I've recently finished a course in differential geometry, and I was having a look at some of my old classical mechanics notes. I found the following written in my notes:
Under the influence of a conservative potential energy $U$, a particle of mass $m$ and position $\underline{\mathrm{r}}$ obeys the equation
$$m\underline{\ddot{\mathrm{r}}}=-\nabla U\tag{1}$$
While the motivation and meaning of this equation is very clear, there is a slight problem. In my differential geometry course, I learned that the $\nabla$ operator, also known as the covariant or semicolon derivative, is defined to act on an $(r,s)$ tensor, in components, as
$$(\nabla\mathbf{T})^{i_1..i_r}_{j_1..j_s~k}=\\ \partial_kT^{i_1..i_r}_{j_1..j_s}\\+\Gamma^{i_1}_{k~l}T^{l~i_2..i_r}_{j_1..j_s}+\dots+\Gamma^{i_r}_{k~l}T^{i_1..i_{r-1}~l}_{j_1..j_s}\\-\Gamma^l_{k~j_1} T^{i_1..i_r}_{l~j_2..j_s}-\dots -\Gamma^l_{k~j_s}T^{i_1..i_r}_{j_1..j_{s-1}~l}$$
In particular the covariant derivative of a scalar $\phi$ should be
$$(\nabla\phi)_i=\partial_i\phi$$
This is a covector. However, going back to (1),
$$m\underline{\ddot{\mathrm{r}}}=-\nabla U$$
The left hand side is a vector! Position is a vector, so its time derivative should be a vector – right??
So what's the deal here? Is equation (1) wrong? It's used quite a lot in classical mechanics, with correct results too. In (1) should I instead write $\nabla^{\mathrm{T}}U$ to indicate the transpose? Or should I just ditch the symbol altogether and write
$$m\underline{\ddot{\mathrm{r}}}=-\sum_{i}\underline{\mathrm{e}_i}\partial_iU$$
Which, although correct, is rather inelegant? I would appreciate some advice here.
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ADDENDUM
This other user seems to define the nabla operator contravariantly. Is this standard outside of differential geometry and topology?
Best Answer
Right, they are different nablas. $\nabla U$ here is actually referring to the gradient, which by definition is $\text{grad}(U):= g^{\sharp}(dU)=g^{\sharp}(\nabla U)$; i.e the vector field associated to the $1$-form $dU=\nabla U$ (here this $\nabla U$ refers to the one from differential geometry; it is a $(0,1)$ tensor field/covector field/one-form).
And if you want to write $F=ma$ in truly differential geometric notation, you'd write (letting $\gamma$ denote the trajectory of the particle) \begin{align} m\nabla_{\dot{\gamma}}\dot{\gamma} &= -\text{grad}(U)=-g^{\sharp}(dU)=-g^{\sharp}(\nabla U). \end{align} "mass times acceleration is force = minus gradient of potential energy"
As an aside:
In terms of a chart, this expression yields \begin{align} m\left(\ddot{x}^i + \Gamma^i_{jk}\dot{x}^j\dot{x}^k\right) &= -g^{ij}\dfrac{\partial U}{\partial x^j} \end{align} (everything evaluated at appropriate places, and of course I really mean $x\circ \gamma$). If you use cartesian coordinates in $\Bbb{R}^3$ this reduces to the simple $m\ddot{x}=-\frac{\partial U}{\partial x}$ likewise for $y$ and $z$.