3.1. Ground state corresponding to a given perturbation

We now turn to the central question in this paper and calculate the energy available to a plasma close to a ground state. To this end, suppose that

(3.1)\begin{equation} f_0(\boldsymbol{x}) = F_0 [\epsilon(\boldsymbol{x}),\boldsymbol{y}], \end{equation}

denotes a ground state and

(3.2)\begin{equation} f(\boldsymbol{x}) = f_0 (\boldsymbol{x}) + \delta f(\boldsymbol{x}), \end{equation}

a nearby state, where $\delta f \ll f_0$. Note that, although $f_0$ is a ground state, it is in general inaccessible to a plasma with the distribution function $f$. We thus need to calculate the ground state corresponding to $f$, which we write as

(3.3)\begin{equation} F[\epsilon(\boldsymbol{x}),\boldsymbol{y}] = F_0[\epsilon(\boldsymbol{x}),\boldsymbol{y}] + \delta F[\epsilon(\boldsymbol{x}),\boldsymbol{y}], \end{equation}

where we expect $\delta F \ll F_0$. (As we shall discuss later, the derivatives of $\delta F$ also need to be much smaller than those of $F_0$.) In the interest of economy, we write $\boldsymbol {x}$ instead of $(\boldsymbol {y},\boldsymbol {z})$ wherever possible. The functions $F_0$ and $\delta F$ are then defined by the integral equations

(3.4)\begin{gather} \int \varTheta[f_0(\boldsymbol{x}) - F_0(w,\boldsymbol{y}) ] \sqrt{g} \, {\rm d} \boldsymbol{z} = \varOmega(w,\boldsymbol{y}), \end{gather}

(3.5)\begin{gather}\int \varTheta[f_0(\boldsymbol{x}) + \delta f(\boldsymbol{x}) - F_0(w,\boldsymbol{y}) - \delta F (w,\boldsymbol{y})] \sqrt{g} \, {\rm d} \boldsymbol{z} = \varOmega(w,\boldsymbol{y}). \end{gather}

The last equation is now expanded to second order, giving

(3.6)\begin{align} & \int \left\{ \varTheta[f_0(\boldsymbol{x}) - F_0(w,\boldsymbol{y}) ] + [\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})] \delta [f_0(\boldsymbol{x}) - F_0(w,\boldsymbol{y}) ] \vphantom{\frac{1}{2}}\right.\nonumber\\ & \quad + \left. \frac{1}{2} [\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})]^2 \delta' [f_0(\boldsymbol{x}) - F_0(w,\boldsymbol{y}) ] \right\} \sqrt{g} \, {\rm d} \boldsymbol{z} = \varOmega(w,\boldsymbol{y}). \end{align}

Since $F_0(w,\boldsymbol {y})$ is a decreasing function of $w$, we have the relation

(3.7)\begin{equation} \varTheta[w-\epsilon(\boldsymbol{x}) ] = \varTheta[f_0(\boldsymbol{x}) - F_0(w,\boldsymbol{y})], \end{equation}

which can be differentiated with respect to $w$ to give

(3.8)\begin{equation} \delta[f_0(\boldsymbol{x}) - F_0(w,\boldsymbol{y})] ={-} \frac{\delta[w-\epsilon(\boldsymbol{x}) ]}{F_0'(w,\boldsymbol{y})}, \end{equation}

where a prime denotes the derivative with respect to the first argument. Differentiating once more gives

(3.9)\begin{equation} \delta'[f_0(\boldsymbol{x}) - F_0(w,\boldsymbol{y})] = \frac{1}{F_0'(w,\boldsymbol{y})} \frac{\partial}{\partial w} \left( \frac{\delta[w-\epsilon(\boldsymbol{x}) ]}{F'_0(w,\boldsymbol{y})} \right). \end{equation}

Substituting these last two equations in (3.6) gives

(3.10)\begin{align} \int \left\{ \delta[w-\epsilon(\boldsymbol{x}) ][\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})] - \frac{1}{2} [\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})]^2 \frac{\partial}{\partial w} \left( \frac{\delta[w-\epsilon(\boldsymbol{x}) ]}{F'_0(w,\boldsymbol{y})} \right) \right\} \sqrt{g} \, {\rm d} \boldsymbol{z} = 0, \end{align}

where we use

(3.11)\begin{equation} \varOmega'(w,\boldsymbol{y}) = \int \delta[w-\epsilon(\boldsymbol{x}) ] \sqrt{g} \, {\rm d} \boldsymbol{z}, \end{equation}

to conclude that

(3.12)\begin{align} \delta F (w,\boldsymbol{y}) & = \frac{1}{\varOmega'(w,\boldsymbol{y})} \int \left\{ \delta f(\boldsymbol{x}) \delta[w-\epsilon(\boldsymbol{x}) ] - \frac{1}{2} [\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})]^2 \right.\nonumber\\ & \quad \left.\frac{\partial}{\partial w} \left( \frac{\delta[w-\epsilon(\boldsymbol{x}) ]}{F'_0(w,\boldsymbol{y})} \right) \right\} \sqrt{g} \, {\rm d} \boldsymbol{z}. \end{align}

The quantity $\varOmega '(w,\boldsymbol {y})$ is equal to the density of states in classical statistical mechanics.

3.2. Available energy

Having thus derived an expression for the ground state that is accurate to second order, we now turn our attention to the available energy

(3.13)\begin{align} A & = \int \epsilon (\boldsymbol{x}) (\delta f(\boldsymbol{x}) - \delta F [\epsilon(\boldsymbol{x}),\boldsymbol{y}]) \sqrt{g} \, {\rm d} \boldsymbol{x}\nonumber\\ & = \int \epsilon (\boldsymbol{x}) \delta f(\boldsymbol{x}) \sqrt{g} \, {\rm d} \boldsymbol{x} - \int\, {\rm d} \boldsymbol{y} \int_0^\infty \delta F (w,\boldsymbol{y}) w \, {\rm d} w \int \delta[w-\epsilon(\boldsymbol{x}) ] \sqrt{g} \, {\rm d} \boldsymbol{z}, \end{align}

which is ostensibly of first order in the smallness of $\delta f$. However, when (3.12) is substituted for $\delta F$, the first-order terms cancel, since in leading order,

(3.14)\begin{equation} \int \delta F(w,\boldsymbol{y}) \delta[w - \epsilon(\boldsymbol{x})]\sqrt{g} \, {\rm d} \boldsymbol{z} \simeq \int \delta f(\boldsymbol{x}) \delta [ w - \epsilon(\boldsymbol{x})] \sqrt{g} \, {\rm d} \boldsymbol{z}. \end{equation}

When substituted in (3.13), this result cancels the first term on the right-hand side, and the available energy thus vanishes in this order. We therefore continue to the next order, where

(3.15)\begin{equation} A = \frac{1}{2} \int \sqrt{g} \, {\rm d} \boldsymbol{x} \int_0^\infty w [\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})]^2 \frac{\partial}{\partial w} \left( \frac{\delta[w-\epsilon(\boldsymbol{x}) ]}{F'_0(w,\boldsymbol{y})} \right) \, {\rm d} w. \end{equation}

Focusing on the integral over $w$, one can make progress by integrating by parts,

(3.16)\begin{align} & \int_0^\infty w [\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})]^2 \frac{\partial}{\partial w} \left( \frac{\delta[w-\epsilon(\boldsymbol{x}) ]}{F'_0(w,\boldsymbol{y})} \right) \, {\rm d} w\nonumber\\ & \quad ={-} \int_0^\infty \left\{ [\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})]^2 - 2 w [\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})] \delta F '(w,\boldsymbol{y}) \right\} \frac{\delta[w-\epsilon(\boldsymbol{x}) ]}{F'_0(w,\boldsymbol{y})} \, {\rm d} w, \end{align}

and the available energy becomes

(3.17)\begin{align} A & ={-} \frac{1}{2} \int_0^\infty \, {\rm d} w \int \frac{{\rm d} \boldsymbol{y}}{F'_0(w,\boldsymbol{y})}\nonumber\\ & \quad \int \left\{[\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})]^2 -2 w [\delta f(\boldsymbol{x}) - \delta F (w,\boldsymbol{y})] \delta F' (w,\boldsymbol{y}) \right\} \delta[w-\epsilon(\boldsymbol{x}) ] \sqrt{g} \, {\rm d} \boldsymbol{z}. \end{align}

Since this expression is of second order, the first-order version of (3.6) may now be used for $\delta F$,

(3.18)\begin{equation} \delta F (w,\boldsymbol{y}) = \frac{1}{\varOmega'(w,\boldsymbol{y})} \int \delta f(\boldsymbol{x}) \delta[w-\epsilon(\boldsymbol{x}) ] \sqrt{g} \, {\rm d} \boldsymbol{z}, \end{equation}

whereupon the last term vanishes as a consequence of (3.14),

(3.19)\begin{equation} \int_0^\infty w \, {\rm d} w \int \frac{\delta F'(w,\boldsymbol{y})}{F'_0(w,\boldsymbol{y})} d\boldsymbol{y} \int [\delta f(\boldsymbol{x}) - \delta F (w ,\boldsymbol{y})] \delta[w-\epsilon(\boldsymbol{x}) ] \sqrt{g} \, {\rm d} \boldsymbol{z} = 0. \end{equation}

We thus obtain the following expression for the available energy:

(3.20)\begin{equation} A ={-} \int \frac{(\delta f(\boldsymbol{x}) - \delta F [\epsilon(\boldsymbol{x}),\boldsymbol{y}])^2}{2 F'_0[\epsilon(\boldsymbol{x}),\boldsymbol{y}]} \sqrt{g} \, {\rm d} \boldsymbol{x}, \end{equation}

where $\delta F$ is given by (3.18). Since $F'_0(w,\boldsymbol {y})$ is negative, $A$ is positive definite. Moreover, it vanishes if and only if $\delta f(\boldsymbol {y},\boldsymbol {z}) = \delta F [\epsilon (\boldsymbol {y},\boldsymbol {z}),\boldsymbol {y}]$, so that $f = f_0 + \delta f$ only depends on $\boldsymbol {z}$ through the energy function $\epsilon$, which is, of course, the condition that $f$ represents a ground state.

Another useful form for the available energy is obtained by expanding the square in (3.20) and noting that, to leading order,

(3.21)\begin{equation} \int \frac{\delta f(\boldsymbol{x}) \delta F [\epsilon(\boldsymbol{x}),\boldsymbol{y}]}{2 F'_0[\epsilon(\boldsymbol{x}),\boldsymbol{y}]} \sqrt{g} \, {\rm d} \boldsymbol{x} \simeq \int \frac{\delta F^2 [\epsilon(\boldsymbol{x}),\boldsymbol{y}]}{2 F'_0[\epsilon(\boldsymbol{x}),\boldsymbol{y}]} \sqrt{g} \, {\rm d} \boldsymbol{x}, \end{equation}

as follows from

(3.22)\begin{equation} \int \frac{\delta f(\boldsymbol{x}) \delta F [\epsilon(\boldsymbol{x}),\boldsymbol{y}]}{2 F'_0[\epsilon(\boldsymbol{x}),\boldsymbol{y}]} \sqrt{g} \, {\rm d} \boldsymbol{x} = \int_0^\infty \, {\rm d} w \int \frac{\delta f(\boldsymbol{x}) \delta F(w,\boldsymbol{y})}{2 F'_0(w,\boldsymbol{y})} \delta[w-\epsilon(\boldsymbol{x}) ]\sqrt{g} \, {\rm d} \boldsymbol{y}\, {\rm d} \boldsymbol{z}, \end{equation}

where substituting equation (3.14) gives the desired result. We thus find, to the same accuracy as (3.20),

(3.23)\begin{equation} A ={-} \int \frac{\delta f^2(\boldsymbol{x}) - \delta F^2 [\epsilon(\boldsymbol{x}),\boldsymbol{y}]}{2 F'_0[\epsilon(\boldsymbol{x}),\boldsymbol{y}]} \sqrt{g} \, {\rm d} \boldsymbol{x}. \end{equation}

3.3. Relation to Helmholtz free energy

Let us temporarily assume that $f_0$ is a Maxwellian with spatially constant density $n_0$ and temperature $T_0$,

(3.24)\begin{equation} f_0 = n_0 \left( \frac{m}{2 {\rm \pi}T_0} \right)^{3/2} \, {\rm e}^{-\epsilon / T_0}, \end{equation}

so that $F'_0 = - F_0/T_0$, and let us take the total number of particles contained in the distribution functions $f_0$ and $f$ to be the same,

(3.25)\begin{equation} \int f_0 \sqrt{g} \, {\rm d} \boldsymbol{x} = \int f \sqrt{g} \, {\rm d} \boldsymbol{x}. \end{equation}

If we define the difference in entropy and energy carried by the distribution functions $f_0$ and $f$ as

(3.26)\begin{gather} \delta S ={-} \int (f \ln f - f_0 \ln f_0) \sqrt{g} \, {\rm d} \boldsymbol{x}, \end{gather}

(3.27)\begin{gather}\delta U = \int \epsilon(\boldsymbol{x}) (f-f_0) \sqrt{g} \, {\rm d} \boldsymbol{x}, \end{gather}

then to second order in $\delta f$,

(3.28)\begin{equation} \delta S = \frac{\delta U}{T_0} - \int \frac{\delta f^2}{2 f_0} \sqrt{g} \, {\rm d} \boldsymbol{x}, \end{equation}

and we conclude that

(3.29)\begin{equation} H = T_0 \int \frac{\delta f^2}{2 f_0} \sqrt{g} \, {\rm d} \boldsymbol{x} = \delta U - T_0 \delta S,\end{equation}

denotes the difference in Helmholtz free energy in the two distribution functions. In plasma physics, this quantity has often been considered in discussions of turbulence, see e.g. Krommes & Hu (Reference Krommes and Hu1993), Brizard (Reference Brizard1994), Sugama et al. (Reference Sugama, Okamoto, Horton and Wakatani1996), Garbet et al. (Reference Garbet, Dubuit, Asp, Sarazin, Bourdelle, Ghendrih and Hoang2005), Schekochihin et al. (Reference Schekochihin, Cowley, Dorland, Hammett, Howes, Quataert and Tatsuno2009), Banon Navarro et al. (Reference Banon Navarro, Morel, Albrecht-Marc, Carati, Merz, Goerler and Jenko2011) and Stoltzfus-Dueck & Scott (Reference Stoltzfus-Dueck and Scott2017). It has recently been used to derive upper bounds on gyrokinetic instabilities that are generally valid irrespective of the geometry of the magnetic field, the number of particle species, collisions, etc. (Helander & Plunk Reference Helander and Plunk2021,  Reference Helander and Plunk2022; Plunk & Helander Reference Plunk and Helander2022)

Equations (3.20) and (3.23) show that the available energy is closely related to, but in general different from, the Helmholtz free energy. If for some reason $\delta F$ vanishes, then these two energies are equal, i.e. $H=A$ if $\delta F = 0$, and otherwise the available energy is smaller than the Helmholtz free energy, i.e. $A < H$ if $\delta F \ne 0$.

This line of thought can be extended to the case of a non-Maxwellian $f_0$ by considering a more general entropy functional

(3.30)\begin{equation} S[f] = \int s(f) \sqrt{g} \, {\rm d} \boldsymbol{x}, \end{equation}

where the function $s$ will be chosen suitably. To second order in $\delta f / f_0$ we then have

(3.31)\begin{equation} \delta S = S[f] - S[f_0] = \int \left[ s'(f_0) \delta f + \frac{s''(f_0)}{2} \delta f^2\right]\sqrt{g} \, {\rm d} \boldsymbol{x}, \end{equation}

and the natural generalisation of the Helmholtz energy is

(3.32)\begin{equation} H = T_0 \int \left[ \left( \frac{\epsilon}{T_0} - s'(f_0) \right) \delta f - \frac{s''(f_0)}{2} \delta f^2\right] \sqrt{g} \, {\rm d} \boldsymbol{x}, \end{equation}

if the function $s$ and the constant $T_0$ are chosen such that the linear term vanishes. If $f_0({\boldsymbol x}) = F_0[\epsilon ({\boldsymbol x})]$, this is accomplished by the choice

(3.33)\begin{gather} s'(F_0) = \frac{\epsilon}{T_0}, \end{gather}

(3.34)\begin{gather}s''(F_0) = \frac{1}{T_0 F_0'(\epsilon)}. \end{gather}

The resulting Helmholtz energy,

(3.35)\begin{equation} H ={-} \int \frac{\delta f^2(\boldsymbol{x})}{2 F'_0[\epsilon(\boldsymbol{x})]} \sqrt{g} \, {\rm d} \boldsymbol{x}, \end{equation}

coincides with (3.20) and (3.23) if $\delta F = 0$. A similar argument was made by Taylor (Reference Taylor1963), see also Kruskal & Oberman (Reference Kruskal and Oberman1958).

This construction shows that, if $F_0$ is chosen in such a way that $\delta F =$(3.18) vanishes, then an appropriate definition of the entropy function $s(f)$ results in a Helmholtz energy that is equal to the available energy. In this sense, the available energy is always equal to the Helmholtz free energy with a suitable choice for the unperturbed state and the entropy. However, in practice, a basic requirement of perturbation theory is that the unperturbed state, in this case $F_0$, should be simple, which will usually not be the case if one insists that (3.18) vanishes.

The reason why it is necessary to include $\delta F$ is that, given a ground state $f_0$ and a nearby state $f = f_0 + \delta f$, the former is generally not accessible from the latter through Gardner restacking. The ground state corresponding to $f$ is instead $F = F_0 + \delta F$, and the available energy (3.20) is equal to the generalised free energy associated with a perturbation from this ground state rather than $f_0$. This circumstance also explains why the available energy cannot exceed the generalised Helmholtz free energy (3.35). The latter energy would be released through Gardner restacking of $f$ into $f_0$, if this were possible. Imposing the extra constraint that the ground state reached by restacking should be accessible from the initial state can only make the released energy smaller.

4.1. Bi-Maxwellian with fluctuating density, temperatures and flow

We begin by considering the case of a bi-Maxwellian distribution function, i.e.

(4.1)\begin{equation} f(\boldsymbol{r}, \boldsymbol{v}) = n \left( \frac{m}{2 {\rm \pi}\bar{T}} \right)^{3/2} \exp \left(-\frac{m (v_x-u_x)^2}{2T_\perp} -\frac{m (v_y- u_y)^2}{2T_\perp} - \frac{m (v_\parallel{-}u_\|)^2}{2T_\parallel}\right),\end{equation}

in Cartesian velocity-space coordinates aligned with the temperature anisotropy. The quantity $\bar {T} = T_\|^{1/3}T_\perp ^{2/3}$ is the geometric mean of the temperatures in the three directions, two of which have been taken to be equal. The density, temperatures and flow velocity, are assumed to be nearly constant,

(4.2)\begin{gather} n = n_0[1 + \nu(\boldsymbol{r})], \end{gather}

(4.3)\begin{gather}T_\| = T_{\|,0}[1 + \tau_\|(\boldsymbol{r})], \end{gather}

(4.4)\begin{gather}T_\perp{=} T_{{\perp},0}[1 + \tau_\perp(\boldsymbol{r})], \end{gather}

(4.5)\begin{gather}\boldsymbol{u} = \boldsymbol{u}_0 + \delta \boldsymbol{u}(\boldsymbol{r}), \end{gather}

where $\boldsymbol {u}_0$ can be made to vanish by a Galilean transformation. Helander (Reference Helander2017) treated the case of constant density and temperatures, and showed that the ground state is an isotropic Maxwellian with temperature $T_{\perp,0}=T_{\|,0}=T_0$. The perturbed distribution function thus becomes $f = f_0 + \delta f$, with

(4.6)\begin{equation} \delta f = \left[\nu + \tau_\perp \left( \frac{m v_\perp^2}{2 T_0} - 1 \right) + \tau_{\|} \left( \frac{m v_\|^2}{2 T_0} - \frac{1}{2} \right) - \frac{m (\boldsymbol{v} \boldsymbol{\cdot} \delta \boldsymbol{u})}{T_0} \right] f_0(\boldsymbol{r}, \boldsymbol{v}), \end{equation}

where $\boldsymbol {v} = (v_x,v_y,v_\|)$ and $v_\perp ^2 = v_x^2 + v_y^2$. The function $f_0$ can be chosen so that the distribution functions $f_0$ and $f$ carry equal amounts of particles and energy,

(4.7)\begin{equation} \langle \nu \rangle = \left\langle \frac{\bar{T}-T_0}{T_0} \right\rangle = 0, \end{equation}

where angular brackets denote an average over the plasma volume $V$,

(4.8)\begin{equation} \langle \cdots \rangle = \frac{1}{V} \int \cdots \, {\rm d} \boldsymbol{r}. \end{equation}

To leading order we must then have $\langle \tau _\| \rangle = -2 \langle \tau _\perp \rangle$, and (3.18) becomes

(4.9)\begin{equation} \delta F(w) = \frac{1}{\varOmega'(w)} \int f_0 \tau_\perp \left(\epsilon - \frac{3}{2}\frac{m v_\|^2}{ T_0} \right) \delta (w - \epsilon)\, {\rm d} \boldsymbol{r}\, {\rm d} \boldsymbol{v} . \end{equation}

One can make progress by introducing the coordinates $v_\perp = v \cos \vartheta$ and $v_\| = v \sin \vartheta$. The volume element becomes $d \boldsymbol {v} = {\rm \pi}(2/m)^{3/2} \cos \vartheta \sqrt {\epsilon } \, {\rm d} \vartheta \, {\rm d} \epsilon$, resulting in

(4.10)\begin{equation} \delta F(w) = \frac{\rm \pi}{\varOmega'(w)} \int f_0 \tau_\perp \left(\frac{2\epsilon}{m} \right)^{3/2} \delta (w - \epsilon) \left( 1 - 3 \sin^2 \vartheta \right) \cos \vartheta \, {\rm d} \vartheta \, {\rm d} \epsilon\, {\rm d} \boldsymbol{r} = 0. \end{equation}

The available energy is thus equal to the Helmholtz free energy and becomes

(4.11)\begin{equation} A = H = \frac{n_0 T_0 V}{2} \left\langle \nu^2 + \frac{1}{2} \tau_\|^2 + \tau_\perp^2 + \frac{m (\delta \boldsymbol{u})^2}{T_0} \right\rangle.\end{equation}

This expression generalises the result found by Helander (Reference Helander2017) to include anisotropic temperature perturbations and a spatially varying flow.

4.2. Kappa distribution in any number of dimensions

We next consider the kappa distribution in any integer number $d$ of dimensions. These functions were originally introduced to model high-energy tails of distribution functions in astrophysical plasmas that exhibit power-law behaviour (Olbert Reference Olbert1968; Vasyliunas Reference Vasyliunas1968). Lazar & Fichtner (Reference Lazar and Fichtner2021) provide a recent review of their use in modelling various plasmas. The kappa distribution function is defined by (Livadiotis & McComas Reference Livadiotis and McComas2013)

(4.12)\begin{equation} f_{\kappa,d}(\boldsymbol{r},\boldsymbol{v}) = n \left(\frac{m}{2 {\rm \pi}T}\right)^{d/2} \frac{\varGamma\left( \kappa_0 + 1 + \dfrac{{\rm d} }{2} \right)}{\kappa_0^{d/2} \varGamma\left( \kappa_0 + 1 \right)} \left( 1 + \frac{1}{\kappa_0} \frac{mv^2}{2T} \right)^{-\kappa_0 - 1 - d/2}, \end{equation}

where $\kappa _0 = \kappa - d/2$, and the normalisation constants have been chosen such that the integral over velocity space returns the number density,

(4.13)\begin{equation} \int f_{\kappa,d} \, {\rm d} \boldsymbol{v} = n, \end{equation}

and the energy density is

(4.14)\begin{equation} \int \frac{mv^2}{2} f_{\kappa,d} \, {\rm d} \boldsymbol{v} = \frac{nTd}{2}, \end{equation}

so that each degree of freedom as usual carries the energy $nT/2$. Note that the tail of the distribution follows a power-law, $f_{\kappa,d} \propto v^{-2\kappa _0 - 2 - d}$. Furthermore, the limit of $\kappa _0 \rightarrow \infty$ corresponds to the usual Maxwellian.

Any spatially constant kappa distribution is trivially a ground state, as it is a monotonically decreasing function of $v^2$. If the density and temperature are perturbed by relative amounts $\nu (\boldsymbol {r})$ and $\tau (\boldsymbol {r})$, respectively, the fluctuating part of the distribution function becomes

(4.15)\begin{equation} \delta f_{\kappa,d} = f_{\kappa,d,0} \left[ \nu + \tau \frac{\dfrac{m v^2}{2 T_0}(1 + \kappa_0) - \dfrac{\kappa_0 d}{2}}{\dfrac{m v^2}{2 T_0} + \kappa_0} \right]. \end{equation}

Imposing the condition $\langle \nu \rangle = \langle \tau \rangle = 0$, we find that the ground state vanishes, $\delta F = 0$. The available energy is thus equal to the Helmholtz free energy and is given by

(4.16)\begin{equation} A = H = \frac{n_0 T_0 V}{2} \left \langle \frac{d ( \nu + \tau )^2 + \kappa_0( 2 \nu^2 +\, d \tau^2 ) }{2 + d + 2 \kappa_0} \right \rangle. \end{equation}

In the limit of a Maxwellian, $\kappa _0 \rightarrow \infty$, the available energy depends on the dimensionality $d$ as

(4.17)\begin{equation} A = \frac{n_0 T_0 V}{2} \left \langle\nu^2 + \frac{d}{2} \tau^2 \right \rangle. \end{equation}

Setting $d=3$ we recover the result of Helander (Reference Helander2017) for the available energy of a three-dimensional Maxwellian plasma with slightly varying density and temperature.

4.3. Magnetic-moment conservation of a bi-Maxwellian

Let us again consider the case of a bi-Maxwellian at rest with slightly varying density and temperatures, where we now impose the condition that $\mu = mv_\perp ^2/2B$ be conserved, so that (4.1) becomes

(4.18)\begin{equation} f(\boldsymbol{r}, \boldsymbol{v}) = n(\boldsymbol{r}) \left( \frac{m}{2 {\rm \pi}\bar{T}(\boldsymbol{r})} \right)^{3/2} \, {\rm e}^{- \epsilon / T_\parallel(\boldsymbol{r}) + \mu B(\boldsymbol{r}) [ 1/T_\parallel(\boldsymbol{r}) - 1/T_\perp(\boldsymbol{r})]}, \end{equation}

where $\bar {T}$ is the geometric mean of the temperatures, as in § 4.1. When the magnetic moment $\mu$ is conserved, any ground state must be of the form $f_0(\boldsymbol {r}, \boldsymbol {v}) = F_0[\epsilon (\boldsymbol {r}, \boldsymbol {v}), \mu (\boldsymbol {r}, \boldsymbol {v})]$ with $F'_0(w,\mu ) \le 0$. (As usual, a prime denotes differentiation with respect to the first argument.) It follows that, in order for the function

(4.19)\begin{equation} f_0=n_0 \left( \frac{m}{2 {\rm \pi}\overline{T}_0} \right)^{3/2} \, {\rm e}^{- \epsilon / T_{{\parallel},0} + \mu B(\boldsymbol{r}) [ 1/T_{{\parallel},0} - 1/T_{{\perp},0}]}, \end{equation}

to be a ground state, the perpendicular and parallel temperatures either must be equal $T_{\perp,0}=T_{\|,0} = T_0$, if $B(\boldsymbol {r})$ is non-constant, or one may have $T_{\perp,0} \neq T_{\|,0}$ if $B(\boldsymbol {r})=B_0$ is a constant. Let us start by investigating the former case.

4.3.1. Spatially varying magnetic field

In the following calculation we have set $T_{\|,0}=T_{\perp,0}$ everywhere. The varying part of the distribution function, $\delta f$, is given in (4.6). The phase-space coordinates are chosen to be $\boldsymbol {x} = (\boldsymbol {z},\mu ) = (\boldsymbol {r},v_\parallel, \mu )$, and the velocity-space volume element becomes

(4.20)\begin{equation} {\rm d} \boldsymbol{v} = \frac{2 {\rm \pi}B}{m} \, {\rm d} v_\|\, {\rm d} \mu. \end{equation}

The density of states is now equal to

(4.21)\begin{equation} \varOmega'(w,\mu) = \int \delta \left( w - \mu B - \frac{mv_\|^2}{2} \right) \frac{2 {\rm \pi}B}{m} \, {\rm d} \boldsymbol{r} \, {\rm d} v_\| = \int_{w>\mu B(\boldsymbol{r})} \frac{4 {\rm \pi}B \, {\rm d} \boldsymbol{r}}{\sqrt{2 m^3 (w - \mu B)}}, \end{equation}

and (3.18) becomes

(4.22)\begin{align} \delta F (w,\mu) = \frac{f_0(w)}{\varOmega'(w,\mu)} \int \left[ \nu + \tau_\| \left( \frac{w - \mu B}{T_{0} } - \frac{1}{2} \right) + \tau_\perp \left( \frac{\mu B}{ T_{0}} - 1 \right) \right] \frac{4 {\rm \pi}B \, {\rm d} \boldsymbol{r}}{\sqrt{2 m^3 (w - \mu B)}}.\end{align}

If the magnetic field varies with $\boldsymbol {r}$ in a way that is correlated with the density or temperature fluctuations, then $\delta F$ is generally non-zero and the available energy will then be smaller than the Helmholtz energy, $A \le H$.

4.3.2. Constant magnetic field

If the magnetic field is constant, $B=B_0$, the perpendicular and parallel temperatures $T_{\perp,0}$ and $T_{\|,0}$ can be different, and will now be taken to be unequal. Under such circumstances, the fluctuating part of the bi-Maxwellian distribution (4.1) with $\boldsymbol {u}=0$ becomes

(4.23)\begin{equation} \delta f = \left[ \nu + \tau_\| \left( \frac{mv_\|^2}{2T_{\|,0} } - \frac{1}{2} \right) + \tau_\perp \left( \frac{\mu B_0}{ T_{{\perp},0}} - 1 \right) \right] f_0, \end{equation}

where we have perturbed the density and temperatures in the usual manner. The ground state simplifies to

(4.24)\begin{equation} \delta F (w,\mu) = f_0 \left[ \langle \tau_\| \rangle \left( \frac{w - \mu B}{T_{\|,0} } - \frac{1}{2} \right) + \langle \tau_\perp \rangle \left( \frac{\mu B}{ T_{{\perp},0}} - 1 \right) \right] , \end{equation}

and we have employed $\langle \nu \rangle = 0$. The available energy can now be calculated, and becomes

(4.25)\begin{equation} A = \frac{n_0 T_{\|,0} V}{2} \left\langle \nu^2 + \frac{1}{2} \left( \tau_\|^2 - \langle \tau_\| \rangle^2 \right) + \tau_\perp^2 - \langle \tau_\perp \rangle^2 \right\rangle. \end{equation}

In order to compare this expression with the case without invariants given in (4.11), we set $T_{\perp,0}=T_{\|,0}=T_0$ in (4.25). Under these conditions, the available energy becomes

(4.26)\begin{equation} A = H - \frac{n_0T_0V}{2}\left( \frac{1}{2} \left \langle \tau_\| \right \rangle^2 + \left \langle \tau_\perp \right \rangle^2 \right). \end{equation}

We thus conclude that the available energy of an isotropic plasma with constant magnetic field, where the perturbations satisfy $\tau _{\perp } = \tau _{\|}=\tau$, is equal to the usual Helmholtz free energy if $\mu$ is conserved, and the introduction of anisotropic perturbations reduces it below this value unless $\langle \tau _\| \rangle = \langle \tau _\perp \rangle = 0$.

4.4. Magnetic-moment conservation of a kappa-Maxwellian with fluctuating density and temperatures

We next consider a plasma distribution function that exhibits non-Maxwellian behaviour in the direction along a constant magnetic field $B=B_0$, but is Maxwellian in the perpendicular directions. In order to model such effects, we employ the kappa-Maxwellian distribution function introduced by Hellberg & Mace (Reference Hellberg and Mace2002),

(4.27)\begin{equation} f_\kappa(\boldsymbol{r},\boldsymbol{v}) = n \left( \frac{m}{2 {\rm \pi}\bar{T}} \right)^{3/2} \frac{\varGamma(\kappa + 1)}{\kappa^{3/2}\varGamma(\kappa - \frac{1}{2})} \sqrt{\frac{\kappa}{\kappa - \frac{3}{2}}} \left( 1 + \frac{m v_\parallel^2}{2 T_\parallel (\kappa - \frac{3}{2})} \right)^{-\kappa} {\rm e}^{ - \mu B_0 / T_\perp}, \end{equation}

which shares many features with the distribution function discussed in § 4.2. In the limit $\kappa \rightarrow \infty$ it becomes bi-Maxwellian, and at high parallel velocities it exhibits power-law behaviour, $f_\kappa \propto v_\|^{-2\kappa }$. The normalisation is chosen such that the particle number density is equal to $n$, and the perpendicular and parallel energy densities are

(4.28)\begin{gather} \int \mu B_0 f_\kappa \, {\rm d} \boldsymbol{v} = n T_\perp, \end{gather}

(4.29)\begin{gather}\int \frac{m v_\|^2}{2} f_\kappa \, {\rm d} \boldsymbol{v} = \frac{n T_\|}{2}. \end{gather}

One may verify that the spatially constant distribution function is a ground state by setting $m v_\|^2/2 = \epsilon - \mu B_0$ and taking the derivative

(4.30)\begin{equation} \left( \frac{\partial f_\kappa}{\partial \epsilon} \right)_{\mu,\boldsymbol{r}} ={-}\frac{f_\kappa \kappa}{\epsilon - \mu B_0 + T_\| ( \kappa - \frac{3}{2} )}.\end{equation}

Hence it is seen that $f_\kappa$ is a decreasing function of particle energy at fixed $\boldsymbol {r}$ for all $\mu$ as long as $\kappa >3/2$. It follows that any spatially constant $f_\kappa$ is then indeed a ground state. We now slightly perturb the distribution function, so that the fluctuating part becomes

(4.31)\begin{equation} \delta f_\kappa = f_{\kappa,0} \left[ \nu + \tau_\perp \left( \frac{\mu B_0}{T_{{\perp},0}} - 1 \right) + \tau_\| \left( \kappa - \frac{1}{2} - \frac{T_{\|,0} \kappa (\kappa - \frac{3}{2})}{\epsilon - \mu B_0 + T_{\|,0}( \kappa - \frac{3}{2} )} \right) \right]. \end{equation}

The ground state may readily be calculated, and becomes

(4.32)\begin{equation} \delta F_\kappa = f_{\kappa,0} \left[ \langle \tau_\perp \rangle \left( \frac{\mu B_0}{T_{{\perp},0}} - 1 \right) + \langle \tau_\| \rangle \left( \kappa - \frac{1}{2} - \frac{T_{\|,0} \kappa (\kappa - \frac{3}{2})}{\epsilon - \mu B_0 + T_{\|,0}( \kappa - \frac{3}{2} )} \right) \right]. \end{equation}

The integrals required for the available energy can be performed analytically, and result in

(4.33)\begin{equation} A = \frac{n_0 T_{\|,0} V}{2 \kappa} \left\langle \frac{1}{2} (\nu + \tilde \tau_\|)^2 + \left( \kappa - \frac{3}{2} \right) \left( \nu^2 + \frac{\tilde \tau_\|^2}{2} \right) + (\kappa - 1) \tilde \tau_\perp^2 \right\rangle, \end{equation}

where $\tilde \tau _\| = \tau _\| - \langle \tau _\| \rangle$ and $\tilde \tau _\perp = \tau _\perp - \langle \tau _\perp \rangle$. As required, this result is positive definite and becomes equal to (4.25) in the Maxwellian limit, $\kappa \rightarrow +\infty$.

4.5. Conservation of magnetic moment and the parallel invariant

For instabilities and turbulence with wavelengths comparable to the ion gyroradius in the direction perpendicular to the magnetic field, the frequency is usually much smaller than that of the motion of electrons along the field. The electron distribution function is then independent of the position along the field line in each trapping well. Moreover, since the bounce frequency of magnetically trapped electrons is much larger than that of the fluctuations, the action integral taken between two consecutive bounce points,

(4.34)\begin{equation} J = \int_{l_1}^{l_2} m v_\| \, {\rm d} l, \end{equation}

is conserved in addition to the magnetic moment. The available energy of the electrons in a thin flux tube aligned with the magnetic field under the constraint of constant $\mu$ and $J$ was recently calculated by Helander (Reference Helander2020) and Mackenbach et al. (Reference Mackenbach, Proll and Helander2022,  Reference Mackenbach, Proll, Wakelkamp and Helander2023). It is interesting to note that their results do not agree with the formulae (3.20) or (3.23).

The reason can be traced back to an implicit assumption made in the previous section, where it was tacitly assumed not only that $\delta f \ll f_0$ but also that all derivatives of $\delta f$ are smaller than the corresponding ones of $f_0$. This assumption is, in general, violated if the number of conserved quantities $\boldsymbol {y}$ is so large that only a single coordinate in phase space is not conserved, i.e. if the vector $\boldsymbol {z}$ only consists of a single component $z$Footnote ². If this is the case, that component can be expressed as a function of the energy and $\boldsymbol {y}$, at least locally, by inverting the function $\epsilon (\boldsymbol {y},z)$, and it must therefore be possible to write $\delta f$ as a function of $\boldsymbol {y}$ and $\epsilon$,

(4.35)\begin{equation} \delta f (\boldsymbol{y},z) = \delta F[\epsilon(\boldsymbol{y},z),\boldsymbol{y}], \end{equation}

where it is expected that $\delta F \ll F_0$. However, since the function

(4.36)\begin{equation} f(\boldsymbol{y},z) =F_0[\epsilon(\boldsymbol{y},z),\boldsymbol{y}] + \delta F[\epsilon(\boldsymbol{y},z),\boldsymbol{y}], \end{equation}

only depends on $z$ through $\epsilon (\boldsymbol {y},z)$, it follows that it actually represents a ground state if

(4.37)\begin{equation} \frac{\partial ( F_0 + \delta F)}{\partial \epsilon} < 0. \end{equation}

The perturbed state $f$ is thus a ground state unless $|\partial \delta F/ \partial \epsilon | > |\partial F_0 / \partial \epsilon |$, which invalidates the perturbation theory developed in §§ 2 and 3. More generally, in order to verify the correctness of the perturbation theory, one could check that that $\partial \delta f / \partial x_i \ll \partial f_0 / \partial x_i$ as required.

As a final observation, note that, in the perturbative framework described in § 3, the available energy is invariant under the substitution $\delta f \rightarrow -\delta f$, as may be verified from (3.18) and (3.20). For a fixed background magnetic field, then, the available energy (3.20) is impervious to the relative sign of the gradient of the magnetic field strength, $\boldsymbol {\nabla } B$, and the gradient of $\delta f$. For instance, if $f=f_0 + \delta f$ is a Maxwellian with a small density gradient contained in $\delta f$, then (3.20) is independent of the sign of $\boldsymbol {\nabla } n \boldsymbol {\cdot } \boldsymbol {\nabla } B$. There is thus no notion of ‘good’ or ‘bad’ curvature, and the expressions (3.20) and (3.23) are oblivious to curvature-driven modes although available energy can, in fact, be used to describe trapped-electron modes if $\mu$ and $J$ are treated as adiabatic invariants (Helander Reference Helander2020; Mackenbach et al. Reference Mackenbach, Proll and Helander2022,  Reference Mackenbach, Proll, Wakelkamp and Helander2023).