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Why is Faraday's law one-sided?

  1. Oct 23, 2015 #1
    Faraday's law states that the curl of E is equal to the negative of the rate of change of a magnetic field. That means that a changing magnetic field causes a curling electric field. But then, shouldn't the opposite be true? That is, shouldn't an electric field with constant circulation generate a changing magnetic flux? This would be similar to the relationship between force and acceleration, with constant mass. If a force is applied, we know that it will cause acceleration. But the opposite is true as well: if an object is accelerating we know that it is under the influence of a force. Shouldn't the law work this way: either field involves the other one?
    Ampere's law states that a curling magnetic field is equal to the rate of change of the electric field, plus the current density. This means that a steady current generates a steady magnetic field. Yet, a steady current means a steady electric field, and according to the above paragraph a steady electric field, hence steady circulation should cause a changing magnetic field. So why is this statement not true?
    Also, since an infinite straight current-carrying wire would produce a magnetic field around it by Ampere's law, would placing a wire in a magnetic field circulating it induce a current? That is, does Ampere's law hold for both sides: a steady magnetic circulation causes a changing electric flux and vice-versa?
    Thanks.
     
  2. jcsd
  3. Oct 23, 2015 #2
    Good thinking!

    Reading about motors and generators might help solidify your understanding. Some helpful search terms are induction motor, slip and counter EMF.
     
  4. Oct 23, 2015 #3
    Thanks a lot for the advice and the search terms. I shall look into them. Again, thanks for the search terms
     
  5. Oct 23, 2015 #4

    Dale

    Staff: Mentor

    Yes. You can switch the left and the right sides of an equation however you like.

    It isn't a good idea to think of Maxwells equations as expressing a cause and effect relationship. Causes always come before effects, so a real cause and effect relationship should have the form ##a(t_E)=b(t_C)## where ##t_E>t_C## and it is said that b is the cause of a or a is the effect of b. Maxwell's equations are not of that form.
     
  6. Oct 23, 2015 #5
    Then if a circulating electric field is around, there will always be a changing magnetic flux? So suppose I have a coil of wire. If the electric circulation around such a wire is zero, then there is no changing magnetic flux. Is it possible then to have a current in the coil of wire without a circulating electric field?
     
  7. Oct 24, 2015 #6

    Dale

    Staff: Mentor

    Yes.
    Certainly. As long as the current is steady you will have that.
     
  8. Oct 25, 2015 #7

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    DaleSpam has a very good point. The Maxwell equations split into two general sorts, namely in the homogeneous ones, i.e.,
    $$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0,$$
    which are constraints in addition to the dynamical equations of motion, which are the inhomogeneous Maxwell equations,
    $$\vec{\nabla} \times \vec{B} -\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho,$$
    where I have used Heaviside-Lorentz (rationalized Gauss) units and wrote down the fundamental Maxwell equations and not the approximate (coarse grained weak-field linear-response) in-medium equations.

    The "constraints" let us introduce the scalar and vector potential (or relativistically the four-vector potential), which however introduces gauge invariance, which makes the explicit causility structure a bit hidden, depending on the choice of gauge, which is arbitrary and made to simplify the problem you want to solve.

    In classical electromagnetism, however we don't need to introduce the potentials but work directly with the electromagnetic field, i.e., ##\vec{E}## and ##\vec{B}## which are gauge invariant and observable quantities, i.e., they should be interpretible in terms of causal equations of motion, and indeed they are.

    The first thing to observe is that the Maxwell equations put a constraint on the possible charge-current-density distributions, which is also a consequence of the underlying gauge invariance. Taking the divergence of the Ampere-Maxwell law gives
    $$-\frac{1}{c} \partial_t \vec{\nabla} \cdot \vec{E}=\frac{1}{c} \vec{\nabla} \cdot \vec{j}.$$
    Now with Gauß's Law we get
    $$\partial_t \rho+\vec{\nabla} \cdot \vec{j}=0,$$
    i.e., even without the equations of motion for the charged particles we know that the electric charge must be strictly conserved, i.e., you have a constraint for the equations of motion of the charges. There must be a (global) symmetry leading to the conservation of the charge you couple the electromagnetic field to. That's a hint of the necessity of gauge invariance of electromagnetism which results from the analysis of the group structure of the proper orthochronous Poincare group, but that's another story.

    The next step is to decouple the equations for ##\vec{E}## and ##\vec{B}## with help of the homogeneous Maxwell equations. To this end it's clear that we should try it with the curl of the Ampere Maxwell Law in order to use Faraday's Law to get rid of ##\vec{E}##:
    $$\vec{\nabla} \times (\vec{\nabla} \times \vec{B})-\frac{1}{c} \partial_t \vec{\nabla} \times \vec{E}=\frac{1}{c} \vec{\nabla} \times \vec{j},$$
    and indeed with Faraday's Law we have
    $$\vec{\nabla} \times (\vec{\nabla} \times \vec{B}) + \frac{1}{c}^2 \partial_t^2 \vec{B}=\frac{1}{c} \vec{\nabla} \times \vec{j}.$$
    Now we work in Cartesian coordinates, and then we can simplify the "double curl" and also use the other homogeneous Maxwell equation (Gauß's Law for the magnetic field, saying that there are no magnetic monopoles):
    $$\left (\frac{1}{c^2} \partial_t^2 - \Delta \right) \vec{B} =\Box \vec{B}=\frac{1}{c} \vec{\nabla} \times \vec{j}.$$
    Now it is clear that the solution of this wave equation is only unique when we invoke the assumption of causility. As any linear differential equation it consists of the full solution of the homogeneous equation, whose concrete form is determined by the initial conditions given by ##\vec{B}## and ##\partial_t \vec{B}## at a time ##t_0##, and the inhomogeneous solution, which relates the sources of the fields, ##\rho## and ##\vec{j}## to the fields and which should obey the causility constraint, i.e., the fields at ##t## should be a functional of the sources at earlier times ##t'<t##. Technically this determines that of all Green's functions of the d'Alembert operator ##\Box## we choose the retarded one, which is very simple since we deal with massless fields and operate in three spactial dimension, where the naive Huygens principle is indeed the correct answer:
    $$\Delta_{\text{ret}}(t,\vec{x})=\frac{1}{4 \pi |\vec{x}|} \Theta(t) \delta(t-|\vec{x}|/c).$$
    Thus the inhomogeneous part (which is the solution for the case where the sources where adiabatically switched on in the remote past) is given by
    $$\vec{B}(t,\vec{x})=\int_{\mathbb{R}^4} \mathrm{d}^4 x' \Delta_{\text{ret}}(t-t',\vec{x}-\vec{x}') \frac{1}{c} \vec{\nabla}' \times \vec{j}(t',\vec{x}').$$
    In the same way we get a wave equation for ##\vec{E}## by taking the time derivative of the Ampere-Maxwell law and eliminate ##\vec{B}## with help of Faraday's Law:
    $$\frac{1}{c} \vec{\nabla} \times \partial_t \vec{B}-\frac{1}{c^2} \partial_t^2 \vec{E}=\frac{1}{c^2} \partial_t \vec{j},$$
    $$-\vec{\nabla} \times (\vec{\nabla} \times \vec{E})-\frac{1}{c^2} \partial_t^2 \vec{E}=\frac{1}{c^2} \partial_t \vec{j}.$$
    Again resolving the double curl for Cartesian components and using Gauß's Law for the electric field gives
    $$\Box \vec{E}=\vec{\nabla} \rho -\frac{1}{c^2} \partial_t \vec{j}.$$
    The causal solution is again given with help of the retarded propagator.

    These two retarded solutions can be simplified by integrating out the ##\delta## distributions, where one must be a bit careful with the derivatives of ##\rho## and ##\vec{j}## occuring in the integrands. The resulting equations are, for some strange reason, named Jefimenko equations, although the retarded solutions of the Maxwell equations reach back to the mid 18's by Lorenz (the Danish physicist, not the Dutch Lorentz):

    http://en.wikipedia.org/wiki/Jefimenko's_equations

    For our argument, the somewhat simpler looking forms with the ##\delta## distribution in the integrals make however clear the conceptual structure behind the Maxwell equations: The homogeneous ones are constraints while the inhomogeneous ones are the equations of motion connecting the sources, ##\rho## and ##\vec{j}## causally to the corresponding "field response".

    You can of course also find other equations, connecting, e.g., the displacement current as a part of the sources of the magnetic field, but as soon as you want the full solution of the problem to derive the electromagnetic field from given charge-current distributions, you'll see that this point of view is utmost more complicated (and I dare say even "ugly"), because you get a pretty nonlocal way to express the magnetic field with the displacement current as a source, which also hides the beautiful "causal structure" of the Jefimenko expressions, which are conceptually much cleaner and last but not least much more in the spirit of relativistic field theory.
     
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