1. Vector Potential & Gauge Transformations in Geometric Algebra.
Setup: Consider a uniform magnetic field \(\vec{B} = B_0 \mathbf{e}_z\text{.}\) In Geometric Algebra (GA), we represent this as the bivector:
\begin{equation*}
\bivec{B} = B_0 (\mathbf{e}_x \wedge \mathbf{e}_y) \, .
\end{equation*}
Note that in 3D, the traditional vector \(\vec{B}\) is the dual of this bivector via \(\vec{B} = -I \bivec{B}\text{,}\) where \(I = \mathbf{e}_x \mathbf{e}_y \mathbf{e}_z\text{.}\)
(a)
Show that vector potentials \(\vec{A}\) for symmetry gauge and Landau gauge satisfy the condition that \(\bivec{B} = \vec{\nabla} \wedge \vec{A}\) for a uniform magnetic field.
Hint.
-
Landau Gauge:\begin{equation*} \vec{A}_{\text{Landau}} = -B_0 y \mathbf{e}_x \end{equation*}
-
Symmetry Gauge:\begin{equation*} \vec{A}_{\text{sym}} = \frac{B_0}{2}(x\mathbf{e}_y - y\mathbf{e}_x) \end{equation*}
Solution 1.
Part 1: Show that the Landau gauge potential satisfies \(\bivec{B}=\vec{\nabla}\wedge\vec{A}\) such that \(\bivec{B} = B_0\mathbf{e}_x\mathbf{e}_y\text{.}\)
Step 1: Rewrite the Landau gauge potential in vector form.
\begin{equation*}
\vec{A}_{\text{Landau}} = -B_0 y \mathbf{e}_x = \begin{bmatrix} -B_0 y \\ 0 \\ 0 \end{bmatrix}
\end{equation*}
Step 2: Evaluate \(\bivec{B} = \vec{\nabla} \wedge \vec{A}_{\text{Landau}}\text{.}\)
\begin{align*}
\vec{\nabla} \wedge \vec{A}_{\text{Landau}} \amp=
\begin{bmatrix} \partial_x \\ \partial_y \\ \partial_z \end{bmatrix}
\wedge
\begin{bmatrix} -B_0 y \\ 0 \\ 0 \end{bmatrix}\\
\amp=
(\partial_x \mathbf{e}_x) \wedge (-B_0 y \mathbf{e}_x + 0\mathbf{e}_y + 0\mathbf{e}_z)\\
\amp\quad + (\partial_y \mathbf{e}_y) \wedge (-B_0 y \mathbf{e}_x + 0\mathbf{e}_y + 0\mathbf{e}_z)\\
\amp\quad + (\partial_z \mathbf{e}_z) \wedge (-B_0 y \mathbf{e}_x + 0\mathbf{e}_y + 0\mathbf{e}_z)\\
\amp=
(\partial_x \mathbf{e}_x) \wedge (-B_0 y \mathbf{e}_x)\\
\amp\quad + (\partial_y \mathbf{e}_y) \wedge (-B_0 y \mathbf{e}_x)\\
\amp\quad + (\partial_z \mathbf{e}_z) \wedge (-B_0 y \mathbf{e}_x)\\
\amp=
\cancelto{0}{(-\partial_x B_0 y)}\cancelto{0}{(\mathbf{e}_x \wedge \mathbf{e}_x)}\\
\amp\quad + (-\partial_y B_0 y)(\mathbf{e}_y \wedge \mathbf{e}_x)\\
\amp\quad + \cancelto{0}{(-\partial_z B_0 y)}(\mathbf{e}_z \wedge \mathbf{e}_x)\\
\amp= -B_0(-\mathbf{e}_x \wedge \mathbf{e}_y)\\
\amp= B_0 (\mathbf{e}_x \wedge \mathbf{e}_y) \quad \proofmark
\end{align*}
Solution 2.
Part 2: Show that the symmetry gauge potential satisfies \(\bivec{B}=\vec{\nabla}\wedge\vec{A}\text{.}\)
(b)
Define a gauge transformation as \(\vec{A}' = \vec{A} + \vec{\nabla}\chi\text{.}\) What is the physical significance of the scalar function \(\chi\text{?}\)
(c)
Determine the specific scalar function \(\chi(x,y)\) that transforms the Landau gauge potential into the Symmetry gauge potential:
\begin{equation*}
\vec{A}_{\text{sym}} = \vec{A}_{\text{Landau}} + \vec{\nabla}\chi
\end{equation*}
Hint.
\(\vec{A}_{\text{sym}} - \vec{A}_{\text{Landau}}\)
(d)
Prove that the magnetic bivector is invariant under this transformation:
\begin{equation*}
\bivec{B}' = \vec{\nabla} \wedge \vec{A}' = \bivec{B}
\end{equation*}
(e)
Why does adding the gradient term \(\vec{\nabla}\chi\) not change the bivector field \(\bivec{B}\text{?}\) Explain this in terms of the properties of the wedge product and second derivatives.
