Appendix I: Electrons’ momentum exchange rate and conductivity in excited dielectric
Electrons transferred to the conduction band exchange momentum in collisions with phonons when the electron density is low. At some electron number density, the Coulomb collisions with electrons and ions become dominant. Let us consider both processes.
(a) Electron–phonon collisions
Temperature-dependent electron–phonon collision rate reads [10]:
$${v_{{\text{e--ph}}}} \approx {n_{\text{a}}}\left( {\frac{{{k_{\text{B}}}T}}{{\hbar {\omega _{{\text{ph}}}}}}} \right) \cdot a_{{\text{B}}}^{3} \cdot {\omega _{{\text{ph}}}} \approx \frac{{{k_{\text{B}}}T}}{\hbar };\quad {n_{\text{a}}} \cdot a_{{\text{B}}}^{3} \approx 1.$$
(11)
For the room temperature \({k_{\text{B}}}T=0.025\;{\text{eV}}\) one gets νe−ph ≈ 3.8 × 1013 s− 1. However, if it is available, it is better to use experimentally measured value introduced in the above formula with the numerical coefficient usually > 1. For Al, the similar estimate was corrected with coefficient 18.8 [27] to match the measured value at the room conditions.
(b) Coulomb collisions
The electron–ion collision rate depends on continuously growing electron number density and electrons’ velocity, which might be approximately considered being practically constant during the ionisation process and equal to ve ≈ (2Δ
g
/me)1/2 = 2 × 108 cm/s. The electron–ion collision rate reads [24]:
$${V_{{\text{e--i}}}} \approx {3.10^{ - 6}}\,1{\text{n}}\Lambda \cdot \frac{{{n_{\text{e}}}Z}}{{\theta _{{{\text{e,eV}}}}^{{3/2}}}};\quad \Lambda =9{N_{\text{D}}}/Z;\quad {N_{\text{D}}}=1.7 \times {10^9}\left( {\frac{{\theta _{{{\text{e,eV}}}}^{3}}}{{{n_{\text{e}}}}}} \right).$$
(12)
We take Z = 1, θe,eV = 10 eV; ε0 = 3.13 for Al2O3. The collision rate grows up to the natural maximum, which is defined by the ratio of the electrons’ velocity to the inter-atomic distance and comprises ~ 5 × 1015 s1 [11].
At εre = 1 [critical density at 800 nm (ω = 2.356 × 1015s−1) \({n_{{\text{cr}}}}={m_{\text{e}}}{c^2}\pi /{e^2}{\lambda ^2}=\) 1.75 × 1021 cm−3], ne = 2.9 × 1021 cm−3; ND = 0.86—thus plasma is NON-IDEAL; lnΛ = 2.05; νe−i = 0.75 × 1015 s−1. In non-ideal plasma, ratio of collision rate to the plasma frequency is \({V_{{\text{ei}}}}/{\omega _{{\text{pe}}}} \approx 1n\Lambda /.10 \times {N_{\text{D}}}\) = 0.238; νei = 0.76 × 1015s−1.
(c) Conductivity
Conductivity of the free electrons in the Drude approximation reads:
$$\sigma =\frac{{{n_{\text{e}}}}}{{4\pi {n_{{\text{cr}}}}}}\frac{v}{{(1+{{\text{V}}^2}/{\omega ^2})}}.$$
(13)
Thus, laser-affected sapphire at εre = 1 has the conductivity \(\sigma \approx \frac{{({\varepsilon _0} - 1)v}}{{4\pi }}\) ≈ 1014s−1, which is three orders of magnitude lower than in good metals (σ ~ 1017 s−1).
(d) Electron temperature dependence on the absorbed fluence
Energy absorbed per unit volume per unit time accounted for boundary conditions (connecting the field inside the laser-affected material and the incident field) reads [10]:
$$\left\langle {{Q_{{\text{abs}}}}} \right\rangle =\frac{{\omega \cdot {\varepsilon _{{\text{im}}}}}}{{8\pi }}\frac{{4E_{0}^{2}}}{{{{\left| {1+{\varepsilon ^{1/2}}} \right|}^2}}}=\frac{{\omega \cdot {\varepsilon _{im}}}}{c}\frac{{4{I_0}(t)}}{{{{\left| {1+{\varepsilon ^{1/2}}} \right|}^2}}}.$$
(14)
The energy absorbed for time t is converted to the electron internal energy as follows:
$${C_{\text{e}}}\,{n_e}\,{k_{\text{B}}}{T_{\text{e}}}=\int\limits_{0}^{t} {\left\langle {{Q_{{\text{abs}}}}} \right\rangle } \,{\text{d}}\tau \;\left[ {\frac{{\text{J}}}{{{\text{c}}{{\text{m}}^3}}}} \right].$$
(15)
The electron temperature can be recovered from this equation taking values for ne, epsilon and laser fluence at moment t:
$${k_{\text{B}}}{T_{\text{e}}}=\frac{{4F(t)}}{{{C_{\text{e}}} \cdot {n_{\text{e}}}}}\frac{{\omega \cdot {\varepsilon _{{\text{im}}}}}}{{{{\left| {1+{\varepsilon ^{1/2}}} \right|}^2}}};\quad F(t)\int\limits_{0}^{t} {{I_0}(\tau ){\text{d}}\tau } .$$
(16)
Appendix II: Electrons’ charge density oscillations (plasma waves) in the laser-excited dielectric
Let us consider the stage of ionisation when ν << ω, ne < < na. Thus, the permittivity (1) takes a form:
$${\varepsilon _{{\text{re}}}} \cong {\varepsilon _0} - \frac{{{n_{\text{e}}}}}{{{n_{{\text{cr}}}}}};\quad {\varepsilon _{{\text{im}}}}=\frac{{{n_{\text{e}}}}}{{{n_{{\text{cr}}}}}}\frac{\nu }{\omega }.$$
(17)
Note that the conditions imposed on the collision rate and electron number density hold when εre = 0. Let us now consider the electronic oscillations due to small perturbations of the electrons’ density (plasma oscillations) driven by the electrostatic field. The fluctuation in the electron density changes the electrostatic equation as follows:
$$\nabla ({\varepsilon _0}E)=4\pi e({n_{\text{e}}}+\Delta {n_{\text{e}}}).$$
(18)
The permittivity is close to the initial value because the number of conduction electrons is still much less than that of the valence electrons. Let us neglect the permittivity gradient on the plasma oscillations. Then the electrostatic equation reduces to the following:
$$\nabla ({\varepsilon _o}E)={\varepsilon _0} \cdot \nabla \cdot E \approx 4\pi e({n_{\text{e}}}+\Delta {n_{\text{e}}}).$$
(19)
Let us consider the geometry of Fig. 3. Then the electrostatic field driving the oscillations in the direction of the field component along the permittivity gradient is as follows:
$${\varepsilon _0} \cdot \nabla \cdot E={\varepsilon _0} \cdot \left( {\frac{{\partial {E_Z}}}{{\partial z}}} \right) \approx 4\pi e{n_{\text{e}}};\quad {E_Z} \approx \frac{{4\pi e{n_{\text{e}}}}}{{{\varepsilon _0}}}z.$$
(20)
The influence of the valence electrons on the electrostatic field gradually diminishes to zero when the real part of the permittivity turns to be unity. The Newton equation for the electrons motion in such a field is:
$${m_{\text{e}}}\frac{{{{\text{d}}^2}z}}{{{\text{d}}{t^2}}}=e{E_Z} \approx \frac{{4\pi {e^2}{n_{\text{e}}}}}{{{\varepsilon _0}}}z.$$
(21)
One can see that the frequency of electron oscillations in the ionised dielectric is lower than that for ideal plasma due to electrostatic field imposed by the positive polarisation of the valence electrons. Thus, the plasma frequency for electrons oscillating in the excited dielectric reads:
$$\omega _{{d,pe}}^{2}=\frac{{4\pi {e^2}{n_e}}}{{{\varepsilon _0}}}.$$
(22)
The condition εre = 0 is achieved when ne ≈ ε0 × ncr. Therefore, at the zero-real-permittivity point, the plasma resonance occurs, \({\omega _{{\text{d,pe}}}}=\omega ,\) and the plasma wave is excited similar to that in the ideal plasma.
One can treat the resonance absorption by dumping the oscillator (electrons’ plasma oscillations) driven by the external periodic force (laser). The collisional dumping is included in the plasma wave equation driven by the external high-frequency field as follows:
$$\frac{{{d^2}z}}{{d{t^2}}}=\omega _{{pe}}^{2}z - v\frac{{dz}}{{dt}}+e{E_0}\exp (i\omega t).$$
(23)
The solution for the amplitude of electrons’ oscillations is straightforward:
$$z=\frac{{e{E_0}/{m_{\text{e}}}}}{{{{\left[ {({\omega ^2} - \omega _{{{\text{pe}}}}^{2})+{\omega ^2}{v^2}} \right]}^{1/2}}}}\exp (i\omega t+\varphi ).$$
(24)
One sees that at the resonance \({\omega _{{\text{d,pe}}}}=\omega ,\) the maximum oscillation amplitude is:
$${z_{\hbox{max} }}=\frac{{e{E_0}}}{{{m_{\text{e}}}\omega \cdot \nu }}.$$
(25)
The maximum amplitude of the electric field in the resonance reads:
$${E_{\hbox{max} }}={E_0}\frac{\omega }{v}.$$
(26)
Maximum velocity (equal to the oscillation velocity in the imposed field) and amplitude of the plasma wave generated near the plasma resonance:
$${v_{{\text{pw}}}}=\frac{{eE({\varepsilon _{{\text{re}}}}=0)}}{{{m_e}\omega }}.$$
(27)
Taking approximately the resonance intensity as I ~ 1013 W/cm2 (E = 2.89 × 105 CGSE) and frequency as for 800 nm one gets vpw = 6.48 × 107 cm/s and amplitude of 2.75 angstrom. The main damping mechanism for plasma wave in considered conditions is collisional damping, as indicated in the main text.