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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?

K.defaoite
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  • It is standard that $∇f$ denotes the gradient of $f$, so that $df(v)=\langle∇f,v\rangle$. In 3 dimensions one then gets the rotation as $∇\times$ and the divergence as $∇·$, with the constant vectors acting geometrically and the partial derivatives as derivatives. – Lutz Lehmann Feb 23 '21 at 20:02

1 Answers1

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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 $z$.

peek-a-boo
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    (by the way if you've ever solved these questions involving polar coordinates for example, you may recall that the radial acceleration is something like $a_r=\ddot{r}-r\dot{\theta}^2$; this extra term is precisely encoded into the $\Gamma$ factors). – peek-a-boo Feb 23 '21 at 20:26
  • Thanks for the response! I guess I should be really careful when I use the $\nabla$ symbol outside of differential geometry. Probably better to write $\operatorname{grad}$ instead. I'm pretty unfamiliar with differential forms and the musical isomorphisms - can you quickly explain what $\mathrm{d}U$ is? What makes you so quick to say $\mathrm{d}U=\nabla U$ ? Are you referring to the covariant derivative or the gradient? Also, as a matter of curiosity, the last equation kind of reminds me of the geodesic equation, albeit in a inhomogeneous form. Is there a good reason for this? – K.defaoite Feb 23 '21 at 22:56
  • @K.defaoite Well if $f:M\to\Bbb{R}$ is a function, then $df$ is a covector field defined by setting for each $p\in M,\xi\in T_pM$ $df_p(\xi):=\xi(f)$ (here I'm using the definition of the tangent space as the space of all derivations). If you're instead more comfortable with the tangent space as the space of equivalence class of smooth curves, then $df_p([\gamma]):=(f\circ \gamma)'(p)$ for any smooth curve $\gamma:(-\epsilon,\epsilon)\to M$ with $\gamma(0)=p$. If you unwind the definitions then $df=\frac{\partial f}{\partial x^i},dx^i$ so its the covector field with components as you'd expect – peek-a-boo Feb 23 '21 at 23:37
  • In the equality $dU=\nabla U$, the RHS refers to the covariant derivative. So both sides are $(0,1)$-tensor fields/covector fields/1-forms (these are all different names for the same thing), and this equality is true almost by definition. In a chart they have the same components $(dU)_i=(\nabla U)_i=\frac{\partial U}{\partial x^i}$ (almost by definition). Regarding the last equation yes: if a curve $\gamma$ satisfies the geodesic equation then it is the straightest possible curve in the manifold (eg straight lines in the plane, or great circles on the sphere etc). – peek-a-boo Feb 23 '21 at 23:50
  • So, newton's first law recast in geometric terms says that "particles under the influence of no force travel along a geodesic in (classical) spacetime"; i.e they satisfy $\nabla_{\dot{\gamma}}\dot{\gamma}=0$. The second law then essentially defines for us what a force is: a force is any "thing" which deviates the motion of a particle from a "striahgt line" i.e a geodesic (up to proportionality factor). So, if the RHS is non-zero $m\nabla_{\dot{\gamma}}\dot{\gamma}=F$, then that is what we call a force. So, it really is just a geometric way of saying Newton's second law. – peek-a-boo Feb 23 '21 at 23:55
  • Regarding musical isomorphisms, this is a simple matter of linear algebra. Let's fix a finite-dimensional vector space $V$ (in geometry, we're interested in the tangent spaces), and a symmetric $(0,2)$ tensor $g:V\times V\to\Bbb{R}$. We say $g$ is non-degenerate if the mapping $g^{\flat}:V\to V^$ defined by $g^{\flat}(v):=g(v,\cdot)$ is injective (or equivalently is an isomorphism...these two notions are equivalent by the rank-nullity theorem) In this case the inverse is denoted $g^{\sharp}:V^\to V$. We call such a $g$ a "pseudo-inner product" or an "indefinite inner product". – peek-a-boo Feb 24 '21 at 00:04
  • Clearly an inner product (which is positive-definite) satisfies this non-degeneracy condition, because if $v\in V$ is such that for all $w\in V$, $\langle v,w\rangle=0$, then by choosing $w=v$, we see $\langle v,v\rangle=0$, hence by positive-definiteness, $v=0$. A usual theme is that anything you can do for vector spaces, you can do on the tangent spaces of the manifold, so given a metric tensor field $g$, for each $p\in M$, you can consider $g_p^{\flat}:T_pM\to T_p^M$ and $g_p^{\sharp}:T_p^M\to T_pM$. This allows you to "go back and forth" between vectors and covectors. – peek-a-boo Feb 24 '21 at 00:07
  • This answer and the following comments really cleared up my confusion. Many thanks @peek-a-boo – K.defaoite Feb 24 '21 at 00:36