hw3.solns

# hw3.solns - Physics 318 Solutions to Homework#3 1 E = A B=A...

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Physics 318: Solutions to Homework #3 February 18, 2008 1 E = −∇ Φ ˙ A B = ∇ × A L ( x , ˙x , t ) = 1 2 m ˙x 2 q Φ( x , t ) + q ˙x · A ( x , t ) (1) a.) Writing out all the components explicitly, the Lagrangian is given by L ( x, y, z, ˙ x, ˙ y, ˙ z, t ) = m 2 ( ˙ x 2 + ˙ y 2 + ˙ z 2 ) q Φ( x, y, z, t )+ q [ ˙ xA x ( x, y, z, t )+ ˙ yA y ( x, y, z, t )+ ˙ zA z ( x, y, z, t )] . We compute d dt L ˙ x L ∂x = 0 , (2) and similarly for y and z , remembering that d dt = ∂t + ˙x · ∇ = ∂t + ˙ x ∂x + ˙ y ∂y + ˙ z ∂z . (3) Using L ˙ x = m ˙ x + qA x (4) leads to m ¨ x = q bracketleftBigg ˙ A x ( ˙ x ∂x + ˙ y ∂y + ˙ z ∂z ) A x Φ ∂x + ∂x ( ˙ xA x + ˙ yA y + ˙ zA z ) bracketrightBigg (5) and similarly for y and z . In vector notation, this becomes m ¨x = q ( −∇ Φ ˙ A ) + q [ ( ˙x · A ) ( ˙x · ∇ ) A ] (6) = q E + q ˙x × ( ∇ × A ) (7) = q E + q ˙x × B (8) We can simplify the computations leading to this result by writing ( x, y, z ) as ( x 1 , x 2 , x 3 ) and intro- ducing the notation x i = ( x 1 , x 2 , x 3 ). We use x i and x i interchangably, there is no significance to the position of the index, up or down. The Lagrangian can be expressed as L ( x i , ˙ x i , t ) = 1 2 m 3 summationdisplay i =1 ˙ x i ˙ x i q Φ( x i , t ) + q 3 summationdisplay i =1 ˙ x i A i ( x i , t ) . (9) 1

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We then compute d dt L ˙ x i L ∂x i = 0 , (10) with d dt = ∂t + 3 summationdisplay i =1 ˙ x i ∂x i . (11) Using L ˙ x i = m ˙ x i + qA i (12) leads to m ¨ x i = q ˙ A i Φ ∂x i 3 summationdisplay j =1 ˙ x j ∂A i ∂x j + ∂x i 3 summationdisplay j =1 ˙ x j A j (13) Note that the indices i match up on the left and right hand side of this equation, and there is no sum on i . We have changed the summation index to be j ; an index that is summed over is just a dummy variable and can always be relabeled. We can see that this Eq. (13) is equivalent to Eq. (8) as follows. The cross product can be written as ( ˙x × B ) i = 3 summationdisplay j,k =1 ǫ ijk ˙ x j B k = 3 summationdisplay j,k =1 ǫ ijk ˙ x j 3 summationdisplay l,m =1 ǫ klm ∂x l A m = 3 summationdisplay j,k,l,m =1 ǫ ijk ǫ klm ˙ x j ∂x l A m (14) Here, ǫ ijk is the antisymmetric Levi-Civita symbol; it is +1 if ( i, j, k ) = (1 , 2 , 3) or any cyclic permu- tation, 1 if ( i, j, k ) = (1 , 3 , 2) or any cyclic permutation, and 0 if any of the indices are the same.
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