A Brief Introduction to Classical Field Theory

Home , Classical field theory, Hamiltonian mechanics, Scalar potential

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, with∞ 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 function 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 approximation for the string problem. Let us try to figure out a  s 2 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  The first step is simple:∞ 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), ∀ i ∈ {i}i=0 , ∀ 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. |qi+1 − qi| 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. ∞ 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, , ∇φ, 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 + δ∇φα d x dt = 0. (13) ∂φ ∂φ˙ ∂∇φ 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 ∇ · (fv) = f∇ · v + v · ∇f, (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 δ∇φα d x = δφα · da dt − δφα d x . (14) ∂∇φ ∂∇φ ∂t ∂φ˙ ZΩ α Zt1 I∂V α ZΩ α ∂L 4 − ∇ · δφα d x . As said earlier, we are considering every possi- ∂∇φ 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 δ∇φα d x = − ∇ · δφα d x . We must then apply a similar argument to the third ∂∇φ ∂∇φ ZΩ α ZΩ α term in Eq. (13). (19)

∂L 4 ∂L 4 δ∇φα d x = ∇(δφα) d x . (16) By implementing Eqs. (15) and (19) into Eq. (13), we ∂∇φ ∂∇φ ZΩ α ZΩ α obtain

∂L ∂ ∂L ∂L 4 − − ∇ · δφα d x = 0. (20) ∂φ ∂t ∂φ˙ ∂∇φ 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. As the equation also must hold for every possible combination of vari- ∂ ∂L ∂L ∂L + ∇ · − = 0. (21) ∂t ∂φ˙ α ∂∇φα ∂φα 4

As an example, let us consider the equations of mo- For a given value of α, if φα is a scalar field it will tion described by the Lagrangian2 for the vibrating hold that ∂µφα is a one-form. In a similar manner, if the string, obtained in SectionI: Lagrangian L is a scalar, then ∂L is a four-vector3 ∂(∂µφα) and the first-term on the Euler-Lagrange Equations is a µ ∂φ2 τ ∂φ2 L = − . (22) scalar. Therefore, if we want the Euler-Lagrange Equa- 2 ∂t 2 ∂x tions to be covariant, we must simply demand L to be a scalar. As dx, the four-dimensional volume element, Thus, we have is invariant under Lorentz Transformations, the action ∂L S = L dx will be a scalar whenever the Lagrangian = 0, ∂φ is a scalar[1,2], and thus the requirement of S to be R  ∂L a scalar is another argument in defense of L being a  = µ∂tφ, (23) ∂(∂ φ) scalar quantity.  t  ∂L = −τ∂xφ, ∂(∂xφ) A. Maxwell’s Electrodynamics   where we denote a partial derivative with respect to a As an example of a Relativistic Field Theory, we may variable u as ∂u. Substitution in the Euler-Lagrange Equations yields consider the Lagrangian for Maxwell’s Electrodynam- ics4: ∂2φ ∂2φ 1 1 µ − τ = 0, (24) L = − F Fµν + JµA , (28) ∂t2 ∂x2 16π µν c µ which is, indeed, the wave equation satisfied by the with string. Fµν = ∂µAν − ∂νAµ. (29) The four-vectors Jµ and Aµ are, in fact, covariant III. RELATIVISTIC FIELD THEORIES forms of familiar quantities. Namely, they are the four- current and four-potential, composed from the charge One might be interested in consider field theories density ρ, vector current density J, scalar potential V under the light of the Theory of Relativity. It would and vector potential A as then be interesting for us to drop description in terms Jµ = (cρ, J) , of time and, instead, consider the coordinate x0 = ct, (30) Aµ = (V, ) . allowing us to treat space-time in a coherent way. A

Notice that the Euler-Lagrange Equations are kept We are now dealing with the fields φα ≡ A , where 0 α unchanged under the scale transformation t → x = ct, α is a four-vector index. Our notation will be simplified for if we write ∂ ∂L 1 ∂ ∂L Aµα ≡ ∂µAα. (31) 0 0 = c · . (25) ∂x ∂(∂φα/∂x ) c ∂t ∂(∂φα/∂t) As ∂L = 1 Jα, the Euler-Lagrange Equations be- ∂Aα c Thus, in terms of x0, we may write the Euler- come Lagrange Equations as ∂L 1 ∂ − Jα = 0. (32) µ ∂A c ∂L ∂L µα ∂µ − = 0, (26) ∂(∂µφα) ∂φα where, as usual, 3 Some texts refer to four-vectors as contravariant (four-)vectors and to one-forms as covariant (four-)vectors. As I am unable to remember ∂ which one is which, and therefore I shall stick to calling them ∂ ≡ . (27) µ ∂xµ four-vectors and one-forms. 4 As you might know from Classical Electrodynamics, the system of units we choose may change the way we express our equations. We are sticking to Gaussian units. Furthermore, as you might know from Relativity, the metric convention we choose may also 2 Sometimes we shall refer to the Lagrangian density as simply “La- change the way we express our equations. We are sticking to the grangian”. The difference should be made clear by the context. (− + ++) metric signature. 5

We then make We know from Classical Electrodynamics[3,4] that τν the electric field, E, is given in terms of the potentials V ∂L −1 ∂Fτν τν ∂F = F + Fτν . (33) and A by ∂Aµα 16π ∂Aµα ∂Aµα 1 ∂A E = −∇V − . (40) But c ∂t τν τν ∂F βγ ∂F Therefore, we obtain Fτν = gτβgνγF , ∂Aµα ∂Aµα τν ∇ · E = 4πρ, (41) ∂F βγ = gβτgγν F , ∂Aµα which is Gauss’s Law. µ = i, i = 1, 2, 3 ∂Fβγ βγ If, on the other hand, we pick , it = F , (34) i follows from the fact that J = Ji, where Ji denotes the ∂Aµα i-th component of the current density vector J, that where gµν is the metric tensor. Therefore, 4π iν Ji = ∂νF , ∂L −1 ∂Fτν τν c = F . (35) = ∂ ∂iAν − ∂νAi , ∂Aµα 8π ∂Aµα ν ∂ ν ν However, we already know that Fµν = Aµν − Aνµ. = ∂ν A − ∂ Ai , ∂xi Thus, ∂ 1 ∂V 1 ∂2A = + ∇ · − ∇2A + i . ∂F A i 2 2 τν µ α µ α ∂xi c ∂t c ∂t = δ τδ ν − δ νδ τ, ∂Aµα (42) ∂F τν Fτν = δµ δα Fτν − δµ δα Fτν, If we join these components in a single vector- ∂A τ ν ν τ µα equation, we get = Fµα − Fαµ, 2 µα 4π 1 ∂V 1 ∂ A = 2F , J = ∇ + ∇ · A − ∇2A + , c c ∂t c2 ∂t2 ∂L 1 = − Fµα, 2 1 ∂ 1 ∂A ∂Aµα 4π = ∇(∇ · A) − ∇ A + + ∇V , c ∂t c ∂t ∂L 1 = Fαµ (36) 1 ∂E ∂A 4π = ∇ × (∇ × A) − , (43) µα c ∂t Substitution in Eq. (32) yields where we used that ∇ × (∇ × A) = ∇(∇ · A) − ∇2A. 4π As we know from Classical Electrodynamics[3,4] ∂ Fµν = Jµ, (37) that the magnetic field B relates to the vector poten- ν c tial A through which is the covariant form of the inhomogeneous Maxwell’s Equations. Indeed, suppose µ = 0. Then, B = ∇ × A, (44) since J0 = cρ (Eq. (30)), it follows then that ∂ F0ν = 4πρ, 1 ∂E 4π ν ∇ × B − = J, (45) 0 ν ν 0 c ∂t c ∂ν ∂ A − ∂ A = 4πρ, 1 ∂ which is the Ampère-Maxwell Law. ∂ Aν + ∂νV = 4πρ. (38) ν c ∂t It appears that the Euler-Lagrange Equations only led us to half of Maxwell’s Equations, and therefore you are Opening up the derivatives, probably wondering where are the others! They come in fact from the definition Fµν = ∂µAν − ∂νAµ. Notice 1 ∂ 1 ∂V 1 ∂2V 4πρ = + ∇ · A + ∇2V − , that c ∂t c ∂t c ∂t2 1 ∂A ∂τFµν = ∂τ∂µAν − ∂τ∂νAµ, = −∇ · + ∇V . (39) c ∂t = ∂τ∂µAν − ∂ν∂τAµ. (46) 6

Thus, it follows that Furthermore,

ijk jik ∂τFµν + ∂µFντ + ∂νFτµ = ∂τ∂µAν − ∂ν∂τAµ ∂i∂jA0 = − ∂i∂jA0, ijk + ∂µ∂νAτ − ∂τ∂µAν = − ∂j∂iA0, + ∂ ∂ A − ∂ ∂ A . ijk ν τ µ µ ν τ (47) = − ∂i∂jA0. (55)

As you can see, every term on the right-hand side of Thus, the expression cancels out and we are left with ijk ijk ijk 0 = ∂0∂iAj + ∂i∂0Aj + ∂i∂jA0 ∂τFµν + ∂µFντ + ∂νFτµ = 0, (48) ijk ijk ijk + ∂0∂jAi + ∂j∂0Ai + ∂i∂jA0. (56) which I claim to be a covariant expression of the homo- geneous Maxwell Equations. After dividing the expression by 2, we obtain Let µ = 1, ν = 2, τ = 3 and i, j, k = 1, 2, 3. Then, if 0 = ijk∂ ∂ A + ijk∂ ∂ A + ijk∂ ∂ A , we open up Eq. (48) in terms of the four-potential (just 0 i j j 0 i i j 0 ijk ijk ijk like in Eq. (47)), we get = ∂0∂iAj − ∂i∂0Aj + ∂i∂jA0, = ijk∂ ∂ A + ∂ A − ijk∂ ∂ A , 0 = ∂ ∂ A − ∂ ∂ A + ∂ ∂ A i j 0 0 j i 0 j 3 1 2 2 3 1 1 2 3 ijk ∂ ∂V 1 ∂Aj ijk ∂ 1 ∂Aj − ∂3∂1A2 + ∂2∂3A1 − ∂1∂2A3, = + − . (57) ∂xi ∂xj c ∂t ∂xi c ∂t = ∂1 (∂2A3 − ∂3A2)

+ ∂2 (∂3A1 − ∂1A3) In vector notation, + ∂3 (∂1A2 − ∂2A1) , 1 ∂A 1 ∂ ijk 0 = ∇ × ∇V + − (∇ × A). (58) = ∂i ∂jAk , (49) c ∂c c ∂t

ijk 123 = 1 ∂A where denotes the Levi-Civita symbol (with If we recall one last time that E = −∇V − c ∂t and 1). B = ∇ × A, it follows that In vector notation, we get 1 ∂B ∇ × E + = 0, (59) ∇ · (∇ × A) = 0. (50) c ∂t

This might seem obvious, but if we recall that B = which is Faraday’s Law. ∇ × A, it follows that

∇ · B = 0, (51) IV. NOETHER’S THEOREM which is Gauss’s Law for Magnetism. We know from Hamiltonian Mechanics that, Finally, let µ = i, ν = j, τ = 0. Once again, we let i, j = 1, 2, 3. Eq. (48) now yields dH ∂L = − , (60) dt ∂t 0 = ∂0∂iAj − ∂j∂0Ai + ∂i∂jA0 where H denotes the Hamiltonian5 of a certain physi- − ∂ ∂ A + ∂ ∂ A − ∂ ∂ A . (52) 0 i j j 0 i i j 0 cal system described by the Lagrangian L. Therefore, This equation holds ∀ i, j = 1, 2, 3. Let now k also whenever the Lagrangian does not depend explicitly on assume the values 1, 2, 3. We might now multiply the time, the Hamiltonian is a constant of motion. Many equation as a whole by a Levi-Civita symbol and sum times, it happens that H = E, where E stands for energy, over all repeated indexes, i.e., and we say that energy is conserved. However, the most remarkable thing is that the rea- ijk ijk ijk 0 = ∂0∂iAj − ∂j∂0Ai + ∂i∂jA0 son for H to be conserved is that L does not depend ijk ijk ijk explicitly on time. Similarly, the canonical momentum − ∂0∂iAj + ∂j∂0Ai − ∂i∂jA0. (53) Notice that

ijk jik 5 You might wonder whether Field Theory can be studied from the ∂0∂iAj = ∂0∂jAi, Hamiltonian point of view. The answer is yes, and more details ijk = − ∂0∂jAi. (54) can be found about that in references [1,2,4]. 7

∂L p = ∂q˙ associated with the generalized coordinate q is where also conserved whenever ∂L . The description of how ∂q 0 0 0 0 0 0 0 4 0 these conservation laws arise (in Classical Field The- S = L (φα(x ), φα;µ(x ), x ) d x , Ω0 ory) from symmetries in the Lagrangian is the main  Z (69)  4 goal of this section.  S = L(φα(x), φα;µ(x), x) d x . We shall consider an inﬁnitesimal transformation ZΩ such that  We must now compute how the Lagrangian is af- xµ → x0µ = xµ + ∆xµ, fected by the transformation. We are going to make 0 0 0 (61) a little notation abuse right now and write L ≡ φα(x) → φα(x ) = φα(x) + ∆φα(x). 0 0 0 0 0 0 L (φα(x ), φα;µ(x ), x ) for simplicity. Taking our ﬁrst We also introduce the notation for the variation due demand into consideration, we see that only to the change in form of φα: 0 0 0 0 0 0 L = L(φα(x ), φα;µ(x ), x ), 0 δφα(x) = φα(x) − φα(x). (62) = L(φα(x) + ∆φα(x), φα;µ(x) + ∆φα,µ(x), x + ∆x), ∂L Notice that = L(φα(x), φα;µ(x), x) + ∆φα ∂φα 0 0 0 0 ∆φα(x) = φα(x ) − φα(x ) + φα(x ) − φα(x), ∂L ∂L ν 0 µ + ∆φα;µ + ν ∆x . (70) = δφα(x ) + ∂µφα(x)∆x . (63) ∂φα;µ ∂x

If we ignore the second-order terms in the infinitesi- If we now consider Eqs. (64) and (65), it follows that mal variations, it follows that 0 ∂L ∂L ∂φα ν µ L = L + δφα + ν ∆x ∆φα(x) = δφα(x) + φα;µ(x)∆x . (64) ∂φα ∂φα ∂x ∂L ∂L ∂φα;µ ν ∂L ν We also notice for further use that, since ∆ involves + δφα;µ + ν ∆x + ν ∆x , ∂φα;µ ∂φα;µ ∂x ∂x evaluating the fields at different points in space-time, ∂L ∂L L ∆∂ 6= ∂ ∆ φ d ν µ µ . Inserting α;ν in Eq. (64) yields = L + δφα + δφα;µ + ν ∆x , ∂φα ∂φα;µ dx µ ∆φα ν(x) = δφα ν(x) + φα ν µ(x)∆x . (65) dL ; ; ; ; ≡ L + δL + ∆xν. (71) dxν Nevertheless, δ simply involves evaluating the field before and after the transformation at the very same In the previous calculation, we write point, and therefore it commutes with ∂µ. dL ∂L ∂φα ∂L ∂φα µ ∂L Finally, the Lagrangian density will transform ac- ≡ + ; + (72) xν ∂φ ∂xν ∂φ ∂xν ∂xν cording to d α α;µ

0 0 0 0 0 0 to denote the partial derivative of the Lagrangian with L(φα(x), φα;µ(x), x) → L (φ (x ), φ (x ), x ). (66) α α;µ respect to xν without keeping the fields constant, i.e., we Assuming space-time is flat, we are going to prove also consider the contribution the fields give to the ν the existence of conserved quantities if this transforma- derivative, as opposed to ∂L/∂x , in which we con- tion satisfies the following properties sider only the explicit dependence. We must now change coordinates on the integral for Form Invariance:: the functional form of the trans- S0 in order to express ∆S as an integral over x. We have formed Lagrangian is equal to the original one, i.e. ∂ x00, x01, x02, x03 4 0 4 0 0 0 0 0 0 0 0 0 0 0 d x = 0 1 2 3 d x . (73) L (φα(x ), φα;µ(x ), x ) = L(φα(x ), φα;µ(x ), x ). ∂ (x , x , x , x ) (67) In order to calculate the Jacobian determinant of the transformation xν → xν + ∆xν, we must use the fact Invariance of the Action:: the numerical value of the that given a matrix J = 1 + A, then action integral is not changed by the transforma- 2 tion, i.e., det J = 1 + tr A + O , (74) 0 ∆S = S − S = 0, (68) which is proved in the Appendix at page 11. 8

The Jacobian matrix for xν → xν + ∆xν can be writ- One should be aware that, although the notation may ten as J = 1 + J0, where J0 is the Jacobian matrix for the suggest the opposite, the indices α and r are not nec- transformation xν → ∆xν. As ∆xν is small, we can ne- essarily tensor indices, but summation over repeated glect higher-order terms when computing tha Jacobian indices is implicit. determinant for xν → xν + ∆xν. Thus, we may apply Due to Eq. (64), we get the aforementioned theorem in order to obtain (r) ν(r) ∂∆xν δφα = Ψα − φα;νX r. (81) d4x0 = 1 + d4x . (75) ∂xν If we define In view of Eqs. (71) and (75), the expression for the µ(r) ∂L (r) ν(r) µ(r) Invariance of the Action becomes Θ = − Ψα − φα;νX − LX , (82) ∂φα;µ dL ∂∆xν L + δL + ∆xν 1 + − L d4x = 0. dxν ∂xν then inserting Eqs. (80) and (81) into Eq. (79) yields ZΩ (76) µ(r) dΘ 4 − r d x = 0, (83) dxµ If we ignore variations of second order and above, ZΩ the expression simplifies to which is valid for every possible combination of r and ∂∆xν dL any integration volume Ω. Due to this freedom, the L + δL + ∆xν 4x = 0, ν ν d integral can only vanish if the integrand vanishes as Ω ∂x dx Z well, and therefore we obtain the following R continuity d δL + (L∆xν) d4x = 0. (77) equations:7 dxν ZΩ µ(r) Notice that, by applying the Euler-Lagrange Equa- ∂µΘ = 0. (84) tions6 It is worth mentioning that these conserved currents ∂L ∂L also may imply the existence of conserved charges. Let δL = δφα + δφα;µ , ∂φα ∂φα;µ Θµ(r) = Θ0(r), Θ(r) . Then we may write Eq. (84) in d ∂L ∂L vector notation as = ν δφα + δφα;µ , dx ∂φα;ν ∂φα;µ ∂Θ0(r) d ∂L (r) 0 + ∇ · Θ = 0. (85) = ν δφα . (78) ∂x dx ∂φα;ν Integration over a volume V yields If we substitute this result in Eq. (77), it follows that d d ∂L Θ0(r) d3x = − ∇ · Θ(r) d3x , ν 4 x0 ν δφα + L∆x d x = 0. (79) d V V dx ∂φα µ Z Z ZΩ ; = − Θ(r) · da . (86) Eq. (79) can be put in a more convenient form if ∂V ν I we express δφα and ∆x in terms of the infinitesimal V parameters that characterize the transformation. Let us If is the entire three-dimensional space and the assume there are R such parameters, each one denoted fields vanish sufficiently fast as the coordinates grow towards infinity, the surface integral vanishes, implying by r with 1 6 r 6 R. Then we can write the global conservation of the quantities ν ν(r) ∆x = X r, (80) (r) 0(r) 3 (r) C = Θ d x , (87) ∆φα = Ψα r. Z which are known as Noether charges.

6 You might be wondering why we are using d/dxµ instead of ∂/∂xµ on the Euler-Lagrange Equations. If you take a look on our derivation of the equations, you will see that the derivatives 7 Since we are not differentiating anything while holding the fields we are considering there, albeit partial, do consider the indirect constant, it is possible to use the notation ∂µ again without any contributions due to the dependence of the fields on xµ. risk of confusion. 9

V. APPLICATIONS OF NOETHER’S THEOREM B. The Conservation of Electric Charge

A. The Stress-Energy Tensor In order to obtain the conservation of electric charge, let us consider the following Lagrangian As a first example, let us consider symmetry under a 1 L = − F Fµν + ψ¯ (iγµ∂ − m)ψ − eψγ¯ µψA . (93) space-time translation, i.e., 4 µν µ µ 0µ µ µ x = x + , (88) This Lagrangian describes the theory of Quantum µ Electrodynamics (QED)[6–8]. The term where is a constant four-vector. The fields remain unchanged. 1 − F Fµν (94) As µ is a constant, the Jacobian of the transforma- 4 µν tion is unity. Since the fields are unchanged, the action is the same we have seen in the Lagrangian for will be kept invariant under this transformation if the Maxwell’s Electrodynamics. The missing 4π is due to Lagrangian does not depend explicitly on the coordi- a new choice of units: when dealing with High Energy nates. Physics it is customary to adopt a system of units such Since that h = c = 1. The term µ µ(r) = X r, µ ψ¯ (iγ ∂µ − m)ψ (95) (r) (89) 0 = Ψ r, α gives rise to the Dirac equation for the motion of a µν µν (r) it follows that X = g and Ψα = 0. In this case, relativistic electron. Finally, the term r does have tensor character. If we substitute these µ −eψγ¯ ψAµ (96) expressions into Eq. (82), we get couples electrons and photons together, allowing them µν ∂L τν µν JµA T = φα;τg − Lg , to interact with each other, and matches the µ we ∂φα;µ had earlier, the sole difference being that we now are µ µ ∂L ∂φα µν dealing with a specific current given by J = −eψγ¯ ψ. = − Lg . (90) µ µ ν ∂φα;µ ∂xν γ are four 4 × 4 matrices satisfying {γ , γ } = −2gµν. A possible representation of these matrices The tensor T µν is called the stress-energy tensor or is the Weyl, or chiral, representation, given by energy-momentum tensor. The Noether charge associated to it is the four-vector 0 1 0 σ γ0 = , γi = i , (97) given by 1 0 −σi 0

where σi are the Pauli matrices. pν = T 0ν d3x . (91) Since γµ are 4 × 4 matrices, it is not that surprising Z that ψ is a 4-component spinor. We define ψ¯ = ψ†γ0. The zeroth component of T 0ν is given by Let now α(x) be some real function depending on the ∂L space-time coordinates. Consider the following trans- T 00 = − φ˙ + L, ∂φ˙ α formation ∂L A0 = A + ∂ α(x), = − φ˙ − L . (92) µ µ µ ∂φ˙ 0 −ieα(x) (98) α ψ = e ψ. This is in fact minus the Hamiltonian density, which Symmetries that are parametrized by a function such is used in the Hamiltonian treatment of Classical Field as α(x) are said to be gauge of local symmetries. On Theory[1,2]. It is then natural to say that T 00 is minus the other hand, if we had a constant α instead, we the energy density of the fields[5]. would call it a global symmetry. Notice that every gauge Therefore, p0 is, up to a signal, the integral of the symmetry implies a global symmetry. energy density of the fields, and therefore represent Notice that F is left unchanged by this transforma- the energy stored in the fields. Due to covariance µν tion: considerations and the knowledge that momentum is 0 0 0 preserved under spatial translations, we conclude that F µν = ∂µA ν − ∂νA µ, pi, i = 1, 2, 3 are the components of the fields’ momen- = ∂µAν + ∂µ∂να(x) − ∂νAµ − ∂ν∂µα(x), tum. This establishes pµ as the fields’ four-momentum and justifies the term energy-momentum tensor. = ∂µAν − ∂νAµ. (99) 10

1 µν Therefore, the term − 4 FµνF of the QED La- yields grangian is left unchanged by the transformation. Xν(r) = 0, In fact, the Lagrangian as a whole is left unchanged (102) Ψa = −ieψa, as well. Indeed, where a stands for the spinorial index of ψ. The Noether current is then given by8 1 L0 = − F0 F0µν + ψ¯ 0(iγµ∂ − m)ψ0 − eψ¯ 0γµψ0A0 , ∂L 4 µν µ µ Θµ = − Ψ, 1 ∂(∂µψ) µν ¯ +ieα(x) µ −ieα(x) = − FµνF + ψe (iγ ∂µ − m)e ψ ∂L 4 = ie ψ, +ieα(x) µ −ieα(x) ∂(∂ ψ) − eψe¯ γ e ψAµ µ +ieα(x) µ −ieα(x) = −eψγ¯ µψ. (103) − eψe¯ γ e ψ∂µα(x),

1 µν µ µ = − FµνF + ψ¯ (iγ ∂µ − m)ψ + eψγ¯ ψ∂µα(x) However, this is simply the electric current density 4 we chose at the beginning of this section, i.e., µ µ − eψγ¯ ψAµ − eψγ¯ ψ∂µα(x), µ µ 1 J = −eψγ¯ ψ. (104) = − F Fµν + ψ¯ (iγµ∂ − m)ψ − eψγ¯ µψA , 4 µν µ µ = L. (100)

Thus, Jµ is the Noether current associated with U(1) gauge symmetry, and we may conclude that Thus, we have found a gauge symmetry in our La- ∂ Jµ = 0. (105) grangian. This also implies a global symmetry by µ α(x) = α ∈ R ψ e−ieα choosing . Since is multiplied by , One should notice that this derivation was made α ∈ R with , the symmetry group associated to this possible when we coupled the electromagnetic field, gauge transformation is U(1)[9]. described by Aµ, to a matter field ψ responsible for carrying the electric charge. Furthermore, we could If we pick an infinitesimal α, we get the following also couple some more fields (corresponding, e.g., to transformation for the fields muons and taus) and obtain a more robust result. The current expression we have yields the conservation law for charge carried by electrons and positrons only and 0 A µ = Aµ, ignores any other charged particles. µ 0 (101) It is also worth mentioning that ψγ¯ ψ is the prob- ψ = (1 − ieα) ψ, ability four-current density for the Dirac Equation[10]. Therefore, the electric four-current density is nothing but the electron’s charge multiplied by its probability with the space-time coordinates left unchanged. This four-current density.

[1] N. Lemos, Analytical Mechanics (Cambridge University [3] J. D. Jackson, Classical Electrodynamics (Wiley, New York, Press, Cambridge, United Kingdom, 2018). 1999). [2] H. Goldstein, C. Poole, and J. Safko, Classical Mechanics [4] J. Frenkel, Princípios de Eletrodinâmica Clássica (EDUSP, (Addison Wesley, San Francisco, 2002). São Paulo, 2005). [5] M. A. M. Aguiar, “Tópicos de mecânica clássica,” (2019), unpublished. [6] M. E. Peskin and D. V.Schroeder, An Introduction to Quan- 8 One might do the calculation considering each spinorial compo- tum Field Theory (Westview Press, New York, 1995). nent of ψ and then see that formally diﬀerentiating with respect [7] M. Schwartz, Quantum Field Theory and the Standard Model to ∂µψ yields the same result. (Cambridge University Press, Cambridge, 2014). 11

[8] D. Tong, “Quantum ﬁeld theory,” (2007), unpublished. Let us write ei for the elements of the canonical basis, [9] L. A. Lessa, “Notes on gauge theories,” (2019), unpub- i.e., ei is the i-th column of 1. In a similar manner, let lished. us denote the i-th column of A by Ai. [10] W. Greiner, Relativistic Quantum Mechanics: Wave Equa- The determinant is a multilinear form on the columns tions (Springer-Verlag, Berlin, 2000). of J. Therefore,

det J = det (e1 + A1, e2 + A2, . . . , eN + AN) , Appendix: Determinant of a Near-Identical Matrix = det (e1, e2, . . . , eN) + det (A1, e2, . . . , eN)

+ det (e1,A2, . . . , eN) + ··· Theorem 1: 2 Let J = 1 + A be a N × N matrix. Then it holds that + det (e1, e2,...,AN) + O , 2 = 1 + (A11 + A22 + ··· + ANN) + O , det J = 1 + tr A + O2. 2 = 1 + tr A + O . Proof: