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Physics
Classical Physics
Electromagnetism
Arriving at the differential forms of Maxwell's equations
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[QUOTE="snoopies622, post: 6303314, member: 112312"] 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. [/QUOTE]
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Physics
Classical Physics
Electromagnetism
Arriving at the differential forms of Maxwell's equations
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