Calculating the cross product of a cross product











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so I really can't see what I am doing wrong. I want to use this formula:



$atimes (btimes c) = b(acdot c) - c(acdot b)$



Calculate the rotation of $v(x,y,z)=(x,y,z)^T times omega$ with $omega in mathbb R^3$



Solution:



$acdot c=nablacdot omega=0$



and



$acdot b = nabla cdot (x,y,z)^T= partial_x x + partial_y y + partial_z z = 3$



so we get $-omega 3=-3omega$



The actual solution (which I do get by direct calculation) is: $-2omega$






cross-product curl







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  • 1




    You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
    – Hans Lundmark
    Oct 23 at 18:32















up vote
1
down vote

favorite












so I really can't see what I am doing wrong. I want to use this formula:



$atimes (btimes c) = b(acdot c) - c(acdot b)$



Calculate the rotation of $v(x,y,z)=(x,y,z)^T times omega$ with $omega in mathbb R^3$



Solution:



$acdot c=nablacdot omega=0$



and



$acdot b = nabla cdot (x,y,z)^T= partial_x x + partial_y y + partial_z z = 3$



so we get $-omega 3=-3omega$



The actual solution (which I do get by direct calculation) is: $-2omega$






cross-product curl







share|cite|improve this question


















  • 1




    You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
    – Hans Lundmark
    Oct 23 at 18:32













up vote
1
down vote

favorite









up vote
1
down vote

favorite











so I really can't see what I am doing wrong. I want to use this formula:



$atimes (btimes c) = b(acdot c) - c(acdot b)$



Calculate the rotation of $v(x,y,z)=(x,y,z)^T times omega$ with $omega in mathbb R^3$



Solution:



$acdot c=nablacdot omega=0$



and



$acdot b = nabla cdot (x,y,z)^T= partial_x x + partial_y y + partial_z z = 3$



so we get $-omega 3=-3omega$



The actual solution (which I do get by direct calculation) is: $-2omega$






cross-product curl







share|cite|improve this question













so I really can't see what I am doing wrong. I want to use this formula:



$atimes (btimes c) = b(acdot c) - c(acdot b)$



Calculate the rotation of $v(x,y,z)=(x,y,z)^T times omega$ with $omega in mathbb R^3$



Solution:



$acdot c=nablacdot omega=0$



and



$acdot b = nabla cdot (x,y,z)^T= partial_x x + partial_y y + partial_z z = 3$



so we get $-omega 3=-3omega$



The actual solution (which I do get by direct calculation) is: $-2omega$






cross-product curl




cross-product curl






share|cite|improve this question













share|cite|improve this question











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asked Oct 23 at 18:16









xotix

36829




36829








  • 1




    You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
    – Hans Lundmark
    Oct 23 at 18:32














  • 1




    You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
    – Hans Lundmark
    Oct 23 at 18:32








1




1




You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
– Hans Lundmark
Oct 23 at 18:32




You can't just replace $a$ with $nabla$ like that, you also need to keep track of what is integrated. See here: math.stackexchange.com/questions/836041/…
– Hans Lundmark
Oct 23 at 18:32










2 Answers
2






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accepted










If you want to generalise formulae with vectors having commuting components, make sure to write the usual formula so as to preserve the order in products. For example, here the $i$th component of the RHS is $sum_j a_j (b_i c_j-b_j c_i)$, once we impose the $abc$ order of the LHS. Now you know the generalisation off the top of your head. For $a_i=partial_i$, the result's $i$th component is $$sum_j partial_j (b_i c_j-b_j c_i)=b_inablacdot c+ccdotnabla b_i-(nablacdot b) c-bcdotnabla c_i.$$The vector, in other words, is $$vec{b}(nablacdotvec{c})-(nablacdotvec{b})vec{c}+(vec{c}cdotnabla)vec{b}-(vec{b}cdotnabla)vec{c}.$$(I've swapped the middle terms to mirror @md2perpe's quoted result, but the letters $b,,c$ still need to be changed to $A,,B$).






share|cite|improve this answer






























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    1
    down vote













    On Wikipedia you can see that the formula for the curl of a cross product is given by
    $$
    nabla times (mathbf{A} times mathbf{B}) = mathbf{A} (nabla cdot mathbf{B}) - mathbf{B} (nabla cdot mathbf{A}) + (mathbf{B} cdot nabla) mathbf{A} - (mathbf{A} cdot nabla) mathbf{B}
    .
    $$



    Applying this on your case gives
    $$begin{align}
    nabla times (mathbf{x} times mathbf{omega})
    &= mathbf{x} (nabla cdot mathbf{omega}) - mathbf{omega} (nabla cdot mathbf{x}) + (mathbf{omega} cdot nabla) mathbf{x} - (mathbf{x} cdot nabla) mathbf{omega} \
    &= mathbf{x} 0 - mathbf{omega} 3 + mathbf{omega} - mathbf{0}
    = -2 mathbf{omega}
    .
    end{align}$$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      If you want to generalise formulae with vectors having commuting components, make sure to write the usual formula so as to preserve the order in products. For example, here the $i$th component of the RHS is $sum_j a_j (b_i c_j-b_j c_i)$, once we impose the $abc$ order of the LHS. Now you know the generalisation off the top of your head. For $a_i=partial_i$, the result's $i$th component is $$sum_j partial_j (b_i c_j-b_j c_i)=b_inablacdot c+ccdotnabla b_i-(nablacdot b) c-bcdotnabla c_i.$$The vector, in other words, is $$vec{b}(nablacdotvec{c})-(nablacdotvec{b})vec{c}+(vec{c}cdotnabla)vec{b}-(vec{b}cdotnabla)vec{c}.$$(I've swapped the middle terms to mirror @md2perpe's quoted result, but the letters $b,,c$ still need to be changed to $A,,B$).






      share|cite|improve this answer



























        up vote
        1
        down vote



        accepted










        If you want to generalise formulae with vectors having commuting components, make sure to write the usual formula so as to preserve the order in products. For example, here the $i$th component of the RHS is $sum_j a_j (b_i c_j-b_j c_i)$, once we impose the $abc$ order of the LHS. Now you know the generalisation off the top of your head. For $a_i=partial_i$, the result's $i$th component is $$sum_j partial_j (b_i c_j-b_j c_i)=b_inablacdot c+ccdotnabla b_i-(nablacdot b) c-bcdotnabla c_i.$$The vector, in other words, is $$vec{b}(nablacdotvec{c})-(nablacdotvec{b})vec{c}+(vec{c}cdotnabla)vec{b}-(vec{b}cdotnabla)vec{c}.$$(I've swapped the middle terms to mirror @md2perpe's quoted result, but the letters $b,,c$ still need to be changed to $A,,B$).






        share|cite|improve this answer

























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          If you want to generalise formulae with vectors having commuting components, make sure to write the usual formula so as to preserve the order in products. For example, here the $i$th component of the RHS is $sum_j a_j (b_i c_j-b_j c_i)$, once we impose the $abc$ order of the LHS. Now you know the generalisation off the top of your head. For $a_i=partial_i$, the result's $i$th component is $$sum_j partial_j (b_i c_j-b_j c_i)=b_inablacdot c+ccdotnabla b_i-(nablacdot b) c-bcdotnabla c_i.$$The vector, in other words, is $$vec{b}(nablacdotvec{c})-(nablacdotvec{b})vec{c}+(vec{c}cdotnabla)vec{b}-(vec{b}cdotnabla)vec{c}.$$(I've swapped the middle terms to mirror @md2perpe's quoted result, but the letters $b,,c$ still need to be changed to $A,,B$).






          share|cite|improve this answer














          If you want to generalise formulae with vectors having commuting components, make sure to write the usual formula so as to preserve the order in products. For example, here the $i$th component of the RHS is $sum_j a_j (b_i c_j-b_j c_i)$, once we impose the $abc$ order of the LHS. Now you know the generalisation off the top of your head. For $a_i=partial_i$, the result's $i$th component is $$sum_j partial_j (b_i c_j-b_j c_i)=b_inablacdot c+ccdotnabla b_i-(nablacdot b) c-bcdotnabla c_i.$$The vector, in other words, is $$vec{b}(nablacdotvec{c})-(nablacdotvec{b})vec{c}+(vec{c}cdotnabla)vec{b}-(vec{b}cdotnabla)vec{c}.$$(I've swapped the middle terms to mirror @md2perpe's quoted result, but the letters $b,,c$ still need to be changed to $A,,B$).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 at 19:40

























          answered Oct 23 at 20:23









          J.G.

          21.1k21933




          21.1k21933






















              up vote
              1
              down vote













              On Wikipedia you can see that the formula for the curl of a cross product is given by
              $$
              nabla times (mathbf{A} times mathbf{B}) = mathbf{A} (nabla cdot mathbf{B}) - mathbf{B} (nabla cdot mathbf{A}) + (mathbf{B} cdot nabla) mathbf{A} - (mathbf{A} cdot nabla) mathbf{B}
              .
              $$



              Applying this on your case gives
              $$begin{align}
              nabla times (mathbf{x} times mathbf{omega})
              &= mathbf{x} (nabla cdot mathbf{omega}) - mathbf{omega} (nabla cdot mathbf{x}) + (mathbf{omega} cdot nabla) mathbf{x} - (mathbf{x} cdot nabla) mathbf{omega} \
              &= mathbf{x} 0 - mathbf{omega} 3 + mathbf{omega} - mathbf{0}
              = -2 mathbf{omega}
              .
              end{align}$$






              share|cite|improve this answer

























                up vote
                1
                down vote













                On Wikipedia you can see that the formula for the curl of a cross product is given by
                $$
                nabla times (mathbf{A} times mathbf{B}) = mathbf{A} (nabla cdot mathbf{B}) - mathbf{B} (nabla cdot mathbf{A}) + (mathbf{B} cdot nabla) mathbf{A} - (mathbf{A} cdot nabla) mathbf{B}
                .
                $$



                Applying this on your case gives
                $$begin{align}
                nabla times (mathbf{x} times mathbf{omega})
                &= mathbf{x} (nabla cdot mathbf{omega}) - mathbf{omega} (nabla cdot mathbf{x}) + (mathbf{omega} cdot nabla) mathbf{x} - (mathbf{x} cdot nabla) mathbf{omega} \
                &= mathbf{x} 0 - mathbf{omega} 3 + mathbf{omega} - mathbf{0}
                = -2 mathbf{omega}
                .
                end{align}$$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  On Wikipedia you can see that the formula for the curl of a cross product is given by
                  $$
                  nabla times (mathbf{A} times mathbf{B}) = mathbf{A} (nabla cdot mathbf{B}) - mathbf{B} (nabla cdot mathbf{A}) + (mathbf{B} cdot nabla) mathbf{A} - (mathbf{A} cdot nabla) mathbf{B}
                  .
                  $$



                  Applying this on your case gives
                  $$begin{align}
                  nabla times (mathbf{x} times mathbf{omega})
                  &= mathbf{x} (nabla cdot mathbf{omega}) - mathbf{omega} (nabla cdot mathbf{x}) + (mathbf{omega} cdot nabla) mathbf{x} - (mathbf{x} cdot nabla) mathbf{omega} \
                  &= mathbf{x} 0 - mathbf{omega} 3 + mathbf{omega} - mathbf{0}
                  = -2 mathbf{omega}
                  .
                  end{align}$$






                  share|cite|improve this answer












                  On Wikipedia you can see that the formula for the curl of a cross product is given by
                  $$
                  nabla times (mathbf{A} times mathbf{B}) = mathbf{A} (nabla cdot mathbf{B}) - mathbf{B} (nabla cdot mathbf{A}) + (mathbf{B} cdot nabla) mathbf{A} - (mathbf{A} cdot nabla) mathbf{B}
                  .
                  $$



                  Applying this on your case gives
                  $$begin{align}
                  nabla times (mathbf{x} times mathbf{omega})
                  &= mathbf{x} (nabla cdot mathbf{omega}) - mathbf{omega} (nabla cdot mathbf{x}) + (mathbf{omega} cdot nabla) mathbf{x} - (mathbf{x} cdot nabla) mathbf{omega} \
                  &= mathbf{x} 0 - mathbf{omega} 3 + mathbf{omega} - mathbf{0}
                  = -2 mathbf{omega}
                  .
                  end{align}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 23 at 20:07









                  md2perpe

                  7,31311027




                  7,31311027






























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