Consider Maxwell's equation in integral form:

$$\int_{\partial V} \pmb E(\pmb r,t)\cdot \ dS =\int_V\frac{\rho(\pmb r,t)}{\epsilon_0}dV$$

$$\int_\Sigma \pmb B(\pmb r,t)\cdot dS=0$$

$$\oint_{\partial \Sigma} \pmb E(\pmb r,t)\cdot dl=-\frac{\partial}{\partial t}\int_\Sigma \pmb B(\pmb r,t)\cdot dS$$

$$\oint_{\partial \Sigma}\pmb B(\pmb r,t)\cdot dl=\mu_o\int_\Sigma \pmb J(\pmb r,t)\cdot dS+\frac{1}{c^2}\frac{\partial}{\partial t}\int_{\Sigma}\pmb E(\pmb r,t)\cdot dS$$

Where $V$ is a volume, $\partial V$ its surface boundary; $\Sigma$ is a surface and $\partial \Sigma$ its line boundary.

My question is this: for the equation to hold, is it necessary that these volumes/surfaces/lines stay fixed in time, or are they allowed to vary along with the fields and the sources?

For example, if I defined a function $\Sigma(t):[t_1,t_2]\rightarrow S(\mathbb{R^3}) $ , where $S(\mathbb{R^3}) $ is the set of all appropriate surfaces in $\mathbb{R^3}$, would it still be true that $$\oint_{\partial \Sigma(t)}\pmb B(\pmb r,t)\cdot dl=\mu_o\int_{\Sigma(t)} \pmb J(\pmb r,t)\cdot dS+\frac{1}{c^2}\frac{\partial}{\partial t}\int_{\Sigma(t)}\pmb E(\pmb r,t)\cdot dS ?$$

(Notice I've added the time dependency at the bottom of the integral signs)


1 Answer 1


In deriving the integral form of the Maxwell equations from the differential form, you have to keep the partial time derivatives of the magnetic and electric field under the integral sign if you have a time dependent integration surface $\Sigma(t)$. Thus the Faraday-Maxwell and the Ampere-Maxwell Laws as written in your question are not correct in the case of a time-varying integration surface.


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