4.1. Isothermal process

In the case where the time scale of the dust process (e.g. expansion/compression), ${\tau _P}$ is larger than the dust neutral collision time scale ${\tau _{\textrm{dn}}}$, the dust is strongly coupled with neutrals. In this case, the dust temperature ${T_d}$ is regulated by the neutral bath and can be taken to be constant ${T_{d0}}({\approx} {T_n})$. This would typically be true in experiments with high gas pressure (Thomas Reference Thomas2010). Thus, in the high gas pressure regime, the equation of state is given by

(4.1)\begin{equation}{P_d} = \frac{{{N_d}{T_{d0}}}}{{V_d^{}}} + \frac{{q_d^2N_d^2{T_e}{T_i}}}{{{q^2}n({T_e} + {T_i})V_d^2}}.\end{equation}

In this case, the thermal energy of the dust component $3{N_d}{T_{d0}}/2$ is constant. However, the internal energy ${U_d}$ which contains ES contributions as well, and is defined in (3.2), is not constant. Processes where ${U_d}$ is constant will be defined later. For quasi-static expansion/contraction of the dust volume $\Delta {V_d}$, the energy conservation can be re-expressed as ${T_d}\Delta {S_d} + \Delta ({T_i}{S_i} + {T_e}{S_e}) = {P_d}\Delta {V_d}$, which shows that the heat taken from the plasma bath maintains an isothermal background with given electron and ion temperatures. The dust then expands/compresses at constant ${T_d}$ in this isothermal background. The change in the dust entropy is given by $\Delta {S_d} = {n_d}\Delta {V_d}$ while the change in the plasma entropy (ES part), in the simple case where ${T_e} = {T_i} = T$ is $\Delta S = q_d^2n_d^2\Delta {V_d}/2{q^2}n$. In terms of coupling parameters ${\varGamma _0},{\kappa _0}$ which, for given ${N_d}$, are related to the two constants ${T_{d0}},{V_{do}}$ by the relations ${\varGamma _0} = q_d^2/4{\rm \pi} {\varepsilon _0}{a_0}{T_{d0}},{\kappa _0} = {a_0}/{\lambda _d}$ where ${a_0} = {(3{V_{d0}}/4{\rm \pi} {N_d})^{1/3}}$, the equation of state is given by

(4.2)\begin{equation}{\bar{P}_d} = \frac{1}{{{{\bar{V}}_d}}} + \frac{3}{2}\frac{{{\varGamma _0}}}{{\kappa _0^2}}\frac{1}{{\bar{V}_d^2}},\end{equation}

where we have normalized pressure with ${n_{d0}}{T_{d0}}$ and ${V_d}$ by ${V_{d0}}$. In the weak correlation regime the ratio of the ES potential energy to the average kinetic of dust is given by $\varGamma _0^\ast = {\varGamma _0}{\rm exp} ( - {\kappa _0}) \lt 1$. In figure  1 we show a dust isotherm for ${\varGamma _0} = 1,{\kappa _0} = 1.5,\varGamma _0^\ast = 0.2$. The isothermal equation of state given in (4.2) has been verified in recent MD simulations (Shukla et al. Reference Shukla, Avinash, Mukherjee and Ganesh2017). In the large ${V_d}$, limit, ${\bar{P}_d} \propto 1/{\bar{V}_d}$ as an ideal gas, however, for smaller dust clouds where ES pressure dominates, ${\bar{P}_d} \propto 1/\bar{V}_d^2$.

Figure 1.

Isothermal process (solid), adiabatic (dash) and constant internal energy (dot-dash) processes of dusty plasma in the ${P_d} - {V_d}$ plane for ${\varGamma _0} = 1,{\kappa _0} = 1.5,\varGamma _0^\ast = 0.2$. In the limit of small ${V_d}$ all the three processes become identical with ${P_d} \propto 1/V_d^2$.

4.2. Adiabatic process

In the case where the expansion/compression time scale of dust ${\tau _P}$ is smaller than the dust neutral collision time scale ${\tau _{\textrm{dn}}}$, the thermal contact of the dust with neutrals is weak. In these cases the dust can regarded as isolated with constant entropy $(\Delta {S_d} = 0)$. This would typically be true in experiments with low neutral gas pressure (Pilch et al. Reference Pilch, Piel, Trottenberg and Koepke2007). As stated earlier, the heat exchanges involved in local plasma perturbations must be taken into account while defining thermodynamic processes in our model. Accordingly, we define the quasi-static adiabatic process involving dust as ‘the expansion/compression processes involving isolated dust that take place in an isothermal plasma background of given temperature’. Hence, in this process, while the dust is isolated and its entropy remains constant in expansion/compression, the background plasma is not and its entropy changes. The equation of state for this process can be calculated from the equation $0 = \Delta {U_d} + {P_d}\Delta {V_d}$. Eliminating ${U_d}$ and ${P_d}$ via (3.4a) we obtain the equation $\Delta {T_d}/{T_d} ={-} (2/3)\Delta {V_d}/{V_d}$ which can be integrated to give ${T_d}V_d^{2/3} = \textrm{const}\textrm{.}$, (like an ideal gas). Eliminating ${T_d}$ using this relation in the expression for ${P_d}$ in (3.4a) finally gives the adiabatic equation of state for gaseous dust in terms one arbitrary constant C (and given ${T_e}$ and ${T_i}$) as

(4.3)\begin{equation}{P_d} = \frac{C}{{V_d^{5/3}}} + \frac{{q_d^2N_d^2{T_e}{T_i}}}{{{q^2}n({T_e} + {T_i})V_d^2}}.\end{equation}

The adiabat/isentrope ${P_d} = {P_d}({V_d})$ can be plotted in the ${P_d} - {V_d}$ plane where C is determined by any point ${P_{d0}},{V_{do}}$ (alternately ${T_{d0}},{V_{do}}$) on the adiabat. In terms of coupling parameters ${\varGamma _0},{\kappa _0}$, defined earlier, the equation of state can be expressed thus

(4.4)\begin{equation}{\bar{P}_d} = \frac{1}{{\bar{V}_d^{5/3}}} + \frac{3}{2}\frac{{{\varGamma _0}}}{{\kappa _0^2}}\frac{1}{{\bar{V}_d^2}}.\end{equation}

In figure 1 we show a dust adiabat for ${\varGamma _0} = 1,{\kappa _0} = 1.5,\varGamma _0^\ast = 0.2$. If the dust volume expands against external pressure from ${\bar{V}_{d1}} \to {\bar{V}_{d2}}$, then the work done is

(4.5)\begin{equation}\bar{W} = \frac{3}{2}\left( {\frac{1}{{\bar{V}_{d1}^{2/3}}} - \frac{1}{{\bar{V}_{d2}^{2/3}}}} \right) + \frac{3}{2}\frac{{{\varGamma _0}}}{{\kappa _0^2}}\left( {\frac{1}{{{{\bar{V}}_{d1}}}} - \frac{1}{{{{\bar{V}}_{d2}}}}} \right).\end{equation}

This equation shows that in expanding from ${\bar{V}_{d1}} \to {\bar{V}_{d2}}$, dust does extra work (second term) compared with the corresponding ideal gas. The extra work is done by the ES pressure to extract requisite amount of heat from the plasma background to maintain uniform electron and ion temperatures. The changes in entropies of electrons and ions $\Delta {S_e},\Delta {S_i}$ can be calculated from expressions of electrons and ion entropies given in (2.3). In the simple case where ${T_e} = {T_i} = T$, the change in the ES part of the total plasma entropy is given by $\Delta S = q_d^2n_d^2\Delta {V_d}/2{q^2}n$.

4.3. Constant internal energy process

In addition to the two processes described above, we can define an additional new process where the internal energy of dust ${U_d}$ is constant. This process will take place under the condition when the dust neutral collision time scale is of the same order as the time scale of the process, i.e. ${\tau _{\textrm{dn}}} \approx {\tau _p}$. Under this condition, the rate at which the dust exchanges heat with the dust heat bath is equal to the work done by/on the dust against the external pressure so that $\Delta {Q_d} = {P_d}\Delta {V_d}$ and $\Delta {U_d} = 0$ in (3.2). The equation of state for this process can be calculated by using (3.2), (3.4a) and the condition $\Delta {U_d} = 0$, and is given by

(4.6a,b)\begin{equation}{T_d} = {T_{d0}} - \frac{{2q_d^2{N_d}{T_e}{T_i}}}{{3{q^2}n{V_d}({T_e} + {T_i})}},\quad {P_d} = \frac{{{N_d}{T_{d0}}}}{{{V_d}}} + \frac{{q_d^2N_d^2{T_e}{T_i}}}{{3{q^2}V_d^2n({T_e} + {T_i})}}.\end{equation}

Thus, in the constant ${U_d}$ process, the dust temperature increases on expansion and decreases on compression. This is an interesting and somewhat counterintuitive consequence which can be verified experimentally. However, the pressure, as usual, decreases with the volume as in the case of isothermal processes, except for a factor of 1/3 in the second term. In figure  1 we show the ${P_d} - {V_d}$ plot for constant ${U_d}$ process. The change in the plasma entropy, as in earlier cases, is given by $\Delta S = q_d^2n_d^2\Delta {V_d}/2{q^2}n$, while the change in the dust entropy is given by $\Delta {S_d} = (1 + 3\varGamma /2{\kappa ^2}){n_d}\Delta {V_d}$. In the large ${\bar{V}_d}$ limit, the effects due to the mean field vanish and the constant ${U_d}$ process becomes identical to isothermal process and $\Delta {S_d} = {n_d}\Delta {V_d}$.

4.4. Specific heat of dust

The specific heats of the dust at constant volume and constant pressure are defined as

(4.7a,b)\begin{equation}{C_{{V_d}}} = {\left. {\frac{{\partial {Q_d}}}{{\partial {T_d}}}} \right|_{{V_d}}},\quad {C_{{P_d}}} = {\left. {\frac{{\partial {Q_d}}}{{\partial {T_d}}}} \right|_{{P_d}}}.\end{equation}

Eliminating $\Delta {Q_d}$ from (2.8a,b) and using (3.2) and (3.4a,b) we get

(4.8a,b)\begin{equation}{C_{{V_d}}} = 3{N_d}/2,\quad {C_{{P_d}}} = 3{N_d}/2 + \frac{{{N_d}{T_d}}}{{{V_d}}}{\left. {\frac{{\partial {V_d}}}{{\partial {T_d}}}} \right|_{{P_d}}}.\end{equation}

Eliminating the partial derivative in (4.8a,b) using (3.4a) we finally obtain the specific heat at constant pressure and the ratio of two specific heats γ as

(4.9a,b)\begin{equation}{C_{{P_d}}} = \frac{3}{2}{N_d} + \frac{{{N_d}}}{{\left( {1 + \dfrac{{3\varGamma }}{{2{\kappa^2}}}} \right)}},\quad \gamma = 1 + \frac{{2/3}}{{\left( {1 + \dfrac{{3\varGamma }}{{2{\kappa^2}}}} \right)}}.\end{equation}

Equation (4.9a,b) shows that γ is no longer constant but a function of dust volume and temperature. In the ideal gas or weak coupling limit (Γ→ 0, κ → ∞), γ = 5/3.

In addition to the two types of specific heats defined in (4.7a,b), we can define a third type of specific heat, i.e. the specific heat at constant internal energy ${C_{{U_d}}}$ defined as

(4.10)\begin{equation}{C_{{U_d}}} = {\left. {\frac{{\partial {Q_d}}}{{\partial {T_d}}}} \right|_{{U_d}}} = {\left. {{P_d}\frac{{\partial {V_d}}}{{\partial {T_d}}}} \right|_{{U_d}}},\end{equation}

where we have used $\Delta {Q_d} = {P_d}\Delta {V_d}$ for constant ${U_d}$. This specific heat has no ideal gas analogue, though it may have a real gas analogue. Evaluating the partial derivative and substituting for pressure from the condition in (4.6a,b), we get

(4.11)\begin{equation}{C_{{U_d}}} = \frac{{3{N_d}}}{2}\left( {\frac{{2{\kappa^2}}}{{3\varGamma }} + 1} \right).\end{equation}

In the limit Γ→ 0, κ→ ∞, ${C_{{U_d}}} \to \infty$.

4.5. Dust acoustic waves

The dust acoustic waves (DAW) are analogues of ion acoustic waves in electron–ion plasma where the inertia is due to the dust mass and the screening is due to electrons and ions. The dispersion relation of these waves can be obtained directly by using equation of states, derived above, in the fluid equations.

The total pressure of the dust ${P_d}$, which is the sum of the kinetic and the ES pressure, drives acoustic modes in dusty plasma. The dispersion of these acoustic modes driven by the total dust pressure is governed by following set of fluid equations:

(4.12a–c)\begin{equation}{\rho _d}\frac{{\textrm{d}{\boldsymbol{v}_d}}}{{\textrm{d}t}} ={-} {\bf \nabla }({P_d}),\quad \frac{{\partial {\rho _d}}}{{\partial t}} + {\bf \nabla}\,\cdot\,{\rho _d}{\boldsymbol{v}_d} = 0,\quad {P_d} = cn_d^\gamma + \frac{{q_d^2{T_e}{T_i}}}{{{q^2}n({T_e} + {T_i})\delta }}n_d^2.\end{equation}

In these equations, ${P_d}$ is the local dust pressure which is the sum of the dust kinetic and the ES pressure and ${\rho _d},{v_d}$, are the local dust mass density and the fluid velocity, respectively. In (4.12c), γ = 5/3 for adiabatic processes, γ = 1 for isothermal or constant ${U_d}$ processes, δ = 1 for adiabatic or isothermal processes and δ = 3 for constant internal energy processes. Performing standard linearization of these equations about a uniform and static equilibrium and using plane wave solutions where ${\bf \nabla } = \textrm{i}k,\partial /\partial t ={-} \textrm{i}\omega$ in (4.12a–c), (ω and k are the frequency and the wave vector of DAW) we obtain following dispersion relation of acoustic waves:

(4.13)\begin{equation}\frac{{{\omega ^2}}}{{{k^2}}} = C_{\textrm{DAW}}^2 = \left[ {\frac{{\gamma {T_d}}}{{{m_d}}} + \frac{{2q_d^2{n_d}{T_e}{T_i}}}{{{q^2}n{m_d}({T_e} + {T_i})\delta }}} \right].\end{equation}

In the low neutral gas pressure regime $({\nu _{\textrm{dn}}}/\omega \ll 1)$, γ = 5/3 and δ = 1 in (4.13) which corresponds to the adiabatic DAW; in the high neutral gas pressure regime $({\nu _{\textrm{dn}}}/\omega \gg 1)$, γ = 1 and δ = 1 which corresponds to the isothermal DAW. While in the intermediate neutral gas pressure regime $({\nu _{\textrm{dn}}}/\omega \approx 1)$, γ = 1, δ = 3 which corresponds to a new mode, i.e. the constant internal energy DAW.

As stated before, in DAW, the inertia is due to dust mass while the screening is due to both electrons and ions. In the usual dispersion relation of ion acoustic modes in electron–ion plasma given by ${\omega ^2}/{k^2} = C_{\textrm{IA}}^2 = (\gamma {T_i} + {T_e})/{m_i}$, the inertia is due to ions and the screening is due to electrons. If we consider screening due to electrons in the dispersion relation of DAW given in (4.13) by taking ${T_i} \gg {T_e}$ and the inertia due to ions by replacing dust with ions, i.e. ${q_d} \to {q_i} = q,{m_d} \to {m_i}$, ${n_d} \to {n_i},{T_d} \to {T_i}$, $2{n_i} \approx n$ (n is plasma density of electron–ion plasma) the dispersion relation of DAW in (4.13) reduces to the dispersion relation of ion acoustic given above.

In dusty plasma experiments, DAW has been observed over a wide range of frequencies and neutral gas pressures. For example, in a 3PDX device (Thomas Reference Thomas2010), DAWs in the range from 7 Hz to 120 Hz were observed. The dust particle velocity distribution function was observed to be close to Maxwellian during the wave motion (Fisher & Thomas Reference Fisher and Thomas2010; Thomas Reference Thomas2010), justifying the quasi-static assumption. The neutral pressure in this device was 72 mTorr and the corresponding dust neutral collision frequency was ${\nu _{\textrm{dn}}} \approx 75\;\textrm{Hz}$. Then, in this experiment, waves observed in the range ω < 75 Hz will correspond to isothermal DAW, waves with ω > 75 Hz will correspond to Adiabatic DAW, while waves with ω in the range ≈75 Hz will correspond to constant internal energy DAW as discussed here. Thus it should be possible to identify the constant ${U_d}$ DAW in this experiment which has the novel and distinctive feature that in crests dust will be cooler and trough it will be hotter.

4.6. Free expansion of dust

In dusty plasma, the hydrodynamic free expansion of Coulomb balls and dust clouds (starting from an equilibrium state) has been examined via MD simulations (Piel & Goree Reference Piel and Goree2013) and analytically using fluid equations (Ivlev Reference Ivlev2013). In experiments (Barkan & Merlino Reference Barkan and Merlino1995; Antonova et al. Reference Antonova, Du, Ivlev, Annaratonne, Hou, Kompaneets, Thomas and Morfill2012), free expansion of dust in dusty plasma has been studied in the afterglow phase of the discharge. In this phase, the confining fields are removed by switching off the anode voltage causing the background plasma to decay. The MD simulations show that at low gas pressure, the background plasma decays rapidly removing the shielding and the cloud explodes due to bare Coulomb repulsion. In contrast, at high gas pressure, the dust cloud expands gradually under the shielded Yukawa repulsion (Ivlev et al. Reference Ivlev, Kretschmer, Zuzic, Morfill, Rothermel, Thomas, Fortov, Molotkov, Nefedov and Lipaev2003; Saxena, Avinash & Sen Reference Saxena, Avinash and Sen2012) and may fission eventually (Merlino et al. Reference Merlino, Meyer, Avinash and Sen2016).

Here, in the last part of our paper, we describe the free, non-quasi-static expansion of isolated gaseous dust in a confined and steady isothermal plasma background. This is different from the free expansion of dust in an unconfined and decaying plasma background of the afterglow phase mentioned above. The non-quasi-static expansion is governed by the equation $\Delta {Q_d} = \Delta {U_d} + {P_{\textrm{ext}}}\Delta {V_d}$. Since the dust is isolated $\Delta {Q_d} = 0$ and is allowed to expand freely, ${P_{\textrm{ext}}} = 0$. Hence the condition for free expansion is given by ${U_d} = \textrm{const}\textrm{.}$ and the variation of temperature with volume is given by (4.6a), i.e.

(4.14)\begin{equation}{T_d} = {T_{d0}} - \frac{{2q_d^2{N_d}{T_e}{T_i}}}{{3{q^2}n{V_d}({T_e} + {T_i})}}.\end{equation}

This is an irreversible process hence the entropy increases. If the dust expands freely from state $({T_{d1}},{V_{d1}}) \to ({T_{d2}},{V_{d2}})$ then the change in entropy is calculated by putting a reversible path between the two states. It is identical to change in entropy of the constant internal energy process and is given by $\Delta {S_d} = (1 + 3\varGamma /{\kappa ^2}){n_d}\Delta {V_d}$. In the ideal gas limit (Γ→ 0, κ → ∞), $\Delta {S_d} = {n_d}\Delta {V_d}$. Of course, as opposed to the constant internal energy process, the work done by dust in free expansion is zero. Since the dust is isolated, the free expansion must occur on time scale faster than the dust neutral time scale, i.e. ${\tau _p} \ll {\tau _{\textrm{dn}}}$, which will be typically true at low neutral pressure. This neutral pressure condition is similar to the adiabatic process, however, the difference in the two processes is that for free expansion ${P_{\textrm{ext}}} = 0$, while for adiabatic processes ${P_{\textrm{ext}}} = {P_d}$. To calculate the rise in dust temperature for typical dusty plasma experimental parameters, we express (4.14) in the following dimensionless form:

(4.15)\begin{equation}{\bar{T}_d} = 1 + \frac{{2{Z_d}p}}{3}\frac{{{{\bar{T}}_e}{{\bar{T}}_i}}}{{({{\bar{T}}_e} + {{\bar{T}}_i})}}(1 - {\bar{n}_d}).\end{equation}

In this equation, the dust temperature, electron and ion temperature are all normalized with initial dust temperature ${T_{d0}}$, the dust density by the initial dust density ${n_{d0}}$, ${Z_d} = {q_d}/q$ and, p is given by $p = {Z_d}{n_{d0}}/n$. Typical parameters of dusty plasma experiments are ${T_e} \approx 3\;\textrm{eV},{T_i} \approx 0.025\;\textrm{eV},n \approx 5 \times {10^{13}}\;{\textrm{m}^{ - 3}},{n_d} \approx 5 \times {10^9}\;{\textrm{m}^{ - 3}},{T_d} \approx 100\;\textrm{eV}$, while for micron sized dust ${Z_d} \approx 3 \times {10^3}$ (Thomas Reference Thomas2010). Then, if we take the values of dust and plasma density given above as initial values, and if in the experiment the dust volume increases (freely) by two times so that ${\bar{n}_d} = 1/2$, then ${\overline T _d} = 1.15$ in (4.15), i.e. the dust temperature increases by 15 % of its initial value. In some dusty plasma experiments higher ion temperatures, ${T_i} \approx 0.1\;\textrm{eV}$, are reported (Trottenberg et al. Reference Trottenberg, Block and Piel2006). In such cases dust temperature increases by 50 %, i.e. ${\bar{T}_d} = 1.5$ for ${\bar{n}_d} = 1/2$.

The reason for heating of dust particles in free expansion is the repulsive nature of the interparticle force. In free expansion of isolated dust particles, the positive potential energy decreases which is converted into thermal energy. The reverse would be true in cases of attractive force between particles. This is similar to the case of free expansion of real gases (sometimes called Joule expansion) where heating observed above inversion temperature ${T_{\textrm{inv}}}$ is attributed to the repulsive part, and cooling below ${T_{\textrm{inv}}}$ is attributed to the attractive part of the Lennard–Jones potential (Goussard & Roulet Reference Goussard and Roulet1993). For example, hydrogen gas with low inversion temperature (~ 200 K) shows heating on free expansion at normal temperatures. For purely repulsive potential like the present case of negatively charged dust, ${T_{\textrm{inv}}} \to 0$, and temperature always increases in free expansion.

In the case of Joule–Thompson effect as well, where the gas is throttled through a constricted passage, the heating observed in the case $T \gt {T_{\textrm{inv}}}$ is due to the repulsive part of the interparticle force. In a separate paper, we will consider a Joule–Thompson-like throttling experiment for dusty plasma where dust particles are electrostatically throttled through a negatively biased mesh. This is expected to further enhance the dust temperature increase.