(a) Show that electric field and magnetic field have units (force)/area or energy/volume.

Transcription

1 Problem. Units (a) how that electric field and magnetic field have units (force)/area or energy/volume. (b) A rule of thumb that you may need in the lab is that coaxial cable has a capcitance of 2 pf/foot. That is why cable length must be kept to a minimum in high speed electronics. The order of magnitude of this result is set by ɛ o = 8.85 pf/m. In the Heavyside- Lorentz system capacitance is still Q HL = C HL V HL. how that C HL has units of meters, and that ( ) CHL C MK = 8.85 pf () meters (c) The impedance of the vacuum is Z o = µ o /ɛ o = 376 Ohms. This is why high frequency antennas will typically have a radiation resistance of this order of magnitude. As this problem will discuss, the unit of resistance is s/m for the Heavyside Lorentz system, and the impedance of the vacuum is /c In Heavyside-Lorentz units Ohm s law still reads, j HL = σ HL E HL, where σ HL is the conductivity, and j is the current per area. how that the conductivity in Heavyside- Lorentz has units [σ HL ] = /seconds and that σ MK = σ HL ɛ o. Then show that a wire of length L and radius R o has resistance What is σ HL for copper? R MK =376 Ohms (R HL c) (2) ( ) Lc =376 Ohms (3) πroσ 2 HL

2 Problem 2. Vector Identities (a) Use the epsilon tensor to prove the analog of b(ac)-(ab)c rule for curls ( V ) = ( V ) 2 V (4) Use this result, together with the Maxwell equations in the absence of charges and currents, to establish that E and B obey the wave equation (b) When differentating /r we write with x = x i e i, and use results like c 2 2 t B 2 B =0 (5) c 2 2 t E 2 E =0 (6) r = (7) xi x i i x j = δ j i i x i = δ i i = d = 3 (8) where d = 3 is the number of spatial dimensions. (It is usually helps to write this as d rather than 3 to get the algebra right). In this way, one finds that field due to a electric charge (monopole) is the familiar ˆr/r 2 j-th component of (/r) = where ˆr n = x/r. ( ) = j = r j xi x i Using tensor notation (i.e. indexed notation) show that 2 (xi δ ji + x i δj) i 3 xk x k = x j r = (ˆr) j 3 r 2 (9) ˆr r 2 = 0 (0) (c) Using the tensor notation (i.e. indexed notation) show that for constant vector p (and away from r = 0) that ( p n ) 3(n p)n p = () 4πr 2 4πr 3 Remark: φ dip = p n/(4πr 2 ) is the electrostatic potential due to an electric dipole p, and Eq. () records the corresponding electric field. Notice the /r 3 as opposed to /r 2 for the monopole, and, taking p along the z-axis, notice how the electric field points at θ = 0 (or n = ẑ) and θ = π/2 (or n = ˆx). How could you derive this using the identities on the front cover of Jackson? 2

3 Problem 3. Easy important application of Helmholtz theorems (a) Using the source free Maxwell equations (i.e. those without ρ and j) and the Helmholtz theorems, explain why E and B can be written in terms of a scalar field Φ (the scalar potential) and a vector field A (the vector potential) B = A (2) E = c ta Φ (3) Thus two of the four Maxwell equations are trivially solved by introducing Φ and A. (b) how that A and Φ are not unique, i.e. A i =(A old ) i + i Λ(t, x) (4) Φ =(Φ old ) c tλ(t, x) (5) gives the same E and B fields. Here Λ(t, x) is any function. This change of fields is known as a gauge transformation of the gauge fields (Φ, A). 3

4 Problem 4. Tensor decomposition (a) Consider a tensor, and define the symmetric and anti-symmetric components = ( 2 + T ji) (6) A = ( 2 T ji) (7) so that = + A. how that the symmetric and anti-symmetric components don t mix under rotation =R i lr j mt lm (8) A =R i lr j mta lm (9) This means that I don t need to know T A if I want to find T in a rotated coordinate system. Remarks: We say that the general rank two tensor is reducable to = into two tensors that dont mix under rotation (b) You should recognize that an antisymmetric tensor is isomorphic to a vector + A V i 2 ɛ ijkt jk A (20) Explain qualitatively the identity ɛ ijk ɛ lmk = δ i l δj m δ j l δi m using ɛ ij3 ɛ lm3 as an example, and use this to show A = ɛijk V k (2) Remark: In matrix form this reads T A = 0 V z V y V z 0 V x (22) V y V x 0 (c) Using the Einstein summation convention, show that the trace of a symmetric tensor is rotationally invariant T i i T i i (23) and that is traceless. 3 δij T l l (24) Remark: A symmetric tensor is therefore reducable to a symmetric traceless tensor and a scalar times δ ij. = + 3 δij T l l where T l 3 lδ ij (25) I don t need to know T l l in order to compute = Ri l Rj mt lm 4

5 Remarks: The results of this problem show that a general second rank tensor is decomposable into irreducable components = T ij + ɛijk V k + T l 3 lδ ij (26) = ( + T ji l l δ ij) + 2 ɛijk ɛ klm T lm + 3 T l l δ ij (27) No further reduction is possible. A general result is that a fully symmetric traceless tensor is irreducable. When this result is applied to the product of two vectors it says E i B j = 2 ( E i B j + B i E j 2 3 E Bδij) + 2 ɛijk (E B) k + 3 E Bδij (28) which expresses the tensor product of two vectors as the sum of an irreducable (traceless and symmetric) tensor, a vector, and a scalar, = 2 0. More physically it says that not all of E i B j is really described by a tensor. Rather, part of E i B j is described by the vector E B, and part is described by the scalar E B. It is for this reason that the tensors we work with in physics (i.e. the moment of inertia tensor, the quadrupole tensor, the maxwell stress tensor) are symmetric and traceless. 5

6 Problem 5. 3d delta-functions A delta-function in 3 dimensions δ 3 (r r o ) is an infinitely narrow spike at r o which satisfies d 3 r δ 3 (r r o ) = (29) In spherical coordinates, where the measure is we must have d 3 r = r 2 dr d(cos θ) dφ = r 2 sin θ dr dθ dφ, (30) δ 3 (r r o ) = r 2 δ(r r o)δ(cos θ cos θ o )δ(φ φ o ) = so that d 3 r δ 3 (r) =. For a general curvlinear coordinate system where u a o are the coordinates of r o. r 2 sin θ δ(r r o) δ(θ θ o )δ(φ φ o ) (3) δ 3 (r r o ) = δ(u a u a g o) (32) (a) What is formula δ 3 (r r o ) for cylindrical coordinates? (b) A uniform ring of charge Q and radius a sits at height z o above the xy plane, and the plane of the ring is parallel to the xy plane. Express the charge density ρ(r) (charge per volume) in spherical coordinates using delta-functions. Check that the volume integral of ρ(r) gives the total Q. a z o y x 6

7 Problem 6. Fourier Transforms of the Coulomb Potential The fourier transfrom takes a function in coordinate space and represents in momentum space F (k) = dx [ e ikx] f(x) (33) The inverse transformation repesents a function as a sum of plane waves F (x) = dk [ ] e ikx F (k) (34) 2π The Fourier transform generalizes the concept of a fourier series to non-periodic, but square integrable functions i.e. dx f(x) 2 should converge. The Fourier transform of a 3D function r = (x, y, z) is: F (k) = d 3 r [ e ik r] F (r) (35) F (r) = d 3 k (2π) 3 [ e ik r ] F (k) (36) To do this problem you will need to know (as dicussed in class) that the integral of a pure phase e ikx is proportional to a delta-fcn. In 3D we have d δ 3 3 k (r) = eik r (37) (2π) 3 (2π) 3 δ 3 (k) = d 3 r e ik r (38) (a) Use tensors notation to show that the Fourier transform of F (r) is ikf (k), (39) and that the Fourier transform of the curl of a vector vector field F (r) is F (r) is ik F (k) (40) (b) The genral rule is to replace ik. What is the Fourier transform of 2 F (r) (c) Prove the Convolution Theorem, i.e. the Fourier Transform of a product is a convolution d d 3 r e i k r F (r) 2 3 k = (2π) F (k)f (k k) (4) 3 making liberal use of the completeness integrals d 3 r e ik r = (2π) 3 δ 3 (k) (42) The notation of putting e ikx in square brackets is not standard, but I have used it in the notes to highlight the similarity between this expansion and other eigenfunction expansions. 7

8 Remark: etting k = 0 we recover Parseval s Theorem d d 3 r F (r) 2 3 k = (2π) F 3 (k) 2 (43) Remark: This is often used in reverse, the fourier transform of a convolution is a product of the fourier transforms F.T. of d 3 r o F (r o ) G(r r o ) = F (k)g(k) (44) (d) The Fourier transform of the Coulomb potential is difficult (try it and find out why!). This is because /(4πr) is not in the space of square integrable functions (Why?). Thus, we will consider the Fourier transform of /(4πr) to be the limit as m 0 of the Fourier transform of a screened Coulomb potential known as the Yukawa potential Φ(x) = e m r 4π r (45) The Yukawa potential is square integrable. how that the Fourier transform of the Yukawa potential is Φ(k) = (46) k 2 + m 2 with k = k 2. Thus, we conclude with m 0 that d 3 x e ik r 4πr = k 2 (47) Note that the inverse transform can be computed by direct integration 4π r r o = d 3 k (2π) 3 e ik (r ro) k 2 (48) (e) In electrostatics the electric field is the negative gradient of the potential, E = Φ. From E = ρ, we derive the Poisson equation 2 Φ = ρ. For a unit charge at the origin, the coulomb potential, /(4πr), satisfies Deduce Eq. (47) by fourier transforming this equation. 2 Φ = δ 3 (r) (49) 8