I was beginning to learn about the vector and scalar potential formulations of classical E & M where you choose a $\phi$ and an $\vec{A}$ such that: \begin{equation} \begin{aligned} &\vec{B} \equiv \vec{\nabla} \times \vec{A} \\ &\vec{E} \equiv-\nabla \phi-\frac{\partial \vec{A}}{\partial t}. \end{aligned} \end{equation} I was curious what would happen if $\phi$ was such that it was simply equal to the emf from Faraday's law, \begin{equation} \phi=-\frac{\partial}{\partial t} \oint \vec{B} \cdot d \vec{A}, \end{equation} So then, \begin{equation} \begin{aligned} \frac{\partial \phi}{\partial t} &=-\frac{\partial^{2}}{\partial t^{2}} \oint \vec{B} \cdot d \vec{S} \\ &=-\frac{\partial^{2}}{\partial t^{2}} \oint \left(\vec{\nabla} \times \vec{A}\right) \cdot d \vec{S}. \end{aligned} \end{equation} Then Stoke's theorem says \begin{equation} \begin{aligned} & \frac{\partial \phi}{\partial t}=-\frac{\partial^{2}}{\partial t^{2}} \int \vec{A} \cdot d \vec{r} \\ \Rightarrow & \vec{\nabla} \frac{\partial \phi}{\partial t}=-\vec{\nabla}\left(\frac{\partial^{2}}{\partial t^{2}} \int\vec{A} \cdot d \vec{r}\right) \\ \Rightarrow &-\frac{\partial}{\partial t}(\nabla \phi)=\frac{\partial^{2} \vec{A}}{\partial t^{2}} \\ \Rightarrow &-\nabla \phi=\frac{\partial A}{\partial t} \end{aligned} \end{equation} Which means: \begin{equation} \vec{E}=-\nabla \phi-\frac{\partial \vec{A}}{\partial t}=0 \end{equation} So in the special case where the scalar potential is equal to the emf, the electric field vanishes.

Is this derivation correct? And if so, what are the implications of this result? I feel that there is some significance here but I'm not sure what it is.


2 Answers 2


You equation $$ \phi=-\frac{\partial}{\partial t}\oint {\bf B}\cdot d{\bf S} $$ makes no sense. The potential $\phi({\bf r},t)$ is a function of position ${\bf r}$ while the RHS is not a function of position. It depends only the curve bounding the surface of integration.

  • $\begingroup$ B is a function of r and t. $\endgroup$
    – Cody Payne
    Commented Feb 24, 2022 at 21:58
  • 1
    $\begingroup$ Yes, but after integration over ${\bf r}$ the RHS is no longer a function of ${\bf r}$. $\endgroup$
    – mike stone
    Commented Feb 24, 2022 at 21:59

To be honest, this doesn't make a lot of sense. The electric potential $\phi(\vec{r},t)$ is a scalar field, meaning it has a real value associated with every point in space and every moment in time. The value of the surface integral $\oint \vec{B} \cdot d \vec{A}$, meanwhile, depends on the values of $\vec{B}$ over an entire surface—in other words, at an infinite number of points.

Also, you appear to be stating that $$ \vec{\nabla} \left( \int \vec{A} \cdot d\vec{l} \right) = \vec{A} $$ which is only true in limited circumstances. Specifically, if $\vec{\nabla} \times \vec{A} = 0$, then we can define a function $U(\vec{r}) = \int^\vec{r}_{\vec{r}_0} \vec{A} \cdot d\vec{l}$, where $\vec{r}$ appears as the endpoint of the path integral. We will then have $\vec{\nabla} U = \vec{A}$ from the fundamental theorem of calculus.

But the vector potential in general does not have $\vec{\nabla} \times \vec{A} = \vec{B} = 0$. What's more, the loop you are integrating over is a closed loop (it's the boundary of a surface, remember), so you can't differentiate it with respect to its "endpoint" because it doesn't have any endpoints.


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