Classical mechanics on a curved background

On a flat background Lagrangians mostly look like

(1)
\begin{align} L = \frac{1}{2} \mathbf{\dot x} \cdot \mathbf{\dot x} - V \left( \mathbf{x} \right) \end{align}

and Hamiltonians look like

(2)
\begin{align} H = \frac{1}{2} \mathbf{p} \cdot \mathbf{p} + V \left( \mathbf{x} \right), \end{align}

where $\mathbf{x}$ is the position vector, $\mathbf{\dot x}$ is its time derivative, $\mathbf{p}$ is the momentum vector, and $\cdot$ represents the standard dot product. On the other hand in a curved background, we have to use the metric to take inner products.

First, we establish a splitting into a time coordinate and three spatial coordinates, and we call the spatial part of the metric $g_{ij} , i,j = 1,2,3$. As usual, we let an overdot denote a derivative with respect to our chosen time coordinate.

So the Lagrangian for a particle moving on a curved background described by the spatial metric $g_{ij}$ and experiencing a potential $V \left( \mathbf{x} \right)$ is

(3)
\begin{align} L = \frac{1}{2} g_{ij} \dot{x}^i \dot{x}^j - V \left( \mathbf{x} \right), \end{align}

the momentum is given by the 1-form $p_i = g_{ij} \dot{x}^j$, and the Hamiltonian is

(4)
\begin{align} H = \frac{1}{2} g^{ij} p_i p_j + V \left( \mathbf{x} \right) \end{align}

Here $g^{ij}$ is defined to be the inverse matrix of $g_{ij}$, so that $g_{ij} g^{jk} = \delta_i^k$ (where we are using Einstein summation convention as always here and $\delta_i^k$ is the Kronecker delta. What we are seeing is just the multiplication of a matrix with its inverse to obtain the identity matrix).

page revision: 0, last edited: 05 Apr 2014 18:52