Two-temperature equation of state for aluminum and gold with electrons excited by an ultrashort laser pulse

Abstract

A short laser pulse converts metal into a two-temperature state with the electron temperature higher than the ion temperature. To describe the electron contributions to the total internal energy and pressure arising as a result of electron heating, we develop the new analytic approximation formulae for two-temperature thermodynamics of metal. Those approximations are based on quantum calculations performed with density functional theory (DFT) packages. DFT calculations provide the internal energies and pressures for densities of the order of solid-state density and for electron temperatures up to 55 kK. The new analytic approximations give a better accuracy in hydrodynamic simulation of laser–matter interaction and should be used instead of the less accurate expressions based on the Fermi model of ideal electron gas, which is widely used for two-temperature states of metal.

Appendices

Appendix 1

In the case of Au, the Vienna ab-initio simulation package (VASP) [25, 26] was used in total energy and electron DoS calculations combined with the projected augmented wave (PAW) potential, the Perdew–Burke–Ernzerhof (PBE) exchange-correlation functional, and a kinetic energy cutoff of 500 eV. A $21 \times 21 \times 21$ Monkhorst-Pack k-point grid was used and electron occupation was treated with a Fermi–Dirac smearing method. The lattice constant for Au in equilibrium was found equal to 4.15 A. The number of empty states was set equal to 30. It is found that this value is enough for total energy calculations at high electron temperatures. Here we used the GGA-PBE approximation because it is more accurate for DoS calculation than LDA, but it gives the less accurate equilibrium lattice parameter (which was not aimed for improving in the present study).

For the Al total energy calculations, we use the FP-LAPW method implemented in the Elk code [27] and the PBE exchange-correlation functional. The muffin-tin radius of aluminum atoms of 2.20 a.u. is kept constant for all calculations; the number of $k$ points is equal to $20 \times 20 \times 20;$ and the product of the muffin-tin radius and the maximum reciprocal space vector is equal to 10. The maximum value for the waves inside the atomic spheres and the largest reciprocal vector in the charge Fourier expansion, Gmax, are set to 10 and 14, respectively. The self-consistent calculation is terminated when the total energy change is less than $10^{-6}$ eV in the case of VASP calculations for Au and less than $10^{-6}$ Ha in the case of Elk calculations for Al. In the latter case the tolerance for total potential change was limited by the value $10^{-6}$ Ha.

Appendix 2

Let us denote as $\mu (T_{\rm e})$ the electron chemical potential at the temperature $T_{\rm e}.$ Then the electron concentration is

$$\begin{aligned} zn=\frac{\sqrt{2}}{\pi ^{2}}\left( \frac{\sqrt{m_{\rm s}}}{\hbar }\right) ^3 \int { \frac{\varepsilon ^{1/2}\text{ d }\varepsilon }{\exp [(\varepsilon -\mu )/(k_{\rm B}T_{\rm e})]+1}}. \end{aligned}$$

($z$ is the number of electrons per atom, and $n$ is the concentration of atoms). Correspondingly internal energy per unit volume, measured from the band bottom, is

$$\begin{aligned} E=\frac{\sqrt{2}}{\pi ^{2}}\left( \frac{\sqrt{m_{\rm s}}}{\hbar }\right) ^3 \int { \frac{\varepsilon ^{3/2}\text{ d }\varepsilon }{\exp [(\varepsilon -\mu )/(k_{\rm B}T_{\rm e})]+1}}. \end{aligned}$$

Values $z n$ and $E$ both are Fermi integrals

$$\begin{aligned} I(\mu )=\int { \frac{F(\varepsilon )\text{ d }\varepsilon }{\exp [(\varepsilon -\mu )/(k_{\rm B}T_{\rm e})]+1}} \end{aligned}$$

with functions $F(\varepsilon )$ correspondingly equal to $F(\varepsilon )=\varepsilon ^{1/2}$ and $F(\varepsilon )=\varepsilon ^{3/2}.$ These integrals in the low-temperature limit in the fourth order in $T_{\rm e}$ are [14]

$$\begin{aligned} I(\mu )&= \int _{0}^{\mu }F(\varepsilon )\text{ d }\varepsilon \\&\quad + \frac{\pi ^{2}}{6}F'(\mu )(k_{\rm B}T_{\rm e})^2+\frac{7\pi ^{4}}{360}F'''(\mu )(k_{\rm B}T_{\rm e})^4 \end{aligned}$$

To derive expressions (9) and (11), first we will find the expression for the chemical potential. Let us introduce the notation

$$\begin{aligned} A=\frac{\sqrt{2}}{\pi ^{2}}\left( \frac{\sqrt{m_{\rm s}}}{\hbar }\right) ^3. \end{aligned}$$

Then for the electron concentration we have

$$\begin{aligned} \frac{zn}{A}&= \frac{2}{3}\mu ^{3/2}+ \frac{\pi ^{2}}{6}\cdot \frac{1}{2\mu ^{1/2}}(k_{\rm B}T_{\rm e})^2\\&\quad +\frac{7\pi ^{4}}{360}\cdot \frac{3}{8\mu ^{5/2}}(k_{\rm B}T_{\rm e})^4. \end{aligned}$$

From the other side at $T_{\rm e}=0$ the same electron concentration can be written as

$$\begin{aligned} zn=A\, \frac{2}{3} \, E_{\rm F}^{3/2}. \end{aligned}$$

Hence we get

$$\begin{aligned} \frac{2}{3}E_{\rm F}^{3/2}&= \frac{2}{3}\mu ^{3/2}+ \frac{\pi ^{2}}{12}\mu ^{-1/2}(k_{\rm B}T_{\rm e})^2\\&\quad + \frac{7\pi ^{4}}{960}\mu ^{-5/2}(k_{\rm B}T_{\rm e})^4 \end{aligned}$$

or

$$\begin{aligned} 1&= \left( \frac{\mu }{ E_{\rm F} }\right) ^{3/2}+ \frac{\pi ^{2}}{8}\left( \frac{\mu }{ E_{\rm F} }\right) ^{-1/2}\left( \frac{k_{\rm B}T_{\rm e}}{E_{\rm F} }\right) ^2\\&\quad +\frac{7\pi ^{4}}{640}\left( \frac{\mu }{ E_{\rm F}}\right) ^{-5/2} \left( \frac{k_{\rm B}T_{\rm e}}{E_{\rm F} }\right) ^4. \end{aligned}$$

When introducing the denotion $k_{\rm B}T_{\rm e}/E_{\rm F}=\tau$, we need to find coefficients $c$ in the expression

$$\begin{aligned} \frac{\mu }{ E_{\rm F} }=1+c_{1}\tau +c_{2}\tau ^2+c_{3}\tau ^3+c_{4}\tau ^4. \end{aligned}$$

It is obvious that the coefficients of odd degrees of $\tau$ are equal to zero, therefore

$$\begin{aligned} \frac{\mu }{ E_{\rm F} }=1+c_{2}\tau ^2+c_{4}\tau ^4. \end{aligned}$$

With the accuracy including the term $\tau ^4$ we have

$$\begin{aligned} 1&= 1+\frac{3}{2}c_{2}\tau ^2+\frac{3}{2}c_{4}\tau ^4 +\frac{3}{8}c_{2}^{2}\tau ^4 \\&\quad + \frac{\pi ^{2}}{8}\left( 1-\frac{1}{2}c_{2}\tau ^2\right) \tau ^2+ \frac{7\pi ^{4}}{640}\tau ^4 \end{aligned}$$

or

$$\begin{aligned}&\left( \frac{3}{2}c_{2}+\frac{\pi ^{2}}{8}\right) \tau ^2 \\&\quad +\left( \frac{3}{2}c_{4}+\frac{3}{8}c_{2}^{2}-\frac{\pi ^{2}}{16}c_{2}+\frac{7\pi ^{4}}{640}\right) \tau ^4=0. \end{aligned}$$

Hence we find

$$\begin{aligned} c_{2}=-\frac{\pi ^{2}}{12}, c_{4}=-\frac{\pi ^{4}}{80}. \end{aligned}$$

Now the asymptotic representation of the internal energy can be obtained. For the internal energy per atom $E$ we have

$$\begin{aligned} \frac{E \, n}{A}&= \frac{2}{5}\mu ^{5/2} +\frac{\pi ^{2}}{6}\cdot \frac{3 \mu ^{1/2}}{2}(k_{\rm B}T_{\rm e})^2\\&\quad -\frac{7\pi ^{4}}{360}\cdot \frac{3}{8 \mu ^{3/2}}(k_{\rm B}T_{\rm e})^4. \end{aligned}$$

Dividing this expression by the value $2 E_{\rm F}^{5/2}/5$ we obtain

$$\begin{aligned} \frac{En}{A}\frac{5}{2E_{\rm F}^{5/2} }&= \left( \frac{\mu }{ E_{\rm F} }\right) ^{5/2} + \frac{5\pi ^{2}}{8}\left( \frac{\mu }{ E_{\rm F} }\right) ^{1/2}\tau ^2 \\&\quad - \frac{7\pi ^{4}}{384}\left( \frac{\mu }{ E_{\rm F} }\right) ^{-3/2}\tau ^4. \end{aligned}$$

Substituting to this expression the expansion $\mu /E_{\rm F}=1+c_{2}\tau ^2+c_{4}\tau ^4$ and taking into account that

$$\begin{aligned} E_{\rm 0}=\frac{1}{n} \, A \, \frac{2}{5}E_{\rm F}^{5/2} = \frac{3}{5}z E_{\rm F} \end{aligned}$$

is the internal energy per atom at $T_{\rm e}=0,$ we have with the accuracy up to terms $\tau ^4$

$$\begin{aligned} \frac{E}{E_{\rm 0}}&= 1+\frac{5}{2}c_{2}\tau ^2+\frac{5}{2}c_{4}\tau ^4 +\frac{15}{8}c_{2}^{2}\tau ^4 \\&\quad + \frac{5\pi ^{2}}{8}\left( 1+\frac{1}{2}c_{2}\tau ^2\right) \tau ^2- \frac{7\pi ^{4}}{384}\tau ^4\\&= 1+\left( \frac{5}{2}c_{2}+\frac{5\pi ^{2}}{8}\right) \tau ^2\\&\quad + \left( \frac{5}{2}c_{4}+\frac{15}{8}c_{2}^{2}+\frac{5\pi ^{2}}{16}c_{2}-\frac{7\pi ^{4}}{384}\right) \tau ^4 \end{aligned}$$

Substituting here the above found coefficients $c_{2}$ and $c_{4}$, we obtain

$$\begin{aligned} E=E_{0}\left( 1+\frac{5\pi ^{2}}{12}\tau ^2- \frac{\pi ^{4}}{16}\tau ^4\right) \end{aligned}$$

Then for thermal energy per atom we obtain

$$\begin{aligned} E_{\rm T}=z E_{\rm F} \left( \frac{\pi ^{2}}{4}\tau ^2- \frac{3\pi ^{4}}{80}\tau ^4\right) \end{aligned}$$

Now the entropy can be found:

$$\begin{aligned} s&= \int _{0}^{T_{\rm e}} { \frac{{\rm d}E_{\rm T}}{T}} \\&= z E_{\rm F} \int _{0}^{T_{\rm e}} { \left( \frac{\pi ^{2}}{4}\cdot \frac{k_{\rm B}^{2}}{ E_{\rm F}^{2}}\cdot 2 - \frac{3\pi ^{4}}{80}\cdot \frac{k_{\rm B}^{4}}{ E_{\rm F}^{4}}\cdot 4T^2 \right) {\rm d}T } \\&= zk_{\rm B}\left( \frac{\pi ^{2}}{2}\tau -\frac{\pi ^{4}}{20}\tau ^{3} \right) \end{aligned}$$

The adiabatic condition $s$ = const therefore means $\tau \,=\,{\rm const}$, and the thermal pressure can be written as

$$\begin{aligned} p_{\rm T}&= -\frac{\partial E_{\rm T}}{\partial v}|_{s}= -z\frac{{\rm d} E_{\rm F}}{{\rm d}v} \left( \frac{\pi ^{2}}{4}\tau ^2- \frac{3\pi ^{4}}{80}\tau ^4\right) \\&= \frac{2}{3}znE_{\rm F} \left( \frac{\pi ^{2}}{4}\tau ^2- \frac{3\pi ^{4}}{80}\tau ^4\right) . \end{aligned}$$