Negative permittivity of metals

In summary, the negative permittivity shown by noble metals in optical frequencies is caused by their behavior as a plasma at high frequencies. The resonance frequency of the metal causes a change in permittivity, resulting in negative values below the resonance and positive values above it. This negative permittivity has a negligible effect on wave propagation due to the high conductivity of metals. However, in lossless materials with negative permittivity, there is a complex refractive index which leads to lossy propagation and reflection of waves at the boundary. Despite the high conductivity of metals, their plasma frequencies are not far below optical frequencies and they are able to reflect most of the visible light, making them good mirrors.
  • #1
krindik
65
1
Hi,

It is seen that noble metals (gold, silver) show a negative permittivity in optical frequencies. Can somebody explain the physical interpretation of this phenomena? How is the negative permittivity measured?

Really appreciate if someone could point me in the correct direction.

Thanks.
 
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  • #2
At high frequencies metals act like a plasma. The conduction electrons are free to flow around while the relatively massive ions remain more or less stationary. When we solve this in a plasma problem with incident electromagnetic waves, what happens is that we get a resonance frequency. Below the resonance frequency, the waves oscillate slow enough for the electrons to follow. Thus, the metal behaves as a good conductor because the currents that can be excited can properly cancel out the incident waves. However, above the resonant frequency, the inertia of the electrons prevents the electrons from oscillating in proper phase with the incident wave. Thus, the currents cannot be excited properly to eliminate the incident wave and now the wave can pass through the metal like it was a vacuum (but this is a dispersive and lossy vacuum).

If you solve for the dispersive curve, you get
[tex]\omega^2 = \omega_p^2+c^2k^2[/tex]
where \omega_p is the plasma frequency (which turns out to be the resonant frequency. Solving for the permittivity assuming vacuum permeability, we find
[tex]\epsilon = \epsilon_0 \left(1-\frac{\omega_p^2}{\omega^2}\right)[/tex]
Thus, we see that the equivalent permittivity that represents the previously described behavior requires that the permittivity become negative below resonance, when above the resonance the permittivity is positive.

However, it is my recollection that metals have their resonance frequencies below optical, in the terahertz range. I certainly know that silver's resonance is in the terahertz range and thus has a positive permittivity in the optical spectrum.
 
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  • #3
Thanks for the reply.

So how can negative permittivity affect wave propagation. We know that a dielectric having a positive permittivity allows EM wave propagation with a reduced velocity than vacuum. Is there a similar explanation for some material that have negative permittivity?
 
  • #4
krindik said:
Thanks for the reply.

So how can negative permittivity affect wave propagation. We know that a dielectric having a positive permittivity allows EM wave propagation with a reduced velocity than vacuum. Is there a similar explanation for some material that have negative permittivity?

For this case it really doesn't matter because there is a conductivity here that is being ignored. For the plasma approximation, below the resonance the conductivity will be more or less infinite. Not a bad approximation for metals either since their conductivities are on the order of 10^8. So the effects of the negative permittivity are negligible since the conductivity prevents any meaningful penetration at anything but the lowest frequencies.

Another way to look at is that the negative permittivity just makes the propagation lossy as well. Since the wave number is related to the square root of the product of the permittivity and permeability, a negative permittivity and positive permeability results in a lossy wave number. Since the plasma frequency is so high this is another way of showing a very high loss. The only other real aspecrts that it may affect would be the refraction and diffraction directions. But once again, the index of refraction is related to the wave number and thus the negative permittivity just results in a complex number.
 
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  • #5
Thanks for the reply.

Assuming a lossless material with a negative permittivity [tex] \epsilon_r < 0[/tex] and a permeability will result in a complex number for refractive index [tex] n = \sqrt{\epsilon_r \mu_r}[/tex] thus lossy propagation.

An EM wave hitting such a medium will decay within the medium and some part would be reflected satisfying boundary conditions.

Hope that's what you mean.

Thanks once again.
 
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  • #6
Yup.
 
  • #7
Born2bwire, there is no difference between permittivity and conductivity, up to an i x frequency factor. Especially in the optical region, conductivity is far from being infinite.
 
  • #8
DrDu said:
Born2bwire, there is no difference between permittivity and conductivity, up to an i x frequency factor. Especially in the optical region, conductivity is far from being infinite.

Conductivity is different than permittivity. Permittivity is the constant that relates the electric field and the electric flux density field (sometimes called the displacement field). Conductivity is the constant used in the general Ohm's Law that relates the electric field with the current density. The permittivity can be rewritten as a combination of the real and imaginary permittivities with the imaginary permittivity being related to the conductivity and frequency of the wave. And as I stated previously, for metals the conductivity can be estimated to be infinite when we are below the plasma frequency for all but the very low frequency range. In addition, as I stated previously, the plasma frequency for most metals is in the terahertz range, below the optical frequencies.
 
  • #9
In the optical region, the real and imaginary part of the permittivity of a metal are usually of the same order of magnitude. E.g. some values are cited here:
http://ucl-4c00-spr-bpatel.tripod.com/
In Gold and Copper, the increase of the imaginary part due to interband transitions has the unusual effect that the reflectivity sinks in the absorption region (blue to green) and some of the light can make it trough very thin films of copper and gold as can be found e.g. as a thin film in some light bulbs. Hence, metals transmit most of the light where they are most strongly absorbing, while other substances (with a positive real part of permittivity) have a minimum of transmittance there.
 
  • #10
I disagree with Born2bwire's statement that plasma frequencies of (common) metals are far below optical frequencies. If you ignore the imaginary component of the permittivity, then a negative (real) permittivity implies a purely imaginary refractive index, which implies a purely imaginary wavenumber. A consequence of this is that power cannot propagate into the plasma; i.e. below the plasma frequency, waves are completely reflected. (This would be a good time to refer to post #2)

Metals such as aluminium and silver are known to make good mirrors. Why? Because they reflect most of the visible light impinging upon them. Based on my foregoing argument, this implies that the plasma frequency lies, at the least, in the short-wavelength visible (aluminium actually has a plasma frequency in the vacuum UV).

Ever wonder why gold looks like "gold?" Apparently the plasma frequency lies somewhere around what we would call yellow/orange.

Regarding the conductivity, there was actually an interesting study done a few years back where they showed that DC conductivities were approximately valid to wavelengths as short as ~25 microns (I'd call that long IR). Shorter than this, the conductivity becomes significantly dispersive. See: Hagan and Rubens, Annalen der Physik, Vol. 11, p. 873, 1903.
 
  • #11
Dear cmos,
you are right that for most metals, the plasma frequency lies in the ultraviolet. This hold true also for gold and copper. The reason why they are coloured is due to interband transitions from the d-band to the conduction band, see my previous post.
 

1. What is negative permittivity of metals?

Negative permittivity of metals refers to the phenomenon where the permittivity (ability to store electric charge) of a metal is less than that of a vacuum. In other words, the metal has a negative dielectric constant, meaning that it can reduce the strength of an electric field within it.

2. How is negative permittivity of metals measured?

Negative permittivity of metals is typically measured using a device called a dielectric spectrometer. This instrument applies an electric field to a sample of the metal and measures the resulting polarization. The permittivity can then be calculated using the relationship between electric field, polarization, and permittivity.

3. What causes negative permittivity of metals?

The negative permittivity of metals is caused by the presence of free electrons within the metal. These electrons are able to move freely in response to an applied electric field, effectively shielding the rest of the material from the field. This results in a lower overall permittivity compared to a vacuum.

4. What are the practical applications of negative permittivity of metals?

Negative permittivity of metals has several practical applications, including in the development of high-frequency electronic devices. It can also be used in the design of metamaterials, which have unique properties not found in natural materials. Additionally, negative permittivity can be used to enhance the resolution of optical microscopes.

5. Are there any disadvantages to negative permittivity of metals?

One potential disadvantage of negative permittivity of metals is that it can lead to energy loss in high-frequency applications. This is due to the fact that the free electrons within the metal dissipate energy as they move in response to the electric field. Additionally, the use of negative permittivity in some applications may be limited by the availability and cost of certain metals.

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