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Coulomb Gauge and Lorentz Gauge 📂Electrodynamics

Coulomb Gauge and Lorentz Gauge

Overview1

A relationship exists between the potential and the charge density, current density as follows.

∇2V+∂∂t(∇⋅A)=−1ϵ0ρ(∇2A−μ0ϵ0∂2A∂t2)−∇(∇⋅A+μ0ϵ0∂V∂t)=−μ0J \begin{align*} \nabla ^2 V +\dfrac{\partial }{\partial t}(\nabla \cdot \mathbf{A}) &= -\frac{1}{\epsilon_{0}}\rho \\ \left( \nabla ^2 \mathbf{A}-\mu_{0}\epsilon_{0} \dfrac{\partial ^2 \mathbf{A} }{\partial t^2} \right) -\nabla\left( \nabla \cdot \mathbf{A} +\mu_{0}\epsilon_{0} \dfrac{\partial V}{\partial t}\right) &= -\mu_{0} \mathbf{J} \end{align*}

Depending on how assumptions about the potential are made, the expression changes.

Coulomb Gauge

As in magnetostatics, the divergence of the vector potential is made 00.

∇⋅A=0 \nabla \cdot \mathbf{A}=0

This allows for the representation of the equation regarding charge density solely in terms of scalar potential, thus resulting in the Poisson equation.

∇2V=−1ϵ0ρ \nabla^{2} V = -\frac{1}{\epsilon_{0}}\rho

The advantage is that it’s easier to calculate the scalar potentialVV, but the downside is that it’s harder to calculate the vector potentialA\mathbf{A}. The vector potentialA\mathbf{A} can be calculated using the equation below.

∇2A−μ0ϵ0∂2A∂t2=−μ0J+μ0ϵ0∇(∂V∂t) \nabla^{2} \mathbf{A} - \mu_{0} \epsilon_{0} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}} = -\mu_{0} \mathbf{J} + \mu_{0} \epsilon_{0} \nabla \left( \frac{\partial V}{\partial t} \right)

Lorenz Gauge

The divergence of the vector potential A\mathbf{A} is set as follows.

∇⋅A=−μ0ϵ0∂V∂t \nabla \cdot \mathbf{A} = -\mu_{0} \epsilon_{0} \frac{\partial V}{\partial t}

Then, the scalar potentialVV and vector potentialA\mathbf{A} are separated and expressed as equations of the same form.

∇2V−μ0ϵ0∂2V∂t2=−1ϵ0ρ \nabla^{2} V - \mu_{0} \epsilon_{0} \frac{\partial^{2} V}{\partial t^{2}} = -\frac{1}{\epsilon_{0}} \rho

∇2A−μ0ϵ0∂2A∂t2=−μ0J \nabla^{2} \mathbf{A} - \mu_{0} \epsilon_{0} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}} = -\mu_{0} \mathbf{J}

At this point, using the d’Alembertian makes it possible to represent it in a simpler form. The d’Alembertian is defined as follows.

□2:=∇2−μ0ϵ0∂2∂t2 \Box^{2} := \nabla^{2} - \mu_{0} \epsilon_{0} \frac{\partial^{2}}{\partial t^{2}}

Using the d’Alembertian,

□2V=−1ϵ0ρ \Box^{2} V = -\frac{1}{\epsilon_{0}}\rho

□2A=−μ0J \Box^{2} \mathbf{A} = -\mu_{0}\mathbf{J}

See Also

  1. David J. Griffiths, 기초전자기학(Introduction to Electrodynamics, 김진승 역) (4th Edition1 2014), p476-478 ↩︎