Einstein's convention and Hamilton's equations in $\Bbb R^3$.

Question

Consider $F = -\nabla U$ a conservative force field in $\Bbb R^3$. Assume we describe the motion of a unit mass particle under this force field by a curve $q(t) = (q^1(t), q^2(t), q^3(t))$. We have Newton's law $$-\frac{\partial U}{\partial q^i}(q(t)) = \ddot{q}^i(t), \quad i=1,2,3,$$which is a second order differential equation in $\Bbb R^3$, and we want to convert that into a first order differential equation in $\Bbb R^6$ by calling $p^i \doteq \dot{q}^i$, so that we have $$\begin{cases} \dot{q}^i(t) = p^i(t) \\[1em] \dot{p}^i(t) = -\dfrac{\partial U}{\partial q^i}(q(t)), \end{cases}$$great.

Question: How can I write indices according to Einstein's summation convention (for psychological reasons, I'll still write summations signs)? The index $i$ in $p^i$ is covariant or contravariant?

If $i$ goes "in the ceiling" as $p^i$, then $\dot{p}^i = -\partial U/\partial q^i$ is bad.

If $i$ goes "in the cellar" as $p_i$, then $p_i = \dot{q}^i$ is bad.

Is there an identification being made? If $M$ is a manifold and $(U, (x^i))$ is a chart, then $(TU, (x^i, \xi_i))$ is a chart in $TM$, and since we see $\Bbb R^6$ as $T\Bbb R^3$, this would indicate to use $p_i$. I'm confused.

Answer 1

In symplectic mechanics, the components of $p$ are local coordinates for the cotangent bundle, so a physicist would "naturally" write them with subscripts. Consequently, you need to invoke the metric $g_{ij}$ to relate $\dot{q}$ and $p$. For instance, Hamilton's equations for the Hamiltonian $$H(q,p)=\tfrac{1}{2m}p_{i}g^{ij}p_j + U(q)$$ are $$\begin{align}\dot{p}_i&=\tfrac{1}{m}p_lg^{lk}\Gamma^j_{ki}p_j-U_{,i}\\ \dot{q}^i&=\tfrac{1}{m}g^{ij}p_j \end{align}$$ where $\Gamma$ is the Christoffel symbol for $g$.

More stuffily, we say that $(\dot{q},\dot{p})$ is the Hamiltonian vector field associated with the function $H$.

Answer 2

Since the question is tagged with symplectic-geometry and the other answer mentions symplectic mechanics, I think there is relevance in what I'm to say (this is just an addition to the existing good answer by K B Dave).

In symplectic mechanics, Hamilton's equations in a symplectic manifold are given by $$\iota_X \omega=dH.*$$ No metric plays a role, and the canonical metric in $\mathbb{R}^n$ is kind of a red herring if we are looking at the issue from an abstract symplectic point of view. The identification of $TM$ with $T^*M$ happens with the symplectic form (and its nondegeneracy). The fact that in the particular case of $\mathbb{R}^{n} \times \mathbb{R}^n$ we can see this coming from the metric is due to the equality $$\omega_{can}=\sum dx_i\wedge dy_i=\langle J\cdot, \cdot\rangle_{can},$$ where $\omega_{can}$ and $\langle \cdot, \cdot \rangle_{can}$ are the canonical symplectic form and inner product, respectively, and $J$ is the standard complex structure $\begin{pmatrix}0 &-I \\ I & 0\end{pmatrix}$ on $\mathbb{R}^{2n}$. Thus, Hamilton's equations translate to their most familiar form since \begin{align*} \omega(X,\cdot)&=\langle JX, \cdot \rangle, \\ dH&=\langle \nabla H, \cdot\rangle \end{align*} imply through Hamilton's equations that $$JX=\nabla H, $$ and thus $$X=-J \nabla H.$$

EDIT: It has occurred to me that since no Hamiltonian is explicitly mentioned in the question or in the answers, this may be a little confusing to the bypasser (particularly due to the fact that $\nabla U$ appears, maybe even suggesting that $H=U$, which is not the case). So I'll elaborate a bit more and, while at it, I changed the sign conventions.

In standard mechanics on $\mathbb{R}^n\times \mathbb{R}^n$ as in the context of OP, our Hamiltonian is $$H(q,p)=\frac{1}{2}p^2+U(q). $$

Thus, $\nabla H=(\nabla U, p)$. We then have that $-J \nabla H=(p,-\nabla U)$, and then the equation $X= -J \nabla H$ obtained above is exactly the one as written in the question.

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*Beware of wildly different sign conventions in literature, including the previous version of this answer!

Also worth of noticing is that this is the symplectic analogue of the gradient vector field, which is given by $$\iota_X g=dH.$$