← Back to Part IV: Electromagnetism & Fields

Chapter 4: Four-Current and Charge Conservation

The four-current unifies charge density and current density into a single four-vector. Charge conservation follows from gauge invariance and is expressed as a covariant continuity equation.

The Four-Current J^μ

\( J^\mu = (c\rho, \vec{J}) \)

ρ = charge density, J = current density

The four-current transforms as a four-vector under Lorentz transformations, mixing charge and current densities between frames.

Charge Conservation

\( \partial_\mu J^\mu = 0 \)

Equivalent to \( \frac{\partial \rho}{\partial t} + \nabla \cdot \vec{J} = 0 \)

This is the relativistic continuity equation. It's automatically satisfied when Maxwell's equations hold, and is a consequence of gauge symmetry (Noether's theorem).

Covariant Lorentz Force

\( \frac{dp^\mu}{d\tau} = qF^{\mu\nu}u_\nu \)

Four-force = charge × field tensor × four-velocity

Interactive Simulations

Four-Current: Charge and Current Density Transformation

Python

script.py58 lines

import numpy as np

print("=" * 70)
print("FOUR-CURRENT J^mu = (rho*c, J_x, J_y, J_z)")
print("Transforms as a four-vector under Lorentz boosts")
print("=" * 70)

c = 1.0  # natural units

def boost_four_current(rho, jx, jy, jz, beta):
    gamma = 1.0 / np.sqrt(1.0 - beta**2)
    rho_p = gamma * (rho - beta * jx)
    jx_p = gamma * (jx - beta * rho)
    jy_p = jy
    jz_p = jz
    return rho_p, jx_p, jy_p, jz_p

# Case 1: Static charge distribution
print("\n--- Case 1: Static charge (rho=1, J=0) ---")
print("A boosted static charge creates a current!")
print(f"{'v/c':>6} {'rho_p':>10} {'jx_p':>10} {'J_mu J^mu':>12}")
print("-" * 42)

for beta in [0.0, 0.1, 0.3, 0.5, 0.7, 0.9, 0.99]:
    rho_p, jx_p, jy_p, jz_p = boost_four_current(1.0, 0.0, 0.0, 0.0, beta)
    norm = -(rho_p**2) + jx_p**2 + jy_p**2 + jz_p**2
    print(f"{beta:6.2f} {rho_p:10.4f} {jx_p:10.4f} {norm:12.6f}")

# Case 2: Neutral current-carrying wire
print("\n--- Case 2: Neutral wire with current (rho=0, jx=1) ---")
print("Boosting creates a net charge density!")
print(f"{'v/c':>6} {'rho_p':>10} {'jx_p':>10} {'Net charge?':>14}")
print("-" * 46)

for beta in [0.0, 0.1, 0.3, 0.5, 0.7, 0.9]:
    rho_p, jx_p, jy_p, jz_p = boost_four_current(0.0, 1.0, 0.0, 0.0, beta)
    charge = "YES" if abs(rho_p) > 1e-10 else "NO"
    print(f"{beta:6.2f} {rho_p:10.4f} {jx_p:10.4f} {charge:>14}")

print("\n" + "=" * 70)
print("CHARGE CONSERVATION: d_mu J^mu = 0")
print("=" * 70)
print("\nContinuity equation: d(rho)/dt + div(J) = 0")
print("This is frame-independent (Lorentz scalar equation)!")
print("Total charge Q = integral(rho dV) is conserved in all frames.")

# Verify invariant
print(f"\nJ_mu J^mu invariance check:")
print(f"{'v/c':>6} {'J_mu J^mu (S)':>16} {'J_mu J^mu (S_p)':>18} {'Match':>8}")
print("-" * 52)
rho0, j0 = 2.0, 1.5
norm0 = -(rho0**2) + j0**2
for beta in [0.1, 0.3, 0.5, 0.7, 0.9]:
    rp, jp, _, _ = boost_four_current(rho0, j0, 0, 0, beta)
    normp = -(rp**2) + jp**2
    print(f"{beta:6.2f} {norm0:16.6f} {normp:18.6f} {'YES' if abs(norm0-normp)<1e-8 else 'NO':>8}")

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Fortran: Four-Current Lorentz Transformation

Fortran

program.f9074 lines

program four_current
  implicit none
  double precision :: beta, gamma_val
  double precision :: rho, jx, jy, jz
  double precision :: rho_p, jx_p, jy_p, jz_p
  double precision :: norm, norm_p
  double precision :: betas(7)
  integer :: i

betas = (/0.0d0, 0.1d0, 0.3d0, 0.5d0, 0.7d0, 0.9d0, 0.99d0/)

print '(A)', repeat('=', 65)
  print '(A)', 'FOUR-CURRENT TRANSFORMATION'
  print '(A)', 'J^mu = (rho, J_x, J_y, J_z) with c = 1'
  print '(A)', repeat('=', 65)

! Case 1: Stationary charge density
  rho = 1.0d0; jx = 0.0d0; jy = 0.0d0; jz = 0.0d0
  norm = -rho**2 + jx**2

print '(A)', 'Case: Static charge rho = 1, J = 0'
  print '(A6, A10, A10, A14)', 'v/c', 'rho''', 'jx''', 'J_mu J^mu'
  print '(A)', repeat('-', 42)

do i = 1, 7
    beta = betas(i)
    if (beta < 1.0d-15) then
      gamma_val = 1.0d0
    else
      gamma_val = 1.0d0 / sqrt(1.0d0 - beta**2)
    end if

rho_p = gamma_val * (rho - beta * jx)
    jx_p = gamma_val * (jx - beta * rho)
    norm_p = -rho_p**2 + jx_p**2

print '(F6.2, F10.4, F10.4, F14.6)', &
      beta, rho_p, jx_p, norm_p
  end do

! Case 2: Neutral wire with current
  print '(/,A)', repeat('=', 65)
  print '(A)', 'NEUTRAL WIRE WITH CURRENT'
  print '(A)', 'rho = 0, jx = 1 (current but no charge)'
  print '(A)', repeat('=', 65)

rho = 0.0d0; jx = 1.0d0

print '(A6, A10, A10, A14)', 'v/c', 'rho''', 'jx''', 'J_mu J^mu'
  print '(A)', repeat('-', 42)

do i = 1, 7
    beta = betas(i)
    if (beta < 1.0d-15) then
      gamma_val = 1.0d0
    else
      gamma_val = 1.0d0 / sqrt(1.0d0 - beta**2)
    end if

rho_p = gamma_val * (rho - beta * jx)
    jx_p = gamma_val * (jx - beta * rho)
    norm_p = -rho_p**2 + jx_p**2

print '(F6.2, F10.4, F10.4, F14.6)', &
      beta, rho_p, jx_p, norm_p
  end do

print '(/,A)', 'A neutral wire appears CHARGED in a moving frame!'
  print '(A)', 'This explains the magnetic force as a relativistic'
  print '(A)', 'Coulomb force in the boosted frame.'

end program four_current

Click Run to execute the Fortran code

Code will be compiled with gfortran and executed on the server

← E and B Field Transformations Next: Lagrangian Formulation →