Four-Vector Formulation of Electrodynamics, Physics tutorial

Transformation Properties of the Differential Operator:

To establish the invariant forms of the differential operators, what we are going to do is to apply the rules of partial differentiation to differentiate the four-dimensional space-time coordinates. First of all, the rule of partial differentiation is as follows:

∂/∂x'1 = (∂x1/∂x'1)(∂/∂x1) + (∂x2/∂x'1)(∂/∂x2) + (∂x3/∂x'1)(∂/∂x3) + (∂x4/∂x'1)(∂/∂x4)

According to Lorentz coordinate transformation

x1 = (x'1 - iβx'4)/√(1 - β2)

x2 = x'2

x3 = x'3

x4 = (x'4 + iβx'1)/√(1 - β2)


(∂x1/∂x'1) = (∂/∂x'1)(x'1 - iβx'4)/√(1 - β2)∂/∂x1 = (1/√1 - β2)(∂/∂x1)

(∂x2/∂x'1) = 0

(∂x3/∂x'1) = 0

(∂x4/∂x'1) = 1/√(1 - β2)(∂/∂x1 + iβ(∂/∂x4)

Therefore, we conclude that ∂/∂xμ is four-vector. We can obtain invariant scalar product of ∂/∂xμ. This provides:

(∂/∂xμ)(∂/∂xμ) = ∂2/∂x21 + ∂2/∂x22 + ∂2/∂x23 + ∂2/∂x24

= ∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2 - (1/c2)(∂2/∂t2) = 2332_d’Alembertian operator.jpg

This four-dimensional operator is called as d'Alembertian and is denoted by 2332_d’Alembertian operator.jpg

Four-vector Form of the Continuity Equation:

The invariant form of differential operator, now show that charge density and current density are four-vectors and then write equation of continuity in invariant form.

Charge is invariant and doesn't depend on relative motion of reference frame. Also, you know that charge is conserved. Statement of conservation of charge is defined quantitatively as

∇.j = -∂ρ/∂t


∇.j + ∂ρ/∂t = 0

where j is current density and ρ volume charge density.

This equation takes on invariant form if current density is expressed in its four-vector, Jμ that is made of current density as space-like part and charge density as time-like part.

Therefore, if ρ is charge density in S- frame in which charges are at rest, then current density four-vector Jμ may be expressed as product of ρ and four-velocity vμ. That is:

2365_Four-vector Form of Continuity Equation.jpg

Here, ρ' = γρ that shows that charge density has increased because of change in volume element (Lorentz contraction).

As you can see, Jμ can be written as Jμ = (J, Jt)

Where J = (Jx, Jy, Jz) represents spatial components or space-like part and Jt = icρ and represents time-like part of four-vector. But still, we could use notation introduced for four-vectors, that is,

Jμ = (J1, J2, J3, J4),

Now write continuity equation in terms of Jμ as follows:

∇.j∇ + ∂ρ/∂t = ∂J1/∂x1 + ∂J2/∂x2 + ∂J3/∂x3 + ∂J4/∂x4 = 0 or simply ∂Jμ/∂xμ = 0

It is rather clear that in this form, equation of continuity is, without doubt, invariant. Also, in S' frame translating at constant speed v along common x-axis relative to S frame, our ρ' must be stated as

ρ' = γ(ρ - vJx/c2)

Therefore, in S frame in which we suppose charges to be at rest, we have Jx = 0 and thus ρ' = γρ as before.

The Four-vector Form of Maxwell's Equations:

For two observers in relative motion, what one sees as the electric field might be seen by other as magnetic field. Electric and magnetic fields are not space-like and time-like parts of the four-vector. Somewhat, they form some components of the quantity known as four-tensor. Therefore Maxwell's equations can be put in Lorentz invariant form by stating them in terms of potentials and formulate equations in four-vector form.

First of all, let us write down Maxwell's the equations in rationalized mks system of units.

∇.E = ρ/ε0

∇.B = 0

∇.E = -∂B/∂t

∇ x B = μ0J  + (1/c2)(∂E/∂t)

Now, electric and magnetic fields can be stated in terms of scalar and vector potentials as follows:

E = -∇Φ = ∂A/∂t

B = ∇ x A

If use vector identities

∇.(∇xA) = 0 and ∇x(∇Φ) = 0

Equation becomes

∇.B = ∇.(∇xA) = 0

∇ x E = ∇x(-∇Φ - ∂A/∂t)

= -∂B/∂t

∇.E = ∇.(-∇Φ - ∂A/∂t)

= ρ/ε0

Use Lorentz condition

∇.A + (1/c2)(∂Φ/∂t) = 0

∇.A = -(1/c2)(∂Φ/∂t)

Write above equation as:

-∇2Φ + (1/c2)(∂2Φ/∂t2) = ρ/ε0

∇ x B - (1/c2)(∂E/∂t)

Applying Lorentz condition the above becomes



-∇2Φ + (1/c2)(∂2Φ/∂t2) = ρ/ε0

-∇2A→ + (1/c2)(∂2A→/∂t2) = μ0J→

Maxwell's equations are Lorentz invariant as needed by theory of special relativity. Besides, they are also in agreement with second postulate of special relativity that needs that all observers measure same speed of light. Aμ and Jμ are four-vectors, so that if have their values one frame of reference, we can work out corresponding values in another reference frame S' in uniform motion translation relative to using Lorentz transformation. For example,

Φ' = γ(Φ - vAx/c2)

A'y = Ay

A'z = Az

A'x = γ(Ax - vΦ/c2)

Inverse of above transformation can be attained in usual way.

Transformation of the Fields:

Now transform the fields, after learning to transform four potential Aμ and differential operator ∂/∂xμ:

Recall that

E = -∇Φ = ∂A/∂t

So that,

E'x = ∂Φ'/∂x' - ∂A'x/∂t'


Aμ = (A, iΦ/c) and ∂/∂xμ = (∇,∂/∂t)


E'x = ic{∂A'4/∂x'1 - ∂A'1/∂x'4}

Since, Aμ and ∂/∂xμ Lorentz invariant, we can write

A'4 = γ(A4 - iβA1)

A'1 = γ(A1 + iβA4)

∂/∂x'1 = γ(∂/∂x1 + iβ(∂/∂x4))


E'x = ic(∂A4/∂x1 - ∂A1/∂x4)

= Ex

Likewise, we can derive other components of E' and B'. which given below:

E'x = Ex

E'y = γ(Ey - vBz)

E'z = γ(Ez + vBy)

B'x = Bx

B'y = γ(By + (v/c2)Ex)

B'z = γ(Bz - (v/c2)Ey)

Electric Field of Point Charge in Uniform Motion:

Consider the point charge q at origin of reference frame S' in uniform translation along common x-axis relative to S frame. Electric field due to this point charge as estimated in S' is

E = q/(4πε0)(r'/r'3)

In terms of its components, we have:

Ex = E'x = (qγy)/(4πε0(x'2 + y'2 + z'2)3/2)

Ex = γE'y = q(x - vt)/(4πε02(x - vt)2 + y'2 + z'2)3/2)

Ez = γE'z = (qγx)/(4πε02(x - vt)2+ y'2 + z'2)3/2)

Bx = 0

By = -γvE'z/c2 = -vEz/c2

Bz = -γvE'y/c2 = -vEy/c2

Or equivalently,

B = 1/c2v x E

The field lines are still straight and radiate from charge but in direction of motion, field pattern is squashed up.

649_Electric Field of a Point Charge in Uniform Motion.jpg

Transformation of Forces:

To complete four-vector formulation of electrodynamics now investigate Lorentz invariance force,

F = q(E + v x B)

You have to remember that and that F = qE transformation of E field is given by equation, that is,

E'x = Ex thus, F'x =qE'x = qEx

E'y = γ(Ey - vBz) thus, F'y = qE'y = qγ(Ey - vBz)

E'z = γ(Ez + vBy) thus, F'z = qE'z = qγ(Ez + vBy)

On the other hand, components of F in S frame are

Fx = qEx = F'x

Fy = qEy = q(Ey - vBz) = F'y

Fz = qEz = q(Ez + vBy) = F'z

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