User talk:Selvasuriyascience

Knowledge of differential forms and identification of the vector field A = (a1, a2, a3), A = ω A = a 1 d x 1 + a 2 d x 2 + a 3 d x 3 {\displaystyle \mathbf {A} =\omega _{\mathbf {A} }=a_{1}dx_{1}+a_{2}dx_{2}+a_{3}dx_{3}} ￼ A = ∗ ω A = a 1 d x 2 ∧ d x 3 + a 2 d x 3 ∧ d x 1 + a 3 d x 1 ∧ d x 2 {\displaystyle \mathbf {A} ={}^{*}\omega _{\mathbf {A} }=a_{1}dx_{2}\wedge dx_{3}+a_{2}dx_{3}\wedge dx_{1}+a_{3}dx_{1}\wedge dx_{2}} ￼ admits a proof similar to the proof using the pullback of ωF. Under the identification ωF = F the following equations are satisfied. ∇ × F = d ω F ψ ∗ ω F = P 1 d u + P 2 d v ψ ∗ ( d ω F ) = ( ∂ P 2 ∂ u − ∂ P 1 ∂ v ) d u ∧ d v {\displaystyle {\begin{aligned}\nabla \times \mathbf {F} &=d\omega _{\mathbf {F} }\\\psi ^{*}\omega _{\mathbf {F} }&=P_{1}du+P_{2}dv\\\psi ^{*}(d\omega _{\mathbf {F} })&=\left({\frac {\partial P_{2}}{\partial u}}-{\frac {\partial P_{1}}{\partial v}}\right)du\wedge dv\end{aligned}}} Here, d stands for exterior derivative of the differential form, ψ∗ stands for pull back by ψ and, P1 and P2 are same as P1 and P2 of the body text of this article respectively.