feos-org · prehner · Feb 24, 2023 · Feb 22, 2023 · Feb 22, 2023 · Feb 24, 2023
diff --git a/docs/theory/dft/derivatives.md b/docs/theory/dft/derivatives.md
@@ -3,29 +3,29 @@ For converged density properties equilibrium properties can be calculated as par
 
 The density profiles are calculated implicitly from the Euler-Lagrange equation, which can be written simplified as
 
-$$\Omega_{\rho_i}(T,\lbrace\mu_k\rbrace,[\lbrace\rho_k(\mathbf{r})\rbrace])=\mathcal{F}_{\rho_i}(T,[\lbrace\rho_k(\mathbf{r})\rbrace])-\mu_i+V^\mathrm{ext}(\mathbf{r})=0$$ (eqn:euler_lagrange)
+$$\Omega_{\rho_i}(T,\lbrace\mu_k\rbrace,[\lbrace\rho_k(\mathbf{r})\rbrace])=F_{\rho_i}(T,[\lbrace\rho_k(\mathbf{r})\rbrace])-\mu_i+V_i^\mathrm{ext}(\mathbf{r})=0$$ (eqn:euler_lagrange)
 
-Incorporating bond integrals can be done similar to the section on the [Newton solver](solver.md) but will not be discussed in this section. The derivatives of the density profiles can then be calculated from the total differential of eq. {eq}`eqn:euler_lagrange`.
+Incorporating bond integrals can be done similar to the section on the [Newton solver](solver.md) but will not be discussed in this section. The derivatives of the density profiles can then be calculated from the total differential of eq. {eq}`eqn:euler_lagrange`, leading to
 
 $$\mathrm{d}\Omega_{\rho_i}(\mathbf{r})=\left(\frac{\partial\Omega_{\rho_i}(\mathbf{r})}{\partial T}\right)_{\mu_k,\rho_k}\mathrm{d}T+\sum_j\left(\frac{\partial\Omega_{\rho_i}(\mathbf{r})}{\partial\mu_j}\right)_{T,\mu_k,\rho_k}\mathrm{d}\mu_j+\int\sum_j\left(\frac{\delta\Omega_{\rho_i}(\mathbf{r})}{\delta\rho_j(\mathbf{r}')}\right)_{T,\mu_k,\rho_k}\delta\rho_j(\mathbf{r}')\mathrm{d}\mathbf{r}'=0$$
 
-Using eq. {eq}`eqn:euler_lagrange` and the shortened notation for derivatives of functionals in their natural variables, e.g., $\mathcal{F}_T=\left(\frac{\partial\mathcal{F}}{\partial T}\right)_{\rho_k}$, the expression can be simplified to
+Using eq. {eq}`eqn:euler_lagrange` and the shortened notation for derivatives of functionals in their natural variables, e.g., $F_T=\left(\frac{\partial F}{\partial T}\right)_{\rho_k}$, the expression can be simplified to
 
-$$\mathcal{F}_{T\rho_i}(\mathbf{r})\mathrm{d}T-\mathrm{d}\mu_i+\int\sum_j\mathcal{F}_{\rho_i\rho_j}(\mathbf{r},\mathbf{r}')\delta\rho_j(\mathbf{r}')\mathrm{d}\mathbf{r}'=0$$ (eqn:gibbs_duhem)
+$$F_{T\rho_i}(\mathbf{r})\mathrm{d}T-\mathrm{d}\mu_i+\int\sum_j F_{\rho_i\rho_j}(\mathbf{r},\mathbf{r}')\delta\rho_j(\mathbf{r}')\mathrm{d}\mathbf{r}'=0$$ (eqn:gibbs_duhem)
 
 Similar to the Gibbs-Duhem relation for bulk phases, eq. {eq}`eqn:gibbs_duhem` shows how temperature, chemical potentials and the density profiles in an inhomogeneous system cannot be varied independently. The derivatives of the density profiles with respect to the intensive variables can be directly identified as
 
-$$\int\sum_j\mathcal{F}_{\rho_i\rho_j}(\mathbf{r},\mathbf{r}')\left(\frac{\partial\rho_j(\mathbf{r}')}{\partial T}\right)_{\mu_k}\mathrm{d}\mathbf{r}'=-\mathcal{F}_{T\rho_i}(\mathbf{r})$$
+$$\int\sum_j F_{\rho_i\rho_j}(\mathbf{r},\mathbf{r}')\left(\frac{\partial\rho_j(\mathbf{r}')}{\partial T}\right)_{\mu_k}\mathrm{d}\mathbf{r}'=-F_{T\rho_i}(\mathbf{r})$$
 
 and
 
-$$\int\sum_j\mathcal{F}_{\rho_i\rho_j}(\mathbf{r},\mathbf{r}')\left(\frac{\partial\rho_j(\mathbf{r}')}{\partial\mu_k}\right)_{T}\mathrm{d}\mathbf{r}'=\delta_{ik}$$ (eqn:drho_dmu)
+$$\int\sum_j F_{\rho_i\rho_j}(\mathbf{r},\mathbf{r}')\left(\frac{\partial\rho_j(\mathbf{r}')}{\partial\mu_k}\right)_{T}\mathrm{d}\mathbf{r}'=\delta_{ik}$$ (eqn:drho_dmu)
 
 Both of these expressions are implicit (linear) equations for the derivatives. They can be solved rapidly analogously to the implicit expression appearing in the [Newton solver](solver.md). In practice, it is useful to explicitly cancel out the (often unknown) thermal de Broglie wavelength $\Lambda_i$ from the expression where it has no influence. This is done by splitting the intrinsic Helmholtz energy into an ideal gas and a residual part.
 
-$$\mathcal{F}=k_\mathrm{B}T\int\sum_im_i\rho_i(\mathbf{r})\left(\ln\left(\rho_i(\mathbf{r})\Lambda_i^3\right)-1\right)\mathrm{d}\mathbf{r}+\mathcal{\hat F}^\mathrm{res}$$
+$$F=k_\mathrm{B}T\int\sum_im_i\rho_i(\mathbf{r})\left(\ln\left(\rho_i(\mathbf{r})\Lambda_i^3\right)-1\right)\mathrm{d}\mathbf{r}+\mathcal{\hat F}^\mathrm{res}$$
 
-Then $\mathcal{F}_{\rho_i\rho_j}(\mathbf{r},\mathbf{r}')=m_i\frac{k_\mathrm{B}T}{\rho_i(\mathbf{r})}\delta_{ij}\delta(\mathbf{r}-\mathbf{r}')+\mathcal{\hat F}_{\rho_i\rho_j}^\mathrm{res}(\mathbf{r},\mathbf{r}')$ and eq. {eq}`eqn:drho_dmu` can be rewritten as
+Then $F_{\rho_i\rho_j}(\mathbf{r},\mathbf{r}')=m_i\frac{k_\mathrm{B}T}{\rho_i(\mathbf{r})}\delta_{ij}\delta(\mathbf{r}-\mathbf{r}')+\mathcal{\hat F}_{\rho_i\rho_j}^\mathrm{res}(\mathbf{r},\mathbf{r}')$ and eq. {eq}`eqn:drho_dmu` can be rewritten as
 
 $$m_i\frac{k_\mathrm{B}T}{\rho_i(\mathbf{r})}\left(\frac{\partial\rho_i(\mathbf{r})}{\partial\mu_k}\right)_T+\int\sum_j\mathcal{\hat F}_{\rho_i\rho_j}^\mathrm{res}(\mathbf{r},\mathbf{r}')\left(\frac{\partial\rho_j(\mathbf{r}')}{\partial\mu_k}\right)_{T}\mathrm{d}\mathbf{r}'=\delta_{ik}$$
 
@@ -39,16 +39,16 @@ $$\mathrm{d}\mu_i=-s_i\mathrm{d}T+v_i\mathrm{d}p$$
 
 which can be used in eq. {eq}`eqn:gibbs_duhem` to give
 
-$$\left(\mathcal{F}_{T\rho_i}(\mathbf{r})+s_i\right)\mathrm{d}T+\int\sum_j\mathcal{F}_{\rho_i\rho_j}(\mathbf{r},\mathbf{r}')\delta\rho_j(\mathbf{r}')\mathrm{d}\mathbf{r}'=v_i\mathrm{d}p$$
+$$\left(F_{T\rho_i}(\mathbf{r})+s_i\right)\mathrm{d}T+\int\sum_j F_{\rho_i\rho_j}(\mathbf{r},\mathbf{r}')\delta\rho_j(\mathbf{r}')\mathrm{d}\mathbf{r}'=v_i\mathrm{d}p$$
 
 Even though $s_i$ is readily available in $\text{FeO}_\text{s}$ it is useful at this point to rewrite the partial molar entropy as
 
-$$s_i=v_i\left(\frac{\partial p}{\partial T}\right)_{V,N_k}-\mathcal{F}_{T\rho_i}^\mathrm{b}$$
+$$s_i=v_i\left(\frac{\partial p}{\partial T}\right)_{V,N_k}-F_{T\rho_i}^\mathrm{b}$$
 
 Then, the intrinsic Helmholtz energy can be split into an ideal gas and a residual part again, and the de Broglie wavelength cancels.
 
 $$\begin{align*}
-&\left(m_ik_\mathrm{B}\ln\left(\frac{\rho_i(\mathbf{r})}{\rho_i^\mathrm{b}}\right)+\mathcal{F}_{T\rho_i}^\mathrm{res}(\mathbf{r})-\mathcal{F}_{T\rho_i}^\mathrm{b,res}+v_i\left(\frac{\partial p}{\partial T}\right)_{V,N_k}\right)\mathrm{d}T\\
+&\left(m_ik_\mathrm{B}\ln\left(\frac{\rho_i(\mathbf{r})}{\rho_i^\mathrm{b}}\right)+F_{T\rho_i}^\mathrm{res}(\mathbf{r})-F_{T\rho_i}^\mathrm{b,res}+v_i\left(\frac{\partial p}{\partial T}\right)_{V,N_k}\right)\mathrm{d}T\\
 &~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+m_i\frac{k_\mathrm{B}T}{\rho_i(\mathbf{r})}\delta\rho_i(\mathbf{r})+\int\sum_j\mathcal{\hat F}_{\rho_i\rho_j}^\mathrm{res}(\mathbf{r},\mathbf{r}')\delta\rho_j(\mathbf{r}')\mathrm{d}\mathbf{r}'=v_i\mathrm{d}p
 \end{align*}$$
 
@@ -60,7 +60,7 @@ and temperature
 
 $$\begin{align*}
 &m_i\left(\frac{\partial\rho_i(\mathbf{r})}{\partial T}\right)_{p,x_k}+\rho_i(\mathbf{r})\int\sum_j\beta\mathcal{\hat F}_{\rho_i\rho_j}^\mathrm{res}(\mathbf{r},\mathbf{r}')\left(\frac{\partial\rho_j(\mathbf{r}')}{\partial T}\right)_{p,x_k}\mathrm{d}\mathbf{r}'\\
-&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-\frac{\rho_i(\mathbf{r})}{k_\mathrm{B}T}\left(m_ik_\mathrm{B}\ln\left(\frac{\rho_i(\mathbf{r})}{\rho_i^\mathrm{b}}\right)+\mathcal{F}_{T\rho_i}^\mathrm{res}(\mathbf{r})-\mathcal{F}_{T\rho_i}^\mathrm{b,res}+v_i\left(\frac{\partial p}{\partial T}\right)_{V,N_k}\right)
+&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-\frac{\rho_i(\mathbf{r})}{k_\mathrm{B}T}\left(m_ik_\mathrm{B}\ln\left(\frac{\rho_i(\mathbf{r})}{\rho_i^\mathrm{b}}\right)+F_{T\rho_i}^\mathrm{res}(\mathbf{r})-F_{T\rho_i}^\mathrm{b,res}+v_i\left(\frac{\partial p}{\partial T}\right)_{V,N_k}\right)
 \end{align*}$$
 
 follow. All derivatives $x_i$ shown here can be calculated from the same linear equation
@@ -73,4 +73,4 @@ by just replacing the right hand side $y_i$.
 |-|-|
 |$\left(\frac{\partial\rho_i(\mathbf{r})}{\partial\beta\mu_k}\right)_T$|$\rho_i(\mathbf{r})\delta_{ik}$|
 |$\left(\frac{\partial\rho_i(\mathbf{r})}{\partial\beta p}\right)_{T,x_k}$|$\rho_i(\mathbf{r})v_i$|
-|$\left(\frac{\partial\rho_i(\mathbf{r})}{\partial T}\right)_{p,x_k}$|$-\frac{\rho_i(\mathbf{r})}{k_\mathrm{B}T}\left(m_ik_\mathrm{B}\ln\left(\frac{\rho_i(\mathbf{r})}{\rho_i^\mathrm{b}}\right)+\mathcal{F}_{T\rho_i}^\mathrm{res}(\mathbf{r})-\mathcal{F}_{T\rho_i}^\mathrm{b,res}+v_i\left(\frac{\partial p}{\partial T}\right)_{V,N_k}\right)$|
+|$\left(\frac{\partial\rho_i(\mathbf{r})}{\partial T}\right)_{p,x_k}$|$-\frac{\rho_i(\mathbf{r})}{k_\mathrm{B}T}\left(m_ik_\mathrm{B}\ln\left(\frac{\rho_i(\mathbf{r})}{\rho_i^\mathrm{b}}\right)+F_{T\rho_i}^\mathrm{res}(\mathbf{r})-F_{T\rho_i}^\mathrm{b,res}+v_i\left(\frac{\partial p}{\partial T}\right)_{V,N_k}\right)$|
diff --git a/docs/theory/dft/enthalpy_of_adsorption.md b/docs/theory/dft/enthalpy_of_adsorption.md
@@ -40,15 +40,22 @@ $$\Delta h^\mathrm{ads}=\sum_ix_i\Delta h_i^\mathrm{ads}=h^\mathrm{b}-\sum_ix_i\
 ## Clausius-Clapeyron relation for porous media
 The Clausius-Clapeyron relation relates the $p-T$ slope of a pure component phase transition line to the corresponding enthalpy of phase change. For a vapor-liquid phase transition, the exact relation is
 
-$$\frac{\mathrm{d}p^\mathrm{sat}}{\mathrm{d}T}=\frac{s^\mathrm{V}-s^\mathrm{L}}{v^\mathrm{V}-v^\mathrm{L}}=\frac{h^\mathrm{V}-h^\mathrm{L}}{T\left(v^\mathrm{V}-v^\mathrm{L}\right)}$$
+$$\frac{\mathrm{d}p^\mathrm{sat}}{\mathrm{d}T}=\frac{s^\mathrm{V}-s^\mathrm{L}}{v^\mathrm{V}-v^\mathrm{L}}=\frac{h^\mathrm{V}-h^\mathrm{L}}{T\left(v^\mathrm{V}-v^\mathrm{L}\right)}$$ (eqn:temp_dep_press)
 
 In this expression, the enthalpy of vaporization $\Delta h^\mathrm{vap}=h^\mathrm{V}-h^\mathrm{L}$ can be identified. If the molar volume of the liquid phase $v^\mathrm{L}$ is assumed to be negligible compared to the molar volume of the vapor phase $v^\mathrm{V}$ and the gas phase is assumed to be ideal, the relation simplifies to
 
 $$\frac{\mathrm{d}p^\mathrm{sat}}{\mathrm{d}T}=\frac{p}{RT^2}\Delta h^\mathrm{vap}$$
 
 which can be compactly written as
 
-$$\frac{\mathrm{d}\ln p^\mathrm{sat}}{\mathrm{d}\frac{1}{RT}}=-\Delta h^\mathrm{vap}$$ (eqn:clausius_clapeyron)
+$$\frac{\mathrm{d}\ln p^\mathrm{sat}}{\mathrm{d}\frac{1}{RT}}=-\Delta h^\mathrm{vap}$$ (eqn:Clausius_Clapeyron)
+
+Without assuming neither ideal gas nor neglecting the volume of the liquid, eq. {eq}`eqn:temp_dep_press` can be rewritten using the compressibility factor $Z$ to
+
+$$\frac{\mathrm{d}p^\mathrm{sat}}{\mathrm{d}T}=\frac{h^\mathrm{V}-h^\mathrm{L}}{T\left(v^\mathrm{V}-v^\mathrm{L}\right)}=\frac{p}{R T^2}\frac{\Delta h^\mathrm{vap}}{Z^\mathrm{V}-Z^\mathrm{L}}$$
+
+Neglecting the compressibility of the liquid phase ($Z^\mathrm{L}=0$) and assuming ideal gas for the vapor phase ($Z^\mathrm{V}=1$) leads to eq. {eq}`eqn:Clausius_Clapeyron`. 
+
 
 A similar relation can be derived for fluids adsorbed in a porous medium that is in equilibrium with a bulk phase. At this point it is important to clarify which variables describe the system
 - The adsorbed fluid and the bulk phase are in equilibrium. Therefore, the temperature $T$ and chemical potentials $\mu_i$ are the same for both phases.
@@ -73,11 +80,10 @@ $$\frac{\mathrm{d}\ln p}{\mathrm{d}\frac{1}{RT}}=-\frac{T}{Z^\mathrm{b}}\left(s^
 
 Finally, using $h^\mathrm{b}=Ts^\mathrm{b}+\sum_ix_i\mu_i$ and $\mathrm{d}U=T\mathrm{d}S+\sum_i\mu_i\mathrm{d}N_i$ leads to
 
-$$\frac{\mathrm{d}\ln p}{\mathrm{d}\frac{1}{RT}}=-\frac{1}{Z^\mathrm{b}}\left(h^\mathrm{b}-\sum_ix_i\left(\frac{\partial U}{\partial N_i}\right)_T\right)=-\frac{\Delta h^\mathrm{ads}}{Z^\mathrm{b}}$$
+$$\frac{\mathrm{d}\ln p}{\mathrm{d}\frac{1}{RT}}=-\frac{1}{Z^\mathrm{b}}\left(h^\mathrm{b}-\sum_ix_i\left(\frac{\partial U}{\partial N_i}\right)_T\right)=-\frac{\Delta h^\mathrm{ads}}{Z^\mathrm{b}}$$ (eqn:deriv_relation_hads)
 
-The relation is exact and valid for an arbitrary number of components in the fluid phase.
+The relation is exact and valid for an arbitrary number of components in the fluid phase. 
 
-(attentive readers are aware of the fact that not assuming ideal gas behavior in the classical Clausius-Clapeyron relation will introduce the same division by the compressibility factor in eq. {eq}`eqn:clausius_clapeyron`.)
 
 ## Calculation of the enthalpy of adsorption from classical DFT
 In a DFT context, the introduction of entropies and internal energies are just unnecessary complications. The most useful definition of the (partial molar) enthalpy of adsorption is
@@ -104,4 +110,4 @@ After multiplying with $T$, the following elegant expression remains
 
 $$0=\sum_j\left(\frac{\partial N_i}{\partial\mu_j}\right)_T\Delta h_j^\mathrm{ads}+T\left(\frac{\partial N_i}{\partial T}\right)_{p,x_k}$$
 
-which is a symmetric linear equation due to $\left(\frac{\partial N_i}{\partial\mu_j}\right)_T=-\left(\frac{\partial^2\Omega}{\partial\mu_i\partial\mu_j}\right)_T$. The derivatives of the particle numbers are obtained by integrating over the respective derivatives of the density profiles which were discussed [previously](derivatives.md).
+which is a symmetric linear system of equations due to $\left(\frac{\partial N_i}{\partial\mu_j}\right)_T=-\left(\frac{\partial^2\Omega}{\partial\mu_i\partial\mu_j}\right)_T$. The derivatives of the particle numbers are obtained by integrating over the respective derivatives of the density profiles which were discussed [previously](derivatives.md).
diff --git a/docs/theory/dft/euler_lagrange_equation.md b/docs/theory/dft/euler_lagrange_equation.md
@@ -1,17 +1,17 @@
 # Euler-Lagrange equation
-The fundamental expression in classical density functional theory is the relation between the grand potential $\Omega$ and the intrinsic Helmholtz energy $F$.
+The fundamental expression in classical density functional theory is the relation between the grand potential functional $\Omega$ and the intrinsic Helmholtz energy functional $F$.
 
 $$\Omega(T,\mu,[\rho(r)])=F(T,[\rho(r)])-\sum_i\int\rho_i(r)\left(\mu_i-V_i^\mathrm{ext}(r)\right)\mathrm{d}r$$
 
-What makes this expression so appealing is that the intrinsic Helmholtz energy only depends on the temperature $T$ and the density profiles $\rho_i(r)$ of the system and not on the external potential $V_i^\mathrm{ext}(r)$.
+What makes this expression so appealing is that the intrinsic Helmholtz energy functional only depends on the temperature $T$ and the density profiles $\rho_i(r)$ of the system and not on the external potentials $V_i^\mathrm{ext}(r)$.
 
-For a given temperature $T$, chemical potentials $\mu$ and external potentials $V^\mathrm{ext}(r)$ the grand potential reaches a minimum at equilibrium. Mathematically this condition can be written as
+For a given temperature $T$, chemical potentials $\mu_i$ and external potentials $V_i^\mathrm{ext}(r)$ the grand potential reaches a minimum at equilibrium. Mathematically this condition can be written as
 
 $$\left.\frac{\delta\Omega}{\delta\rho_i(r)}\right|_{T,\mu}=F_{\rho_i}(r)-\mu_i+V_i^{\mathrm{ext}}(r)=0$$ (eqn:euler_lagrange_mu)
 
-where $F_{\rho_i}(r)=\left.\frac{\delta F}{\delta\rho_i(r)}\right|_T$ is short for the functional derivative of the intrinsic Helmholtz energy. In this context, eq. (1) is commonly referred to as the Euler-Lagrange equation, an implicit nonlinear integral equation which needs to be solved for the density profiles of the system.
+where $F_{\rho_i}(r)=\left.\frac{\delta F}{\delta\rho_i(r)}\right|_T$ is short for the functional derivative of the intrinsic Helmholtz energy. In this context, eq. (1) is commonly referred to as the Euler-Lagrange equation, an implicit nonlinear integral equation which needs to be solved for the equilibrium density profiles of the system.
 
-For a homogeneous (bulk) system, $V^\mathrm{ext}=0$ and we get
+For a homogeneous (bulk) system, $V_i^\mathrm{ext}=0$ and we get
 
 $$F_{\rho_i}^\mathrm{b}-\mu_i=0$$ (eqn:euler_lagrange_bulk)
 
@@ -50,8 +50,8 @@ For chain molecules that do not resolve individual segments (essentially the PC-
 
 $$\beta F^\mathrm{chain}=-\sum_i\int\rho_i(r)\left(m_i-1\right)\ln\left(\frac{y_{ii}\lambda_i(r)}{\rho_i(r)}\right)\mathrm{d}r$$
 
-Here, $m_i$ is the number of segments (i.e., the PC-SAFT chain length parameter), $y_{ii}$ the cavity correlation function at contact in the reference fluid, and $\lambda_i$ a weighted density.
-The presence of $\rho(r)$ in the logarithm poses numerical problems. Therefore, it is convenient to rearrange the expression as
+Where $m_i$ is the number of segments (i.e., the PC-SAFT chain length parameter), $y_{ii}$ is the cavity correlation function at contact in the reference fluid, and $\lambda_i$ is a weighted density.
+The presence of $\rho_i(r)$ in the logarithm poses numerical problems. Therefore, it is convenient to rearrange the expression as
 
 $$\begin{align}
 \beta F^\mathrm{chain}=&\sum_i\int\rho_i(r)\left(m_i-1\right)\left(\ln\left(\rho_i(r)\Lambda_i^3\right)-1\right)\mathrm{d}r\\

diff --git a/docs/theory/dft/functional_derivatives.md b/docs/theory/dft/functional_derivatives.md
@@ -25,7 +25,7 @@ F_{\rho_\alpha}(r)&=\int\sum_\gamma f_{n_\gamma}(r')\frac{\delta n_\gamma(r')}{\
 &=\sum_\gamma\int f_{n_\gamma}(r')\omega_\gamma^\alpha(r'-r)\mathrm{d}r
 \end{align}$$
 
-At this point the parity of the weight functions has to be taken into account. By construction scalar and spherically symmetric weight functions (the standard case) are even functions, i.e., $\omega(-r)=\omega(r)$. In contrast, vector valued weight functions, as they appear in fundamental measure theory, have odd parity, i.e., $\omega(-r)=-\omega(r)$. Therefore, the sum over the weight functions needs to be split into two contributions, as
+At this point the parity of the weight functions has to be taken into account. By construction scalar and spherically symmetric weight functions (the standard case) are even functions, i.e., $\omega(-r)=\omega(r)$. In contrast, vector valued weight functions, as they appear in fundamental measure theory, are odd functions, i.e., $\omega(-r)=-\omega(r)$. Therefore, the sum over the weight functions needs to be split into two contributions, as
 
 $$F_{\rho_\alpha}(r)=\sum_\gamma^\mathrm{scal}\int f_{n_\gamma}(r')\omega_\gamma^\alpha(r-r')\mathrm{d}r-\sum_\gamma^\mathrm{vec}\int f_{n_\gamma}(r')\omega_\gamma^\alpha(r-r')\mathrm{d}r\tag{2}$$
 

diff --git a/docs/theory/dft/solver.md b/docs/theory/dft/solver.md
@@ -52,7 +52,7 @@ The linear integral equation has to be solved for the step $\Delta\rho(r)$. Expl
 
 $$\int\sum_\beta\frac{\delta\mathcal{F}_\alpha\left(r;\left[\rho(r)\right]\right)}{\delta\rho_\beta(r')}\Delta\rho_\beta(r')\mathrm{d}r'\approx\frac{\mathcal{F}_\alpha\left(r;\left[\rho(r)+s\Delta\rho(r)\right]\right)-\mathcal{F}_\alpha\left(r;\left[\rho(r)\right]\right)}{s}$$
 
-However this approach requires the choice of an appropriate step size $s$ (something that we want to get away from in $\text{FeO}_\text{s}$) and also an evaluation of the full residual in every step of the linear solver. The solver can be sped up by doing parts of the functional derivative analytically beforehand. Using the definition of the residual in the rhs of eq. {eq}`eqn:newton` leads to
+However this approach requires the choice of an appropriate step size $s$ (something that we want to avoid in $\text{FeO}_\text{s}$) and also an evaluation of the full residual in every step of the linear solver. The solver can be sped up by doing parts of the functional derivative analytically beforehand. Using the definition of the residual in the rhs of eq. {eq}`eqn:newton` leads to
 
 $$\begin{align*}
 q_\alpha(r)&\equiv-\int\sum_\beta\frac{\delta\mathcal{F}_\alpha\left(r;\left[\rho(r)\right]\right)}{\delta\rho_\beta(r')}\Delta\rho_\beta(r')\mathrm{d}r'\\