Sign of permittivity/permeability

  1. Hi.

    I have a book called "Electrodynamics of Continuous Media" where it is stated that for a passive, lossy material, the imaginary part of the permeability and permittivity must always be positive. But I don't see why. Why is it not OK for one of them to be negative, if the other one is positive and much larger?
     
  2. jcsd
  3. olgranpappy

    olgranpappy 1,273
    Homework Helper

    In theory, what you propose could happen... at least, I see no obvious reason why not.

    In practice a great many materials can be treated as if
    [tex]
    \mu\approx 1
    [/tex]
    and so
    [tex]
    Im(\mu)\approx 0\;.
    [/tex]

    Then, with the above approximation, the imaginary part of the dielectric function must be positive.
     
  4. marcusl

    marcusl 2,100
    Science Advisor
    Gold Member

    Positive imaginary part is associated with the transfer of energy from the fields to the medium. This describes attenuation of an EM wave through a material, for example. If the imaginary part is negative, the field absorbs energy from the medium, which is highly unusual. It can be made to happen through special effort (by optically pumping a three state quantum system where one state is metastable, for instance, to create a population inversion--this is how laser amplification is created), but this cannot occur in a passive system in equilibrium .
     
  5. Hello.

    My book reads

    [tex]Q=\frac{\omega}{8\pi} (\epsilon''|\mathbf E|^2+\mu''|\mathbf H|^2)[/tex]

    where [tex]\epsilon''[/tex] is the imaginary part of the permittivity. And then he goes on to say that because Q>0 then [tex]\epsilon''>0[/tex] and [tex]\mu''>0[/tex], but I don't understand why both MUST be positive always.
     
  6. olgranpappy

    olgranpappy 1,273
    Homework Helper

    ...but, I think you misunderstand the question...

    For a medium with both a complex permittivity *and* permeability the energy lost into the medium (which, of course must be positive if we consider materials which are equilibrated--I.e., the material heats up when irradiated, it doesn't cool down) is roughly proportional to

    [tex]
    \omega Im(\epsilon)|E|^2+\omega Im(\mu)|H|^2\;,
    [/tex]
    where E is the electric field, H is the magnetic field, \omega is the angular frequency of the field, \epsilon is the electric "permitivity", and mu is the magnetic "permeability".

    I believe the OP's point is that the electric and magnetic field cannot be taken as independent and thus there is no apparent reason why *both* Im(\epsilon) *and* Im(\mu) must simultaneously be positive. Indeed, for the case of electromagnetic wave is seems that one only has the condition
    [tex]
    |\mu|Im(\epsilon)+|\epsilon|Im(\mu)>0\;.
    [/tex]
     
  7. marcusl

    marcusl 2,100
    Science Advisor
    Gold Member

    Sorry, you are right that I didn't catch that. Thanks for correcting me!
     
  8. Andy Resnick

    Andy Resnick 5,744
    Science Advisor

    Is that Landau and Lifgarbagez? I quickly skimmed through and didn't see that comment.

    I did see some discussion about the analytical properties of the permittivity, but the book does note that analytic continuation is possible on the frequency as well- that's what leads to the Hamaker constant. Finally, it's possible to generate a negative index of refraction using engineered materials (metamaterials), so in that case, both the permittivity and the permeability are negative.
     
  9. marcusl

    marcusl 2,100
    Science Advisor
    Gold Member

    Double negative meta-materials are those where the real parts are negative. I think the comments still apply to the imaginary parts.
     
  10. olgranpappy

    olgranpappy 1,273
    Homework Helper

    I think it's section 80.

    I think the OP has a valid point. L+L are usually very good, but they do on occasion make a mistake. Or, usually, it is the case that they make a very general statement that does not actually hold in complete generality.

     
  11. Andy Resnick

    Andy Resnick 5,744
    Science Advisor

    Got it- thanks.

    Ok, Q is the heat, not the total energy. I would say the restriction (Eq. 80.7 in my edition) holds always because Q represents energy *lost* from the electromagnetic field. The change in entropy of the system TdS = dQ must always be positive, so the energy loss of each component eE^2 and uH^2 must be positive.

    The footnote at the bottom of p.274 indicates that for nonequilibrium conditions, the situation may be different.
     
  12. olgranpappy

    olgranpappy 1,273
    Homework Helper

    still... the question is: Why for each component eE^2" and uH^2 *individually*?

    It seems as if only the *sum* must be positive, why should each term in the sum individually be positive. I don't think it is true in general that one can treat E and H as independent.
     
  13. marcusl

    marcusl 2,100
    Science Advisor
    Gold Member

    Maybe we shouldn't argue the electromagnetic case. What if we access each parameter individually by applying a slowly changing E or B field? If either imaginary part is negative, then the resulting loss tangent or hysteresis loss (or whatever the magnetic loss is called) would have the wrong sign and, again, the applied field would (individually) gain energy from the medium. Hence each must be positive in equilibrium.
     
  14. olgranpappy

    olgranpappy 1,273
    Homework Helper

    But, we are not interested only in a slowly varying field. We are interested in any given frequency [itex]\omega[/itex].

    Anyways, I believe that there are different ways to prove that each Im(\epsilon) and Im(\mu) individually are positive (for positive frequencies). But, I don't think this fact follows from the single positivity condition on the heat flowing into the medium.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook