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Maxwell Equations when current density is time independent and divergence free

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data

    If the current density is time independent and divergence free, show that the Maxwell Equations separate into independent equations for [tex]\vec{E}[/tex] and [tex]\vec{B}[/tex].

    2. Relevant equations

    Maxwell's equations

    3. The attempt at a solution

    The only Maxwell equation with [tex]\vec{j}[/tex] in it is the Maxwell-Ampere law, so that seems like the right place to start. By taking the partial derivative with respect to time of this equation and using the fact [tex]\vec{j}[/tex] is time independent, Faraday's Law and Gauss' Law for [tex]\vec{B}[/tex] I can get a wave equation for [tex]\vec{B}[/tex].

    What is confusing me is how to use the fact [tex]\vec{j}[/tex] is divergence free. If I take the divergence of the Maxwell-Ampere equation I get:

    ∇⋅∇x[tex]\vec{B}[/tex] = εµ (∂∇⋅[tex]\vec{E}[/tex]/∂t) + µ ∇[tex]\vec{j}[/tex]

    The LHS = 0 (vector identity), ∇[tex]\vec{j}[/tex] = 0 as given, and then using Gauss' Law for [tex]\vec{E}[/tex] I simply get:

    (∂ρ)/(∂t) = 0

    But this isn't surprising as the equation of charge conservation would have given me this anyway. How do I get the equation for [tex]\vec{E}[/tex]? Thanks.
     
  2. jcsd
  3. Nov 22, 2008 #2
    In the presence of a current, the charge conservation equation is actually ∇⋅[tex]\vec{j}[/tex] + (∂ρ)/(∂t) = 0. So you need the fact that the divergence of the current is zero to obtain (∂ρ)/(∂t) = 0. You did that, so you are halfway there.

    Now you do the same thing you did for B. You have a condition on ρ; find which Maxwell's equation involves ρ, and go from there.
     
  4. Nov 22, 2008 #3
    I think I get what you're saying. Is this correct:

    (∂ρ/∂t) = 0 => ρ = constant

    Take the curl of Faraday's Law:

    ∇x∇x[tex]\vec{E}[/tex] = -(∂∇x[tex]\vec{B}[/tex]/∂t)

    From Maxwell-Ampere Law can substitute into RHS and use identity for LHS:

    ∇(∇⋅[tex]\vec{E}[/tex]) - ∇²[tex]\vec{E}[/tex] = -∂/∂t[ εµ(∂[tex]\vec{E}[/tex]/∂t) + µ[tex]\vec{j}[/tex] ]

    Subbing in Gauss' Law for [tex]\vec{E}[/tex] and using the fact ρ = constant cancels the first part of LHS and fact [tex]\vec{j}[/tex] is time independent fixes last part of RHS, thus get wave equation for E to add to the one for B above.

    Thanks.
     
  5. Nov 22, 2008 #4
    Ooh you need to be careful. (∂ρ/∂t) = 0 just means that the partial derivative of ρ with time is zero, i.e. ρ doesn't change with time. It doesn't mean that ρ is the same everywhere in space. In fact, you should get a different wave equation for E than for B.
     
  6. Nov 22, 2008 #5
    So is the right approach to not cancel ρ, but instead leave the answer as:

    εµ (∂²[tex]\vec{E}[/tex]/∂t²) - ∇²[tex]\vec{E}[/tex] = ∇(ρ/ε)

    as the second equation? If this is so then was (∂ρ/∂t) = 0 even needed?

    I'm really struggling with manipulating these equations :S. I think I just need to get more practice.

    This one question in a series on Maxwell's Equations. The next ones says show that the two equations above (so the inhomogeneous wave equation for E and the homogeneous wave equation for B) are solved in the Coulomb gauge when E is written as the electrostatic potential and B is written as the vector potential, and both φ and [tex]\vec{A}[/tex] satisfy Poisson's equation.

    I sub everything in, but I can't see at all where to start with this one, none of the identities I have allow me to manipulate the equations so that I can introduce the Coulomb gauge or the fact the two potentials satisfy Poisson's equation. Any suggestions? Thanks
     
  7. Nov 22, 2008 #6
    You're right, ∂ρ/∂t = 0 wasn't really needed I guess. Yes, this stuff does become easier with practice.

    Write down the fields in the Coulomb gauge. It might be easier to use these in Maxwell equations rather than the wave equations, since you know that they are equivalent.
     
  8. Nov 22, 2008 #7
    The first question is to derive the two waves equations, and then the following question is to show that these equations are consistent with the potentials, Coulomb gauge and Poisson's equation.

    I agree that the wave equations are essentially equivalent to Maxwell's equations, but I think the question wants you to specifically work with the wave equation.

    If that is what is required I have no idea where to start. For the homogeneous case (which would seemingly be the simpler of the two) you get:

    εµ (∂²∇x[tex]\vec{A}[/tex]/∂t²) - ∇²(∇x[tex]\vec{A}[/tex]) = 0

    I have no idea how to simplify this any further, and the inhomogeneous case for φ runs me into the same problems.
     
  9. Nov 22, 2008 #8
    Hmm, can you state the problem verbatim please?
     
  10. Nov 22, 2008 #9
    There are a couple of introductory questions that aren't relevant to the last parts then it says:

    (d) If the current density is time independent and divergence free, show that the Maxwell Equations separate into independent equations for [tex]\vec{E}[/tex] and [tex]\vec{B}[/tex].
    [This was the one I asked originally].

    (e) Express [tex]\vec{E}[/tex] in terms of the electrostatic potential φ, and [tex]\vec{B}[/tex] in terms of the vector potential [tex]\vec{A}[/tex]. Show that the equations in question (d) are satisfied when (in Coulomb Gauge) φ and [tex]\vec{A}[/tex] satisfy Poisson equations. Write down those equations.

    I can obviously write down the Poisson equations and the fields in terms of the potentials but that is about as far as I can get with (e).
     
  11. Nov 22, 2008 #10
    I just don't understand (e), sorry :confused: Maybe someone else can shed more light. You cannot plug in [tex]\vec{E} = -\nabla\phi[/tex] into the wave equation, that will just not give you anything. You can start from Maxwell's equations, and derive "wave equations" for the two potentials, but that doesn't seem to be what the question is.

    Also, the equations in (d) will be satisfied no matter what gauge you choose for the potentials. So I'm really confused. Sorry :frown:
     
  12. Nov 23, 2008 #11
    No worries mate, it's always comforting to know it's not just me who is confused! Subbing in the potentials just led to a mess for me.

    I might post it as a separate question and see if I have any luck. Thanks for your help.
     
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