kolahal_b
16th May 2007 - 02:54 PM
we are to show a=(1/2) closed loop integral over [r x dl]
I suppose this can be done formally from the alternative form of Stokes' theorem that can be obtained by replacing the vector field in curl theorem by VxC where C is a constant vector
The identity is :
surface int [(da x grad) x V]=closed loop integral over [dl x V]
The RHS matches.But how to show that LHS leads to the required value?
AlphaNumeric
18th May 2007 - 11:56 PM
Work in suffix notation
(da x ▼)_i = ε_(ijk) da_j ▼_k
[(da x ▼) x r]_m = ε_(min) ( ε_(ijk) da_j ▼_k ) r_n
= ε_(min)ε_(ijk) da_j (▼_k r_n)
= ε_(min)ε_(ijk) da_j δ_kn
= ε_(inm)ε_(ijn) da_j
= (δ_nj δ_mn - δ_nn δ_mj ) da_j
= (δ_mj - 3δ_mj) da_j
= -2 da_m
Therefore the left hand side is Int_s (-2 da) = -2 Int_s da = -2 a
Therefore
-2a = Int_c (dl x r)
a = 1/2 Int_c (r x dl)
Done.
kolahal_b
19th May 2007 - 01:07 AM
Thank you very m,uch.I did this in another method.
You just dot int [r x dl] with a constant vector.Insert it within the integral.Use Stokes's theorem and the result follows...
AlphaNumeric
19th May 2007 - 08:58 AM
True, but I got the impression from your first post you might have wanted it done another way.
Also, you'll find that plenty of questions like this can be trivialised if you work in suffix notation so it's a very good idea to be competant at it.