For a second order differential equation (many physical systems) in one variable, I know "procedures" to compute the energy. Given $$q''(t)=f(q(t),q'(t)),\ \ q(0)=q_0,\ \ q'(0)=v_0,$$ if we're lucky we can read off the related Lagrangian $L$, introduce $p=\frac{\partial L}{\partial q}$, do a Legendre transform and we got the Hamiltonian function $H(q,p)$ for which $\frac{\text d}{\text dt}H(q(t),p(t))=0$ for solutions of the differential equation.

We can be more exact and give all the conditions for Noethers theorem to hold and the result is that along the flow $X(t)=\langle q(t),p(t)\rangle$ in phase space given by a solution with initial conditions $X(0)$, the function $H(q,p)$ always takes the same values. It defines surfaces in phase space indexed by initial conditions $X(0)$.

I wonder how to view this for a priori first order systems $$q'(t)=f(q(t)),\ \ q(0)=q_0,$$ where I think this must be even simpler. E.g. I thing some functional $$q(t)\mapsto q(t)-\text{an integral over}\ f(q(t))\ \text{something},$$ should exists which will be constant for solutions of the equation, i.e. only depend on $q_0$. For each $f$, the functional dependence of this "energy" on "$q_0$" will be different.

However, I can't seem to find a general relation. What's the theory behind this, is there an energy'ish time integral of motion? What is the functional dependence on the intial condition, for a suitable constant of motion for for first order systems. And what would be the interpretation, given that we speak of a situation with only one initial condition?

If it is that case that the system is too restricted so that there is no meaningful geometrical interpretation, then let's think about a system of first order differential equations. This is like the one we generated from phase space, except that it doesn't really come from a second order situation and so the intial conditions aren't really e.g. intial position and velocity. I'm pretty sure there are situation where one considers such a directly generated flow (the equation might be more complicated than exponential flow $\dot x\propto x$), but I don't recall any talk about the time-constant of motion in these systems, or how to interpret it.


Assuming the first order system is Hamiltonian, for one degree of freedom, we may construct the hamiltonian by integrating the pair of ode's using the relation:

$ q'= \frac{\partial H}{\partial p}\qquad p'= -\frac{\partial H}{\partial q} $

Higher order systems have the same "symplectic" structure, and the integrals could have no closed forms. So we would have:

$ \frac{\partial H}{\partial p} = f(q),\quad p'=??$

Things seem to get notationally ambiguous. The point is that we must look at variables as conjugated pairs when working in the Hamiltonian formulation of mechanics. As a definitive answer to the OP, if we knew that the system was Hamiltonian, we could in principle integrate the above relations to generate the "Hamiltonian". Also, energy and time are considered to be conjugated in this framework. May I suggest the Poisson bracket as a starting point.


Consider the autonomous ODE system

$$ \dot{x} = f(x) $$

You have a first integral of motion if there is a function $K$ such that $K(x(t))=const$. Therefore,

$$ d K/dt = \dot{x} \cdot \nabla K = f(x)\cdot \nabla K =0 $$

This tells you that the field $f$ that defines the motion must be orthogonal to the direction that defines the level set of $K$ (i.e. the trajectories are on the level sets of $K$).

So the problem is to find $K$ (if it exists), but this $K$ has (in general) nothing to do with the "concept of Hamiltonian":

  • Trivial point: even in the case you have an even number of $x$ components it is not guaranteed that you can see some of them as $q$ and others as $p$ and recover the Hamiltonian formulation.

  • The Hamiltonian is not just a constant of motion: it fully defines the motion.

  • The conservation of energy is a drawback of the symplectic structure. If you remove this structure you should expect to loose the fact that your "tentative-Hamiltonian" is conserved.

  • Given the points above and loosely speaking: the Hamiltonian has "more to do" with $f$ than with $K$.

It may be interesting to have a look at what happens in the "symplectic case" and in the general case when you formally split the variables as $x = (q,p)$.

In the "symplectic" case the field $f$ can be written in terms of a single function $H$, a genuine Hamiltonian, and the ODE system is

$$ \dot{q} = \nabla_p H \qquad \quad \dot{p} = -\nabla_q H $$

This immediately tells you that

$$ d/dt = \dot{x}\nabla = \dot{q} \nabla_q + \dot{p} \nabla_p = (\nabla_p H ) \nabla_q - (\nabla_q H )\nabla_p $$

so that $\dot{H}=0$ and the Hamiltonian is automatically a constant of motion.

Now consider a non-necessarily symplectic case with an even number of variables $x$ that you formally split into $x=(q,p)$:

$$ \dot{q} = A(q,p) \qquad \qquad \dot{p} = B(q,p) $$

This system is too generic and nothing can be said. However, the point of mechanics is that the dynamics of the system will be determined by a function on the manifold called "Hamiltonian". So we may "downgrade" the genuine-Hamiltonian concept from the symplectic theory to the non-symplectic one by asking that

"we want to turn a smooth function $H$ into a vector field $f=(A,B)$... then the dynamics consists of the flow along integral curves of $f$, namely $\dot{x}=(A,B)$."

A natural way to turn a function into a field is to consider $\nabla_x H$, so that we may impose $$ \dot{q} = A(q,p) = \nabla_q H \qquad \qquad \dot{p} = B(q,p) = \nabla_p H $$

But now $H$ is not conserved: the "particles" slide along the level sets of $H$!

In the end, the conservation of $H$ is really a drawback of the symplectic structure: a sort of "rotation of 90-degrees" makes the whole difference,

$$ (\dot{q} , \dot{p}) = (\nabla_q H , \nabla_p H) \qquad VS \qquad (\dot{q} , \dot{p}) = (\nabla_p H , -\nabla_q H) = Rot_{90} (\nabla_q H , \nabla_p H) $$ Not a surprise: in two dimensions $Rot_{90}$ this is really a genuine rotation of 90-degrees, and this rotation turns the gradient of $H$ in a way that it is parallel to the level sets of $H$: hence, $H$ is a constant of motion.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.