Maxwell’s Equations and Parity Invariance
One of the most beautiful features of Maxwell’s equations is their symmetry. In this post, we will prove that Maxwell’s equations are invariant under a parity transformation (spatial inversion).
1. What is a Parity Transformation?
A parity transformation (spatial inversion) is defined by:
\[ \mathbf{r} \;\longmapsto\; \mathbf{r}' = -\mathbf{r}, \qquad t' = t. \]Transformation rules:
- Polar vector \( \mathbf{V} \): \(\mathbf{V}'(\mathbf r',t) = -\mathbf{V}(\mathbf r,t)\)
- Axial vector \( \mathbf{W} \): \(\mathbf{W}'(\mathbf r',t) = +\mathbf{W}(\mathbf r,t)\)
- Scalar \( \rho \): \(\rho'(\mathbf r',t) = \rho(\mathbf r,t)\)
Gradient operator: \[ \nabla' = \frac{\partial}{\partial \mathbf r'} = -\nabla. \]
In Maxwell’s theory:
- \( \mathbf{E}, \mathbf{J} \) are polar vectors
- \( \mathbf{B} \) is an axial vector
- \( \rho \) is a scalar
2. Maxwell’s Equations in SI Units
\[ \begin{aligned} &(1) \quad \nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}, \\[6pt] &(2) \quad \nabla \cdot \mathbf{B} = 0, \\[6pt] &(3) \quad \nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}, \\[6pt] &(4) \quad \nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}. \end{aligned} \]3. Applying Parity
(1) Gauss’s Law for \( \mathbf{E} \)
\[ \nabla' \cdot \mathbf{E}' = (-\nabla) \cdot (-\mathbf{E}) = \nabla \cdot \mathbf{E}. \] \[ \rho'/\varepsilon_0 = \rho/\varepsilon_0. \] ✅ Invariant.(2) Gauss’s Law for \( \mathbf{B} \)
\[ \nabla' \cdot \mathbf{B}' = (-\nabla) \cdot (+\mathbf{B}) = -\nabla \cdot \mathbf{B}. \] Since RHS = 0, equation is unchanged. ✅ Invariant.(3) Faraday’s Law
\[ \nabla' \times \mathbf{E}' = (-\nabla) \times (-\mathbf{E}) = \nabla \times \mathbf{E}. \] \[ -\frac{\partial \mathbf{B}'}{\partial t} = -\frac{\partial \mathbf{B}}{\partial t}. \] ✅ Invariant.(4) Ampère–Maxwell Law
\[ \nabla' \times \mathbf{B}' = (-\nabla) \times (+\mathbf{B}) = -(\nabla \times \mathbf{B}). \] \[ \mu_0 \mathbf{J}' + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}'}{\partial t} = \mu_0 (-\mathbf{J}) + \mu_0 \varepsilon_0 \frac{\partial (-\mathbf{E})}{\partial t}. \] \[ = -\left(\mu_0 \mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\right). \] Both sides gain the same minus sign. ✅ Invariant.4. Continuity Equation
\[ \frac{\partial \rho'}{\partial t} + \nabla' \cdot \mathbf{J}' = \frac{\partial \rho}{\partial t} + (-\nabla) \cdot (-\mathbf{J}) = \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0. \]5. Conclusion
By treating \( \mathbf{E} \) as a polar vector, \( \mathbf{B} \) as an axial vector, and assigning the correct transformation rules to sources, we find that all four Maxwell equations remain unchanged under parity transformation.
This symmetry is one of the many reasons Maxwell’s theory is so elegant: it respects both the structure of space and the distinction between vectors and pseudovectors.