homework_3_solution

homework_3_solution - 1 ∂µ T µν = − F νλ Jλ c...

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Unformatted text preview: 1 ∂µ T µν = − F νλ Jλ c ∂λ M µνλ = ∂λ (xµ T νλ − xν T µλ ) µ ν = δλ T νλ − δλ T µλ + xµ ∂λ T νλ − xν ∂λ T µλ = xµ ∂λ T νλ − xν ∂λ T µλ 1 = (xν F µα Jα − xµ F νρ Jρ ) c 1µ = (x Jα F αν − xν Jα F αµ ) c M µνλ ∂λ M µνλ = 0 jν ∂0 ˆ d3 xj 0 = − ˆ d3 x∂ i j i = − ˆ ￿j dS · ￿ = 0 0 Q= ˆ d3 xj 0 M µνλ Q µν = M T µν T µν E 1 Sx c = 1 Sy c 1 c Sz E= ˆ d3 xM µν 0 1 c Sx −σxx −σyx −σzx 1 c Sy −σxy −σyy −σzy 1 c Sz −σxz −σyz −σzz 1 ￿2 ￿ (|E | + |B |2 ) 8π c￿ ￿ ￿ S= E×B 4π M ij 0 = xi T j 0 − xj T i0 = Qij M = 1 ij 1 (x S − S i xj ) = ￿ijk (￿ × S )k x￿ c c 1 d3 x ￿ijk (￿ × S )k x￿ c ˆ 1 t￿ M i00 = xi T 00 − x0 T i0 = xi E − x0 S i = (￿ E − S )i x c c Q i0 M = ˆ t￿ d3 x(￿ E − S )i x c x=0 x=0 ￿ E = −4πσ ex (0 < x < a) ￿ ϕ(x) = −4πσ x(0 < x < a) 2πσ ex ￿ ￿ f = −2πσ 2 ex ￿ T ij ￿ − 81 |E |2 π = 0 0 0 1 ￿2 |E | 8π 0 0 −2πσ 2 = 0 0 1 ￿2 0 |E | 8π 0 2πσ 2 0 0 0 2πσ 2 x = a/2 ￿ f = f i /A · ￿i = e ˆ dAj T ij · ￿i = −2πσ 2 ex e ￿ (0, L/2, 0) (x, 0, 0) (0, −L/2, 0) ￿ E ( x) = ( x2 2q | x | ex ￿ 2 + L ) 3 /2 4 (x, 0, 0) T ij = ￿ − 81 |E |2 π 0 0 ￿ f = f i ￿i = ( e S= ˆ ˆ 0 1 ￿2 8π | E | 0 dtd3 rψ ∗ (i∂t + 0 = 0 ￿ |E |2 1 8π dAj T ij )￿i = ey e ￿ ˆ ∞ 0 ￿2 ∇2 −V )ψ = m2 − 81 · π 4q 2 x 2 2 ( x 2 + L )3 4 1 8π 0 dtd3 r(ψ ∗ i∂t ψ − ˆ 0 0 4q 2 x 2 0 2 (x 2 + L ) 3 4 0 ∞ (r 2 1 8π ˜ δL =0 δ (∂t ψ ∗ ) ˜ δL ￿2 ∇ 2 =− 2 ψ ∗) δ (∂i ψ m ψ∗ · 4q 2 x 2 2 (x 2 + L )3 4 r2 q2 dr2 = 2 ey ￿ r2 3 L + 4) ￿2 ￿ ∗ ￿ ( ∇ψ ) · ( ∇ψ ) − V ψ ∗ ψ ) = m2 ˜ δL = i∂ t ψ − V ψ δψ ∗ ∂i · 0 1 4q 2 r2 q2 2π rdr = ey ￿ 2 + L2 ) 3 8π ( r 2 4 ˆ 0 ˆ ˜ dtd3 rL ˜ ˜ δL δL = ∂i ∗ δψ δ (∂i ψ ∗ ) i∂ t ψ = − ￿2 ∇2 ψ+Vψ m2 µ0 I ￿ B= eφ ￿ 2π r E= B2 µ0 I 2 = 2µ 0 8π r2 El = ˆ R δ /2 µ0 I 2 = 4π E · 2π rdr ˆ R δ /2 dr r µ0 I 2 2R = ln( ) 4π δ El = µ0 I 2 2R ln( ) 4π δ = (10−7 ) · (20)2 · ln( = 2.7 × 10−4 J/m 2·5 )J/m 10 × 10−3 ...
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This note was uploaded on 12/29/2011 for the course PHYSICS 606 taught by Professor Bedaque during the Spring '11 term at Maryland.

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