Contents. 1 Introduction to Vector and Tensor Analysis 7. 2 Foundations of Theory of Relativity Relativistic Dynamics 34

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1 X I N TA O C L A S S I C A L E L E C T R O DY N A M I C S P U B L I S H E R

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3 Contents 1 Introduction to Vector and Tensor Analysis 7 2 Foundations of Theory of Relativity 19 3 Relativistic Dynamics 34 4 Charges in a Given Electromagnetic Field 41 5 The Electromagnetic Field Equations 53 6 General Electrostatics 66 7 Electrostatics in Matter and Boundary Value Problems 75 8 General Magnetostatics 94 9 Magnetic Fields in Matter Time-varying Electromagnetic Fields Fields of Moving Charges Radiation of Electromagnetic Waves 145

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5 讲义说明 2018 版说明 : 一个主要变化是重新加入了从分析力学的角度推导 Maxwell 方程的章节, 因为这一章节非常有助于从场论的角度理解电动力学, 同时也强调了场作为另一种描述物理现象的方式与粒子的同等重要地位 与 2013 年的讲义不同的是, 这次的场的分析力学部分为选修章节, 不会计入最终成绩, 也不会参加正常的考试部分 2015 版说明 : 这本讲义主要是我自己讲课时参考所用, 其内容主要了参考了 Landau 和 Lifshitz 的 The Classic Theory of Fields, David J. Griffiths 的 Introduction to Electrodynamics (4th Edition) 其余部分还参考了 Jackson 的 Classical Electrodynamics 在这三本书中,Landau 和 Jackson 的书并不太适合一般本科生阅读 这个讲义里对 Landau 的书作了大幅简化, 而 Jackson 的书则只是参考了一小部分数学证明和一些个别有意思的章节 正常情况下, 我推荐的阅读书是 Griffiths 的电动力学导论 这个讲义第一版的使用是在 2014 年春季学期, 当时采用的单位制是 Gauss 单位制, 因为我觉得 Gauss 单位制更适合理论电动力学 但在实行过程中, 不少同学向我建议换回国际单位制, 以求更好地和之前的课程衔接 因此从 2015 年春季学期开始, 我对这个讲义作了比较大幅度的改写, 一是增加更多的基础数学知识, 二是去除场的分析力学部分, 三是将 Gauss 单位制换为 SI 单位制 在这里要特别感谢金泽宇, 王贤瞳, 和吴昊楠三位同学的协助, 他们在改单位制的过程中付出了大量劳动, 并且修正了 2014 年讲义中的很多错误 2014 年春季学期班上还有好几位其他同学在 14 年教学过程中也陆续地指出讲义中的错误, 但遗憾当时并没有记录名单, 在此一并感谢

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7 1 Introduction to Vector and Tensor Analysis 1.1 Basic Concepts How do you define a vector? From your elementary math classes: A Vector is an object with both magnitude and direction; e.g., displacement, velocity, etc. A more general way of defining a vector is related to how they transform when one changes coordinates. A formal definition of a vector is that any set of three (in ordinary space) components that transforms in the same manner as the displacement vector when one changes coordinates. You can see that, the displacement vector is the model for all vectors. This definition of a vector is very important to understand four-vectors in special relativity. You can see that some sets of numbers do look like a vector, but they do not transform like the model vector (the displacement vector), therefore, they are not vectors Einstein Summation Convention Einstein considered this to be his most important invention. It essentially contains three rules: 1. Repeated indices are implicitly summed over. 2. Each index can appear at most twice in any term. 3. Each term must contain identical non-repeated indices. The repeated indices are called dummy indices, and the nonrepeated indices are called free indices. The actual use of this summation convection is actually much more straightforward than the above three rules. Consider two

8 8 classical electrodynamics vectors, A and B. In Cartesian coordinates, A = A 1 e 1 + A 2 e 2 + A 3 e 3 = B = B 1 e 1 + B 2 e 2 + B 3 e 3 = 3 A i e i, (1.1) i=1 3 B i e i. (1.2) i=1 We know that A B = 3 i=1 A ib i. With the Einstein notation, we can write That is, A B = A i B i. (1.3) A i B i = 3 A i B i. (1.4) i=1 You do not need the summation sign. Index i is a dummy index. This can greatly simplify the notation of vector analysis. You ll see this when you do more and more vector or tensor analysis. Another example, (A B)C = j ( i A i B i )C j e j = (A i B i )C j e j. (1.5) Both i and j are dummy indices. You should now see how the Einstein notation can simplify the vector/tensor expression. One more example, M ij A j = M ij A j, (1.6) j is the i-th component of M A, where M is a second-rank tensor. Here i is a free index and j is a dummy index. Rule 2 and 3 can easily help to identify wrong expressions. For example, A i B i C i cannot be correct because of rule 2: the index i appears three times in this term. Expression A i B j C j + E p is also wrong, because of rule 3: The non-repeated index in A i B j C j is i, while in E p is p, and they are not identical. On the other hand, A i B j C j + E i is correct. This expression means the i-th component of the vector A(B C) + E. One more useful notation, the Einstein notation for A B, A B = (A 2 B 3 A 3 B 2 )e 1 + (A 3 B 1 A 1 B 3 )e 2 + (A 1 B 2 A 2 B 1 )e 3. So how do you write this using Einstein notation? Let s introduce the Levi-Civita symbol ɛ ijk given by +1 if i,j,k form an even permutation of 1, 2, 3. ɛ ijk = 1 if i,j,k form an odd permutation of 1, 2, 3. 0 otherwise (1.7)

9 introduction to vector and tensor analysis 9 Hence, ɛ ijk = ɛ jki, ɛ ijk = ɛ jik, and ɛ iik = 0. A very useful identity of ɛ ijk is ɛ ijk ɛ imn = δ jm δ kn δ jn δ km, (1.8) where δ jm = 0 if j = m and 1 if j = m. With the Levi-Civita symbol, you can easily see that A B = ɛ ijk A j B k e i. (1.9) The Del Operator The gradient of a scalar is T. In Cartesian coordinates, T T = e x x + e T y y + e T z z. (1.10) We can define x 1 x, x 2 y, x 3 z, and write T = e i T x i. (1.11) We say that is a vector operator. In Cartesian coordinates, = e i. (1.12) x i A few expressions related to can be quickly given. For example, the divergence of a vector, A = x i A i = A i x i. (1.13) The curl of a vector A is A A = ɛ ijk e i A x k = ɛ ijk e k i. (1.14) j x j 1.2 Vector Algebra In this section, I will teach you how to memorize/derive commonly used vector algebra without referring to a handbook. We will use a

10 10 classical electrodynamics lot of vector analysis and identities in this class. A B C = B C A = C A B = A B C A (B C) = (C B) A = (A C)B (A B)C ( f g) = f g + g f ( f A) = f A + A f ( f A) = f A + f A (A B) = B A A B (A B) = A ( B) + B ( A) + (A )B + (B )A ( A) = ( A) 2 A A = 0 f = 0 A = A B B A ˆB A = ˆB ( ˆB A) In the last two identities, the direction is w.r.t. ˆB. There are also some useful identities involving r, the radial vector: r = 3 r = 0 r = I r = r r = e r [ f (r)r] = 0. First, for A (B C), you can interchange the dot and cross. A (B C) = A B C. Or you can cyclically permute the order of vectors, like A (B C) = B C A. Of course, since B C = C B, A (B C) = A C B. Second, for A (B C), you can use the middle-outer rule 1. A ( }{{} B }{{} C ) = B middle outer }{{} middle ( } A {{ C } other two dotted ) C }{{} outer ( } B {{ A } other two dotted ). 1 Fundamentals of Plasma Physics by Paul M. Bellan ( B }{{} outer C }{{} middle ) A = C }{{} middle ( } B {{ A } other two dotted ) B }{{} outer ( } A {{ C } other two dotted I ve found this middle-outer rule quite convenient. Third, for identities involving, you should treat it as being both a vector and a differential operator. Applying the calculus rule (ab) = a b + ab, we know ( f g) = g f + f g ).

11 introduction to vector and tensor analysis 11 I find it not difficult to see ( f A) = f A + A f ( f A) = f A + f A For more complicated identities involving, we introduce two mathematical tricks. The first one involves 3 steps. I ll use (A B) as an example. Step 1: Treat as an operator; it applies to both A and B (adding subscripts), and add the resulting terms. (A B) = A (A B) + B (A B). Step 2: Treat A and B as vectors, applying vector rules, and making necessary adjustments to put A right before A and B right before B. Adjustments should also satisfy vector rules. A (A B) = ( A B)A B( A A) Similarly, we have middle-outer rule = (B A )A B( A A) adjustment B (A B) = ( B B)A ( B A)B = ( B B)A (A B )B Step 3: Dropping subscripts of, we have (A B) = (B )A B( A) + ( B)A (A )B Now let s use this trick to derive (A B). 1. (A B) = A (A B) + B (A B). 2. A (A B) = A A B = B A A B (A B) = B B A = A B B 3. (A B) = B A A B Now let s try another example, A ( B). 1. A ( B) = A ( B B) 2. A ( B B) = B (A B) (A B )B Making adjustment ( B only operates on B) = ( B B) A (A B )B. 3. A ( B) = ( B) A (A )B Note: you might ignore adding subscripts after you get familiar with the method. The second trick for vector/tensor analysis uses the property that a vector/tensor identity is valid in all coordinate systems. Using this property, we can perform vector-tensor analysis in the following way: 1. write out the components in Cartesian coordinates

12 12 classical electrodynamics 2. do necessary analysis 3. from components, restore the original variables Usually the last step is the most difficult one, it requires some experience and practice. Example 1: Prove r = In Cartesian coordinates, r = xe x + ye y + ze z 2. r = (e x x + e y y + e z z ) (xe x + ye y + ze z ) = x x + y y + z z = 3. or simply in Cartesian coordinates, r = e i x i x j e j = x j x i δ ij = x i x i = 3. (1.15) Example 2: Prove r = 0 1. ( r) i = ɛ ijk xj x k = 0. Let s try more complicated examples. First, prove (A B) = B ( A) A B. (1.16) Here it goes: (A B) = e i (ɛ x jkl A k B l )e j = (ɛ i x ikl A k B l ) (1.17) ( ) i Ak B = ɛ ikl B x l + A l A k = ɛ k B i x ikl B i x l + ɛ ikl A l k i x i (1.18) = ɛ lik A k x i B l A k ɛ kil B l x i (1.19) = ( A) B A ( B) (1.20) = B ( A) A ( B) (1.21) One more example. Let s now use the method to prove (A B) = (B )A B( A) + ( B)A (A )B. The steps are (A B) = ɛ ijk (A B) x k e i = ɛ ijk ɛ j x klm A l B m e i j ( ) A = ɛ ijk ɛ l klm B m + B m A x j x l e i j ( ) A = ɛ kij ɛ l klm B m + B m A x j x l e i j ( ) A = (δ il δ jm δ im δ jl ) l B m + B m A x j x l j e i

13 introduction to vector and tensor analysis 13 Take a break and let s continue... (A B) = (δ il δ jm δ im δ jl ) = δ il δ jm ( A l x j B m + B m x j A l ( A l B m + B m A x j x l j ) ( ) A δ im δ l jl B m + B m A x j x l j ( A = B i B j A j j + A x i B j x i j = ( B A i x j j x j A B i j + B j A x j x i A ) j B B j x i A i j j x j = (B )A + ( B)A ( A)B (A )B e i e i ) ) e i e i e i 1.3 Curvilinear coordinates I do not talk about general vector analysis in curvilinear coordinates in this class; it s quite complicated. If you are interested, see books in my reference list. I will only very briefly review spherical and cylindrical coordinates, since you should have learned this before. In spherical coordinates (e r, e θ, e φ ), Gradient of a scalar f = f r e r + 1 f r θ e θ + 1 f r sin θ φ e φ, Divergence A = 1 ) (r 2 r 2 A r + 1 r r sin θ θ (sin θ A θ) + 1 A φ r sin θ φ, Curl of a vector ( A) r = 1 ( ) 1 sin θaφ r sin θ θ r sin θ ( A) θ = 1 A r r sin θ φ 1 ( ) raφ, r r ( A) φ = 1 r r (ra θ) 1 A r r θ. A θ φ Laplacian 2 f = 1 ( r 2 r 2 f ) + 1 ( r r r 2 sin θ f ) + sin θ θ θ 1 r 2 sin 2 θ 2 f φ 2.

14 14 classical electrodynamics In cylindrical coordinates (e r, e φ, e z ), Gradient of a scalar f = f r e r + 1 f r φ e φ + f z e z, Divergence A = 1 r r (ra r) + 1 A φ r φ + A z z, Curl of a vector ( A) r = 1 r A z φ A φ z ( A) φ = A r z A z r, ( A) z = 1 r r ( raφ ) 1 r A r φ. Laplacian 2 f = 1 r ( r f ) f r r r 2 φ f z The Dirac Delta Function The one-dimensional Dirac Delta Function The 1D Dirac delta function, δ(x), is defined as 0, if x = 0, δ(x) =, if x = 0, and δ(x)dx = 1. Strictly speaking, δ(x) is not a function, since its value is not finite at x = 0. It is a distribution. Note that we also have b a δ(x)dx = 1, as long as 0 [a, b]. Suppose f (x) is a ordinary continuous function, then it immediately follows from the definition of δ(x) that f (x)δ(x) = f (0)δ(x).

15 introduction to vector and tensor analysis 15 Correspondingly, f (x)δ(x)dx = f (0) δ(x)dx = f (0). A slightly more general form of 1D Dirac delta function is of course δ(x a), 0, if x = a, δ(x a) =, if x = a, and δ(x a)dx = 1. And it is straightforward to show that f (x)δ(x a)dx = f (a). We almost always think that expressions involving δ(x) should be used under an integral sign. In particular, two expressions involving delta functions, D 1 (x) and D 2 (x), are considered equal if f (x)d 1 (x)dx = for all ordinary functions f (x). For example, δ(kx) = 1 k δ(x), f (x)d 2 (x)dx, from which you can see that δ(x) is even; i.e., δ( x) = δ(x). Another frequently used expression is δ(g(x)) = i δ(x x i ) g (x i ), where g(x) is a continuously differentiable function and x i are all roots of g(x) = 0 and of course g (x) = The three-dimensional Dirac delta function The generation of δ(x) to 3D is very easy. δ(r) = δ(x)δ(y)δ(z), where r = xe x + ye y + ze z is the displacement vector. You see that 0, if r = 0, δ(r) =, if r = 0, and all space δ(r)dv = 1.

16 16 classical electrodynamics Of course, you can have f (r)δ(r a)dr = f (a). Questions: Now, try to express the charge density ρ(x) of an ideal point charge q located at x 0 using the Dirac delta function. The charge density satisfies, ρdv = q. 1.5 A very brief introduction to tensor The simplest tensor, a dyad ( 并矢 ), can be introduced via the dotproduct between vectors 2. A dyad AB is defined by X AB (X A)B, 2 W.D.D haeseleer, W.N.G.Hltchon, J.D.Callen, and J.L.Shohet, Flux coordinates and magnetic field structure. or AB Y A(B Y). The notation for a dyad is a simple juxtaposition of the two vectors making up the dyad, without a symbol in between. From the definition of the dyad, you can see that in general AB = BA, since AB C = BA C. The component of a dyad can be written, noting that A = A i e i and B = B i e i, AB = A i e i B j e j = A i B j e i e j. Of course, we can use a single notation for a dyad; e.g., D AB, then D ij = A i B j. In this class, the most common form of tensors is dyad, so you d better get familiar with this one. A general second-order tensor is F = F ij e i e j. A general second-order tensor cannot be written as a dyad, but it can be written as a summation of dyads; i.e., F = ab + cd + e f + In 3-D space, you need at least three dyads to represent a general second-order tensor. A vector can be considered to be a first-order tensor, and a scalar is a zeroth-order tensor. If you define a second-order tensor using transformation properties, then a tensor transforms under coordinate transformations like the products of components of two vectors. As you can see from the definition of a dyad. The dot-product between a second-order tensor and a vector is a vector. For example, F A = F ij e i e j A k e k = F ij A k e i δ jk = F ij A j e i.

17 introduction to vector and tensor analysis 17 So a dot product reduces the rank of a tensor by one. The dotproduct between a tensor and two vectors gives a scalar, like A F B = A i e i F jk e j e k B l e l = A i F jk B l δ ij δ kl = A i F ik B k. This dot product can also be written as A F B BA : F F : BA. It s easy to see that in the component of a tensor, there are two free indices; e.g., F ij. In a vector, there is one; e.g., A i. In a scalar, there is zero; e.g., C. So the number of free indices in the component of an object equals the rank of the tensor. A special second-order tensor is the unit tensor I, which has the property that I A = A I = A, I : AB = A B, (ϕi) = ϕ, I : B = B. Here are two examples involving tensors. Example 1: Prove that Solution: ( f g r) = [ ( f g)] r + g f. ( f g r) = e i x i f k (g r) l e k e l = Example 2: Prove that Solution: x i [ f i (g r) l ] e l = ɛ lkm ( f x i g k r m )e l i = ɛ lkm ( f x i g k )r m e l + ɛ lkm f i g k (r m )e i x l i = ɛ lkm [ ( f g)] k r m e l + ɛ lkm g k ( f r) m e l = [ ( f g)] r + g ( f I) = [ ( f g)] r + g f ( f gh) = ( f )gh + ( f g)h + g( f h). 1. ( f gh) = f ( f gh) + g ( f gh) + h ( f gh) 2. f ( f gh) = ( f f )gh g ( f gh) = ( g f )gh = ( f g )gh = ( f g g)h h ( f gh) = ( h f )gh = ( f h )gh = g( f h h) 3. ( f gh) = ( f )gh + ( f g)h + g( f h)

18 18 classical electrodynamics 1.6 Some useful references about vector analysis Note: See the course homepage for links to these documents. 1. Curvilinear coordinates and tensors, Chapters 1,2,and 3, Flux and Coordinates and Magnetic Field Structure, by D haeseleer, HItchon, Callen, and Shohet. 2. 向量张量运算符号法胡友秋, 电动力学讲义 ( 课程主页上可以下载 ). 3. 张量分析, 黄克智, 薛明德, 陆明万

19 2 Foundations of Theory of Relativity 2.1 The principle of relativity In Newtonian mechanics, the Galileo s principle of relativity (PR) is that the laws of mechanics are identical in all interval systems of reference under Galileo transformations. Suppose there are two inertial reference frames K and K ; K moves with V relative to K. Then r = r + Vt, (2.1) t = t. (2.2) Note that time is absolute in classical mechanics. It can be shown, however, that Maxwell equations do not satisfy Galileo s principle of relativity under Galileo transformation. (Left as Ex.) There are three possible ways to solve this problem. 1. The Maxwell equations were wrong. The proper theory of EM was invariant under Galileo transformations. 2. Galileo PR only applied to classical mechanics. EM is not mechanics. 3. There exists a general principle of relativity for both classical mechanics and electromagnetism. Einstein did some really hard thinking and chose 3. He then propose the following two postulates based on experiments done by other people and lots of his own thinking. The principle of relativity. All the laws of nature are identical in all inertial systems of reference. The constancy of the speed of light. The speed of light (c) is independent of the motion of its source; its numerical value is c = cm/s. It can be immediately shown that time being absolute is not consistent with Einstein s PR. From Galileo s PR, the velocity transforms like v = v + V. (2.3)

20 20 classical electrodynamics This equation directly follows from that t being invariant. However, this leads to that v can be larger than c, not consistent with Einstein PR. 2.2 Intervals in spacetime For convenience of presentation, we ll first introduce a few concepts. Event: An event is described by the place it occurred and the time when it occurred. We also introduce a fictitious four-dimensional space; marked by 3 space coordinates and 1 time coordinates. For an idealized particle, an event is defined by three coordinates and the time when the event occurs. We consider two inertial reference frames K and K, with parallel axes. Suppose K moves in V relative to K. Now define two events in K system. Event 1: sending out a light signal from (x 1, y 1, z 1 ) at t 1. Event 2: the arrival of the signal at (x 2, y 2, z 2 ) at t 2. The signal traveled c t, or ( x 2 + y 2 + z 2 ) 1/2, with definition f = f 2 f 1. So we have c 2 t 2 ( x 2 + y 2 + z 2 ) = 0 (2.4) In K, noting that time is not absolute, the coordinates of the same events are Event 1, (x 1, y 1, z 1 ) and t 1 Event 2, (x 2, y 2, z 2 ) and t 2 Because of the constancy of light speed c, we have in K system c 2 t 2 ( x 2 + y 2 + z 2 ) = 0. (2.5) If x 1 y 1 z 1 t 1 and x 2 y 2 z 2 t 2 are the coordinates of any two events, we define s by s = [c 2 t 2 x 2 y 2 z 2 ] 1/2, (2.6) and call it the interval between these two events. From the previous analysis, we reach an important conclusion that if s = 0 in K, then s = 0 in any K. If two events are infinitely close to each other, then the interval ds is ds = [c 2 dt 2 dx 2 dy 2 dz 2 ] 1/2. (2.7)

21 foundations of theory of relativity 21 The form of ds allows us to regard it as the distance between two world points in the fictitious four-dimensional space. This space is called Minkowski space (axes: x, y, z, and ct), it s pseudo-euclidean. If Euclidean, the distance would be l = [c 2 t 2 + x 2 + y 2 + z 2 ] 1/2. (2.8) Now the question: What s the relationship between ds in K and ds in K in general (valid for ds = 0)? We have two constraints: 1. If ds = 0, then ds = ds and ds are infinitesimal of the same order. From these two conditions, we have ds = ads (2.9) We now derive the factor a in ds = ads. For an inertial reference frame, space and time are homogeneous. If a = a(t), Equation (2.9) violates that time being homogeneous. If a = a(x), Eq (2.9) depends on the location of the origin of K in K; it is a violation of space being homogeneous. If a = a(v), that means Eq (2.9) depends on the direction of K moving in K; a violation of space being homogeneous. Therefore we conclude a = a( V ) or a = a(v 2 ), and ds = a(v 2 )ds. If K moves relative to K in V, then K moves relative to K in V. We should have ds = a(v 2 )ds. (2.10) These two equations immediately lead to a = 1 and ds = ds OR ds 2 = ds 2. (2.11) The interval between two events is invariant under transformation from one inertial frame to another. This is the mathematical formulation of the invariance of c. From the invariance of ds, we can immediately reach the following conclusion: If a particle moves with v < c in K, then v < c in all other K. Proof: ds 2 = c 2 dt 2 dx 2 = (c 2 v 2 )dt 2. The invariance of ds 2 finishes the proof. With the invariance of ds, time is no longer absolute. Statements like two events occur simultaneously do not necessarily hold if we transform to another reference frame. Let s now discuss this type of problem. Our first question is, if two events occur at two different times in K ( t = 0), can we find a reference frame K in which t = 0? Suppose we can find a K so that t = 0. From the invariance of s 2, we have s 2 = c 2 t 2 x 2 = x 2 < 0. (2.12)

22 22 classical electrodynamics Hence if s 2 < 0 or if s is imaginary, it is possible to find a reference frame where t = 0. Imaginary intervals are said to be space-like. For space-like intervals, the concepts of simultaneous, earlier, and later are relative. Note that c 2 t 2 < x 2, meaning that the two events are so separated that no signal can propagate from one point to the other point within t. Following the previous question, it is natural to ask another one: if two events occur at two different times in K ( t = 0), what s the condition for t = 0 in all K? From the invariance of s 2, we have s 2 = c 2 t 2 x 2 = c 2 t 2 x 2. (2.13) The minimum value for c 2 t 2 to take is when x 2 = 0, and s 2 = c 2 t 2 x 2 = c 2 t 2 > 0. (2.14) Hence if s 2 > 0 or if s is real, it is not possible to find a K so that t = 0. Note that c 2 t 2 > x 2. Real intervals are said to be time-like. Note that for time-like intervals, it is possible to connect two events using a signal with propagation speed less than c, since x/ t < c. For time-like intervals, the concepts of future and past are absolute. To see this, let s assume the interval between event 1 and event 2 is time-like, then s 2 = c 2 t 2 x 2 = c 2 t 2 x 2 > 0. (2.15) Solving for t from this equation, we have t = ± t (1 ( x 2 x 2 )/c 2 t 2 ) 1/2. (2.16) Which sign should we take for t? To see this, we know that t t as V 0, where V is the velocity of K relative to K. Hence t = t(1 ( x 2 x 2 )/c 2 t 2 ) 1/2. (2.17) Therefore for time-like intervals, we must have 1. If t > 0, then t > 0 in all K. 2. If t < 0, then t < 0 in all K. That is, the concepts of future or past are absolute for time-like intervals. The concept of time-like or space-like intervals are important if two events are causally related. For event 1 to be the reason of event 2, event 1 must occur before event 2 in all reference frames; i.e., event 1 is in the absolute past of event 2. Therefore the interval must be time-like. On the other hand, if event 1 is the reason of event 2, a signal has to propagate from event 1 to event 2. The propagation speed of signal is then x 2 / t 2, which is less than c 2, since the interval is time-like. This is the same statement as that c is the maximum speed of propagation of interaction. Finally we point out that because ds is an absolute concept, the time-like or space-like property of an interval is also absolute.

23 foundations of theory of relativity Proper time Since time is not absolute, it is convenient to introduce the concept of proper time. Proper time is the time read by a clock moving with the object; it is normally denoted by τ. Let s now derive the relationship between dt observed in laboratory frame K and the proper time dτ of the object. In the laboratory frame K, the object moves with a constant velocity v. Let the inertial reference frame moving with the object be called K. From the invariance of ds 2 and dx = 0 (because the object is still in K ), we have From this equation, ds 2 = c 2 dτ 2 = c 2 dt 2 dx 2. (2.18) dτ = ds/c = dt (1 dx 2 /c 2 dt 2 ) 1/2. (2.19) But dx/dt = v, therefore we have dτ = dt (1 v 2 /c 2 ) 1/2 dt/γ, (2.20) where γ = 1/(1 β 2 ) 1/2 with β = v/c; β 1 and γ 1. Note that dτ = dt/γ; the fact that γ 1 leads to Or by integrating this equation, we have τ 2 τ 1 = dτ dt. (2.21) t2 t 1 dt/γ < t 2 t 1 (2.22) Conclusion: Moving clocks go more slowly than those at rest. There are lots of experiments proving this conclusion. Here I list a few: 1. The lifetime of mu-mesons (muons) 1. 1 Ref: Feynman s Lecture on Physics, Vol 1, Average lifetime of muons is or about 2 µs. If they travel at v c, the Galilean distance 600 m. Muons can travel from the top of atmosphere to the surface of Earth. 2. A similar example: The lifetime of pions 2. 2 Ref: Jackson, Classical Electrodynamics, 2nd Edition, pp 520. On the other hand, the concept is new and confusing. See the following two paradoxes: 1. If K moves relative to K in V, then dt = dt/γ. From K, however, K moves relative to K in V. then dt = dt /γ. How to understand this? 3. 3 Ref: 郭硕鸿, 电动力学, 第三版, 第 204 页 2. The twin paradox 4 : Identical twins; one of them makes a jour- 4 Ref: Feynman s Lectures on Physics, Vol 1, 16-2 ney into space in a high-speed spaceship and returns home. Which twin should have aged more?

24 24 classical electrodynamics 2.4 The Lorentz transformation With Galileo PR, r = r + Vt, t = t, and we know this is not correct in general. Now time is not absolute, how do we make coordinate transformations? Again we consider two IRF s K and K ; K moves relative to K in V. For an event with coordinate (t, r), what s the corresponding coordinate (t, r ) in K? Basic requirement: the transformation should not change any s, the distance between any two points in Minkowski space. There are basically two choices for the transformation: 1. Parallel displacement 2. Rotation of the coordinate system. However, a parallel displacement only leads to a change in r or t reference point, therefore, the required transformation = a rotation of the 4D space. In 3D space, each rotation can be resolved into three rotations in the planes xy, yz, and xz. In 4D space, each rotation can be resolved into six rotations in the planes xy, zy, xz, tx, ty, tz. Rotations in xy, zy, xz only changes space coordinates; they are ordinary space rotations. We now consider rotations involving time t. We want to know the transformation equation for rotations in planes, e.g., tx. Let s first review how to derive the transformation equations in an Euclidean space with Cartesian coordinates. Rotations in xy plane need to conserve r 2 = x 2 + y 2 (2.23) To derive the rotation equation, we define an angle θ so that x = r cos θ, (2.24) y = r sin θ. (2.25) A rotation would change the angle from θ to θ + φ, so that r 2 = x 2 + y 2 is always conserved. Changing θ to θ + φ, in the rotated coordinate, x = r cos(θ + φ) = r cos θ cos φ r sin θ sin φ, (2.26) y = r sin(θ + φ) = r cos θ sin φ + r sin θ cos φ. (2.27) Note x = r cos θ and y = r sin θ, we have x = x cos φ y sin φ, (2.28) y = x sin φ + y cos φ. (2.29) We ll derive similarly the transformation equation in the Minkowski space. In the Minkowski space, a rotation in tx needs to conserve s 2 = c 2 t 2 x 2, (2.30)

25 foundations of theory of relativity 25 with all y = y and z = z. Since we will derive the transformation from K to K system, we start from The constant of s 2 describes a hyperbola. s 2 = c 2 t 2 x 2. (2.31) Y cosh a x² y² = 1 a/2 sinh a X From the definition of hyperbolic functions 5, if we define a 5 cosh x = (e x + e x )/2, sinh x = parameter θ and let (e x e x )/2. ct = s cosh θ, (2.32) x = s sinh θ, (2.33) then c 2 t 2 x 2 = s 2 is always satisfied. In the rotated system K, the angle parameter becomes θ + φ, ct = s cosh(θ + φ), (2.34) x = s sinh(θ + φ). (2.35) It can be easily seen that c 2 t 2 x 2 = s 2. Noting ct = s cosh θ and x = s sinh θ, we have ct = s cosh(θ + φ) = s cosh θ cosh φ + s sinh θ sinh φ (2.36) = ct cosh φ + x sinh φ, (2.37) x = s sinh(θ + φ) = s sinh θ cosh φ + s cosh θ sinh φ (2.38) = x cosh φ + ct sinh φ. (2.39) We rewrite the equations as x = x cosh φ + ct sinh φ, (2.40) ct = x sinh φ + ct cosh φ. (2.41) These are the needed transformation equations for rotation in 4D space. Let s see what the rotation in xt plane corresponds to. Starting from the origin of the K system, we have r = 0. The rotation

26 26 classical electrodynamics equations are then x = ct sinh φ, (2.42) ct = ct cosh φ. (2.43) We thus have V/c = tanh φ with V = x/ t, the speed of the origin of the K system in K. Because y = y = 0, z = z = 0, V y = V z = 0. In short, the rotation in xt plane corresponds to the coordinate transformation for K moving relative to K with V = Ve x. Now from V/c = tanh φ = (e φ e φ )/(e φ + e φ ), we have V + c e φ = V c. (2.44) Therefore sinh φ = eφ e φ 2 cosh φ = eφ + e φ 2 = = V/c βγ 1 V 2 /c2 (2.45) 1 1 V 2 /c γ 2 (2.46) Substituting the expressions of sinh φ and cosh φ into rotation equations, we have, x = x + Vt 1 V 2 /c 2, t = t + (V/c 2 )x 1 V 2 /c 2. y = y, z = z. These equations are the needed transformations; they are called Lorentz transformations. The Lorentz transformation corresponds to the rotation in the 4D Minkowski space involving t-axis. The inverse transformation can be obtained by noting that K moves relative to K in V. Therefore we have x = x Vt 1 V 2 /c 2, (2.47) t = t (V/c2 )x 1 V 2 /c 2. (2.48) This inverse Lorentz transformation can also be directly obtained by solving for (t, r ) from previous equations. If V/c 1 or equivalently c, (1 V 2 /c 2 ) 1/2 1. The Lorentz transformation is reduced to the Galileo transformation, x = x + Vt, (2.49) t = t. (2.50) This means that classical mechanics work well because c V, where V means the characteristic speed of our daily life. From the Lorentz transformation, we derive the proper length, which is the length of an object in an inertial reference frame in

27 foundations of theory of relativity 27 which it is at rest. We denote proper length by l 0. Suppose we have a rod parallel to the x -axis; it s at rest in K system. In K system, the length of the rod is l 0 = x 1 x 2. In K system, to measure the length of the rod, we have x 1 at t 1 and x 2 at t 2, and t 1 = t 2. From the inverse Lorentz transformation, we have, x 1 = x 1 Vt 1 1 V 2 /c, (2.51) 2 x 2 = x 2 Vt 2 1 V 2 /c. (2.52) 2 Noting t 1 = t 2, we have x 1 x 2 = x 1 x 2 1 V 2 /c. (2.53) 2 Or l 0 = l/ 1 V 2 /c 2 or l = l 0 1 V 2 /c 2. Since l < l 0, this is called Lorentz contraction. Similarly, we can introduce proper volume, which is the volume of an object in an IRF in which it is at rest. Since transverse dimensions do not change, we can immediately have V = V 0 1 V 2 /c 2, (2.54) where V 0 is the proper volume of the object. Using the Lorentz transformation, we can re-derive the equation for proper time. Suppose a clock to be at rest in K, two events occurred at r at t 1 and t 2. Then τ = t 2 t 1 and in K, t 1 = t 1 + (V/c2 )x 1 V 2 /c 2, (2.55) t 2 = t 2 + (V/c2 )x 1 V 2 /c 2. (2.56) Therefore, t = τ/ 1 V 2 /c 2, the same as we obtained before. 2.5 Transformation of velocities To completely determine the state of a particle, we need both its space coordinates and velocity. From the Lorentz transformation, we can easily derive the needed formulas for the transformation of velocities. Suppose a particle has a velocity v in IRF K and v in IRF K ; K moves with V = Ve x relative to K. The differentiation of the Lorentz transformation gives dx = (dx + Vdt )/ 1 V 2 /c 2, (2.57) dy = dy, (2.58) dz = dz, (2.59) dt = (dt + V/c 2 dx )/ 1 V 2 /c 2. (2.60) The velocity transformation can be obtained easily as v = dr/dt,

28 28 classical electrodynamics v = dr /dt, etc. The resulting velocity transformations are v x = dx + Vdt v dt + (V/c 2 )dx = x + V 1 + (V/c 2 )v (2.61) x v y v y = 1 + (V/c 2 )v 1 V2 x c 2 (2.62) v v z = z 1 + (V/c 2 )v 1 V2 x c 2 (2.63) In case of c, the above equations recovers the familiar Galileo velocity transformation. 2.6 Vectors and Tensors in Spacetime Minkowski once said: Space of itself, and time of itself will sink into mere shadows, and only a kind of union between them shall survive. Indeed, in the framework of the special theory of relativity, time and space are put on equal footing: both time and space are not absolute and are transformed together from one reference system to another reference system Four-vectors It s convenient to discuss time and space together in terms of vectors in a pseudo-euclidean Minkowski space. For example, the coordinates of an event (ct, x, y, z) can be considered as the components of a radius four-vector, x 0 = ct, x 1 = x, x 2 = y, x 3 = z. (2.64) The radius four-vector satisfies the Lorentz transformation when changing reference system from K to K. Note that with this representation, the Lorentz transformation can be written as (easier to remember) x 0 = γ(x 0 + βx 1 ), (2.65) x 1 = γ(x 1 + βx 0 ), (2.66) where γ = 1/ 1 V 2 /c 2, and β = V/c. Like the case in 3D, we model four-vectors after the radius fourvector. Definition: any set of four quantities A 0, A 1, A 2, A 3, which transform like the radius four-vector x α under the change from K to K is called a four-vector. This definition essentially means that when changing from K to K, all four-vectors are transformed using Lorentz transformation. Therefore for K with axes parallel to K and moves relative to K in

29 foundations of theory of relativity 29 V = Ve x, A 0 = A 0 + (V/c)A 1 1 (V/c) 2, (2.67) A 1 = A 1 + (V/c)A 0 1 (V/c) 2, (2.68) and A 2 = A 2 and A 3 = A 3. The component A 0 is called the time component, and A 1, A 2, A 3 the space components. We sometimes write the four-vector as A α = (A 0, A), (2.69) where A is a three-dimensional vector. In this course, we use Greek letters (α, β, ) to represent 0, 1, 2, 3, and use Latin letters (i, j, ) to represent 1, 2, 3. Also we use prefix four- to specifically denote variables in the Minkowski space. Ordinary three-dimensional variables will just be called scalars, vectors, or tensors. The square magnitude of the radius four-vector x α is defined as (x 0 ) 2 (x 1 ) 2 (x 2 ) 2 (x 3 ) 2 = s 2, (2.70) which is the square of s, the spacetime interval (the magnitude of the radius four-vector in Minkowski space). Similarly, the square magnitude of an arbitrary four-vector A α is (A 0 ) 2 (A 1 ) 2 (A 2 ) 2 (A 3 ) 2. (2.71) To more conveniently represent the dot-product between fourvectors, we introduce two kinds of components of four-vectors: 1. The contravariant ( 逆变 ) components, denoted by A α. 2. The covariant ( 协变 ) components, denoted by A α. so that A α and A α are related by A 0 = A 0, A 1 = A 1, A 2 = A 2, A 3 = A 3. (2.72) By using contravariant and covariant components, the square magnitude of a four-vector A α can now be compactly written as 3 A α A α = A 0 A 0 + A 1 A 1 + A 2 A 2 + A 3 A 3 A α A α. (2.73) α=0 The dot-product between two four-vectors A α and B α is A α B α or A α B α. In these expressions, we have used the Einstein summation rule. Another example, the square magnitude of the radius fourvector, the interval, can be written as s 2 = x α x α. Note the locations of indices in these expressions. There are simple ways to memorize the location of sub/super scripts: 1. 中文 : 上逆下协

30 30 classical electrodynamics 2. English: There is a r in both contra- (contra-variant) and super- (super-script). To convert between the contravariant and the covariant components, one needs to use the metric coefficients g αβ. From the definitions of contra- and co-variant components above, you should be able to see that for the four-vectors in Minkowski space, (g αβ ) = (g αβ ) = (2.74) Therefore we can convert between A α and A α by A α = g αβ A β or A α = g αβ A β. (2.75) The metric tensor g plays an very important role in the general theory of relativity (not covered by this course). In your textbook, the authors represented the Minkowski space in a different way. Let x 0 = ct, x 1 = ix, x 2 = iy, and x 3 = iz, then s 2 = (x 0 ) 2 + (x 1 ) 2 + (x 2 ) 2 + (x 3 ) 2. (2.76) There are two reasons why I wouldn t use this in my class: 1. Hard to understand why one would need the contra- and covariant components of a vector. 2. Hard to generalize this to the General Theory of Relativity (the metric g, curved or flat spacetime). The concept of contra- and co-variant components of a vector is important in using curvilinear coordinates. Curvilinear coordinates are very convenient in certain cases (like fusion and space plasmas, general relativity, etc). For more information about curvilinear coordinates, see my reference list near the end of the previous Chapter. Here are a few rules regarding contra- and co-variant components of four-vectors. 1. The dot product between two vectors is A α B α or A α B α. 2. The squared length of A α is A α A α or A α A α. 3. You can convert between A α and A α by A α = g αβ A β and A α = g αβ A β. Regarding Einstein notation of four-vectors and four-tensors, here are two simple rules that you should keep in mind. 1. For a dummy index, one index must be superscript and one must be subscript; e.g., A α B α. Terms like A α B α do not make sense, as you can see from the definitions of contra and covariant components of a four-vector given above. 2. Equations do not change the position of a free index, like A α = g αβ A β. Note the location of the free index α.

31 foundations of theory of relativity Four-scalars Definition: Like 3D case, we define four-scalars as invariants under the transformation of coordinates, here the Lorentz transformation. Because they are invariant under Lorentz transformations, the four-scalars play important roles in the study of special relativity and electrodynamics. You should learn how to construct fourscalars from four-vectors and four-tensors. For example, like the 3D case, the dot product between two four-vectors is a four-scalar. In the Minkowski space, the dot product between two four-vectors A α and B α is A α B α or A α B α, and it is a four-scalar. Note that there is no free index in this term. Example: the interval s 2 = x α x α is a four-scalar. We can construct more invariants after we learn more four-vectors. 2.7 Four-tensors Definition: A second-order four-tensor is a set of 16 quantities F αβ, which under coordinate transformations, transform like the products of components of two four-vectors. A four-tensor can be written in different forms: F αβ, F αβ, F α β, and F β α. Different positions of indices represent different kind of components of a four-tensor. In case of a four-vector, there are two different kinds of components (co- and contra-variant components); in the case of a four-tensor, there are four different kinds of components. Again one use the metric tensor g to convert between different kinds of representations of a four-tensor. One just needs to remember the rules of Einstein notation, and you can immediately figure out that, F µν = g µα F αβ g βν (2.77) F β µ = g µα F αβ (2.78) (2.79) Noting that in the special relativity, 1 if α = β = 0 g αβ = 1 if α = β = 0 0 otherwise (2.80) From F µν = g µα F αβ g βν, we have the following simple rules: 1. raising or lowering index (1, 2, 3) changes the sign, 2. raising or lowering index (0) does not change the sign.

32 32 classical electrodynamics Example: F 00 = F 00, F 01 = F 01, F 11 = F 11, F 0 0 = F 00, F 1 0 = F 01, F 1 0 = F 01, F 0 1 = F10, From four-vectors A α and B α, we know A α A α and A α B α are fourscalars; i.e., they are invariants under Lorentz transformations. The trick to construct 4-scalars from 4-vectors/tensors is that all indices are dummy indices; there is no free index in the final expression. Similarly, we can also construct four-scalars from 4-tensors. For example, for F αβ, we know that F αβ F βα = inv. (2.81) Also it s possible to construct four-scalars from four-vectors and four-tensors, like A α B β F βα is a four-scalar, since there is no free index in the final expression. This operation is called contraction Basic four-vector/tensor differential calculus. Let s now introduce four-vector calculus. The four-gradient of a scalar φ is the four-vector ( ) φ 1 x α = φ c t, φ From the location of the index, we know they are the covariant components of a four-vector. To make this more obvious, some people write the four-gradient of φ as, φ x α αφ. (2.82) Similarly, you might have the contra-variant components of a fourvector, φ x α α φ. The 4-divergence of a 4-vector A α = (A 0, A) is a four-scalar, A α x α = αa α = α A α = 1 c A 0 t It s a scalar under coordinate transformation. + A. (2.83) 2.8 Four-velocity and four-acceleration Using the knowledge of four vectors, we can construct the fourvelocity by u α = dx α /dτ, where dτ = dt/γ, u 0 = cdt γ = γc, dt (2.84) u 1 = dx dt γ = γv x, (2.85) u 2 = γv y, (2.86) u 3 = γv z. (2.87)

33 foundations of theory of relativity 33 Therefore, the four-velocity is u α = (γc, γv x, γv y, γv z ) = (γc, γv). (2.88) The contraction of the four-velocity is, noting that dτ = ds/c, or it can be calculated directly by u α u α = dxα dx α dτdτ = ds2 ds 2 /c 2 = c2, (2.89) u α u α = γ 2 (c 2 v 2 ) = c 2, (2.90) which is a scalar and an invariant. One can construct the four-acceleration similarly by w α = du α /dτ, Without calculating the components of the four-acceleration, one can show that u α w α = 0. (2.91) This shows that the four-acceleration is always perpendicular to the four-velocity.

34 3 Relativistic Dynamics 3.1 A brief review of Lagrangian mechanics We ll use some Lagrangian mechanics in this class, so let me first briefly give it a review, using materials from Landau s book Mechanics. In classical mechanics, the action is defined by the integral S = t2 t 1 L(q, q, t)dt. (3.1) Suppose q(t 1 ) = q (1) and q(t 2 ) = q (2), there could be infinite number of paths connecting t 1 and t 2. It is required that the actual path minimizes S; i.e., δs = 0. (3.2) This is called the principle of least action, with L the Lagrangian. To obtain equations of motion from the principle of least action, note that if q = q(t) is the actual path, then for a given variation of q, the change in S is δs = t2 t 1 The constraints we have are q(t) + δq(t), (3.3) t2 L(q + δq, q + δ q, t)dt L(q, q, t)dt = 0. (3.4) t 1 δq(t 1 ) = 0, δq(t 2 ) = 0, (3.5) since q(t 1 ) = q (1) and q(t 2 ) = q (2) are given. Keeping only the leading (first-order) terms in δs, we have t2 ( ) L L δq + q q δ q dt = 0. (3.6) t 1 Noting that δ q is, because of the variation in q, δ q = d dt (q + δq) d dt (q) = d δq. (3.7) dt

35 relativistic dynamics 35 Therefore, δs = [ ] L t 2 t2 ( L q δq + t 1 t 1 q d ) L δqdt = 0. (3.8) dt q Because δq(t 1 ) = 0 and δq(t 2 ) = 0, we have t2 ( L δs = q d ) L δqdt = 0, (3.9) dt q Since δs must vanish for all δq, we have t 1 L q d L = 0, (3.10) dt q these equations are called Lagrange s equations. If we know L of a system, then we can write down its equations of motion. Note that the Lagrangian is not uniquely determined for a given system. For if we have two Lagrangians L (q, q, t) and L(q, q, t), and L = L + d dt f (q, t), (3.11) then S = t2 t 1 L dt = t2 t 1 Ldt + t2 t 1 d f dt dt = S + f (q(2), t) f (q (1), t). (3.12) Therefore δs and δs would lead to the same equations of motion. Now we apply the concept to solving a free body problem in classical mechanics. To consider mechanical problems, it s necessary to choose a reference frame. In principle, we can choose any reference system we like. However if we choose one where time or space is inhomogeneous, then the description of nature laws would become unnecessarily difficult. The simplest one is thus the so-called inertial reference frame so that space is homogeneous and isotropic,and time is homogeneous. In such a frame, a free body would move with a constant velocity or is at rest forever. Let s find the the Lagrangian for a free particle in an inertial reference system. Note that time is homogeneous so L does not depend on t. Space is homogeneous so L does not depend on r. Therefore, L = L(v). Space is isotropic, therefore L must not depend on the direction of v. Conclusion: L = L( v ) or L = L(v 2 ). Applying the Lagrange s equation (3.10), noting that L/ v must be a function of v only, the equation of motion is v = 0. (3.13) This equation states that a free body in an inertial reference frame moves with a constant velocity. Of course, this velocity can equal 0, which means the free body is at rest all the time. This is the laws of inertia. To find the form of L(v 2 ), we need to use Galileo s principle of relativity. Consider two IRF s K and K, K moves in ɛ relative to K. v = v + ɛ. (3.14)

36 36 classical electrodynamics Because of Galileo s PR, the leading terms of the L are L = L(v 2 ) = L(v 2 + 2v ɛ + ɛ 2 ), (3.15) L(v 2 ) = L(v 2 ) + L 2v ɛ (3.16) v2 We know from Galileo s PR that L and L should lead to the same equations of motion, therefore, L = L + d dt f (r, t), (3.17) where f (q, t) is a function of r and t, Hence Because v = dr/dt, we must have where α is some constant. From we have for a free particle L v 2 2v ɛ = d f (r, t). (3.18) dt L = α, (3.19) v2 L = α, (3.20) v2 L = αv 2. (3.21) Normally we write α = m/2, where m is called mass, and L = 1 2 mv2. (3.22) From previous derivations, we have learned that either ds is an invariant under Galileo transformation, or ds and ds differ by a d f (r, t). We will use this property of action in relativistic dynamics. Of course, there the action is an invariant under the Lorentz transformation (a four-scalar) The Lagrangian for a particle moving in a given external field can be obtained by adding to L = mv 2 /2 a quantity describing the interaction between the field and the particle, so that the equations of motion become L = 1 2 mv2 U(r, t), (3.23) m dv dt = U. (3.24) Note that now the total action S becomes t2 ( ) 1 S = S p + S f = 2 mv2 U(r, t) dt. (3.25) t 1

37 relativistic dynamics 37 Using Lagrangian mechanics, it is very convenient to discuss conservation properties. Here we discuss the conserved quantities from the isotropy and homogeneity of space and time. If time is homogeneous, then for a closed system L = L(q i, q i ). The total time derivative of L is dl dt = q L i + dq ( ) i L d L = q q i dt q i + dq i L (3.26) i dt q i dt q i = d ( ) L q dt i (3.27) q i d ( ) L q dt i L = 0 (3.28) q i We define total energy E of the system by E q i L q i L. (3.29) Therefore, if time is homogeneous, E is a constant. Equation (3.29) defines the energy of the system. If, however, we consider a free particle, then it is the definition of energy for the particle. We will use this to define particle energy in relativistic dynamics. If space is homogeneous; i.e., if we replace r i by r i + ɛ, then Therefore δl = i From Lagrange s equation, i L L δr r i = ɛ i = 0. (3.30) r i i i d L = d dt v i dt i If space is homogeneous, for a closed system, L r i = 0. (3.31) P i L v i = 0. (3.32) L v i (3.33) remains constant. This is called the momentum of the system Similarly, we will use this to define the momentum of a free particle in relativistic dynamics. 3.2 Relativistic action for a free particle I ll now apply what I ve introduced to obtain the relativistic action for a free particle, S = The relativistic action must satisfy b a ds. (3.34)

38 38 classical electrodynamics 1. ds is invariant under Lorentz transformation, or differed by a d f (x α ) when transformed from K to K. 2. It should recover non-relativistic dynamics if c (v/c 0). From the previous chapter, we know that for a free particle, the event interval ds is an invariant. Try S = b a αds, (3.35) where α is a 4-scalar. Let s see whether S can recover non-relativistic dynamics if c. If it can, then it suits our need and can be used as the relativistic action for a free particle. To recover the non-relativistic dynamics, we first write S in the normal 3D form; i.e., S = Ldt form. Because ds = cdτ = c 1 v 2 /c 2 dt, we write b b S = αds = αc 1 v2 dt. (3.36) c2 From this form of S, we have the Lagrangian In case of c, we have a L = αc a 1 v2 c 2. (3.37) L = αc αv2 2c. (3.38) Therefore, if α = mc, then S satisfies the two constraints. The relativistic action for a free particle is then And the corresponding Lagrangian is b S = mc ds. (3.39) a L = mc 2 or L = mc 2 /γ, with γ = 1/ 1 v 2 /c 2. 1 v2 c 2, (3.40) 3.3 Energy and momentum The momentum of a particle is given by Using L = mc 2 1 v 2 /c 2, we have p = mc p = L v. (3.41) ) 1 (1 v2 2 ( 1c ) c 2 2 2v,

39 relativistic dynamics 39 or p = mv 1 v 2 /c 2 = γmv. If c, p = mv, the non-relativistic (NR) momentum. The energy of a particle is defined from L by Using p = γmv, we have E = v L L = p v L. (3.42) v E = mc 2 1 v 2 /c 2 = γmc2 (3.43) If v 0, E mc 2, the rest energy of the particle. In case of c, E = mc mv2. This definition of energy leads to that mass is not a conserved quantity anymore in relativistic mechanics. If we have a body, with mass m, consisting of n particles. In classical mechanics, we have m = i m i. In relativistic mechanics, however, the energy of a body at least contains 1. the rest energies of its constituent particles i m i c 2 2. the kinetic energy of particles 3. the interaction energy 4. therefore, mc 2 = i m i c 2 and m = i m i. Only the law of conservation of energy holds. The energy and momentum are closely related. Note that the 4-velocity we introduced is u α = (γc, γv). Multiplying u α by m, a 4-scalar, we have the four-momentum vector ( ) E p α = mu α = (γmc, γmv) = c, p. (3.44) Since p α is a four-vector, we can apply the knowledge we know about four-vector. First, we can immediately obtain their transformation equation, since any four-vector transforms like the radius four-vector x α ; i.e., E = E + Vp x 1 V 2 /c 2, (3.45) p x = p x + (V/c 2 )E 1 V 2 /c 2, (3.46) p y = p y, (3.47) p z = p z. (3.48) Here (p x, p y, p z ) = p. Also the dot product of a four-vector with itself is a Lorentz invariant. Since four-momentum p α = mu α, we immediately have p α p α = m 2 c 2. (3.49)

40 40 classical electrodynamics Substituting p α = (E/c, p) and p α = (E/c, p), we have or E 2 c 2 p2 = m 2 c 2, (3.50) E 2 = p 2 c 2 + m 2 c 4, and p = E v. (3.51) c2 this is the energy-momentum relation. If v c, both p and E become infinite unless m 0. Actually, for a photon (m = 0), we have p = E c. (3.52) This equation is also useful for the case E E 0, then p E/c. 3.4 The Covariant Equation of Motion We can derive the equations of motion in covariant form by viewing S as a function in the Minkowski space, S = mc b a ds = mc b Then if we vary the actual path x α by δx α, b δs = mc δ ds = mc δ a from the principle of least action. Let s first calculate δ dx α dx α. a b a dxα dx α. (3.53) dxα dx α = 0, (3.54) δ dx α dx α = 1 2ds δ(dx αdx α ) = 1 2ds (δdx αdx α + dx α δdx α ) (3.55) Note that we can switch the upper- and lower- indices, δdx α dx α = δdx α dx α. (3.56) Therefore, δ dx α dx α dx α /ds = u α /c, and = dx α δdx α /ds. Noting that ds = cdτ, δ dx α dx α = u α δdx α /c = u α dδx α /c. (3.57) With δ dx α dx α = u α dδx α /c, we have b b δs = m u α dδx α = mu α δx α b a + m a a δx α du α ds. (3.58) ds If we want to find the actual path, then we fix x α (a) and x α (b), or the four-velocity u α is constant. δs = 0 duα ds = 0 (3.59)

41 4 Charges in a Given Electromagnetic Field 4.1 Four-potentials of a field As in classical mechanics, the total action for a particle in a given field is S = S p + S p f (4.1) Here S p = mc ds is the action for the particle, and S p f is the action characterizing the interaction. Experiments suggest that the interaction is determined by the charge q, and the four-vector potential A α characterizing the field. To make the integral S p f a Lorentz invariant, we can construct b S p f = q A α dx α, (4.2) a where A α = (φ/c, A) or A α = (φ/c, A). Therefore the total action is In 3D coordinates, and S = S = b a S p f = t2 The total Lagrangian is L = t 1 = ( mcds qa α dx α ). (4.3) b a b a ( mc2 γ (q A dr qφdt) (4.4) (q A v qφ) dt, (4.5) ) + q A v qφ dt. (4.6) mc2 + q(a v φ). (4.7) γ }{{}}{{} interaction with the field free particle 4.2 Equations of motion of a charge in a field We derive here the equations of motion of a charge. For simplicity, we assume that the charge is small, so its effects on the field can be

42 42 classical electrodynamics neglected. The field potential do not depend on the position and velocity of the charge. The Lagrange s equation is d dt From the total Lagrangian, Noting that L r ( ) L L v r = 0 (4.8) = L = q (A v) q φ. (4.9) (A v) = (A )v +(v )A + v ( A) + A ( v). }{{}}{{} =0 =0 (4.10) On the other hand, L = p + qa. (4.11) v Substituting Equations (4.9)-(4.11) into the Lagrange s equation (4.8) gives d (p + qa) = q(v )A + qv ( A) q φ (4.12) dt However, q d dt A = q [ A t ] + (v )A. (4.13) Correspondingly, dp dt A = q t q φ }{{} +qv ( A). (4.14) no dependence on v The right hand side of the equation can be divided into two parts; one depends on velocity v and the other not. If we define the electric field intensity to be E = A t Similarly, we define the magnetic field intensity to be φ. (4.15) B = A. (4.16) Then the equations of motion of a charge can be simply written as dp dt = qe + qv B. (4.17) This is the Lorentz force felt by a particle in a given field.

43 charges in a given electromagnetic field Gauge invariance Note that we can describe the field in two ways. We can use A α as in the action S p f. Or as in the equations of motion, we can use only E and B, dp dt = qe + qv B. (4.18) However, a given E and B do not allow the unique determination of A and φ. For example, it s easy to show that the transformation A = A + f (4.19) φ = φ f t (4.20) do not change E and B. Therefore all equations must be invariant under the transformation of the potentials; this invariance is called gauge invariance ( 规范不变性 ). This is the first (and maybe the last) gauge invariance you have ever seen and it is a very important property of fields in more advanced field theories describing other types of interactions. The gauge invariance can also be clearly seen from the action, If we replace where f is a function of t and r, then b S p f = q A α dx α, (4.21) a A α A α f x α, (4.22) b S p f = q = q a b a b A α dx α + q A α dx α + q a b a f x α dxα (4.23) d f. (4.24) The total differential d f has no effect on the equations of motion. The gauge invariance allows one to choose one 1 auxiliary condition for A and φ. Two popular gauges are Coulomb gauge and Lorenz gauge. One is 1 This can be seen from that a scalar function is involved in Equation (4.22). A = 0, the Coulomb gauge (4.25) This gauge is popular in some problems. The Lorenz gauge is very popular in theoretical analysis. In the Lorenz gauge, we require or in 4D form A + 1 c 2 φ t = 0. the Lorenz gauge (4.26) A α = 0. (4.27) xα Note that if A and φ satisfy the Lorenz gauge in one inertial reference frame, then they satisfy the Lorenz gauge in all other inertial reference frames.

44 44 classical electrodynamics 4.4 Static electromagnetic fields If potentials do not have time dependence, then B = A, (4.28) E = φ. (4.29) Therefore, φ determines E, while A determines B for static electromagnetic fields. In case of static fields, we can add to φ an arbitrary constant. Usually, we choose φ( ) = 0. Also it is obvious that E and B fields are completely decoupled in case of static fields. For a closed system in a static EM field, we have L t = 0. (4.30) From Mechanics, we know that the energy is conserved, E = v L v L = γmc2 + qφ = const. (4.31) }{{}}{{} kinetic potential If the field is constant and uniform, then simple forms of A and φ can be obtained. From E = φ, From B = A, we can choose A to be The choices of φ and A can be easily proved by φ = E r. (4.32) A = 1 B r. (4.33) 2 φ = (E r) = E r = E I = E, (4.34) A = 1 2 (B r) = 1 [B r (B )r] = B. (4.35) 2 Note that we have used the fact that E and B are uniform in the proof. 4.5 Motion in uniform static EM fields First, let s consider the motion in a uniform static E field. The equations of motion are dp dt Assuming E = Ee x, p x (t = 0) = 0, we have = qe. (4.36) dp x = qe p x = qet dt (4.37) dp y = 0 p y = const dt (4.38) dp z = 0 p z = const dt (4.39)

45 charges in a given electromagnetic field 45 Assuming p y = p 0, and p z = 0, then p x = qet, p y = p 0. The kinetic energy 2 of the particle is 2 Here kinetic energy means γmc 2, where E k = m 2 c 4 + p 2 c 2 = E0 2 + (qect)2, (4.40) E 0 = m 2 c 4 + p 2 yc 2 + p 2 zc 2 = m 2 c 4 + p 2 0 c2, (4.41) is the energy at t = 0. To find the trajectory of the particle, we need to solve dr dt = v = The components of the equation are dx dt = p γm = pc2 E k. (4.42) qec 2 t E (qect)2, (4.43) dy dt = p 0 c E 2. (4.44) (qect)2 as opposed to the potential energy qφ. See Equation (4.31) Note that as t, v x c and v y 0. Integrate dx/dt and dy/dt, assuming r(t = 0) = 0, x = 1 qe y = p 0c qe sinh 1 E0 2 + (qect)2, (4.45) ( ) qect, (4.46) From y = y(t), we obtain t = t(y). From x = x(t), we have E 0 x = E 0 qey cosh qe p 0 c. (4.47) Now let s consider the motion of a particle in a uniform static B field, assuming B = Be z. The equations of motion are dp dt Using p = γmv that γ is constant in a B field, = qv B. (4.48) dv dt = qv B = v Ω, (4.49) γm where Ω = qb/γm = Ωe z. The components of dv/dt are v x = Ωv y, v y = Ωv x, v z = 0, (4.50) from which v z = v z0. To solve for v, let v = v x + iv y, then from v x = Ωv y, (4.51) i v y = Ωiv x, (4.52)

46 46 classical electrodynamics we have v = iω v, from which v = v 0 e iωt. Here v 0 = v 0 e iα is a complex constant with α a constant. From v = v 0 e i(ωt+α) and v = v x + iv y, we have v x = v 0 cos(ωt + α), (4.53) v y = v 0 sin(ωt + α), (4.54) It s clear that v = v 2 x + v 2 y = const = v 0. We see from v x and v y that Ω is the angular frequency. The trajectory can be obtain by integrating r = dv/dt, which gives x = x 0 + ρ sin(ωt + α), (4.55) y = y 0 + ρ cos(ωt + α), (4.56) z = z 0 + v z0 t. (4.57) Here ρ = v /Ω is the gyro-radius (or ρ ). In case of v/c 0, γ 1, Ω qb/m. To make life more complicated, let s now consider the motion of a charge in a static uniform E and B field. We consider only nonrelativistic case here (v c). Assume B = Be z, E = E y e y + E z e z. For a non-relativistic particle, p = mv. The components of the equation are m v = qe + qv B. (4.58) m v x = qv y B, (4.59) m v y = qe y qv x B, (4.60) m v z = qe z, (4.61) from which we immediately have v z = qe z t/m + v z0. To solve for v, we again let v = v x + iv y. Then The general solution of the equation is Or letting v 0 = v 0 e iα, d v dt + iω v = i qe y m. (4.62) v = v 0 e iωt + qe y mω. (4.63) v x = v 0 cos(ωt + α) + qe y mω = v 0 cos(ωt + α) + E y B, (4.64) v y = v 0 sin(ωt + α). (4.65) The x-velocity consists of two parts, v x = v 0 cos(ωt + α) + }{{} E y /B }{{} = ṽ x + v x (4.66) oscillatory non-oscillatory

47 charges in a given electromagnetic field For cases like this, it s frequently useful to average over the fast oscillation to obtain the average motion by vx = Z t+ T 1 T t v x (t0 )dt0 = 1 T Z t+ T t v 0 cos(ωt0 + α)dt0 + Ey /B, (4.67) or v x = Ey /B. Note that vy = 0. The motion in x-y plane is the very famous E B drift. In vector form, the drift velocity is E B/B v vx vy time Fig: v x and vy from Equations (4.64) and (4.65). To find the trajectory, we integrate v x, vy and vz and x = ρ sin(ωt + α) + v E t + x0 (4.68) y = ρ[cos(ωt + α) cos α] + y0 (4.69) z= qez 2 t + vz0 t + z0 2m (4.70) Here v E = Ey /B is the E B drift velocity. 1.0 v =0.5,vE =0.1,y0 =0.5,x0 =0.0,T =1 0.8 y x Fig: Trajectories in the x-y plane for different v E and v = ρω. 47

48 48 classical electrodynamics 1.0 v =0.1,vE =0.5,y0 =0.5,x0 =0.0,T =1 0.8 y x Fig: Trajectories in the x-y plane for different v E and v = ρω. 1.0 v =0.5,vE =0.5,y0 =0.5,x0 =0.0,T =1 0.8 y x Fig: Trajectories in the x-y plane for different v E and v = ρω. 4.6 The electromagnetic field tensor To obtain the Lorentz equation of motion in covariant form, we consider the action in four-dimensional form. Recall the action of a particle in a given field is S= Z b a ( mcds qaα dx α ) (4.71) The principle of least action leads to δs = δ Z b a ( mcds qaα dx α ) = 0. (4.72) Recall that δds = dxα dδx α /ds = uα dδx α /c, δs = Z b a (muα dδx α + qaα dδx α + qδaα dx α ) = 0. (4.73)

49 charges in a given electromagnetic field 49 Integrating by parts, δs becomes δs = b Because δx α (a) = δx α (b) = 0, δs = a (mδx α du α + qδx α da α qδa α dx α ) [mu α + qa α ] δx α b a. (4.74) b a (mδx α du α + qδx α da α qδa α dx α ). (4.75) Noting da α = ( A α / x β )dx β and δa α = ( A α / x β )δx β, b ( δs = mdu α δx α + q A α a x β dxβ δx α q A ) α x β δxβ dx α (4.76) b ( = mdu α δx α + q A α a x β dxβ δx α q A ) β x α δxα dx β (4.77) b [ = m du ( α a dτ Aβ q x α A ) ] α x β u β dτδx α. (4.78) If we define a four-tensor F αβ by then δs = b a F αβ = A β x α A α x β, (4.79) [ m du ] α dτ qf αβu β dτδx α. (4.80) The tensor F αβ is called the electromagnetic field tensor; the tensor is anti-symmetric, F αβ = F βα. (4.81) To obtain the equations of motion, we apply the principle of least action. From δs = 0, We can also move indices up or down, m du α dτ = qf αβu β. (4.82) m duα dτ = qfαβ u β. (4.83) Equations (4.82) or (4.83) are the covariant form of the equations of motion of a charge in a given field. Now let s find out the components of the electromagnetic field tensor F αβ ; they can be obtained by noting that A α = (φ/c, A), In SI units, e.g., F 10 = A 0 x 1 A 1 x 0 = 1 c φ x + 1 A x c t = E x /c. (4.84) 0 E x /c E y /c E z /c E F αβ = x /c 0 B z B y E y /c B z 0 B x. (4.85) E z /c B y B x 0

50 50 classical electrodynamics From F αβ, we can immediately obtain using the rule 0 E x /c E y /c E z /c F αβ E = x /c 0 B z B y E y /c B z 0 B x, (4.86) E z /c B y B x 0 1. Moving the time index (0) No change in sign. 2. Moving space indices (1, 2, 3) change sign. Using F αβ, we obtain components of the covariant equations of motion, Equation (4.83), rewritten here du α /dτ = q m Fαβ u β. (4.87) For example, the time component (α = 0) is ( m du0 dτ = qf0β u β = q E ) ( γv). (4.88) c Noting that u 0 = γc and dτ = dt/γ, we have γm dγc dt or de k dt = γ q E v, (4.89) c = qe v, with E = γmc 2 (4.90) The space components of du α /dt can be obtained similarly; they are just the Lorentz equations of motion (Equation (4.17)). 4.7 Lorentz transformation of the EM fields The Lorentz transform of potentials φ and A can be obtained easily, because A α = (φ/c, A) is a four-vector. Any 4-vector transforms like x α ; therefore φ = φ + VA x 1 V 2 /c 2, (4.91) A x = A x + (V/c 2 )φ 1 V 2 /c 2, (4.92) A y = A y, (4.93) A z = A z. (4.94) Of course, this is for K move with V = Ve x in K. The transformation of E and B can be obtained from the transformation of components of F αβ. For the E field, E x = E x, E y = E y + VB z 1 V 2 /c 2 E z = E z VB y 1 V 2 /c 2. (4.95) For the B field, B x = B x, B y = B y (V/c 2 )E z 1 V 2 /c 2 B z = B z + (V/c 2 )E y 1 V 2 /c. (4.96) 2

51 charges in a given electromagnetic field 51 The non-relativistic case V/c 1 is very important, E x = E x, E y = E y + VB z, E z = E z VB y. (4.97) B x = B x, B y = B y (V/c 2 )E z, B z = B z + (V/c 2 )E y. (4.98) Or in vector form, E = E + B V, B = B 1 c 2 E V. or E = E + cb β, B = B (E /c) β, (4.99) (4.100) with β = V/c. Note that if B = 0 in K system, then in K system, If E = 0 in K system, then in the K system, B = β (E/c). (4.101) E = cb β. (4.102) In both cases, E B in K. This point can be obtained more clearly by considering the invariants of E and B. From the field tensor F αβ, we can obtain two important and useful invariants by constructing four-scalars. First, we can easily see that F αβ F βα = inv. Because F αβ is anti-symmetric, normally we write it as which leads to F αβ F αβ = inv, (4.103) c 2 B 2 E 2 = inv. (4.104) To construct the second invariant, we first introduce e αβµν the completely anti-symmetric unit tensor of fourth-order. The tensor e αβµν has the following properties, 1. e αβµν changes sign if interchanging any pair of indices 2. e αβµν = 0 if any two indices are the same 3. normally we choose e 0123 = 1. You can see that e αβµν is the 4D version of Levi-Civita symbol ɛ ijk. An extra note: e αβµν is actually a pseudo-tensor. Read Pg of Landau s book about this. Basically a pseudo-tensor transforms similarly as real tensors in rotations of coordinates. They differ only in those transformation that cannot be reduced to rotation of coordinates. Since Lorentz transformation is a rotation of fourdimensional space, we can use pseudo-tensors to construct Lorentz invariants. Using e αβµν, we can construct the following invariant, F αβ e βαµν F νµ = inv. (4.105)

52 52 classical electrodynamics Because e αβµν and F αβ are anti-symmetric, we write e αβµν F αβ F µν = inv, (4.106) which leads to E B = inv. (4.107) In short summary, the two important invariants of the field are c 2 B 2 E 2 = inv E B = inv Using two invariants, we draw the following conclusions: 1. If E > cb in K, then E > cb in all other K, and vice versa. 2. If E B in K, then we can find a K so that E = 0 if c 2 B 2 E 2 > 0. B = 0 if c 2 B 2 E 2 < If E B in K, then E B in all other K if E = 0 and B = 0.

53 5 The Electromagnetic Field Equations 5.1 Electrodynamics before Maxwell From your class on electromagnetism, you should have learned the following laws E = ρ/ɛ 0, Gauss s law. B = 0, no name E = B t, B = µ 0 j. Faraday s law Ampere s law There are inconsistencies in these equations that cannot not be resolved by these equations themselves. Taking the divergence of Ampere s law, you will get While is always correct, ( B) = µ 0 j. ( B) = 0, j = 0, only holds if the current is steady. This implies that Ampere s law only holds for static case. 5.2 The charge conservation law To see how Maxwell fixed these laws, we need to first describe the conservation of charge. We first prove the equation of continuity in a more traditional way, then we express it using the current density 4-vector. The conservation of charge means t ρdv = ρv ds, (5.1)

54 54 classical electrodynamics i.e., the time rate change of the total charge in volume V equals minus the total amount of charge leaving the system. Using j = ρv, t ρdv = ρ t dv = j ds = jdv. (5.2) V Since this integration must hold for any volume V, the above equation leads to the equation of continuity, ρ + j = 0. (5.3) t This equation of continuity can also be directly proved using the definition of ρ. For simplicity, we note that the charge density ρ for a single particle 1 is defined as, From this equation, we see ρ(t, r) = qδ[r r 0 (t)]. (5.4) 1 How do you derive the equation of continuity for a system of particles? Hint: Consider what is the appropriate definition of v. Then ρ t = q δ dr 0 r 0 dt ( = q δ r ) v = (ρv), since v = 0. Therefore, with j ρv, ρ + j = 0. t Now we introduce the current four-vector. While the charge q is a Lorentz invariant, ρ is not, since dq = ρdv and dv is not an invariant. Multiplying dq = ρdv by dx α leads to dqdx α }{{} four-vector = ρdvdx α = dvdt }{{} ρ dxα dt. (5.5) four-scalar Therefore ρdx α /dt must be a four-vector, we define it to be the current density four-vector; i.e., j α = ρ dxα dt, (5.6) or j α = (ρc, j) with j = ρv the normal current density vector you re familiar with. Since j α is a four-vector, its divergence is a four-scalar; i.e., j α = inv. (5.7) xα Writing this out with j α = (ρc, j) and using the equation of continuity, we have j α x α = ρ + j = 0. (5.8) t That is the charge conservation law or the equation of continuity can be expressed as that the the-divergence of the current density 4-vector is 0 (the current density 4-vector is divergence-free).

55 the electromagnetic field equations How Maxwell fixed Ampere s law Ampere s law gives ( B) = µ 0 j, Because the left hand side (LHS) is always right, one might try to fix this by adding a term to the right hand side (RHS). Well, we ve just learned the equation of continuity, therefore, we can fix Ampere s law using 2 ( ( B) = µ 0 j + ρ ). t 2 Note that in actual history, Maxwell added the extra term for a reason based on ether, a model which is now considered to be incorrect. Considering Gauss s law that E = ρ/ɛ 0, the above equation is written as ( ) E ( B) = µ 0 j + µ 0 ɛ 0 E = µ 0 j + µ 0 ɛ 0. t t Therefore, the Ampere s law can be fixed to be B = µ 0 j + µ 0 ɛ 0 E t. The new extra term is called by Maxwell the displacement current, This new term means that j d = ɛ 0 E t. A changing electric field induces a magnetic field. Now the new electrodynamics has a certain kind of symmetry in it. Of course, theoretical convenience and consistency are only suggestive. The confirmation of Maxwell s law came later with Hertz s experiments on electromagnetic waves after Maxwell s death. 5.4 Maxwell s Equations The whole set of Maxwell s equations are E = ρ/ɛ 0, Gauss s law. B = 0, no name E = B t, B = µ 0 j + µ 0 ɛ 0 E t. Faraday s law Ampere s law These are the fundamental equations of the EM field theory. The Maxwell equations and the Lorentz force law, dp dt = q(e + v B),

56 56 classical electrodynamics are the entire content of the theoretical classical electrodynamics. *5.5 Deriving Maxwell equations using the principle of least action The homogeneous Maxwell equations From definitions of E and B using potentials φ and A, we have B = A, (5.9) E = φ A t, (5.10) B = 0, (5.11) E = B t. (5.12) These are the two homogeneous Maxwell equations The action of the electromagnetic field To obtain the other two equations of fields, we introduce the action of the electromagnetic field. Previously we considered the action for charges in a given field S = S p }{{} + S p f }{{}. (5.13) free particle interaction The action for the whole system is, by adding S f to the total action S, S = S p }{{} + S p f }{{} + S f }{{}. (5.14) free particle interaction field only To derive the field action S f, we note the following three constraints, 1. The field satisfies the principle of supposition. Equations are linear in E and B; therefore, S f must be quadratic in E and B; i.e., E 2 or B S f must be a 4-scalar. Considering these constraints, other people have figured out S f Fαβ F αβ dω. In SI units, S f = 1 4µ 0 F αβ F αβ dω, where dω = cdtdv (5.15) In 3D form, since F αβ F αβ = 2(B 2 E 2 /c 2 ), S f = c (E 2 /c 2 B 2 )dvdt, (5.16) 2µ 0

57 the electromagnetic field equations 57 or the Lagrangian for the field is L f = c (E 2 /c 2 B 2 )dv. (5.17) 2µ 0 Putting all pieces together, the total action is S = mc ds qa α dx α 1 4µ 0 c F αβ F αβ dω, (5.18) where the summation is over all particles. Note that now charges are not assumed to be small, so fields include external fields and the field produced by the particles themselves. Using the charge density 4-vector j α, the interaction action S p f becomes S p f = = qa α dx α = ρa α dv dxα dt dt = 1 j α A α dω. c ρa α dvdx α Substituting this form of S p f into the Equation (5.18) S = mc ds 1 c j α A α dω 1 4µ 0 c F αβ F αβ dω. This is the total action expressed using the current density 4-vector The inhomogeneous Maxwell equations In this subsection, we use the principle of least action to determine equations obeyed by fields. Deriving field equations using the principle of least action is similar to deriving particle equations. Recall that to obtain particle trajectory, we vary x α x α + δx α, and then we obtain equations satisfied by x α. Now we want to obtain equations obeyed by field A α, we do it in the following three steps. 1. Particles move along their actual path 2. Vary A α A α + δa α. 3. The field equations can be obtained from δs = 0. First, noting that particles all move along their actual path (δs p = 0), the variation of action due to δa α is δs = δs p 1 }{{} c δ = 0 A α j α dω 1 4µ 0 c δ F αβ F αβ dω. (5.19) Since particle move along their actual trajectories, δj α = 0, δs = 1 c δa α j α dω 1 4µ 0 c δ F αβ F αβ dω. (5.20) The integrand of δs p f is in the form of ( )δa α, so we do not need to do anything about it. We just need to write the integrand of δs f in a similar form.

58 58 classical electrodynamics Now let s calculate δs f = 1 4µ 0 c δ(f αβ F αβ )dω. (5.21) The variation of the action of field is δs f = 1 δ(f 4µ 0 c αβ F αβ )dω (5.22) = 1 (δf 4µ 0 c αβ F αβ + F αβ δf αβ )dω (5.23) = 1 F αβ δf 2µ 0 c αβ dω. (5.24) Substituting F αβ = A β x α A α x β into Equation (5.24) leads to δs f = 1 2µ 0 c = 1 2µ 0 c = 1 2µ 0 c = 1 2µ 0 c = 1 µ 0 c δf αβ = δa β x α δa α x β (5.25) F αβ δf αβ dω (5.26) ( F αβ δa β x α ( F αβ δa β x α ( F αβ δa β x α δa ) Fαβ α x β dω (5.27) Fβα δa β x α + Fαβ δa β x α ) dω (5.28) ) dω (5.29) F αβ δa β dω, (5.30) xα where we have used F αβ = F βα. Perform the integration in Equation (5.30) by parts, δs f = 1 µ 0 c = 1 µ 0 c F αβ δa β dω (5.31) xα [ ( ) F x α F αβ αβ ] δa β dω x α δa βdω. (5.32) The first term can be proved to be 0 using Gauss s theorem in four-dimensional space, because ( ) x α F αβ δa β dω = Therefore Equation (5.32) becomes F αβ δa β ds α = 0, (5.33) δa β = 0 for t = t 0 and t = t 1. (5.34) F αβ = 0 for r =. (5.35) δs f = 1 µ 0 c F αβ x α δa βdω. (5.36)

59 the electromagnetic field equations 59 The variation of the total action is δs = δs p f + δs f, δs = ( 1 c jβ + 1 µ 0 c ( 1 = c jβ + 1 µ 0 c F αβ ) x α δa β dω (5.37) F βα x α The principle of least action δs = 0 gives The components of these equations are 1 c ) δa β dω. (5.38) F βα x α = µ 0j β. (5.39) E x x 1 E y c y 1 E z c z = µ 0ρc (5.40) 1 E x c 2 B z t y + B y z = µ 0j x (5.41) 1 E y c 2 + B z t x B x z = µ 0j y (5.42) 1 E z c 2 B y t x + B x y = µ 0j z (5.43) In vector form, these equations give, using c 2 = 1/µ 0 ɛ 0, the inhomogeneous Maxwell equations. E = ρ/ɛ 0, (5.44) B = µ 0 ɛ 0 E t + µ 0j. (5.45) 5.6 Integral forms of Maxwell equations We now obtain useful laws by integrating the differential Maxwell equations. First, V B dv = S B ds. Gauss s theorem (5.46) The integral S B ds is the flux through surface S, B = 0 B ds = 0, (5.47) means the flux of B through any closed surface is 0. S We now integrate the E equation. First, From the differential equation, S E ds = E dl. Stokes theorem (5.48) l E = B t, (5.49)

60 60 classical electrodynamics we have the conclusion E ds = E dl = S l t S B ds = Φ t. (5.50) In the equation l E dl = Φ t, (5.51) the two terms mean 1. Φ: the magnetic flux through surface S. 2. E dl: the electromotive force in the contour. l The electromotive force thus equals minus the time rate change of Φ. Third, the integral form of Gauss s law is EdV = 1 ρdv, (5.52) ɛ 0 E ds = 1 ρdv = Q/ɛ 0. (5.53) ɛ 0 The flux of E through a closed surface equals 1/ɛ 0 times the total charge inside the volume. The integral form of Ampere s law is B ds = µ 0 ɛ 0 E ds + µ 0 j d (5.54) t ( ) E B dl = µ 0 j + ɛ 0 ds (5.55) t The circulation of B around any contour equals µ 0 times the sum of the displacement current and the true current through the surface. 5.7 Energy density and energy flux of the EM field Now we discuss some conservation properties of electromagnetic field regarding their energy and momentum. Recall that field is similar to a continuous media, therefore, when talking about field energy and momentum, we almost always talk about energy density and momentum density in the context of classical electrodynamics. Field carries energy; and the total energy is conserved.

61 the electromagnetic field equations 61 To figure out the energy density of the field, we see that the work done by EM field on a charge q in interval dt is dε k = F dl = q(e + v B) vdt = qe vdt (5.56) If we consider a continuous distribution of charge, dε k dt = ρe vdv = j EdV. (5.57) Through conservation of energy, dε k must equal minus the energy change of the field in time dt. To see the energy variation of the field, we express j E using field quantities only. The assumption here is that the only other way of changing energy is through the field. j E = 1 ( ) E B µ 0 ɛ 0 E µ 0 t = 1 E B E ɛ 0 µ 0 t E = 1 µ 0 ( B E E B) + 1 µ 0 E B ɛ 0 2 = 1 (E B) 1 B 2 µ 0 2µ 0 t = 1 µ 0 (E B) 1 2 t ɛ 0 E 2 2 t ) (ɛ 0 E 2 + 1µ0 B 2. Reorganizing the equation as ( 1 t 2 ɛ 0E ) ( ) 1 B 2 = j E E B. (5.58) 2µ 0 µ 0 Denoting the Poynting flux vector by E 2 t P = 1 µ 0 E B, (5.59) the equation can be written as t ( 1 2 ɛ 0E µ 0 B 2 ) + j E = P. (5.60) To see what the equation (5.60) means, we apply dv to it. t ( 1 2 ɛ 0E ) B 2 dv + 2µ 0 j EdV = PdV. or using Gauss theorem ( 1 t 2 ɛ 0E ) B 2 dv + 2µ 0 j EdV = P ds (5.61) If we extend the integration to all space, the field is 0 at, Equation (5.61) becomes ( 1 t 2 ɛ 0E ) B 2 dv + j EdV = 0. (5.62) 2µ 0

62 62 classical electrodynamics Noting that dε k dt Equation (5.61) then becomes t ( 1 2 ɛ 0E ) B 2 dv 2µ 0 }{{} field energy = j EdV, (5.63) + ε k }{{} particle kinetic energy = 0, (5.64) from which we conclude that w = 1 2 (ɛ 0 E 2 + 1µ0 B 2 ) (5.65) is the field energy density. Yes, the field has energy. Expressed using w, Equation (5.61) is ( ) wdv + ε t k = P ds. (5.66) The left hand side (LHS) is the total energy variation inside a given finite volume, thus P, the Poynting flux, is the field energy flux density the amount of energy passing through unit area per unit time. 5.8 Momentum density and the Maxwell stress tensor We derive the momentum of the field in a similar way as that of the field energy density. That is, Field carries momentum; the total momentum is conserved. The time rate change of momentum of a particle q, dp m dt = F = q(e + v B). (5.67) The subscript m denotes mechanical quantity. If we consider a continuous distribution of charges, then dp m dt = ρ(e + v B)dV = (ρe + j B) dv. (5.68) We now express ρ and j using E and B, and the purpose is of course to express the final equation in the form d ) (p dt m + p f + GdV = 0, (5.69) where p f is the field momentum, and G is the field momentum flux, which is a tensor (since momentum is a vector).

63 the electromagnetic field equations 63 We now express the right hand side of Equation (5.69) using field quantities only. From Maxwell equations, ρ = ɛ 0 E and j = 1 ( ) E B µ 0 ɛ 0 (5.70) µ 0 t The first term ρe = ɛ 0 E( E). The second term is Using j B = 1 µ 0 ( B) B ɛ 0 E t B. (5.71) B E t = B (E B) + E t t, (5.72) and adding B( B), which is 0, to equation (5.68) gives, ρe + j B = ɛ 0 (E B) [ t + ɛ 0 E( E) + 1 B( B) ɛ 0 E ( E) 1 ] B ( B). µ 0 µ 0 (5.73) Substituting Equation (5.73) into equation (5.68) gives dp m + d ( ) ɛ 0 E BdV = dt dt [ ɛ 0 E( E) + 1 B( B) ɛ 0 E ( E) 1 ] B ( B) dv µ 0 µ 0 Define (5.74) g = ɛ 0 E B, (5.75) which will be shown later to be the field momentum density, and p f = gdv, (5.76) which is the total field momentum in the chosen volume, then Equation (5.74) is dp m + dp f = dt dt [ ɛ 0 E( E) + 1 B( B) ɛ 0 E ( E) 1 ] B ( B) dv, µ 0 µ 0 (5.77) Let s now figure out what s in the right hand side. Recall that in Homework 2, you have proved ( E ( E) = ( E)E EE 1 ) 2 E2 I, (5.78) so ( E( E) E ( E) = EE 1 ) 2 E2 I (5.79)

64 64 classical electrodynamics similarly, ( B( B) B ( B) = BB 1 ) 2 B2 I (5.80) Define the Maxwell stress tensor T to be T = [ ɛ 0 EE + 1 BB 1 ) ] (ɛ 0 E 2 + 1µ0 B 2 I, (5.81) µ 0 2 Equation (5.77) becomes d ) (p dt m + p f = TdV = T ds. (5.82) Or written in the form of Equation (5.69), d ) (p dt m + p f + G dv = 0, (5.83) where G = T is the field momentum flux tensor. To actually understand the meaning of p f and T (or G), let V. Then T = 0 since field is 0 at infinity. Equation (5.82) becomes d ) (p dt m + p f = 0. (5.84) From the conservation of momentum, p f thus represents the momentum of the field. To see the meaning of T or G, let V be finite, then d ) (p dt m + p f = T ds = G ds. (5.85) Therefore T represents the field momentum flux flowing into the volume through surface. Or G, the field momentum flux tensor, represents the field momentum flux flowing out of the volume through surface. In a Cartesian coordinate system, let the normal direction of ds to be n = (n 1, n 2, n 3 ), d (p dt m + p f )i = T ij n j ds (5.86) Therefore, T ij n j is the ith component of the momentum per unit area per unit time into the volume. On the other hand, d ) (p dt m + p f = force on the field and particles, (5.87) T ij n j is the ith component of the force per unit area through S. If the field is static, then p f is constant, or dp m dt d(p m ) i dt = = T ds (5.88) T ij n j ds (5.89)

65 the electromagnetic field equations 65 On the other hand, if p m is constant, then or or dp f dt = g t g t dv = TdV (5.90) = T = G, (5.91) g + G = 0, (5.92) t further proving that the field momentum flux is G.

66 6 General Electrostatics 6.1 General considerations The Maxwell equations are, E = B t, (6.1) E = ρ ɛ 0, (6.2) E B = µ 0 ɛ 0 t + µ 0j, (6.3) B = 0. (6.4) In this chapter, we consider static EM fields; i.e., The corresponding static EM field equations are = 0. (6.5) t E = 0 (6.6) E = ρ ɛ 0, (6.7) B = µ 0 j, (6.8) B = 0. (6.9) Notes that E = φ A t = φ, B = A. That is, E is completely determined using φ, and E and B are decoupled. So we separate static electromagnetic fields into two parts: one about the electrostatic field and the other about magnetostatic field. In this chapter, we discuss electrostatics. 6.2 Electrostatic fields In case of a electrostatic field, E = φ. Using φ can significantly simplify analysis, since it s a scalar. One exception is if the problem possesses some symmetry, then you can apply Gauss s theorem to obtain E.

67 general electrostatics 67 The Coulomb s law expressed using φ is E = ρ ɛ 0 2 φ = ρ ɛ 0. (6.10) If ρ = 0 (e.g., vacuum), the equation becomes 2 φ = 0. (The Laplace equation.) (6.11) E and φ apparently satisfy the principle of superposition. We first calculate the electrostatic field of a point charge. The ES field of a distribution of charges can be obtained from the principle of superposition. This is an example of applying Gauss s law to a problem with symmetry. For a point charge, because of symmetry, E = E(r)e r, where r is the distance to the charge, and e r = r/r. From the Coulomb s law, EdV = E ds = EdV = ρ dv = q, (6.12) ɛ0 ɛ 0 where we have used ρ = qδ(r). Choose the volume to be a sphere centering at the charge, E ds = 4πr 2 E = q ɛ 0. (6.13) Therefore, the ES field of a point charge is The potential of the field is clearly since E = 1 4πɛ 0 q r 2 e r. φ = 1 4πɛ 0 q r, φ = 1 4πɛ 0 q r 2 r = 1 4πɛ 0 q r 2 e r = E. (6.14) Using this expression of φ, we can obtain an useful identify for δ(r). Putting φ = q/4πɛ 0 r into 2 φ = ρ/ɛ 0, for a point charge at r = 0, ρ = qδ(r), therefore, or ( ) q 2 4πɛ 0 r = qδ(r) ɛ 0 (6.15) ( ) 1 2 = 4πδ(r) (6.16) r 6.3 Electrostatic field multipole moments One electrostatic problem we frequently encountered is to calculate the potential or the E field of a system of charges. Normally the calculation is tedious and is best done using a computer. However, in lots of cases, we are only interested in the field far far away

68 68 classical electrodynamics from the charges. For example, we might be interested in the field produced by a group of molecules at a distance much larger compared to the spatial scale of system molecules. In this case, we can use Taylor expansion to obtain leading fields of the charge system. This technique and the corresponding fields are introduced in this section. Let s introduce a coordinate system where the origin is within the system of charges, the radius vector of charge a is x a. The ES potential at an arbitrary point x is φ(x) = 1 q a 4πɛ 0 a x x a. (6.17) We now consider field at large distances x x a ; i.e., at distances large compared to the dimensions of the system. The method is to expand φ using the small parameter x a /r, with r x, and find out the dominant terms. We will also drop script a with understanding that the summation is over all charges. field point R = x-x x q a x a O Illustration of the coordinates used in this chapter. Using x x, we can Taylor expand a function f (x x ) by f (x x ) = f (x) x f (x) x i x j f (x) + (6.18) x i x j In our case, f (x x ) = φ(x x ), and φ = 1 q 4πɛ 0 a x x = φ(0) + φ (1) + φ (2) +, (6.19) where φ (i) denotes the ith order term. If you compare, e.g., φ (1) and φ (0), φ (1) φ (0) x φ x φ/ x x 1. (6.20) φ φ x If we denote ɛ x / x, then φ (1) /φ (0) O(ɛ). Similarly, you can show that φ (2) /φ (0) O(ɛ 2 ). 2

69 general electrostatics 69 Applying Equation (6.18) to Equation (6.19), you can immediately arrive at the conclusion that φ (0) = φ(x) = 1 q 4πɛ 0, with r x, (6.21) a r φ (1) = 1 qx a 1 r, (6.22) 4πɛ 0 φ (2) = 1 8πɛ 0 qx i x j x i x j a a 2 ( ) 1. (6.23) r We ll now treat these terms one by one. The 0 th order term φ (0), normally called the monopole term, is φ (0) = φ(x) = 1 i q i. (6.24) 4πɛ 0 r This is equivalent to ignoring the internal distribution of charges and treating them as a single point charge with Q = a q. Apparently φ (0) dominates unless Q = 0, which is common, e.g., when calculating fields produced by water molecules. The 0 th term is also consistent with our intuition: if we are far far away from a distribution of charges, we can approximate the charges by a point charge. If Q = 0 (φ (0) = 0), we then need higher order terms φ (1), or even φ (2) if φ (1) = 0, to better approximate φ. Let s first discuss φ (1). Introducing the electrostatic dipole moment p, p = qx, (6.25) a we can write φ (1), the electrostatic dipole potential, as φ (1) = 1 4πɛ 0 qx 1 r = p 1 4πɛ 0 r = a p x 4πɛ 0 r 3. (6.26) This is the potential due to the dipole moment of the charges. The field E = φ. Note here = / x; i.e., with respect to the field coordinate, therefore the dipole moment p is treated as a constant when we perform. Performing the gradient operation, E = p x 4πɛ 0 r 3 = 1 1 (p x) (p x) 1 4πɛ 0 r3 4πɛ 0 r 3 (6.27) = 1 4πɛ 0 r 3 p x p x 3 r 4πɛ 0 r4 (6.28) = p 4πɛ 0 r 3 + 3p x x 4πɛ 0 r 4 r (6.29) or finally with n = x/r, E (1) = 3(n p)n p 4πɛ 0 r 3. (6.30) Note that E (1) r 3. You can also directly Taylor expand E to obtain E (0), E (1),.

70 70 classical electrodynamics The components of the electric field E can be obtained easily from Equation (6.30). Assuming p = pe z, then, with θ = n, e z, e e z. E z = E = 3(n p)n p 4πɛ 0 r 3 e z = p 3 cos2 θ 1 4πɛ 0 r 3, (6.31) 3(n p)n p 3 cos θ sin θ 4πɛ 0 r 3 e = p 4πɛ 0 r 3. (6.32) Or if using spherical coordinates, E r = E n = 3(n p)(n n) p n 4πɛ 0 r 3 E θ = E e θ = 3(n p)(n e θ) p e θ 4πɛ 0 r 3 = p 2 cos θ 4πɛ 0 r 3 = p cos θ 2πɛ 0 r 3, (6.33) = p sin θ 4πɛ 0 r 3. (6.34) Two useful facts about the dipole moment are worth mentioning. First, if the total charge a q = 0, then p does not depends on the choice of the origin of coordinates. Consider two systems, x = x + a, then p = qx = qx + qa = qx = p. (6.35) On the other hand, if Q a q = 0, then p = p + Qa. Second, a simple and frequently used dipole moment is for two point charges q at x + and q at x. p = q x + q x = q l, with l = x + x. (6.36) Let s now continue to the second order term φ (2) ; this term will dominate if both Q and p equal 0. Rewrite Equation (6.23) here, φ (2) = 1 8πɛ 0 qx i 1 x j x i x j r. (6.37) Define temporarily the quadrupole moment of the system then D = a a 2 3qx x, or D ij = 3qx i x j, (6.38) a φ (2) = D ij 24πɛ 0 2 x i x j 1 r. (6.39)