📂Electrodynamics

Gauge Transformation

Overview¹

A given scalar potential $V$ and vector potential $\mathbf{A}$ uniquely determine the electric field $\mathbf{E}$ and the magnetic field $\mathbf{B}$, but the converse is not true. In other words, there are multiple potentials $V$, $\mathbf{A}$ that can represent a single electromagnetic field $\mathbf{E}$, $\mathbf{B}$. Therefore, within the changes that do not alter $\mathbf{E}$ and $\mathbf{B}$, $V$ and $\mathbf{A}$ can be changed freely.

Gauge Transformation

There exist two pairs of potentials ($V$, $\mathbf{A}$), ($V^{\prime}$, $\mathbf{A}^{\prime}$) that generate the same electric and magnetic field. Let us refer to them by $\mathbf{A}^{\prime}=\mathbf{A} + \mathbf{\alpha}$, $V^{\prime}=V+\beta$. Since the two vector potentials $\mathbf{A}$, $\mathbf{A}^{\prime}$ produce the same magnetic field $\mathbf{B}$,

$$\mathbf{B} = \nabla \times \mathbf{A} =\nabla \times \mathbf{A}^{\prime}$$

Using $\mathbf{A}^{\prime}=\mathbf{A} + \mathbf{\alpha}$, it follows that

$$\nabla \times \mathbf{A}^{\prime} = \nabla \times (\mathbf{A}+\mathbf{\alpha}) = \nabla\times \mathbf{A} + \nabla \times \mathbf{\alpha}  = \nabla \times \mathbf{A}$$

$$\implies \nabla \times \mathbf{\alpha}=0$$

The curl of $\mathbf{\alpha}$ is $0$, so $\alpha$ can be expressed as the gradient of some scalar function $\lambda$.

$$\mathbf{\alpha}=\nabla \lambda$$

The same calculation can be performed for the two scalar potentials $V$, $V^{\prime}$. Since the two pairs of potentials ($V$, $\mathbf{A}$), ($V^{\prime}$, $\mathbf{A}^{\prime}$) generate the same electric field $\mathbf{E}$,

$$\begin{align*} \mathbf{E}^{\prime} &= -\nabla V – \frac{\partial \mathbf{A}}{\partial t} \\ &= -\nabla V^{\prime} -\frac{\partial \mathbf{A}^{\prime}}{\partial t} \\ &= -\nabla(V+\beta)-\frac{\partial \mathbf{A}}{\partial t} +\frac{\partial \mathbf{\alpha}}{\partial t} \\ &= -\nabla V – \frac{\partial \mathbf{A}}{\partial t} -\nabla \beta –\frac{\partial \mathbf{\alpha}}{\partial t} \end{align*}$$

Therefore, $-\nabla \beta –\dfrac{\partial \mathbf{\alpha}}{\partial t}= 0$, and because $\mathbf{\alpha}=\nabla \lambda$,

$$\nabla \left( \beta + \frac{\partial \lambda}{\partial t} \right)=0$$

$\nabla$ involves differentiation with respect to position, and since the result is $0$, the quantity inside the parentheses is independent of position. Therefore, referring to it as a function of time $k(t)$, we get $\beta = -\dfrac{\partial \lambda}{\partial t}+k(t)$. Applying a transformation to the equation gives a new equation for $\Lambda$ as follows: $$-\frac{\partial \lambda}{\partial t}+k(t)=-\frac{\partial}{\partial t} \left[ \lambda - \int_{0}^t{k(t^{\prime})}dt^{\prime} \right]=-\frac{\partial \Lambda}{\partial t}$$ The second equation uses the fundamental theorem of calculus. The reason this representation is possible is because the gradients of $\lambda$ and $\Lambda$ are identical.

$$\nabla \Lambda = \nabla \left\{ \lambda -\int_{0}^t{k(t^{\prime})}dt^{\prime} \right\} =\nabla \lambda - \nabla \int_{0}^t{k(t^{\prime})}dt^{\prime} =\nabla \lambda = \mathbf{\alpha}$$

To summarize, it is as follows.

$$\mathbf{A}^{\prime} =\mathbf{A} +\nabla \Lambda \\ V^{\prime}=V-\frac{\partial \Lambda}{\partial t}$$

The implications of the above equation are as follows.

There exists a pair of potentials ($V$, $\mathbf{A}$) that generate an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$. In this case, adding $\nabla \Lambda$ to $\mathbf{A}$ and subtracting $\dfrac{\partial \Lambda}{\partial t}$ from $V$, to produce a new potential $(V^{\prime}, \mathbf{A}^{\prime}) = (V-\dfrac{\partial \Lambda}{\partial t},\ \mathbf{A}+\nabla \Lambda)$, also generates the same electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$. That is, the electromagnetic field does not change if the conditions are properly met when modifying the potentials.. This modification of potentials is called gauge transformation. Using transformations makes it possible to simplify complex equations for easier calculation.

See Also

• Coulomb Gauge and Lorenz Gauge