# Free particles in space

### Notation convention

When dealing with Euclidean coordinates, the convention is to use the notation, \begin{align*}q_i & \to x_i\\ \dot{q_i} & \to \dot{x_i} \text{ or } v_i\\ \pi_i & \to p_i\end{align*}

### Single particle

The Lagrangian for a particle of mass $m$ in free space is the classical one, \begin{align*}L(x, \dot{x}) &= T - U\\& = \tfrac{1}{2} m \dot{x}^2 - U(x)\end{align*}

Now, the momentum conjugate to $x$ and the generalised force are just the standard momentum and force, \begin{align*}p &= \frac{\partial L}{\partial \dot{x}} \quad ({= m \dot{x} })\\ \frac{\partial L}{\partial x} &= -\frac{\partial U}{\partial x}\end{align*} and we recover Newton's Law, $$\dot{p} = -\frac{\partial U}{\partial x}$$

### Systems of free particles

Consider a system of $N$ particles of mass $m_i$ in free space. The Lagrangian generalises to a seemingly more complicated expression $$L(x_i, \dot{x_i}) = \sum_{i = 1}^{3N} \left(\tfrac{1}{2} m_i \dot{x_i}^2 \right) - U(x_1, \dots, x_{3N})$$ but the conjugate momenta and generalised forces are just indexed versions of the single particle case, \begin{align*}p_i &= \frac{\partial L}{\partial \dot{x_i}} \quad ({= m \dot{x_i}})\\ \frac{\partial L}{\partial x_i} &= -\frac{\partial U}{\partial x_i}\end{align*}

So we get the multidimensional version of Newton's Law, $$\dot{p_i} = -\frac{\partial U}{\partial x_i}$$

### Example of a 2-particle system

Figure 3.1 - Two masses on a line

Consider a system of two particles, with coordinates $x_1, x_2$ and masses $m_1, m_2$, constrained to move on a straight line, and where the force between them is governed by a potential function that only depends on the distance between them, $$U(x_1, x_2) = U(x_1 - x_2)$$

Nb. If we had more than 2 particles on the line then this potential would generalise to a sum of potentials of the same form, one for every pair of particles.

The kinetic energy is $$T = \tfrac{1}{2} m \dot{x_1}^2 + \tfrac{1}{2} m \dot{x_2}^2$$ so the Lagrangian is $$L = T - U = \tfrac{1}{2} m \dot{x_1}^2 + \tfrac{1}{2} m \dot{x_2}^2 - U(x_1 - x_2)$$

Writing $l = x_1 - x_2$, we see that \begin{align*}\dot{p_1} = -\frac{\partial U}{\partial x_1} = -\frac{\partial U}{\partial l} \frac{\partial l}{\partial x_1} &= -\frac{\partial U}{\partial l}\\ \dot{p_2} = -\frac{\partial U}{\partial x_2} = -\frac{\partial U}{\partial l} \frac{\partial l}{\partial x_2} &= +\frac{\partial U}{\partial l}\end{align*} hence, as expected, the forces acting on each of the masses are equal in magnitude and opposite in direction.

Another way to say that is to say the sum of the internal forces is zero, hence the momentum is conserved.