Revista Brasileira de Ensino de F´ısica, vol. 43, e20210075 (2021) Articles www.scielo.br/rbef c b DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0075 Licenc¸aCreative Commons

An alternative way to present a relativistic Lagrangian definition

M.A. De Andrade1, C. Neves*1

1Universidade do Estado do Rio de Janeiro, Faculdade de Tecnologia, Resende, RJ, Brasil.

Received on February 26, 2021. Revised on June 25, 2021. Accepted on June 28, 2021. As long we known, the non-relativistic Lagrangian definition, L = T − U with T and U as being, respectively, kinetic and potential energy, it is not valid at the relativistic level. This issue partially obstructs the college student’s comprehension of special theory of relativity from the Lagrangian formalism. Further, it is well known that the space-time world demands essentials principles of special theory of relativity: four-vectors and Lorentz invariance. In this work, we departure from these basics concepts of space-time world in order to get a relativistic Lagrangian definition parallel to the non-relativistic one and, as a consequence, this can help the college student to better understand special theory of relativity [1]. Keywords: Special Relativity, Lorentz Invariance, Four-vector in space-time, Lagrangian and Hamiltonian formalism.

1. Introduction how these concepts help to formulate the relativistic Lagrangian in a straightforward and transparent way It is well known that the Lagrangian formalism, as well in order to facilitate the college student to understand as, Hamiltonian one are important to describe classical special theory of relativity from the Lagrangian point systems, because these formalisms reveal many impor- of view. tant features, symmetries [2, 3] for example, and allow In order to get a self-contained work, it is organized to quantize them through some quantization processes: as follows: in section 2 the basics of four-vectors and canonical [4] and path integral [5] quantization methods the importance of Lorentz invariance are reviewed. for instance. However, when it is considered relativistic After that, departing from this review, the relativistic systems, the usual non-relativistic Lagrangian definition Lagrangian of the free particle is obtained in a very clean (L = T − U) is not extended to the relativistic one, and straightforward way. In section 3, the steps that as the Hamiltonian formalism is, i.e., H = T + U. led us to obtain the relativistic Lagrangian of the free It happens because, unlike Hamiltonian, Lagrangian is particle given in the end of the section 2 are extended defined in the configuration space. This issue embraces to explore the dynamics of a generic interacting system. some concepts that are difficulty to the college student In section 4, it is considered a particle in a background to understand. Among others things, this is a point that electromagnetic field in order to illustrate how this pre- we want to clarify in this work. viously alternative relativistic Lagrangian formulation The relativistic Lagrangian formalism is usually pre- is more effective than the usual one. Consequently, it sented, through fundamental books in college, in an is possible to explore a parallel between the relativistic opaque way due to some assumptions considered and it and non-relativistic Lagrangian definitions, indeed, it is is not possible to do a parallel between non-relativistic obtained a relativistic Lagrangian definition parallel to and relativistic Lagrangian definitions. Precisely, the L = T −U. At the end, some conclusions and discussions non-relativistic Lagrangian definition, L = T − U, is are presented. not valid at the relativistic level. Consequently, this obliterates the student to better comprehend the special theory of relativity and, indeed, some gaps arise in the 2. Basics on Four-Vectors of the Special student comprehension of the special theory of relativity Relativity in the Lagrangian framework. In order to fill this gap and to promote a better way to teach special theory We start this Section describing the notation used here. of relativity from the Lagrangian formalism, at college The Greek indices indicate the four-dimensional space- level, we start to discuss the concepts and importance of time with the coordinates xµ = x0, xi
while the Latin four-vector and Lorentz invariance. After that, we show indices indicate the standard three-dimensional space with i = 1, 2, 3. So the x0 and xi refer to the timelike * Correspondence email address: cliff[email protected] and the spacelike components, respectively.

Copyright by Sociedade Brasileira de F´ısica.Printed in Brazil. e20210075-2 An alternative way to present a relativistic Lagrangian definition

We would like to do a brief comment about the metric Whenever the motion of the particle is taken in definition. Consider a N-dimensional space (V N ) with consideration, space-time comes into play, then we must no special structure in which tensors play a role are use the metric tensor ηµν . With the intervention of µ metric spaces, i.e., they possess a rule which assigns the metric tensor, an additional condition for Λ ν , or “distances” to pairs of neighboring points. In particular, for the corresponding transformation matrix Λ, can be one calls a space (pseudo-)Riemannian if there exists an found in a way to provide invariant scalar forms in invariant quadratic differential form, namely, four-dimensional space-time, like the distance in the three-dimensional space. With the aim to simplify the 2 a b ds = gabdx dx , (1) procedure, we can employ the usual notation in which the space-time metric tensor (η ) and its inverse (ηµν ) with a, b = 1, 2,...,N and the g’s are generally func- µν can be used to lower or raise indices. Thus, using the tions of position and are subject only to the following inverse of the metric tensor to write the four-differentials restriction: det(g ) 6= 0. As it is required that ds2 is ab on both sides of the Eq. (5) with their indices lowered, invariant, it follows that g must be a tensor, which ab as well as in dx0µ = ηµαdx0 , it can be verified that is called the metric tensor, while Eq. (1) is the metric. α In pseudo-Riemannian spaces such as Minkowski space, 0 −1 ρ dxν = (ηΛη )ν dxρ. (10) the square of a vector, given in Eq. (1), can be a positive 0 or a negative real number and Eq. (1) is rewritten as So that dxν is covariant four-differential, whose contrac- tion with its contravariant partner dx0ν results in an 2 µ ν ds = ηµν dx dx , (2) invariant scalar quantity, the transformation matrix in Eq. (10) must be the same as that found for the four- with the metric tensor ηµν , that reflects the nature of derivative operator in Eq. (9), therefore the space-time, as well as its inverse ηµν , reads as ηΛη−1 = Λ−1 T or ΛT ηΛ = η. (11) µν ηµν = η = diagonal (+, −, −, −), (3) The transformation of the four-differential performed by µ and, consequently, it is get the following identity, the Λ ν , which satisfies the Eq. (11), is called a Lorentz transformation1. Therefore, it assures the invariance of the quadratic form, i.e., the invariance of the metric, µν µ η ηνλ = δλ , (4) namely: µ 0µ 0 µ with δλ as the Kronecker delta. dx dxµ = dx dxµ = invariant. (12) The transformation law of a contravariant tensor, in particular a component four-differential dxρ, from A generic contravariant four-vector must have the one coordinate system {xµ} to another, {x0µ}, is transformation matrix satisfying the Eq. (11) and its expressed by transformation law, as that of the four-differential in Eq. (5), then 0µ µ ρ dx = Λ ρdx , (5) 0µ µ ν A = Λ ν A . (13) where, by the chain rule of differentiation, we have A generic covariant four-vector must have the transfor- ∂x0µ mation matrix satisfying the Eq. (11) and its transfor- Λµ = . (6) mation law, as that of the four-derivative in Eq. (9), ρ ∂xρ 0 −1
ρ By using, again, the chain rule of differentiation, we can Bµ = Λ µBρ. (14) take the following trivial identity In such a way that the contraction of a contravariant 0µ ρ 0µ ρ ∂x ∂x ∂x four-vector with other covariant one, and vice-versa, Λµ Λ−1
= = = δµ. (7) ρ ν ∂xρ ∂x0ν ∂x0ν ν results in a Lorentz invariant scalar: 0µ 0 ν From which, we can read A Bµ = A Bν . (15)

ρ ∂x ρ Any four-vector can be split into timelike and spacelike = Λ−1
. (8) ∂x0ν ν components and the metric tensor can be used to lower the index as After the contraction of both sides of the Eq. (8) with the ∂ Aµ = A0, A
and A = η Aν = A0, −A
, (16) four-derivative operator , we get the transformation µ µν ∂xρ law of the four-derivative operator as being 1 Lorentz transformation encompasses a variety of transformations like spatial rotations [3], Lorentz boosts [6], discrete symmetries ∂ ρ ∂ ρ = Λ−1
or ∂0 = Λ−1T
∂ , (9) transformations, let it be said that not all Lorentz-invariant laws ∂x0ν ν ∂xρ ν ν ρ are expressible as relations between 4-vectors and scalars; some require 4-tensors and, in quantum mechanics, spinors transforma- where the superscript T means the transpose operation. tions, which is a representation of Lorentz-invariant laws [7].

Revista Brasileira de Ensino de F´ısica, vol. 43, e20210075, 2021 DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0075 Andrade and Neves e20210075-3 where A0 is the timelike component and A is the vector obtained with the help of Eq. (20), Eq. (21) and Eq. (22). assembled with the spacelike components Ai. If we go Now, consider another reference frame S0, where this 0 0 0 ahead and write that Aµ = (A0,Ai) then A0 = A and particle is momentarily at rest(v = 0). In S the Lorentz i 0 Ai = −A . Thus, the invariant scalar of the Eq. (15) factor is γ = 1 and, due to Lorentz invariance of the takes the following form Eq. (23), the Einstein second postulate [1, 6], v0 = v00 = c, is justified. 0 0 A0 B0 − A0 · B0 = A0B0 − A · B. (17) The four-momentum of a free particle with rest mass m is defined by The four-derivative operator (∂µ) is split into time- like and spacelike components according the identity pµ = muµ = γmvµ. (24) µ µ ∂ν x = δν , which can be verified without the interven- tion of the metric tensor in a similar way to Eq. (7) with Similar to what happens to the four-velocity uµ, a the interchange of Λ and Λ−1. So, for the four-derivative general Lorentz boost on the four-momentum pµ changes operator, it’s worth from one reference frame S, where the particle has an ordinary momentum p = γmv, to another S0, where the µ µν ∂µ = (∂0, ∇) and ∂ = η ∂ν = (∂0, −∇) . (18) particle is momentarily at rest p0 = 0. Thus, we have at our disposal two different Let dxµ to be the infinitesimal four-displacement of four-vectors associated with the particle, the four- the particle, which obviously behaves as a contravariant displacement dxµ and the four-momentum pµ that can vector under Lorentz transformation. Splitting dxµ into be combined to form three Lorentz invariant scalars: timelike and spacelike components as follows µ µ µ (i) dx dxµ; (ii) p pµ; (iii) p dxµ. dxµ = dx0, dr
, (19) We can assemble the first Lorentz invariant quadratic scalar by combining the four-displacement dxµ = vµdt where dx0 = cdt, with c as being the speed of light in the with its covariant partner: vacuum and dt as the infinitesimal time interval spent by 2 2 the particle to move along the infinitesimal displacement 2 µ µ 2 c dt ds = dx dxµ = v vµdt = 2 , (25) dr, whose components are dxi. The time derivative of the γ four-position of the particle will be represented by vµ, where it was used the identity given in Eq. (21). The dxµ quantity ds is the infinitesimal interval [6], which can be vµ = = (c, v) , (20) dt written as r where v = dr/dt is the ordinary velocity of the particle; cdt v2 ds = = cdt 1 − = cdτ. (26) vµ isn’t a four-vector because its contraction with its γ c2 2 partner vµ depends on v ≡ v ·v, that isn’t a Lorentz invariant quantity: So, from the Lorentz invariance of ds, we get 0  v2  c2 dt dt vµv = c2 − v2 = c2 1 − = , (21) = = dτ, (27) µ c2 γ2 γ γ0 where γ is the Lorentz factor where dτ is the infinitesimal time interval that is the numerator of a supposed fraction whose denominator is 1 γ00 = 1. Thus, this time interval was measured in the γ = r . (22) v2 reference frame where the particle is momentarily at rest. 1 − c2 So, dτ is called the infinitesimal proper time. As it can be understood from Eq. (27), time intervals have different Obviously, a genuine four-vector with the dimension of values if measured in different reference frames, therefore velocity which undergoes a Lorentz transformation like time is a relative quantity, that is the essence of the µ a contravariant vector is γv , because its contraction special relativity. with its covariant partner, according to Eq. (21), yields After that, combining pµ = γmvµ with its covariant 2 the Lorentz invariant quantity c . So we can define the partner, using the Eq. (21), we can write the second µ four-velocity u as Lorentz invariant quadratic scalar as

uµ = γvµ ⇒ uµu = c2. (23) µ 2 µ p pµ = (mc) . (28)

Here, we would like to point out that the notation The four-momentum of the particle split into timelike for four-velocity follows two basic text books [3, 8]. and spacelike components can be read as Moreover, we would to explore the relation given in Eq. (23). First, consider a particle in a reference frame H  pµ = free , p , (29) S with a non null arbitrary velocity v, then Eq. (23) is c

DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0075 Revista Brasileira de Ensino de F´ısica,vol. 43, e20210075, 2021 e20210075-4 An alternative way to present a relativistic Lagrangian definition

2 where Hfree = γmc is the energy of the free particle. can always express a genuine four-vector in terms of So, we can get a Lorentz invariant quadratic scalar as “non-four-vectors”, namely: uµ = γvµ, where uµ is an authentic four-velocity and they can evolve during 2 µ Hfree 2 calculation in a way that, sometimes, is useful or subtle, p pµ = − p . (30) c2 for example, as in the expression obtained in Eq. (33) Then, by comparing Eq. (28) with Eq. (30), the energy- for the Lagrangian of the free particle. momentum relation is obtained as being The technique used here is to assume that the timelike and spacelike components of vµ are variable quantities H2 and, in the end, to submit its timelike component to the free − p2 = (mc)2. (31) c2 Einstein’s second postulate [1]: v0 = c = constant. In order to extend the procedure described in the It has already been calculated the quadratic Lorentz µ µ previous section for an interacting system, it is con- invariants by combining dx and p with their respective sidered a particle in a generic background field. This covariant partners. The next task will be obtain the third interacting system has its dynamics governed by a Lorentz invariant scalar by combining the two unique Lagrangian (L), which must have among its terms different four-vectors available for the free particle, µ µ the one that corresponds to the interaction in which p = mu with dxµ = uµdτ. Then using the Eq. (23), µ 2 the particle is subjected. Therefore, the timelike and u uµ = c , and Eq. (26), expressed as ds = cdτ, spacelike components of the generalized four-momentum we obtain P µ of an interacting system [6] are defined by the 2 µ following operation p dxµ = mcds = −dSfree = −Lfreedt, (32)

0 ∂L ∂L where dSfree is the Lorentz invariant infinitesimal action P = − 0 and P = . (35) for the free particle [6], which means that the Physics ∂v ∂v can not depend on the reference frame. The Lfree is the By inspection of Eq. (35) and with P µ = (P 0, P ) and corresponding Lagrangian for the free particle, which by 0 vµ = (v , −v), the generalized four-momentum of an the use of dxµ = vµdt in Eq. (32), it renders to interacting system is written as µ Lfree = −p vµ. (33) ∂L P µ = − . (36) µ ∂vµ Therefore, eliminating p in the expression for Lfree in Eq. (33) through the use of the four-momentum of µ µ With the aim to approach the dynamics of interacting the free particle given in Eq. (24), p = γmv , then system, we can start from the Legendre transformation employing the Eq. (21), we get which, in general, links the Lagrangian L to the Hamil- mc2 tonian H [6], namely, Lfree = − . (34) γ ∂L H = · v − L. (37) As γvµ = uµ and from Eq. (33), it is obtained that ∂v µ γLfree = −p uµ, which is a Lorentz invariant scalar From Eq. (29), it can be inferred the relation H = µ free because both the four-momentum p and four-velocity p0c and, consequently, the energy-momentum relation. uµ are genuine four-vectors, the same result can be Then, based on covariance, in general, the Hamiltonian verified from Eq. (34). So, Lfree changes with the of the interacting system can be defined as being changing of the reference frame. This result is valid for any relativistic Lagrangian, i.e., a relativistic Lagrangian H = P 0c = P 0v0. (38) never will be a Lorentz invariant scalar. On the other hand, a non-relativistic Lagrangian is always a Galilean Therefore, from Eq. (35) and Eq. (38), the Legendre invariant scalar, because the time interval dt is a Galilean transformation in Eq. (37) can be expressed in a compact invariant scalar. way as

0 0 3. Dynamics of Interacting System L = −P v + P · v, (39) by Four-Vectors I that is,

In general, when we assemble a four-quantity in a naive µ L = −P vµ. (40) way, gathering the ordinary vector of three-dimensional space in the simplest way, as an example, vµ = (v0, v), ∂L where v is the ordinary three-dimensional velocity, this 2 The symbol represents the vector whose components are the ∂v one obviously will not transform like a Lorentz vector. derivatives of L with respect to the corresponding components of However, it can be useful for calculations, because we v [6].

Revista Brasileira de Ensino de F´ısica, vol. 43, e20210075, 2021 DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0075 Andrade and Neves e20210075-5

Therefore, with the use of the four-vectors, the relation 4. Dynamics of Interacting System between the Lagrangian of the interacting system and by Four-Vectors II the generalized four-momentum takes a very simple form, which resembles Lfree, given in Eq. (33). The vµ In order to illustrate the previous proposed procedure, derivative of both sides of the Eq. (40) is an interacting system is assumed. Consider, as example, a particle with rest mass (m) and electric charge (e) in  ν  ∂L ∂P µ a background electromagnetic four-potential (Aµ) split = − vν − P . (41) ∂vµ ∂vµ in timelike and spacelike component as

Then, in the relativistic case, the compatibility between Aµ = (A0, A), (46) the formal definition of generalized four-momentum 0 given in Eq. (36) and the Legendre transformation in in which A and A are, respectively, the scalar and the compact four-dimensional form given in Eq. (40) is vector potentials, whose connection with the three- achieved through the validity of the following constraint dimensional electric (E) and magnetic (B) fields are given by  ν  ∂P 1 ∂A vν = 0. (42) E = −∇A0 − and B = ∇ × A, (47) ∂vµ c ∂t Note that Eq. (36) is restored when Eq. (42) is replaced which can be explored in some basics references [3, 6–8]. in Eq. (41). The standard relativistic Hamiltonian for this system, On the other hand, we can departure from the such as the one found in Ref. [6], is expressed in anal- compact Legendre transformation given in Eq. (40), ogy to the non-relativistic Hamiltonian by adding the 0 multiplying both its sides with the infinitesimal time contribution of potential energy (eA ) to the relativistic 3 interval dt to obtain an expression, similar to the one Hamiltonian of free particle, namely: for the free particle given in Eq. (32), in this case for the H = γmc2 + eA0, (48) Lorentz invariant infinitesimal action of the interacting system, which can be interpreted as being energy only for static situations4, i.e., electromagnetic field does not depend µ dS = Ldt = −P dxµ, (43) on the time and, consequently, the Lagrangian for the electric charge particle also does not depend explicitly where dxµ = vµdt is the infinitesimal four-displacement. on the time and the energy is the Hamiltonian given in Note that the Lagrangian L is a non-Lorentz invariant Eq. (48). scalar, as well as the infinitesimal time interval dt. How- With the intention to obtain the relativistic ever the product of both provides a Lorentz invariant Lagrangian for an interacting system, we must follow scalar, because Ldt = (γL) dτ, where both γL and the recipe described in the end of the Section 3; for an proper time interval dτ are Lorentz invariant scalars. interacting system the relativistic Hamiltonian is given Thus, from Eq. (43), the generalized four-momentum by H = P 0c, thus the division of both sides of Eq. (48) can be also expressed in terms of the action functional by c results in S as e P 0 = γmc + A0. (49) ∂S c P µ = − , (44) ∂xµ As P 0 is the timelike component of the four-vector P µ and by the covariance of all the components of P µ, it whose timelike and spacelike components can be read as can be induced that ∂S µ µ e µ P 0 = − and P = ∇S, (45) P = γmv + A , (50) ∂x0 c which can be used together with the equation for 3 On the relativistic Lagrangian L, which must always be expressed Hamiltonian, Eq. (38), to obtain the Hamilton-Jacobi in terms of the ordinary, time, position and velocity, we must use the standard Lorentz factor given in Eq. (22). However, for the equation for an interacting system [6]. Hamiltonian H to be well defined, it must always be expressed in The recipe to obtain the relativistic Lagrangian L term of the ordinary, time, position and momentum; in this case, through the Hamiltonian H is straightforward: by know- the alternative form of the Lorentz factor, written in terms of the ing that H = P 0c, we can identify P 0 as a function of v0 momentum of a free particle (p), reads as r which, by the covariance of all the components, leads us p2 γ = 1 + . to P µ as a function of vµ. The Lagrangian can then be (mc)2 found through the expression given in Eq. (40), which 4 Vide chapter 3, section 19, of Ref. [6] and for a nice discussion can be expressed in the usual shape after the use of the about how Hamiltonian plays the role of energy, see chapter 2, identity given in Eq. (21). section 2 and 3, of Ref. [8].

DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0075 Revista Brasileira de Ensino de F´ısica,vol. 43, e20210075, 2021 e20210075-6 An alternative way to present a relativistic Lagrangian definition where P µ is the generalized four-momentum of the one as electromagnetic field. Note that the first term in the T v right hand size of the previous expression is the four- L = −mc2 + − U + eA · , (56) momentum pµ of the free particle. Inserting Eq. (50) γ c into Eq. (40), we get the relativistic Lagrangian for the where the potential energy was represented by U = interacting particle, eA0. In contrast with the relativistic Hamiltonian of e interacting particle, given in Eq. (48), we get L = −γmvµv − Aµv . (51) µ c µ H = mc2 + T + U. (57) Note that γL is a Lorentz invariant scalar. The Lagrangian in Eq. (51) can be simplified by using In the extreme non-relativistic regime, time is an abso- Eq. (21), which renders to lute quantity, as in Galilean relativity and Newtonian Mechanics. Consequently, all the Lorentz factors in the mc2 e L = − + A · v − eA0. (52) denominators of Eq. (27) must be always equal to one, γ c independently of the velocity (v). Therefore, from the Eq. (27) and Eq. (22), we get5 An alternative and simplest way to obtain the gener- alized four-momentum via Lagrangian is to express the 1 γ = 1, c = ∞ and T = mv2. (58) Lagrangian with all its terms multiplied by the common 2 factor vµ, as in Eq. (51), and to obtain the generalized four-momentum through the comparison with Eq. (40). In this case, we must perform an operation like a “renor- For the same task, we can also employ the derivative malization” in order to eliminate the constant infinity 2 in Eq. (36), which seems more familiar, but it can only energy mc , i.e., absorbing it into the Lagrangian, Eq. (56), and Hamiltonian, Eq. (57), in such a way that be done considering that v0 is a variable quantity, as it was assumed in the beginning of the Section 3. With the corresponding redefined non-relativistic Lagrangian the aim to justify the validity of these two procedures and Hamiltonian can be written as [2] to obtain the generalized four-momentum in the scope L = T − U, (59) of electromagnetism, it appears that from Eq. (50), it is possible to get an equation similar to Eq. (32) H = T + U. (60) when interacting particle is considered; if there is an electromagnetic field interacting with a particle, then We would like to emphasize that any relativistic Lagrangian can never be presented in the simple shape µ e µ like that of the non-relativistic one in Eq. (59). In fact, P dxµ = mcds + A dxµ = −dS = −Ldt, (53) c they have very different shapes, because in the latter, we where dS and L are, respectively, the infinitesimal action should subtract the rest energy, consider the extra factor and Lagrangian for the particle in the electromagnetic 1/γ on the kinetic energy and add a term that consists of field [6]. a scalar product of vectors to make it equal to that of the Next, to confirm the equivalence of both methods to Eq. (56). On the other hand, the relativistic Hamiltonian get the generalized four-momentum, one through the in Eq. (57), except for the addition of the rest energy, Eq. (36) and the other by Eq. (40), the constraint given has its shape very similar to the non-relativistic one in in Eq. (42) can be verified in the case of the electro- Eq. (60). magnetic field, vide Appendix A. So, the compatibility between the Eq. (36) and Eq. (40) is consolidated. There- 5. Conclusion fore, both ways of proceeding are equally legitimate and they provide the same result for P µ, which is the one It is well known that the Lagrangian, at the non- that is induced by covariance in Eq. (50). relativistic level, is defined as being L = T − U. On the other hand, the kinetic energy T is known as However, this definition is not valid at the relativistic being level, indeed, it was shown in this work. According to Eq. (56), the part of the relativistic Lagrangian that T = H − mc2 = (γ − 1)mc2. (54) T free corresponds to the non-relativistic one is − U. Due γ In order to evidence the kinetic energy in the expression to this, it is also well known that this issue is not of the Lagrangian in Eq. (52), we can write it as so clear to the college student and, consequently, the understanding of special theory of relativity from the (γ − 1)mc2 e L = −mc2 + + A · v − eA0. (55) Lagrangian formalism is jeopardized. In this work it γ c was presented an alternative procedure, even when an So, the relativistic Lagrangian of the interacting particle can be expressed in a similar way to the non-relativist 5 For the kinetic energy T vide Appendix B.

Revista Brasileira de Ensino de F´ısica, vol. 43, e20210075, 2021 DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0075 Andrade and Neves e20210075-7 interacting system is considered, to teach special theory Then, we get again, for the specific case of the electro- of relativity from the Lagrangian formalism based on magnetic field, the constraint of the Eq. (42), the four-vectors and Lorentz invariance. This demon-  ν  strated, once again, the profound importance of these ∂P vν = 0. (A7) fundamental concepts of special theory of relativity. ∂vµ Here, it was explained in a detailed way the meaning and the origin of proper time, the energy-momentum relation and action; all of these three quantities are due B: The relativistic kinetic energy to the four-vector framework and Lorentz invariance. In order to illustrate this procedure, it was considered The relativistic kinetic energy can be algebraically the free particle and a charged particle in a background manipulated as electromagnetic field. Consequently, it was possible to 2 make a parallel between the definitions of relativistic T = (γ − 1) mc (A8) and non-relativistic Lagrangian definitions. γ2 − 1 = mc2 γ + 1 Acknowledgments  2   2  γ γ − 1 2 = 2 mc C. Neves and M. A. De Andrade thank to Brazilian γ + 1 γ Research Agencies (CNPq and FAPERJ) for partial  γ2   1  = 1 − mc2 financial support. Further, we would like to thank the γ + 1 γ2 Reviewers for a careful reading of the manuscript and  γ2    v2  the valuable suggestions. = 1 − 1 − mc2, γ + 1 c2

Supplementary material  γ2  T = mv2. (A9) γ + 1 A: The four-momentum of the electromagnetic field In the extreme non-relativistic regime, when we take γ = 1 and mc2 = ∞ in Eq. (A8), it results in 0 · ∞, which is a value that can’t be calculated. However, we Starting from the standard generalized four-momentum can change from this indeterminate form to a determined of the electromagnetic field one through the algebraic manipulation shown above. e P ν = γmvν + Aν . (A1) Therefore, if we take γ = 1 in the equivalent Eq. (A9), we c will obtain the usual expression for the non-relativistic Knowing that the electromagnetic four-potential, Aν , kinetic energy. depends only on the four-position, the vµ derivative of the generalized four-momentum is References ∂P ν  ∂γ  = mvν + γmηµν . (A2) [1] H.A. Lorentz, A. Einstein and H. Minkowski, The Prin- ∂v ∂v µ µ ciple of Relativity (Dover Publications, New York, 1952). It follows that [2] L. Landau and E. Lifshitz, Mechanics (Butterworth-  ν    Heinemann, Oxford, 1976). ∂P ∂γ ν µ vν = mv vν + γmv . (A3) [3] H. Goldstein, C. Poole and J. Safko, Classical Mechanics ∂vµ ∂vµ (Pearson Education, Essex, 2011), 3a ed. c2 [4] P.M. Dirac, The Principles of Quantum Mechanics ν (Clarendon Press, Gloucestershire, 1982), 4a ed. However, from Eq. (21) it is known that v vν = 2 , thus γ [5] R.P. Feynman and A. Hibbs, Quantum Mechanics and ∂P ν   ∂γ  mc2 Path Integrals (McGraw-Hill, New York, 1965). v = + γmvµ. (A4) ∂v ν ∂v γ2 [6] L. Landau and E. Lifshitz, The Classical Theory of Fields µ µ (Butterworth Heinemann, Oxford, 1975), 4a ed. ν −1/2 [7] J.J. Sakurai, Advanced Quantum Mechanics (Addison- Since γ = c(v vν ) , then its vµ derivative can be obtained as Wesley Publishing Company, Reading, 1967). [8] A.O. Barut, Electrodynamics and Classical Theory of µ ∂γ µ ν −3/2 3 v Fields & Particles (Dover publications, New York, 1980). = −cv (v vν ) = −γ 2 . (A5) ∂vµ c Using the previous result in Eq. (A4),

 ν   µ  2 ∂P 3 v mc µ vν = −γ 2 2 + γmv . (A6) ∂vµ c γ

DOI: https://doi.org/10.1590/1806-9126-RBEF-2021-0075 Revista Brasileira de Ensino de F´ısica,vol. 43, e20210075, 2021