This post considers an aspect of time-reparametization invariance in classical Hamiltonian mechanics. Specifically, it concerns the use of Lagrange multipliers to rewrite the action for a classical system in a time-reparametization-invariant way.


Suppose we have a system with a single degree of freedom $q(t)$ with conjugate momentum $p$, and action $$I = \int dt\ L. \tag{8.1}$$ The Hamiltonian is the Legendre transform $$H(p,q) = p\dot{q}-L(q,\dot{q})|_{p=\partial L/\partial\dot{q}}.\tag{8.2}$$ The independent variable $t$ is special. It labels the dynamics but does not participate as a degree of freedom.

Time-reparametrization symmetry:

Let us introduce a fake time-reparameterization symmetry by labelling the dynamics by an arbitrary parameter $\tau$ and introducing a physical 'clock' variable $T$, treating it as a dynamical degree of freedom. So we consider the system of variables and conjugate momenta $$q(\tau),\qquad p(\tau),\qquad T(\tau), \qquad \Pi(\tau)\tag{8.4}$$ where $\Pi$ is the momentum conjugate to $T$. This is equivalent to the original original system if we use the 'parameterized' action $$I' = \int d\tau\ (pq'+\Pi T'-NR), \qquad R \equiv \Pi + H(p,q),\tag{8.5}$$ where prime $= d=d/d\tau$. Here $N(\tau)$ is a Lagrange multiplier, which enforces the 'constraint equation' $$\Pi + H(p,q) = 0.\tag{8.6}$$

My difficulty lies with the introduction of the Lagrange multipliers. How do you show that the action

$$I' = \int d\tau\ (pq'+\Pi T'-NR), \qquad R \equiv \Pi + H(p,q),\tag{8.5}$$

with Lagrange multiplier $N(\tau)$, reduces to the action

$$I' = \int d\tau\ (pq' - H(p,q)T')~?\tag{8.8}$$

Edit to question:

OkThen's answer suggests that, when an action is extremised under a set of constraints, and the constraints are implemented using Lagrange multipliers, each Lagrange multiplier becomes a dynamical variable which satisfies its own equation of motion.

Why must each Lagrange multiplier be a dynamical variable which satisfies its own equation of motion?

  • 1
    $\begingroup$ Comment to the post (v5). The block quotes are apparently taken from pages 83-84 of Tom Hartman's lectures on Quantum Gravity and Black Holes. Please always include reference for quotes. $\endgroup$
    – Qmechanic
    Commented Jun 25, 2017 at 11:44

1 Answer 1


Lagrange multipliers don't have derivatives in the action. This means that their equation of motion is algebraic -- it is not a differential equation.

So, you can calculate the equation of motion for N, solve it, and plug back into the action.

It is straightforward since you get just $R=0$.

  • $\begingroup$ Please see my edit to the post. $\endgroup$ Commented Apr 18, 2017 at 2:26
  • 1
    $\begingroup$ You can find an explanation for this in any classical mechanics book. I recommend looking into it. They will have more detailed explanations. In any case, when you have constraints, you cannot perform arbitrary variations on your fields. These must respect your constraints. The Lagrange multiplier is an additional degree of freedom that soaks up the constraint allowing you to perform your variations freely. Answering your question: it is so by construction! $\endgroup$
    – OkThen
    Commented Apr 18, 2017 at 14:19

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.