Physical cause of a negative real part of the permittivity

[BlackHole213]

I originally posted the following question on physics.stackexchange, but no one was able to answer it. I did find this answer on PhysicsForums, but I was already aware of the oscillation of electrons in response to an external electric field.

What is the physical cause behind a material having a negative real part of its dielectric function? Given the complex permittivity, \epsilon(\omega)=\epsilon(\omega)'+i\epsilon(\omega)'', the Drude model gives
\begin{align}
\epsilon'=1-\frac{\omega_{P}^2}{\omega^2+\omega_{\tau}^2}
\end{align}
where \omega is the frequency of the incoming light, \omega_{P}=\sqrt{\frac{Ne^2}{m\epsilon_0}}is the plasma frequency, N is the electron density, m is the electron's mass, e is the electronic charge, and \omega_{\tau} is the frequency of collisions between conduction electrons and the ion lattice.

If \omega is small enough, then \epsilon'<0. But this is just the mathematics behind negative permittivity. How does it physically happen?

 

[tech99]

I originally posted the following question on physics.stackexchange, but no one was able to answer it. I did find this answer on PhysicsForums, but I was already aware of the oscillation of electrons in response to an external electric field.

What is the physical cause behind a material having a negative real part of its dielectric function? Given the complex permittivity, \epsilon(\omega)=\epsilon(\omega)'+i\epsilon(\omega)'', the Drude model gives
\begin{align}
\epsilon'=1-\frac{\omega_{P}^2}{\omega^2+\omega_{\tau}^2}
\end{align}
where \omega is the frequency of the incoming light, \omega_{P}=\sqrt{\frac{Ne^2}{m\epsilon_0}}is the plasma frequency, N is the electron density, m is the electron's mass, e is the electronic charge, and \omega_{\tau} is the frequency of collisions between conduction electrons and the ion lattice.

If \omega is small enough, then \epsilon'<0. But this is just the mathematics behind negative permittivity. How does it physically happen?

No one answered. I am guessing that an electron moves in response to an incident E-field, and as it accelerates it re-radiates some of the energy. In the vicinity of the plasma frequency, I imagine there will be a 180 degree phase shift in the re-radiated E-field as the frequency is changed from below to above. Maybe the dielectric has an inductive effect above the plasma frequency?

 

[metatrons]

Hi BlackHole213:

You have asked a very 'big' question, and a sound explanation may require a very good fundamentals of the material and electromagnetic-dynamic science. I will try to provide my answer while avoiding the very complicated mathematics and theoretical terminlogies here. But please do study it in more details in a formal textbook (e.g. Fundamentals of Semiconductors: Physics and Materials Properties, by YU et, al) if you are really devoted into it.

In general, the permittivity of a material is a measure of its response to an (usually oscillated) external electric field. The real part of the permittivity described the 'resistance' of the material to the external field (say, what is the excited field inside the materials compare to the external field). If the real part of the permittivity has a negative sign, it means that the excited field is in opposite direction to the external field.

For a good metallic material, we may consider its Electric-model as small balls (as the free electrons) oscillated while attached to a spring (as the Coulomb force of the atom) with the other end fixed in a periodic box (as the crystalline structure of the atom nucleus) under external perturbation (as the external field). The oscillation of the small ball will induce an 'excited' field of the material, which is depended on both external field and the material' structure. If the frequency of the external field is too low, the oscillation of the ball can fully catch its pace, so in every instantaneous of time, the electron will build an exact 'reaction field' counter to the external field, much like the spring produces the reaction to the external force with equal and opposite sign. In this case, the material has a negative real part of the permittivity, and resembles a 'mirror'.

On the other hand, if the external field oscillate too fast for the excited field to keep pace, the excited field can not build in a full-resistance field in the material (but it will somehow partially counter the external field), which will result in a 'damping' oscillation, resulting a non-unity positive real part of permittivy. And, if the frequency of the external field is EXACTLY coincide to the spatial frequency of the crystalline structure of the material, a perfect electric resonance will occur and the material becomes totally transparent to the external field. This frequency is called 'plasmon frequency'.

It is very important to understand that the drude model is only an approximation of a good metal in a LOW FREQUENCY range. The drude-lorentz model will be more realistic when including higher frequency. Again, please refer to the formal textbook for a complete details.

Hope this helps!

 

[tech99]

Hi BlackHole213:

You have asked a very 'big' question, and a sound explanation may require a very good fundamentals of the material and electromagnetic-dynamic science. I will try to provide my answer while avoiding the very complicated mathematics and theoretical terminlogies here. But please do study it in more details in a formal textbook (e.g. Fundamentals of Semiconductors: Physics and Materials Properties, by YU et, al) if you are really devoted into it.

In general, the permittivity of a material is a measure of its response to an (usually oscillated) external electric field. The real part of the permittivity described the 'resistance' of the material to the external field (say, what is the excited field inside the materials compare to the external field). If the real part of the permittivity has a negative sign, it means that the excited field is in opposite direction to the external field.

For a good metallic material, we may consider its Electric-model as small balls (as the free electrons) oscillated while attached to a spring (as the Coulomb force of the atom) with the other end fixed in a periodic box (as the crystalline structure of the atom nucleus) under external perturbation (as the external field). The oscillation of the small ball will induce an 'excited' field of the material, which is depended on both external field and the material' structure. If the frequency of the external field is too low, the oscillation of the ball can fully catch its pace, so in every instantaneous of time, the electron will build an exact 'reaction field' counter to the external field, much like the spring produces the reaction to the external force with equal and opposite sign. In this case, the material has a negative real part of the permittivity, and resembles a 'mirror'.

On the other hand, if the external field oscillate too fast for the excited field to keep pace, the excited field can not build in a full-resistance field in the material (but it will somehow partially counter the external field), which will result in a 'damping' oscillation, resulting a non-unity positive real part of permittivy. And, if the frequency of the external field is EXACTLY coincide to the spatial frequency of the crystalline structure of the material, a perfect electric resonance will occur and the material becomes totally transparent to the external field. This frequency is called 'plasmon frequency'.

It is very important to understand that the drude model is only an approximation of a good metal in a LOW FREQUENCY range. The drude-lorentz model will be more realistic when including higher frequency. Again, please refer to the formal textbook for a complete details.

Hope this helps!

My understanding is that permittivity can be measured at low or zero frequency and is ordinarily positive. I presume this to arise from the "springiness" of the system. At frequencies above the plasmon frequency, I presume the inertia of the electrons (inductance and mass) is predominant. This is where I am guessing that the permittivity becomes negative.

 

[metatrons]

My understanding is that permittivity can be measured at low or zero frequency and is ordinarily positive. I presume this to arise from the "springiness" of the system. At frequencies above the plasmon frequency, I presume the inertia of the electrons (inductance and mass) is predominant. This is where I am guessing that the permittivity becomes negative.

tech99：

For a good metal, at low frequency （lower than plasmon frequency）, its permittivity (real part) is negative, and at high frequency (higher than plasmon frequency) the permittivity (real part) becomes positive.

It is very important to understand the permittivity as the 'overall material response to external field'. When a EM field illuminates a material, the total field at a certain position will be the superposition of the external excitation field and the material response field. A negative (real part) permittivity means the response field is in opposite sign with the excitation. Take DC (Zero-frequency) field on bulk metals for example, in this case, the electrons in the metal is completely 'pulled' from the nucleus by the external field into equilibrium. The coulomb field induced by the pulled electron and the nucleus is exactly the same (but in opposite direction) of the external field as a 'electric shield'. The two field completely canceled each other, and no field shall pass through the metal.

On the other hand, when the frequency of the external field is high enough, and the electron in the metal is too 'dull' to catch the speed. The electric shield can not be built, and the external field can pass through the metal. Some energy of the transmitted field is lost by scattering during the collision of the electrons to the nucleus, causing a damping oscillation and a positive (larger than unity) real part of the permittivity. And in the case of the perfect frequency match (external field has exactly the same frequency as the material's plasmon frequency), a perfect timing of oscillation in the metal is occur, with no collision occurs, which induces a total transparent material and unity permittivity.

Hope this helps.

 

[Andy Resnick]

I<snip>
What is the physical cause behind a material having a negative real part of its dielectric function? <snip>

Negative values of the permittivty and permeability are associated with 'anomalous dispersion'- no new physics are involved.

http://www.phy.duke.edu/~rgb/Class/phy319/phy319/node50.html

Anomalous dispersion occurs in the region of resonant interaction, and requires the medium to be absorptive. This effect can now be engineered (metamaterials, photonic crystals, 'left handed materials', etc.)

 

[metatrons]

Negative values of the permittivty and permeability are associated with 'anomalous dispersion'- no new physics are involved.

http://www.phy.duke.edu/~rgb/Class/phy319/phy319/node50.html

Anomalous dispersion occurs in the region of resonant interaction, and requires the medium to be absorptive. This effect can now be engineered (metamaterials, photonic crystals, 'left handed materials', etc.)

Usually, this 'anamalous dispersion' is required when acquiring BOTH negative permittivity and permeability, or single negative permeability (as natural materials have no magnetic response at visible wavelength). The negative permittivity is actually very common in natural metals at optical range. For example the Ag, Au, Al... check John and Christine's materials handbook..

As a matter of fact, although Ag and Au have a very negative permittivity in near infared (approximately 750nm~1500nm), they are nearly lossless, and resemble 'perfect mirror' at this range (complete reflection).