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Maxwell's curl equations

  1. Sep 8, 2015 #1
    I first learned Maxwell's equations in their integral form before I was introduced to the differential form, i.e. w/curl & divergence.

    As I understand, in order to derive the curl form from the integral form, apply Stokes Theorem to the integral form of

    ∫(closed)E⋅dl=-d/dt[∫(closed)B⋅dA],

    and you first need to commute the 'd/dt' through the integrand, where it becomes it a partial derivative:

    ∫(closed)E⋅dl=-∫(closed)[∂B/∂t⋅dA]

    It is this commutation which gives it the form of Stoke's Theorem, which is how we derive

    curl E = -∂B/∂t.

    What I don't understand is how this commutation is permissible unless we know the limits of integration on the right side intregand are constant with respect to time... and ... without this commutation through the integrand, it doesn't really take the form of Stoke's Theorem, does it?

    In short, I wonder of this curl equation has some limitations:

    Imagine the B vector is constant everywhere with respect to time, and there is a shrinking loop of wire in the area. Well, according to the relation

    curl E = -∂B/∂t,

    we ought to expect that curl E is zero everywhere ... but it (intuitively?) simply cannot be zero everywhere ... at least not if there is a non-zero integral of E around the loop, correct?




     
  2. jcsd
  3. Sep 8, 2015 #2
    If I may make a quick correction, I should have presented the integral form as

    ∫(closed)E⋅dl=-d/dt[∫B⋅dA],

    and not as

    ∫(closed)E⋅dl=-d/dt[∫(closed)B⋅dA]

    like I did, because the surface integral on the right is NOT around a closed surface. (sorry).
     
  4. Sep 9, 2015 #3

    vanhees71

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    You hit the nail on the head, as we say in Germany, and you are very right in questioning the exchange of spatial integration and time differentiation.

    The first message is that the integral form of the laws you wrote down, are not precise without stating that they are valid only for all domains of integration at rest. The local laws, i.e., Maxwell's equations in differential form are always valid, and they are the form which is most natural from the point of view of relativistic classical field theory, which is underlying classical electromagnetism. So let's take Faraday's Law as an example. Its local form, which is always valid, reads (in the obviously used SI units, which I don't like, but anyway):
    $$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.$$
    Now you are right in saying that you can bring this in an integral form by applying Stokes's theorem of vector calculus. This is triggered by the occurance of the curl operation on the left-hand side, because if you take an arbitrary surface ##F## with surface-normal vectors ##\mathrm{d} \vec{F}## and its closed boundary curve ##\partial F##, which is orientied relative to the surfrace-normal vectors according to the right-hand rule, you get via Stokes's theorem
    $$\int_{F} \mathrm{d} \vec{F} \cdot (\vec{\nabla} \times \vec{E})=\int_{\partial F} \mathrm{d} \vec{x} \vec{E}(t,\vec{x}).$$
    In the final step, I've written out the space-time arguments of the electric fields. It's very clear that you integrate here along the surface at a fixed time ##t## (in the inertial reference frame under consideration!).

    Now the integral form of Faraday's law reads
    $$\int_{\partial F} \mathrm{d} \vec{x} \cdot \vec{E}(t,\vec{x})=-\int_F \mathrm{d} \vec{F} \cdot \partial_t \vec{B}(t,\vec{x}).$$
    Now it's clear that also on the right-hand side you integrate the time derivative of the magnetic field at constant time ##t##, and that's the form that is always valid also for a moving surface and boundary.

    Now the usual integral form of Faraday's Law involves the magnetic flux, which depends on the magnetic field and the surface:
    $$\Phi_F(t)=\int_{F} \mathrm{d} \vec{F} \cdot \vec{B}(t,\vec{x}).$$
    Again, we integrate at a fixed time ##t##.

    If (AND ONLY IF) the surface is not moving in the here considered frame of reference, you have
    $$\frac{\mathrm{d}}{\mathrm{d} t} \Phi_F(t) \stackrel{*}{=} \int_F \mathrm{d} \vec{F} \cdot \partial_t \vec{B}(t,\vec{x}),$$
    because then time is just a parameter in the function under the integral and (provided ##\vec{B}## is well-behaved enough, and this we assume as physicists always but should be aware that one must be careful, if there are singularities of ##\vec{B}##) then you can indeed interchange the time derivative and the spatial integration. I wrote ##\stackrel{*}{=}## to alert the readers to keep in mind that the equation only holds under the constraint that the surface and its boundary is not moving. Then (AND ONLY THEN) the simple form of Faraday's Law holds:
    $$\int_{\partial F} \mathrm{d} \vec{x} \cdot \vec{E}(t,\vec{x}) \stackrel{*}{=}-\frac{\mathrm{d}}{\mathrm{d} t} \Phi_F(t).$$
    If your surface and its boundary is time dependent, you must be more careful, because then
    $$\frac{\mathrm{d}}{\mathrm{d} t} \Phi_F(t)=\lim_{\Delta t \rightarrow 0} \frac{1}{\Delta t} \left [\int_{F(t+\Delta t)} \mathrm{d} \vec{F} \cdot \vec{B}(t+\Delta t,\vec{x})-\int_{F(t)} \mathrm{d} \vec{F} \cdot \vec{B}(t,\vec{x}) \right ].$$
    A careful analysis, also using Gauß's Law for the magnetic field,
    $$\vec{\nabla} \cdot \vec{B}=0,$$
    which tells you that there are no magnetic monopoles, then gives the complete Faraday Law in integral form
    $$\mathcal{E}=\int_{\partial F} \mathrm{d} \vec{x} \cdot [\vec{E}(t,\vec{x})+\vec{v}(t,\vec{x}) \times \vec{B}(t,\vec{x})]=-\frac{\mathrm{d}}{\mathrm{d} t} \Phi_F(t),$$
    i.e., under the line integral you must have the additional ##\vec{v} \times \vec{B}## term with ##\vec{v}(t,\vec{x})## describing the momentaneous velocity of the surface and its boundary. The line integral is the correct and complete electromotive force ##\mathcal{E}## along the boundary of the surface, and it is taken again at a fixed time ##t##.

    For an excellent derivation of the correct Faraday Law, see Wikipedia:

    https://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof_of_Faraday.27s_law
     
  5. Sep 9, 2015 #4

    ChrisVer

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    Hmm isn't that form very similar to the Lorentz force? Is there some connection that allowed you to write it down fast, since you considered a moving surface?



    PS
    Since I'm studying in Germany it would be fun to know how you say that :biggrin:
     
  6. Sep 9, 2015 #5
    ∂∂∂
    Thank you very much for your reply!

    Getting back to my shrinking loop of wire in an area of time-constant B field, would the following analysis be correct, in light of what you've said? (Let us assume either no induced current, or negligible induced current in my shrinking loop, so it won't affect the analysis.)

    1) Because B is constant everywhere in the area of my shrinking loop, ∇ x E is indeed zero everywhere inside -- and outside -- the shrinking loop since, as you've said, the local form of Faraday's law, i.e., curl E = -∂B/∂t, is ALWAYS correct, even when evaluating moving material.

    2) And because ∇ x E is zero everywhere inside the shrinking loop, by Stokes's theorem ∫E⋅dx must be zero around the perimeter of the shrinking loop, as intuitively strange as this seems for me to grasp!

    3) But there IS, however, an electromotive force around the loop, due ... not to E ... but, rather, to v x B; the 'v' being the velocities of the differential-length segments of the shrinking loop as it shrinks.

    4) The integral form of Faraday with which I'm most familiar,
    ∫(closed loop)E⋅dx = -d/dt∫(surface)B⋅dA
    in which there's a TOTAL time derivative OUTSIDE the integrand, cannot be relied upon when evaluating moving closed loops, because if we did adhere to it, we would, for example, incorrectly conclude that there IS non-zero ∫E⋅dx around my shrinking loop, when in fact, there is none ... only ∫(v x B)⋅dx.

    Am I at least close? Thanx.
     
    Last edited: Sep 9, 2015
  7. Sep 10, 2015 #6

    vanhees71

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    Of course, you get the Lorentz force from the Maxwell equations by considering the energy-momentum balance of the electromagnetic-field energy-momentum-stress tensor. For a closed system you can infer the mechanical energy-momentum tensor from the conservation of total energy and momentum. Unfortunately I'm far from this point in my SRT article, but it's crucial for the resolution much unnecessary confusion like "hidden momentum" and also the obstacle with the radiation-reaction problem, where the idea of classical point particles within classical electrodynamics clearly comes to an end of applicability in a strict sense.

    The German saying is: "Du triffst den Nagel auf den Kopf!"
     
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