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Conservation of energy (wave in a conductor)

  1. Sep 26, 2006 #1

    quasar987

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    I need to prove conservatino of energy in a conductor of real condictivity [itex]\sigma[/itex]. I need opinion on how to proceed. My best guess was to use Poynting's theorem in differential form, which says that the rate at which work is done per unit time per unit volume on the charges in the metal by the EM wave is given by

    [tex]-\frac{\partial u}{\partial t}-\nabla \cdot \vec{S}[/tex]

    where u is the E-M energy density and S the Poynting vector. Alright, so since this amount of work is done by unit time by unit volume, it must be that the energy in the wave decreases at the opposite rate! An expression of the decrease in wave energy per unit time per unit volume I found is

    [tex]\frac{\partial u}{\partial z} v_{ph}[/tex]

    where v_ph is the phase velocity.

    We made the computation by plugging E = a damped monochormotic plane wave and B = the associated B field, but the two sides were pretty far from equal. In particular, the phase difference btw E and B was particularly bugging.

    By the way, I need to find the solution today (it's 18h30 here) absolutely, so help is eminently needed! Heeeelp!
     
  2. jcsd
  3. Sep 26, 2006 #2
    If I remember correctly, doesn't the "real conductor" part give you a clue to the wave vector, i.e. this gives rise to an approximation, which simplifies the calculation. What surface are you constructing to consider the energy flow through? My understanding is a little weak, but its coming back. I don't know if this helps, but I remember a problem I had similar to this, and those two things helped me.
     
  4. Sep 27, 2006 #3
    I now have an idea to tackle this problem. Consider the Maxwell equations

    [itex] \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon} [/itex]
    [itex] \nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t }[/itex]
    [itex] \nabla \cdot \mathbf{B} = 0 [/itex]
    [itex] c^2 \nabla \times \mathbf{B} = \frac{\mathbf{j}}{\epsilon} + \frac{\partial \mathbf{E}}{\partial t } [/itex]

    But since the considered piece of matter is a real conductor, [itex] \mathbf{j} = \sigma \mathbf{E} [/itex] and [itex] \rho=0 [/itex].

    [itex] \nabla \cdot \mathbf{E} = 0 [/itex]
    [itex] c^2 \nabla \times \mathbf{B} = \frac{\sigma \mathbf{E}}{\epsilon} + \frac{\partial \mathbf{E}}{\partial t } [/itex]

    With this you can derive a wave-equation for the damped wave in the conductor. If you can (as you said above) show that the work done by the field is equal to the loss to the "friction term", you've shown the conservation of energy. Perhaps this is what you already did ...
     
    Last edited: Sep 27, 2006
  5. Sep 27, 2006 #4
    Something that still confuses me: I think you have to know how much heat a real conductor "produces" if it has the current density [itex] \mathbf{j} [/itex], since you can't derive that from the Maxwell equations (as far as I see), should be something like [itex]P=RI^2[/itex] in "local" form. So maybe it's sufficient to know this and calculate the work done by the field as you did above?
     
    Last edited: Sep 27, 2006
  6. Sep 27, 2006 #5

    quasar987

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    P=RI² in local form would be J.E, or as you said, (sigma)E². This is the rate at which work is done on the charges in the metal per unit volume by the fields. It is, if you will, the left hand side of the Poynting thm. The right hand side is my first equation in the original post. This result is known.

    What we need to do to show conservation of energy is to craft an expression for the rate of local loss in E-M energy density and show that it is equal to either side of Poynting.

    Someone else from my class found a solution, but it is not the most general solution in that it does not show instantaneous conservation of energy but rather conservation in the mean. The solution consists of taking the mean value (over time) of the energy density u. Set z=vt (i.e. put yourself in the wave referential) and differentiate that wrt time. If energy is conserved, this should be equal to the rate at which mechanical energy density increases, i.e. <[itex]\sigma[/itex]E²>(z=vt,t). And it works.
     
  7. Sep 27, 2006 #6
    I'll show you some calculations I did, at the moment I don't know what they mean myself. :smile:

    First, you can consider one side of "Poynting":

    [itex] \nabla \cdot \vec S = \nabla \cdot (\vec B \times \vec E) [/itex]
    Knowing

    [itex] \nabla \cdot (\vec C \times \vec D) = \vec C \cdot (\nabla \times \vec D) - \vec D \cdot (\nabla \times \vec C) [/itex],
    this gets
    [itex]\nabla \cdot (\vec B \times \vec E) = \vec B \cdot (\nabla \times \vec E) - \vec E \cdot (\nabla \times \vec B) [/itex].

    But now you can insert the "modified" Maxwell equations from above

    [itex]\nabla \cdot (\vec B \times \vec E) = - \vec B \cdot \frac{\partial \vec B}{\partial t } - \vec E \cdot \frac{1}{c^2}(\frac{\sigma \vec E}{\epsilon} + \frac{\partial \vec E}{\partial t }) [/itex].

    This looks very much like the (negative) derivative of the energy density, plus an additional term, and we get the result
    [itex] \nabla \cdot \vec S = -\frac{\partial }{\partial t} \frac{\vec B^2 + \vec E^2}{2} - \vec E \cdot \frac{1}{c^2} \frac{\sigma \vec E}{\epsilon} = - \frac{\partial }{\partial t} u - \frac{1}{c^2} \frac{\sigma \vec E^2}{\epsilon}[/itex]

    edit: Latex typos are freaking me out. :D
     
    Last edited: Sep 27, 2006
  8. Sep 27, 2006 #7

    quasar987

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    Actually what you did is rederive Poynting's theorem. There are a few errors in the constants but if you fix them, you find exactly

    [tex]\sigma \vec{E}^2 =-\frac{\partial u}{\partial t}-\nabla \cdot \vec{S}[/tex]

    There are 3 constant-related errors:
    1° You forgot to include the 1/c² coefficient of E² when you wrote your last expression for the divergence of S ([itex] -\frac{\partial }{\partial t} \frac{\vec B^2 + \vec E^2}{2} - \vec E \cdot \frac{1}{c^2} \frac{\sigma \vec E}{\epsilon} [/itex]). With it, your last term becomes

    [itex] -\frac{\partial }{\partial t} \frac{\vec B^2 + \frac{1}{c^2}\vec E^2}{2} - \vec E \cdot \frac{1}{c^2} \frac{\sigma \vec E}{\epsilon} [/itex]

    2° When in matter, you cannot use c² in Maxwell's equation. 1/c² is [itex]\epsilon_0 \mu_0[/itex]. But in matter, the permitivity and permeability become [itex]\epsilon[/itex] and [itex]\mu[/itex] respectively. And while [itex]\epsilon \mu[/itex] is the inverse square of the velocity for a wave in a dielectric ([itex]\sigma = 0[/itex]), in a conductor, the velocity is more complicated. So if I rewrite your last term, it becomes

    [itex] -\frac{\partial }{\partial t} \frac{\vec B^2 + \epsilon \mu \vec E^2}{2} - \vec E \cdot \mu \sigma \vec E [/itex]

    3° The way S is defined is (E x B)/mu, so the equations becomes

    [itex] \nabla \cdot \vec{S} = -\frac{\partial }{\partial t} \frac{ \frac{1}{\mu}\vec{B}^2 + \epsilon \vec{E}^2}{2} - \vec{E} \cdot \sigma \vec E =\frac{\partial u}{\partial t}-\sigma \vec{E}^2[/itex]
     
    Last edited: Sep 27, 2006
  9. Sep 27, 2006 #8
    Yeah, thx. Well, if you get the solution (by yourself or from the classes ;) ), I'd like to hear it. :smile:
     
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