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Not all equations of motion admit a Hamiltonian. Several questions and answers on this site concern this correspondence, for example Hamiltonian or not?, When can an autonomous system be written using a Hamiltonian?, and What are the necessary/sufficient conditions for a system to be Hamiltonian/non-Hamiltonian?.

My question is who was the first to establish that not all equations of motion can be described in a Hamiltonian fashion? Hamilton himself? Someone else? In which paper or book was this lack of total correspondence first exposed? Who has then developed the details of this correspondence? What is the current state of affairs? (From Hamiltonian or not? I gather that it is still a matter of trial and error to obtain a Hamiltonian from the equation of motions?)

Ricardo
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  • Comment to the post (v1): It is notoriously difficult to formulate No-Go theorems without loopholes, cf. the mentioned links. For the question to be well-posed, OP should consider to be more precise on what counts as a Hamiltonian system & what doesn't. – Qmechanic Oct 21 '16 at 17:09
  • Although I haven't used the term Hamiltonian system, a Hamiltonian system would be a set of equations of motion that can be derived from a Hamiltonian in the usual way. – Ricardo Oct 21 '16 at 17:58

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The conditions under which a function admits a Lagrangian or Hamiltonian are known as "conditions of variational self-adjointness." Whatever violates these conditions thus do not admit a Lagrangian or Hamiltonian.

The conditions are studied in the context of the Inverse Problem, which

formulates as:

Given the totality of solutions $y(x) = \left\{y^1(x), \ldots, y^n(x)\right\}$ of a system of $n$ ordinary differential equations of order $r$, $$F_k\left(x, y^{(0)}, y^{(1)}, \ldots, y^{(r)}\right)=0\qquad\text{(I.23)}$$ $$y^{(i)}=\frac{d^iy}{dx^i},\qquad i=1,\ldots,r,\qquad k=1,2,\ldots,n,$$ determine whether there exists a functional $$A(y)=\int_{x_1}^{x_2}dxL\left(x,y^{(0)},\ldots,y^{(r-1)}\right)\qquad\text{(I.24)}$$ which admits such solutions as extremals.

It appears Helmholtz was the first to study the Inverse Problem (ibid., p. 12):

The necessary and sufficient conditions for the existence of a solution $L$ of system (I.29)* were apparently formulated for the first time by Helmholtz (1887)26 on quite remarkable intuitional grounds. In essence, Helmholtz's starting point was the property of the self-adjointness of Lagrange's equations, i.e., their system of variational forms coincides with the adjoint system (see Chapter 2 and following). This is a property which goes back to Jacobi (1837).27 Without providing a rigorous proof, Helmholtz indicated that the necessary and sufficient condition for the existence of a solution $L$ of system (I.28)** is that the system $F_k = 0$ be self-adjoint.


26. Helmholtz did not consider an explicit dependence of the equations of motion on time. Subsequent studies indicated that his findings were insensitive to such a dependence.
27. The equations of variations of Lagrange's equations or, equivalently, of Euler's equations of a variational problem, are called *Jacobi's equations* in the current literature of the calculus of variations. We shall use the same terminology for our Newtonian analysis.

*(I.29) is the Euler-Lagrange equation corresponding to (I.24) when $n>1$, $r=2$.
**(I.28) is the case when $n=r=1$.

This same analysis of conditions for variational self-adjointness of a Lagrangian can be applied to Hamiltonians, as Hamiltonians are simply the Legendre transform of Lagrangians (cf. Callen's Thermodynamics and an Introduction to Thermostatistics §5.2, pp. 137-145, for a good introduction to Legendre transforms).

Helmholtz (1887) is:

Geremia
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    Your answer is amazing! Thank you so much! You don't know how long I've been looking for it and only getting unhelpful answers like AGML's back. I wonder when people will stop repeating the mantra that Hamilton equations are a restatement of Newton's laws. – Ricardo Oct 21 '16 at 20:34
  • @Ricardo Yes, as Santilli says (p. 7): "Lagrange and Hamilton were fully aware that the Newtonian forces are generally not derivable from a potential. Therefore, to avoid an excessive approximation of the physical reality, they formulated their equations with external terms"—i.e., with a third, non-zero term in the Euler-Lagrange or Hamiltonian equations, what Goldstein 1980 (§1.5 p. 21) calls the "dissipation function." – Geremia Oct 21 '16 at 22:24
  • @Ricardo Also, non-Lagrangian/non-Hamiltonian functions (i.e., those that don't satisfy the Euler-Lagrange / Hamiltonian equations without the "external term," as they're most commonly written today) lead to velocity-dependent potentials (like Neumann's potential for Weber's electrodynamics force law). – Geremia Oct 21 '16 at 22:24
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    @Ricardo Also check out vol. 2 of Santilli's book. He shows that there are Hamiltonians that are not derivable from a corresponding Lagrangian. – Geremia Feb 02 '18 at 03:59
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The Hamilton equations are a restatement of Newton's laws, so any classical physical system will obey them.

However, one typically uses the word "Hamiltonian system" to describe the case in which the Hamiltonian is furthermore time-independent.

The Hamiltonian measures the energy, so real physical systems including e.g. friction will obviously not be "Hamiltonian" in this sense. It thus seems to follow pretty immediately from the definition that there will exist systems even within physics that are "non-Hamiltonian", and it's hard to imagine anyone ever thought otherwise.

AGML
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My friend, who doesn't have a StackExchange account, provided another answer which may help:

Ricardo, I apologize if I do not completely understand the specifics of your question. You are correct however to complain that the “Hamiltonian equations” are not just “a restatement of Newton’s laws”.

As Geremia indicated, the Hamiltonian $H\left(\left\{q_i\right\},\left\{p_i\right\},t\right)$ is mathematically generated from the Lagrangian by the Legendre transform, that is, by:

$$H\left(\left\{q_i\right\},\left\{p_i\right\},t\right)=\sum_iq_ip_i-L\left(\left\{q_i\right\},\left\{p_i\right\},t\right).\qquad\text{(1)}$$

Thus (1) simply is an algebraic process expressing the physical equivalence of $H$ and $L$. The Hamiltonian formulation of classical mechanics is irreducible to “Newton’s laws” because we precisely add that which makes it irreducible to the latter, i.e. the generalized impulses/conjugate momenta $p_i$ to the generalized coordinates $q_i$ according to:

$$p_i=\frac{\partial L\left(\left\{q_i\right\},\left\{p_i\right\},t\right)}{\partial q_i}.\qquad\text{(2)}$$

The Hamiltonian equations of motion are given by:

$$q_i=\frac{\partial H}{\partial p_i},\qquad-p_i=\frac{\partial H}{\partial q_i}.\qquad\text{(3 and 4)}$$

However in the Hamiltonian format, moving from second order differential equations to first order ones may not always make life mathematically that much easier. Starting from the generalized coordinates and the Lagrangian, you need to calculate the Hamiltonian, formulate the generalized velocities as a function of generalized impulses, substitute the velocities in the definition of the Hamiltonian, and solve the $2n$ first order equations for the trajectories of the system.

Thus while Hamilton’s method laid out in his equations (3 and 4) is related to Lagrange’s by (3) via (2), it also offers an entirely distinct perspective from that Lagrange, which by the same token guarantees its irreducibility to Newton’s formulation.

Geremia
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