end{eqnarray}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ill have another look but I think statement is the bit im not seeing
            $endgroup$
            – Permian
            Dec 5 '18 at 14:07










          • $begingroup$
            Yeah I cant see how (1) is so obvious to you
            $endgroup$
            – Permian
            Dec 5 '18 at 20:43










          • $begingroup$
            @Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
            $endgroup$
            – caverac
            Dec 5 '18 at 20:45












          • $begingroup$
            where are the C's in 1a)?
            $endgroup$
            – Permian
            Dec 5 '18 at 20:47






          • 1




            $begingroup$
            It is a very fascinating subject, congratulations!
            $endgroup$
            – caverac
            Dec 7 '18 at 20:02














          1












          1








          1





          $begingroup$

          Remember that the torsion in basically the antisymmetric part of $C$



          $$
          T_{abc} = C_{a[bc]} tag{1}
          $$



          So that



          begin{eqnarray}
          color{blue}{C_{bac} - C_{bca}} &=& color{blue}{C_{b[ac]} = T_{bac}} \
          color{red}{C_{abc} - C_{acb}} &=& color{red}{C_{a[bc]} = T_{abc}} \
          color{magenta}{C_{cab} + C_{cba}} &=& color{magenta}{2C_{c(ba)}} tag{2}
          end{eqnarray}



          With this in mind ($nabla g = 0$)



          begin{eqnarray}
          tilde{nabla}_ag_{bc} + tilde{nabla}_bg_{ac} - tilde{nabla}_cg_{ab} &=& [nabla_a g_{bc} + C_{cab} + C_{bac}] + [nabla_{b}g_{ac} + C_{cba} + C_{abc}] \
          && -[nabla_cg_{ab} + C_{bca} + C_{abc}] \
          &=& (color{magenta}{C_{cab} + C_{cba}}) + color{blue}{C_{bac} - C_{bca}} + (color{red}{C_{abc} - C_{acb}}) \
          &stackrel{(2)}{=}& color{magenta}{2C_{c(ba)}} + color{blue}{C_{b[ac]}} + color{red}{C_{a[bc]}} \
          &stackrel{(1)}{=}& color{magenta}{2C_{c(ab)}} + color{blue}{T_{bac}} + color{red}{T_{abc}}
          end{eqnarray}






          share|cite|improve this answer









          $endgroup$



          Remember that the torsion in basically the antisymmetric part of $C$



          $$
          T_{abc} = C_{a[bc]} tag{1}
          $$



          So that



          begin{eqnarray}
          color{blue}{C_{bac} - C_{bca}} &=& color{blue}{C_{b[ac]} = T_{bac}} \
          color{red}{C_{abc} - C_{acb}} &=& color{red}{C_{a[bc]} = T_{abc}} \
          color{magenta}{C_{cab} + C_{cba}} &=& color{magenta}{2C_{c(ba)}} tag{2}
          end{eqnarray}



          With this in mind ($nabla g = 0$)



          begin{eqnarray}
          tilde{nabla}_ag_{bc} + tilde{nabla}_bg_{ac} - tilde{nabla}_cg_{ab} &=& [nabla_a g_{bc} + C_{cab} + C_{bac}] + [nabla_{b}g_{ac} + C_{cba} + C_{abc}] \
          && -[nabla_cg_{ab} + C_{bca} + C_{abc}] \
          &=& (color{magenta}{C_{cab} + C_{cba}}) + color{blue}{C_{bac} - C_{bca}} + (color{red}{C_{abc} - C_{acb}}) \
          &stackrel{(2)}{=}& color{magenta}{2C_{c(ba)}} + color{blue}{C_{b[ac]}} + color{red}{C_{a[bc]}} \
          &stackrel{(1)}{=}& color{magenta}{2C_{c(ab)}} + color{blue}{T_{bac}} + color{red}{T_{abc}}
          end{eqnarray}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 13:38









          caveraccaverac

          14.5k31130




          14.5k31130












          • $begingroup$
            Ill have another look but I think statement is the bit im not seeing
            $endgroup$
            – Permian
            Dec 5 '18 at 14:07










          • $begingroup$
            Yeah I cant see how (1) is so obvious to you
            $endgroup$
            – Permian
            Dec 5 '18 at 20:43










          • $begingroup$
            @Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
            $endgroup$
            – caverac
            Dec 5 '18 at 20:45












          • $begingroup$
            where are the C's in 1a)?
            $endgroup$
            – Permian
            Dec 5 '18 at 20:47






          • 1




            $begingroup$
            It is a very fascinating subject, congratulations!
            $endgroup$
            – caverac
            Dec 7 '18 at 20:02


















          • $begingroup$
            Ill have another look but I think statement is the bit im not seeing
            $endgroup$
            – Permian
            Dec 5 '18 at 14:07










          • $begingroup$
            Yeah I cant see how (1) is so obvious to you
            $endgroup$
            – Permian
            Dec 5 '18 at 20:43










          • $begingroup$
            @Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
            $endgroup$
            – caverac
            Dec 5 '18 at 20:45












          • $begingroup$
            where are the C's in 1a)?
            $endgroup$
            – Permian
            Dec 5 '18 at 20:47






          • 1




            $begingroup$
            It is a very fascinating subject, congratulations!
            $endgroup$
            – caverac
            Dec 7 '18 at 20:02
















          $begingroup$
          Ill have another look but I think statement is the bit im not seeing
          $endgroup$
          – Permian
          Dec 5 '18 at 14:07




          $begingroup$
          Ill have another look but I think statement is the bit im not seeing
          $endgroup$
          – Permian
          Dec 5 '18 at 14:07












          $begingroup$
          Yeah I cant see how (1) is so obvious to you
          $endgroup$
          – Permian
          Dec 5 '18 at 20:43




          $begingroup$
          Yeah I cant see how (1) is so obvious to you
          $endgroup$
          – Permian
          Dec 5 '18 at 20:43












          $begingroup$
          @Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
          $endgroup$
          – caverac
          Dec 5 '18 at 20:45






          $begingroup$
          @Permian It is actually in the solution you attach: Problem 1a. I assumed since you were asking about part c, you already solved part a
          $endgroup$
          – caverac
          Dec 5 '18 at 20:45














          $begingroup$
          where are the C's in 1a)?
          $endgroup$
          – Permian
          Dec 5 '18 at 20:47




          $begingroup$
          where are the C's in 1a)?
          $endgroup$
          – Permian
          Dec 5 '18 at 20:47




          1




          1




          $begingroup$
          It is a very fascinating subject, congratulations!
          $endgroup$
          – caverac
          Dec 7 '18 at 20:02




          $begingroup$
          It is a very fascinating subject, congratulations!
          $endgroup$
          – caverac
          Dec 7 '18 at 20:02


















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