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During a meeting yesterday, a colleague of mine stated that a potential cannot depend upon arbitrary powers of the canonical momentum. So I am curious, is the potential limited to: $$ \Phi = \Phi\left( \mathbf{q}_{0}, \mathbf{q}_{1}, ..., \mathbf{q}_{N}; \ \mathbf{p}_{0}, \mathbf{p}_{1}, ..., \mathbf{p}_{N} \right) $$ or can it be represented as: $$ \Phi = \Phi\left( \mathbf{q}_{0}^{i}, \mathbf{q}_{1}^{j}, ..., \mathbf{q}_{N}^{k}; \ \mathbf{p}_{0}^{l}, \mathbf{p}_{1}^{m}, ..., \mathbf{p}_{N}^{n} \right) $$ where $i$, $j$, $k$, $l$, $m$, and $n$ are arbitrary real numbers? I did not quite follow why they were so certain this could not be so.

I am okay with the answers being purely classical and non-relativistic if that makes the explanation easier/simpler.

Questions

  • Can someone explain why, for instance, $\Phi\left( \mathbf{q}^{i}; \ \mathbf{p}_{0}^{2} \right)$ or $\Phi\left( \mathbf{q}^{i}; \ \dot{\mathbf{q}}_{0}^{2} \right)$ would not be allowed?
  • Is there any physical reason to limit the order of derivatives of $\mathbf{q}$ in the explicit dependence of $\Phi$? Meaning, could $\Phi$ depend upon $\tfrac{d^{K} \mathbf{q}}{dt^{K}}$, where $K$ is an integer $\geq$2?
  • The first subquestion (v2) does not seem clear: E.g. what stops us from absorbing the real powers $i$, $j$, $k$, $l$, $m$, and $n$ into a redefinition of the function $\Phi$? – Qmechanic Mar 21 '17 at 15:34
  • See also http://en.wikipedia.org/wiki/Ostrogradsky_instability – AccidentalFourierTransform Mar 21 '17 at 16:09
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    The last subquestion (v2) is essentially a duplicate of http://physics.stackexchange.com/q/4102/2451 – Qmechanic Mar 21 '17 at 16:26
  • @Qmechanic - Yes, you are correct. Should I just remove the last subquestion then or leave it in place since there is already an answer that addresses it? – honeste_vivere Mar 22 '17 at 05:29
  • Interestingly, I have seen people redefine the potential (for radiation belt dynamics) as a function of distance along a dipole magnetic field away from the magnetic equator, $s$, the parallel (to $\mathbf{B}{o}$) momentum, $p{\parallel}$, and the magnetic moment of the particles, $\mu$, where the magnetic moment for a single particle is $\propto \ p_{\perp}^{2}/B_{o}$. – honeste_vivere Mar 22 '17 at 05:36
  • FYI: 1. SE discourages radical changes to the parts of a post that have received answers. 2. SE also discourages asking more than one subquestion per post. – Qmechanic Mar 22 '17 at 14:02
  • @Qmechanic - I thought as much, but I thought I would ask for clarification. I asked the second sub-question as a sort of clarification for the first. Thanks for pointing the similar question. I tried several variations of the question title looking for similar posts to see if others had already answered my question but to no avail. I see I should have looked a little longer. – honeste_vivere Mar 22 '17 at 14:50

1 Answers1

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NB Your first question is improperly stated as Qmechanic pointed out in his comment. I interpret it in a precise sense: If there is a reason why $\Phi$ is supposed to depend at most linearly on the first derivatives of Lagrangian coordinates.

I guess you are considering generalized Lagrangians of the form $$L(t,q, \dot{q})= T(t,q, \dot{q}) - \Phi(t,q, \dot{q})\:, \tag{-1}$$ for classical systems described in a generalized coordinate system and also taking holonomous ideal constraints into accounts if any. In this case the kinetic energy $T$ takes the form $$T(t,q, \dot{q}) = \sum_{i,j=1}^n A(t,q)_{ij} \dot{q}_i\dot{q}_j + \sum_{j=1}^n B(t,q)_j\dot{q}_j + C(t,q)\:. \tag{0}$$ It turns out that the matrix $A(t,q) = [ A(t,q)_{ij}]_{i,j=1,\ldots, n}$ is symmetric an positively defined and in particular is invertible. Suppose that $$\Phi= \Phi(t,q)$$ If you write down the E-L equations, $$\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} - \frac{\partial L}{\partial q_j}=0\:, \quad \frac{dq_j}{dt} = \dot{q}_j\:, \quad j=1,\ldots, n \tag{1}$$ using the fact that $A$ is invertible you see, with a tedious computation, that it is possible to re-write these equations into the precise form $$\frac{d^2q_j}{dt^2} = F_j(t,q, \frac{dq}{dt}) \quad j=1,\ldots,n\:.\tag{2}$$ where in particular, for some functions $G_k$ we have $$F_j(t,q, \frac{dq}{dt}) = \sum_{k=1}^nA(t,q)^{-1}_{jk} G_k(t,q, \frac{dq}{dt})\:. \tag{3}$$ The form (2) of Euler-Lagrange's equations is said to be normal. This is a general notion in the theory of ordinary differential equation systems of order $n$ and just means that

the derivatives of highest order $n$ can be separated, and inserted in the left-hand side, from the derivatives of other orders $n-1, n-2,\ldots, 0$ which appear in the right-hand side in any functional form.

If the right-hand side is sufficiently regular (jointly continuous and locally Lipschitz in the variables $(q, \dot{q})$), the existence and uniqueness theorem establishes that any system of 2nd-order differential equations of the normal form (1) admits a unique (local and global) solution as soon as you fix the state of the system at initial time: $$q(t_0) = Q\quad \dot{q}(t_0) = \dot{Q}\:.$$ This property is the mathematical translation of the determinism principle of classical physics.

The crucial facts to pass from (1) to (2) are that (a) the first time-derivatives $\dot{q}_j$ appear quadratically in (0), (b) they do not appear in $\Phi$ and (c) that $A$ in (0) in invertible.

The same result can be obtained if $\Phi$ is also function of the $\dot{q}_j$, but they appear therein linearly.

Any different dependence, in particular a quadratic dependence of $\dot{q}_j$ in $\Phi$ could give rise to an obstruction to reach the normal form of the Euler-Lagrange equations, so that the principle of physical determinism may fail to be satisfied.

It is worth stressing that linearity in $\dot{q}_j$ appearing in $\Phi$ is only a sufficient condition to fulfill standard hypotheses for the existence and uniqueness theorem. So one may construct physical systems respecting the determinism principle, but described with Lagrangians including potentials with non-linear dependence on $\dot{q}$.

However the only two cases of generalized potentials $\Phi(t,q, \dot{q})$ known in classical physics, the potential of electromagnetic (Lorentz) forces and the potential of general inertial forces respect this linearity constraint.

Inserting time derivatives of order greater that $1$ in $\Phi$ gives rise to the same type of problems regarding the determinism principle, though these higher order derivatives are not completely forbidden and they are used in some semi-classical models, to describe the self-acceleration of an electric charge in particular.

  • Thanks, this is a great answer. I was confused by my colleague because several people working in radiation belt dynamics construct Hamiltonians that have potentials that depend upon the particle magnetic moment (i.e., it's "orbital" magnetic moment, not the intrinsic quantum mechanical property), which depends upon the square of the perpendicular momentum. However, I see from your answer that such higher order derivatives can make life difficult, but they are not forbidden. Correct? – honeste_vivere Mar 22 '17 at 15:02
  • Yes, correct, as far as I know at least – Valter Moretti Mar 22 '17 at 15:04