# What's the physical interpretation of hamiltonian in calculus of variation?

I'm studying calculus of variation in this semester. I have difficulties in understanding the conditions for a curve to be an extremal.

Specifically, when we are deriving the general variation of a functional of the form $$\int_{a}^{b}F(x,y,y')\mathrm{d}x$$, we define $$p$$ to be $$\frac{\partial F}{\partial y'}$$ and $$H$$ to be $$y'\frac{\partial F}{\partial y'}-F$$. If we don't put any restrictions on the two endpoints, we must have a) Euler-Lagrange equation must be satisfied; b) $$p\delta_y|_a^b - H\delta_x|_a^b=0$$.

Our professor mentioned that $$p$$ and $$H$$ represent momentum and hamiltonian respectively. But I can't understand the relationship between these two quantities and the variational problem. Because I've never seen any variational problem which involves momentum $$p$$ and hamiltonian $$H$$. An example from physics which involves momentum and hamiltonian would be very helpful.

--------The followings are edited on Apr 4, 2020---------

Moreover, I would like to know the intuition of the quantity $$\delta J = p\delta_y|_{a}^{b}-H\delta_x|_{a}^{b}$$ where $$J[y] = \int_{a}^{b}F(x,y,y')$$. This corresponds to the situation where the endpoints are not fixed and the curve $$y$$ is taken to be the extremal (satisfies the Euler-Lagrange equations). How could this quantity be interpreted as the variation when the endpoints are perturbed? Need a concrete example to illustrate this.

• What's the definition of $\delta_x$ and $\delta_y$? – Qmechanic Feb 15 at 12:56
• They represent the perturbations of positions of endpoints. – Chris Jing Feb 15 at 23:59
• Consider to spell out more clearly the logic. Condition (b) seems to be part of Noether's theorem but the post (v2) has no mentioning of a symmetry per se. – Qmechanic Feb 16 at 8:20

Let's discuss a generalization of @Botond's example,$$L=\frac12m\dot{x}^2-V(x)\implies p=m\dot{x}\implies H=\frac12m\dot{x}^2+V(x)$$(this even works in multiple dimensions). Then $$L$$ is kinetic energy minus potential energy, while $$H$$ is kinetic plus potential. In other words, $$H$$ is the total energy. Indeed, $$H$$ is conserved much more generally, namely whenever $$\partial_tL=0$$, because$$\frac{dH}{dt}=\frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t}.$$So $$H$$ is a popular definition of energy.
But let me show you a simpler example: Imagine a point particle at the end of an ideal spring (simple harmonic oscillator). The mass of the particle is $$m$$, and the spring constant is $$k$$. We know that the energy of the system will be $$E=\frac{1}{2}mv^2+\frac{1}{2}kx^2$$ with momentum $$p=mv$$, but we can get it directly from the Lagrange function: $$L=\frac{1}{2}mv^2-\frac{1}{2}kx^2$$ The generalised momentum (which happens to equal to the momentum this time) is $$p=\frac{\partial L}{\partial v}=mv$$ So the Hamiltonian is $$H=pv-L=pv-\frac{1}{2}mv^2+\frac{1}{2}kx^2=p\frac{p}{m}-\frac{1}{2}m\frac{p^2}{m^2}+\frac{1}{2}kx^2=\frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2$$ Where we defined $$\omega$$ as $$\omega=\sqrt{\frac{k}{m}}$$ (It's nit neccessary, but people usually do this).