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Consider a scalar potential $\phi$, a vector potential $\mathbf{A}$, an electric field satisfying $\mathbf{E}=-\mathbf{\nabla}\phi-\dfrac{\partial}{\partial t}\mathbf{A}$, and a magnetic field satisfying $\mathbf{B}=\mathbf{\nabla}\times\mathbf{A}$.

I need to prove that $\mathbf{E}$ and $\mathbf{B}$ satisfy Maxwell's first equation in the differential form $$\nabla\times\mathbf{E}=-\partial_t \mathbf{B}$$

So I started off just by rearranging to obtain that:

$\nabla\times\mathbf{E}+\partial_t \mathbf{B}=0$

But

$\mathbf{\nabla} \times \mathbf{E} + \partial_t \mathbf{B}$

$=\mathbf{\nabla}\times\left(-\mathbf{\nabla}\phi-\dfrac{\partial}{\partial t}\mathbf{A}\right)+\partial_t \mathbf{B} $

$=-\mathbf{\nabla}(\mathbf{\nabla}\phi)-\mathbf{\nabla}\partial_t\mathbf{A} + \partial_t\mathbf{B}$

But I'm unsure as of what to do next...and if what I've done is actually correct? Thanks for any help

symplectomorphic
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Sarah Jayne
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1 Answers1

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Your last line isn't quite correct.

You didn't distribute the curl correctly: it should be $\nabla\times\nabla\phi$ and $\nabla\times\partial_t\mathbf{A}$. But the curl of a gradient is zero, and for $\nabla\times \partial_t\mathbf{A}$, you can exchange the operators, because $\nabla$ only encodes derivatives with respect to spatial coordinates, whereas $\partial_t$ only deals with time. So you get $$-\underbrace{\nabla\times\nabla\phi}_{0}-\partial_t\nabla\times\mathbf{A}+\partial_t\mathbf{B}=-\partial_t\mathbf{B}+\partial_t\mathbf{B}=0$$ where I've used the definition of the vector potential: $\nabla\times\mathbf{A}=\mathbf{B}$.

symplectomorphic
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