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On what empirical laws are Maxwell equations based?

  1. Aug 20, 2014 #1
    For example, it could be said that the equation ∇*B= 0 is based on the observation that there are no magnetic monopoles.

    But for Faraday's law of induction, it is easy to derive it from other equations but it's hard to say on what empirical law it is based. Could it be said that it is itself an empirical law and definition of it shall suffice?
  2. jcsd
  3. Aug 20, 2014 #2
    Hi Earthland, The simple answer is that Maxwell determined the mathematical form of Faraday's law of induction himself. There was no pre-existing law or mathematical expression that fit. Faraday expressed his experimental results and theoretical ideas in a sort of quasi-mathematical text description. Maxwell quantified Faraday's description and brought it into agreement with other EM dynamical expressions.

    Of course, Maxwell used all of the physical and mathematical skills and knowledge at his disposal (such as Lagrangian analysis) to ensure that all of EM could be described consistently. So Maxwell's equations are not purely empirical.
  4. Aug 20, 2014 #3
    The equations kind of suggest themselves in the lagrangian formulation, Einstein gravity by A. Zee has a really nice section on this.
    Essentially when you are coupling a vector field to a particle you need to find a lagrangian to describe the dynamics of the vector field.Taking gauge invariance into account you have to form a "curl" on the four potential to get a gauge invariant object, this is the faraday tensor. Then you just do what comes natural and contract the faraday tensor on itself. Varying this will give you maxwells equations.
  5. Aug 20, 2014 #4


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    Staff: Mentor

    For some comments on Faraday's "lines of force" and their connection with Maxwell's theory:

  6. Aug 20, 2014 #5
    in my opinion a very coerent way is to derive everything by coulomb force, lorentz force, biot savart and faraday-neumann. For example, using only biot savart you can see that you can express B with a vector potential:
    [tex]\vec B = \nabla\wedge\vec A[/tex]
    and you know that the div of a rotor is alwais 0, so you can obtain the third maxwell equation:
    [tex] \nabla\cdot \vec B =0[/tex]
    The first equation of maxwell [tex] \nabla\cdot \vec D=\rho[/tex] is a consequence of Gauss theorem, that come by the fact that E is poportional to the inverse of the square of the distance by the source.
    The second is [ŧex] \nabla \wedge \vec E=-\frac{\partial B}{\partial t}[/tex], and come by the lorentz force plus and faraday induction's law.
    The fourth is derivable by these experimental laws (biot savart, coulomb...) only in the stationary case. The correction with [tex]\frac{\partial D}{\partial t}[/tex] looks as experimental valid, but i don't know by what it derives (but i'd like really know!). I know only that the [tex]\nabla \wedge H =J[/tex] is valid only in the stationary case, as you can see taking the divergence of both the sides and using the eq. of the divergence of D.
    Last edited: Aug 20, 2014
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