I have the following issue with understanding. A light ray traveling from $q(\tau_1)$ to $q(\tau_2)$ minimizes the integral $\int\limits_{\tau_1}^{\tau_2} n(q(\tau))|\dot{q}(\tau)| d\tau$, so the corresponding Lagrangian of the light ray is $$n(q(\tau))|\dot{q}(\tau)| .$$ Then the Legendre transform yields $$H = \dot{q}\frac{\partial L}{\partial \dot{q}} - L.$$ Now, the derivative of $L$ works out to $$\frac{\partial L}{\partial \dot{q}} = n(q) \frac{\dot{q}}{|\dot{q}|}, $$ so the Hamiltonian is simply identical to zero. Surely this is now right. Can someone clarify my confusion? Thanks in advance!


1 Answer 1


That the Hamiltonian is zero is completely correct.

The system is time-reparametrization invariant - changing $\tau$ to $\xi(\tau)$ transforms $$ n(q(\tau))\mapsto n(q(\xi)),\quad \dot{q}\mapsto \frac{\mathrm{d}\xi}{\mathrm{d}\tau}q', \quad \mathrm{d}\tau\mapsto \frac{\mathrm{d}\tau}{\mathrm{d}\xi}\mathrm{d}\xi$$ and the action is invariant under this transformation (up to a sign we will ignore). It is a general fact that time-reparametrization-invariant actions where also $q$ and $p = \frac{\partial L}{\partial \dot{q}}$ are invariant lead to a zero Hamiltonian.

To develop a working Hamiltonian formalism for such systems - or more generally for systems where the Legendre transform is non-invertible - requires the notions of gauge theory and constrained Hamiltonian systems.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.