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Why does (current density) dJ/dt=0?

  1. Sep 1, 2014 #1
    could someone explain to me why the the time derivative of current density J would be zero?

    more specifically..

    $$\mu_0\frac{∂\vec{J}}{∂t}=0$$


    trying to derive maxwell's wave equation for E-field and i guess one of the assumptions i have to make is the above. but if i'm making a wave, i thought i had to have accelerating charges in which case why would the above be true?
     
  2. jcsd
  3. Sep 1, 2014 #2

    rude man

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    The Maxwell equation you are referring to is ∇ x H = j + ∂D/∂t

    where

    H = B0
    j = current density vector
    D0E.

    Fact: e-m waves are created by virtue of the ∂D/∂t term, not the j term.

    BTW I am not aware that ∂j/∂t has to be zero.
     
  4. Sep 1, 2014 #3
    yea that's the equation i'm referring to (the corrected ampere's law).


    I agree that a constant J would not produce a wave, however ∂J/∂t should. and here's where i get ∂J/∂t:


    if we set this equation equal to faraday's law in differential form:

    $$\frac{∂}{∂t}(\vec{\nabla} \times \vec{B})= \frac{∂}{∂t}(\mu_0(\vec{J}+\epsilon_0\frac{∂\vec{E}}{∂t}))$$

    $$\nabla \times \frac{∂\vec{B}}{∂t}=-\vec{\nabla} \times (\vec{\nabla} \times E)$$


    $$\frac{∂}{∂t}(\mu_0(\vec{J}+\epsilon_0\frac{∂\vec{E}}{∂t}))=-\vec{\nabla} \times (\vec{\nabla} \times E)$$

    this is the place i'm referring to. why does the [itex]\mu_0 \frac{∂\vec{J}}{∂t}[/itex] term disappear?
     
  5. Sep 1, 2014 #4

    ehild

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    [tex]\vec{\nabla} \times \vec{B}= \mu_0\left(\vec{J}+\epsilon_0\frac{∂\vec{E}}{∂t}\right) [/tex]

    and

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

    Take the curl of the second one,

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

    and substitute the first one
    --->

    [tex]\vec{\nabla}\times\vec{\nabla} \times \vec{E}= -\frac{∂}{∂t}\mu_0\left(\vec{J}+\epsilon_0\frac{∂\vec{E}}{∂t}\right) [/tex]

    ehild
     
  6. Sep 1, 2014 #5
    exactly what i did..

    still don't see how that term disappears
     
  7. Sep 1, 2014 #6

    ehild

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    What do you mean that J disappears? It is still there.

    ehild
     
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