A Brief Introduction to Classical Field Theory Níckolas Alves∗ Department of Mathematical Physics, Institute of Physics, University of São Paulo, São Paulo, SP, Brazil (Dated: February 15, 2020) In this work, I present an introduction to classical field theory by exploring the limiting case of N coupled harmonic oscillators as N ! in order to obtain the equation of motion for a vibrating string. The Euler-Lagrange Equations for a collection of N three-dimensional fields are presented and, finally, Noether’s Theorem is proved, with1 the stress-energy tensor and the conservation of electric charge due to gauge invariance in QED being given as examples of application. Keywords: classical field theory, Noether’s theorem m I. VIBRATING STRINGS m m m Suppose you want to describe the motion of a vi- m brating string of length L and linear mass density µ m with clamped ends. A simple model for this situation is considering a series of coupled harmonic oscillators m q1 q2 qN confined to movement in a single direction.Suppose x x x you want to describe the motion of a vibrating string of 1 2 a N length L and linear mass density µ with clamped ends. A simple model for this situation is considering a series Figure 1. A simple model for a vibrating string of length L of coupled springs, with spring constant k, confined to and linear mass density µ consists of N coupled harmonic L movement in a single direction. Such oscillators should oscillators, each a distance a = N apart from its neighbors L with mass m = µa each have mass m = µa, where a = N+1 is the distance between two consecutive oscillators. The height of the oscillator located at the position xi is given by a func- tion qi(t). As the string should be clamped at its ends, we impose that q0(t) = qN+1(t) = 0 at all times. We still must discover what is the value of l. We know The kinetic energy Ki stored in the i-th oscillator is that L0 = (N + 1)l is the rest length of the string, while L = (N + 1)a m 2 it’s current length is (roughly1) . If a ten- Ki = q˙ : (1) 2 i sion τ is being applied in the string, then we know from k Hooke’s Law that τ = - N+1 ((N + 1)l -(N + 1)a) (the On the other hand, the potential energy Ui stored in k factor N+1 comes from associating springs in series) the spring between positions xi and xi+1 is given by and it follows that k 2 Ui = (li - l) ; (2) 2 τ l = a - l; where stands for the rest length of the spring (which k is equal for every spring here considered) and li is the τ l = a - : (4) length of the spring between positions xi and xi+1, k which can be obtained through trigonometry. If the angle the spring makes with the horizontal line is θ, then we know that li = a·sec θ. Furthermore, we know qi+1-qi 2 2 that tan θ = a . Thus, as tan θ + 1 = sec θ, it Thus, we have that the energy stored in each spring follows that s 2 qi+1 - qi li = a 1 + : (3) a 1 Don’t bother with this approximation right now, we are going to ∗ Correspondence email address: [email protected] justify it in a minute or two. 2 is given by So far, we have only dealt with a discrete approxima- tion for the string problem. Let us try to figure out a 0 s 12 way to take the limit N ! properly and obtain a full k q - q 2 τ U = a 1 + i+1 i + - a ; description of the string. i 2 @ a k A The first step is simple:1 we are currently using a ! 2 2 finite number of degrees of freedom through the gen- k 2 qi+1 - qi k τ = a 1 + + a - eralized coordinates qi. If we want to describe a con- 2 a 2 k tinuous string, we should drop this description in fa- s 2 vor of something that accepts a continuous indexation. qi+1 - qi τ - ka 1 + a - : (5) Thus, we are going to define a field φ(x; t) such that a k N+1 φ(xi; t) = qi(t); 8 i 2 figi=0 ; 8 t. Furthermore, we are going to write ∆x ≡ a from now on, since this is the We will now make the assumption that the string spatial variation between two oscillators and our plan is vibrating under a regime of small oscillations, i.e., is to take such a value to zero in the limiting case. jqi+1 - qij a. Without this approximation, even Recalling that m = µa, our Lagrangian should now though the springs are being treated as linear, we would be written as obtain a nonlinear behaviour, which is more appropri- " # ate for a deeper analysis of the problem. For simplicity, µ ∂φ(x; t)2 we are going to assume the string is being held really L = ∆x 2 @t tight and we can only disturb it slightly. X " # This assumption justifies the approximation we τ φ(x + ∆x, t)- φ(x; t)2 - ∆x : (8) made earlier that the length of the string is L = (N+1)a 2 ∆x (which is, in fact, also the definition of a we provided). X If the disturbances are small, indeed the length of string By taking the limit as N ! , and recalling that is very close to L = (N + 1)a and our treatment is justi- L = N∆x remains constant, we obtain fied. 1 By Taylor expanding the square root in Eq.5, it µ ∂φ2 τ ∂φ2 L = - dx : (9) follows that 2 @t 2 @x Z ! 2 2 k 2 qi+1 - qi k τ Ui = a 1 + + a - This motivates us to define a new function, named 2 a 2 k Lagrangian density and denoted by L, such that ! 1 q - q 2 τ - ka 1 + i+1 i a - ; 2 a k L = L dx : (10) k k k τ 2 Z = a2 + (q - q )2 + a - + aτ 2 2 i+1 i 2 k τ k II. EULER-LAGRANGE EQUATIONS - ka2 + (q - q )2 - (q - q )2 ; 2a i+1 i 2 i+1 i 2 k 2 k τ τ 2 If we want the description in terms of the Lagrangian = - a + a - + aτ + (qi+1 - qi) : 2 2 k 2a density to be useful, we must obtain the equations of (6) motion described by this quantity. The procedure is fairly similar to the way one obtains the Euler-Lagrange We might now calculate the Lagrangian L = K - U Equations for usual Lagrangians: apply Hamilton’s by summing Eqs. (1) and (6) over every mass and Principle considering every possible field under the spring. Recalling that constant terms may be dropped, given boundary conditions. for they leave the Euler-Lagrange Equations unaltered, t2 As the action is defined as S = L dt, in terms of it follows that t1 the Lagrangian density it becomes R N N 2 L = K - U ; t2 x ∂φ ∂φ i i S = L(φ, ; ; x; t) dx dt : (11) i=1 i=0 @x @t X X t1 x1 N N Z Z m 2 τ 2 = q˙ - (qi+1 - qi) : (7) In a more general way, let us consider a Lagrangian 2 i 2a i=1 i=0 N X X density describing fields, each of them denoted by 3 φi and collectively described as φ, in three spatial di- written as mensions and one time dimension. The action is then ∂φ S = L φ, ; rφ, x; t d4x : (12) @t ZΩ Hamilton’s Principle states that δS = 0. From this knowledge, we will be able to obtain the Euler- Lagrange Equations for our field theory. With implicit summation over repeated indexes, we have t2 @L @L @L 3 δS = δφα + δφ˙ α + δrφα d x dt = 0: (13) ∂φ @φ˙ @rφ Zt1 ZV α α α Let us first deal with the second term (the first term If we recall that is already as simple as we want). Notice that r · (fv) = fr · v + v · rf; (17) @L 4 @L @ δφα 4 δφ˙ α d x = d x ; @φ˙ @φ˙ @t ZΩ α ZΩ α t2 then it follows from Gauss’s Theorem that @L 3 = δφα d x ˙ V @φα t1 Z t2 @L 4 @L @ @L 4 δrφα d x = δφα · da dt - δφα d x : (14) @rφ @rφ @t @φ˙ ZΩ α Zt1 I@V α ZΩ α @L 4 - r · δφα d x : As said earlier, we are considering every possi- @rφ ZΩ α ble field under the given boundary conditions. Thus, (18) t2 δφα = 0. Thus, t1 Once again, since we have fixed the boundary condi- δφ = 0 tions, α @V . Thus, @L 4 @ @L 4 δφ˙ α d x = - δφα d x : (15) ˙ ˙ Ω @φα Ω @t @φα Z Z @L 4 @L 4 δrφα d x = - r · δφα d x : We must then apply a similar argument to the third @rφ @rφ ZΩ α ZΩ α term in Eq. (13). (19) @L 4 @L 4 δrφα d x = r(δφα) d x : (16) By implementing Eqs. (15) and (19) into Eq. (13), we @rφ @rφ ZΩ α ZΩ α obtain @L @ @L @L 4 - - r · δφα d x = 0: (20) ∂φ @t @φ˙ @rφ ZΩ α α α As the equation must hold for every domain of inte- ations in the fields δφα we finally conclude that gration, the integrand must be zero.