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Question regarding Curl of Charge Flow, and the vanishing constants in Faraday's law.

  1. Nov 10, 2008 #1
    I am trying to derive the electromagnetic wave equations from Faraday's law of Induction, and the Ampere-Maxwell law.

    But, I am having a problem with the 1/c^2 disappearing.

    This is what I am using:

    [tex]\nabla\times\vec{B}=\mu\vec{J}+\mu\epsilon\stackrel{\partial\vec{E}}{\partial t}[/tex]

    Taking the cross product of both sides, I eventually end up with...

    [tex]\nabla (\nabla\cdot\vec{B}) - \nabla^{2} (\vec{B}) = \mu\nabla\times\vec{J}+\mu\epsilon\stackrel{\partial(\nabla\times\vec{E})}{\partial t}[/tex]

    Divergence of B is 0, so that goes away.

    I'm stuck, though, when it comes to the curl of the charge flow. I know that the charge flow is as follows:

    [tex]\vec{J}=\sigma\vec{E}[/tex]



    Does the curl of the charge flow vanish? Faraday's law of Induction would imply that I'd be taking the curl of the negative time rate of change of the magnetic field. I'd ideally like to see the curl of the charge flow disappear, unless something else appears when talking about the charge flow through a medium. I understand that space is essentially non-conductive, and if you are deriving these equations in the vacuum of space, you won't have a charge flow. I'd like to not make that assumption, though, and it would be great if I ended up with an extremely generalized form of the wave equation, which could show the speed of light as being different through different media.


    Why doesn't mu and epsilon show up in Faraday's law of Induction? I'd imagine that the relationship between the Displacement field and Magnetization field would still end up producing constants. And, if it does, then I'm in big trouble, because my 1/c^2 disappears due to the [tex]\mu\epsilon[/tex] canceling out with the [tex]\mu\epsilon[/tex] in the curl of Ampere-Maxwell's law.

    Maybe after more discussion, I can clear my mind a bit on why I'm hung up on this concept.
     
  2. jcsd
  3. Nov 11, 2008 #2

    atyy

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    Re: Question regarding Curl of Charge Flow, and the vanishing constants in Faraday's

    There can be a current in "empty space" (not really, by definition, but you know what I mean). The most common form of the wave equation is derived assuming no currents since a wave can travel far from the charges that caused it.

    The speed of light in materials is different because the permittivity/permeabilities/dielectric constants are different from vacuum. Also in materials some sort of spatially avergaed field is considered, since the true microscopic field is very inhomogeneous and unknown.
     
  4. Nov 11, 2008 #3
    Re: Question regarding Curl of Charge Flow, and the vanishing constants in Faraday's

    So, what does the curl of the charge flow reduce to in this case?

    Also, why aren't there constants [tex]\mu\epsilon[/tex] in Faraday's law?
     
  5. Nov 12, 2008 #4
    Re: Question regarding Curl of Charge Flow, and the vanishing constants in Faraday's

    Does anyone have any idea what my error in thought is?

    I basically want to know where the constants in Faraday's law are, and what the curl of a charge flow is.
     
  6. Nov 12, 2008 #5

    gabbagabbahey

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    Re: Question regarding Curl of Charge Flow, and the vanishing constants in Faraday's

    In classical electrodynamics, "in vacuum" means a region devoid of all charges....so, the charge density and the current density are both zero in vacuum.

    Assuming you are deriving the wave equation in a vacuum, then Faraday's Law is actually in terms of the permittivity [itex]\epsilon_0[/itex] and permeability of free space [itex]\mu_o[/itex]:

    [tex]
    \nabla \times \vec{B} =\mu_0 \vec{J}+\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}
    [/tex]

    When deriving the wave equation in a general medium on the other hand, you need to use Faraday's Law in terms of the Auxiliary and displacement fields instead:

    [tex]
    \nabla\times\vec{H}=\vec{J_f}+ \frac{\partial \vec{D}}{\partial t}
    [/tex]

    In this case, all that matters is that there is no free charge or free current inside the medium, so that [itex]\vec{J}_f=0[/itex]

    In either case, when taking the curl of zero you naturally get zero!
     
  7. Nov 13, 2008 #6
    Re: Question regarding Curl of Charge Flow, and the vanishing constants in Faraday's

    That's Ampere's circuital law with the Maxwell correction, relating the induced magnetomotive force to the charge flux and electric flux. I understand how this law works, but I still don't quite understand why the curl of the charge flow is 0. You're saying that the charge flow represents free charges? How do you know that there is no free charge flow through the medium?

    I can assume it's zero and deriving the EM wave equation becomes easy, but I'm still not sold on why J is 0, plus it wouldn't be a generalized proof. For instance, how would it be 0 if there is indeed charge flow inside the medium?

    Additionally, once J is 0, I still end up with trying to convince myself why there are no constants in Faraday's law, which directly relates the electric field to the magnetic field, without any sort of scaling.

    [tex]\nabla\times\vector{E}=-\frac{\partial \vec{B}}{\partial t}[/tex]

    If D = epsilon*E, and B = mu*H, then why isn't Faraday's law actually...

    [tex]\nabla\times\vector{E}=-\mu\epsilon\frac{\partial \vec{B}}{\partial t}[/tex]?
     
  8. Nov 13, 2008 #7

    jtbell

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    Re: Question regarding Curl of Charge Flow, and the vanishing constants in Faraday's

    What you're looking for looks like it's discussed in most intermediate-level E&M textbooks, under the subject of "Electromagnetic Waves in Conductors." Two examples that I have on my bookshelf are Griffiths (section 9.4) and Pollak & Stump (section 13.3).

    I don't want to try to reproduce their derivations here because I'm not familiar with this topic myself, so I'm not in a good position to answer questions about it. Do you have a library nearby?

    [added] I just remembered a set of online lecture notes which includes this topic:

    http://farside.ph.utexas.edu/teaching/em/lectures/node102.html
     
    Last edited: Nov 13, 2008
  9. Nov 13, 2008 #8

    tiny-tim

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    It's the dimensions …

    E has the same dimensions as cB

    and ∂/∂t has the same dimensions as c∇

    So [tex]\nabla \times \vec{B} =\mu_0 \vec{J}+\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}[/tex]

    but [tex]\nabla\times\vector{E}\ =\ -\frac{\partial \vec{B}}{\partial t}[/tex] :smile:
     
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