Explicit Hamiltonian without legendre transform in symplectic geometry

Question

Most textbooks derive the Hamiltonian through the Legendre transform, so for simple mechanical systems one lands at $H = T + V$. I am interested in whether the time-evolution generator $H$ can be fixed directly from Newton’s laws together with the canonical symplectic form; without ever introducing a Lagrangian, even in situations where the result is not $T+V$.

Answer 1

The Hamiltonian formalism is more general and you cannot always use the Lagrangian one as a starting point. Indeed, In the latter, you start with a manifold $M$, consider its tangent bundle $TM$ and then by duality introduce $T^*M$ as phase space. This means that it necessarily has vector bundle structure and the symplectic 2-form is exact, being the derivative of the tautological 1-form.

You can consider more general phase spaces with symplectic structures. In particular, the 2-form only needs to be closed, so if the second cohomology is non trivial, you can have Hamiltonian systems with no Lagrangian counterparts.

The simplest example for this is when the phase space is the sphere $\mathbb S^2$ with the usual area form giving the symplectic structure. Explicitly, at $s\in\mathbb S^2$ and $ds_1,ds_2\in T_s\mathbb S^2$: $$ \omega(ds_1,ds_2)=s\cdot(ds_1\times ds_2) $$ This is even relevant physically, as it describes angular momentum. For example: $$ H=-\gamma B\cdot s $$ describes a gyromagnetic interaction, or $$ H=\frac12s\cdot(Is) $$ is an inertia term.

Hope this helps.

Answer 2

1. More generally, a Hamiltonian formulation consists of a (possibly degenerate) Poisson manifold $(M,\{\cdot,\cdot\})$ equipped with a Hamiltonian function $H:M\to \mathbb{R}$ generating time evolution.

   It may or may not be formulated outside (and may or may not have any relation to) the framework of Newtonian mechanics.

2. OP seems to be essentially asking:

   If we are given an arbitrary physical system in Newtonian mechanics (say, without dissipation), can we in general define a corresponding Hamiltonian system without introducing a Lagrangian?

   Well, for starters, it is unclear how in general to directly identify the Hamiltonian canonical momentum with an appropriate notion of canonical momentum in Newtonian mechanics without effectively using the Lagrangian canonical momentum at an intermediate stage as a "go-between".

   See also e.g. this related Phys.SE post.

Answer 3

We can formulate the stationary action principle directly in terms of a Hamiltonian $H(x,p)$ and two independent variables $x$ and $p$ (we consider just one "degree of freedom" for presentational simplicity): $$ S = \int_{t_1}^{t_2}dt\left(p\dot x - H(x,p)\right)\tag{1}\;, $$ where Eq. (1) might be reducible to an action principle in one independent variable $x$ and a Lagrangian $L(x,\dot x)$, but we don't necessarily have to make such a reduction. Further, we can still determine interesting properties of the Hamiltonian $H$, such as its status as a generator of time translations, as discussed below.

From Eq. (1) alone we can see that the thing denoted by $H(x,p)$ is the generator of time translations, since under an infinitesimal change of the time parameter $t\to t + \eta(t)$, we have $$ S \to \int_{t_1}^{t_2} dt (1+\dot \eta)(p(1-\dot \eta)\dot x - H(x,p)) $$ and so $$ \delta S = -\int_{t_1}^{t_2} dt H(x,p)\dot \eta $$ $$ -\int_{t_1}^{t_2} dt \frac{d}{dt} \left(\eta H(x,p)\right) + \int_{t_1}^{t_2} dt \eta (t) \dot H\;, $$ where the first term above, by definition, provides us with the generator of time translation, by considering the variation of the classical action $S_{cl}$ with respect to the endpoint times: $$ \delta S_{cl,\;\text{time endpoint variation}} = -\eta(t_2) H(t_2) + \eta(t_1) H(t_1)\;. $$

In other words: $$ \frac{\partial S_{cl}}{\partial t_1} = H(t_1) $$ and $$ \frac{\partial S_{cl}}{\partial t_2} = - H(t_2)\;. $$

If someone just "gives you" a Hamiltonian $H(x,p)$, you can use Eq. (1) with no explicit reference to a "Lagrangian" $L(x,\dot x)$, if you would like. And, as we have just shown, the Hamiltonian is the generator of time translations.

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As another example, suppose that you are simply "given" a classical action as an explicit function of the endpoints: $$ S_{cl} = S_{cl}(t_1, x_1, t_2, x_2)\;. $$

You can use such a function directly to determine the "generator" of time translation by taking the derivative with respect to the initial time: $$ \frac{\partial S_{cl}}{\partial t_1} = H(t_1)\;. $$

For example, for a free particle $$ S_{cl}(t_1, x_1, t_2, x_2)=\frac{m}{2}\frac{(x_2 - x_1)^2}{(t_2 - t_1)} $$ and $$ \frac{\partial S_{cl}}{\partial t_1} = \frac{m}{2}\frac{(x_2 - x_1)^2}{(t_2 - t_1)^2} $$

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I am interested in whether the time-evolution generator H can be fixed directly from Newton’s laws

As another explicit example, consider a system that obeys the following equation of motion, derived from Newton's laws: $$ \ddot x = -kx -\beta \dot x\;. $$

Define $$ v \equiv \dot x $$ and $$ \vec y \equiv \left(\begin{matrix}x\\v\end{matrix}\right)\;. $$

So, now we have reduced our second-order equation of motion to a first-order linear equation: $$ \dot{\vec y} = {\bf H}\cdot \vec{y}\;, $$ where $$ {\bf H}\equiv \left(\begin{matrix}0 & 1\\-k/m & -\beta/m\end{matrix}\right)\;. $$

You can call the matrix ${\bf H}$ the generator of time translations, since it takes you from one time step to the next: $$ \vec y \underbrace{\to}_{t\to t+\delta t} \vec y + \delta\vec y = \left(1 + \delta t {\bf H}\right)\cdot\vec y\;. $$

So, we have a generator of time translation and we didn't ever write the word "Lagrangian."

Answer 4

$\def \c {\boldsymbol} \def \p {\dot} \def \pp {\ddot} \def \b {\mathbf} \def \e {\rm e}$

Given the Newton equations of motion in generalized coordinates:

$$\b M(\b q)\,\b{\ddot{q}}=\b f(\b q)+\b f_p(\b q~,\b{\dot{q}})$$

and assuming the force $~\b f(\b q)~$ is conservative, and $~\b f_p(\b q~,\b{\dot{q}})~$ is pseudo force due to coordinates transfer, we can derive the total mechanical energy of the system, which consists of kinetic and potential energy.

Kinetic Energy

The kinetic energy $T$ is defined as: $$T=\frac 12 \b{\dot{q}}^T\,\b M(\b q)\,\b{\dot{q}}$$

Potential Energy

Since the force is conservative, there exists a potential energy function $V(\b q)~$ such that:

$$\b f(\b q)=-\nabla_{\b q}\,V(\b q)$$

Total Energy

the total energy E is :

$$E(\b q~,\b{\dot{q}})=T+V=\frac 12 \b{\dot{q}}^T\,\b M(\b q)\,\b{\dot{q}}+V(\b q)$$

In this case: $~H=E~$

• The Lagrangian is: $$L=T-V$$
• The canonical momentum is: $$\b p=\frac{\partial L}{\partial \b{\dot{q}}}=\b M(\b q)\, \b{\dot{q}}$$
• The Hamiltonian becomes: $$H(\b q~,\b p)=\b p^T\,\b{\dot{q}}-L=T+V=E$$

Answer 5

whether the time-evolution generator can be fixed directly from Newton’s laws together with the canonical symplectic form; without ever introducing a Lagrangian,

Even the Lagrange equations do not follow from Newton directly. You first must specify what the ideal holonomic constraints are. Besides this the Legendre transform takes a dynamical system (Lagrange equations) defined on the tangent bundle to one (Hamilton equations) defined on the cotangent bundle. This is a very important story by itself