# Why does the Hamiltonian vanish for $\int f(x,y)\sqrt{\dot{x}^2+\dot{y}^2} \text{ d}{t}$?

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.

• 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/… Commented Jan 17, 2021 at 19:05

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 $$t \to \tau$$.