Vector Calculus Subscript Notation

  • #1
Hi there is there a tutorial or post explaining vector calculus subscript notation please?
e.g. Eijk Kklm

dil djm etc etc

is there a tutorial explaining these thoroughly and how these can convert into div grad and curl??
i've used the search engine but cant seem to find them. thnx
 

Answers and Replies

  • #3
thanks a lot for that. it does help but there is far too much information there. i was just looking for a general explanation of the fundamentals and how the notations can be used to solve questions such as proving that:

grad(A.B) = (B.Delta)A + (A.Delta)B + BX(CurlA) + AX(CurlB)

etc etc... any suggestions please?
 
  • #4
im actually getting the hang of it now. but stuck on dot product of cross products..

e.g. show... (AXB) . (CXD) = (A.C)(B.D) - (A.D)(B.C)

here's what i have done but its partially correct.

[tex] (AXB)i = \varepsilon_{ijk}A_jB_k[/tex].
[tex] (CXD)i = \varepsilon_{ijk}C_jD_k[/tex]

so (AXB) . (CXD) = [tex] (\varepsilon_{ijk}A_jB_k)_i . (\varepsilon_{ijk}C_jD_k)_i [/tex]

= [tex]\varepsilon_{ijk}\varepsilon_{ijk}A_jC_jB_kD_k[/tex]

= [tex]( \delta_{ij} \delta_{jk} - \delta_{ik}\delta_{jj})(A.C)_j(B.D)k [/tex]

= [tex] [ \delta_{ij}(A.C)_j ][ \delta_{jk}(B.D)_k ] - [ \delta_{ik}(B.D)_k ][ \delta_{jj}(A.C)_j ] [/tex]

= [tex](A.C)_i(B.D)j - (B.D)_i(A.C)_j[/tex]

=(A.C)(B.D) - (B.D)(A.C)

the first part of the answer (in red) i got right.. but the 2nd part is wrong as you can see

how am i meant to get -(A.D)(B.C)???
thanks guys :) please help out
 
  • #5
Is there anybody who knows how to calculate Eijk(ijk is sbuindex) times itself. The value is 6 but I need to prove that. Thanks
 

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