# The Unapologetic Mathematician

## Maxwell’s Equations (Integral Form)

It is sometimes easier to understand Maxwell’s equations in their integral form; the version we outlined last time is the differential form.

For Gauss’ law and Gauss’ law for magnetism, we’ve actually already done this. First, we write them in differential form:

\displaystyle\begin{aligned}\nabla\cdot E&=\frac{1}{\epsilon_0}\rho\\\nabla\cdot B&=0\end{aligned}

We pick any region $V$ we want and integrate both sides of each equation over that region:

\displaystyle\begin{aligned}\int\limits_V\nabla\cdot E\,dV&=\int\limits_V\frac{1}{\epsilon_0}\rho\,dV\\\int\limits_V\nabla\cdot B\,dV&=\int\limits_V0\,dV\end{aligned}

On the left-hand sides we can use the divergence theorem, while the right sides can simply be evaluated:

\displaystyle\begin{aligned}\int\limits_{\partial V}E\cdot dS&=\frac{1}{\epsilon_0}Q(V)\\\int\limits_{\partial V}B\cdot dS&=0\end{aligned}

where $Q(V)$ is the total charge contained within the region $V$. Gauss’ law tells us that the flux of the electric field out through a closed surface is (basically) equal to the charge contained inside the surface, while Gauss’ law for magnetism tells us that there is no such thing as a magnetic charge.

Faraday’s law was basically given to us in integral form, but we can get it back from the differential form:

$\displaystyle\nabla\times E=-\frac{\partial B}{\partial t}$

We pick any surface $S$ and integrate the flux of both sides through it:

$\displaystyle\int\limits_S\nabla\times E\cdot dS=\int\limits_S-\frac{\partial B}{\partial t}\cdot dS$

On the left we can use Stokes’ theorem, while on the right we can pull the derivative outside the integral:

$\displaystyle\int\limits_{\partial S}E\cdot dr=-\frac{\partial}{\partial t}\Phi_S(B)$

where $\Phi_S(B)$ is the flux of the magnetic field $B$ through the surface $S$. Faraday’s law tells us that a changing magnetic field induces a current around a circuit.

A similar analysis helps with Ampère’s law:

$\displaystyle\nabla\times B=\mu_0J+\epsilon_0\mu_0\frac{\partial E}{\partial t}$

We pick a surface and integrate:

$\displaystyle\int\limits_S\nabla\times B\cdot dS=\int\limits_S\mu_0J\cdot dS+\int\limits_S\epsilon_0\mu_0\frac{\partial E}{\partial t}\cdot dS$

Then we simplify each side.

$\displaystyle\int\limits_{\partial S}B\cdot dr=\mu_0I_S+\epsilon_0\mu_0\frac{\partial}{\partial t}\Phi_S(E)$

where $\Phi_S(E)$ is the flux of the electric field $E$ through the surface $S$, and $I_S$ is the total current flowing through the surface $S$. Ampère’s law tells us that a flowing current induces a magnetic field around the current, and Maxwell’s correction tells us that a changing electric field behaves just like a current made of moving charges.

We collect these together into the integral form of Maxwell’s equations:

\displaystyle\begin{aligned}\int\limits_{\partial V}E\cdot dS&=\frac{1}{\epsilon_0}Q(V)\\\int\limits_{\partial V}B\cdot dS&=0\\\int\limits_{\partial S}E\cdot dr&=-\frac{\partial}{\partial t}\Phi_S(B)\\\int\limits_{\partial S}B\cdot dr&=\mu_0I_S+\epsilon_0\mu_0\frac{\partial}{\partial t}\Phi_S(E)\end{aligned}

February 2, 2012 -

1. [...] law, we’re already done, since it’s exactly the third of Maxwell’s equations in integral form. So far, so [...]

Pingback by Deriving Physics from Maxwell’s Equations « The Unapologetic Mathematician | February 3, 2012 | Reply

2. who is maxwell and why did he decide to come with his equations that are difficult to understand

Comment by TPL | May 17, 2012 | Reply

• Blame the universe’s electromagnetism for being most easily calculable with vector calculus.

Comment by Jose | February 12, 2013 | Reply

3. That note is clear and good prapared

Comment by Habamenshi Pierre Claver | November 23, 2012 | Reply

4. why are these equations called maxwell’s equations while none of them are derived or proved by mexwell ,all of them were alread present

Comment by abdur rahim | December 26, 2012 | Reply

5. where on earth is maxwell from?

Comment by Uc | June 20, 2013 | Reply