Electromagnetism Meets Relativity: The Charge and Current Four-Vector
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You should be comfortable with the definition of the four-velocity \( U^\mu \) and understand that the Lorentz transformation mixes space and time coordinates. Familiarity with the continuity equation from classical electromagnetism is assumed.
Why this matters
Electric charge is not just a property of matter; it is the source of the electromagnetic field, and its conservation must hold in all inertial frames.
In classical physics, charge conservation is expressed by the continuity equation, which relates the change in charge density to the flow of current. However, this equation is not invariant under Galilean transformations. If we demand that the laws of physics, including charge conservation, hold for all inertial observers, we must reformulate it using the tools of special relativity. This leads to the concept of the four-current, a single geometric object that unifies charge density and current density, ensuring that charge conservation is a Lorentz-scalar statement.
The Geometric Unification
Imagine a crowd of people moving through a stadium. The number of people in a specific section (charge density) and the rate at which they flow out of that section (current) are two different ways of describing the same underlying reality: the distribution and motion of the crowd. In relativity, space and time are intertwined, so a quantity that looks like a "density" in one frame may look like a "flow" in another. Charge density \( \rho \) and current density \( \mathbf{J} \) are not separate entities; they are the time and space components of a single four-vector. This four-vector, called the four-current, transforms according to the Lorentz transformation, just like position and time. This ensures that if charge is conserved in one frame, it is conserved in all frames.
Constructing the Four-Current
To build the four-current, we start with the proper charge density \( \rho_0 \), which is the charge density measured in the rest frame of the charge distribution. We then multiply this scalar by the four-velocity \( U^\mu \) of the charge element. The four-velocity is defined as \( U^\mu = \gamma (c, \mathbf{v}) \), where \( \gamma = 1/\sqrt{1 - v^2/c^2} \). The resulting four-current is:
We identify the time component as \( J^0 = \rho c \) and the spatial components as \( \mathbf{J} = \rho \mathbf{v} \). Here, \( \rho = \rho_0 \gamma \) is the charge density in the lab frame. This shows that charge density is not invariant; it increases by a factor of \( \gamma \) in the frame where the charges are moving. This is consistent with length contraction: the same amount of charge is compressed into a shorter length, increasing the density.
Covariant Charge Conservation
The continuity equation in its classical form is \( \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0 \). We can rewrite this equation in a manifestly covariant form by recognizing that \( \frac{\partial}{\partial t} \) and \( \nabla \) are parts of the four-gradient operator \( \partial_\mu = \left( \frac{1}{c} \frac{\partial}{\partial t}, \nabla \right) \). The four-divergence of the four-current is then:
Since \( \partial_\mu J^\mu \) is a Lorentz scalar (the contraction of a covariant and a contravariant vector), the equation \( \partial_\mu J^\mu = 0 \) holds in all inertial frames if and only if it holds in one. This is the relativistic statement of local charge conservation. It is more fundamental than the classical continuity equation because it explicitly respects the symmetry of spacetime.
Physical Interpretation
The relation \( \rho = \gamma \rho_0 \) has a direct physical interpretation. Consider a line of charges at rest in their own frame, with proper density \( \rho_0 \). In a frame where they move with velocity \( v \), their length is contracted by \( 1/\gamma \). Since the total charge in a segment is invariant, the density must increase by \( \gamma \). The current density is then \( \mathbf{J} = \rho \mathbf{v} = \gamma \rho_0 \mathbf{v} \). This shows that current is not an independent quantity but is directly tied to the motion of the charge density. The four-current encapsulates this relationship geometrically.
The derivation
The Four-Current and Charge Conservation
The charge density ρ and the three-current density J are the time and space parts of a single four-vector, the four-current Jμ = (ρc, J), built as Jμ = ρ0Uμ from the proper charge density and four-velocity. Its vanishing four-divergence, ∂μJμ = 0, is a Lorentz-scalar equation equivalent in every frame to the continuity equation ∂ρ/∂t + ∇·J = 0, which states that electric charge is locally conserved.
A common misconception is that charge density is invariant. It is not. Only the total charge \( Q = \int \rho \, d^3x \) is invariant (for a closed system). The density \( \rho \) depends on the frame because volume elements contract. When transforming between frames, you must transform both \( \rho \) and \( \mathbf{J} \) simultaneously using the Lorentz transformation of the four-current. Treating them separately leads to violations of charge conservation in moving frames. C
Worked example
Problem. A beam of protons has a proper charge density \( \rho_0 = 1.0 \times 10^{-5} \ \mathrm{C/m^3} \) in its rest frame. The beam moves with a speed \( v = 0.8c \) relative to the laboratory frame. Calculate the charge density \( \rho \) and the current density \( J_x \) in the laboratory frame.
- Calculate the Lorentz factor \( \gamma \): \[ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - 0.8^2}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = \frac{5}{3} \approx 1.667 \]
- Calculate the lab frame charge density \( \rho \): \[ \rho = \gamma \rho_0 = 1.667 \times 1.0 \times 10^{-5} \ \mathrm{C/m^3} = 1.67 \times 10^{-5} \ \mathrm{C/m^3} \]
- Calculate the current density \( J_x \): \[ J_x = \rho v = (1.67 \times 10^{-5} \ \mathrm{C/m^3}) (0.8 \times 3.0 \times 10^8 \ \mathrm{m/s}) \] \[ J_x = 1.67 \times 10^{-5} \times 2.4 \times 10^8 \ \mathrm{A/m^2} = 4.0 \times 10^3 \ \mathrm{A/m^2} \]
Units check. \( \rho \) is in \( \mathrm{C/m^3} \). \( J_x \) is \( \mathrm{C/m^3} \times \mathrm{m/s} = \mathrm{C/(m^2 s)} = \mathrm{A/m^2} \). The units are consistent.
Three quick tests
- Is the four-current \( J^\mu \) a Lorentz scalar or a Lorentz vector?
Answer
A Lorentz vector. It transforms according to the Lorentz transformation, mixing its time and space components. - If a charge distribution is at rest in frame S, what is the current density in that frame?
Answer
Zero. Since \( \mathbf{v} = 0 \), \( \mathbf{J} = \rho \mathbf{v} = 0 \). - Why is \( \partial_\mu J^\mu = 0 \) preferred over \( \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0 \) in relativistic contexts?
Answer
Because \( \partial_\mu J^\mu = 0 \) is a Lorentz-scalar equation, ensuring that charge conservation holds in all inertial frames without needing to specify the frame explicitly.
Problem set for PU-105 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.