So I have this integral $$ \int_{t_1}^{t_2} f(x,y)\sqrt{\dot{x}^2+\dot{y}^2} \text{ d}t, $$

where $x,y$ are functions of $t$ and $f$ is a function of just $x,y$. Also $\dot{x}$ denotes the first derivative of $x$ with respect to $t$. I am trying to study this type of integral and look for its extremas, depending on what $f$ is. The first thing I am wondering though is what is the reason for its vanishing Hamiltonian?

First we have the Lagrangian $L = f(x,y)\sqrt{\dot{x}^2+\dot{y}^2}$ which is just the integrand. Then the generalized momenta we get from that are $$ p_x = \frac{f(x,y)\dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}} \quad\text{and}\quad p_y = \frac{f(x,y)\dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}}. $$ And now we calculate the Hamiltonian to see that it is zero $$ H = p_x\dot{x} + p_y\dot{y} - L = 0. $$ I want to know what is the reason for this? I think it has something to do with the integrand's independence from $t$. But I'm not sure where to go from there.

  • 1
    $\begingroup$ The integral's explicit independence of $t$ implies that the Hamiltonian is conserved, $dH/dt = 0$, so $H$ is a constant as you find: en.wikipedia.org/wiki/… $\endgroup$
    – Winther
    Commented Jan 17, 2021 at 19:05

1 Answer 1


This is due to the reparametrization invariance of the Lagrangian. Specifically, your Lagrangian has the property that if we rewrite it in terms of a new "time function" $\tau(t)$, then we have \begin{align*} \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dx}{dt}\right)^2 } \, dt &= \left[ \frac{d\tau}{dt} \sqrt{ \left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dx}{d\tau}\right)^2 } \right] \left[ \frac{dt}{d\tau} \, d\tau \right] \\&= \sqrt{ \left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dx}{d\tau}\right)^2 } \, d\tau, \end{align*} since $(dt/d\tau)(d\tau/dt) = 1$. Thus, the Lagrangian does not change under this reparametrization $t \to \tau$.

The Hamiltonian for any Lagrangian which is parametrization-invariant is identically equal to zero. A nice proof of this can be found in this answer over on Physics.SE.

  • $\begingroup$ Gotchu, Thank you very much! 🙏 $\endgroup$
    – Jean
    Commented Jan 17, 2021 at 19:10

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