Ampère-Maxwell's Law, in it's integral form, is $$ \oint_{\partial \Sigma}\vec{B}\cdot d \vec{l}=\mu_0\left(\iint_{\Sigma}\vec{J}\cdot d \vec{A}+\epsilon_0\frac{d }{d t}\iint_{\Sigma}\vec{E}\cdot d\vec{A}\right), $$ let's take $\vec{J}=0$ to make things easier. We can, then, write it as $$\iint_{\Sigma}\vec{\nabla}\times\vec{B}\cdot d \vec{l}=\mu_0\epsilon_0\frac{d }{d t}\iint_{\Sigma}\vec{E}\cdot d\vec{A},$$

and since $\Sigma$ is arbitrary, we get

$$\vec{\nabla}\times\vec{B}=\mu_0\epsilon_o\frac{\partial}{\partial t}\vec{E}.$$

Here's my question: is there a quick way to know if can we write $\partial/\partial t$ as $d/dt$ besides evaluating $\sum_i(\partial\vec{E}/\partial x_i)(dx_i/dt)$?


1 Answer 1


It;s a variant of the notation in Leibnitz' formula which says $$ \frac d{dt} \int^{b(t)}_{a(t)} f(x,t) dx= \frac{db}{dt} f(b(t))- \frac{da}{dt}f(a(t)) + \int_{a(t)}^{b(t)} \frac{\partial f }{\partial t}(x,t)dx. $$ On the LHS the expression only depends on $t$ so there is no need for a partial derivative. But in the integral on the RHS the expression $f(x,t) $ depends of both $x$ and $t$ so you need to use the partial derivative symbol for the time derivative because $x$ is being understood as being fixed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.