PHY331: Advanced Electrodynamics & MagnetismPart I: Electrodynamics Contents 1 Field theories and vector calculus 3 1.1 Field theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Vector calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Second derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.5 Index notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2 Electromagnetic forces, potentials, Maxwell’s equations 13 2.1 Electrostatic forces and potentials . . . . . . . . . . . . . . . . . . . . . . . . 13 2.2 Magnetostatic forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.3 Electrodynamics and Maxwell’s equations . . . . . . . . . . . . . . . . . . . 16 2.4 Charge conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.5 The vector potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 3 Electrodynamics with scalar and vector potentials 21 3.1 Scalar and vector potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.2 Gauge transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.3 A particle in an electromagnetic field . . . . . . . . . . . . . . . . . . . . . . 22 4 Electromagnetic waves and Poynting’s theorem 25 4.1 Electromagnetic waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 4.2 Complex fields? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 4.3 Poynting’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 5 Momentum of the electromagnetic field 31 6 Multipole expansion and spherical harmonics 34 6.1 Multipole expansion of a charge density . . . . . . . . . . . . . . . . . . . . 34 6.2 Spherical harmonics and Legendre polynomials . . . . . . . . . . . . . . . 36 6.3 Multipole expansion of a current density . . . . . . . . . . . . . . . . . . . 37 7 Dipole fields and radiation 41 7.1 Electric dipole radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 7.2 Magnetic dipole radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 7.3 Larmor’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 0 8 Electrodynamics in macroscopic media 47 8.1 Macroscopic Maxwell equations . . . . . . . . . . . . . . . . . . . . . . . . 47 8.2 Polarization and Displacement fields . . . . . . . . . . . . . . . . . . . . . . 47 8.3 Magnetization and Magnetic induction . . . . . . . . . . . . . . . . . . . . 49 9 Waves in dielectric and conducting media 52 9.1 Waves in dielectric media . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 9.2 Waves in conducting media . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 9.3 Waves in plasmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 10 Relativistic formulation of electrodynamics 59 10.1 Four-vectors and transformations in Minkowski space . . . . . . . . . . . . 59 10.2 Covariant Maxwell equations . . . . . . . . . . . . . . . . . . . . . . . . . . 62 10.3 Invariant quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 A Special coordinates and vector identities 67 A.1 Coordinate systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 A.2 Integral theorems and vector identities . . . . . . . . . . . . . . . . . . . . . 69 A.3 Levi-Civita tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 B Unit systems in electrodynamics 70 Literature The lecture notes are intended to be mostly self-sufficient, but below is a list of recom- mended books for this course: 1. R.H. Good, Classical Electromagnetism, Saunders College Publishing (1999). Most of the material in this course is covered in this book. It is very accessible and probably should be your first choice to look something up. 2. J.D. Jackson, Classical Electrodynamics, Wiley (1998). This is the standard work on classical electrodynamics, and it has everything that is covered in this course. The level is quite high, but it will answer your questions. 3. R.P. Feynman, Lectures on Physics, Addison-Wesley (1964). Probably the best gen- eral books on physics ever. The emphasis is on the physical intuition, and it is written in a very accessible narrative. 4. J. Schwinger et al., Classical Electrodynamics, Westview Press (1998). This is quite a high-level textbook, containing many topics. It has detailed mathematical deriva- tions of nearly everything. After each lecture there is a detailed “further reading” section that points you to the relevant chapters in various books. 2 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 1 1 Field theories and vector calculus 1.1 Field theory Michael Faraday (1791-1867) Electrodynamics is a theory of fields, and all matter enters the theory in the form of densities. All modern physical theories are field theories, from general relativity to the quantum fields in the standard model and string theory. Therefore, apart from learning some important topics in electromagnetism, in this course you will aquire an under- standing of modern field theories without having to deal with the strangeness of quantum mechanics or the math- ematical difficulty of general relativity. Fields were intro- duced by Michael Faraday, who came up with “lines of force” to describe magnetic phenomena. You will be familliar with particle theories such as clas- sical mechanics, where the fundamental object is char- acterised by a position vector and a momentum vector. Ignoring the possible internal structure of the particles, they have six degrees of freedom (three position and three momentum components). Fields, on the other hand, are characterised by an infinite number of degrees of freedom. Let’s look at some examples: A vibrating string: Every point x along the string has a displacement r, which is a degree of freedom. Since there are an infinite number of points along the string, the displacement r(x) is a field. The argument x denotes a location on a line, so we call the field one-dimensional. Landscape altitude: With every point on a surface (x, y), we can associate a number that denotes the altitude h. The altitude h(x, y) is a two-dimensional field. Since the altitude is a scalar, we call this a scalar field. Temperature in a volume: At every point (x, y, z) in the volume we can measure the temperature T, which gives rise to the three-dimensional scalar field T(x, y, z). Mathematically, we denote a field by F(r, t), where the value of the field at position r and time t is given by the quantity F. This quantity can be anything: if F is a scalar, we speak of a scalar field, and if F is a vector we speak of a vector field. In quantum field theory, the mathematical object that makes the quantum field are operators acting on a vacuum state. In this course we will be mostly dealing with scalar and vector fields, but occasion- ally we will encounter tensor fields. A common example of a tensor field that you may have encountered is the stress in a material. We will discuss the difference between tensors and ordinary matrices in a moment. 3 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 1 1.2 Vector calculus In mechanics, a particle does not randomly jump around in phase space, but follows equations of motion determined by the laws of mechanics and the boundary conditions. These equations of motion typically involve derivatives, namely the velocity and accel- eration of the particle. The derivatives tell you how much a quantity changes. Likewise, fields obey equations of motion, and we need to define the derivatives of fields. First, take the scalar field. The interesting aspect of such a field is how the values of the field change when we move to neighbouring points in space, and in what direction this change is maximal. For example, in the altitude field (with constant gravity) this change determines how a ball would roll on the surface, and for the temperature field it determines how the heat flows. It is easy to see that both a rolling ball and heat flow have a magnitude and a direc- tion. The measure of change of a scalar must therefore be a vector. Since the change is defined at every point r (and time t), it is a vector field. Let the scalar field be denoted by f (x, y, z, t). Then the change in the x direction (denoted by ˆ i) is given by lim h→0 f (x + h, y, z, t) − f (x, y, z, t) h ˆ i = ∂ f (x, y, z, t) ∂x ˆ i . (1.1) A x (r + l ˆ i) A x (r) A y (r) A y A z (r) A z (r + l ˆ k) l l l r Figure 1: The divergence of a vector field A can be found by considering the total flux of Athrough the faces of a small cube of volume l 3 at point r = x ˆ i + y ˆ j + z ˆ k. Similar expressions hold for the change in the y and z direction, and in general the spa- tial change of a scalar field is given by ∂ f ∂x ˆ i + ∂ f ∂y ˆ j + ∂ f ∂z ˆ k = ∇f ≡ gradf , (1.2) called the gradient of f . The “nabla” or “del” symbol ∇ is a differential operator, and it is also a vector: ∇ = ˆ i ∂ ∂x + ˆ j ∂ ∂y + ˆ k ∂ ∂z . (1.3) Clearly, this makes a vector field out of a scalar field. Note that we do not include changes over time in the gradient. We first study static fields. When we want to describe the behaviour of vector fields, there are two main concepts: the divergence and the curl. The divergence is a measure of flow into, or out of, a volume element. Consider a volume element dV = l 3 at point r = x ˆ i + y ˆ j + z ˆ k and a vector field A(x, y, z), as shown in figure 1. We assume that l is very small. The difference of the flux of the field going into the volume 4 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 1 A(x −l, y −l) A(x + l, y −l) A(x + l, y + l) A(x −l, y + l) y x x −l x + l y −l y + l Figure 2: The curl of a vector field A can be found by considering the change of the vector field around a closed loop. and the flux coming out of the volume in the x direction l 2 A x (r +l ˆ i) −l 2 A x (r) is related to the change of the component A x in the x direction. From the Taylor expansion we get the approximation (l ≪1, and ultimately infinitesimal) A x (x + l, y) = A x (x, y) + l ∂A x ∂x (x, y) . (1.4) This leads to l 2 A x (x + l, y, z) −l 2 A x (x, y, z) = l 3 ∂A x ∂x . In order to find the total change in the vector field we have to add the changes in the y and z components as well, which then leads to ∂A x ∂x + ∂A y ∂y + ∂A z ∂z = ∇ A ≡ divA, (1.5) where we removed the common factor l 3 in order to consider the divergence per unit volume. By taking l infinitesimally small, we can properly define the divergence (and later the curl) at a single point. The divergence is a measure of how the vector field A(r) spreads out at position r. The curl of a vector field is a measure of the vorticity of the field. In two dimensions (here meaning the xy plane), consider a four-part infinitesimal closed loop starting at point (x − l, y − l), going to (x + l, y − l), (x + l, y + l) and via (x − l, y + l) back to (x − l, y − l), as shown in figure 2. The area of the square is 4l 2 . The accumulated change of the vector field around this loop is given by the projection of A along the line elements. We evaluate all four sides of the infinitesimal loop in Fig. 2. For example, the line element that stretches from x −l to x + l at y −l is oriented in the x direction, and the corresponding A dl is _ A x ˆ i + A y ˆ j + A z ˆ k _ _ 2l ˆ i + 0 ˆ j + 0 ˆ k _ = 2l A x (x, y −l) , (1.6) 5 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 1 where we evaluate A x in the middle of the line element at the point (x, y −l). We repeat the same procedure for the other three line elements. Make sure you get the orientation right, because the top horizontal part is directed in the −x direction. The corresponding term gets a minus sign. Now add them all up and get A dl around the loop: _ A dl = 2l A x (x, y −l) + 2l A y (x + l, y) −2l A x (x, y + l) −2l A y (x −l, y) . (1.7) Just like for the divergence, we can use the first two terms in Taylor’s expansion of A j (x + l, y) to evaluate this formally, since l will be again an infinitesimal length. For example: A x (x, y + l) = A x (x, y) + l ∂A x ∂y (x, y) . (1.8) This leads to _ A dl = 4l 2 _ ∂A y ∂x − ∂A x ∂y _ . (1.9) For a three-dimensional vector space, we not only have to take the contribution of a loop in the xy plane, but also in the xz and the yz planes. Since there are three orthogonal loops, they can be thought of as the components of a vector. For a three-dimensional vector field we thus have ˆ i _ ∂A z ∂y − ∂A y ∂z _ + ˆ j _ ∂A x ∂z − ∂A z ∂x _ + ˆ k _ ∂A y ∂x − ∂A x ∂y _ = ∇A ≡ curlA, (1.10) the curl of a vector field A per unit surface 4l 2 . We again take l →0 to define the curl of A(r) at a point r. In four dimensions (which is relevant when we do relativity) we have loops not only in the xy, xz, and yz planes, but also in the xt, yt, and zt planes. As you can see, there are now six loops, which can no longer be regarded as components of a three-dimensional vector. Therefore the cross product (and hence the curl operator) as a vector is special to three-dimensional space. In compact matrix notation, the curl can be written as a determinant ∇A = ¸ ¸ ¸ ¸ ¸ ¸ ˆ i ˆ j ˆ k ∂ x ∂ y ∂ z A x A y A z ¸ ¸ ¸ ¸ ¸ ¸ . (1.11) Sometimes, when there are a lot of partial derivatives in an expression or derivation, it saves ink and space to write the derivative ∂/∂x as ∂ x , etc. The curl and the divergence are in some sense complementary: the divergence mea- sures the rate of change of a field along the direction of the field, while the curl measures the behaviour of the transverse field. If both the curl and the divergence of a vector field A are known, and we also fix the boundary conditions, then this determines Auniquely. 6 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 1 It seems extraordinary that you can determine a field (which has after all an infinite number of degrees of freedom) with only a couple of equations, so let’s prove it. Sup- pose that we have two vector fields A and B with identical curls and divergences, and the same boundary conditions (for example, the field is zero at infinity). We will show that a third field C = A−B must be zero, leading to A = B. First of all, we observe that ∇C = ∇A−∇B = 0 ∇ C = ∇ A−∇ B = 0 , (1.12) so C has zero curl and divergence. We can use a vector identity (see Appendix A) to show that the second derivative of C is also zero: ∇ 2 C = ∇(∇ C) −∇(∇C) = 0 −0 = 0 . (1.13) This also means that the Laplacian ∇ 2 of every independent component of C is zero. Since the boundary conditions for A and B are the same, the boundary conditions for C must be zero. So with all that, can C be anything other than zero? Eq. (1.13) does not permit any local minima or maxima, and it must be zero at the boundary. Therefore, it has to be zero inside the boundary as well. This proves that C = 0, or A = B. Therefore, by determining the divergence and curl, the vector field is completely fixed, up to boundary conditions. Looking ahead at the next lecture, you now know why there are four Maxwell’s equations: two divergences and two curls for the electric and magnetic fields (plus their time derivatives). These four equations and the boundary conditions completely deter- mine the fields, as they should. 1.3 Second derivatives In physics, many properties depend on second derivatives. The most important exam- ple is Newton’s second law F = ma, where a is the acceleration, or the second derivative of the position of a particle. It is therefore likely that we are going to encounter the sec- ond derivatives of fields as well. In fact, we are going to encounter them a lot! So what combinations can we make with the gradient, the divergence, and the curl? The div and the curl act only on vectors, while the grad acts only on scalars. More- over, the div produces a scalar, while the grad and the curl produce vectors. If f is a scalar field and A is a vector field, you can convince yourself that the six possible combinations are ∇ (∇f ) ∇(∇ A) ∇(∇f ) ∇ (∇A) ∇(∇A) ∇ 2 A Of these, the first (∇ (∇f )) and the last (∇ 2 A) are essentially the same, since in the latter case the Laplacian acts on each component of A independently, and a component of a vector is a scalar. 7 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 1 Another simplification is that two of the six expressions above are identically zero for any f or A: ∇ (∇A) = 0 and ∇(∇f ) = 0 . (1.14) Since these identities tend to simplify equations a lot, it is a good idea to learn them by heart. The remaining derivatives are related by the vector identity ∇(∇A) = ∇(∇ A) −∇ 2 A, (1.15) which is also used very often. In fact, we just used it in Eq. (1.13). All these relations, and more, can be found in Appendix A. 1.4 Tensors Let’s talk about vectors and tensors. You know that a vector is a quantity with a mag- nitude and a direction. There is, however, also a more formal definition, which makes it easier to generalise the concept of a vector to higher rank objects such as matrices (re- member, a scalar has rank 0, a vector has rank 1, a matrix has rank 2, etc.). These objects are called tensors. A tensor of rank 0 is just a scalar, and a tensor of rank 1 is just a vector. A tensor of rank 2 is indeed a matrix, but is every matrix a tensor of rank 2? The answer is no, because tensors must obey certain transformation properties. The formal definition of a tensor says that it is an object that transforms in a special way under coordinate transformations. To see what we mean by this, consider again the case of a vector. We can write a vector a in Cartesian coordinates as a list of components (a x , a y , a z ) or, in our notation, a x ˆ i + a y ˆ j + a z ˆ k. But the coordinate system is something that we choose, and has nothing to do with the vector itself. In particular, we can rotate our coordinates around the z axis over an angle θ according to x ′ = cos θ x + sinθ y y ′ = −sinθ x + cos θ y z ′ = z . (1.16) Since the vector a does not change, its description in the new coordinate system must change, so a = (a x , a y , a z ) = (cos θ a ′ x −sinθ a ′ y , sinθ a ′ x + cos θ a ′ y , a ′ z ) . (1.17) You can view this geometrically as follows: If we rotate our coordinate system over an angle θ, then in the new coordinate system we must rotate the vector back over an angle −θ in order to describe the same vector. This means that you cannot choose any old function of x, y and z as your vector field, because it must obey this transformation rule. If a j = f (x, y, z), then a ′ j = f (x ′ , y ′ , z ′ ), that is, the same function f . For example, you can check using Eq. (1.16) that A = (y, x, 0) is not a proper vector field. 8 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 1 a a x a ′ x a y a ′ y θ Figure 3: Rotation of coordinate frame. In general, we can write a linear coordinate transformation (such as Eq. (1.16)) as a matrix equation: _ _ x ′ y ′ z ′ _ _ = _ _ cos θ sinθ 0 −sinθ cos θ 0 0 0 1 _ _ _ _ x y z _ _ , (1.18) or, more compactly: x ′ i = 3 ∑ k=1 R ik x k , (1.19) where we now write the general (non-Cartesian) coordinates as x 1 , x 2 , and x 3 , and R is an orthogonal real matrix. Note that from this equation we can write R ik = ∂x ′ i ∂x k . (1.20) Our vectora can then be written in component form as a ′ i = 3 ∑ k=1 ∂x ′ i ∂x k a k or a ′ i = ∂x ′ i ∂x k a k , (1.21) where we introduced Einstein’s summation convention: When two indices are repeated (here k), the sum is implied. This saves a lot of writing, but it also hides the complexity to some extent. Make sure that in the beginning you write every expression with and without sums, until you develop an intuition for the summation convention. At last, we are ready to define the tensor: A tensor of rank 2 is a matrix T ik that transforms under coordinate transformations according to T i ′ k ′ = 3 ∑ i=1 3 ∑ k=1 ∂x i ′ ∂x i ∂x k ′ ∂x k T ik = ∂x i ′ ∂x i ∂x k ′ ∂x k T ik . (1.22) We have chosen to differentiate between coordinate systems by putting the primes on the components. This is good practice, because the tensor T does not change, only the 9 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 1 coordinate systems change. A general tensor of rank n then obeys T i ′ 1 ...i ′ n = ∂x i ′ 1 ∂x i 1 . . . ∂x i ′ n ∂x i n T i 1 ...i n . (1.23) This includes the regular vector in Eq. (1.21). We will not make use of tensors very often in this course, but they are indispensible when talking about the momentum of the electromagnetic field, and when we give the relativistic description of Maxwell’s equations. You should therefore be aware that a tensor exists, and that it is different from a matrix in its transformation properties. 1.5 Index notation It is instructive to see how we can write the gradient, the divergence, and the curl in index notation (using the summation convention). The gradient produces a vector, so we can write this as (∇f ) i = (gradf ) i = ∂ i f . (1.24) Note that the i th component of the field is given by the derivative ∂ i , as it should. The divergence is written as ∇ A = divA = ∂ i A i , (1.25) where divA does not carry an index because it is a scalar. The repeated index i on the right hand side is summed over, so it does not show up on the left hand side. The curl, finally, makes use of the Levi-Civita tensor ǫ i jk , which returns 1 if the sequence i jk is an even permutation of the index numbers 1, 2, and 3, and it returns −1 if the sequence i jk is an odd permutation of 1, 2, and 3. When some indices are repeated (say i = j), then the Levi-Civita tensor returns 0. Verify that the curl of a vector field can be written as (∇A) i = (curlA) i =ǫ i jk ∂ j A k . (1.26) You can always do an immediate check on equations like this, because the unpaired indices on the left must match the unpaired indices on the right. Further reading – D. Mowbray, Mathematics for Electromagnetism, PHY205 (http://www.david-mowbray.staff.shef.ac.uk/mathematics for electromagnetism.htm). – H.M. Schey, Div, Grad, Curl, and all that, Norton & Co. (1973). – G. Weinreich, Geometrical Vectors, Chicago Lectures in Physics (1998): An unorthodox but very insightful treatment of vector calculus. – R.H. Good, Classical Electromagnetism, Saunders College Pub. (1999): Ch. 1, pp 1-32. – R.P. Feynman, Lectures on Physics, volume II, Addison-Wesley (1964): Ch. 2. – W.J. Duffin, Electricity and Magnetism, McGraw-Hill (1990): App. A & B, pp 390-405. 10 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 1 – B.I. Bleaney & B. Bleaney, Electricity and magnetism, Vol. 1, Oxford University Press (1976): App. A, pp A1-A12. – J.R. Reitz, F.J. Milford, & R.W. Christy, Foundations of Electromagnetic Theory, fourth edition, Addison-Wesley (1993): Ch. 1, pp 1-25. Exercises 1. Let f = 1 √ x 2 +y 2 +z 2 . Calculate (a) ∇f , (b) ∇ ∇f . 2. Let A = 1 2 B(−y, x, 0). (a) Calculate B = ∇A, (b) Sketch A and B. 3. Prove that ∇ (∇A) = 0 and ∇(∇f ) = 0. 4. Let Q = (−y, x, 0). Can you find a function T such that Q = ∇T? If T is supposed to be a temperature field, what does Q represent? Interpret your result physically. 5. Calculate _ S A dS, where S is the surface of the unit sphere centered around the origin, and A is (Hint: use Gauss’ theorem): (a) A = (−y, x, 0), (b) A = 1 3 (x, y, z). 6. Calculate _ C A dl, where the closed loop C is the unit circle centered at the ori- gin in the xy plane. It is traversed in anti-clockwise direction (Hint: use Stokes’ theorem): (a) A = (−y, x, 0), (b) A = 1 3 (x, y, z). 7. Using arrows of the proper magnitude, sketch (a) (x, y) (b) (1, 1)/ √ 2 (c) (0, x) (d) ( y √ x 2 +y 2 , x √ x 2 +y 2 ) 11 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 1 8. An object moves in the xy plane such that its position at time t is r = (a cos ωt, b sinωt) with a, b, and ω constants. (a) How far is the object from the origin at time t? (b) Find the velocity and the acceleration as a function of time. (c) Show that the object moves in an elliptical path _ x a _ 2 + _ y b _ 2 = 1 9. Find a unit vector normal to each of the following surfaces (a) z = 2 −x − y (b) z = x 2 + y 2 (c) z = √ 1 −x 2 10. Calculate the divergence of F(x, y, z) = ( f (x), f (y), f (−2z)) and show that it be- comes zero at the point (c, c, −c/2). 11. Let F(r) = ˆ r f (r) and ˆ r = (x, y, z)/r. Determine f (r) such that divF = 0. 12. Calculate the curl of the following functions: (a) (z 2 , x 2 , −y 2 ) (b) (3xz, 0, −x 2 ) (c) (e −y , e −z , e −x ) (d) (yz, xz, xy) (e) (−yz, xz, 0) (f) (x, y, x 2 + y 2 ) 13. Show that curl 1 2 Ar = A for r = (x, y, z) and A constant. 14. Verify Stokes’ theorem when F = (z 2 , −y 2 , 0), the closed loop C is given by the square (0, 0, 0) → (0, 0, 1) → (1, 0, 1) → (1, 0, 0) → (0, 0, 0), and S is the surface of unit cube between (0, 0, 0) and (1, 1, 1) without the side enclosed by C. 15. Verify the vector identities for the first and second derivatives. Use index notation. 12 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 2 + + (a) (b) Figure 4: Electric fields and potentials. (a) The electric field lines (solid) and the equipotential surfaces (dashed) due to a point charge. (b) The electric field lines and equipotential surfaces due to two opposite charges. The equipotential surfaces are always perpendicular to the field lines. 2 Electromagnetic forces, potentials, Maxwell’s equations In this lecture we review the laws of electrodynamics as you have learned them previ- ously, and write them in the form of Maxwell’s equations. We start with electrostatics and magnetostatics, and then we include general time-dependent phenomena. We also introduce the scalar and vector potential. The theory of electrodynamics is mathematically quite involved, and it will get very technical at times. It is therefore important to know when we are being mathematically rigorous, and when we are just putting equations together to fit the observed phenom- ena. First, we postulate the laws. In fact, it was Coulomb, Biot and Savart, Gauss, Fara- day, etc., who did measurements and formulated their observations in mathematical form. There is nothing rigorous about that (although the experiments were amazing). However, when Maxwell put all the laws together in a consistent mathematical frame- work the rules of the game changed: In order to find out what are the consequences of these postulated laws, we have to be mathematically rigorous. 2.1 Electrostatic forces and potentials We start our journey to the Maxwell equations with the electrostatic force: The electric force on a particle with charge q is proportional to the field E at the position of the particle: F = qE . (2.1) The field E itself must be generated by some charge density ρ, and Gauss’ law relates the flux of the electric field lines through a closed surface to the charge density inside the surface: _ S E dS = _ V ∇ E dr = _ V ρ ǫ 0 dr = Q ǫ 0 , (2.2) 13 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 2 (a) (b) (c) dl I I θ r r B B D Figure 5: (a) The B field due to a current I. (b) The same B field as in (a) but now seen along the axis of the current, pointing into the page. (c) The magnetic field lines are always closed, and therefore ∇ B = 0: the flux through the surface of a closed volume D is always zero. where Q is the total (net) charge enclosed by the surface, andǫ 0 = 8.85 . . . 10 −12 Fm −1 is the electric permittivity of free space. This leads to the first Maxwell equation: ∇ E = ρ ǫ 0 (2.3) It determines the divergence of E. The static electric field can be expressed as the gradient of a scalar function Φ, called the scalar potential. This implies that the curl of E is zero: E = −∇Φ → ∇E = 0 . (2.4) Combining Eqs. (2.3) and (2.4), we obtain the Poisson equation ∇ 2 Φ = − ρ ǫ 0 , (2.5) which, in vacuum (ρ = 0) becomes the Laplace equation ∇ 2 Φ = 0 . (2.6) 2.2 Magnetostatic forces The force of a magnetic field B on a charged particle is more complicated than the elec- tric force, because it depends on the velocity v of the particle. The Lorentz force can be stated as F = q(E +vB) or dF = ρ dV E + IdlB = (ρE +JB) dV , (2.7) where I is a current, dl is a line element, and J is a current density. Magnetostatics means that the work done by the magnetic part of the force is zero. 14 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 2 The Biot-Savart law states that the magnetic field due to a current density J is given by B(r) = µ 0 4π _ J(r ′ )(r −r ′ ) [r −r ′ [ 3 dr ′ , (2.8) where µ 0 = 4π10 −7 Hm −1 is the magnetic permeability of free space. Look at Fig. 5a to read this equation: First, forget the integral sign. The B field due to an infinitesimally small current Idl points into the page and is proportional to sinθ. This is consistent with a cross product Idlr. Next, you take the contribution to B(r) of all infinitesimal currents, which gives you the integral. And rather than taking a wire through a current, we generalize to the current density J(r ′ ). From Figs. 5b and 5c you see that the magnetic field lines must be closed. This leads to ∇ B = 0, which we will now prove. Using the vector identity ∇ (ab) = (∇a) b −a (∇b) (2.9) we get ∇ B(r) = µ 0 4π _ _ r −r ′ [r −r ′ [ 3 (∇J(r ′ )) −J(r ′ ) _ ∇ r −r ′ [r −r ′ [ 3 __ dr ′ = 0 . (2.10) The first term in the integrand is zero because the curl takes derivatives to x, y, and z, while the current density J is a function of the variables x ′ , y ′ , and z ′ . The second term is zero because (r −r ′ )/[r −r ′ [ 3 is the gradient of −1/[r −r ′ [, and the curl of a gradient field must be zero. This gives us the second Maxwell equation: ∇ B = 0 (2.11) which is sometimes called Gauss’ law for magnetism. It determines the divergence of the magnetic field. We find Amp` ere’s law by taking the curl of the magnetic field: ∇B(r) = µ 0 4π _ _ J(r ′ )∇ r −r ′ [r −r ′ [ 3 −(J(r ′ ) ∇) r −r ′ [r −r ′ [ 3 _ dr ′ = µ 0 J(r) . (2.12) So for electrostatics and magnetostatics we have ∇ E = ρ ǫ 0 , ∇E = 0 , ∇ B = 0 , ∇B = µ 0 J . (2.13) These are valid only when the E and B fields are constant in time. We consider time- dependent fields in the next section. 15 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 2 2.3 Electrodynamics and Maxwell’s equations C Φ B I Figure 6: Lenz’ law. So far we have considered only electric and magnetic fields that do not change in time. Now, let’s consider general elec- tromagnetic fields. Lenz’ law states that the electromotive force c on a closed wire C is related to the change of the mag- netic flux Φ B through the loop: c = _ C E dl = − dΦ B dt and Φ B = _ S B(r, t) dS . (2.14) Do not confuse Φ B with the scalar potential Φ: they are two completely different things! In general, we can define the general (motional) electromotive force as c = _ C F dl q = _ C E dl + _ C vB dl , (2.15) which has both a purely electric component, as well as a magnetic component due to vB. This is the motional e.m.f. For a strictly electric e.m.f., we can use Stokes’ theorem to write _ C E dl = _ S (∇E) dS = − _ S ∂B(r, t) ∂t dS , (2.16) or ∇E + ∂B(r, t) ∂t = 0 (2.17) This is Faraday’s law, and our third Maxwell equation. 2.4 Charge conservation Charges and currents inside a volume V bounded by the surface S can be a function of time: Q(t) = _ V ρ(r, t) dr and I(t) = _ S J dS = _ ∇ J dr . (2.18) Conservation of charge then tells us that the change of the charge in V is related to the flow of charge through S: dQ(t) dt + I(t) = 0 or ∂ρ ∂t +∇ J = 0 . (2.19) As far as we know today, charge conservation is strictly true in Nature. 16 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 2 The fourth, and last, Maxwell equation is a modification of Amp` ere’s law ∇B = µ 0 J. As it turns out, Amp` ere’s law, as stated in Eq. (2.12) was wrong! Or at least, it does not have general applicability. To see this, let’s take the divergence of Eq. (2.12): ∇ J = 1 µ 0 ∇ (∇B) = 0 . (2.20) But charge conservation requires that ∇ J = − ∂ρ ∂t , (2.21) James Clerk Maxwell (1831-1879) So Amp` ere’s law in Eq. (2.12) is valid only for static charge distributions. We can fix this by adding the relevant term to Amp` ere’s law. Using Gauss’ law of Eq. (2.3), you see that this is the time derivative of the E field. This was the great insight of James Clerk Maxwell (1831-1879) when he unified the known laws of electrodynamics in his four equations. The final Maxwell equation therefore becomes ∇B −µ 0 ǫ 0 ∂E ∂t = µ 0 J (2.22) and the complete set of (microscopic) Maxwell equations is ∇ B = 0 and ∇E + ∂B ∂t = 0 (2.23) ∇ E = ρ ǫ 0 and ∇B −µ 0 ǫ 0 ∂E ∂t = µ 0 J . (2.24) The top two equations are the homogeneous Maxwell equa- tions (they are equal to zero), and the bottom two are the inhomogeneous Maxwell equations (they are equal to a charge or current density). For every Maxwell equation we have the behaviour of the fields on the left hand side, and the source terms on the right hand side. This is gener- ally how field equations are written. In general relativity, the tensor field describing the curvature of space-time is related to the energy-momentum density, which includes all masses. This way, it is clear how different source terms affect the fields. 2.5 The vector potential Since the electric field is the gradient of a scalar field (the scalar potential), it is natural to ask whether the magnetic field is also some function of a potential. As the Lorentz force shows, the magnetic field is a good deal more complicated than the electric field, 17 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 2 and this is reflected in the magnetic potential. Rather than a scalar potential, we need a vector potential to make the magnetic field. The proof is easy: Since ∇ B = 0, we can always write B as the curl of a vector field A because ∇ B = ∇ (∇A) = 0 , (2.25) by virtue of a well-known vector identity. At this point, the vector potential A, as well as the scalar potential Φ, are strictly mathematical constructs. The physical fields are E and B. However, you can see imme- diately that whereas E and B have a total of six components (E x , E y , E z , B x , B y , and B z ), the vector and scalar potentials have only four independent components (Φ, A x , A y , and A z ). This means that we can get a more economical description of electrodynamics phenomena using the scalar and vector potential. When the current is constant, we can choose a vector potential of the following form: A(r) = µ 0 4π _ J(r ′ ) [r −r ′ [ dr ′ . (2.26) We can prove this by taking the curl on both sides ∇A(r) = µ 0 4π ∇ _ J(r ′ ) dr ′ r ′′ = µ 0 4π _ ∇ _ J(r ′ ) dr ′ r ′′ _ , (2.27) with r ′′ = [r −r ′ [. Using the chain rule we can rewrite this as ∇A(r) = µ 0 4π _ ∇J(r ′ ) dr ′ r ′′ + µ 0 4π _ ∇ _ 1 r ′′ _ J(r ′ ) dr ′ . (2.28) The first term is zero because the derivative are to r and J is a function of r ′ . The second term can be evaluated using ∇(1/r ′′ ) = −ˆ r ′′ /r ′′ 2 . We therefore have ∇A(r) = µ 0 4π _ J(r)r ′′ r ′′ 2 dV = µ 0 4π _ J(r)(r −r ′ ) [r −r ′ [ 3 dV = B(r) . (2.29) Further reading – R.H. Good, Classical Electromagnetism, Saunders College Pub. (1999): Ch. 2-6, pp 33-164. – J.D. Jackson, Classical Electrodynamics, Wiley (1998): Sec. 1.1-1.7 & 5.1-5.4, pp 24-35, 174- 181. – R.P. Feynman, Lectures on Physics, volume II, Addison-Wesley (1964): Ch. 15 & 18. – J.R. Reitz, F.J. Milford, & R.W. Christy, Foundations of Electromagnetic Theory, fourth edition, Addison-Wesley (1993): Ch. 2, 3, 7, 8, 11, & 16, pp 26-96, 162-217, 271-288, 386-411. – C.A. Brau, Modern Problems in Classical Electrodynamics, Oxford University Press (2004): Sec. 0.1-0.5., pp 1-20. – J. Schwinger, L.L. DeRaad, K.A. Milton, &W. Tsai, Classical Electrodynamics, The Advanced Book Program, Westview Press (1998): Ch. 1-2, pp 1-20. 18 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 2 Exercises 1. A line charge of constant linear charge density λ Cm −1 extends along the positive y axis from 0 to L. Find the field at the position a along the positive x axis. 2. Calculate the pressure (force per area) between two parallel plates that carry op- posite charge densities ±σ. 3. Calculate the magnetic field at a distance z above the center of a circular loop of radius r carrying a current I. Evaluate this expression for z = 0, a = 1 cm, and I = 1 A. 4. Show that charge conservation is implicit in Maxwell’s equations. 5. Give the expression of the magnetic flux for both B and the vector potential A. What does the vector potential look like outside an infinite solenoid of radius R that confines a homogeneous field B in the z direction pointing along the symmetry axis? 6. Write Maxwell’s equations in index notation 7. Assessed Homework Exercise: A device that can generate large currents is the so-called Faraday disk generator (see figure). It consists of a conducting thin disk, rotating with angular velocity ω in a homogeneous magnetic field in the z direc- tion (B = B ˆ k). The rim of the disk and the axis are connected via a wire with Ohmic resistance R. R ω B I (a) Argue qualitatively that a current will flow in the direction indicated in the figure. (2 points) (b) Suppose the resistance is a light bulb. Where does the energy come from that is dissipated in R? (2 points) (c) Calculate the (motional) electromotive force between the rim and the center of the disk. What is the current in the wire? (6 points) 19 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 2 (d) Explain why, when the rotation is not driven, the disk will stop spinning. (Hint: calculate the Lorentz force due to the surface current on the disk. 5 points) (e) Calculate the torque rF on the disk. (5 points) 8. An electron travels in a circular orbit around a proton that is fixed in space with angular momentum ¯ h (the Bohr model). Show that the magnetic field at the loca- tion of the proton is given by B = µ 0 e¯ h 4πma 3 0 , (2.30) where e is the elementary charge, m is the mass of the electron, and a 0 is the Bohr radius. Calculate the numerical value of B. 20 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 3 3 Electrodynamics with scalar and vector potentials 3.1 Scalar and vector potential We can also write Maxwell’s equations in terms of the scalar and vector potential, but we have to modify our relation E = −∇Φ: In electrodynamics ∇E ,= 0, so E can’t be a pure gradient. Since we have ∇ B = 0, the construction B = ∇A still works. According to Faraday’s law ∇E + ∂B ∂t = ∇E + ∂ ∂t ∇A = 0 or ∇ _ E + ∂A ∂t _ = 0 . (3.1) Therefore E + ˙ A is a gradient field: E + ∂A ∂t = −∇Φ or E = − ∂A ∂t −∇Φ . (3.2) We should re-derive Poisson’s equation (2.5) from ∇ E = ρ/ǫ 0 using this form of E: ∇ 2 Φ = − ρ ǫ 0 − ∂ ∂t (∇ A) (3.3) Similarly, we can write ∇B = ∇(∇A) = ∇(∇ A) −∇ 2 A = µ 0 J +µ 0 ǫ 0 ∂ ∂t _ −∇Φ− ∂A ∂t _ , (3.4) or ∇ 2 A−µ 0 ǫ 0 ∂ 2 A ∂t 2 = −µ 0 J +µ 0 ǫ 0 ∇ _ ∂Φ ∂t _ +∇(∇ A) (3.5) Eqs. (3.3) and (3.5) are the Maxwell equations in terms of the scalar and vector potential. This looks rather a lot worse than the original Maxwell equations! Can we simplify these equations so that they look a bit less complicated? The answer is yes, and involves so- called gauge transformations. 3.2 Gauge transformations The electric and magnetic fields can be written as derivative functions of a scalar poten- tial Φ and a vector potential A. We have already seen that we can add a gradient field to the vector potential without affecting the field equations for B. In general, we can apply a gauge transformation to the scalar and vector potentials without changing any of the physical content of the theory. 21 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 3 From B = ∇A and ∇(∇Λ) = 0 for all Λ we see that we can always add a gradient field ∇Λ to the vector potential. However, we established that the electric field E depends on the time derivative of the vector potential. If E is to remain invariant, we need to add the time derivative of Λ to the scalar potential: E ′ = −∇Φ− ∂A ∂t − ∂∇Λ ∂t +∇ ∂Λ ∂t = E , (3.6) since ∇(∂ t Λ) = ∂ t (∇Λ). It is clear that the full gauge transformation is Φ(r, t) → Φ ′ (r, t) = Φ(r, t) − ∂Λ(r, t) ∂t , A(r, t) → A ′ (r, t) = A(r, t) +∇Λ(r, t) . (3.7) On the one hand, you may think that it is rather inelegant to have non-physical de- grees of freedom, because it indicates some kind of redundancy in the theory. However, it turns out that this gauge freedom is extremely useful, because it allows us to simplify our equations, just by choosing the right gauge Λ. In the Coulomb gauge (which is sometimes also called the radiation gauge), we set ∇ A = 0. We will encounter this when we discuss electromagnetic waves. Another useful gauge is the Lorenz gauge, in which we set ∇ A +µ 0 ǫ 0 ∂Φ ∂t = 0 . (3.8) This leads to the following Maxwell equations: _ ∇ 2 −µ 0 ǫ 0 ∂ 2 ∂t 2 _ Φ(r, t) = − ρ(r, t) ǫ 0 , (3.9) _ ∇ 2 −µ 0 ǫ 0 ∂ 2 ∂t 2 _ A(r, t) = −µ 0 J(r, t) . (3.10) Gauge transformations are important in field theories, particularly in modern quantum field theories. The differential operator in brackets in Eqs. (3.9) and (3.10) is called the d’Alembertian, after Jean le Rond d’Alembert (1717–1783), and is sometimes denoted by the symbol . The differential equations (3.9) and (3.10) are called d’Alembert equa- tions. 3.3 A particle in an electromagnetic field Often we want to find the equations of motion for a particle in an electromagnetic field, as given by the vector potential. There are several ways of doing this, one of which involves the Hamiltonian H(r, p), where r and p are the position and momentum of the particle, respectively. You know the Hamiltonian from quantum mechanics, where it is 22 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 3 the energy operator, and it is used in the Schr¨ odinger equation to find the dynamics of the wavefunction of a quantum particle. However, the Hamiltonian was first constructed for classical mechanics as a regular function of position and momentum, where it also completely determines the dynamics of a classical particle. The idea behind this is that the force on an object is the spatial derivative of some potential function. Once the Hamiltonian is known, the equations of motion become ∂r i ∂t = ∂H ∂p i and ∂p i ∂t = − ∂H ∂r i . (3.11) These are called Hamilton’s equations, and it is clear that the second equation relates the force (namely the change of momentum) to a spatial derivative of the Hamiltonian. William Rowan Hamilton (1805-1865) The derivation of the Hamiltonian is beyond the scope of this course, but in vector notation it becomes: H(r, p) = 1 2m [p −qA(r, t)] 2 −qΦ(r, t) . (3.12) This expression can be further simplified for the specific problem at hand by choosing the most suitable gauge. In the weak field approximation, we set [A[ 2 = 0, and the Hamiltonian is that of a free particle with a coupling term 2q p A. As mentioned, the Hamiltonian completely deter- mines the dynamics of a system, and as such is a very use- ful quantity to know. In quantum mechanics, we replace the position vector r by the operator 1 ˆ r and the momentum vector p by the operator ˆ p. This makes the Hamiltonian an operator as well. The Schr¨ odinger equation for a particle in an electromagnetic field is therefore i¯ h d dt [Ψ) = _ 1 2m ( ˆ p −qA) 2 −qΦ _ [Ψ) . (3.13) The vector potential A and the scalar potential Φare classical fields (not operators). The full theory of the quantized electromagnetic field is quantum electrodynamics. Further reading – R.H. Good, Classical Electromagnetism, Saunders College Pub. (1999): Ch. 6, pp 131-163. – J.D. Jackson, Classical Electrodynamics, Wiley (1998): Sec. 6.3, pp 240-243. – C.A. Brau, Modern Problems in Classical Electrodynamics, Oxford University Press (2004): Sec. 2.4, pp 110-116; this is quite an advanced text. 1 Note that the hat now denotes an operator, rather than a unit vector. In the rest of the lecture notes the hat denotes unit vectors. 23 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 3 Exercises 1. Derive Eqs. (3.3) and (3.5). 2. Construct a homogeneous B field in the z direction by two vector potentials A 1 and A 2 (with B = ∇A 1 = ∇A 2 ), one of which points in the x direction, and one in the y direction. Find the gauge transformation that connects both poten- tials. 3. For a charge q in a vector potential A due to some current, show that the charge receives a momentum kick m∆v = −qA when the current is suddenly switched off. 4. Find the vector potential for a static charge in the gauge where Φ = 0. 24 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 4 4 Electromagnetic waves and Poynting’s theorem In field theories we can often find nontrivial solutions to the field equations (meaning that the field is not zero everywhere), even when there are no sources present. For electrodynamics, in the absence of charges and currents we can still have interesting effects. So interesting in fact, that it covers a whole discipline in physics, namely optics. 4.1 Electromagnetic waves We obtain Maxwell’s equations in vacuum by setting ρ = J = 0 in Eq. (2.23): ∇ B = 0 and ∇E = − ∂B ∂t (4.1) ∇ E = 0 and ∇B = µ 0 ǫ 0 ∂E ∂t . (4.2) Taking the curl of ∇E and substituting ∇B = µ 0 ǫ 0 ∂ t E we find ∇ _ ∇E + ∂B ∂t _ = ∇(∇ E) −∇ 2 E + ∂ ∂t _ µ 0 ǫ 0 ∂E ∂t _ = 0 , (4.3) which, with ∇ E = 0, leads to the wave equation _ ∇ 2 −µ 0 ǫ 0 ∂ 2 ∂t 2 _ E(r, t) = 0 . (4.4) This differential equation has the well-know plane wave solutions E = E 0 e i(kr−ωt) , (4.5) where E 0 is a constant vector, k is the wave vector pointing in the propagation direction, and ω is the frequency. By virtue of Eq. (4.4), the wave vector and frequency obey the relation (k 2 = k k): k 2 −µ 0 ǫ 0 ω 2 = 0 . (4.6) Clearly, in SI units 2 µ 0 ǫ 0 has the dimension of inverse velocity-squared, and substitut- ing the numerical values for µ 0 and ǫ 0 reveals that this velocity is c, the speed of light. Another great triumph of Maxwell’s theory was the identification of light as electromag- netic waves. For the right frequencies, the above equation is therefore the dispersion relation of light propagating through vacuum: k 2 = ω 2 c 2 . (4.7) 2 See Appendix B. 25 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 4 x y z Figure 7: An electromagnetic wave with linear polarization. The E field is pointed in the x direction, while the B field points in the y direction. The Poynting vector therefore points in the z direction, the direction of propagation. The orientation of the E and B fields determine the polarization of the wave. So not only does Maxwell’s theory unify electric and magnetic phenomena, it also en- compasses the whole of optics. On top of that, it predicts a range of new types of radi- ation from radio waves and infrared in the long wavelengths, to ultraviolet, X-rays (or R¨ ontgen rays), and gamma rays in the short wavelength. Prior to Maxwell’s discovery of electromagnetic waves, the known types of radiation other than light were infrared, discovered by William Herschel in 1800, and ultraviolet, discovered in 1801 by Johann Wilhelm Ritter. After Maxwell, Heinrich Hertz discovered radiowaves and microwaves in 1887 and 1888, respectively. Wilhelm Conrad R¨ ontgen discovered X-rays in 1895, and Paul Ulrich Villard discovered gamma rays in 1900. However, it was not until the work of Ernest Rutherford and Edward Andrade in 1914 that gamma rays were understood as electromagnetic waves. Similarly, Maxwell’s equations in vacuum give rise to a wave equation for the mag- netic field, yielding the solution B = B 0 exp[i(k r −ωt)]. Since ∇ E = ∇ B = 0, the electric and magnetic fields are perpendicular to the direction of propagation, k. Elec- tromagnetic waves are thus transverse waves. It is left as a homework question to prove that E ⊥ B. 4.2 Complex fields? You may have noticed that in the previous section the solution to the wave equation gave us the complex field of Eq. (4.5). But the electric field must be a real quantity, oth- erwise we would get complex forces, momenta, velocities, etc. This is clearly nonsense. The usual way out of this is that we take the real part of Eq. (4.5) as the physical solution, and we discard the imaginary part. We calculate things this way because it is easier to deal with exponentials than with trigonometric functions. However, it is instructive to delve a little deeper into this: why does the maths allow 26 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 4 complex solutions when all the quantities we put in (r, t, µ 0 , and ǫ 0 ) are real? Notice that a different solution of the wave equation is E = E 0 e −i(kr−ωt) , (4.8) where the exponential has an overall minus sign. We can understand this as a wave travelling in the −k direction, rather than the +k direction. However, the frequency ω must be a positive number, so that leaves us with a wave travelling in the −t direction, or backwards in time! We conclude that Maxwell’s equations are symmetric under time reversal, just like Newton’s laws and quantum mechanics. The story is not finished, though, because we can superpose two solutions to a linear differential equation to obtain a third solution. When we do this we can construct real solutions for the field: E = E 0 e i(kr−ωt) +E 0 e −i(kr−ωt) = 2E 0 cos(k r −ωt) . (4.9) More generally, we can take E 0 to be complex, in which case the cosine picks up a phase shift (check this!). Every real field can be written as a sum of two complex fields, one travelling forward in time, and one travelling backward in time. You may have heard slogans like “anti-particles are particles moving backward in time”, and this is where that comes from. In quantum field theory, the fields are usually written in terms of the two components where one is the complex (actually, Hermitian) conjugate of the other. The so-called positive frequency part corresponds to regular parti- cles, while the negative frequency part corresponds to the anti-particles. 4.3 Poynting’s theorem John Henry Poynting (1852-1914) Electromagnetic waves carry energy. For example, the Earth gets all of its energy from the Sun in the form of elec- tromagnetic radiation. We therefore want to know what is the energy density of the fields. We first determine the en- ergy density of the electric and magnetic fields separately. Suppose we have a charge q 1 at position r 1 , and we bring in a second charge q 2 from infinity to the position r 2 . The work done on the second charge by the field is W = q 2 _ r 2 −∞ E dl = q 2 Φ 1 (r 2 ) = 1 2 [q 1 Φ 2 (r 1 ) + q 2 Φ 1 (r 2 )] , (4.10) where the last equality follows from the fact that it does not matter whether we bring q 1 from infinity to r 1 before or after we bring q 2 from infinity to r 2 . We can symmetrize 27 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 4 c B I Figure 8: Energy in B. the work function. In general, the work needed to assem- ble a charge distribution ρ(r) is then W = 1 2 _ V ρ(r)Φ(r) dr = − ǫ 0 2 _ V Φ∇ 2 Φdr = ǫ 0 2 _ V ∇Φ ∇Φdr = ǫ 0 2 _ V E 2 dr . The energy density of the electric field in vacuum is therefore given by U e = 1 2 ǫ 0 E 2 , where E 2 = E E. The energy density of the magnetic field is slightly more involved. Since the Lorentz force always acts perpendicular to the magnetic field, the work done on a particle by a static magnetic field is zero. The work done on a particle therefore occurs when the magnetic field is changing, say in a closed loop C with a current I, and it creates an electric field, called the electromotive force c. The rate of work needed to keep the current going can be found by Lenz’ law: dW dt = −cI = − _ C E I dl = − _ V E J dV . (4.11) Now we use that here J = ∇B/µ 0 and ∇E = −∂ t B, and ∇ (EB) = B (∇E) −E (∇B) . (4.12) This leads to dW dt = − 1 µ 0 _ V E (∇B) dr = 1 µ 0 _ V [B (∇E) −∇(EB)] dr = − 1 µ 0 _ V B ∂B ∂t dr − _ S EB µ 0 dS = − 1 µ 0 _ V B ∂B ∂t dr = − 1 2µ 0 d dt _ V B 2 dr , (4.13) where in the second line we have used that we can take the volume to be the entire space, and the surface integral becomes zero. The magnetic energy density is therefore U m = 1 2 B 2 /µ 0 and the total energy density of the field is U = ǫ 0 E 2 2 + B 2 2µ 0 . (4.14) 28 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 4 Continuing the manipulation of Eq. (4.12), from Maxwell’s equations we have E (∇B) −B (∇E) = µ 0 J E + ∂ ∂t _ B 2 2 + E 2 2c 2 _ . (4.15) Using the vector identity in Eq. (4.12) we find ∇ (BE) = µ 0 J E +µ 0 ∂ ∂t U , (4.16) where U = U e +U m . A little rearrangement of the terms will reveal Poynting’s theorem: ∂U ∂t +∇ S +E J = 0 , (4.17) where S = µ −1 0 EB is the Poynting vector, after John Henry Poynting (1852-1914). To interpret this theorem properly, let’s integrate the expression over the volume V bounded by the surface S (now not at infinity): d dt _ V U dV + _ S S n dS + _ V J E dV = 0 . (4.18) We again used Gauss’ theorem to relate a volume integral to a surface integral. The first term is the rate of change of the field energy, the second term is the rate of the energy flow through the surface, and the third term is the rate of work done on the charge density. Poynting’s theorem is a statement of conservation of energy. The energy can leave the fields in a region of space, but it must then be either transported through the sur- face (measured by the Poynting vector), or it must be used to do work on the charges in the volume. When we make the volume infinitesimally small, you see that energy conservation is not only true globally, but it is true locally, at every point in space. Further reading – R.H. Good, Classical Electromagnetism, Saunders College Publishing (1999): Ch. 14, pp 336-370. – J.D. Jackson, Classical Electrodynamics, Wiley (1998): Sec. 6.8 & 7.1, pp 262-264-566, 295-300. – R.P. Feynman, Lectures on Physics, volume II, Addison-Wesley (1964): Ch. 20. – H.J. Pain, The Physics of Vibrations and Waves, Wiley (1983): Ch. 7, pp 187-218. – B.I. Bleaney & B. Bleaney, Electricity and magnetism, Vol. 1, Oxford University Press (1976): Ch. 8, pp 225-257. – J.R. Reitz, F.J. Milford, & R.W. Christy, Foundations of Electromagnetic Theory, fourth edition, Addison-Wesley (1993): Ch. 16, pp 386-411. – C.A. Brau, Modern Problems in Classical Electrodynamics, Oxford University Press (2004): sec. 0.5-0.6, pp 19-28. 29 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 4 Exercises 1. Show that E ⊥ B for electromagnetic waves. Argue why the E and B fields must be out of phase by π/2. 2. The Maxwell equations can be written in terms of E, A, and Φ as follows (in this exercise we set c =ǫ 0 = µ 0 = 1): E + ∂A ∂t +∇Φ = 0 , and ∂E ∂t −∇(∇A) + J = 0 , and ∇ E = ρ . Furthermore, when a vector field V is zero at infinity, we can separate it into two components V = V r + V g , such that ∇ V r = 0 and ∇V g = 0. V r is the solenoidal part of the field, while V g is the irrotational part. (a) Show that the three equations above are equivalent to the Maxwell equations. (b) Show that ∂A r ∂t +E r = 0 , and ∂A g ∂t +E g +∇Φ = 0 . (c) The current density J can also be separated into J r and J g (J = 0 at infinity). Use this to write down the solenoidal and irrotational parts of the second Maxwell equation above. (d) From the above results, derive the wave equation for A r (Since J ,= 0, this is a driven wave equation). (e) Using the Lorenz gauge, derive the wave equation for A g . (f) Derive the wave equation for the scalar potential Φ. (g) Finally, derive the wave equations for the scalar potential Φand the complete vector potential A directly from the Maxwell equations above and the Lorenz gauge. 3. The Poynting vector for complex E and B fields is given by S = µ −1 0 sinθ Re(E)Re(B), where θ is the angle between E and B. For an electromagnetic wave with E(t) = E 0 e −iωt and B(t) = B 0 e −iωt+iφ , show that the time averaged Poynting vector can be written as S av = Re(EB ∗ ) 2µ 0 . 30 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 5 5 Momentum of the electromagnetic field It is not surprising that the electric and magnetic fields carry energy. After all, you’re quite familiar with the notion that it is the potential energy in the fields that make charges move. The general behaviour of the energy in the fields is described by Poynt- ing’s theorem, discussed in the previous lecture. The electromagnetic field also has momentum. There are several immediate reasons why this must be the case: 1. The theory of electromagnetism is a relativistic theory, and we know that energy in one frame of reference gives rise to momentum in another reference frame. There- fore, to have energy is to have momentum 3 . 2. Last year you learned about radiation pressure, so you know that electromagnetic waves can exert a force on objects, and therefore transfer momentum. 3. When two charged particles fly close part each other with small relative velocity (v ≪ c), Newton’s third law holds, and the momentum of the two particles is conserved. However, when the relative velocity becomes large, the particles will each experience a B field due to the other particle’s motion. We can configure the situation such that the total momentum of the two charges is not conserved (see exercise), and in order to save Newton’s third law, the field must be imbued with momentum. Now let’s study the properties of the momentum of the electromagnetic field. Since momentum is closely related to force, we first consider the Lorentz force. We can write the Lorentz force ∆F on a small volume ∆V as ∆F = (ρE + JB)∆V . (5.1) we now define the force f on a unit volume exerted by the electromagnetic field as f = ∆F/∆V = ρE + JB. Since we are interested in the fields, and not the charge or current densities, we want to eliminate ρ and J from the expression for f. We use the inhomogeneous Maxwell equations for this: f =ǫ 0 (∇ E)E + _ 1 µ 0 ∇B −ǫ 0 ∂E ∂t _ B. (5.2) Next, we wish to rewrite the term ˙ EB using the chain rule: ∂E ∂t B = ∂(EB) ∂t −E ∂B ∂t . (5.3) Also, the term (∇B)B can be rewritten as 1 µ 0 (∇B)B = 1 µ 0 (B ∇)B − 1 2µ 0 ∇B 2 = 1 µ 0 (B ∇)B + 1 µ 0 (∇ B)B − 1 2µ 0 ∇B 2 , (5.4) 3 Even though the actual value of the momentum in specific frames may be zero. 31 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 5 where we used the fact that ∇ B = 0, always. We put this term in to make the expres- sion more symmetric. This leads to a horrible mess: f =ǫ 0 [(∇ E)E + (E ∇)E] + 1 µ 0 [(∇ B)B + (B ∇)B] −∇U − 1 c 2 ∂S ∂t , (5.5) where U =ǫ 0 E 2 /2 +B 2 /2µ 0 is the potential energy per unit volume, and S is the Poynt- ing vector. By making the following clever substitution, this expression will simplify consider- ably: T i j =ǫ 0 _ E i E j − δ i j 2 E 2 _ + 1 µ 0 _ B i B j − δ i j 2 B 2 _ . (5.6) This is the Maxwell stress tensor, which we will interpret in due course. Since it is a rank two tensor, for all practical purposes it behaves like a matrix. We are going to take the divergence of T, which now will yield a vector, rather than a scalar. This comes down to multiplying a vector with a matrix (where ∇ T = ∑ i ∂ i T i j ): ∇ T =ǫ 0 _ (∇ E)E + (E ∇)E − 1 2 ∇E 2 _ + 1 µ 0 _ (∇ B)B + (B ∇)B − 1 2 ∇B 2 _ . (5.7) It is then easy to see that f = ∇ T − 1 c 2 ∂S ∂t . (5.8) When we integrate this over the total volume, the total Lorentz force is F = _ V _ ∇ T − 1 c 2 ∂S ∂t _ dτ = _ S T da − d dt _ V S c 2 dτ , (5.9) where da is the surface increment on the boundary surface S of volume V. The first term on the right-hand side is the pressure and shear on the volume, while the second term is the time derivative of the electromagnetic momentum _ S dτ/c 2 . The Lorentz force F is moving around particles, and changing the amount of mechanical energy. The tension in the fields (or field lines) also explains why some configurations of charges and magnets lead to repulsion, and some to attraction. The Maxwell stress tensor makes precise the notion of tension and pressure in the field lines. Finally, now that we have an expression for the momentum of the electromagnetic field per unit volume p = S c 2 = EB µ 0 c 2 =ǫ 0 EB, (5.10) we can define the angular momentum per unit volume of the field: l = rp =ǫ 0 r(EB) . (5.11) However, we will not investigate this further here. 32 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 5 Further reading – R.H. Good, Classical Electromagnetism, Saunders College Pub. (1999): Sec. 14.2, pp 343-347. – J.D. Jackson, Classical Electrodynamics, Wiley (1998): Sec. 6.7, pp 258-262. – R.P. Feynman, Lectures on Physics, volume II, Addison-Wesley (1964): Ch. 27. – C.A. Brau, Modern Problems in Classical Electrodynamics, Oxford University Press (2004): Sec. 0.6, pp 24-28. – J. Schwinger, L.L. DeRaad, K.A. Milton, &W. Tsai, Classical Electrodynamics, The Advanced Book Program, Westview Press (1998): Ch. 3, pp 21-32. Exercises 1. Write out Maxwell’s stress tensor in matrix form. What does the trace correspond to? 2. Two positive charges move with velocities vA and v B as indicated in the figure below. Give a qualitative estimate of the force on charge a and on charge b. Is the A B v A v B Figure 9: Two moving charges. total momentum conserved in this situation? 3. A photon with energy ¯ hω crosses a (ficticious) surface area A in time t. Calculate the momentum of the photon given that the momentum per unit volume of the field is given by S/c 2 . 4. Write the derivation starting with Eq. (5.1) to Eq. (5.5) in index notation. 33 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 6 6 Multipole expansion and spherical harmonics Charge and current distributions in a small region of space may be very complicated, and the potentials will generally not be simple. For example, a water molecule is elec- trically neutral as a whole, but one side of the molecule is slightly positive (+q), while the other side is slightly negative (−q). If the two charges are separated by a distance d, the potential of each is Φ + (r) = − q 4πǫ 0 [r[ and Φ − (r) = q 4πǫ 0 [r −d[ . (6.1) Using the superposition principle to find the scalar potential of the combined charges Φ, we find (r ≫d) Φ(r) = − q[r −d[ 4πǫ 0 [r[[r −d[ + q[r[ 4πǫ 0 [r[[r −d[ ≃ qd cos θ 4πǫ 0 r 2 , (6.2) where θ measures the angle between d and r. This is the so-called dipole potential. Note that it is no longer a 1/r potential but falls off faster, with 1/r 2 . If, for some reason the potentials in Eq. (6.1) had different charges q 1 and q 2 = −q 1 −δ, then the scalar potential would be Φ(r) = − q 1 4πǫ 0 [r[ + q 1 +δ 4πǫ 0 [r −d[ ≃ δ 4πǫ 0 r + q 1 d cos θ 4πǫ 0 r 2 . (6.3) Now the scalar potential is a superposition of a monopole and a dipole contribution. In general, the scalar potential of a charge distribution will have a multipole expansion, and the same will apply to the vector potential. Let’s look at this in a bit more detail. 6.1 Multipole expansion of a charge density Suppose that we have a localized (meaning enclosed in a volume V) static charge den- sity ρ(r), which gives rise to a scalar potential Φ(r) = 1 4πǫ 0 _ ρ(r ′ ) dr ′ [r −r ′ [ . (6.4) Far away from the charge density, the potential looks mainly like that of a point charge, but with some higher-order corrections. These corrections are the multipoles, and we can find them by looking at the Taylor expansion of Eq. (6.4). Let f (r −r ′ ) = 1 [r −r ′ [ = 1 _ (x − x ′ ) 2 + (y − y ′ ) 2 + (z −z ′ ) 2 . (6.5) Now we make a Taylor expansion of f (r −r ′ ) around r ′ = 0: f (r −r ′ ) = f (r) − 3 ∑ i=1 r ′ i ∂ f ∂r i ¸ ¸ ¸ ¸ r ′ =0 + 1 2 3 ∑ i, j=1 r ′ i r ′ j ∂ 2 f ∂r i ∂r j ¸ ¸ ¸ ¸ ¸ r ′ =0 + . . . (6.6) 34 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 6 dρ r r ′ r −r ′ O Figure 10: The field at position r far away from a charge distribution ρ(r) can be expressed conveniently in terms of a multipole expansion. O denotes the origin of the coordinate system. This can be rewritten in compact form as 1 [r −r ′ [ = 1 r −r ′ ∇ 1 r + 1 2 (r ′ ∇) 2 1 r + . . . = 1 r + 3 ∑ i=1 r i r ′ i r 3 + 1 2 3 ∑ i, j=1 r i (3r ′ i r ′ j −r ′ 2 δ i j )r j r 5 + . . . (6.7) where r = _ r 2 1 + r 2 2 + r 2 3 = [r[ is the magnitude of the vector with components r i . After substituting this into the generic form of the scalar potential in Eq. (6.4), we find Φ(r) = 1 4πǫ 0 _ ρ(r ′ ) dr ′ r + 1 4πǫ 0 3 ∑ i=1 _ r ′ i r i r 3 ρ(r ′ ) dr ′ + 1 8πǫ 0 3 ∑ i, j=1 _ 3r ′ i r ′ j −r ′ 2 δ i j r 5 r i r j ρ(r ′ ) dr ′ + . . . (6.8) = 1 4πǫ 0 _ Q r + 3 ∑ i=1 Q i r i r 3 + 1 2 3 ∑ i, j=1 r i Q i j r j r 5 + . . . _ . (6.9) The multipole moments of the charge distribution are the total charge Q, the dipole mo- ment Q i , the quadrupole moment Q i j , etc. They are defined as follows: Q = _ ρ(r ′ ) dr ′ (6.10) Q i = _ r ′ i ρ(r ′ ) dr ′ (6.11) Q i j = _ (3r ′ i r ′ j −r ′ 2 δ i j ) ρ(r ′ ) dr ′ (6.12) . . . (6.13) 35 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 6 The quadrupole moment has nine components, but it is easy to see that Q i j = Q ji , and 3 ∑ i=1 Q ii = 3 ∑ i=1 _ (3r i r i −r 2 )ρ(r) dr = _ __ 3 3 ∑ i=1 r i r i _ −3r 2 _ ρ(r) dr = 0 . (6.14) The quadrupole moment is therefore characterised by five independent variables. 6.2 Spherical harmonics and Legendre polynomials It is clear from the previous section that we can continue the multipole expansion to octupoles and higher, but it is also pretty obvious that the polynomials in the definitions of Q i jk... get quite unwieldy very quickly. Luckily, there is a more systematic approach based on spherical harmonics. These are functions of the spherical coordinates θ and φ of a vector r, and the function f can then be written as 1 [r −r ′ [ = ∞ ∑ l=0 +l ∑ m=−l r ′ l r l+1 _ 4π 2l + 1 Y lm (θ, φ) _ 4π 2l + 1 Y ∗ lm (θ ′ , φ ′ ) (6.15) with Y lm (θ, φ) = ¸ 2l + 1 4π (l + m)! (l −m)! e imφ sin m θ _ d d cos θ _ l−m (cos 2 θ −1) l 2 l l! . (6.16) This is still rather complicated, but the advantage is that this is valid for all l. Note also that the spherical harmonics are complex, but Eq. (6.15) is still real due to the sum over m. You should think of the Y lm as basis functions that can be used to write arbitrary functions (of θ and φ) as a series, just like any polynomial can be written as a series ∑ n a n x n , or a periodic function as a Fourier series. Like any proper set of basis functions, the Y lm obey an orthogonality relation: _ dΩ Y ∗ lm (θ, φ)Y l ′ m ′ (θ, φ) ≡ _ π 0 sinθ dθ _ 2π 0 dφ Y ∗ lm (θ, φ)Y l ′ m ′ (θ, φ) = δ ll ′ δ mm ′ , (6.17) where we introduced the integration over the sold angle dΩ. We can now define the Legendre polynomials as P l (w) = _ d dw _ l (w 2 −1) l 2 l l! , (6.18) with normalisation P l (1) = 1. In the spherical harmonics, we have set w = cos θ. The Legendre polynomials also obey an orthogonality relation _ 1 −1 dw P l (w)P l ′ (w) = 2 2l + 1 δ ll ′ . (6.19) 36 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 6 Figure 11: The fluctuations of the cosmic microwave background radiation as measured by WMAP. On the right is the plot of the amplitude of Y lm for different values of l. The plot is going all the way up to l = 1 000. The Legendre polynomials are closely related not only to the multipole moments Q i jk... , but also to the quantum mechanical wavefunction of a particle in a central poten- tial. The index l is then the quantum number for the orbital angular momentum of the particle, and m is the quantum number of its z-component. The most famous example of this is the wavefunction of the electron in a hydrogen atom. The more general spher- ical harmonics also have many applications outside electrodynamics. One example is the description of the temperature fluctuations of the cosmic microwave background. In figure 11 the relative strength of the harmonics are plotted against l. 6.3 Multipole expansion of a current density In order to find the magnetic multipole expansion due to complicated stationary current desities, we use the time-independent form of the Maxwell equation (3.5) in either the Lorentz gauge or the Coulomb gauge (that makes no difference here): ∇ 2 A(r) = −µ 0 J(r) . (6.20) The three components of this vector equation are independent of each other, so they are effectively three scalar Poisson equations. If the source J vanishes at infinity (in other words, it is localized), then we can write this as A(r) = µ 0 4π _ J(r ′ ) dr ′ [r −r ′ [ . (6.21) If we assume that the total charge in the system does not change (dρ/dt = 0), then we can use the fact that ∇ J = 0. We can again insert the Taylor expansion for the 37 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 6 function 1/[r −r ′ [ given by Eq. (6.6), which leads to A i (r) = µ 0 4π _ 1 r _ J i (r ′ ) dr ′ + 1 r 3 3 ∑ j=1 _ r j r ′ j J i (r ′ ) dr ′ + O(r −5 ) _ . (6.22) The first term in square brackets is the magnetic monopole term, which we expect to vanish for magnetostatics, and the second term is the magnetic dipole moment. To show that the magnetic monopole contribution vanishes for all J(r), note that _ J i (r ′ ) dr ′ = 3 ∑ j=1 _ J j (r ′ ) ∂r ′ i ∂r ′ j dr ′ . (6.23) Integration by parts of the right-hand side of this expression gives _ J i (r ′ ) dr ′ = 3 ∑ j=1 _ J j (r ′ )r ′ i d 2 r ′ k,=j ¸ ¸ ¸ ¸ r ′ j =∞ r ′ j =−∞ − 3 ∑ j=1 _ r ′ i ∂J j (r ′ ) ∂r ′ j dr ′ . (6.24) The first termof the right-hand side is zero because the current density at infinity is zero: we consider a localized current density. The differential in the second term (together with the sum) is ∇ J, which, we already determined, is zero. The magnetic dipole moment is therefore the lowest order term in Eq. (6.22). Let’s separate it into the symmetric and anti-symmetric parts: r ′ j J i = 1 2 _ r ′ j J i + r ′ i J j _ + 1 2 _ r ′ j J i −r ′ i J j _ . (6.25) The integral over the symmetric part is zero. You can see this by integration by parts: 3 ∑ j=1 r j _ _ r ′ j J i + r ′ i J j _ dr ′ = 3 ∑ j,k=1 r j _ _ r ′ j ∂r ′ i ∂r ′ k J k + r ′ i ∂r ′ j ∂r ′ k J j _ dr ′ = 3 ∑ j,k=1 r j _ _ r ′ j r ′ i J k d 2 r l,=k ¸ ¸ ¸ ¸ r ′ k =∞ r ′ k =−∞ − _ r ′ i ∂J j (r ′ ) ∂r ′ j dr ′ _ = 0 . (6.26) In the first line we again used the Kronecker delta representation ∂r i /∂r j = δ i j , and in the second line we used the same tricks to show that the monopole moment is zero. The lowest order moment of the vector potential is therefore A i (r) = µ 0 8πr 3 3 ∑ j=1 _ r j _ r ′ j J i (r ′ ) −r ′ i J j (r ′ ) _ dr ′ . (6.27) It is very tempting to write the antisymmetric part r ′ j J i −r ′ i J j as a cross product. This leads to the definition of the magnetic dipole moment of the current distribution m = 1 2 _ rJ(r) dr . (6.28) 38 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 6 However, we can’t just plug the expression for m in Eq. (6.28) into Eq. (6.27), because the components do not match up: The cross product (r j J i −r i J j ) in Eq. (6.27) makes use of component i, which according to the equation should contribute to A i . This is not possible unless another cross product is involved. If we call 1 2 _ (r j J i − r i J j )dr ′ = m k with k ,= i, j, then we have the equation A i (r) = µ 0 4πr 3 3 ∑ j=1(,=i) ∑ k,=j,i r j m k . (6.29) You see that the A i component depends on terms r j and m k , with neither j nor k equal to i. This is crying out for another cross product, and it will involve m and r. So let’s see if we can massage mr such that we obtain Eq. (6.27): (mr) i = 3 ∑ j=1 1 2 _ _ r j r ′ j J i −r j r ′ i J j _ dr ′ . (6.30) The integrand involves the antisymmetric part of r ′ j J i . However, since we have just proved that the integration over the symmetric part is zero, we can add this to the integrand without affecting the integral. But adding the symmetric and anti-symmetric part of r ′ j J i is just r ′ j J i , and the cross product becomes (mr) i = 3 ∑ j=1 1 2 _ r j r ′ j J i (r ′ ) dr ′ . (6.31) This leads immediately to the vector potential of a magnetic dipole: A(r) = µ 0 4π mr r 3 . (6.32) Further reading – R.H. Good, Classical Electromagnetism, Saunders College Pub. (1999): Ch. 7, pp 164-182. – J.D. Jackson, Classical Electrodynamics, Wiley (1998): Sec 4.1 & 5.6, pp 145-150, 184-188. – B.I. Bleaney & B. Bleaney, Electricity and magnetism, Vol. 1, Oxford University Press (1976): Sec. 1.4, 2.3 & 4.2, pp 11-14, 38-43, 101-107. – J.R. Reitz, F.J. Milford, & R.W. Christy, Foundations of Electromagnetic Theory, fourth edition, Addison-Wesley (1993): Sec. 2.9, pp 46-48. – J. Schwinger, L.L. DeRaad, K.A. Milton, &W. Tsai, Classical Electrodynamics, The Advanced Book Program, Westview Press (1998): Ch. 22, pp 257-264. 39 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 6 Exercises 1. Let the charge distribution ρ(r, t) consist of two static charges ρ(r, t) = qδ(r) − qδ(r −d). Calculate the total charge Q, the dipole moment Q i , and the quadrupole moment Q i j by evaluating Eqs. (6.10), (6.11), and (6.12). How does the position of the origin affect the outcome? 2. A sphere of radius R rotates with angular velocity ω around a symmetry axis, and carries a uniformly distributed surface charge Q. Give the charge and current densities ρ and J, and verify that ∇ J = 0. 3. The magnetic moment of an electron is one Bohr magneton, µ B = e¯ h/2m. Suppose we model this as a small ring current of radius r, which can be considered the size of the electron. What is the smallest possible size of the electron according to this model? Experimentally, the electron behaves as if it is a point particle. Is this a problem for this model? 4. Verify Eqs. (6.7) and (6.8). 40 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 7 7 Dipole fields and radiation 7.1 Electric dipole radiation s θ +q −q ∆r r Figure 12: The Hertzian dipole: Radiation from an electric dipole current. So far we have looked at properties of electromagnetic waves (radiation) without asking how they are pro- duced. Since radiation is an electromagnetic effect, we expect that it is created by electric charges. How- ever, both static and uniformly moving charges cannot produce radiation (a uniformly moving charge is the same as a static charge in a different frame) because there is no characteristic time scale ω −1 in the phys- ical system. We therefore need to look at accelerating charges, and to this end we construct an electric dipole in which the charge q oscillates between two points separated by a distance s (see figure 12). The charge at the top of the dipole can be written as Q = q e −iωt , and we have a dipole p = sq e −iωt in the z direction. This is equivalent to a sinusoidal current I 0 e −iωt . We now calculate the scalar and vector potentials in spherical coordinates for this situation. At some distance r the scalar potential due to the top of the dipole is given by Φ(r, t ′ ) = q e −iωt ′ 4πǫ 0 r = q e −iω(t−r/c) 4πǫ 0 r = q e −iωt+ikr 4πǫ 0 r = Φ R (r, t) , (7.1) where t ′ = t −r/c is the retarded time: It takes a while for the change in the potential at r = 0 to propagate to r. To take this into account, we introduced the retarded potential Φ R (r, t), and we used the dispersion relation for radiation in free space k = ω/c. Next, we add the two retarded potentials of the two charges Φ (+) R and Φ (−) R = −Φ (+) R in the dipole. They almost cancel, but for the small separation s in the z direction, which gives rise to δr in the r direction: Φ R (r, t) = Φ (+) R +Φ (−) R = Φ (+) R (r, t) −Φ (+) R (r +δr, t) ≃ −δr ∂Φ (+) R ∂r (7.2) Substituting Eq. (7.1) and δr = s cos θ we obtain Φ R (r, t) = −s cos θ ∂ ∂r _ q e i(kr−ωt) 4πǫ 0 r _ = qs e i(kr−ωt) k cos θ 4πǫ 0 r _ 1 kr −i _ 41 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 7 = p R k cos θ 4πǫ 0 r _ 1 kr −i _ , (7.3) where p R is the retarded dipole moment. The vector potential of this oscillating dipole in spherical coordinates is A R (r, t) = µ 0 4π _ J R (r, t) r dV = 1 4πǫ 0 c 2 r I 0 e i(kr−ωt) s = 1 4πǫ 0 c 2 r I R s , (7.4) where s is the dipole vector s(cos θ ˆ r − sinθ ˆ θ), parallel to the z axis, and I R is the re- tarded current in the dipole. Using the relations Q = q e −iωt and I = dQ/dt, we can write the dipole moment p R in the scalar potential in terms of I R : p R = is ω I R . (7.5) We therefore have Φ R (r, t) = I R s cos θ 4πǫ 0 cr _ 1 + i kr _ and A R (r, t) = I R s(cos θ ˆ r −sinθ ˆ θ) 4πǫ 0 c 2 r . (7.6) The electric field E is given by E = −∇Φ R − ˙ A R . Using the derivatives in spherical coordinates we find E r (r, t) = − ∂Φ R ∂r −( ˙ A R ) r = kI R s cos θ 4πǫ 0 cr _ 2 kr + 2i (kr) 2 _ , E θ (r, t) = − 1 r ∂Φ R ∂θ −( ˙ A R ) θ = kI R s sinθ 4πǫ 0 cr _ 1 kr + i (kr) 2 −i _ . (7.7) We already know that radiation carries energy, so the total energy flux through a closed surface around the source is constant. If we take the surface to be that of a sphere with radius r, the energy flux 4πr 2 (ǫ[E[ 2 /2 +[B[ 2 /2µ 0 ) must be constant for all r. The radiating part of the E and B fields are therefore proporional to 1/r. Looking at Eq. (7.7), we see that the only contribution to the radiation is due to the i term in E θ . Similarly, the only contributing term of the B field is (see exercise) B φ (r, t) = −i kI R s sinθ 4πǫ 0 c 2 r = E θ (r, t) c , (7.8) where E θ (r, t) now denotes the radiating part of the E field. Poynting’s vector is given by S e = ReE θ ReB φ µ 0 = Re EB ∗ µ 0 = 1 16π 2 ǫ 0 c _ ksI sinθ r _ 2 ˆ r . (7.9) Note that we can calculate the Poynting vector by taking the complex conjugate of the B field and take the real part of S afterards. Again, it is often much easier to calculate with complex phase factors of the form e iϕ than with the trigonometric counterparts sinϕ, cos ϕ and tanϕ. 42 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 7 E B Figure 13: The directions of the E and B fields of a spherical wave radiated by a dipole in the centre. After half the oscillation period the fields point in the opposite direction. 7.2 Magnetic dipole radiation s s θ I r Figure 14: Radiation from a mag- netic dipole current. Now suppose we have a small current going in a square loop of area s 2 , leading to a magnetic dipole moment m = s 2 I 0 e −iωt . The current is changing in time in a periodic way. Again, we can define the re- tarded magnetic dipole by substituting t → t −r/c. If the current has a low impedance the scalar potential is zero, while the vector potential is A R (r, t) = µ 0 m R k sinθ 4πr _ − 1 kr + i _ ˆ φ. (7.10) We can again field the radiating parts of the E and B fields via E = − ˙ A and B = ∇A: E φ = µ 0 m R k 2 c sinθ 4πr = −cB θ , (7.11) and the Poynting vector is S m = Re EB ∗ 2µ 0 = 1 16π 2 ǫ 0 c _ (ks) 2 I sinθ r _ 2 ˆ r . (7.12) The difference between S e and S m is a factor (ks) 2 . 7.3 Larmor’s formula Clearly, radiation is generated by accelerated charges and changing currents. In the previous sections we have looked at periodically changing charges and currents, but 43 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 7 in many situations the charges and currents change in an aperiodic way. For example, when a charge is kicked, it will emit a burst of radiation. We will now consider the question of how much power P is radiated when a charge is accelerated by an amount a. This will lead to Larmor’s formula. Joseph Larmor (1857-1942) Suppose a particle with charge q follows a trajectory r q (t). We can formally associate a dipole moment p to this particle: p(t) = q r q (t) . (7.13) The current due to the motion of the charge is clearly re- lated to the velocity v q , so we can write I = dp dt = q dr q dt = q v q (t) . (7.14) The acceleration a q of the particle is proportional to the second derivative of the dipole moment: d 2 p dt 2 = q a q (t) . (7.15) Next, we calculate the vector potential and the magnetic field B of the moving charge. First, we note that the current I is the spatial integration over the current density, so we have A(r, t) = µ 0 4πr _ J(r ′ , t −r/c) dr ′ = µ 0 4πr I R = µ 0 4πr dp dt (t −r/c) , (7.16) evaluated at the retarded time, because the field needs time to propagate from the charge to the point where the vector potential is calculated. The magnetic field is found using B = ∇A. We need to keep in mind two things: (1) We can ignore all the contributions that do not scale like 1/r, because they do not propagate. (2) We must remember that the time variable depends on r because we consider the retarded time. The magnetic field is then B(r, t) = ∇A(r, t) = µ 0 4π _ ∇ _ 1 r _ dp dt + 1 r ∇ dp dt _ = µ 0 4πr ∇ dp dt , (7.17) where the last equality holds because ∇(1/r) leads to a non-propagating contribution to the field. The calculation of the curl of dp/dt is a bit tricky, so we’ll do it here in some detail. In particular, it is convenient to use index notation: _ ∇ dp dt _ i =ǫ i jk ∂ j _ dp dt _ k . (7.18) Using the chain rule, we can write (t = t q −r/c with ) ∂ j = ∂t ∂x j d dt = − 1 c x j r d dt , (7.19) 44 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 7 which allows us to rewrite the cross product: _ ∇ dp dt _ i = − 1 c ǫ i jk x j r _ d 2 p dt 2 _ k . (7.20) In vector notation, this becomes ∇ dp dt = − 1 cr r d 2 p dt 2 = − 1 c ˆ r d 2 p dt 2 = − q c ˆ ra q , (7.21) and therefore the B field can be written as B(r, t) = µ 0 4πr ∇ dp dt = − µ 0 4πcr ˆ r d 2 p dt 2 = − qµ 0 4πcr ˆ ra q , (7.22) evaluated at time t, and due to an accelerated charge at the retarded time t −r/c. The radiating part of the electric field E must be perpendicular to the radiating part of the magnetic field that we just calculated, and they differ by a factor c. Keeping the correct handedness, the radiating E field can be written as a cross product between B and ˆ r: E(r, t) = c Bˆ r = − qµ 0 4πr _ ˆ ra q _ ˆ r . (7.23) Now that we know the E and B fields, we can calculate the Poynting vector S, which gives the energy flux through a surface. If we take the surface to be a sphere with radius R surrounding the accelerating charge, we can calculate the total radiated power P = _ S S ˆ R R 2 dΩ, (7.24) where dΩ is a solid angle. The Poynting vector is straightforward to calculate: S = EB µ 0 = µ 0 q 2 16π 2 cr 2 __ ˆ ra q _ ˆ r ¸ _ ˆ ra q _ . (7.25) Using the vector identity A(BC) = B(A C) −C(A B) we rewrite this as S = q 2 16π 2 ǫ 0 c 3 r 2 _ a q − ˆ r(a q ˆ r) ¸ _ ˆ ra q _ = q 2 16π 2 ǫ 0 c 3 r 2 _ ˆ r _ [a q [ 2 −(a q ˆ r) 2 _ −a q _ a q ˆ r −a q ˆ r _ _ = q 2 16π 2 ǫ 0 c 3 r 2 _ a 2 q −(a q ˆ r) 2 _ ˆ r = q 2 a 2 q sin 2 θ 16π 2 ǫ 0 c 3 r 2 ˆ r , (7.26) 45 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 7 where we defined a q = [a q [ and we oriented our coordinate system such that the accel- eration is along the z direction. The inner product a q ˆ r is therefore equal to a q cos θ. The radiated power through a sphere surrounding the accelerating charge is now straightforward to calculate: P = _ S S ˆ R R 2 dΩ = _ S q 2 a 2 q sin 2 θ 16π 2 ǫ 0 c 3 R 2 R 2 dΩ = q 2 a 2 q 16π 2 ǫ 0 c 3 _ π 0 _ 2π 0 sin 2 θ dθ dφ. (7.27) The integral over sin 2 θ is equal to 8π/3, so we arrive at P = 1 4πǫ 0 2q 2 a 2 q 3c 3 . (7.28) This is the celebrated Larmor formula, derived by Joseph Larmor in 1897. Further reading – R.H. Good, Classical Electromagnetism, Saunders College Publishing (1999): Sec. 15.1 & 15.2, pp 371-379. – J.D. Jackson, Classical Electrodynamics, Wiley (1998): Sec. 9.1-9.3 pp 407-416. – R.P. Feynman, Lectures on Physics, volume II, Addison-Wesley (1964): Ch. 21. – J.R. Reitz, F.J. Milford, & R.W. Christy, Foundations of Electromagnetic Theory, fourth edition, Addison-Wesley (1993): Ch. 20, pp 525-544. – J. Schwinger, L.L. DeRaad, K.A. Milton, &W. Tsai, Classical Electrodynamics, The Advanced Book Program, Westview Press (1998): Ch. 32-33, pp 351-366. Exercises 1. Atoms can have both an electric and a magnetic dipole that produces radiation. Which of these typically two produces stronger radiation? 2. Consider again the Bohr model, in which an electron orbits a proton. The size of the orbit is given by the Bohr radius (5.29 10 −11 m), and the angular momentum of the ground state is ¯ h = 1.05 10 −34 Js. According to classical electrodynamics, how strong is the radiated power? Give an order-of-magnitude estimate of the lifetime of the atom. 46 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 8 8 Electrodynamics in macroscopic media 8.1 Macroscopic Maxwell equations So far, we have considered the fields of (moving) charges and the interactions between the charges. In principle, this covers all classical electromagnetic phenomena, since all matter is made up of charged particles that interact with each other. However, if we are dealing with 10 23 atoms we don’t want to solve the Maxwell equations for every individual atom. Especially when the time and length scales relevant to our problem are large compared to that of the atoms, we much rather average over the atomic fields and have effective fields in the media. The electric and magnetic fields in Maxwell’s equations are now average fields ¸E) and ¸B), and the charge and current densities ¸ρ) and ¸J) now consist of free charges and currents as well as bound charges and currents. Maxwell’s equations in macroscopic media then become ∇ ¸B) = 0 and ∇¸E) + ∂¸B) ∂t = 0 (8.1) ∇ ¸E) = ¸ρ) ǫ 0 and ∇¸B) −µ 0 ǫ 0 ∂¸E) ∂t = µ 0 ¸J) . (8.2) The charge and current densities are ¸ρ) = ρ f +ρ b and ¸J) = J f +J b (8.3) where the subscripts f and b denote “free” and “bound”, respectively. The fields in the materials are no longer the bare E and B fields due to freely moving charges and currents (as was the case for the microscopic Maxwell equations), but they are modified by the presence of the bound charges and currents. We want to relate the modified fields to material properties (the bound charges and currents) and modify Maxwell’s equations such that they involve the macroscopic fields and the free charges and currents. 8.2 Polarization and Displacement fields First, we consider a material that has bound charges. These are, for example, electrons that cannot move freely through the material, but that are bound to the parent molecule. When an external electric field is applied, the negative electrons move with respect to the positive nuclei, creating a small dipole moment. This dipole moment is aligned against the external field, because the external field pulls the charges such that the local field becomes smaller. All these little dipole moments create a macroscopic field P, which is related to the bound charge density via ρ b = −∇ P . (8.4) 47 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 8 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Uniform polarization Nonuniform polarization Figure 15: Polarization in a medium This is in some sense the definition of P. We must now relate the P to the dipole moments of the molecules. Let V be the volume of the medium, and the dipole density in the i direction is ρ b r i . Then _ V r i ρ b dr = − 3 ∑ j=1 _ V r i ∂P j ∂r j dr = − 3 ∑ j=1 _ S r i P j d 2 r k,=j ¸ ¸ ¸ ¸ r j onS r j onS + _ V P i dr , (8.5) where the first equality comes from Eq. (8.4), and the second comes from partial inter- gration. The surface term is zero when we take the surface just outside the medium. In vector notation, we therefore have _ V rρ b dr = _ V Pdr = ∑ V p n , (8.6) where the last term is the sum of all microscopic dipole moments in the volume V. Outside V we have P = 0. Now that we know what P means, we can rewrite the Maxwell equation ∇ ¸E) = ¸ρ)/ǫ 0 using Eqs. (8.3) and (8.4)as ∇ ¸E) = ρ f +ρ b ǫ 0 = ρ f ǫ 0 − 1 ǫ 0 ∇ P . (8.7) Considering we want the macroscopic field in terms of the free charges, we can define the macroscopic field as the displacement field D: D =ǫ 0 ¸E) +P (8.8) All the material properties are determinedby P. The corresponding macroscopic Maxwell equation is then ∇ D = ρ f (8.9) Often the polarization field is directly proportional to E, and we write P = χǫ 0 E, where χ is the susceptibility of the medium. The displacement field then becomes D =ǫ 0 (1 +χ)E ≡ǫ 0 ǫ r E ≡ǫE . (8.10) 48 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 8 Assuming for the moment that the permeability µ 0 does not change, one can see imme- diately that the wave equation yields a propagation velocity c 2 /ǫ r . In ordinary materi- als, we therefore have ǫ r ≥ 1. As we will see in the next lecture, this is directly related to the index of refraction. Some materials have a non-isotropic response to the electric field, for example P i =ǫ 0 3 ∑ j=1 χ i j E j , (8.11) where the polarization in the i direction depends in some way on the E field in the x, y, and z direction, while the polarization in the k ,= i direction depends on the same external E field in a different way. The susceptibility is therefore in general a tensor. Clearly, such a medium has an intrinsic orientation. When the medium is transparent, a medium like this gives rise to birefringence, where light with different polarizations (not to be confused with P) is bent differently when it enters the medium. Other materials have a polarization field P that depends on the external field E in a nonlinear way: P i =ǫ 0 _ χ (0) i + 3 ∑ j=1 χ (1) i j E j + 3 ∑ j,k=1 χ (2) i jk E j E k + 3 ∑ j,k,l=1 χ (3) i jkl E j E k E l + . . . _ . (8.12) The susceptibility has different orders χ (n) i...l corresponding to the n th power in the exter- nal field. In practice, we consider only the effects up to third order, since higher order effects tend to be extremely weak. The χ (2) term is responsible for frequency doubling materials, among other things, and the χ (3) term can be used to induce optical phase shifts that are proportional to the intensity of a second field. It is also called the Kerr nonlinearity. 8.3 Magnetization and Magnetic induction Now we will construct the macroscopic magnetic field using the free and bound cur- rents. Apart from the free currents in the material, there may be bound currents that are confined to the molecules. In quantum mechanics these bound currents can be the spins of the electrons and nuclei, but let’s keep things classical for now. The bound currents may be due to a change of bound charges, which can be written as ∂P/∂t. Alternatively, they may correspond to little current loops, which give rise to a microscopic magnetic dipole moment. Under the influence of an external field B, these little magnets can line up to form a macroscopic field that we shall denote by ∇M. We then have the following expression for the bound current: J b = ∂P ∂t +∇M (8.13) 49 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 8 Uniform magnetization Nonuniform magnetization Figure 16: Magnetization in a medium The divergence of the bound charge is ∇ J b = ∂∇ P ∂t +∇ (∇M) = − ∂ρ b ∂t = 0 , (8.14) since the bound charge is conserved. Now let’s consider the second inhomogeneous Maxwell equation ∇¸B) −µ 0 ǫ 0 ∂¸E) ∂t = ∇¸B) −µ 0 ǫ 0 ∂ ∂t _ D ǫ 0 − P ǫ 0 _ = µ 0 ¸J) = µ 0 _ J b +J f _ . (8.15) This leads to ∇¸B) −µ 0 ∂D ∂t +µ 0 ∂P ∂t = µ 0 ∂P ∂t +µ 0 ∇M +µ 0 J f . (8.16) We can now define the magnetic field H in terms of the magnetic induction B and the macroscopic magnetization M as H = 1 µ 0 ¸B) −M, (8.17) yielding the Maxwell equation ∇H− ∂D ∂t = J f (8.18) The homogeneous Maxwell equations remain unchanged, since here E and B are fundamental (that is, no material properties are present in these equations). Also, for the macroscopic case it is understood that all fields are averages, and we drop the brackets ¸.). The macroscopic Maxwell equations then become ∇ B = 0 and ∇E + ∂B ∂t = 0 (8.19) ∇ D = ρ f and ∇H− ∂D ∂t = J f . (8.20) 50 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 8 Further reading – R.H. Good, Classical Electromagnetism, Saunders College Publishing (1999): Ch. 9 & 10, pp 204-277. – J.D. Jackson, Classical Electrodynamics, Wiley (1998): Sec. 4.3-4.6 & 5.8, pp 151-165, 191-194. – R.P. Feynman, Lectures on Physics, volume II, Addison-Wesley (1964): Ch. 11 & 36. – W.J. Duffin, Electricity and Magnetism, McGraw-Hill (1990): Ch. 11, 12, pp 282-335. – J.R. Reitz, F.J. Milford, & R.W. Christy, Foundations of Electromagnetic Theory, fourth edition, Addison-Wesley (1993): Ch. 4 & 9, pp 97-126, 218-255. – C.A. Brau, Modern Problems in Classical Electrodynamics, Oxford University Press (2004): Sec. 6.1, pp 282-293. – J. Schwinger, L.L. DeRaad, K.A. Milton, &W. Tsai, Classical Electrodynamics, The Advanced Book Program, Westview Press (1998): Ch. 4, pp 33-44. Exercises 1. A slab of linear dielectric material is placed inside a capacitor. The dielectric does not fill the space inside the capacitor completely. Determine the field lines of E, P, and D between the capacitor plates. 2. A slab of linear dielectric material is placed inside a capacitor. The dielectric slab and the plates of the capacitor are square with side w, and the thickness of the slab is s. The plates are held at constant potential V. w x s F V (a) Write down the energy in a capacitor. (b) How does the charge on the plates change? Does this change the energy of the capacitor? (c) Calculate the force on the dielectric slab. 3. A one-dimensional dielectric has a nonlinear polarization field given by P(t) = χ (1) E(t) +χ (2) E(t) 2 . If the incoming wave is described by E(t) = E 0 cos ωt show that the dielectric gives rise to a wave with double the frequency. What happens when the incoming wave is a superposition of two frequencies ω 1 and ω 2 ? (Hint: write the waves in exponential notation.) 51 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 9 9 Waves in dielectric and conducting media From our everyday experience it is clear that electromagnetic radiation (in particular light) is able to propagate through some media, but not through others. Light has no problem travelling through air and glass, but when it hits a conductor, most of it will be reflected. In this lecture we investigate why this is, and what are the characteristic features of different media. Georg Ohm (1789-1854) When waves penetrate a medium, the molecules in the medium respond to the electric field and the magnetic induction to form polarization and magnetization fields. Let’s assume that the medium is electrically neutral, so that there are no free charges (ρ = 0), but there may be electrons in a conduction band that can produce free cur- rents J. These electrons presumably respond directly to the electric field, such that the free current is propotional to E: J =σE (Ohm’s law). (9.1) This is Ohm’s law, and the constant of proportionality σ is the conductance. This law has a wide range of applicability, but you should be aware that there are many interesting situations where Ohm’s law does not hold. There are essentially three regimes of interest for σ: (1) σ is small, and the medium is a dielectric; (2) σ is large, and the medium is a conductor; or (3) σ is imaginary, and the medium is a plasma. We will derive the index of refraction for dielectrics, the skin depth for conductors, and the plasma frequency for plasmas. We will also derive the reflectivity of these three types of materials. 9.1 Waves in dielectric media First, we rederive the wave equations from the Maxwell equations, but this time we keep the J term. We find the following differential equations for E and B: ∇ 2 E −µǫ ∂ 2 E ∂t 2 = µ ∂J ∂t , (9.2) ∇ 2 B −µǫ ∂ 2 B ∂t 2 = −µ∇J , (9.3) where ǫ =ǫ r ǫ 0 and µ = µ r µ 0 . Using Ohm’s law, this leads to ∇ 2 E −µǫ ∂ 2 E ∂t 2 −µσ ∂E ∂t = 0 , (9.4) ∇ 2 B −µǫ ∂ 2 B ∂t 2 −µσ ∂B ∂t = 0 . (9.5) 52 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 9 Air Dielectric Air Dielectric (a) (b) E fields B fields Figure 17: Waves incident on a dielectric. The waves propagate in the dielectric. (a) The electric field is mostly transmitted, and the reflected wave experiences a phase shift. (b) The magnetic field is mostly transmitted, and the reflected wave does not experience a phase shift. We substitute again the plane wave solutions for waves in the positive z direction E = E 0 e i(kz−ωt) and B = B 0 e i(kz−ωt) , and we immediately find that k = _ ǫµω 2 + iωµσ = k 0 √ µ r ǫ r _ 1 + iσ ωǫ , (9.6) where k 0 = ω/c. You can see that the wave vector k is complex! Now we can look at the three different regimes for σ. If the conductance is small (σ/ωǫ ≪ 1), the medium is a dielectric and the imaginary root is approximately equal to 1. The wave vector is real, so the waves propagate freely. However, the value of the wave vector is modified by the factor √ ǫ r µ r . It turns out that this is indeed the index of refraction, at least for apolar materials 4 . The reflectivity of a dielectric can be calculated as well. For ordinary dielectrics where D = ǫE and B = µH, the ratio between the electric field and the magnetic induction is given by the velocity of the wave in the medium: E B = c √ ǫ r µ r . (9.7) In air, this leads to approximately E = cB, and the (average) Poynting vector in a medium is then S av = nǫ 0 cE 2 rms , (9.8) where n is the index of refraction, and E 2 rms is the average strength to the electric field squared. 4 In apolar media the distribution of the electrons in the molecules is fast enough to follow the oscillat- ing fields. In polar materials, however, the response of the medium is slow since the polar molecules have to re-orient themselves. This is the principle behind the microwave: The water molecules try to realign themselves with an oscillating microwave field, and as a consequence the water gets heated. 53 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 9 In order to calculate the reflectivity, we need to figure out what the boundary con- ditions are for the different fields at the separation of the medium and the air. We will consider the difference between the fields just outside the medium and the field just inside the medium. From ∇ D = ρ we find that ∆D n = D 1,n −D 2,n =σ f , (9.9) where the subscript n denotes the normal component with respect to the surface, and the 1 and 2 denote the field outside and inside the medium, respectively. σ f is the surface charge. From ∇ B = 0 we find that ∆B n = B 1,n −B 2,n = 0 . (9.10) Air Medium l l ∆z Figure 18: Boundary condi- tions on a dielectric surface. The boundary condition for ∆E t is more complicated. From ∇E = − ˙ B we have (see Fig. 18) _ E dl = − _ ∂B ∂t da , (9.11) where the contour is taken as a rectangle perpendicular to the surface and sticking half-way into the surface. The parallel sides of the loop have length l, and the sides per- pendicular to the surface have length ∆z. We therefore have l(E 1,t −E 2,t ) = −l∆z ∂B ∂t = 0 , (9.12) where the last equality follows from taking the limit ∆z → 0. We therefore have the third boundary condition ∆E t = E 1,t −E 2,t = 0 . (9.13) Finally, we establish the boundary condition for the transverse magnetic field ∆H t . There may be infinite bound surface current densities J b , so we use ∇H = J + ˙ D, which ignores J b . The free surface current will be denoted by K f , and the last boundary condition reads ∆H t = H 1,t −H 2,t = K f . (9.14) Now we can state the relations between the incident (I), reflected (R), and transmitted fields (T): E I −E R = E T and B I + B R = B T µ r . (9.15) Note that the reflected B field picks up a relative minus sign. This is because the Poynt- ing vector must point in the opposite way to the incident wave, and from Eq. (9.7) we know that the magnetic induction must gain in relative strength in the medium. 54 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 9 Now assume that the medium is such that µ r = 1. We then have B = nk ω E , (9.16) which leads to E I + E R = nE T . (9.17) Together with E T = E I −E R , this yields the ratio E R E I = n −1 n + 1 . (9.18) The reflectivity R of a surface can be defined as the ratio of the reflective and incident Poynting vectors. Since the Poynting vector is proportional to E 2 , the reflectivity of a dielectric medium becomes R = _ n −1 n + 1 _ 2 . (9.19) For glass, the index of refraction is n = 1.6, so the normal reflection is about ten percent. 9.2 Waves in conducting media In a conducting mediumσ is very large, and we can ignore the +1 term in Eq. (9.6). The wave vector k becomes k = k 0 ¸ iµ r σ ωǫ 0 = ± _ ωσµ 2 (1 + i) ≡ ± _ 1 δ + i δ _ . (9.20) Since the wave propagates in the positive z direction, we choose the positive sign. We defined the skin depth δ as δ = ¸ 2 ωσµ . (9.21) When we substitute this value for k into the solution for the E field we obtain E(z, t) = E 0 exp _ i _ z δ −ωt __ exp _ − z δ _ , (9.22) that is, there is a propagating term and an exponentially decaying term, which falls off to e −1 when the wave has reached the skin depth. The magnetic field is given by B = kE/ω, where now k is complex. With the time dependence made explicit, we find that B = _ σµ ω E 0 e i(z/δ−ωt+π/4) e −z/δ . (9.23) 55 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 9 Air Metal Air Metal (a) (b) E fields B fields Figure 19: Waves incident on a metal. (a) The electric field is mostly reflected with a phase shift. (b) The magnetic field is mostly reflected, and does not have a phase shift. There is no propaga- tion of the wave in the metal, and the exponential fall-off of the amplitudes is characterized by the skin depth. Note the π/4 phase lag of B with respect to E. Inside the metal we have B 0 = _ σµ ω E 0 e iπ/4 (9.24) and the Poyning vector is S av = 1 2 _ σ 2ωµ E 2 0 e −2z/δ ˆ k . (9.25) This shows that the power falls off twice as fast as the fields. The reflectivity of metals is high, and consequently the transmittance T is low (R + T = 1 from energy conservation). Using again B = kE/ω, and the boundary conditions of the previous section, we once more use E I −E R = E T and B I + B R = B T µ r , (9.26) and find (n is a large complex number) E T E I = 2 1 + n ≃ 2 n , (9.27) where n is now a complex index of refraction n = ¸ iµ r σ ωǫ 0 . (9.28) 56 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 9 In addition, we can solve Eq. (9.26) for the magnetic induction, which yields B T B I = 2 1 + 1/n ≃ 2 . (9.29) The ratio of the Poynting vectors for the transmitted and incident waves is therefore S T S I = Re 4 n . (9.30) This leads to the transmittance T = _ 8ωǫ 0 σ . (9.31) Usually, the transmitted waves will be absorbed completely unless the metal is suffi- ciently thin. Whenσ →∞such as in a superconductor, the transmittance drops to zero and everything is reflected. That is why high quality cavities are made with supercon- ducting mirrors. 9.3 Waves in plasmas Finally, consider a plasma in which the electrons are disassociated from the atomic nu- clei. When an electric field is applied the electrons accelerate. The current density J therefore lags the electric field, and the phase happens to be e iπ/2 = i. The nuclei are much heavier than the electrons, so they do not accelerate as fast and make a negligable contribution to the conductivity. We can find the velocity v of the electrons by integrat- ing F = ma = eE, where all vector quantities are in the z direction. Using the electron density Ne, the current density is J = Nev = Ne ieE ωm , (9.32) and the conductivity is σ = J E = ie 2 N ωm . (9.33) We substitute this into our expression for the complex wave vector k k = k 0 ¸ 1 − e 2 N ω 2 mǫ 0 = k 0 _ 1 − _ ω p ω _ 2 , (9.34) where we defined the plasma frequency ω p = ¸ e 2 N mǫ 0 . (9.35) 57 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 9 Clearly, when the frequency of the wave ω is larger than the plasma frequency, k is real and the plasma acts as a dielectric. When ω < ω)p the plasma acts as a conductor. The ionosphere of the Earth is a plasma, and consequently for low enough frequen- cies (long wavelengths) it acts as a mirror. Since the earth is also a conductor, low frequency radio waves (AM) can bounce between the ionosphere and the Earth to reach much further than high frequency (FM) radiowaves. The plasma frequency of the iono- sphere is about 3 MHz. Further reading – R.H. Good, Classical Electromagnetism, Saunders College Publishing (1999): Ch. 16, pp 390-412. – J.D. Jackson, Classical Electrodynamics, Wiley (2003): Sec. 7.1-7.6, pp 295-319. – R.P. Feynman, Lectures on Physics, volume II, Addison-Wesley (1964): Ch. 32 & 33. – H.J. Pain, The Physics of Vibrations and Waves, Wiley (1983): Ch. 7, pp 187-218. – J.R. Reitz, F.J. Milford, & R.W. Christy, Foundations of Electromagnetic Theory, fourth edition, Addison-Wesley (1993): Ch. 17, pp 412-440. – J. Schwinger, L.L. DeRaad, K.A. Milton, &W. Tsai, Classical Electrodynamics, The Advanced Book Program, Westview Press (1998): Ch. 41-42, pp 427-448. Exercises 1. Calculate the skin depth of copper for 60 Hz, 10 kHz, and 10 MHz. The conduc- tivity of copper is 5.8 10 7 Ω −1 m. 2. Calculate the speed of a 1 MHz wave in copper and in glass. 58 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 10 10 Relativistic formulation of electrodynamics Albert Einstein (1879-1955) One of the remarkable things about the Maxwell equa- tions is that they give rise to a correct description of all sorts of radiation, most notably light. In particular, it pre- dicts the correct velocity of light in vacuum via the rela- tion c 2 = (µ 0 ǫ 0 ) −1 , as we have seen in Lecture 5. Ex- perimentally it was found by Michelson and Morley in 1887 that the velocity of light is independent of the veloc- ity of the source. This caused Lorentz quite a headache, which ultimately resulted in the Lorentz transformations for moving bodies: Objects that move with respect to an inertial observer are seen to experience a length contrac- tion in the direction of their motion, and their clocks seem to run slower. There are similar rules for the electromagnetic field, meaning that what one inertial observer calls the E field, a second observer may describe in terms of both E and B fields. We therefore want to know two things: (1) How do the fields transform from one coordinate system to another, and (2) What do Maxwell’s equations look like when written in covariant form, that is, independent of the coordinate system? We will first revise the basics of special relativity and Minkowski space, and then we will con- struct Maxwell’s equations from the vector and scalar potentials, and the fields. This leads naturally to the transformation laws for the E and B fields. We end the lecture with a look at some invariant properties. This lecture will rely heavily on the transfor- mation properties of vectors and tensors, and everything will be in index notation. 10.1 Four-vectors and transformations in Minkowski space Remember that covariant descriptions take place in four-dimensional Minkowski space. There is the position four-vector x µ , where µ = 0, 1, 2, 3, r = (x 1 , x 2 , x 3 ) or r = x ˆ i + y ˆ j + z ˆ k, and x 0 = ct, and the momentum four-vector p µ with p = (p 1 , p 2 , p 3 ) and p 0 = U/c. Minkowski space is a strange place in that it has a non-trivial metric: Ordinarily, if you want to find the length of a (short) interval, you add all the components squared, according to Pythagoras: ds 2 = dx 2 + dy 2 + dz 2 . (10.1) However, in relativity the length changes in different reference frames, and we need to include the change in time coordinate as well. It is tempting to just add c 2 dt 2 to Eq. (10.1), but that would be wrong! The correct distance between two events is ds 2 = −c 2 dt 2 + dx 2 + dy 2 + dz 2 , (10.2) 59 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 10 with a minus sign in front of c 2 dt 2 ! This is the definition of the distance in Minkowski space, and we can rewrite this as a vector equation (and also using Einstein’s summation convention): ds 2 = (ct, x, y, z) _ _ _ _ −1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 _ _ _ _ _ _ _ _ ct x y z _ _ _ _ ≡ x µ g µν x ν = x µ x µ = x µ x µ , (10.3) where the four-vector x µ can be written as (ct, x, y, z). Notice that the matrix g µν has a “−1” on the diagonal, which is responsible for the minus sign in the metric. You also see that there is a difference between upper and lower indices. The four-vectors with upper indices are called contravariant, and the four-vectors with lower indices are called covariant. You can verify from Eq. (10.3) that the covariant four-vector has a relative minus sign in the first component: x µ = (−ct, x, y, z) . (10.4) We are cheating a bit here, because the x µ on the left is a component, while the quantity on the right is a proper vector. However, this expression is to emphasize the effect of the vertical position of the index; it is not a proper equation. As you can see from Eq. (10.3), the metric raises or lowers the index, and g µν = g µν in the case of special relativity. We now amend the Einstein summation convention: the sum is implied over two re- peating indices, one of which is upper, and the other is lower (convince yourself that this distinction is important). Carrying out the sum is called contraction. We can construct invariant quantities (scalars) by contracting contravariant four-vectors with covariant four-vectors. In general we write the components of a four-vector a µ as (a 0 , a 1 , a 2 , a 3 ), and we have: a µ b µ = a µ g µν b ν = ∑ µ,ν a µ g µν b ν = −a 0 b 0 + 3 ∑ j=1 a j b j = −a 0 b 0 +a b , (10.5) where greek indices sum over all four components, while roman indices sum only over the spatial part 5 . As an example, let’s look at the four-momentum of a particle: p = (U/c, p 1 , p 2 , p 3 ), and p 2 = p µ p µ can be written as p 2 = −U 2 /c 2 +[p[ 2 . Since this is true for all inertial frames, it is true for the frame where the particle is at rest, so [p[ 2 = 0. Using U = mc 2 with m the rest mass of the particle, we find p 2 = −m 2 c 2 (yes, in Minkowski space 5 You should be aware of the fact that the metric g µν can also be written as a diagonal matrix with three −1 entries for the spatial part and one +1 for the temporal part. This does not lead to observable differences in the theory, so it is a convention. Unfortunately, both conventions are used regularly, so make sure you know what the metric is before you copy a relativistic formula from a book! 60 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 10 a square can be negative without complex numbers!), and the energy of a particle is therefore U = _ m 2 c 4 +[p[ 2 c 2 . (10.6) This is one of the most important formulas in physics. Hendrik Antoon Lorentz (1853-1928) Any four-vector a µ is transformed into a µ ′ due to a Lorentz transformation Λ µ ′ µ via a µ ′ = Λ µ ′ µ a µ = ∑ µ Λ µ ′ µ a µ . (10.7) Note that the primed frame of reference is indicated by a primed index µ ′ . The Lorentz transformation involves a contraction over the old (unprimed) coordinates in order to remove them from the equation. In general, a tensor transforms as T µ ′ ν ′ = Λ µ ′ µ Λ ν ′ ν T µν . (10.8) Every component is transformed with a separate Lorentz transformation Λ µ ′ µ . The actual form of the Lorentz trans- formation, for example for a boost in the z direction, is Λ µ ′ µ = _ _ _ _ γ 0 0 −γβ 0 1 0 0 0 0 1 0 −γβ 0 0 γ _ _ _ _ , (10.9) with γ = 1/ _ 1 −β 2 and β = v/c. In component notation the Lorentz transformation x µ ′ = Λ µ ′ µ x µ becomes x 0 ′ = γx 0 −γβx 3 = 1 _ 1 −v 2 /c 2 _ ct − vz c _ , x 1 ′ = x 1 = x , x 2 ′ = x 2 = y , x 3 ′ = −γβx 0 +γx 3 = 1 _ 1 −v 2 /c 2 (z −vt) . (10.10) Note that the Lorentz transformation is a proper coordinate transformation: Λ µ ′ µ = ∂x µ ′ ∂x µ . (10.11) Using these rules it is straghtforward (but somewhat lengthy) to find the transformation rule for any tensor. 61 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 10 10.2 Covariant Maxwell equations Just like the position and momentum four-vectors, we would like to construct elec- tric and magnetic four-vectors. Unfortunately, it is not that simple: There are six field components (three for E and three for B), and we cannot force them into four-vectors. On the other hand, we can combine the vector potential and the scalar potential into a fourvector A µ : A µ (x) = _ Φ(x) c , A(x) _ , (10.12) where x is again the position four-vector. The second four-vector that we can construct straight away is the current density j µ : j µ (x) = _ cρ(x), J(x) _ . (10.13) Since the Maxwell equations are differential equations, we also need to construct a four-vector out of the differential operators. Again, they come in two variations, covari- ant ∂ µ and contravariant ∂ µ : ∂ µ = ∂ ∂x µ = _ − 1 c ∂ ∂t , ∇ _ and ∂ µ = ∂ ∂x µ = _ 1 c ∂ ∂t , ∇ _ . (10.14) These operators transform just like ordinary vectors (but note the vertical position of the indices). We can construct an invariant with these operators if we contract them: ∂ µ ∂ µ = ∂ 2 x + ∂ 2 y + ∂ 2 z − 1 c 2 ∂ 2 t = . (10.15) This is the d’Alembertian, which we first encountered in lecture 3. The continuity equation (conservation of charge) then becomes particularly com- pact: ∂ µ j µ = 0 . (10.16) Compare this with Eq. (2.19); the two equations say exactly the same thing! The Maxwell equations in Eqs. (3.9) can be written as ∂ µ ∂ µ A ν = µ 0 j ν (10.17) Often people say that this encompasses all of electrodynamics. However, you should remember that we also need to know how charges respond to the fields, that is, we need to know the Lorentz force. Since this force depends explicitly on the velocity of the particle, it is not in covariant form. We will consider the covariant Lorentz force later. We also see explicitly that the Maxwell equations are already relativistically correct: We did not change anything in Eqs. (3.9), we just rewrote everything. The compactness of Eq. (10.17) indicates that Lorentz invariance is a symmetry of electrodynamics (The more symmetric the object, the more compact we can make its description: For example, 62 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 10 compare the description of a perfect sphere with the description of a sponge). Lorentz invariance is an important symmetry in quantum field theory and the standard model. Maxwell thought that the electric fields were carried by some substance that per- vades all space, called the æther. Such a substance defines a natural reference frame, which would break Lorentz invariance. It was Einstein who disposed of the æther and made all inertial reference frames equivalent. Thus he uncovered that electrodynamics was a relativistic theory all along. In fact, his famous paper of 1905 on special relativity was titled “On the electrodynamics of moving bodies”. In addition to the Maxwell equations in Eq. (10.17), we need to specify a particular gauge for the four-vector potential A µ if we actually want to calculate anything. The relativistically invariant gauge is the Lorenz gauge ∂ µ A µ = 0 . (10.18) Nowthat we have constructed the four-vector potential for the electromagnetic field, we can ask what the E and B fields look like in a relativistic setting. We can use the relations E = −∇Φ− ∂A ∂t and B = ∇A. (10.19) If we write out a few components we get, for example, E x = − ∂Φ ∂x − ∂A x ∂t = −c _ ∂ 1 A 0 −∂ 0 A 1 _ , B x = ∂A z ∂y − ∂A y ∂z = −(∂ 2 A 3 −∂ 3 A 2 ) . (10.20) Obviously we can only fit all six components of the electric and magnetic field into a rel- ativistic object if we have something with two indices. We call this the (anti-symmetric) field-strength tensor F: F µν = ∂ µ A ν −∂ ν A µ = _ _ _ _ 0 E x /c E y /c E z /c −E x /c 0 −B z B y −E y /c B z 0 −B x −E z /c −B y B x 0 _ _ _ _ . (10.21) The relativistic transformation of the fields then becomes F µ ′ ν ′ = Λ µ ′ µ Λ ν ′ ν F µν . (10.22) For a boost v in the z direction given by Eq. (10.9) we find E ′ x = γ _ E x + vB y _ , E ′ y = γ _ E y −vB x _ , E ′ z = E z (10.23) B ′ x = γ _ B x − v c 2 E y _ , B ′ y = γ _ B y + v c 2 E x _ , B ′ z = B z . (10.24) 63 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 10 + + (a) (b) v Figure 20: The field lines of an electric charge. (a) The charge is stationary. (b) The charge is moving with velocity v. In terms of the parallel and perpendicular components of the fields with respect to the direction of motion this becomes E ′ [[ = E [[ E ′ ⊥ = γE ⊥ +γ vB B ′ [[ = B [[ B ′ ⊥ = γB ⊥ −γ vE c 2 . (10.25) A physical situation with only a static charge distribution and no currents obviously does not involve B fields. The magnetic fields arise when the charge distribution is viewed by a moving observer. In particular, the electrostatic force F = qE aquires a term proportional to vB, so that F ′ = qE +qvB (verify this!). This is of course the familiar Lorentz force. In other words, when an observer sees a moving electric charge, the E field is not spherically symmetric due to Lorentz contraction in the direction of motion. As a result, there is a component of the electric force that acts only on moving charges. This is the magnetic field. This argument shows that you can think of the magnetic field as just the relativistic part of the electric field of moving charges. Whenever there are B fields, there must be moving charges, whether it be macroscopic currents or spinning charges in atoms. The field lines of a stationary and moving charge are shown in Fig. 20. Finally the Lorentz force must be formulated in a covariant way. The force experi- enced by a particle with charge q, as viewed from the lab frame, is proportional to the velocity, so we need to define a four-velocity u. We can use the relation p µ ≡ mu µ = mγ(c, v) (10.26) such that u µ u µ = −c 2 (verify this!). Rather than deriving the relativistic Lorentz force, we will state it here and confirm that it behaves the way it should. Suppose that τ is the proper time of a particle, as recorded by a co-moving clock. The force will cause a mo- mentum transfer dp µ /dτ. This must be proportional to E + vB. The cross product is built in the field strength tensor F µν , so we guess that the covariant form of the Lorentz force is m d 2 x µ dτ 2 = dp µ dτ = qu ν F µν = qF µν dx µ dτ . (10.27) 64 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE 10 These are the equations of motion for a relativistic particle in an electromagnetic field. Let’s see what the different components of f look like: f x = dp 1 dτ = qF 1ν u ν = γq _ c E x c + v x 0 + v y B z −v z B y _ f y = dp 2 dτ = qF 2ν u ν = γq _ c E y c −v x B z + v y 0 + v z B x _ f z = dp 3 dτ = qF 3ν u ν = γq _ c E x c + v x B y −v y B x −v z 0 _ (10.28) These are indeed the three components of the Lorentz force, with an extra factor γ. This is correct, because the term dp µ /dτ also has an implicit γ factor in the relativistic mass. Remains the question of what is f t : c f t = c dp 0 dτ = qcF 0ν u ν = γq _ v x E x + v y E y + v z E z _ = γq v E. (10.29) This is the relativistic work done on the particle by the fields. So you see that it all works out. 10.3 Invariant quantities When the fields and potentials change when viewed in different inertial frames, it is important to know what the invariant quantities of a theory are. These quantities are the same in all reference frames, so you can evaluate them in whichever frame makes the calculation easiest. There are many invariants, most notably the magnitude of four- vectors a 2 = a µ a µ . You can use the differential operator as well in the construction of invariants (see Eq. (10.13) for an important example). A quantity is invariant when it does not depend on the coordinates, so in index notation the quantity has no indices. The quantity 1 4 F µν F µν is invariant (why?), and can be evaluated as 1 4 F µν F µν = 1 4 g ρµ g σν F ρσ F µν = B 2 2 − E 2 2c 2 . (10.30) This is the Lagrangian (density) of the electromagnetic field, which plays a pivotal role in particle physics. We can also construct the dual of the field-strength tensor T µν = 1 2 ǫ µνρσ F ρσ = _ _ _ _ 0 −B x −B y −B z B x 0 E z /c −E y /c B y −E z /c 0 E x /c B z E y /c −E x /c 0 _ _ _ _ . (10.31) which is a pseudo-tensor. Here we used the Levi-Civita (pseudo) tensor ǫ µνρσ , which returns 1 if the indices make an even permutation, −1 if they make an odd permutation, 65 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE and 0 otherwise. The invariant product F µν T µν is F µν T µν = 4E B/c . (10.32) Finally, the microscopic Maxwell equations can be written in terms of the fields as ∂ µ F µν = µ 0 j ν and ∂ µ T µν = 0 (10.33) Further reading – R.H. Good, Classical Electromagnetism, Saunders College Publishing (1999): Ch. 18, pp 430-490. – J.D. Jackson, Classical Electrodynamics, Wiley (1998): Ch. 11, pp 514-566. – R.P. Feynman, Lectures on Physics, volume II, Addison-Wesley (1964): Ch. 25, 26 & 31. – J.R. Reitz, F.J. Milford, & R.W. Christy, Foundations of Electromagnetic Theory, fourth edition, Addison-Wesley (1993): Ch. 22, pp 558-578. – C.A. Brau, Modern Problems in Classical Electrodynamics, Oxford University Press (2004): Ch. 1, pp 29-72. – J. Schwinger, L.L. DeRaad, K.A. Milton, &W. Tsai, Classical Electrodynamics, The Advanced Book Program, Westview Press (1998): Ch. 10, pp 111-124. Exercises 1. Calculate the determinant of the Lorentz transformation Λ µ ′ µ in Eq. (10.9) 2. Calculate the expression for the energy of a relativistic particle as in Eq. (10.6), but now with the metric g µν = _ _ _ _ 1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1 _ _ _ _ . 3. Show that g µν is a proper transformation that can be written as g µν = ∂x µ ∂x ν . 4. Prove that E B is Lorentz invariant. 5. Show that the phase of an electromagnetic wave is a Lorentz invariant quantity. What exactly is invariant here? 6. Calculate F µν , F µν F µν , T µν , and verify Eq. (10.32). 66 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE A A Special coordinates and vector identities A.1 Coordinate systems Often it is much more convenient to use cylindrical or spherical coordinates, rather than cartesian coordinates. The differential operators change accordingly. For cartesian coordinates: AB = (A y B z − A z B y ) ˆ x + (A z B x − A x B z ) ˆ y + (A x B y − A y B x ) ˆ z (A.1) ∇f = ∂ f ∂x ˆ x + ∂ f ∂y ˆ y + ∂ f ∂z ˆ z (A.2) ∇ A = ∂A x ∂x + ∂A y ∂y + ∂A z ∂z (A.3) ∇A = _ ∂A z ∂y − ∂A y ∂z _ ˆ x + _ ∂A x ∂z − ∂A z ∂x _ ˆ y + _ ∂A y ∂x − ∂A x ∂y _ ˆ z (A.4) Cylindrical coordinates: ∇f = ∂ f ∂ρ ˆ ρ + 1 ρ ∂ f ∂φ ˆ φ+ ∂ f ∂z ˆ z (A.5) ∇ A = 1 ρ ∂(ρA ρ ) ∂ρ + 1 ρ ∂A φ ∂φ + ∂A z ∂z (A.6) ∇A = _ 1 ρ ∂A z ∂φ − ∂A φ ∂z _ ˆ ρ + _ ∂A ρ ∂z − ∂A z ∂ρ _ ˆ φ+ 1 ρ _ ∂(ρA φ ) ∂ρ − ∂A ρ ∂φ _ ˆ z (A.7) ∇ 2 f = 1 ρ ∂ ∂ρ _ ρ ∂ f ∂ρ _ + 1 ρ 2 ∂ 2 f ∂φ 2 + ∂ 2 f ∂z 2 (A.8) x y z Figure 21: Cartesian coordinates (x, y, z). 67 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE A x y z φ ρ Figure 22: Cylindrical coordinates (ρ, φ, z). Spherical coordinates: ∇f = ∂ f ∂r ˆ r + 1 r ∂ f ∂θ ˆ θ + 1 r sinθ ∂ f ∂φ ˆ φ (A.9) ∇ A = 1 r 2 ∂(r 2 A r ) ∂r + 1 r sinθ ∂(A θ sinθ) ∂θ + 1 r sinθ ∂A φ ∂φ (A.10) ∇A = 1 r sinθ _ ∂(A φ sinθ) ∂θ − ∂A θ ∂φ _ ˆ r (A.11) + 1 r _ 1 sinθ ∂A r ∂φ − ∂(rA φ ) ∂r _ ˆ θ (A.12) + 1 r _ ∂(rA θ ) ∂r − ∂A r ∂θ _ ˆ φ (A.13) ∇ 2 f = 1 r 2 ∂ ∂r _ r 2 ∂ f ∂r _ + 1 r 2 sinθ ∂ ∂θ _ sinθ ∂ f ∂θ _ + 1 r 2 sin 2 θ ∂ 2 f ∂φ 2 (A.14) x y z φ r θ Figure 23: Spherical coordinates (r, φ, θ). 68 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE A A.2 Integral theorems and vector identities Having defined the divergence and the curl of vector fields, we can state the important integral theorems (without proof): _ V ∇ AdV = _ S A dS (Gauss’ theorem) (A.15) _ S (∇A) dS = _ C A dl (Stokes’ theorem) (A.16) _ b a ∇f dl = f (b) − f (a) (A.17) _ V _ f ∇ 2 g − g∇ 2 f _ dV = _ S ( f ∇g −g∇f ) dS (Green’s theorem) (A.18) and some important vector identities, such as relations between 3-vectors: A (BC) = B (CA) = C (AB) (A.19) A(BC) = (A C)B −(A B)C (A.20) (AB) (CD) = (A C)(B D) −(A D)(B C) (A.21) first derivatives: ∇ (φA) = φ(∇ A) +A (∇φ) (A.22) ∇(φA) = φ(∇A) + (∇φ)A (A.23) ∇(A B) = A(∇B) +B(∇A) + (A ∇)B + (B ∇)A (A.24) ∇ (AB) = B (∇A) −A (∇B) (A.25) ∇(AB) = (B ∇)A−(A ∇)B +A(∇ B) −B(∇ A) (A.26) and second derivatives: ∇ (∇A) = 0 (A.27) ∇(∇φ) = 0 (A.28) ∇(∇A) = ∇(∇ A) −∇ 2 A (A.29) A.3 Levi-Civita tensor Finally, we may sometimes use the so-called Levi-Civita symbol ǫ i jk , which returns a value of 1 when (i jk) is an even permutation of the indices, −1 if (i jk) is an odd per- mutation, and 0 otherwise (for example when two indices are the same). It obeys the identity 3 ∑ k=1 ǫ i jk ǫ klm = 3 ∑ k=1 ǫ i jk ǫ lmk = δ il δ jm −δ im δ jl , (A.30) where δ i j is of course the Kronecker delta, which returns 1 if i = j and 0 if i ,= j. 69 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE B B Unit systems in electrodynamics Carl Friedrich Gauss (1777-1855) Traditionally, electrodynamics is plagued by the use of many different sets of units. In these lecture notes we use SI units (Syst` eme International) exclusively. However, many people also use Gaussian units, so it is worthwhile to spend some time exploring the difference betwen these units. In particular, you should be able to read the set of units used off the Maxwell equations. The rule of thumb is: if they involve factors of 4π, then they most likely use Gaussian units. Carl Friedrich Gauss was a German mathematician who in the 1830s did (among very many other things) worldwide measurements of the Earth’s magnetic field. Not only are the Gaussian units named after him, the unit of magnetic induction in Gaussian units is also called the Gauss (G). In SI units the unit of magnetic induction is the Tesla (T), after the Serbian inventor Nikola Tesla (1853- 1943). There is a certain arbitrariness in Maxwell’s equations, the Lorentz force, and the continuity equation, in the sense that we can add constants k 1 , k 2 , . . . to the equations that affect only the units, and not the physics: k 1 ∇ J + ∂ρ ∂t = 0 and F = k 2 ρE + k 3 JB (B.1) ∇ E = k 4 ρ and ∇ B = 0 (B.2) ∇E + k 5 ∂B ∂t = 0 and ∇B = k 6 J + k 7 ∂E ∂t (B.3) Not all the constants k 1 to k 7 are independent. Some manipulation of the above equa- tions will give the identity k 6 ∇ J + k 4 k 7 ∂ρ ∂t = 0 (B.4) so we have the restriction that k 6 = k 1 k 4 k 7 . Similarly, by rederiving the wave equations we find that k 5 k 7 = c −2 . More confusion arises when we consider macroscopic media. Gauss’ law in terms of the polarization field becomes ∇ (E + k 4 P) = k 4 ρ free . (B.5) which defines the displacement field D = k 8 (E + k 4 P) . (B.6) 70 PHY331 PART I: ADVANCED ELECTRODYNAMICS LECTURE B Similarly, with a bound current density J bound = ∇M + ˙ P the Maxwell-Amp` ere law becomes ∇(B −k 6 M) = k 6 J free + k 7 ∂ ∂t (E + k 4 P) . (B.7) The magnetic field is then defined as H = k 9 (B −k 6 M) . (B.8) The different constants k 1 . . . k 9 in both SI and Gaussian units are given by k 1 k 2 k 3 k 4 k 5 k 6 k 7 k 8 k 9 SI units 1 1 1 ǫ −1 0 1 µ 0 c −2 ǫ 0 µ −1 0 Gaussian units 1 1 c −1 4π c −1 4πc −1 c −1 1 1 As you know, in the SI system the fundamental units are meter (m), kilogram (kg), second (s), and amp` ere (A). The unit of force is the Newton (N), which is of course kg m/s 2 . Just like the second and the kilogram, the amp` ere is defined operationally. One amp` ere is the amount of current needed to create a force of 2 10 −7 N per meter length between two infinitely long parallel wires separated by a distance of one meter. In the Gaussian system, the fundamental units of mass, length, and time are the gram (g), centimeter (cm), and again the second (s). Instead of the amp` ere, the other fundamental unit is the statcoulomb (statC) for the unit of charge. The statvolt (statV) is the unit of potential, and obeys 1 statV = 1g cm/statC s 2 . charge current E B µ 0 ǫ 0 SI C = As A N/C = V/m T = N/Am Tm/A C 2 /Nm 2 Gauss statC statA = statC/s statV/cm G(statV/cm) – – In Gaussian units the permittivity and the permeability are dimensionless. Further reading – W.J. Duffin, Electricity and Magnetism, McGraw-Hill (1990): App. C, pp 406-413. – J.R. Reitz, F.J. Milford, & R.W. Christy, Foundations of Electromagnetic Theory, fourth edition, Addison-Wesley (1993): App. III, pp 599-602. – C.A. Brau, Modern Problems in Classical Electrodynamics, Oxford University Press (2004): Appendix, pp 581-589. – J. Schwinger, L.L. DeRaad, K.A. Milton, &W. Tsai, Classical Electrodynamics, The Advanced Book Program, Westview Press (1998): App. A, pp 555-559. 71