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Electromagnetic continuity equation

  1. Feb 26, 2016 #1
    Hi, this looks stupid and simple, but I just can't get my head around it.
    Assuming a homogeneous medium.
    The electromagnetic continuity equation goes as
    ∇⋅J + ∂ρ/∂t = 0
    since J = σE, ρ = ɛ∇⋅E, and assuming the time dependence exp(-iωt)
    we have
    σ∇⋅E - iωɛ∇⋅E = 0
    (σ - iωɛ)∇⋅E = 0
    So, σ - iωɛ = 0 or ∇⋅E = 0

    Is it right? Thank you.
     
  2. jcsd
  3. Feb 26, 2016 #2

    BvU

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    Hello eadanlin, :welcome:

    Interesting, but hard to follow for me. Could you explain what exactly has time dependence ##e^{-i\omega t}##, and in what circumstances this reasoning should hold ?
     
  4. Feb 26, 2016 #3
    I don´t quite understand the time dependency you are using either. exp(-iωt) seems the time dependence of a complex magnitude in compact electrical notation, with both active and reactive components.I think that it does not make sense to use i here, because it is only a mathematical tool to ease the calculation in electrical engineering to represent time derivatives of cos (ωt) that appear in impedances, time shifts between different voltages, currents, etc. but there does not exist a complex magnitude. The current function (so ρ) depends only on ωt. Additionally, I believe that the assumption that ∂ρ/∂t=∂(ɛ∇⋅E)/∂t=ɛ.(∂E/∂t)∇⋅E is not correct or at least not always correct, but I´ve nearly forgotten the little Maths I ever knew. Following your supposed time dependence (complex exponential), the right time derivative of E(t) in any case should be multiplied by E peak value. By the way, there is also an inconsistency in your calculation since J is current density (then charge per surface unit and per unit time) and dif(rho)/dif(t)is the total charge/time
     
    Last edited: Feb 26, 2016
  5. Feb 26, 2016 #4
    Thank you for your replies. Sorry for the confusion.
    So this is a plane wave in a homogeneous medium characterized by finite conductivity σ and permittivity ɛ.
    Therefore, σ and ɛ are not a function of space;
    just think of them as constants (hence, ∇⋅ɛE = ɛ∇⋅E, ∇⋅σE = σ∇⋅E).

    I followed the symbol using in wiki (https://en.wikipedia.org/wiki/Continuity_equation)
    ρ is charge density, not total charge.

    I don't see the problem using exp(iωt).
    I follow the standard approach dealing with time harmonic oscillating fields,
    as described in CH9 in http://www.ece.rutgers.edu/~orfanidi/ewa/ch01.pdf, for example.

    I think there is a more fundamental physics concept here;
    just wanted to check how fields, charge density, and current distribute in a bulk metal.
    I will keep researching and post anything I find.
     
  6. Feb 26, 2016 #5
    You cannot use i in this case the way you are doing. Supposing your solution were right, what would you get, imaginary physics constants or resistances/conductivities? Again you are mixing concepts picking things here and there. Be it a sinusoidal field or another kind of function, remember that complex quantites or harmonics are mathematical tools to represent, respectively, time shifts and not sinusoidal magnitudes, but they have no real meaning. If you have a current with 50 Hz fundamental component and a 150 Hz harmonic, there exists only a current. Harmonics are a tool. If you have a current I=50j A ( in engineering j is used to prevent confusions between currents and imaginary i ), it only means that the current has a rms value of 50A and 90deg ahead of the reference angle. The current would stiil be 50*sq(2)*sin (wt), shifted these 90 deg. In the field calculations, you must use wave equations and not complex or polar notation. In wikipedia's article about Maxwell's equations rho is volume charge density, not surface charge density. However, I think the most important thing here is that you understand what is the complex notation used for. I find your question very interesting. Sorry for my typing, I'm on a cell phone.For real current distributions, please consider also the magnetic induction and the related phenomena, like skin effect.
     
    Last edited: Feb 26, 2016
  7. Feb 26, 2016 #6
  8. Feb 26, 2016 #7
    Nope, it is wrong. To begin with, the real part of exp(-iωt) is...0. Maybe you refer to be the module of the imaginary component, which is (-sin(wt)), that is, in the time dominion as I said before. Besides, in the continuity equation, ρ is the amount of the quantity q per unit volume. But in Maxwell equation, ρ = ɛ∇⋅E, equivalent to ∇⋅D=ρ , ρ is the total free charge.
     
  9. Feb 26, 2016 #8
  10. Feb 26, 2016 #9

    Charles Link

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    There are two ways of writing the linear response to an electric field in a medium: 1) J (x,t)=integral(sigma(x-x', t-t')*E(x',t')d^3x'dt' or 2) P(x,t)=integral(chi(x-x',t-t')*E(x',t')d^3x'dt' (sigma and chi are linear response functions to the electric field-the one giving polarization current, the other giving polarization). The second way is essentially D(x,t)=integral(e(x-x',t-t')*E(x',t')d^3x'dt' where "e" is a dielectric response function. Taking fourier transforms gives J(k,w)=sigma(k,w)*E(k,w) where these quantities are now fourier transforms, and similarly for the other equations. Assuming J is actually current from polarization charge i.e. J=Jp=dP/dt, and taking -div P=rho p where "rho p" is the polarization charge density. (sorry, I don't know how to do the greek letters yet), and also doing the other F.T.'s e.g. P(k,w)=chi(k,w)*E(k,w) you can show a consistency between these two linear methods. Using the continuity equation, I think you will find the results give your result that sigma=i*w*e. i.e. sigma(k,w)=i*w*e(k,w). See also the textbook Plasma Physics by Ichimaru where they present these linear equations. Ichimaru does not show the complete algebra described here, but his final results are in agreement with this. And, yes, I do think your result is correct. the sigma is related to the chi, and thereby the sigma is related to the "e", as the above calculations will show.
     
    Last edited: Feb 26, 2016
  11. Feb 26, 2016 #10
    Yes, you are right, I forgot the cos(wt). Anyway,the calculation is wrong (at least as I see it) because of the confusion with the rhos. It makes it inconsistent in terms of unit dimensions. A
    s you said, Re (exp(-iwt) would be needed to cope with the i.
    I had understood from first posts that ɛ was constant, but is is time dependant with the exp(-iwt).
     
  12. Feb 26, 2016 #11
    Hey, thank you guys for your time and thoughts.
    There are some interesting thought; I need some time to sink in.
    anyway, thanks!
     
  13. Feb 26, 2016 #12
    OK. I think I know the answer (probably)
    If it is a plane wave, then yes, ∇⋅E = 0. One can simply write E as a plane wave and see if it is 0.
    In fact, there is actually no charge or current whatsoever.

    If it is not a plane wave, and one finds ∇⋅E ≠ 0, then σ - iωɛ = 0 has to be true.
    As some ppl have pointed out above, the effective permittivity ɛ* = ɛ(1+iσ/ωɛ)
    One finds that σ - iωɛ = 0 gives ɛ* = 0.
    This is exactly the condition for bulk plasmons in the metal.
    Current and charge vibrate along with the propagation direction of waves.

    see (5.2) (5.3) in
    http://theses.ulaval.ca/archimede/fichiers/24879/ch05.html
     
    Last edited: Feb 26, 2016
  14. Feb 26, 2016 #13

    Charles Link

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    Just a quick follow-on: When you have an equation Jp=dP/dt in real space, that of course gives you Jp(k,w)=-iwP(k,w) in F.T. space. Similarly, -div P=rho p in real space gives -ik*P(k,w)=rho p(k,w) (* =dot product) in F.T. space. The continuity equation div Jp +d (rho p)/dt=0 in real space gives ik*Jp(k,w) -(iw)(rho p(k,w)) =0 in F.T. space.
     
  15. Feb 26, 2016 #14

    Charles Link

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    A little algebra, using the Jp and P equations, etc. gives the result sigma(k,w)=-iw*chi(k,w). You may get some slight differences if you consider the J and/or "rho" to represent and/or include free charges, but this is in reasonably good agreement to the original post above. A subsequent follow-on: If we define a system to contain both free charges and polarization charges, and can assume linear responses in all cases, we can write the equations as follows: J free(k,w)=sigma free (k,w)*E(k,w). The div E=rho free/e equation is
    ik*E(k,w)=rho free (k,w)/e(k,w) in F.T. space. The continuity equation (for the free charges) in F.T. space becomes
    ik*J free(k,w)-iw*rho free (k,w)=0 . A little algebra gives sigma free(k,w)=iw*e(k,w) where the "sigma free(k,w)" used here has a slightly different definition than the "sigma(k,w)" contained in my previous comments above. (And note the e(k,w) dielectric function is a polarization type screening of the E field from the free charges)
     
    Last edited: Feb 26, 2016
  16. Feb 28, 2016 #15

    Charles Link

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    Additional comment on the above: What apparently is the case is that a sinusoidal (in space and time) electric charge density distribution , whether from free charges or polarization charges, goes hand-in-hand with a sinusoidal (in space and time) current distribution (with a phase change, etc.). Together with these will be a sinusoidal (in space and time) electric field. Notice also, that the sinusoidal electric field is longitudinal (rather than transverse) because the equations contain the dot product of the wave vector with the electric field. The result is that the conductivity sigma free (k,w) is related to the dielectric function e(k,w). Also of interest is something I have read in at least a couple of places, that the designation of what is free charges vs. polarization charges is somewhat arbitrary. The choice of what is designated as free vs. polarization is to simplify the mathematics, but as long as all of the electrical charges are accounted for, the calculations should yield very similar answers.
     
    Last edited: Feb 28, 2016
  17. Feb 28, 2016 #16

    Charles Link

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    ## \sigma_f(k,\omega)=i*\omega*\epsilon(k,\omega) ## (Taking a couple of minutes to practice with the Latex. It's my first try at it on Physicsforums) A google of the topic showed the derivation of the dispersion relation for the Langmuir longitudinal plasma waves (from which the velocity of propagation can be computed). The google also showed that electrostatic waves are necessarily longitudinal. Some of these derivations are easier to read upon knowing how the conductivity ## \sigma_f(k,\omega) ## and the dielectric function ## \epsilon(k,\omega) ## are related.
     
    Last edited: Feb 28, 2016
  18. Feb 29, 2016 #17
    But if the E field is sinusoidal, ∂ρ/∂t=∂ (ɛ∇⋅E)/∂t is not equal to (∂ɛ/∂t)∇⋅E as in the calculation which originated this thread (leaving apart the rho issue).
     
  19. Feb 29, 2016 #18

    Charles Link

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    In the original post, the Fourier Transforms were not done in a complete manner, but I think his results are still correct. The second equation is div D=## \rho ## Each divergence operation picks up an "ik" in F.T. space and each time derivative picks up a " -i ## \omega ## ". The dielectric response function ## \epsilon(x,t) ## is in the convolution integral with E(x,t) and in F.T. space this reads ## D(k,\omega)=\epsilon(k,\omega)*E(k,\omega) ## . When you convert d (div D)/ dt to F.T. space, it becomes ## -i \omega*ik*D(k,\omega)=\omega*\epsilon(k,\omega)*k*E(k,\omega) ## I do think if you work out te F.T.'s in detail, you will agree with the result of the original post. Also consult Ichimaru's Plasma Physics for additional details. It shows the convolution integrals with the dielectric response function ## \epsilon(x-x',t-t') ## and ## E(x',t') ## in full detail.
     
    Last edited: Feb 29, 2016
  20. Feb 29, 2016 #19

    Charles Link

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    To put into Latex one of my previous comments above: (I just learned how to use Latex) ## D(x,t)=\int \int \epsilon(x-x',t-t')*E(x',t') \,dx' \,dt' ## Taking the Fourier transform (in space and time) of this convolution equation gives ## D(k,w)=\epsilon(k,\omega)*E(k,\omega) ## where the functions are now all Fourier transforms. Alternatively, a linear response type equation can be written for the free current density ## J_f(x,t) ## in the medium: ## J_f(x,t)=\int \int \sigma(x-x',t-t')*E(x',t') \,dx' \,dt' ## The Fourier transform (in space and time) of this equation yields ## J_f(k,\omega)=\sigma(k,\omega)*E(k,\omega) ## where again each of these functions are all Fourier transforms. With the Fourier transform of the continuity equation along with the Fourier transform of the ## d(div D)/dt= d(\rho_f)/dt ## equation , the result is ## \sigma(k,\omega)=i*\omega*\epsilon(k,\omega) ## that is found in the very original post at the top. As previously mentioned, Ichimaru's Plasma Physics textbook shows much of this linear response theory in the first couple chapters of his text.
     
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