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I'm looking for the proof that the 1st Maxwell equation is valid also on non conservative electric field.

When we are talking about a electrostatic field, the equation is ok. We can apply the Gauss (or Flux) theorem and get Gauss' law:

$$\mathbf{\nabla} \cdot \mathbf{E} ~=~ \frac{1}{\epsilon _0} \rho (x,y,z).$$

The question is, why when there is a time dependent magnetic field, and then a time dependent (non conservative) induced electric field, the 1st Maxwell equation is the same?

How we can prove that?

Waffle's Crazy Peanut
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EmarJ
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2 Answers2

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As you've said, and just to be completely clear, in vacuum (neglecting, in other words, effects in macroscopic media like polarization), Gauss' law is the full, time-dependent expression of what you're calling the "first Maxwell equation."

The "derivation" of the Maxwell equations were originally formulated as differential (local) versions of the well known empirically observed laws of Ampere, Faraday, and Gauss. This is discussed some in Jackson's book ("Classical Electrodynamics"). Also see Griffith's book ("Intro to Electrodynamics").

The Maxwell equations aren't really derived from more fundamental considerations. Their integral form (the "laws" cited above) were deduced from observation, compared with phenomena not originally used in the determination of the empirical "laws," and found, in some regimes, to work.

In the regime of atomic physics, Planck found that the assumed continuous radiation of an accelerating charge predicted a black-body spectrum at large frequency in contradiction with that observed. And this led to a modification of the classical electrodynamics and the advent of the quantum theory.

The form of the Maxwell equations is, however, tightly constrained by invariance under Lorentz transformations. Jackson discusses this in Chapter 11.

MarkWayne
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  • Then the fact that 1st Maxwell law is the same with electrostatic field and also with induced ones, is an experimental fact? We cannot prove it with computation? – EmarJ Oct 23 '12 at 19:03
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    Yes, in the sense that I attempted to describe above. In fact, the induced fields (say by a loop external to the field point) will not contribute to $\nabla\cdot\vec{E}$ since there is no local charge associated with that field. – MarkWayne Oct 23 '12 at 20:20
  • @MarkWayne : You seem to be missing the point that the absence of local charges result in zero electric flux because of the inverse square variation if electric field..had it not been there even absence of local charges would have given rise to non zero electric flux that would have been coming from charges not inside the volume enclosed by your local gaussian surface . – Primeczar Feb 23 '15 at 10:14
  • I'm not sure how you gather that I "seem to be missing the point" of a trivial case of Gauss' law. Elaborate, if you want but please be specific. – MarkWayne Mar 03 '15 at 21:31
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Great question. Almost all authors don't show that further justification is is needed to get Gaus's law for induced(time dependent) electric fields

The third Hertz’s equation for electrostatic field is a generalization of the Gauss law for electrostatic fields arrived at as follows:

$$\nabla \cdot E_{static} = \frac{Q}{e}$$ - Gaus's law for electrostatic(time independent) fields.

From the fact that induced electric field does not have sources(think of Faradays coil induction experiment with a centrally placed straight round core in the induction coil to avoid distracting asymmetries. The induced E field is radially symmetric - another way it is usually stated is: the induced EMF is distributed), it follows at once that

$$\nabla \cdot E_{induced} = 0$$ (Use the divergence theorem to convince yourself)

Summing these two equations, we get the differential form of the third Maxwell-Hertz’s equation:

$$\nabla \cdot E_{total} = \nabla \cdot [E_{static} + E_{induced}] = 0$$

in the absence of charges

Pius
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