The Lagrange function is defined as $\mathcal{L}(q,\dot{q}) = T(q,\dot{q}) – V(q,\dot{q})$ where $T$ defines the kinetic energy and $V$ the potential energy.
The equations of motion are given by
$\frac{\partial \mathcal{L}}{\partial q_i} – \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q_i}} = 0$.
In the $n$-body problem we have $n$ planets with masses $m_1, \dots, m_n \in \mathbb{R}_+$. The kinetic and potential energy is given by
$T = \sum_i \frac{1}{2} m_i \Vert \dot{q_i} \Vert_2^2$ and $V = G \cdot \sum_{i<j} \frac{m_i m_j}{\Vert q_i – q_j \Vert_2}$ where $G$ denotes a gravitational constant. Furthermore, $q_i(t) \in \mathbb{R}^3$ decribes the position of the $i$-th planet at time $t$.
Now I need to calculate the equations of motions.
But now I do not understand how to deal with $\frac{\partial \mathcal{L}}{\partial q_i}$. The first thing which confuses me is that $q_i$ is a three-dimensional vector. The second thing would be the derivative of the norm because in calculus we have learned that the norm is not differentiable.
Could anyone explain this problem to me? Any help is really appreciated.
Best Answer
$ \let\ss\scriptstyle \let\sss\scriptscriptstyle \let\ds\displaystyle \renewcommand{\+}{\hspace{1mu}} \renewcommand{\bs}[1]{\boldsymbol{#1}} \renewcommand{\dt}[1]{\overset{\sss \bullet}{#1}} \renewcommand{\ddt}[1]{\overset{\sss \bullet \bullet}{#1}} \renewcommand{\pow}[1]{\raise{.8ex}{\ss{#1}}} \renewcommand{\a}{\alpha} \renewcommand{\p}{\partial} \renewcommand{\r}{\bs{r}} \renewcommand{\rdot}{\dt{\r}} \renewcommand{\qdot}{\dt{q}} \renewcommand{\F}{\bs{F}} \renewcommand{\L}{\mathcal{L}} \renewcommand{\deriv}[3]{\frac{{#1}{#2}}{{#1}{#3}}} \renewcommand{\ddx}[2]{\deriv{d}{#1}{#2}} \renewcommand{\pdx}[2]{\deriv{\partial}{#1}{#2}} \renewcommand{\lagrange}[1]{\ddx{}{t} \pdx{#1}{\qdot_\a} \+ - \+ \pdx{#1}{q_\a}} $
Let me change your notation slightly. Let $q_\a$ be the set of coordinates we use to specify the positions $\bs{r}_i = \boldsymbol{r}_i(q)$ of the planets. Starting with Newton's law, if we introduce virtual work and change our philosophy we arrive at d'Alembert's principle, for which the equations of motion take the form \begin{equation} \sum_i (\F_i - m \,\bs{a}_i) \cdot \delta \r_i \; = \; 0 \end{equation} where $\F_i$ is the force-on and $\bs{a}_i$ is the acceleration-of the $i$th planet. Setting the virtual displacements to $\delta \r_i = \sum_\a (\p\r_i \+ / \+ \p q_\a) \; \delta q_\a $ leads us to the set of $3n$ Lagrange's equations (see Goldstein sec. 1-4) \begin{equation} \L_\a[T] = \; F_\a \end{equation} where the generalized force $F_\a = \sum_i \F_i \! \cdot \! \pdx{\r_i}{q_\a}$, the kinetic energy $T = \sum_i \frac{1}{2} m_i \bs{v}_i \! \cdot \! \bs{v}_i$, and the Lagrange operator $\L_\a = \lagrange{}$. Let us use a non-script $L$ to denote the Lagrangian. In writing down the equations of motion, we can freely switch between using the forces \begin{equation} \F_i = \sum_{j \; : \; j \neq i} \frac{ G \, m_i m_j \, (\r_j - \r_i)}{|\r_j - \r_i|^3} \end{equation} and using the potential \begin{equation} V = \sum_{i,j \; : \; i<j} \frac{-G \, m_i m_j}{|\r_j - \r_i|} \end{equation} The equivalence stems from the fact that the forces can be written as the gradient $\F_i = -\nabla_i V$, which can be used to equate the generalized force to $\L[V]$ \begin{equation} F_\a \;\; = \;\; \sum_i -\nabla_i V \! \cdot \! \pdx{\r_i}{q_\a} \;\; = \;\; -\pdx{V}{q_\a} \;\; = \;\; \L_\a[V] \end{equation} Since $\L$ is a linear operator, we can combine $T$ and $V$ into the Lagrangian $L=T-V$ to arrive at the Euler-Lagrange equations of motion \begin{equation} \L_\a[L] = 0 \end{equation} I often see people claim that Lagrange's or Hamilton's equations are inapplicable when there are forces present that cannot be written as a potential (eg. non-conservative forces). But there is nothing stopping us from leaving the corresponding generalized forces on the RHS. Now, if we want a more explicit expression for the equations of motion we need to choose coordinates. For simplicity, let's use Cartesian coordinates and assume that there are no constraints on the system $\{q_1, q_2, q_3, q_4, \ldots, q_{3n}\} \equiv \{x_1, y_1, z_1, x_2, \ldots, z_n \}$. Non-Cartesian coordinates are best handled with tensor notation -- which I'd rather not introduce in a Stack Exchange post. It is useful to see the vectors expanded out. \begin{equation} \begin{array}{rcl} \r_i &=& q_{3i-2} \+ \bs{i} + q_{3i-1} \+ \bs{j} + q_{3i} \+ \bs{k} \\ \bs{v}_i &=& \qdot_{3i-2} \+ \bs{i} + \qdot_{3i-1} \+ \bs{j} + \qdot_{3i} \+ \bs{k} \\ \F_i &=& F_{i1} \+ \bs{i} + F_{i2} \+ \bs{j} + F_{i3} \+ \bs{k} \\ \end{array} \end{equation} The positions only depend on three coordinates so the terms $\pdx{\r_i}{q_\a}$ are nonzero for only three values of $\alpha$ (for which they become $\bs{i}, \bs{j},$ or $\bs{k}$). It is not hard to see that $\L_\a[T]$ are the coordinate accelerations and that $F_\a$ are the force components. Thus, Lagrange's equations $\L_\a[T] = F_\a$ mirror Newton's law \begin{equation} m_{i} \ddt{q}_{3i+j-3} = F_{ij} \end{equation} where we identified $\a = 3i+j-3$ for convenience. We can condense these $3n$ equations down to $n$ vector equations \begin{equation} m_{i} \bs{a}_i = \F_i \end{equation} We've come full circle. Indeed, there is really no reason to introduce the Lagrangian at all because: one there are no constraints, two we are using Cartesian coordinates, and three we have explicit expressions for the forces.