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The Gradient Operator Copyright C by Joel A Last Latexed: June 30, 2011 at 15:50 1 The gradient operator Copyright c by Joel A. Shapiro We can define the gradient operator ∂ ~ e . ∇ = X ˆi ∂x (1) i i While this looks like an ordinary vector, the coefficients are not numbers Vi but are operators, which do not commute with functions of the coordinates xi. We can still write out the components straightforwardly, but we must be careful to keep the order of the operators and the fields correct. The gradient of a scalar field Φ(~r) is simply evaluated by distributing the gradient operator ∂ ∂ ~ e ~r e Φ . ∇Φ=(X ˆi ∂x )Φ( )= X ˆi ∂x (2) i i i i ∂AB ∂A B A ∂B Because the individual components obey the Leibnitz rule ∂xi = ∂xi + ∂xi , so does the gradient, so if A and B are scalar fields, ∇~ AB =(∇~ A)B + A∇~ B. (3) The general application of the gradient operator ∇~ to a vector A~ gives an object with coefficients with two indices, a tensor. Some parts of this tensor, however, can be simplified. The first (which is the trace of the tensor) is called the divergence of the vector, written and defined by ∂ ∂B ∂B ~ A~ e e B e e j δ j ∇· = (X ˆi ∂x ) · (X ˆj j)= X ˆi · ˆj ∂x = X ij ∂x i i j ij i ij i ∂B i . = X ∂x (4) i i In asking about Leibnitz’ rule, we must remember to apply the divergence operator only to vectors. One possibility is to apply it to the vector V~ = ΦA~, with components Vi = ΦAi. Thus ∂ A ∂ ∂A ~ A~ (Φ i) Φ A i ∇· (Φ ) = X ∂x = X ∂x i + Φ X ∂x i i i i i i = (∇~ Φ) · A~ + Φ∇·~ A.~ (5) 615: Intro to Noether’s Theorem Last Latexed: June 30, 2011 at 15:50 2 We could also apply the divergence to the cross product of two vectors, ∂ A~ B~ ∂ ǫ A B ∂ A B ~ A~ B~ ( × )i (Pjk ijk j k) ǫ ( j k) ∇· ( × ) = X ∂x = X ∂x = X ijk ∂x i i i i ijk i ∂A ∂B ǫ j B ǫ A k . = X ijk ∂x k + X ijk j ∂x (6) ijk i ijk i This is expressible in terms of the curls of A~ and B~ . The curl is like a cross product with the first vector replaced by the differential operator, so we may write the i’th component as ∂ ~ A~ ǫ A . (∇ × )i = X ijk ∂x k (7) jk j We see that the last expression in (6) is ∂A ∂B ǫ j B A ǫ k ~ A~ B~ A~ ~ B~ . X(X kij ∂x ) k − X j X jik ∂x =(∇ × ) · − · (∇ × ) (8) k ij i j ik i where the sign which changed did so due to the transpositions in the indices on the ǫ, which we have done in order to put things in the form of the definition of the curl. Thus ∇·~ (A~ × B~ )=(∇~ × A~) · B~ − A~ · (∇~ × B~ ). (9) Vector algebra identities apply to the curl as to any ordinary vector, except that one must be careful not to change, by reordering, what the differential operators act on. In particular, the “bac-cab” equation becomes ∂B~ A~ ~ B~ A ~ B A . × (∇ × )= X i∇ i − X i ∂x (10) i i i.
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