Arriving at the differential forms of Maxwell's equations

[snoopies622]

In college I learned Maxwell's equations in the integral form, and I've never been perfectly clear on where the differential forms came from. For example, using \int _{S} and \int _{V} as surface and volume integrals respectively and \Sigma q as the total charge enclosed in the given volume, we can express Gauss's law as
<br /> \int _{S} E \cdot dS = \frac {1}{\epsilon_0} \Sigma q<br /> <br />
and with the divergence theorem we can replace
<br /> \int _{S} E \cdot dS<br /> with <br /> \int _{V}<br /> \nabla \cdot E<br />
and of course the total enclosed charge can be thought of as the volume integral of charge density, so
<br /> \frac {1}{\epsilon_0} \Sigma q = \frac {1}{\epsilon_0} \ \int _{V} \rho<br />

giving us
<br /> \int _{V}<br /> \nabla \cdot E<br /> =<br /> \frac {1}{\epsilon_0} \ \int _{V} \rho =<br /> \int _{V} \frac {\rho}{\epsilon_0}<br /> <br />
Since the differential form is
<br /> <br /> \nabla \cdot E<br /> =<br /> \frac {\rho}{\epsilon_0}<br /> <br />

does that mean it's mathematically valid to simply drop those integral symbols? I understand conceptually why that would be true, since the volume can be any non-zero size or at any location in space, it's just something I haven't seen before. Something similar happens with the other equations and the Kelvin Stokes theorem.

Thanks.

 

[PeroK]

does that mean it's mathematically valid to simply drop those integral symbols? I understand conceptually why that would be true, since the volume can be any non-zero size or at any location in space, it's just something I haven't seen before. Something similar happens with the other equations and the Kelvin Stokes theorem.

Thanks.

A rough mathematical argument is: if two functions are not equal at some point then one must be larger than the other. Assuming they change continuously, they one must be larger than the other on some finite volume, hence the integrals over this volume would not be equal.

If the integrals are equal on all volumes, therefore, the functions must be equal at every point.

 

[vanhees71]

You can also think of the differential operators div and curl being defined as limits of integrals, which has the advantage that the definitions are manifestly covariant, i.e., independent on the coordinates choosen to evaluate them. Then your example becomes a triviality, because assuming that Gauss's Law in integral form holds for all volumes and boundaries of these volumes by just shrinking a volume to a point leads to the differential form of the equation.

As a theoretical physicist I'd also consider the differential (local) form of Maxwell's equations the fundamental laws. The integral forms are much more delicate to handle right, which is shown by the great confusion they provide due to sloppy treatment in many (otherwise maybe good) textbooks. The prime example for this is Faraday's Law, which is mostly quoted for the special case of areas and boundaries at rest and then applied without further thinking to moving areas and boundaries, leading to a lot of confusion.

The resolution of this confusion BTW finally leads to the only adequate framework for electrodynamics, which is (special) relativity.

 

[snoopies622]

Thanks both PeroK and vanhees71, that was very helpful.

. . which has the advantage that the definitions are manifestly covariant, i.e., independent on the coordinates choosen to evaluate them.

Ah! I was in fact also wondering about the usefulness of the differential forms. For example, with
<br /> \int _{C} B \cdot dl = \mu _{0} i
I can easily arrive at the magnetic field strength at a given distance r from a wire by drawing a circle around it of radius r and then dividing \mu _{0} i by the circumference of that circle. While the differential form \nabla X B = \mu _{0} J only tells me that the curl of the magnetic field at that distance is zero, which is the case for countless magnetic fields, including none at all.