User:Bob K31416/TF

The fraction of electrons at $$\vec{r}$$ that have momentum between p and p+dp is,


 * $$\begin{align}

F_\vec{r} (p) dp & = \frac{4 \pi p^2 dp} {\frac{4}{3} \pi p_f^3(\vec{r})} \qquad \qquad p \le p_f(\vec{r}) \\ & = 0 \qquad \qquad  \qquad \quad otherwise \\ \end{align} $$

Using the classical expression for the kinetic energy of an electron with mass m, the kinetic energy per unit volume at $$\vec{r}$$ for the electrons of the atom is,


 * $$\begin{align}

t(\vec{r}) & = \int \frac{p^2}{2m} \  n(\vec{r}) \ F_\vec{r} (p) \ dp \\ & = n(\vec{r}) \int_{0}^{p_f(\vec{r})} \frac{p^2}{2m} \ \ \frac{4 \pi p^2 } {\frac{4}{3} \pi p_f^3(\vec{r})} \ dp \\ & = C_F \ [n(\vec{r})]^{5/3} \end{align} $$


 * where a previous expression relating $$n(\vec{r})$$ to $$p_f(\vec{r})$$ has been used and,


 * $$C_F=\frac{3h^2}{10m}\left(\frac{3}{8\pi}\right)^{\frac{2}{3}}.$$

Integrating the kinetic energy per unit volume $$t(\vec{r})$$ over all space, results in the total kinetic energy of the electrons,


 * $$T=C_F\int [n(\vec{r})]^{5/3}\ d^3r \ .$$

This result shows that the total kinetic energy of the electrons can be expressed in terms of only the spatially-varying electron density $$n(\vec{r})$$, according to the Thomas–Fermi model.

$$n(\vec{r})$$$$F_\vec{r} (p)$$

The potential energy of the electrons due to the electric attraction of the positively charged nucleus is,
 * $$U_{eN} = \int n(\vec{r}) \ V_N(\vec{r}) \ d^3r \, $$

where $$V_N(\vec{r}) \, $$ is the potential energy of an electron at $$\vec{r} \, $$ that is due to the positively charged nucleus.

The potential energy of the electrons due to their mutual electric repulsion is,
 * $$U_{ee} = \frac{1}{2} \ e^2 \int \frac{n(\vec{r}) \ n(\vec{r} \, ')} {\left\vert \vec{r} - \vec{r} \, ' \right\vert } \ d^3r \ d^3r' $$

where e is the absolute value of an electron's charge.

The total energy of the electrons is the sum of their kinetic and potential energies,
 * $$ \begin{align}

E & = T_{\text{TF}} \ + \ U_{eN} \ + \ U_{ee} \\ & = C_F\int [n(\vec{r})]^{5/3}\ d^3r \ + \int n(\vec{r}) \ V_N(\vec{r}) \ d^3r \ + \ \frac{1}{2} \ e^2 \int \frac{n(\vec{r}) \ n(\vec{r} \, ')} {\left\vert \vec{r} - \vec{r} \, ' \right\vert } \ d^3r \ d^3r'  \\ \end{align} $$