The Unapologetic Mathematician

Mathematics for the interested outsider

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}

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February 2, 2012 - Posted by | Electromagnetism, Mathematical Physics

6 Comments »

  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


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