Lagrangian Dynamics

When is \(L = K - V\)?

I make a big deal about how the prescription of forming the Lagrangian \(L = K-V\) as the difference between kinetic and potential energies is only valid for Newtonian mechanics. How does one know if this is appropriate?

The way I think about it is to imagine writing the full Lagrangian for all of the particles in terms of their Cartesian inertial coordinates \(\vect{r}_i = (x_i, y_i, z_i)\), with all other complications inserted as constraints with Lagrange multipliers etc. If, in this case, the kinetic energy is quadratic in the velocities, then you can use the \(L = K - V\) prescription after transforming to whatever (non-singular) coordinates make the problem easiest to solve:

\[\begin{gather*} K = \frac{1}{2} \vect{v}^T\cdot \mat{M} \cdot \vect{v}, \qquad \vect{v} = \begin{pmatrix} \dot{x}_0\\ \dot{y}_0\\ \dot{z}_0\\ \dot{x}_1\\ \dot{y}_2\\ \dot{z}_3\\ \vdots\\\ \dot{x}_{N-1}\\ \dot{y}_{N-1}\\ \dot{z}_{N-1} \end{pmatrix}. \end{gather*}\]

Note: the “mass matrix” \(\mat{M}\) does not need to be diagonal.

To see how this fails, consider special relativity. Here, it is natural to start from the Hamiltonian formulation with the relativistic energy-momentum relation (here, with no potential \(V(x)=0\))

\[\begin{gather*} H(x, p) = E(p) = \sqrt{p^2c^2 + m^2c^4}, \qquad v = \dot{x} = E'(p) = \frac{pc^2}{\sqrt{p^2c^2 + m^2c^4}}. \end{gather*}\]

Solving for \(p(v)\) we have the relativistic kinetic energy in terms of the time-dilation factor \(\gamma(v/c)\):

\[\begin{gather*} p = mv\sqrt{1 - \frac{v^2}{c^2}}, \qquad E(v) = mc^2\overbrace{\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}}^{\gamma(v/c)}. \end{gather*}\]

However, with a little bit of work, one can find that the Lagrangian for a relativistic particle should actually be:

\[\begin{gather*} L[x, \dot{x}] = p\dot{x} - H = -mc^2\sqrt{1-\frac{\dot{x}^2}{c^2}} - V(x), \end{gather*}\]

which can be obtained via the Legendre transform, but is not of the form \(E(v) - V(x)\).