Introduction and Review Lecture 1

Transcription

1 Introduction and Review Lecture 1 1 Fields 1.1 Introduction This class deals with classical electrodynamics. Classical electrodynamics is the exposition of electromagnetic interactions between the develoment of Maxwell s equations, introduced in 1860, and quantum field theory initially begun in 1920 by Dirac and completed by Feynmann and Schwinger in It is remakable that this exposition remained virtually intact during the revolutionary changes to Newtonial physics in the first quartile of the 20th century. The interpretation of the physics obviously changed, particularly after the introduction of special relativity, but the mathematical expression of the theory remained. Of course, this was because the theory already described photons as waves and the equations were relativistically invarient. 1

2 1.2 Fields Most importantly we describe the interaction of charges through the concept of a field. A field connect charge (currents) to a geometry of space-time. Thus charge modifies surrounding space such that the space affects another charge. While the abstraction of an interaction to include an intermediate step seems irrelevant to those approaching the subject from a Newtonian viewpoint, it is the only way to develop a consistent method to include relativity in the interaction. The concept of action-at-a-distance is not consistent with relativity. Most of you come to the class thinking in terms of force, and forces acting directly between objects. You should now think in terms of fields, i.e. a modification to space-time geometry which affects a charge that is positioned at a geometric point. In fact, it is not the fields obtained from the force equations, but the potentials (which may be more closely aligned with energy) that are fundamental to the dynamics of interactions. 2

3 2 The Electric and Magnetic Field 2.1 The Electric Force Field Experimentally the force between STATIC charges is determined to be; F = q 1q 2 4πǫ 0 1 r 1 r 2 2 ˆr 12 with the parameters defined in figure 1. The unit vectors ˆr i point in opposite directions, demonstrating the direction of the force on q i, and r 1 r 2 = r 12 is the distance between the charges. The constant 1 4πǫ is used 0 to connect the strength of the electromagnetic force to mechanical units (coupling constant). In mks units E is given in volts/m, and the permittivity constant, ǫ 0, is in units of C 2 /Nm 2. The force written in terms of the force field is; F = q 1 E2 E 2 = q 2 4πǫ 0 1 r 1 r 2 2 ˆr 12 We consider an element of distributed charge, dq = ρ dτ, where ρ(x,y, z) is the charge density at a partic- 3

4 ^ r 12 q 1 r 12 r 1 r 2 q 2 ^ r 21 Figure 1: The geometry of the Coulomb force ular spatial position, and dτ = dxdy dz is the volume element. We assume that the interaction is linear, i.e. that the law of superposition holds, so that the force on a charge is the linear sum of all the forces due to all charges. d F i = q i j d E ij = 1 4πǫ 0 d E ij ρ j dτ j r i r j 2 ˆr ij Using superposition the total force is the vector sum; F = d F i = q i d Ei = g i ( 1 4πǫ 0 ) ρ ˆr i dτ j r i r j 2 ˆr ij 4

5 2.2 The Electric Potential Now we can relate this to a scalar potential. For the force between point charges, q i, the force is F = q 1 V 12, where V 12 is the scalar potential function. It is the energy per unit charge for a unit charge placed at that point in space. W/q 1 = F dl = ( V12 ) dl = dv 12 Now it is clear that; E 12 = V 12 = ( 1 4πǫ 0 1 r 1 r 2 ) For a distributed charge we obtain the equation below which represents integration as shown in figure 2. V ( r 1 ) = 1 4πǫ 0 dτ2 ρ( r 2 ) r 1 r 2 ) 2.3 The Magnetic Field The magnetic force is more complicated that the electric force, as it depends on the velocity of the charges and does not act along the line between the charges. We begin by assuming static currents, i.e. charges in motion 5

6 dq j r ij q i Figure 2: The Coulomb force for a distributed charge at constant speed and direction. We define a current element, I dl = qd v, where I is the current in the direction of dl, q is the charge that flows, and d v the elemental velocity. Eventually we extend this to include movement of an elemental volume of charge. Experimentally the force on a small current element by another element of current is given by (a form of) the Biot-Savart law. F = µ 0I 1 I 2 4π dl1 dl 2 r 12 r 1 r 2 3 As with the coupling constant for the electric force, the magnetic coupling is µ 0 /4π. The above force equation can be written: 6

7 Idl 1 Idl 2 r 12 Figure 3: The geometry for the magnetic force F = µ 0I 1 I 2 4π [ dl2 ( dl 1 r r 3 ) ( dl 1 dl 2 ) r r 3] Figure 3 illustrates the geometry. Suppose one integrates completely over both currents, closing the current loop so that the currents in each loop remain static. Upon integration over the loops, the first term in the integral vanishes. This results in; F = µ 0I 1 I 2 4π dl1 dl 2 r r 3 This shows the expected symmetry, F 12 = F 21, but note that this is true ONLY for static currents. The 7

8 time dependent case is considered below. From the above equations we rewrite this force equation in terms of a vector field. Thus the force on a charge, q, moving with velocity, v, due to a current which produces a magnetic field B is : ( B is more correctly called the magnetic induction and H is the magentic field. We ignore this distinction here.) F 12 = q 1 v 1 B 12 B 12 = µ 0I 2 4π dl2 ˆr 12 r 2 12 The mks units of B are Tesla or Weber/m 2 (cgs units are Gauss). The premeability constant, µ 0, has units of Newtons/A 2. Now consider the flow of a volume charge dendity, ρ. This produces a current density, J = ρ v. The flow of charge through a surface area, da, and aligned with the velocity gives the current through the area. di = J da Therefore the above equation for the field is; 8

9 B = µ 0 4π dτ[ J ˆr r 2 ] where dτ is the volume element, and the radial distances and unit vectors were previously defined. 2.4 The vector potential Suppose we consider B. Write this as; B = µ 0 4π dt[ J ˆr r 2 ] B = µ 0 4π [ ˆr r 2 ( J)dτ J ( ˆr/r 2 )dτ] The current density J involves only source coordinates while is the gradient with respect to the field point. Thus J = 0. Then ; ˆr/r 2 = r/r 3 = 0 Thus B = 0 and we may define a vector A such that B = A. B = µ 0 4π J ˆr r 2 dτ = µ 0 4π [ (1/r) J] dτ 9

10 B = µ 0 4π ( J/r)dτ = µ 0 4π J r 1 r 2 dτ Here operates on the field coordinates. Therefore; A = µ 4π J r 1 r 2 dτ + Λ Here Λ is an arbitrary scalar function. The B field and the force are independent of the value of Λ 3 Maxwell s equations 3.1 Equations for a static field We first consider the divergence of the electric field. E = 1 4πǫ 0 ρ 2 ˆr dτ r 1 r 2 The divergence operation results in a delta function δ( r 1 r 2 ) on the left hand side of the equation. We rewrite this by shifting the origin to the point where the arguement of the delta function is zero. E = 4πǫ 1 ρ ( hatr 0 r 2 dτ Then apply Gauss law; 10

11 E = 1 4πǫ 0 ρ ˆr r 2 da Noting that the area has the form r 2 dω, we find that; E = ρ/ǫ 0 This is the differential form of Gauss law. It assumes Coulomb s law and superposition. In integral form it is expressed as ; E da = Q/ǫ 0 where Q is the charge enclosed by the area, da. In the previous section we observed that B = 0. In analogy with the above equation for the electric field, this implies that here is no free magnetic charge. Now we consider the operation of the curl on the electric and magnetic fields. E = ( V ) = 0 Also B = A = ( A) 2 A 2 11

12 Then A = µ 4π 1 ( J r 1 r 2 )dτ 2 Allow the divergence to operate on the integration variable (2). 1 ( J r 1 r 2 ) = 1 J r 1 r 2 + J 1 ( 1 r 1 r 2 ) 1 ( J r 1 r 2 ) = J 2 2 ( 1 r 1 r 2 ) Integration by parts gives J2 2 (( 1 r 1 r 2 dτ 2 = J 2 r 1 r 2 surface + 2 J 2 r 1 r 2 dτ 2 The surface term vanishes. Now suppose a charge density ρ, is enclosed within a volume. Each element of charge has a velocity, v. We have the equation of continuity; da (ρ v) = dτ (ρ v) = dτ ( ρ t ) 12

13 J = ρ t For a static current the charge density in a volume remains constant (inflow = outflow). Thus A = 0. For a static charge; B = 2 A 2 2 A 2 = µ 0 4π dτ2 J2 2 1( 1 r 1 r 2 ) Now 2 (1/r) = 0 unless r = 0. This is a representation of a δ function, to be discussed in more retail below. Suppose r 0 i.e. r 1 r 2. 2 A 2 µ 0 4π J 1 dτ2 2 [ 2 (1/r)] 2 A 2 µ 0 4π J 1 dτ2 [ 2 (1/r) da] = µ 0 4π J 1 (4π) 2 A 2 = µ 0 J This is Ampere s law. For the static force field; B = µ 0 J 13

14 and in integral form; ( B) da = B dl = µ0 J da = µ0 I 3.2 The δ function The δ function is not a proper mathematical function, but a short-hand method for obtaining limiting values of a mathematical sum or integral. The function is singular. The definition of δ(x x 0 ) is b a dxf(x) δ(x x 0 ) = f(x 0 ) a < x 0 < b 1/2f(x 0 ) a,b = x 0 (1) 0 a < 0 or b > 0 It is perhaps easiest to understand the operation of a δ function through several examples. Although not all δ functions are expressed by integrals of sums of complete sets of functions, these provide the best illustrations of the δ function features. Thus consider the Fourier transforms of a function. These are developed through the completeness of the complex harmonic functions, e ikx. 14

15 F(k) = 1 2π f(x) = 1 2π dxe ikx f(x) dk e ikx F(k) We note by substitution that; f(x) = 1 2π dk e ikx [ 1 2π After re-arranging the integration; dx e ikx f(x )] f(x) = dx f(x ) [ 1 2π dk e ik(x x) ] Identify; δ(x x) = [ 1 2π dk e ik(x x) ] Note that the integral for the δ function may be evaluated as δ(x x) = 1 π lim k sin(k[x x]) x x Figure 4 shows this function for a finite value of k. Then for any complete set of functions, g n (x), where 15

16 8 k Figure 4: A representation of the δ function which occurs as k x < x < b we assume; f(x) = n a n g n (x) Here a n are the expansion coefficients. Using the orthonormality of the set of complete functions; a m = b a dxf(x) g m (x) Substitution and re-arrangement of the sum and integral ; f(x) = b a dx f(x ) n g n (x) g n (x ) 16

17 The δ function is then; δ(x x) = n g n (x) g n (x ) At a singularity, the δ function approaches the function value in the mean at this point. Thus this provides a definition for δ(x x 0 ) valid for a x 0 b when a = x 0 or b = x Faraday s Law We have previouisly obtained E = 0 for a static charge. In integal form this is ; E dl = ( E) da which is obtained by Stokes theorem. Thus we note that E = 0. The electromotove force EMF, around a current look is E dl, and by Faraday s law this is connected to the time change of the magnetic flux through the loop. EMF = E dl = φ t 17

18 where the magnetic flux is defined as φ = B da Thus in integral form E dl = B t da The differential form is ; E = B t Now look at the induced EMF in a rotating, conducting disk as shown in figure 5. We have the Lorentz force acting on charges in the disk F = q( E + v B) where q is a charge and v its velocity. We obtain; v = ωrˆθ F = qωrb ˆr Note that the magnetic field is stationary and the circuit is moving, so it would seem that the magnetic flux is 18

19 ω B δθ r v EMF Figure 5: The geometry for the EMF generated in a rotating disk also stationary. However, in this case the EMF is balanced by the electrostatic force so that F = 0. q E = q v B In integral form we must integrate over a wedge as shown in figure 5. Thus ; EMF = E dl = ( v B) EMF = ωb a 0 r dr = ωba 2 /2 19

20 The flux through the wedge is ; dφ = [b r dr]dθ = (Ba 2 /2) dθ Since dθ dt = ω dφ dt = ωba2 /2 as one might have expected. However to better understand what is happening, we note that if the circuit is at rest the E field is at rest and the magnetic field which is generated by an external current is in the same rest frame. If the circuit moves however the E field moves. In that case we need the convective derivative; d dt = t + v Therefore we should write; d B dt = B t + ( v ) B Now for constant v; B B = ( v ) B ( B) v = ( v ) B 20

21 By substitution; d dt d dt d dt B da = ( B ) da + ( v ) B da t B da = [ ( B v)] da + ( B B da = ( B v) dl + ( B t ) da ) da t Collecting this information we understand that Faraday s law needs to be applied to fields in the same frame of reference. Thus ( Emoving v B) dl = B rest t da and we note that; E moving = E rest + v B Again Faraday s law requires that E and B are in the same frame of reference. 21

22 3.4 Displacement current Ampere s law violates the equation of continuity; J + ρ t = 0 Remember that this equation represents conservation of charge. The divergence is the flux out of a volume per unit volume which is balanced by the change in charge density within the volume. Therefore if; B = µ 0 J then ( B) = µ J = 0 This equation should have the form; ( B) = µ[ J + ρ t ] = 0 Now we have that E = ρ/ǫ 0 so that ǫ 0E E is the displacement current. This means that there is an additional current density to be added to Faraday s law; 22

23 B = µ o J + µ 0 ǫ 0 E t We keep B = Maxwell s equations (mks units) Collecting all the equations that define the electromangnetic field; E = ρ/ǫ 0 B = 0 E = B t B = µ 0 J + µ0 ǫ 0 E t (2) Mathematically these equations express the following laws in the above order 1. Coulomb s law and the law of supperposition 2. Magnetic Coulomb s law and the Biot-Savart law 3. Faraday s law 23

24 4. Ampere s law and the equation of continuity 24