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Is it possible to derive the equations of motion from the energy of a system alone, without knowing canonical coordinates or the Lagrangian?

I'm confused about which parts of the fundamental specification of a physical system can be derived, vs. have to be known in advanced. Given a specification of a system where total energy (the quantity that will be conserved in time) is defined as a function of some coordinates, is this enough to derive the other components of the system, like canonical coordinates, the Lagrangian, equations of motion etc? I have in mind systems like fluid dynamics, where the energy depends only on the coordinates and their first time derivatives, and sometimes only on the coordinates, as in the 2d point vortex energy, where for $n$ point vortices with locations $x_i$ in the 2d plane, the energy is the sum of the logarithms of all pairwise distances times the product of the vorticities ($\Gamma_i$): $$H(x_1,...,x_n)=\sum_{i,j} \Gamma_i\Gamma_j\log|x_i-x_j|$$

I am aware that for simple systems where you can "recognize" part of the energy as being potential energy, you can negate this in the energy to get the Lagrangian, and then use the Euler-Lagrange equation to derive the equations of motion. Also, if you happen to know a transformation to canonical coordinates, then you can use Hamilton's equations. But are canonical coordinates derivable? In short, given just the energy of a system, what can I do? (Or do I need to know more than just the energy? If so, why is the Lagrangian so much more useful than the energy? Is there anything I can do with the Poisson bracket? In cases like fluid dynamics, the Lagrangian introduces a weird additional term $\sum_i \Gamma_i |x_i\times \dot{x}_i|$, which seems hard to guess for other related systems, and doesn't seem to have much physical intuition.)

Qmechanic
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poyi
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1 Answers1

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TL;DR: It is not possible in general to determine the EOMs from the energy function$^1$ $h(q,\dot{q},t)$.

Example: Consider e.g. a non-relativistic 2D point charge in a constant $B$-field$^2$

$$L= T+qBx\dot{y},\qquad T~=~\frac{m}{2}(\dot{x}^2+ \dot{y}^2).\tag{1}$$

The energy function

$$ h~:=~\left(\dot{x}\frac{\partial}{\partial \dot{x}}+\dot{y}\frac{\partial}{\partial \dot{y}}-1 \right) L~\stackrel{(1)}{=}~T\tag{2}$$

is just the kinetic energy $T$, which carries no information about the $B$-field, the EOMs and the classical trajectories of the point charge.

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

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$^1$ The (Lagrangian) energy function $h(q,\dot{q},t)$ should not be confused with the Hamiltonian $H(q,p,t)$ although they take the same value. The latter determines EOMs via Hamilton's equations.

$^2$Here we have used the gauge $(A_x,A_y)=(0,Bx)$. The choice of gauge does not affect the EOMs.

Qmechanic
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