Derivative of a curve on tangent bundle $TQ$ and second-order equations (where is the acceleration?)

Question

This is very basic question that I should have resolved long ago but didn't and it still plagues me.

Let $TQ$ be the tangent bundle of some (configuration) manifold, $Q$, and let $(\pmb{q},\pmb{v}) = (q^1,\dots q^n,v^1,\dots v^n):TQ\to \mathbb{R}^{2n}$ be any local natural/induced/tangent-lifted coordinates on $TQ$ (many write $v^i=\dot{q}^i$). Let $\mathbf{X}\in\mathfrak{X}(TQ)$ be a vector field and $u_t\in TQ$ some integral curve which I will regard as $u_t = ( r_t , \mathbf{u}_t ) \in [r_t] \times T_r Q$. I am confused about the left-hand-side of the equations of motion defined by

$$ \tfrac{d}{dt}u_t = \mathbf{X}_{u_t} $$

I will use $\dot{(\cdot)}=\frac{d(\cdot)}{dt}$. Lets make it specific to second order equations such that $\mathbf{X}=v^i \partial_{q^i} + F^i\partial_{v^i}$ for $F^i\in\mathcal{F}(TQ)$. For example, the Euler-Lagrange equations can be expressed as such a field:

$$ \mathbf{X} = v^i \partial_{q^i} - g^{ij}( \tfrac{\partial L}{\partial q^k \partial v^j} v^k - \tfrac{\partial L}{\partial q^j} ) \partial_{v^i} $$

The integral curves of such fields are then of the form $u_t = (r_t,\dot{r}_t)$ such that $\dot{u}_t = (\dot{r}_t, \ddot{r}_t) $. However, it is clear from every source I've encountered that, while $\dot{r}$ is just the velocity along the curve in the usual sense, $\ddot{r}$ does NOT mean as the acceleration along the curve (we need the covariant derivative for that I believe). Instead, everybody seems to imply that $\ddot{r} = \ddot{q}^i \partial_{v^i}$ and so $\dot{u}_t$ is given simply by $\dot{u} =\dot{q}^i\partial_{q^i} + \ddot{q}^i \partial_{v^i}$ (where, for short, I'm using $\dot{q}^i$ to mean $\dot{q}^i = \frac{d}{dt}(q^i\circ r_t)\equiv v^i(\dot{r})$). I have no idea why we can say this. Clearly, $\tfrac{d}{dt}$ does not mean what I am used to when it is applied to a curve on $TQ$. I am used to it simply meaning the (total) time derivative so that if $\dot{r}=\dot{q}^i\partial_{q^i}$ then I would expect $\ddot{r}=\ddot{q}^i\partial_{q^i}+ \dot{q}^i \frac{d}{dt}\partial_{q^i}$ where the second term would then involve the covariant derivative of $\partial_{q^i}$ along $r_t$. But that is not what is going on.

short version: If $u_t=(r_t,\dot{r}_t)\in TQ$ why is the derivative of this curve given by $\dot{u}_t=\dot{q}^i\partial_{q^i} + \ddot{q}^i\partial_{v^i}$ instead of involving the actual acceleration via covariant derivative. What is the actual formula I should use to express $\dot{u}_t\in T_{u_t}(TQ)$ in a coordinate basis?

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Note: I am asking this question with classical mechanics being my primary concern such that $Q$ is usually embedded in some higher dimensional flat space and where $r_t\in Q$ is a curve describing the evolution of the configuration of some mechanical system. I should maybe post this in the physics SE. I never know whether to post here or there.

Edit: part of my confusion may stem from the fact that, while I'm comfortable-ish with the "position-level" coordinate basis vectors, $\partial_{q^i}$, I still don't really know what the make of the velocity-level coordinate basis vectors, $\partial_{v^i}$. I asked a question about this at this link.

Answer 1

When in doubt, simplify notation. The second tangent bundle is indeed tricky (as are tangent bundles of a given vector bundle). For now, forget about the fact that $TQ$ itself came about as the tangent bundle of an $n$-dimensional manifold $Q$. Just call $M=TQ$. The manifold $M$ is $2n$-dimensional and locally it has coordinates $(z^1,\dots, z^{2n})$, so a curve $\Gamma:I\subset\Bbb{R}\to M$ in $M$ has at each time $t\in I$ a tangent vector $\dot{\Gamma}(t)\in T_{\Gamma(t)}M$ whose expression in local coordinates is \begin{align} \dot{\Gamma}(t)&=\sum_{i=1}^{2n}(z^i\circ\Gamma)’(t)\frac{\partial}{\partial z^i}\bigg|_{\Gamma(t)}\in T_{\Gamma(t)}M. \end{align} Notice the sum goes up to $2n$. The primes on the RHS are the usual derivatives for functions $I\to\Bbb{R}$. That’s it. The way I got this formula is simply because $M$ is a smooth manifold, and I used the definition of $TM$ being its tangent bundle.

The above formula is very general and holds for all smooth curves $\Gamma:I\to M$ and all coordinate charts $(z^1,\dots, z^{2n})$ on $M$. Now at this stage, you may want to remember that $M$ itself was the tangent bundle of $Q$. So, starting with base coordinates $(x^1,\dots, x^n)$ on $Q$, this lifts to adapted coordinates $(q^1,\dots, q^n,v^1,\dots, v^n)$ on $M=TQ$. Next, rather than considering arbitrary curves $\Gamma$, you may want to start with a curve $\gamma:I\to Q$ in the base manifold, and consider its velocity $\dot{\gamma}:I\to TQ$. So, applying the above formula with $\Gamma=\dot{\gamma}$ gives us a curve $\ddot{\gamma}:I\to T(TQ)$ whose local coordinate formula with respect to the adapted coordinates $(z^1,\dots, z^n,z^{n+1},\dots, z^{2n})=(q^1,\dots, q^n,v^1,\dots, v^n)$ is \begin{align} \ddot{\gamma}(t)&= \sum_{i=1}^n(q^i\circ\dot{\gamma})’(t)\frac{\partial}{\partial q^i}\bigg|_{\dot{\gamma}(t)}+\sum_{i=1}^n(v^i\circ\dot{\gamma})’(t)\frac{\partial}{\partial v^i}\bigg|_{\dot{\gamma}(t)}\in T_{\dot{\gamma}(t)}(TQ). \end{align} To simplify further, we should understand the functions $q^i\circ\dot{\gamma}:I\to \Bbb{R}$ and $v^i\circ\dot{\gamma}:I\to \Bbb{R}$. Well, essentially by definition of the phrase “$(q^1,\dots, q^n,v^1,\dots, v^n)$ are adapted coordinates on $TQ$ induced by $(x^1,\dots, x^n)$ on $Q$”, it follows that $q^i\circ\dot{\gamma}=x^i\circ\gamma$ and $v^i\circ\dot{\gamma}=(x^i\circ\gamma)’$ (review the definition and properties of tangent bundle charts if this is not clear). So, plugging this into the above equation, we get \begin{align} \ddot{\gamma}(t)&= \sum_{i=1}^n(x^i\circ\gamma)’(t)\frac{\partial}{\partial q^i}\bigg|_{\dot{\gamma}(t)}+\sum_{i=1}^n(x^i\circ\gamma)’’(t)\frac{\partial}{\partial v^i}\bigg|_{\dot{\gamma}(t)}\in T_{\dot{\gamma}(t)}(TQ). \end{align} This is exactly the formula you asked about, and this is certainly correct.

Finally, one doesn’t really refer to $\ddot{\gamma}(t)\in T_{\dot{\gamma}(t)}(TQ)$ as the acceleration vector, because usually we’d like the acceleration vector $A(t)$ of $\gamma:I\to Q$ to be an element of $T_{\gamma(t)}Q$. However, if all you have at your disposal is a smooth manifold $Q$, then unfortunately, you cannot create a definition for $A(t)$ which captures your usual intuition of acceleration as “second derivative” yet still lands in the first tangent bundle. For this you need a connection. Then, you can talk about $\nabla_{\dot{\gamma}(t)}\dot{\gamma}\in T_{\gamma(t)}Q$. Of course its coordinate expression is simply \begin{align} \nabla_{\dot{\gamma}(t)}\dot{\gamma}&=\left[(x^i\circ\gamma)’’(t)+\Gamma^i_{\,jk}(\gamma(t))\cdot(x^j\circ\gamma)’(t)\cdot(x^k\circ\gamma)’(t)\right]\frac{\partial}{\partial x^i}\bigg|_{\gamma(t)}\in T_{\gamma(t)}Q. \end{align}

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Throughout, I’ve restrained myself from abusing notation in anyway, so I kept all the compositions explicit, and I also distinguished between where the various coordinate functions live. So, that should hopefully clarify where everything belongs.

Also, a few extra remarks: to define the covariant derivative, the extra $\Gamma^{i}_{\,jk}$ terms are there because abstractly speaking, in order to convert an element of $T_{\dot{\gamma}(t)}(TQ)$ to an element of $T_{\gamma(t)}Q$, Ehresmann’s very geometric definition of a connection says that you need to subtract the “horizontal component” in order to get a vertical vector in $V_{\dot{\gamma}(t)}(TQ)$, which is equal to the tangent space of the fiber $T_{\dot{\gamma}(t)}(T_{\gamma(t)}Q)$, and since the fiber is a vector space, you can identify this with an element of $T_{\gamma(t)}Q$. In other words, the extra terms you add are precisely there so that you can go from the second tangent bundle to the first.

Btw, here are some answers you may find helpful: Second derivatives, Hamilton and tangent bundle of tangent bundle TTM and On which tangent bundles of $\Bbb{R}^2$ does position, velocity, acceleration live?

Answer 2

@peek-a-boo answered my question but I realized what my fatal error was and thought I'd post in case anyone else has been confused about this. For some velocity curve $u_t=(r_t,\dot{r}_t)\in TQ$, it is indeed correct that $\dot{u}_t = \dot{q}^i\partial_{q^i} + \ddot{q}^i\partial_{v^i}\in T_{u_t}TQ$ (I'm abusing notation a bit, see peek-a-boo's answer for the strictly correct expressions). My error was in also writing this as $\dot{u}_t=(\dot{r}_t,\ddot{r}_t)$ and then trying to make the identifications $\dot{r}_t=\dot{q}^i\partial_{q^i}\in T_{u_t}TQ$ and $\ddot{r}_t=\ddot{q}^i\partial_{v^i}\in T_{u_t}TQ$. That is wrong. If I want to view $\dot{u}_t$ as containing a velocity term and an acceleration term, I need to use the horizontal-vertical decomposition $T_{u_t}TQ=H_{u_t}TQ\oplus V_{u_t}TQ$ given by the Levi-Cita connection $\nabla$ for the Riemannian base manifold $(Q,g)$. Briefly, the relation is as follows. Let $\dot{r}_t\in T_{r_t} Q$ and $\nabla_{\dot{r}_t} \dot{r}_t \in T_{r_t} Q$ be the velocity and acceleration along $r_t$ the way one would usually think of those things in physics. For the curve $u_t=(r_t,\dot{r}_t)\in TQ$, its derivative $\dot{u}_t \in T_{u_t}TQ$ can indeed be expressed in terms of this velocity and acceleration as $\dot{u}_t = (\dot{r}_t)^H + (\nabla_{\dot{r}_t} \dot{r}_t)^V$, where $(\cdot)^H$ and $(\cdot)^V$ denote the "horizontal lift'' and ``vertical lift" of tangent vectors on $Q$. One could also write this as $\dot{u}_t = \dot{r}_t \oplus \nabla_{\dot{r}_t} \dot{r}_t$ where it is understood that the direct sum is with respect to the horizontal-vertical decomposition. It is still true that $\dot{u}_t = \dot{q}^i\partial_{q^i} + \ddot{q}^i\partial_{v^i}$ but it is not true that $\dot{q}^i\partial_{q^i}$ corresponds to $\dot{r}_t$ lifted to $T_{u_t}TQ$ and it is not true that $\ddot{q}^i\partial_{v^i}$ corresponds to $\nabla_{\dot{r}_t} \dot{r}_t$ lifted to $T_{u_t}TQ$. That is: $$ \dot{u}_t = \dot{q}^i\partial_{q^i} + \ddot{q}^i\partial_{v^i} = (\dot{r}_t)^H + (\nabla_{\dot{r}_t} \dot{r}_t)^V = \dot{r}_t \oplus \nabla_{\dot{r}_t} \dot{r}_t $$

but $\dot{q}^i\partial_{q^i} \neq (\dot{r}_t)^H$ and $\ddot{q}^i\partial_{v^i}\neq (\nabla_{\dot{r}_t} \dot{r}_t)^V $. We can, however, define so-called $\nabla$-adapted basis vectors $D_i:=\partial_{q^i} - v^j\Gamma^k_{ji}\partial_{v^k}\in\mathfrak{X}(TQ)$ such that

$$ u_t=(r_t,\dot{r}_t) : \qquad \dot{u}_t = \dot{q}^i\partial_{q^i} + \ddot{q}^i \partial_{v^i} = \dot{q}^i D_i + (\ddot{q}^i + \Gamma^i_{jk}\dot{q}^j\dot{q}^k)\partial_{v^i} = \dot{r}_t \oplus \nabla_{\dot{r}_t} \dot{r}_t $$

where now, finally, the term $\dot{q}^i D_i=(\dot{r}_t)^H$ is indeed the velocity lifted to $T_{u_t}TQ$ and $(\ddot{q}^i + \Gamma^i_{jk}\dot{q}^j\dot{q}^k)\partial_{v^i} = (\nabla_{\dot{r}_t} \dot{r}_t)^V $ is the acceleration lifted to $T_{u_t}TQ$. They are already contained in the expression $\dot{u}_t = \dot{q}^i\partial_{q^i} + \ddot{q}^i\partial_{v^i}$! We just have to transform this expression to the "correct'' basis in order to see it. Lastly, none of this required that $u_t$ be of the form $u_t=(r_t,\dot{r}_t)$. It applies the same way for arbitrary $u_t=(r_t,\pmb{u}_t)$ :
$$ u_t=(r_t,\pmb{u}_t) : \qquad \dot{u}_t = \dot{q}^i\partial_{q^i} + \dot{u}^i \partial_{v^i} = \dot{q}^i D_i + (\dot{u}^i + \Gamma^i_{jk}\dot{u}^j\dot{q}^k)\partial_{v^i} = \dot{r}_t \oplus \nabla_{\dot{r}_t} \pmb{u}_t $$

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I have not seen my above claims stated anywhere but I'm reasonably confident It is correct. Please let me know if anything is wrong (other than some hopefully clear abuses of notation).