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The principle of superposition (ie fields add vectorially) means that P ~ r) = E(~ 1 X qi (~r − ~ri ) . 4πǫ0 i |~r − ~ri |3 ~r i ~ri O In the limit of (infinitely) many charges this becomes P ~ r) = 1 E(~ 4πǫ0 Z V ~r − ~r′ (~r − ~r′ ) . dV ′ ρ(~r) |~r − ~r′ |3 ~r ~r′ O To return to a single charge at position ~r1 set ρ(~r) = q1 δ(~r − ~r1 ) which gives the original result. Finally as ∂φ ρ(~r′ ) dV ≡ − , |~r − ~r′ | ∂ri V ~ = −∇φ. ~ where we have defined a potential, φ by E So Z 1 ρ(~r′ ) φ(~r) = . dV ′ 4πǫ0 V |~r − ~r′ | ∂ − E(~r)i = ∂ri 4.3.2 1 4πǫ0 Z ′ The multipole expansion First expand 1 |~ r +~a| for r ≫ a. We have ∞ X 1 1 ~ r )n 1 = (~a · ∇ |~r + ~a| n! r n=0 1 1 1 1 1 + ai ∂i + (ai ∂i aj ∂j ) + · · · r 1! r 2! r 2 2 2 1 1 ~a · ~r 3(~a · ~r) − a r . − 3 + +O = 5 r r 2r r4 = Thus Z 1 ρ(~r′ ) φ(~r) = dV ′ 4πǫ0 V |~r − ~r′ | Z 1 ~r′ · ~r 3(~r′ · ~r)2 − r 2 r ′2 1 ′ ′ dV ρ(~r ) + 3 + + ... , = 4πǫ0 V r r 2r 5 45 (upon using the Taylor expansion). This gives φ(~r) = 1 ri pi 1 ri rj Qij 1 Q + + + ... . 3 4πǫ0 r 4πǫ0 r 4πǫ0 2r 5 where Q = pi = Qij = Z dV ′ ρ(~r′ ) ZV ZV (total) charge or monopole term dV ′ ri′ ρ(~r′ ) dipole moment dV ′ (3ri′ rj′ − r ′2 δij )ρ(~r′ ) quadrupole moment V This expansion is valid in the far zone, ie r ≫ r0 . • If Q 6= 0 the monopole term dominates φ(~r) = 1 Q . 4πǫ0 r ~ field emanates from a point charge at the origin. The E • If Q = 0 the dipole term dominates φ(~r) = 1 p~ · ~r . 4πǫ0 r 3 If the charge density is given by two equal opposite charged particles close ~ − δ(~r′ )] then together, ie ρ(~r′ ) = q[δ(~r′ − d) Z ~ − δ(~r′ )] = q d~ , ~p = dV ′ ~r′ q[δ(~r′ − d) this justifies the name. • If Q = 0, p~ = 0, the quadrupole term dominates φ(~r) = 1 ri rj Qij . 4πǫ0 2r 5 Qij is symmetric (Qij = Qji ) and traceless (Tr Q = 0). A simple example is given by the charge distribution of two opposite dipoles ρ(~r) = q[δ(~r − ~a) − δ(~r)−δ(~r −~a−~b)+δ(~r −~b)] which gives p~ = 0 and Qij = 2q[~a·~bδij − 23 (ai bj +aj bi )]. Similar considerations also hold for the gravitational potential. Simply replace 1 → G, 4πǫ0 q → m. 46 Chapter 5 Some Physical Examples of Tensors: I We now consider some examples 5.1 Polarisation of Media • Electric ~ = ǫ0 E ~ + P~ , D ~ is the (total) electric field, D ~ is the external electric field (from free where E charges) called ‘electric displacement field’ and P~ is the polarisation vector ~ · P~ . We have from (averaged) bound charges, ρb = −∇ ~ = χij Ej + . . . , Pi = Pi (E) for a linear, anisotropic field. • Magnetic ~ = µ0 H ~ + µ0 M ~ B ~ is the external magnetic field (due to external currents), B ~ is (total) where H ~ is the magnetic polarmagnetic field, called ‘magnetic induction field’ and M ~ ×M ~ isation vector from (averaged) bound charges/currents ~jb = ∂ P~ /∂t + ∇ ~ = χij Hj + . . . , Mi = Mi (H) again for a linear, anisotropic field. For both cases χ is the susceptibility tensor (tensor by the quotient theorem). 47 5.2 Ohm’s Law Current density: ~j def = electrical charge transported through a unit area ⊥ to ~j per unit time, ~j and so ~j = j~n , ~n2 = 1 , where j = total charge to cross unit area in direction ~n per unit time. The current is def i = Z ~, ~j · dS (or alternatively I for the current). The units are Cs−1 = A (amps). Ohm’s Law: ~ applied to the material, Current density is proportional to the electric field E ~, ~j = σ E where σ is the conductivity. ~ = ~0 and σ = 0 is a (perfect) insulator σ = ∞ is a (perfect) conductor where E (for example a vacuum). We shall usually consider good conductors, such as copper wire. ~ E φ1 A φ2 The potential difference is V = φ1 − φ2 (measured in volts, V ). Now E = V /l (potential gradient) so j = σV /l and hence Z ~ = jA = σ V A , i = ~j · dS l A 48 or i= V R with R = l . σA R is the resistance, measured in V /A or Ohms, Ω. We can generalise Ohm’s Law by writing ji = σij Ej , where σ is the conductivity tensor (by the quotient theorem). • Layered material of conductor and insulator where the current can only flow in the layers. 1111111111111 0000000000000 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 0000000000000 1111111111111 z y x 1111111111111 0000000000000 0000000000000 1111111111111 0000000000000 1111111111111 σ0 0 0 σ = 0 σ0 0 0 0 0 j1 σ0 E1 j2 = σ0 E2 . j3 0 giving ~ • Uniform material, no prefered direction, isototropic, so σij = σ0 δij and ~j = σ0 E ~ are always parallel to each other. (the original Ohm’s Law when ~j and E 5.3 5.3.1 The Inertia Tensor Angular momentum and kinetic energy Suppose a rigid body of arbitrary shape rotates about a fixed axis with angular velocity ~ω = ω~n. Consider a blob of matter dm at point P , with position vector ω P O ~r 49 dm ~r relative to O. The velocity of a little element dm = ρdV where ρ ≡ ρ(~r) is the (mass) density is ~v = ~ω × ~r . Repeating previous results: • Simple proof δr = δθr sin φ = |δθ~n × ~r| and so ~v = δ~r/δt = ω ~ × ~r where ~ω δθ ~r φ ~ω ~ω = δθ/δt~n. • More complicated proof: Use previous result for rotation matrix R(θ, ~n), δri = Rij (δθ, ~n)rj − ri = (−ǫijk nk δθ + O((δθ)2 ))rj , δθ δri = (~n × ~r)i , δt δt or ~v = ~ω × ~r again. Angular momentum ~ of an element of mass dm = ρdV at ~r is d~h = The angular momentum d~h (or dL) ρ(~r)dV ~r × ~v giving Z ~h = ρ~r × (~ω × ~r)dV , body giving hi = Z ρǫijk rj ǫklm ωl rm dV Z ρ(δil δjm − δim δjl )rj ωl rm dV . body = body Thus hi = Iij ωj , Iij = Z ρ(r 2 δij − ri rj )dV . body 50 The geometric factor I(O) (where O is the origin) is the moment of Inertia tensor. It is a tensor because ~h is pseudovector, ~ω is a pseudovector and hence from the quotient theorem I is a tensor. [Furthermore note that Iij is symmetric and independent of the ~n axis chosen.] Kinetic energy For the kinetic energy, T , we have dT = 21 (ρdV )(~ω × ~r)2 or Z 1 ρǫijk ωj rk ǫilm ωl rm dV T = 2 body Z 1 = ρ(δjl δkm − δjm δkl )ωj rk ωl rm dV 2 body Z 1 = ρ(ω 2 r 2 − (~r · ~ω )2 )dV 2 body Z 1 ρ(r 2 δij − ri rj )dV ωi ωj , = 2 body or 1 1 T = Iij ωi ωj ≡ ~ω I~ω . 2 2 Alternative (more familiar) forms Often write L = T = I (n) ω 1 (n) 2 ω 2I with L = ~h · ~n , I (n) = Iij ni nj . where I (n) is now the moment of inertia with respect to ~n, Z Z (n) 2 2 2 I = Iij ni nj = ρ(r − (~r · ~n) ) dV ≡ ρr⊥ dV , body body with r⊥ being the perpendicular distance from the ~n-axis. Simpler if only interested in one ~n, but for a different ~n must re-compute; Iij contains all the information. Sometimes write I (n) = Mkn2 , where kn is called the radius of gyration (ie distance of a point mass giving the same result). 51 An example For example, a cube of side a of constant density, ρ with M = ρa3 , z y a x O Iij (O) = Z ρ(r 2 δij − ri rj )dV . This gives So by symmetry a dxdydz (x2 + y 2 + z 2 ) − x2 10 3 a = ρ 3 y xz + 13 z 3 xy 0 = 23 ρa5 = 23 Ma2 , Z a = ρ dxdydz(−xy) 0 a = −ρ 21 x2 21 y 2 z 0 = − 41 ρa5 = − 14 Ma2 . I11 = ρ I12 Z I(O) = Ma2 5.3.2 2 3 − 14 − 14 − 41 − 14 2 1 . 3 −4 1 2 −4 3 Parallel Axes Theorem Often simpler to find the moment of inertia tensor about the centre of gravity, G, rather than an arbitrary point, O. There is, however, a simple relationship between ~ =R ~ we have ~r = R ~ + ~r′ , them. Taking O to be the origin, and OG ω P dm ~r′ O ~r ~ R G 52