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In case of electrostatics, the work done in moving a point charge from one point to another point is independent of path. This leads to concept of electric potential \(\phi(\vec{x},t\) which is related to the electric field by \(\vec{E}=-\nabla \phi\). The electric potential, as explained above for electrostatic case, does not generalize to the case when time varying fields are present because the force due to electric field is no longer path independent. In other words, the force due to time dependent electric field on a point charge is not a conservative force. The generalization to time varying situations comes from Maxwell's equations. The Maxwell's equations also allow us to introduce a vector potential \(\vec{A}\) related to the field \(B\).
Two results from vector calculus:
Before we come to Maxwell's equations, we recall two theorem from vector calculus.
Theorem I:
If curl of a vector field is zero, \(\nabla \times \vec{F}=0\), then there exists a scalar function \(C(x)\) such that \(\vec{F}=\nabla C(\vec{x})\).
Theorem II:
Also if divergence of a vector field is zero, \(\nabla \cdot \vec{F}=0\), then there exists a vector field \(\vec{D}(\vec{x})\), such that \( \vec{F}=\nabla \times D(x)\)
Potentials in electromagnetic theory:
The four Maxwell's equations in absence of a medium are
\begin{eqnarray} \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0} &\qquad& \text{curl} \vec{E}=-\frac{\partial\vec{B}}{\partial t}\\ \nabla \cdot \vec{B} = 0,&\qquad& \text{curl} \vec{B}=\frac{\mu_0}{4\pi} \vec{J} +\mu_0\epsilon_0 \frac{\partial \vec{E}]{\partial t}. \end{eqnarray} Noting the \(\text{curl} \vec{E}=0\) equation, and using Theorem-I, we see that there exists a function \(\phi(\vec{x},t)\) such that \(\vec{E}=-\nabla \phi\).
Application of Theorem-II to magnetic field, \(\text{div} \vec{B}=0\), implies that there exists a field \(A(\vec{x},t)\) such that \(\vec{B}=\nabla \times \vec{A}\). The functions \(\phi(\vec{x},t)\) and \(A(\vec{x},t)\) introduced above are known as scalar and vector potentials. It may be remarked that the potentials are not determined uniquely from the fields, In other words if \(\Lambda(\vec{x},t)\) is any function of \(\vec{x},t\), the potentials \(A, \phi\) and \(A', \phi'\) related by \begin{equation} \vec{A}'(\vec{x},t) = \vec{A}(\vec{x},t) - \nabla \Lambda(\vec{x},t),\qquad \phi'(\vec{x},t) = \phi(\vec{x},t) + \Lambda(\vec{x}, t) \end{equation} give the same \(\vec{E}, \vec{B}\) field because, curl of gradient of an arbitrary function vanishes.\\ More details can be found in standard text books.
