Consider an electron with mass $1$ moving in $\mathbb R^n$ in under the influence of a static electromagnetic field. Up to identifying vector fields with differential forms, Maxwell's equations state that the electric force is given by a closed one-form and the magnetic field is given by a closed two-form. Since $\mathbb R^n$ is contractible, we can pick antiderivatives of each of these: let $C$ be the electric potential (a function on $\mathbb R^n$) and $B$ the magnetic potential (a one-form on $\mathbb R^n$). Again identifying vectors with covectors, the equations of motion of the electron are: $$ \ddot q = dB\cdot \dot q + dC \quad\quad \text{(EOM)} $$ where $q(t)$ is the position at time $t$, I have identified one-forms with vector fields, and $\cdot$ is the pairing that takes the two-form $dB$ and the vector $\dot q$ to a covector. (Pick your favorite sign, perhaps swapping $B$ for $-B$ below.)

Pick an open set $U \subseteq \mathbb R^n$ and let $Q: U \times [0,1] \to \mathbb R^n$ satisfy: $Q(x,0) = 0$, $Q(x,1) = x$, and $Q(x,-)$ is a solution to (EOM) for each $x \in \mathbb R^n$; i.e. $Q$ is a family of solutions to (EOM) starting at $0$ and parameterized by the value at $t=1$. Define the Hamilton principal function $S$ on $U$ by $$ S(x) = \int_0^1 \left( \frac12 \left( \frac{\partial Q}{\partial t}(x,t)\right)^2 + B\bigl(Q(x,t)\bigr) \cdot \frac{\partial Q}{\partial t}(x,t) + C\bigl(Q(x,t)\bigr) \right) dt $$ $S$ depends on the choice of antiderivatives $B,C$ from the first paragraph. It satisfies: $$ \frac{\partial S}{\partial x}(x) = \frac{\partial Q}{\partial t}(x,1) + B(x) $$ up to identifying vectors and covectors. So the differential $dS = \frac{\partial S}{\partial x}$ does not depend on the choice $C$ of antiderivative of the electric field. By differentiating again, the Hessian $\frac{\partial^2 S}{\partial x^2}$ does not depend on the choice $B$ of antiderivative of the magnetic field.

My question

I know how to prove that the Hessian of $S$ is nondegenerate on the open set $U$. I would like to know whether the Hessian is necessarily positive-definite?

Suppose that $B = 0$. (See my answer below when $B \neq 0$; sorry about bringing it into play before, I was confused.) Does it follow that the Hessian of $S$ is positive-definite?

Some examples

I can work very few examples. In particular, I know the answer when $B = 0$ and $C(q)$ is homogeneous quadratic. Pick a basis in which $C(q)$ is diagonalized, and let the eigenvalues be $C_1,\dots,C_n$. Then the Hessian is diagonalized in the same basis. If $C_i = 0$, then the $i$th eigenvalue of the Hessian is $1$; if $C_i > 0$, then the $i$th eigenvalue of the Hessian is $C_i / \sinh^2\!\! \sqrt{C_i}$. If $C_i < 0$, then the $i$th eigenvalue is $|C_i| / \sin^2\!\! \sqrt{|C_i|}$. (Actually, this is true even if $C(q)$ is quadratic but not homogeneously so; let $C_i$ be the eigenvalues of the homogeneous part.)

But the other examples I've thought of seem to require either coupled ODEs or elliptic integrals, so I haven't solved them directly.

Bonus question

I used the metric on $\mathbb R^n$ twice: once in (EOM) to identify the vector $\ddot q$ with the covector on the RHS, and once in the definition of $S$ to square the vector $\frac{\partial Q}{\partial t}$. But consistently changing the metric in both places allows everything to be defined still. Up to changing bases, the only way to change a metric is to change its signature. Anyway, in the quadratic case, changing the signature of the metric changes the signature of the Hessian in exactly the same way. So I expect that the correct statement is that the ratio of the Hessian to the metric is positive-definite. But I'm not sure.

Bonus bonus question

Experts know that one can equations of motion and Hamilton principal functions for much more general Lagrangian functions $L: \mathbb R^{2n} \to \mathbb R$ — suppose that the matrix $\frac{\partial^2 L}{\partial v^2}(v,q)$ is invertible for every $(v,q)$, so that the equations of motion define a nondegenerate second-order ODE. Then is the Hessian of the action necessarily positive-definite?


The answer to your bonus bonus question is negative, I'm afraid. The Euler-Lagrange equations only extremise the action in general, not minimise it. Nondegeneracy of the lagrangian and positive-definiteness of the Hessian are two separate notions: the former indicating only the absence of constraints. Just take a massive particle moving on the line with a potential which is not bounded below. The lagrangian $$L = \frac12 m v^2 - V(q)$$ is nondegenerate, since $\frac{\partial^2 L}{\partial v^2} = m$ is nonzero, but the Hessian depends on the second derivative of the potential.

  • $\begingroup$ I certainly don't deny that "the Hessian depends on the second derivative of the potential". But you're too quick in your answer. As I said in the examples section, I can explicitly solve EOM when the potential V(q) is at most quadratic, with arbitrary sign, including when it is bounded below, and so in this case (and only this case) I can explicitly write the action and its Hessian. And unless I made a mistake, if V(q) is quadratic, then the Hessian is a perfect square depending on V''(q). $\endgroup$ – Theo Johnson-Freyd Nov 25 '09 at 6:11
  • $\begingroup$ Yes, sorry -- I must have misunderstood. I thought your last question referred to a general lagrangian. If that is the case, I stand by what I said. $\endgroup$ – José Figueroa-O'Farrill Nov 25 '09 at 14:00

Shortly after posting my question, I realized that the answer to the original question is that the Hessian is not necessarily positive-definite. In particular, consider picking a different antiderivative to the magnetic term. This corresponds to changing $B$ to $B + d\beta$, where $\beta$ is an arbitrary function on $\mathbb R^n$. Then the Hessian of $S$ changes by the Hessian of $\beta$, which can be very negative.

  • $\begingroup$ @fpqc: I changed the question as soon as posting my answer. $\endgroup$ – Theo Johnson-Freyd Nov 25 '09 at 6:06

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