Relativistic Dynamics

15 Relativistic Dynamics 15.1 Lagrange Equations Let us consider a free particle, that is, a particle that is not under the influence of any forces. The equations describing the motion of this particle are obtained using the variational principle. The action integral is defined by (11.4), that is,  I=

SB

SA

   SB   dxn ds = L xn , L xn , un ds, ds SA

(15.1)

where un is the four-velocity of the particle. The equations of motion of the particle are obtained using the variational principle, that is, the condition that the variation of the action integral is equal to zero, δI = 0. The Lagrange equations of motion of the particle are given by (11.12) in the form ∂L d − n ∂x ds



∂L ∂un



=0

(n = 0, 1, 2, 3).

(15.2)

Let us now define the form of the Lagrangian function for a free particle. Equations (15.2), derived using the variational principle applied to the action integral (15.1), are equal in all inertial systems of reference and the action integral is a scalar invariant with respect to the Lorentz transformations (13.29). Thus, the Lagrange function L itself is also a scalar invariant. Due to the homogeneity of the four-dimensional space-time in the inertial frames of reference, the Lagrange function cannot be a function of space-time coordinates xn . Thus, we may write L = L(un )

(n = 0, 1, 2, 3).

(15.3)

The only invariant which can be created using the four-velocity vector un is given by un un = gkn uk un = 1

(k, n = 0, 1, 2, 3).

(15.4)

Thus, the action integral of a free particle is simply proportional to the arc length in the four-dimensional space-time manifold, that is,  I = −m c

SB SA

 ds = −m c

SB

SA



gkn

dxk dxn ds, ds ds

(15.5)

where m is a mass parameter of the free particle. From (15.5) we see that the Lagrangian function is given by the expression L = −m c

 gkn uk un .

Tensors, Relativity, and Cosmology. http://dx.doi.org/10.1016/B978-0-12-803397-5.00015-7 Copyright © 2015 Elsevier Inc. All rights reserved.

(15.6)

111

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TENSORS, RELATIVITY, AND COSMOLOGY

Using the Lagrangian (15.6) we may write



1/2 ∂L ∂ k j = u u −m c g kj ∂un ∂un

−1/2

 1 = − m c gkj uk uj gkn uk + gnj uj . 2

(15.7)

Using here gkj uk uj = 1 and the symmetry of the metric tensor gnj = gjn we obtain ∂L 1 = − m c 2 gkn uk = −m c gkn uk = −m c un . ∂un 2

(15.8)

Thus, we may write −

d ∂L dun . = mc n ds ∂u ds

(15.9)

Using the Lagrangian (15.6) we may also write ∂L = 0. ∂xn

(15.10)

Substituting (15.9) and (15.10) into the Lagrange equations (15.2), we obtain dun d = (m c un ) = 0. ds ds

mc

(15.11)

The equations of motion of a free particle are then given in the following form: dun =0 ds

wn =

(n = 0, 1, 2, 3).

(15.12)

From (15.12) we see that a free particle moves along a straight line with constant velocity. Let us now investigate the non-relativistic limit of the action integral (15.5) when the particles are moving with low velocities (v  c). If we substitute (13.37) into (15.5) we obtain  I = −m c2



tB

1−

tA

v2 dt = c2



tB

  LT xα , vα dt,

(15.13)

tA

where LT is a Lagrangian defined with respect to the non-scalar time variable t, given by 

α

LT x , v

α





= −m c

2



v2 1 − 2 = −m c2 c

1+

gαβ vα vβ c2

(15.14)

In the non-relativistic limit of particles moving with low velocities (v  c), we may use the approximation 

1−

v2 v2 ≈ 1− . 2 c 2 c2

(15.15)

Substituting (15.15) into (15.13) we obtain  I=

tB

tA

  1 −m c2 + m v2 dt. 2

(15.16)

Chapter 15 • Relativistic Dynamics

113

Thus, the non-relativistic approximation of the Lagrangian LT can be written as follows:   1 LT xα , vα ≈ m v2 − m c2 . 2

(15.17)

As the non-relativistic Lagrangian is defined up to an arbitrary additive constant, the second term in (15.17) does not contribute to the non-relativistic equations of motion and can be dropped in the non-relativistic applications. Its physical significance in the special theory of relativity will be discussed later in this chapter. Thus, the action integral (15.5) has the proper non-relativistic limit.

15.2 Energy-Momentum Vector 15.2.1 Introduction and Definitions In the non-relativistic mechanics there is a number of constants of motion. These include the energy and the momentum of the particle. The objective of this section is to define the energy and the momentum of a particle in the framework of the special theory of relativity. Let us start with the action integral (15.13), in the form  I=

tB tA

  LT xα , vα dt.

(15.18)

The Lagrange equations of motion of a particle , analogous to (11.12), defined in terms of the Lagrangian LT with the non-scalar time variable t as the parameter, are given by ∂LT d − ∂xα dt



∂LT ∂vα



=0

(α = 1, 2, 3).

(15.19)

Let us now calculate the total time derivative of the Lagrangian LT as follows: ∂LT dxα ∂LT dvα dLT = + α . α dt ∂x dt ∂v dt

(15.20)

Using here the equations of motion (15.19) and the definition of the three-dimensional velocity (13.33), we obtain dLT d = dt dt



∂LT ∂vα



vα +

d ∂LT dvα = ∂vα dt dt

The result (15.21) can be rewritten in the form d dt



∂LT α v − LT ∂vα



 ∂LT α v . ∂vα

(15.21)

 = 0.

(15.22)

Thus, the quantity E=

∂LT α v − LT = Constant ∂vα

(15.23)

is a constant of motion. The quantity (15.23) can be recognized from the non-relativistic mechanics as the total energy of the particle. Let us also define the momentum pα , conjugate to the coordinate xα , as follows:

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= p

where

∂LT ∂LT ∂LT ⇒ pα = −g αβ ⇒ pα = − α ∂ v ∂vβ ∂v

  pα = px , py , pz ,

(15.24)

  pα = −px , −py , −pz

(15.25)

Substituting (15.24) into (15.23) we obtain  · v − LT . E = −pα vα − LT = −gαβ pα vβ − LT = p

(15.26)

Using the definition (15.14) of the Lagrangian LT , we can calculate pα = −

∂ ∂LT = m c2 α ∂vα ∂v

 1/2 gνβ vν vβ 1+ , c2

(15.27)

or pα = m c2

1 2



gνβ vν vβ c2

1+

−1/2

 1 ∂  gνβ α vν vβ 2 ∂v c

(15.28)

Using the symmetry of the metric tensor gνβ = gβν , we obtain pα = m

1 2

 −1/2 gνβ vν vβ 1+ 2 gαβ vβ . c2

(15.29)

Using vα = gαβ vβ , we finally obtain m vα m vα pα =  ⇒ pα =  . 2 2 1 − vc2 1 − vc2

(15.30)

Substituting (15.30) and (15.14) into (15.26), we obtain the total energy of the particle as follows: m vα vα E = − + m c2 2 1 − vc2



v2 m v2 1− 2 =  + m c2 2 c 1 − vc2



1−

v2 c2

   m c 2 − v2 m c2 v2 =  −  + m c2 1 − 2 2 2 c 1 − vc2 1 − vc2 m c2 =  − m c2 v2 1 − c2





v2 1 − 2 + m c2 c

1−

v2 m c2 =  . 2 2 c 1 − vc2

(15.31)

Thus, the total energy of the particle is given by m c2 E=  . 2 1 − vc2

(15.32)

The particle at rest with v = 0 has the so-called rest energy E0 given by E0 = m c 2 .

(15.33)

Chapter 15 • Relativistic Dynamics

115

The kinetic energy of the particle is obtained if the rest energy (15.33) is subtracted from the total energy (15.32). Thus, we obtain ⎛

EK = E − E0 = m c 2 ⎝ 

⎞

1 1−

v2 c2

− 1⎠ .

(15.34)

For the particles moving with low velocities (v  c), we may write 1  1−

≈ 1+

v2 c2

v2 . 2 c2

(15.35)

Substituting (15.35) into (15.34) we obtain EK =

1 m v2 . 2

(15.36)

In the same approximation (v  c), the components of the momentum of the particle are given by pα = m vα .

(15.37)

The approximate results (15.36) and (15.37) are the well-known non-relativistic expressions for the kinetic energy and the momentum of a moving particle, respectively.

15.2.2 Transformations of Energy-Momentum The results for the momentum of the free particle (15.30) and the energy of the free particle (15.32) were derived from the Lagrangian LT , defined with respect to the non-scalar time variable t, in a non-covariant way. Now we want to define the energy and momentum as the constants of motion using the covariant Lagrangian L given by (15.6). In analogy with the definitions (15.24), we now define the energy-momentum four-vector pn with following covariant and contravariant components: pn = −

∂L ∂L = m c un ⇒ pn = − = m c un ∂un ∂un

(15.38)

From (15.11) we see that the four components of the energy-momentum of a free particle satisfy the equations dpn =0 ds

(n = 0, 1, 2, 3).

(15.39)

Thus, for a free particle the four components of the energy-momentum four-vector are conserved. Using here (13.39), we obtain the temporal zeroth component of the energy-momentum vector, which is proportional to the energy E of the particle, in the form E mc p0 = m c u0 =  = , c v2 1 − c2

(15.40)

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TENSORS, RELATIVITY, AND COSMOLOGY

where the energy of the particle is given by (15.32). Furthermore, using (13.40), we obtain the three spatial components of the energy-momentum four-vector, which are equal to the components of the three-dimensional momentum of the particle pα given by (15.30), that is, m vα pα = m c uα =  . 2 1 − vc2

(15.41)

From the results (15.40) and (15.41), we see that the energy and momentum of a particle are not two independent quantities as in the non-relativistic mechanics. In the relativistic mechanics they are components of the same four-vector. It should be noted that the constant of motion analogous to the non-relativistic result (15.23), is identically equal to zero ∂L n u − L = −pn un − L ≡ 0. ∂un

(15.42)

The energy-momentum vector as a four-vector, transforms with respect to the transformations from one inertial system of reference to another, according to the transformation law pk = kn pn ,

(15.43)

where pk are the components of the energy-momentum tensor in the inertial system of reference K , while pk are the components of the energy-momentum tensor in the inertial system of reference K  moving along the common x-axis with a velocity V with respect to K . Using here the explicit form of the tensor kn given by (13.29), we obtain 1 E=  1−

V2 c2

   E + Vpx ,  V  + 2E , c

(15.44)

pn pn = m2 c2 un un = m2 c2 .

(15.45)

E2 + pα pα = m2 c2 . c2

(15.46)

1

px =  1−

V2 c2



px

py = py , pz = pz .

Using (15.38) with (13.41) we obtain

From (15.45) we may write

From the results (15.25) for the components of the three-dimensional momentum vector, we may write  = −p2 . pα pα = gαβ pα pβ = − p·p

(15.47)

Chapter 15 • Relativistic Dynamics

117

Substituting (15.47) into (15.46), we obtain E2 − p2 = m2 c2 . c2

(15.48)

The relation between the energy and the three-dimensional momentum of a particle then becomes E=c

 p2 + m2 c2 .

(15.49)

The function (15.49) is usually called the Hamiltonian of the particle and it is denoted by H. Thus, we may write H =c

  p2 + m2 c2 = p2 c2 + m2 c4 ,

(15.50)

being the most usual definition of the Hamiltonian of a particle.

15.2.3 Conservation of Energy-Momentum The conservation law of the momentum of a particle is a consequence of the homogeneity of space and the conservation law of the energy of a particle is a consequence of the homogeneity of time. Due to the homogeneity of space-time, the mechanical properties of a free particle remain unchanged after the translation of a particle from a point with coordinates xn to a point with coordinates xn + λn , where λn is an infinitesimally small constant four-vector. Thus, the Lagrange function must be invariant with respect to this translation and its variation must be zero. Therefore, we may write δL =

∂L k ∂L δx = k λk = 0. ∂xk ∂x

(15.51)

Using here the equations of motion (15.2) we obtain d ∂L λk = 0. δL = ds ∂uk

(15.52)

Since λn is a non-zero infinitesimal four-vector, the expression (15.52) gives d ∂L = 0. ds ∂uk

(15.53)

d ∂L dun dpn = mc = = 0. ds ∂un ds ds

(15.54)

−

Using now (15.38) we obtain −

The expression (15.54) then gives dpn 1 =  ds c 1−

V2 c2

dpn = 0 ⇒ pn = Constant. dt

(15.55)

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TENSORS, RELATIVITY, AND COSMOLOGY

Thus, we obtain the three-dimensional momentum conservation law d p  = Constant, =0 ⇒ p dt

(15.56)

dE = 0 ⇒ E = Constant. dt

(15.57)

and the energy conservation law

15.3 Angular Momentum Tensor In the relativistic mechanics the angular momentum tensor is defined by the expression Mnk = xn pk − xk pn

(k, n = 0, 1, 2, 3).

(15.58)

Only the spatial components of the angular momentum tensor with (k, n = 1, 2, 3) have a physical meaning and coincide with the usual definition of the angular momentum in the non-relativistic mechanics. In the non-relativistic mechanics it is customary to form an axial three-dimensional angular momentum vector Mν =

1 ναβ e Mαβ = eναβ xα pβ 2

or in the classical vector notation

    = M  

1 x1 p1

2 x2 p2

3 x3 p3

(α, β = 1, 2, 3)

    = r × p .  

(15.59)

(15.60)

The conservation law of the angular momentum tensor (15.58) is a consequence of the isotropic nature of the four-dimensional space-time. The three-dimensional angular momentum conservation law is a consequence of the isotropic nature of the threedimensional space. Due to the isotropic nature of the four-dimensional space-time, the mechanical properties of a free particle remain unchanged after rotations in the four-dimensional space-time. Let us consider the special case of a rotation for some angle θ about the 3-axis in the three-dimensional space. Then the relation between the coordinates zm in the rotated inertial system of reference K  and the coordinates xk in the original inertial system of reference K is given by zn = nj xj

(j, n = 0, 1, 2, 3).

(15.61)

Using (5.40) we may rewrite (15.61) in the matrix form ⎡

⎤ ⎡ z0 1 0 0 0 1 ⎢ z ⎥ ⎢ 0 cos θ sin θ 0 ⎢ ⎥ ⎢ ⎣ z2 ⎦ = ⎣ 0 − sin θ cos θ 0 0 0 0 1 z3

⎤⎡

⎤ x0 1 ⎥⎢ x ⎥ ⎥⎢ ⎥ ⎦ ⎣ x2 ⎦ . x3

(15.62)

If we now consider the rotation for some infinitesimal angle δθ ≈ 0, then we have cos δθ ≈ 1,

sin δθ ≈ δθ.

(15.63)

Chapter 15 • Relativistic Dynamics

119

Substituting (15.63) into (15.62) we obtain ⎡

⎤ ⎡ z0 1 0 0 ⎢ z1 ⎥ ⎢ 0 1 δθ ⎢ ⎥=⎢ ⎣ z2 ⎦ ⎣ 0 −δθ 1 0 0 0 z3

⎤⎡ 0 x 0 ⎢ 1 0⎥ ⎥⎢ x 0 ⎦ ⎣ x2 1 x3

⎤ ⎥ ⎥, ⎦

(15.64)

or zn = xn + δ nj xj = xn + δ nk gkj xj = xn + δ nk xk ,

(15.65)

where δ nj is a mixed tensor defined by the following antisymmetric matrix: ⎡



δ nj



0 0 0 ⎢ 0 0 δθ =⎢ ⎣ 0 −δθ 0 0 0 0

⎤ 0 0⎥ ⎥. 0⎦ 0

(15.66)

Using here (13.9) and (13.12), we obtain the covariant coordinates xk of the fourdimensional space-time in matrix form ⎡ 1   ⎢0     j ⎢ xk = gjk x = ⎣ 0 0

⎤⎡ 0 x 0 0 0 ⎢ 1 −1 0 0 ⎥ ⎥⎢ x 0 −1 0 ⎦ ⎣ x2 0 0 −1 x3

⎤

⎡

⎤ x0 ⎥ ⎢ −x1 ⎥ ⎥=⎢ ⎥ ⎦ ⎣ −x2 ⎦ . −x3

(15.67)

We can also calculate the components of the antisymmetric contravariant tensor δ nk in matrix form as follows: 

or

    δ nk = g jk δ nj ,

⎡ 1 0 0 0   ⎢ 0 −1 0 0 δ nk = ⎢ ⎣ 0 0 −1 0 0 0 0 −1

⎤⎡

0 0 0 ⎥ ⎢ 0 0 δθ ⎥⎢ ⎦ ⎣ 0 −δθ 0 0 0 0

(15.68) ⎤ 0 0⎥ ⎥. 0⎦ 0

(15.69)

Thus, we finally obtain the matrix form of the antisymmetric contravariant tensor δ nk as follows: ⎡ 0   ⎢0 nk ⎢ δ

=⎣ 0 0

⎤ 0 0 0 0 −δθ 0 ⎥ ⎥. δθ 0 0 ⎦ 0 0 0

(15.70)

For an arbitrary infinitesimal rotation, the contravariant tensor δ nk has a more complex form compared to the simple matrix (15.70), but it is always an antisymmetric tensor. Thus, we always have δ nk = −δ kn

(k, n = 0, 1, 2, 3),

(15.71)

and the most general infinitesimal rotation of the coordinates is given by the following transformation relations:

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zn = xn + δxn ,

δxn = δ nk xk .

(15.72)

The most general infinitesimal rotation of four-velocity vector is given by the following transformation relations: un = un + δun ,

δun = δ nk uk .

(15.73)

The Lagrange function of a particle L(xn , un ) must be invariant with respect to this infinitesimal rotation, that is, we must have δL = 0. Let us now calculate the variation of the Lagrangian L with respect to the infinitesimal rotations (15.72) and (15.73), that is, ∂L n ∂L δx + δun = 0, ∂xn ∂un

(15.74)

∂L ∂L δ nk xk + δ nk uk = 0. ∂xn ∂un

(15.75)

δL =

or δL =

Substituting the equations of motion (15.2) into (15.75) we obtain

δL = δ nk

d ds



∂L ∂un



xk +

∂L u = 0. k ∂un

(15.76)

Using the definition of the energy-momentum four-vector (15.38) and the definition of the four-velocity (13.34), we obtain

δL = −δ nk

dx dpn xk + pn k ds ds



=−

 d nk δ pn xk = 0. ds

(15.77)

Using here the antisymmetry of the tensor δ nk , (15.77) becomes  1 1 d  dMnk pn xk − pk xn = δ nk = 0. δL = − δ nk 2 ds 2 ds

(15.78)

From (15.78), as a direct consequence of the isotropic nature of the space-time, we obtain the conservation law of the angular momentum tensor, that is, dMnk dMnk =0 ⇒ = 0 ⇒ Mnk = Constant ds dt

(15.79)

The spatial part of this conservation law for (n, k = 1, 2, 3), that is, M ν = eναβ xα pβ = Constant

is the usual conservation law of the three-dimensional angular momentum vector.

(15.80)