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The Four Maxwell's equation that are given by $$\nabla . \mathbf{E}=\frac{\rho}{\epsilon_0}$$ $$\nabla.\mathbf{B}=0$$ $$\nabla\times \mathbf{E}+\frac{\partial \mathbf{B}}{\partial t}=0$$ $$\nabla \times \mathbf{B}-\frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}=\mu_0 \mathbf{J}$$ For an vector $\mathbf{V}$ , you can define a matrix $$\partial _i \mathbf{V}_j = \left( \begin{array}[ccc] \partial \mathbf{V_1} /\partial x_1 & \partial \mathbf{V_1} /\partial x_2 &\partial \mathbf{V_1} /\partial x_3 \\ \partial \mathbf{V_2} /\partial x_1 & \partial \mathbf{V_2} /\partial x_2 & \partial \mathbf{V_2} /\partial x_3 \\ \partial \mathbf{V_3} /\partial x_1 & \partial \mathbf{V_3} /\partial x_2 & \partial \mathbf{V_3} /\partial x_3 \\ \end{array} \right)$$

In Maxwell equation He take some special combination from this matrix , Why it is so? How do you know that How do you know that only these special combination are needed? Is there any deep meaning ?

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  • $\begingroup$ I agrees with experiment. $\endgroup$ – lalala Jan 30 '20 at 7:30
  • $\begingroup$ You don’t mean $\partial_iV_i$. $\endgroup$ – G. Smith Jan 30 '20 at 7:32
  • $\begingroup$ Look up the Helmholtz theorem. en.m.wikipedia.org/wiki/Helmholtz_decomposition. "Any sufficiently smooth, rapidly decaying vector field in three dimensions can be resolved into the sum of an irrotational (curl-free) vector field and a solenoidal (divergence-free) vector field;" $\endgroup$ – StudyStudy Jan 30 '20 at 9:29
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Yes, there is (somewhat) deep meaning. The divergence is the combination that is a scalar under rotations. The curl is composed of three combinations that form a vector under rotations. These are the only combinations that have these transformation properties. That’s why the divergence and the curl use particular combinations of elements from your matrix, and that’s why they’re “so different”.

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  • $\begingroup$ Small correction: curl is pseudovector. $\endgroup$ – A.V.S. Jan 30 '20 at 8:31
  • $\begingroup$ @A.V.S. When I said “under rotations” I had proper rotations in mind. It’s a vector under proper rotations, right? $\endgroup$ – G. Smith Jan 30 '20 at 8:35
  • $\begingroup$ I think I would prefer to explain what a proper rotation is than explain what a pseudovector is. $\endgroup$ – G. Smith Jan 30 '20 at 8:36
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There are physical interpretations of divergence and curl.

For example, the flux through any closed surface in an electric field is proportional to the charge inside of that surface. The natural way to express this in a vector field is with divergence.

The deeper question might be why do we conceptualize these forces as fields in the first place? Sure, we can picture the vectors, but what is really going on?

An interesting answer to this question lies in the Standard Model! The electromagnetic force operates through the exchange of virtual photons. Forces in this model have a corresponding boson that mediates the force.

I'd invite you to read more about the Standard Model because I can't do it justice. I'd also recommend the second volume of the Feynman Lectures. I believe a satisfying answer to your question is given there!

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If we look at the eigenvalue equation of divergence ( sum of diagonal)  we find that it gives component of vector field along it's propagation vector (wave vector).

Curl(sum of difference of non diagonal) gives component perpendicular to propagation direction. Let A be a vector.

div(A exp(ik.r))=i A.k exp(ik.r)

Curl(A exp(ik.r))=iA×k exp(ik.r)

Hence vector field is completely determined.

Also you can see Helmholtz theorem

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