I am a bit confused as to how to solve this problem, in particular due to the way that the force term is expressed in this formulation of the second law. The way that it is formulated is this:
$$ m_i \frac{d^2\mathbf{Q}_i}{dt^2}=-\nabla_i \mathrm{V}(\mathbf{Q}_1,\:...\:,\mathbf{Q}_N) $$
Now, I know that Newton's second law should be separable, and that should give me a square term that makes both $q(t)$ and $q(-t)$ solutions. I have tried making the substitution $\frac{d\mathbf{Q}_i}{dt}=R$. However, I am at a loss about how to handle the right-hand side of the equation. The author of the book (Tumulka, 2022) also says that this equation is an ordinary differential equation; but that only reinforces my confusion, since how could an expression that involves a gradient be an ODE?
Thank you all.
Let’s reduce the clutter and focus only on one particle, of positive mass $m\in (0,\infty)$, moving in 3-dimensions under the influence of a force-field generated by a potential $V:\Bbb{R}^3\to\Bbb{R}$. So, Newton’s second law says that the particle’s trajectory (which by definition is a curve $\gamma:I\subset \Bbb{R}\to\Bbb{R}^3$, where $I$ is an open interval) is such that for all $t\in I$, we have \begin{align} m\ddot{\gamma}(t)&=-(\nabla V)(\gamma(t))\tag{$*$} \end{align} Recall that $V:\Bbb{R}^3\to\Bbb{R}$ is a function of three variables, so its gradient vector field is again a vector-valued function of three variables, $\nabla V:\Bbb{R}^3\to\Bbb{R}^3$. The point is we have to evaluate the gradient along the trajectory. We can thus write $(*)$ more succinctly as: \begin{align} m\ddot{\gamma}&=-(\nabla V)\circ \gamma.\tag{$**$} \end{align} Notice that both sides of this equation are functions $I\to\Bbb{R}^3$. Since the domain of both these functions is $I\subset\Bbb{R}$ (not $\Bbb{R}^2$ or $\Bbb{R}^3$ or something else), and since we’re involving the derivatives of $\gamma$ (on the LHS), that’s why the equations $(*)$ (or equivalently $(**)$) are called ordinary differential equations.
You are given $V$ and $m$. Your task, when solving Newton’s equations, is to find a curve $\gamma:I\to\Bbb{R}^3$ such that it satisfies $(*)$, or equivalently $(**)$ (along with appropriate initial conditions, but lets gloss over that now). We’re dealing with equations involving derivatives of a function of one-variable, hence ODEs.
As an explicit example, consider $V:\Bbb{R}^3\to\Bbb{R}$, $V(x,y,z)=x^2+y^2+3z^2$ (this is known as a 3-dimensional anisotropic harmonic potential). Then, $(\nabla V)(x,y,z)=(2x,2y,6z)$. So, your task is to find a curve $\gamma:I\to\Bbb{R}^3$ (i.e three single-variable real-valued functions) such that for all $t\in I$, \begin{align} m(\ddot{\gamma}_1(t),\ddot{\gamma_2}(t),\ddot{\gamma}_3(t))=-(\nabla V)(\gamma_1(t),\gamma_2(t),\gamma_3(t))=-\left(2\gamma_1(t),2\gamma_2(t),6\gamma_3(t)\right) \end{align} Or writing this as a system of scalar equations,rather than a single vector-valued equation, the above is equivalent to \begin{align} \begin{cases} m\ddot{\gamma}_1(t)&=-2\gamma_1(t)\\ m\ddot{\gamma}_2(t)&=-2\gamma_2(t)\\ m\ddot{\gamma}_3(t)&=-6\gamma_3(t) \end{cases}\tag{$***$} \end{align} Clearly, this is a system of three ODEs (in this case, they are very easily solved using sines and cosines, so the domain is in fact $I=\Bbb{R}$).
As for proving the time-reversal property, it has all to do with the fact that second derivatives appear in Newton’s law. Let us assume $\gamma$ is a curve which satisfies Newton’s second law (*). Let us now define the time-reversed curve $\rho(t)=\gamma(-t)$; Our goal is to show that $\rho$ also satisfies Newton’s second law. Well, we have \begin{align} m\ddot{\rho}(t)&=m\cdot (-1)^2\ddot{\gamma}(-t)=m\ddot{\gamma}(-t)=-(\nabla V)(\gamma(-t))=-(\nabla V)(\rho(t)). \end{align} The first equal sign is by the chain rule, the second equal sign is obvious, the third is since $\gamma$ satisfies Newton’s second law, and the last is by definition of $\rho$. Hence, we have shown that the time-reversed curve $\rho$ also satisfies Newton’s second law. The key is to just be careful with evaluating things at the right places, and using the chain rule correctly. If this is not clear, then work this out explicitly in the special case $(***)$.
Finally, the case with more particles is just extra notation, an index $i$ for each particle.