How Can the Refractive Index Be Less Than One?

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SUMMARY

The discussion centers on the Lorentz model for calculating the refractive index of a dielectric, specifically the formula n^2(ω) = 1 + (ω²_p / (ω²_0 - ω²)). It is established that when the frequency ω exceeds the resonance frequency ω₀, the refractive index n can indeed be less than one, a phenomenon supported by experimental results. This situation does not violate the principle that nothing can travel faster than light in a vacuum, as the phase velocity may exceed c, but no energy is transmitted at that speed. Historical context is provided through references to Sommerfeld and Brillouin's work, which clarifies that signal propagation remains consistent with Einstein's causality.

PREREQUISITES
  • Understanding of the Lorentz model in optics
  • Familiarity with plasma frequency (ωₚ) and resonance frequency (ω₀)
  • Knowledge of wave propagation and phase velocity concepts
  • Basic principles of Einstein's theory of relativity
NEXT STEPS
  • Study the derivation of the Lorentz model for dielectric materials
  • Explore Sommerfeld's Lectures on Theoretical Physics, vol. 4 for historical insights
  • Investigate the implications of complex integrals and the theorem of residues in optics
  • Learn about the concept of anomalous dispersion and its effects on signal propagation
USEFUL FOR

Physicists, optical engineers, and students of electromagnetism seeking to understand the nuances of refractive index behavior in dielectrics and the implications for signal propagation in relation to the speed of light.

DannyJ108
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Homework Statement
The Lorentz model to calculate the refraction index of a dielectric, in the simplest of terms states the following equation: (see relevant equations)
Relevant Equations
##n^2(\omega) = 1 + \frac {\omega^2_p} {\omega^2_0 - \omega^2}##
Hello fellow users,

I've been given the Lorentz model to calculate the refraction index of a dielectric, the formula in its simplest way states that:
##n^2(\omega) = 1 + \frac {\omega^2_p} {\omega^2_0 - \omega^2}##

Where ##\omega_p## is the plasma frequency and ##\omega_0## is the resonance frequency.

If ##\omega > \omega_0## the refraction index ##n## can be smaller than 1 and experimental results verify this. How does this result reconcile with the fact that "nothing can travel faster than light in a vacuum"?

I need to make a bibliographical search and give an explanation for this, but I can't find an exact answer to this question or the same formula I'm given.
I need your help please! Thank you in advance!
 
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I think it is just that the phase velocity is faster than light; no energy is being propagated at that speed.
 
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Thank you
tech99 said:
I think it is just that the phase velocity is faster than light; no energy is being propagated at that speed.
 
This is a very old apparent problem. It was asked Sommerfeld by Willy Wien in 1907, and Sommerfeld wrote a very short article showing by using a complex integral and the theorem of residues that there's nothing violating Einstein causality and that there's nothing propagating faster than the speed of light that is not supposed to do so in the region of anomalous dispersion. The details have been worked out by Sommerfeld and Brillouin in two papers in the Annalen der Physik in 1914.

You find a summary in Sommerfeld's Lectures on Theoretical Physics vol. 4 (Optics) as well as in Jackson's Classical electrodynamics. The upshot is that in the frequency realm around the resonance group velocity doesn't make sense as an approximate signal-propagation velocity as it does in regions of "normal dispersion", because the transient signal is deformed so much that you cannot consider it a smoothly moving wave packet. In this model the front velocity of the wave is exactly ##c_{\text{vac}}##, and thus there's no signal which propagates faster than light, and everything is in accordance with relativity as you expect from Maxwell's equations, which is the paradigmatic example of a classical relativistic field theory :-).
 
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