3.10.1 Energy-Momentum Tensor of a Free Point-Like Particle
We have seen in the previous section thatT0μs can be interpreted as the components of the 4-momentum density of the system. For a free point-like particle, the 4- momentum is pμ=muμ, wherem is the particle mass anduμ is the particle 4- velocity. For a point-like particle the mass density is
ρ=mδ3
x− ˜x(τ)
, (3.98)
where δ3 is the 3-dimensional Dirac delta function,x(τ)˜ =(x(τ),y(τ),z(τ))are the space coordinates of the particle, and we assume Cartesian coordinates. We can thus write
T0μ=mδ3
x− ˜x(τ)
cuμ. (3.99)
Ti js are proportional to the components of the momentum flux density and there- fore we should add one more velocity. Sinceu0=γc, Eq. (3.99) can be rewritten as
T0μ=mδ3
x− ˜x(τ)u0uμ
γ . (3.100)
Eventually we have the following expression for the energy-momentum tensor of a free point-like particle in Cartesian coordinates
Tμν=mδ3
x− ˜x(τ)uμuν
γ . (3.101)
As we will see in Sect.7.4, Eq. (3.101) can be derived from the Lagrangian density in a straightforward way.
3.10.2 Energy-Momentum Tensor of a Perfect Fluid
A perfect fluid can be described as a gas of quasi-free point-like particles in which the fluid pressure comes from the particle collisions. Assuming that all particles have the same massmfor simplicity, the energy-momentum tensor can be written as
Tμν = N k=1
mδ3
xk− ˜xk(τ)uμkuνk
γ (3.102)
where k=1, . . . ,N is the label of every particle. In the rest-frame of the fluid, T00 =ε, whereεis the energy density (not the mass density!) of the fluid.T0i =0, becauseuμk =0, where·indicates the average over the sample and we assume that N → ∞.Ti j =0 for i = j, because uikukj =0. Ti j =P for i= j, where Pis the pressure. Eventually, the energy-momentum tensor of a perfect fluid in its rest-frame in Cartesian coordinates should be
||Tμν|| = ||Tμν|| =
⎛
⎜⎜
⎝
ε 0 0 0
0 P 0 0
0 0 P 0
0 0 0 P
⎞
⎟⎟
⎠. (3.103)
66 3 Relativistic Mechanics At this point we have to writeTμν in terms of vectors/tensors and such that we recover the expression in Eq. (3.103) when we are in the fluid rest-frame. It turns out that the correct form is
Tμν=(ε+P)UμUν
c2 +Pημν. (3.104)
whereUμis the 4-velocity of the fluid (uμi was the 4-velocity of the particlei) and in the fluid rest-frameUμ=(c,0,0,0).
Problems
3.1 Write the 4-momentum pμ and pμ for a free point-like particle in spherical coordinates.
3.2 Find the constants of motion for a free point-like particles in spherical coordi- nates.
3.3 Repeat Problems3.1and3.2for cylindrical coordinates.
3.4 The counterpart of reaction (3.57) for cosmic ray photons is
γ+γC M B →e++e− (3.105)
namely a high-energy photon (γ) collides with a CMB photon (γC M B) and there is the production of a pair electron-positron (e+e−). Calculate the threshold energy of the high-energy photon that permits the reaction.me+c2 =me−c2=0.5 MeV.
3.5 Iron-56 is the most common isotope of iron. Its nucleus is made of 26 protons and 30 neutrons. Calculate the binding energy per nucleon, namely the binding energy divided the number of protons and neutrons. The mass of a nucleus of iron- 56 isMc2=52.103 GeV. The masses of a proton and of a neutron are, respectively, mpc2 =0.938 GeV andmnc2=0.940 GeV.
3.6 Write the field equations of the following Lagrangian density L
φ, ∂μφ
= − 2gμν
∂μφ
(∂νφ)−1 2
m2c2
φ2. (3.106)
3.7 Write the field equations in Problem3.6in Cartesian and spherical coordinates.
3.8 Let us consider a Cartesian coordinate system. Write the energy-momentum tensor associated with the Lagrangian in Problem3.6.
3.9 Eqs. (3.103) and (3.104) are the expressions in Cartesian coordinates. Find their expressions in spherical coordinates and compare the result with that found in Prob- lem2.1.
Electromagnetism
In this chapter, we will revise the pre-relativistic theory of electromagnetic phenomena, namely the theory of electromagnetism formulated before the advent of special relativity that the reader is expected to know, and we will write a fully relativistic and manifestly covariant theory. For manifestly covariant (or, equiva- lently,manifestly Lorentz-invariant) formulation, we mean that the theory is written in terms of tensors; that is, the laws of physics are written as equalities between two tensors or equalities to zero of a tensor. Since we know how tensors change when we move from a Cartesian inertial reference frame to another Cartesian inertial ref- erence frame, it is straightforward to realize that the equations are invariant under Lorentz transformations. In this and the following chapters, we will always employ Gaussian units to describe electromagnetic phenomena. In the present chapter, unless stated otherwise, we will use Cartesian coordinates; the generalization to arbitrary coordinate systems will be discussed in Sect.6.7.
The basic equations governing electromagnetic phenomena are the Lorentz force law and Maxwell’s equations. The equation of motion of a particle of massmand electric chargeemoving in an electromagnetic field is
mx¨ =eE+e
cx˙×B, (4.1)
wherecis the speed of light andEandBare, respectively, the electric and magnetic fields. Equation (4.1) is just Newton’s Second Law,mx¨=F, whenFis the Lorentz force. The equations of motion (field equations) for the electric and magnetic fields are the Maxwell equations
∇ ·E=4πρ , (4.2)
∇ ·B=0, (4.3)
∇ ×E= −1 c
∂B
∂t , (4.4)
∇ ×B= 4π c J+1
c
∂E
∂t , (4.5)
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67
68 4 Electromagnetism whereρis the electric charge density andJis the electric current density.
From Eq. (4.3), we see that we can introduce thevector potentialAdefined as
B= ∇ ×A. (4.6)
Indeed the divergence of the curl of any vector field vanishes (see AppendixB.3). If we plug Eq. (4.6) into Eq. (4.4), we find
∇ ×
E+1 c
∂A
∂t
=0. (4.7)
We can thus introduce thescalar potentialφas E+1
c
∂A
∂t = −∇φ , (4.8)
because∇ ×(∇φ)=0 (see AppendixB.3). The electric field can thus be written in terms of the scalar potentialφand the vector potentialA
E= −∇φ−1 c
∂A
∂t . (4.9)