1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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..


    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

    User Avatar
    Homework Helper
    Gold Member

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


    H = B0
    j = current density vector

    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


    User Avatar
    Homework Helper

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


    [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]

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

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


    User Avatar
    Homework Helper

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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted