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Formulae involving ∇ Vector Identities with Proofs: Nabla Formulae ...

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=<br />

⎡⎛<br />

∂A<br />

⎤ ⎡<br />

⎤<br />

3 ∂A2<br />

⎞ ⎛ ∂A3<br />

∂A1<br />

⎞ ⎛ ∂A2<br />

∂A1<br />

⎞ ⎛ ∂B3<br />

∂B2<br />

⎞ ⎛ ∂B3<br />

∂B1<br />

⎞ ⎛ ∂B2<br />

∂B1<br />

⎞<br />

⎢⎜<br />

− ⎟i<br />

− ⎜ − ⎟ j + ⎜ − ⎟k⎥<br />

+ ⎢⎜<br />

− ⎟i<br />

− ⎜ − ⎟ j + ⎜ − ⎟k⎥<br />

⎣⎝<br />

∂y<br />

∂z<br />

⎠ ⎝ ∂x<br />

∂z<br />

⎠ ⎝ ∂x<br />

∂y<br />

⎠ ⎦ ⎣⎝<br />

∂y<br />

∂z<br />

⎠ ⎝ ∂x<br />

∂z<br />

⎠ ⎝ ∂x<br />

∂y<br />

⎠ ⎦<br />

=<br />

i j k<br />

∂ ∂ ∂ +<br />

∂x<br />

∂y<br />

∂z<br />

A A A<br />

∴ <strong>∇</strong>x(A + B) = <strong>∇</strong>xA + <strong>∇</strong>xB<br />

(7) Prove <strong>∇</strong>.(φA) = (<strong>∇</strong>φ).A + φ(<strong>∇</strong>.A)<br />

⎛ ∂ ∂ ∂ ⎞<br />

<strong>∇</strong> .( φ ) = ⎜ i + j + k ⎟ 1 2 + φ<br />

⎝ ∂x<br />

∂y<br />

∂z<br />

⎠<br />

1<br />

2<br />

3<br />

i<br />

∂<br />

∂x<br />

B<br />

. ( φAi<br />

+ φA<br />

j A k)<br />

A 3<br />

= ∂( φ A1<br />

) ∂(<br />

φA2<br />

) ∂(<br />

φA3<br />

) = LHS<br />

∂x<br />

+<br />

∂y<br />

+<br />

∂z<br />

⎛ ∂φ<br />

∂φ<br />

∂φ<br />

⎞<br />

⎡⎛<br />

∂ ∂ ∂ ⎞<br />

⎤<br />

( <strong>∇</strong>φ ). A + φ(<br />

<strong>∇</strong>.<br />

A)<br />

= ⎜ i + j + k ⎟.<br />

( A1<br />

i + A2<br />

j + A3k<br />

) + φ ⎢⎜<br />

i + j + k ⎟.<br />

( A1i<br />

+ A2<br />

j + A3k<br />

)⎥<br />

⎝ ∂x<br />

∂y<br />

∂z<br />

⎠<br />

⎣⎝<br />

∂x<br />

∂y<br />

∂z<br />

⎠<br />

⎦<br />

=<br />

1<br />

j<br />

∂<br />

∂y<br />

B<br />

2<br />

k<br />

∂<br />

∂z<br />

B<br />

⎛ ∂φ<br />

∂φ<br />

∂φ<br />

⎞ ⎛ ∂A1<br />

∂A2<br />

∂A3<br />

⎞<br />

⎜ A1<br />

+ A2<br />

+ A3<br />

⎟ + φ⎜<br />

+ + ⎟<br />

⎝ ∂x<br />

∂y<br />

∂z<br />

⎠ ⎝ ∂x<br />

∂y<br />

∂z<br />

⎠<br />

= ⎛ ∂φ<br />

∂A1<br />

⎞ ⎛ ∂φ<br />

∂A2<br />

⎞ ⎛ ∂φ<br />

∂A3<br />

⎞<br />

⎜ A1<br />

+ φ ⎟ + ⎜ A2<br />

+ φ ⎟ + ⎜ A3<br />

+ φ ⎟<br />

⎝ ∂x<br />

∂x<br />

⎠ ⎝ ∂y<br />

∂y<br />

⎠ ⎝ ∂z<br />

∂z<br />

⎠<br />

= ∂( φ A1<br />

) ∂(<br />

φA2<br />

) ∂(<br />

φA3<br />

) = RHS<br />

∂x<br />

+<br />

∂y<br />

LHS = RHS<br />

+<br />

∂z<br />

∴ <strong>∇</strong>.(φA) = (<strong>∇</strong>φ).A + φ(<strong>∇</strong>.A)<br />

(8) Prove <strong>∇</strong>x(φA) = (<strong>∇</strong>φ)xA + φ(<strong>∇</strong>xA)<br />

<strong>∇</strong> x<br />

( φA)<br />

i<br />

∂<br />

=<br />

∂x<br />

φA<br />

1<br />

j<br />

∂<br />

∂y<br />

φA<br />

2<br />

k<br />

∂<br />

∂z<br />

φA<br />

3<br />

= ⎛ ∂( φ A3<br />

) ∂(<br />

φA2<br />

) ⎞ ⎛ ∂(<br />

φA3<br />

) ∂(<br />

φA1<br />

) ⎞ ⎛ ∂(<br />

φA2<br />

) ∂(<br />

φA1<br />

) ⎞<br />

⎜ − ⎟i<br />

− − j + ⎜ − ⎟k<br />

⎝<br />

∂y<br />

∂z<br />

⎠<br />

⎜<br />

⎝<br />

∂x<br />

∂z<br />

⎟<br />

⎠<br />

⎝<br />

∂x<br />

= ⎛ ∂A3 ∂φ<br />

∂A2<br />

∂φ<br />

⎞ ⎛ ∂A3<br />

∂φ<br />

∂A1<br />

∂φ<br />

⎞ ⎛ ∂A2<br />

∂φ<br />

∂A1<br />

∂φ<br />

⎞<br />

⎜ + A −φ<br />

− A ⎟i<br />

− φ + A −φ<br />

− A j + ⎜φ<br />

+ A −φ<br />

− A ⎟k<br />

=<br />

φ 3<br />

2<br />

3<br />

1<br />

2<br />

1<br />

⎝<br />

∂y<br />

∂y<br />

∂y<br />

∂y<br />

⎠<br />

⎜<br />

⎝<br />

∂x<br />

∂x<br />

⎡⎛<br />

∂φ<br />

∂φ<br />

⎞ ⎛ ∂φ<br />

∂φ<br />

⎞ ⎛ ∂φ<br />

∂φ<br />

⎞ ⎤<br />

⎢⎜<br />

A3<br />

− A2<br />

⎟i<br />

− ⎜ A3<br />

− A1<br />

⎟ j + ⎜ A2<br />

− A1<br />

⎟k⎥<br />

⎣⎝<br />

∂y<br />

∂y<br />

⎠ ⎝ ∂x<br />

∂z<br />

⎠ ⎝ ∂x<br />

∂y<br />

⎠ ⎦<br />

3<br />

3<br />

∂y<br />

∂z<br />

⎠<br />

⎟<br />

∂z<br />

⎠<br />

⎝<br />

∂x<br />

∂x<br />

∂y<br />

∂y<br />

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