- #1

snoopies622

- 813

- 25

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 [itex] \int _{S} [/itex] and [itex] \int _{V} [/itex] as surface and volume integrals respectively and [itex] \Sigma q [/itex] as the total charge enclosed in the given volume, we can express Gauss's law as

[tex]

\int _{S} E \cdot dS = \frac {1}{\epsilon_0} \Sigma q

[/tex]

and with the divergence theorem we can replace

[tex]

\int _{S} E \cdot dS

[/tex] with [tex]

\int _{V}

\nabla \cdot E

[/tex]

and of course the total enclosed charge can be thought of as the volume integral of charge density, so

[tex]

\frac {1}{\epsilon_0} \Sigma q = \frac {1}{\epsilon_0} \ \int _{V} \rho

[/tex]

giving us

[tex]

\int _{V}

\nabla \cdot E

=

\frac {1}{\epsilon_0} \ \int _{V} \rho =

\int _{V} \frac {\rho}{\epsilon_0}

[/tex]

Since the differential form is

[tex]

\nabla \cdot E

=

\frac {\rho}{\epsilon_0}

[/tex]

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.

[tex]

\int _{S} E \cdot dS = \frac {1}{\epsilon_0} \Sigma q

[/tex]

and with the divergence theorem we can replace

[tex]

\int _{S} E \cdot dS

[/tex] with [tex]

\int _{V}

\nabla \cdot E

[/tex]

and of course the total enclosed charge can be thought of as the volume integral of charge density, so

[tex]

\frac {1}{\epsilon_0} \Sigma q = \frac {1}{\epsilon_0} \ \int _{V} \rho

[/tex]

giving us

[tex]

\int _{V}

\nabla \cdot E

=

\frac {1}{\epsilon_0} \ \int _{V} \rho =

\int _{V} \frac {\rho}{\epsilon_0}

[/tex]

Since the differential form is

[tex]

\nabla \cdot E

=

\frac {\rho}{\epsilon_0}

[/tex]

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.

Last edited: