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What is the refraction index of ions in plasma ?

And why people do not discuss it in books ? Is it that it is not important for plasma phenomena ?

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    $\begingroup$ What books have you searched? Which of them give the index for electrons? $\endgroup$ – grochmal Jun 24 '16 at 0:39
  • $\begingroup$ it says plasma refractive index, but there are no ions inside the equation $\endgroup$ – Anonymous Jun 24 '16 at 0:52
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Cold Plasma Index of Refraction

We start by deriving the expression for the dielectric tensor, as shown at https://physics.stackexchange.com/a/138460/59023.

If we consider the case of a cold uniform plasma with only linear waves, then one can show the dielectric tensor has the form: $$ \begin{align} \overleftrightarrow{\mathbf{K}} & = \left[ \begin{array}{ c c c } S & -i \ D & 0 \\ i \ D & S & 0 \\ 0 & 0 & P \end{array} \right] \tag{0} \end{align} $$ where the $S$, $D$, and $P$ terms are defined as: $$ \begin{align} S & = 1 - \sum_{s} \frac{\omega_{ps}^{2}}{\omega^{2} - \Omega_{cs}^{2}} \tag{1a} \\ & = \frac{1}{2} \left( R + L \right) \tag{1b} \\ D & = \sum_{s} \frac{ \Omega_{cs} \ \omega_{ps}^{2} }{\omega \left( \omega^{2} - \Omega_{cs}^{2} \right)} \tag{1c} \\ & = \frac{1}{2} \left( R - L \right) \tag{1d} \\ P & = 1 - \sum_{s} \frac{ \omega_{ps}^{2} }{ \omega^{2} } \tag{1e} \\ R & = 1 - \sum_{s} \frac{ \omega_{ps}^{2} }{ \omega \left( \omega + \Omega_{cs} \right) } \tag{1f} \\ L & = 1 - \sum_{s} \frac{ \omega_{ps}^{2} }{ \omega \left( \omega - \Omega_{cs} \right) } \tag{1g} \end{align} $$ where $\Omega_{cs}$ is the gyrofrequency (or cyclotron frequency) of species $s$, $\omega_{ps}$ is the plasma frequency of species $s$, and $\omega$ is the wave frequency. These parameters are defined as: $$ \begin{align} \Omega_{cs} & = \frac{ Z_{s} \ e \ B_{o} }{ \gamma \ m_{s} } \\ \omega_{ps} & = \sqrt{\frac{ n_{s} \ Z_{s}^{2} \ e^{2} }{ \varepsilon_{o} \ m_{s} }} \end{align} $$ where $Z_{s}$ is the charge state of species $s$ (e.g., +1 for protons), $e$ is the elementary charge, $B_{o}$ is the quasi-static magnetic field magnitude, $\gamma$ is the relativistic Lorentz factor, $m_{s}$ is the mass of species $s$, $n_{s}$ is the number density of species $s$, and $\varepsilon_{o}$ is the permittivity of free space.

Note that the $S$, $D$, $P$, $R$, and $L$ terms do have physical significances (e.g., $R$ and $L$ terms correspond to right- and left-hand polarized modes, respectively) but here we rely upon them mostly for brevity.

The dispersion relation, $D(\mathbf{k}, \omega)$, is derived from the equation: $$ \mathbf{n} \times \left( \mathbf{n} \times \mathbf{E} \right) + \overleftrightarrow{\mathbf{K}} \cdot \mathbf{E} = 0 $$ where we rewrite this in the tensor form $\overleftrightarrow{\mathbf{D}} \cdot \mathbf{E} = 0$. If $\overleftrightarrow{\mathbf{D}}$ has a determinant that goes to zero, then there is a non-trivial solution for $\mathbf{E}$. This solution is the dispersion relation, $D(\mathbf{k}, \omega)$, which can be simplified down if we assume the index of refraction, $\mathbf{n}$, is parallel to the wave vector, $\mathbf{k}$, then: $$ D\left( \mathbf{k}, \omega \right) = A \ n^{4} - B \ n^{2} + R \ L \ P = 0 \tag{2} $$ where the terms $A$ and $B$ are defined by: $$ \begin{align} A & = S \ \sin^{2}{\theta} + P \cos^{2}{\theta} \tag{3a} \\ B & = R \ L \ \sin^{2}{\theta} + P \ S \ \left( 1 + \cos^{2}{\theta} \right) \tag{3b} \end{align} $$ where $\theta$ is the angle between $\mathbf{k}$ and the quasi-static magnetic field. The dispersion relation in Equation 2 has the unique solutions of: $$ n^{2} = \frac{B \pm F}{2 \ A} \tag{4} $$ where $F$ is defined by: $$ F = \left( R \ L - P \ S \right)^{2} \sin^{2}{\theta} + 4 \ P^{2} \ D^{2} \ \cos^{2}{\theta} \tag{5} $$ We can see that $\Im{ [F] } = 0$, always. Since the terms $A$ and $B$ are real, then we can say that $n^{2}$ must either be purely real ($n^{2}$ $>$ 0) or purely imaginary ($n^{2}$ $<$ 0). If $n^{2}$ $<$ 0, then the wave becomes evanescent.

Warm Plasma Index of Refraction

Note that including a finite temperature for each species complicates things significantly, but does not result in a neglect of ions (often the opposite). There are numerous articles on the warm plasma dispersion relation so I won't go into detail here [e.g., Seough and Yoon, 2009]

Answers

What is the refraction index of ions in plasma ?

I am not sure why you think only electrons are included in the index of refraction of plasmas. As you can see from my Equation 4 above, the index of refraction for a plasma includes electrons and all ion species.

And why people do not discuss it in books ?

Gurnett and Bhattacharjee, [2005] have several chapters on the index of refraction of various types of plasmas, and all of them discuss both ions and electrons. In some instances one can neglect ion contributions because, for instance, $\omega \gg \Omega_{ci}$ and $\omega \gg \omega_{pi}$ which results in all the ion contributions in the $S$ (Equation 1a), $D$ (Equation 1c), $P$ (Equation 1e), $R$ (Equation 1f), and $L$ (Equation 1g) terms being much much smaller than the electron contributions.

In summary, good plasma physics books will not ignore the ion contributions unless they are genuinely negligible for very specific circumstances.

Is it that it is not important for plasma phenomena ?

No, quite the opposite. For instance, the dispersion relation for electromagnetic ion cyclotron (EMIC) waves is almost entirely dominated by the ion contributions. There are numerous plasma modes that rely heavily upon the ion contributions.

So as you can see, ion contributions to the index of refraction in plasmas can be very important or negligible, it just depends upon the specific circumstance.

References

  • Gurnett, D.A. and A. Bhattacharjee Introduction to Plasma Physics, Cambridge, UK: Cambridge University Press, ISBN:0521364833, 2005.
  • Seough, J.J. and P.H. Yoon "Analytic models of warm plasma dispersion relations," Phys. Plasmas 16(9), doi:10.1063/1.3216459, 2009.
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  • $\begingroup$ $-1$ for the use of unexplained notation, however standard it may be in the field. $\endgroup$ – Emilio Pisanty Jun 24 '16 at 20:34
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    $\begingroup$ @EmilioPisanty - To which part exactly are you referring? I can clarify if necessary. $\endgroup$ – honeste_vivere Jun 25 '16 at 0:02
  • $\begingroup$ Is there magnetic field in these derivations ? $\endgroup$ – Anonymous Jun 25 '16 at 3:15
  • $\begingroup$ @Anonymous - Yes of course, the cyclotron frequency depends upon the quasi-static magnetic field (see my edits). $\endgroup$ – honeste_vivere Jun 25 '16 at 15:16
  • $\begingroup$ @EmilioPisanty - Have my edits helped to clarify the notation I used or do I need more details? $\endgroup$ – honeste_vivere Jun 25 '16 at 15:17
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There is no such thing as refractive index of electrons or ions. Although @honeste_vivere has given a very detailed answer but I am answering again thinking that his answer is too much for you. When the electromagnetic radiation interact with plasma the electrons respond quickly and ions are too slow.

The plasma refractive index is

$$n=\sqrt{\left(1-\frac{\omega_p^2}{\omega^2}\right)}$$

where $\omega_p$ is the natural oscillation frequency of electrons

$$\omega_p^2=\frac{ne^2}{\epsilon_0m}$$

where n is the electron density inside plasma and m is electron mass. Put the ion mass here and you will see that the ion natural frequency is too low and ions do not respond to electromagnetic waves.

You can also see that if $\omega<\omega_p$ refractive index become imaginary and the electromagnetic wave can not travel into plasma. Hence if you use very low frequency wave such that ions could respond to them they will be completely blocked by electrons.

I hope this will give you some idea why ions are not present in the equations of refractive index.

Try to read Introduction to plasma physics and controlled fusion by FF Chen, and Fundamentals of plasma physics by J. A. Bittencourt, for introductory level plasma physics.

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  • $\begingroup$ Neutral atom have refractive index. Ions are mostly same electron configuration just charged. Why would their refractive index not matter? $\endgroup$ – Anonymous Jun 24 '16 at 18:57
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    $\begingroup$ @Anonymous That's perfectly well explained in the answer as is. Since the ion mass is bigger (by >2000), the contribution to the plasma at high frequencies is correspondingly smaller by $m_e/m_\mathrm{ion}$. The contribution only begins to matter at low frequencies, when $\omega_p/\omega$ is big, but this threshold is reached much sooner for electrons than for ions. $\endgroup$ – Emilio Pisanty Jun 24 '16 at 20:40
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    $\begingroup$ @Anonymous The refractive index of neutral atoms is due to dipole moment induced by the em field, which is also due to electron oscillation induced by em field. In the case of ions when one or two electrons leave the atomic core the remaining electrons become more tightly bound with the ion, which results in reduced dipole moment compared to the atom. In the case of plasmas and metals the contribution of atoms/ions in refractive index is usually much smaller than the free electrons. It may be noted that the refractive index by free electrons is <1 and the from neutral atoms/ions is >1. $\endgroup$ – hsinghal Jun 25 '16 at 7:57

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