2.2. LORENTZ INVARIANCE 51
3) The 4-D current density:(2.2.17)
Jμ= (J 0 ,J 1 ,J 2 ,J 3 ),
Jμ= (J^0 ,J^1 ,J^2 ,J^3 ) =gμ νJν,whereJ^0 =−J 0 =cρ,ρis the charge density, andJ~= (J 1 ,J 2 ,J 3 )is the current density
field.4) The 4-D energy-momentum vector:(2.2.18)
Eμ= (E,cP 1 ,cP 2 ,cP 3 ),
Eμ=gμ νEν= (−E,cP 1 ,cP 2 ,cP 3 ),whereEis the energy, andP= (P 1 ,P 2 ,P 3 )is the momentum vector,5) The 4-D gradient operators:(2.2.19)
∂μ=(
∂
∂x^0,
∂
∂x^1,
∂
∂x^2,
∂
∂x^3)
=
(
1
c∂
∂t,∇
)
,
∂μ=gμ ν∇μ=(
−
∂
∂x^0,
∂
∂x^1,
∂
∂x^2,
∂
∂x^3)
=
(
−
1
c∂
∂t,∇
)
,
transform as the first-order Lorentz tensors.2.2.3 Relativistic invariants
All Lagrange actions in relativistic physics are Lorentz invariants. The most common Lorentz
invariants are contractions of 4-D tensors. For example, let
Aμ= (A 0 ,A 1 ,A 2 ,A 3 ), Bμ= (B^0 ,B^1 ,B^2 ,B^3 )be the covariant and contra-variant vectors. Then the contraction
(2.2.20) AμBμ=A 0 B^0 +A 1 B^1 +A 2 B^2 +A 3 B^3
is a Lorentz invariant. In fact, under the Lorentz transformation (2.2.8),AμandBμsatisfy
A ̃μ=LμνAν, B ̃μ=lνμBν.It follows that
(2.2.21) A ̃μB ̃μ=LαμlμβAαBβ.
Since(lνμ) = (Lνμ)−^1 is the inverse matrix of(Lνμ), then
Lαμlμβ=δβα.Thus, (2.2.21) becomes
A ̃μB ̃μ=AμBμ,
which shows that the contraction (2.2.20) is a Lorentz invariant. In fact, the following theorem
can be easily verified in the same fashion.