Matter Modeling Asked on December 16, 2021
In classical density functional theory, one traditionally calculates the chemical potential by taking the variational derivative,
begin{equation}
mu_{i} = frac{delta F}{delta rho_{i}}tag{1}
end{equation}
of the Helmholtz free energy
begin{equation}
F[rho] = int dtextbf{r} f(rho, nabla rho, …) textrm{.}tag{2}
end{equation}
However, this is not directly analogous to the chemical potential in classical thermodynamics. In the latter theory, the chemical potential is defined as a partial derivative with respect to the number of moles,
begin{equation}
hat{mu}_{i} = frac{partial A}{partial n_{i}}
end{equation}
where $A$ is the homogeneous Helmholtz free energy analogous to $F$.
Importantly, $n_{i}$ is an extensive quantity (e.g. $n_{i} = rho_{i} V$, where $V$ is the system volume). This means that $mu_{i}$, defined in DFT is actually analogous to the derivative
begin{equation}
mu_{i} = frac{partial A}{partial rho_{i}}tag{3}
end{equation}
How then does one obtain the actual analogue,
begin{equation}
hat{mu}_{i} = frac{delta F}{delta n_{i}}tag{4}
end{equation}
to the traditional chemical potential?
Is this generalization correct? If so, how does one go about computing such a quantity when the number of moles $n_{i}$ is now itself a functional of the density,
begin{equation}
n_{i} = int dtextbf{r} rho_{i}(textbf{r})tag{5}
end{equation}
Aside:
Related:
Seeing that this question has gathered attention but no replies, I will give it a stab. Note that I am not an expert on DFT or functional calculus, so take this with a grain of salt. As usual, suggestions to the post will be welcome!
Using an approach I saw here, we can use a chain rule and obtain the following:
$$frac{delta F[rho(boldsymbol{r})]}{delta n_i[rho_i(boldsymbol{r})]} = int frac{frac{delta F[rho(boldsymbol{r})]}{delta rho(boldsymbol{r})}}{frac{delta n_i[rho_i(boldsymbol{r})]}{delta rho(boldsymbol{r})}} dboldsymbol{r} = int frac{frac{delta F[rho(boldsymbol{r})]}{delta rho_i(boldsymbol{r})}}{frac{delta n_i[rho_i(boldsymbol{r})]}{delta rho_i(boldsymbol{r})}} dboldsymbol{r}tag{1}$$
where the last equality stems from the fact that the integrand will vanish for any $rho_k, kneq i$. It is straightforward to see that:
$$frac{delta n_i[rho_i(boldsymbol{r})]}{delta rho_i(boldsymbol{r})} = 1tag{2}$$ so the above integral reduces to:
$$frac{delta F[rho(boldsymbol{r})]}{delta n_i[rho_i(boldsymbol{r})]} = int frac{delta F[rho(boldsymbol{r})]}{delta rho_i(boldsymbol{r})} dboldsymbol{r}tag{3}$$
which is what I assume you mean by writing $frac{partial F}{partial rho_i}$, since this will be a function of $boldsymbol{r}$, unless you integrate over it, and coordinate-dependent chemical potentials don't make much sense to me! Also, note that the functional derivative is only equal to $frac{partial f}{partial rho_i}$ if your free energy functional doesn't depend on any derivatives of the density. In this case, you will need higher-order terms as well.
Edit: I will give the osmotic pressure a try as well, but this will definitely need to be checked for some non-obvious errors. Use at your own discretion.
You can express $frac{delta F[rho(boldsymbol{r})]}{delta V}$ as $frac{delta F[rho(boldsymbol{sr})]}{delta s^3}Big|_{s=1} = frac{1}{3s^2}frac{delta F[rho(sboldsymbol{r})]}{delta s}Big|_{s=1}$ for some scaling factor $s$. In this case, the chain rule tells us that:
begin{align}frac{1}{3s^2}frac{delta F[rho(sboldsymbol{r})]}{delta s}Bigg|_{s=1} &= frac{1}{3s^2} int frac{delta F[rho(sboldsymbol{r})]}{delta rho(sboldsymbol r)} frac{partial rho(sboldsymbol{r})}{partial s} d(sboldsymbol{r})Bigg|_{s=1}tag{4}\ &= frac{1}{3} int frac{delta F[rho(boldsymbol{r})]}{delta rho(boldsymbol{r})} (nablarho(boldsymbol{r})cdotboldsymbol{r}) dboldsymbol{r}tag{5} end{align}
Answered by Godzilla on December 16, 2021
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