Homework 7-8 Solutions. Problems

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1 Homework 7-8 Solutions Problems 26 A rhombus is a parallelogram with opposite sides of equal length Let us form a rhombus using vectors v 1 and v 2 as two adjacent sides, with v 1 = v 2 The diagonals of the rhombus are d 1 = v 1 + v 2 and d 2 = v 1 v 2 The angle θ between the two diagonals is obtained from cos θ = d 1 d 2 / d 1 d 2 We can obtain this angle from d 1 d 2 = ( v 1 + v 2 ) ( v 1 v 2 ) = v v 2 v 1 v 1 v 2 v 2 2 = v 1 2 v 2 2 = 0 So cos θ = 0 and the diagonals are orthogonal 27 ɛ ijk ɛ ijk = (ɛ 123 ) 2 + (ɛ 132 ) 2 + (ɛ 213 ) 2 + (ɛ 231 ) 2 + (ɛ 312 ) 2 + (ɛ 321 ) = 6 (All other ɛ ijk s are 0) Or ɛ ijk ɛ ijk = δ il δ jm (ɛ ijk ɛ lmk ) = δ il δ jm (δ il δ jm δ im δ jl ) = = 6 28 (a) [(A )B + (B )A + A ( B) + B ( A)] i = A j j B i + B j j A i + ɛ ijk A j (ɛ klm l B m ) + ɛ ijk B j (ɛ klm l A m ) = A j j B i + B j j A i + ɛ ijk ɛ klm (A j l B m + B j l A m ) = A j j B i + B j j A i + (δ il δ jm δ im δ jl )(A j l B m + B j l A m ) = A j j B i + B j j A i + (A j i B j + B j i A j A j j B i B j j A i ) = A j i B j + B j i A j = i (A j B j ) (chain rule for derivatives) = [ (A B)] i (b) V ( U) U ( V) = V i ɛ ijk j U k U i ɛ ijk j V k = ɛ kij V k i U j ɛ jik U j i V k (just re-labeling indices) = ɛ ijk V k i U j + ɛ ijk U j i V k (antisymmetry of ɛ ijk ) = ɛ ijk i (V k U j ) (chain rule for derivatives) = i (ɛ ijk U j V k ) = (U V) 29 We can write the charge distribution as ρ( r) = qδ (3) ( r) + q δ(z)δ(r R) The δ (3) ( r) is defined so δ (3) d 3 r( r) if the integral includes the origin, and the second term puts the remaining charge in the xy plane (δ(z)) at radius R (δ(r R)) with appropriate normalization I am using cylindrical coordinates (r, θ, z) We calculate the first 3 electric multipole moments (a) Monopole: Q = d 3 r ρ( r) = q + q dz r dr dθ δ(z)δ(r R) = q + q () = 0 This verifies that we have the correct normalization for the second term of ρ( r)

2 (b) Dipole: By symmetry, the only possible direction that this could point in is the z direction But Q 3 = d 3 r ρ( r)z = 0, since both terms in ρ( r) contain δ(z) (All of the charge lies in the plane z = 0) Thus, we expect (Q 1, Q 2, Q 3 ) = (0, 0, 0) Let us check for Q 1 : Q 1 = d 3 r ρ( r)x = d 3 r ρ( r)r cos θ = = qr 2π 2π 0 dθ cos θ = 0 So we have Q 1 = Q 2 = Q 3 = 0 q dz r dr dθ δ(z)δ(r R)r cos θ (c) Quadrupole: Use Q ij 2 d 3 r ρ( r)(3x i x j r 2 δ ij ) By symmetry of ρ( r) and the fact that Q ij is symmetric and traceless, we expect Q 12 = Q 21, Q 13 = Q 31 = Q 23 = Q32, Q 11 = Q 22, and Q 33 = Q 11 Q 22 Thus, we only need to calculate three components: Q 12 q 2 dz r dr dθ δ(z)δ(r R)3(r cos θ)(r sin θ) = 3 qr 2 2π 2 2π 0 dθ sin θ cos θ = 0 Q 13 q 2 dz r dr dθ δ(z)δ(r R)3(r cos θ)(z) = 0 Q 11 q 2 dz r dr dθ δ(z)δ(r R) [3(r cos θ) 2 (r 2 + z 2 )] qr 2 2π 2 2π 0 dθ [3 cos 2 θ 1] qr 2 (π) 2 2π = qr2 4 Although we know that Q 33 = 2Q 11 = 2Q 22 by symmetry and tracelessness, let us calculate it anyway as a check: q dz r dr dθ δ(z)δ(r R) [3z 2 (r 2 + z 2 )] Q 33 2 = 1 qr 2 2π 2 2π 0 dθ = qr2 2 Thus, we get Q 11 = Q 22 = qr2, Q 4 33 = qr2, and all other Q 2 ij = 0 30 We need the most general four-index, symmetric and traceless tensor that we can build out of x i, δ ij, and ɛ ijk Of course, since it is symmetric, we can t use ɛ ijk So there are only three possible types of terms: x i x j x k x l, δ ij x k x l, and δ ij δ kl Of these, the first term is already symmetric under interchange of any indices To make the other terms symmetric, we must add all non-identical permutations of indices In this way our tensor must be of the form: X ijkl = x i x j x k x l + b [δ ij x k x l + δ ik x j x l + δ il x k x j + δ jk x i x l + δ jl x i x k + δ kl x i x j ] +c [δ ij δ kl + δ ik δ jl + δ il δ jk ], where b and c are coefficients to be determined, and an overall normalization factor a has already been pulled out Now we just need to make this traceless, by requiring δ ij X ijkl = 0 (Tracing on any two indices will give the same answer, since it has already been made symmetric) This gives 0 = r 2 x k x l + b [3x k x l + x k x l + x k x l + x k x l + x k x l + δ kl r 2 ] + c [3δ kl + δ kl + δ kl ] = x k x l ( r 2 + 7b) + δ kl ( r 2 b + 5c) Since this must be zero for any r, we must have the coefficients of x k x l and δ kl both equal to zero We get b = r 2 /7 and c = +( r 2 ) 2 /35, so X ijkl = x i x j x k x l 1 7 r2 [δ ij x k x l + δ ik x j x l + δ il x k x j + δ jk x i x l + δ jl x i x k + δ kl x i x j ] ( r2 ) 2 [δ ij δ kl + δ ik δ jl + δ il δ jk ]

3 31 In cylindrical coordinates, (q 1, q 2, q 3 ) = (ρ, θ, z), a position in three space is r = (x, y, z) = (ρ cos θ, ρ sin θ, z) (a) The curvilinear unit vectors are obtained from ρ = (cos θ, sin θ, 0) = ê ρ θ = ( ρ sin θ, ρ cos θ, 0) = ρê θ z = (0, 0, 1) = ê z, so we have h ρ, h θ = ρ, h z and ê ρ = (cos θ, sin θ, 0) ê θ = ( sin θ, cos θ, 0) ê z = (0, 0, 1) (b) In cylindrical coordinates: V [ (V ρ ρ ρ 1) + (V θ θ 1 1) + (V z z 1 ρ) ] ρ (ρv ρ) ρ 1 ρ ρ V θ θ + Vz z (c) Now [ for] V : V = [ 1 (1 V ρ 1 ρ θ z) (ρv z θ) ] [ Vz ρ [ ] [ V (1 V z ρ) (1 V ρ z) ] = [ V ρ [ V ] z 1 1 z 1 ρ [ ρ (ρv θ) θ (1 V ρ) ] ρ (ρv ] θ) θ z ] Vz z ρ [ ] (ρvθ ) Vρ ρ θ (d) Fluid flows in a pipe of radius R with a velocity field given by V = V 0 (1 ρ 2 /R 2 ) ê z So V ρ = V θ = 0 and V z = V 0 (1 ρ 2 /R 2 ) The divergence of this field is V = Vz = 0 z The curl of this field is V V z ρ θ êρ Vz ρ êθ = 2V 0ρ ê R 2 θ 32 The position vector r(ξ, η, φ) = (x, y, z) = (u(ξ, η) cos φ, u(ξ, η) sin φ, v(ξ, η)) (a) = u = u (cos φ, sin φ, 0) + (0, 0, 1) (cos φ, sin φ, 0) + (0, 0, 1) = u( sin φ, cos φ, 0) (b) The vector Thus we only need to check is easily seen to be orthogonal to the other two vectors = u u + If u and v satisfy the Cauchy-Riemann equations, u = u and, then = ( ) u + ( u The vectors are orthogonal ) = 0 (c) Now let u(ξ, η) = ξη and v(ξ, η) = (ξ 2 η 2 )/2 (parabolic coordinates) Then

4 = η(cos φ, sin φ, 0) + ξ(0, 0, 1) = ξ(cos φ, sin φ, 0) η(0, 0, 1) = ξη( sin φ, cos φ, 0) From this we find h 1 = = η 2 + ξ 2, h 2 = = η 2 + ξ 2, h 3 = = ξη (d) Now 2 f = 1 (η 2 +ξ 2 ) ξη(η 2 +ξ 2 ) [ 1 ξ [ ( ξ ( ξη η 2 +ξ η 2 2 +ξ 2 ) ( + 1 η η ) + )] f ξ 2 η 2 2 ( ) ξη η 2 +ξ η ξ 2 ( (η 2 +ξ 2 ) ξη ) ] 33 We want to prove that x 1 = (1, 1, 1,, 1, 1) x 2 = (0, 1, 1,, 1, 1) x 3 = (0, 0, 1,, 1, 1) x n = (0, 0, 0,, 0, 1) are linearly independent To do this we must show that n i=1 α i x i = 0 implies all α i = 0 So, let s assume n i=1 α i x i = 0 This means ni=1 α i x i = (α 1, α 1 + α 2, α 1 + α 2 + α 3,, n 1 i=1 α i, n i=1 α i ) = (0, 0, 0,, 0, 0) Looking at the first position of the n-tuple gives α 1 = 0 The second position gives α 1 + α 2 = 0, which leads to α 2 = 0 Continuing by induction, we see that each of the α i = 0 Thus, the n-tuples are linearly independent 34 If the set of n vectors {x i } form a basis for V, then we want to show that the set {c i x i } is also a basis for V, where c i are arbitrary nonzero scalars To show that {c i x i } is a basis, we must show they are linearly independent and they span V 1) Linear independence: Assume n i=1 α i (c i x i ) = 0 This also can be written n i=1 (α i c i )x i = 0 Since the set {x i } are linearly independent, this must imply each of the α i c i = 0 Since we are given that each of the c i 0, this also implies that each of the α i = 0 Therefore, n i=1 α i (c i x i ) = 0 only if all α i = 0 Therefore, the set {c i x i } must be linearly independent 2) It spans V : Assume x V Then we can write x = n i=1 β i x i for some scalars β i Since the c i 0, we can multiply and divide each term by c i to get

5 x = n i=1 β i(c i x i ), where β i = β i /c i Thus, every vector can be written as a linear combination of the {c i x i } The set spans V A geometrical interpretation is that the length of the basis vectors can be scaled, and it will still be a good basis