# Changing between Maxwell equations in differential and integral formats?

It takes me a long time to think about the equations even in one format and also to deduce things with Stokes. So how can you swap between the equations? I am looking more on the lines that suppose that I know one of the format -- how can I swap to the other?

• ...not my professor and by the description I wanted to dramatize that I am somehow supposed to fast change format between one another, cannot do it yet. It takes me to long time do the partial derivates etc, there must be some tricks I am missing. How do you do it?
– hhh
May 17 '11 at 21:11
• You have to use both forms, depending on the application, e.g. the Kirchhoff's laws of a circuit can be derived from the integral form of the Maxwell equations, while the electromagnetic wave equation is derived from the differential (local) form. May 17 '11 at 22:10

$\newcommand{\vect}{\boldsymbol{#1}}$ $\newcommand{\vH}{\vect{H}}$ $\newcommand{\vE}{\vect{E}}$ $\newcommand{\vj}{\vect{j}}$ $\newcommand{\vx}{\vect{x}}$ $\newcommand{\vr}{\vect{r}}$ $\newcommand{\vnu}{\vect{\nu}}$ The integral form of a PDE usually comes from the physical definition of a quantity, say, for example curl and div. The curl operator $\nabla \times$ and the div operator $\nabla \cdot$ are both defined in a physical way in Physics, which are line integral and surface integral respectively. But for Mathematician, we always view this process in a reverse way: say, the differential equation is already there, and it holds in the sense of weak derivative, multiplying both side by a testing function which is infinitely smooth and vanishes on the boundary $\partial \Omega$(variational formulation); or more intuitively, since the equation holds for every points in the region(classical sense), we integrate both sides in an infinitesimal region(surface or volume), the equation also holds. ie, we usually deduce the integral form from differential form, not the reverse.

For example, this is the differential form of Maxwell's equations describing how electromagnetic waves propagating through a linear isotropic material, where electric field intensity denoted as $\vE = \vE(\vx,t)$ and the magnetic field intensity denoted as $\vH = \vH(\vx,t)$.

\left\{ \begin{aligned} \frac{\partial (\mu \vH)}{\partial t} + \nabla \times \vE &= 0 \quad &\text{ in } \Omega\times (0,T) \\ \frac{\partial(\epsilon \vE)}{\partial t} + \sigma \vE - \nabla\times \vH &= \vj \quad &\text{ in } \Omega\times (0,T) \\ \nabla \cdot (\epsilon \vE) &= q \quad &\text{ in } \Omega\times (0,T) \\ \nabla \cdot (\mu \vH) &= 0 \quad &\text{ in } \Omega\times (0,T) \end{aligned} \right.

For example, we would like to get the integral form for the Faraday's law of the electric field $\vE$, integrate the first equation on any surface $S$ w.r.t. its surface measure $d S$: $$- \int_{S} \frac{\partial (\mu \vH)}{\partial t}\cdot \vnu\,d S= \int_S (\nabla \times \vE)\cdot \vnu \,d S = \oint_{\partial S} \vE \cdot d\vr$$

where

1. $\vnu$ is the unit outward normal vector to the surface $S$.

2. if you parametrize $\partial S$ as $\vr(t)$, then $d\vr = \vr'(t) dt$, where $\vr'(t)$ essentially is the tangential vector of the curve $\partial S$ at every point.

3. For the left side of that equation, if we assume the enough continuity of $\mu \vH$ w.r.t $t$ and $\vx$(which is implied from the divergence free condition at every $t$), then we have $$\int_{S} \frac{\partial (\mu \vH)}{\partial t}\cdot \vnu\,d S = \frac{\partial}{\partial t} \int_{S}\mu \vH \cdot \vnu\,d S$$ since $\vnu$ is the outward normal vector to $S$, you could think the surface integral as "how much the magnetic field is flowing along the outer normal direction of the surface", which we called this quantity magnetic flux.

This is basically the derivation from differential form to integral form. The reverse is similar, just the integral equations you have at first are all physical laws, for example Faraday's Law from wikipedia:

The induced electromotive force (EMF) in any closed circuit is equal
to the time rate of change of the magnetic flux through the circuit.


Then simply write this out using definition which is integral form, apply Stokes theorem in the reverse way, then you will end up with two equal integrals, for eg, use above example, you have $$- \int_{S} \frac{\partial (\mu \vH)}{\partial t}\cdot \vnu\,d S= \int_S (\nabla \times \vE)\cdot \vnu \,d S$$ for any surface $S$, then you argue that the arbitrariness of $S$ implies $$-\frac{\partial (\mu \vH)}{\partial t} = \nabla \times \vE$$ and there you go.

Here mostly I state my own intuitive understanding of this subject while I was learning Physics as a mathematics student, will appreciate if anybody points out where I was wrong.

• +1 very nice! In electrical engineering both forms are used, depending on the application. May 17 '11 at 22:19
• can you elaborate on $\int_{S} (\nabla \times \bar{E} ) \cdot \bar{v} dS = \oint_{\partial S} \bar{E} \cdot d\bar{r}$? The left is a surface integral which you are changing to closed path integral, or is that right and is that allowed?
– hhh
May 18 '11 at 15:24
• @hhh: This is Stokes theorem in three dimensions, please refer to "special cases" on this page: en.wikipedia.org/wiki/Stokes'_theorem, it's essentially using fundamental theorem of calculus for each component of the curl vector field May 18 '11 at 19:07
• Very nice, I was looking for this. THe only step I was missing was: −∫S∂(μH)∂t⋅νdS=∫S(∇×E)⋅νdS for any surface S, then you argue that the arbitrariness of S implies −∂(μH)∂t=∇×E What do you mean by artitrariness of S? And why can you admit that in this example? Thanks
– user11473
May 28 '11 at 18:11