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Is a faraday Field an electric field?

  1. Jun 11, 2012 #1
    A changing magnetic field induces a field, G:

    ∇×G = -∂B/∂t (1)

    Now my book then says, since ∇×G≠0 while ∇×E=0 and ∇[itex]\cdot[/itex]G=0 while ∇[itex]\cdot[/itex]E ≠0 we might aswell see G and E as the same field, and thus see G as an induced electric field.
    I can understand that, I just don't understand how one finds that ∇[itex]\cdot[/itex]G≠0. My book says it is because there exists no source charges, but I find that a bad argument since the fact that E is not divergenceless comes from the mathematical nature of Coulombs law rather than the source charges.

    So my question:
    How does one show that the above law (1) means that E is divergenceless?
     
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  3. Jun 11, 2012 #2

    vanhees71

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    I don't know, what you (or your book) means by [itex]\vec{G}[/itex]. Faraday's Law is one of the fundamental equations of electromagnetism (alltogether known as Maxwell's equations). There is no additional field. There is only one electromagnetic field, which in the notation of the three-dimensional formalism, is given by the electric and magnetic field components [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex].

    There are the two homogeneous Maxwell equations

    [tex]\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E} + \frac{1}{c} \frac{\partial}{\partial t}\vec{B}=0[/tex]

    and the two inhomogeneous ones, coupling the em. field to its sources, which are electric charge and current densities,

    [tex]\vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\vec{j},[/tex]

    all written in the Heaviside-Lorentz system of units.
     
  4. Jun 11, 2012 #3
    I think you misunderstand what I am saying, so I will refer directly to my book (Attached picture), which is Griffiths.
    My question is: How does he conclude that the field induced by the changing magnetic field has zero divergence?
     

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  5. Jun 11, 2012 #4
    It is defined that way, so E+G will satisfy the ordinary Maxwell equations, which we know to be true. We know that the source term in the divergence equation represents electric charges, so the splitting of the electric field that is mentioned in the book is simply defined so as to have a vanishing source term in the divergence equation for G, so we can say that one part of the field is caused by electric charges, while the other part does not...
     
  6. Jun 11, 2012 #5
    You should conclude that [itex]\nabla \cdot G = 0[/itex] because [itex]\nabla \cdot (\nabla \times A) = 0[/itex] for any vector field [itex]A[/itex].
     
  7. Jun 11, 2012 #6
    But it is not stated in the assumptions on that page that G is the curl of some vector field A. As far as I can see, this all boils down to a convenient definition for a splitting of the electric field.
     
  8. Jun 11, 2012 #7

    vanhees71

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    Well, in Griffiths [which is a quite good book as far as I can say, but I haven't studied it very carefully. My favorites are Sommerfeld's Lectures on Theoretical Physics 3, Schwinger, Classical Electrodynamics and of course Jackson, Classical Electrodynamics] he correctly writes Faraday's Law with the electric field and not some new field, [itex]\vec{G}[/itex].

    You can of course not conclude that [itex]\vec{\nabla} \cdot \vec{E}=0.[/itex] In general this is wrong, because [itex]\vec{\nabla} \cdot \vec{E}=\rho[/itex], where [itex]\rho[/itex] is the density of electric charge.

    Of course, with the other homogeneous Maxwell equation (see my previous posting) tells you that [itex]\vec{\nabla} \cdot \vec{B}=0[/itex]. From Helmholtz's fundamental theorem of vector calculus you can conclude that the magnetic field is always a solenoidal field, i.e., there is a vector potential [itex]\vec{A}[/itex] such that

    [tex]\vec{B}=\vec{\nabla} \times \vec{A}.[/tex]

    Substituting this into Faraday's Law, which is the other homogeneous Maxwell equations, gives

    [tex]\vec{\nabla} \times \left (\vec{E}+\frac{1}{c} \frac{\partial \vec{A}}{\partial t} \right )=0.[/tex]

    Here, I've interchanged the spatial derivatives with the time derivatives (assuming the appropriate smoothness conditions for the fields). These equations tell you, again according to Helmhotz's theorem that the field in parentheses is a potential field, i.e., there exists a scalar potential [itex]\Phi[/itex] such that

    [tex]\vec{E}=-\vec{\nabla} \Phi-\frac{1}{c} \frac{\partial \vec{A}}{\partial t}.[/tex]

    The scalar and vector potential for a given electromagnetic field [itex](\vec{E},\vec{B})[/itex] is only determined up to a gauge transformation, i.e., with [itex]\Phi[/itex] and [itex]\vec{A}[/itex] also the fields

    [tex]\Phi'=\Phi-\frac{1}{c} \partial_t \chi, \quad \vec{A}'=\vec{A}+\vec{\nabla} \chi[/tex]

    with an arbitrary scalar field [itex]\chi[/itex] gives the same [itex](\vec{E},\vec{B})[/itex].
     
  9. Jun 11, 2012 #8
    Okay, I'm not good enough to understand all of that unfortunately. But from what I can get you are using the fact, that you can always make a vector potential divergenceless - or putting it another way: The Faraday law specifies the curl of E, not the divergence, to which we are at liberty to assign any value, zero being the smartest.
     
  10. Jun 12, 2012 #9

    vanhees71

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    No! There's another Maxwell equation specifying the divergence to be the charge distribution (Gauss's Law!).
     
  11. Jun 12, 2012 #10
    I am talking about the field G, which is a pure faraday field.
     
  12. Jun 13, 2012 #11

    vanhees71

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    If you introduce non-standard notation and non-standard naming, you have to clearly define the quantities. What's a "pure Farayday field"? I've never heard this notion before, and there is no such thing in classical Maxwell theory. Neither is there a field commonly called [itex]\vec{G}[/itex]. You may also point to a page in Griffiths. I'm not so sure anymore, whether this is really a well-understandable book, if it introduces strange fields into the anyway not so simple physics of electromagnetism :-(.
     
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