Can you prove $F=ma$ mathematically? So, it is one of Hilbert's famous problems to "axiomatize physics". 
In these attempts to establish physics as a subfield of mathematics, 
I wonder where does the Newton's law $F=ma$ stand. 
To be more specific, I ask the following questions. 
(1) How do we define the force $F$, and the mass $m$ in a mathematical way?
(2) Then, is $F=ma$ an axiom, or a theorem? 
 A: 1) I assume you have knowledge of the energy-momentum tensor, $T_{\mu\nu}$, and GR. $T^{00}$, in natural units, is the energy/mass density. So, in an object of volume $V$, the mass $m=V\cdot T^{00}$. So, I believe that mathematically, one can define the force $\mathbf{\vec{F}}$ as $\dfrac{d\mathbf{\vec{p}}}{dt}$, where if $S[\phi]=\int\mathcal{L}\mbox{d}^dx$ is the action of the theory ($d=\dim(M)$, $M$ is spacetime), and $q^j$ are canonical coordinates, then $p_j\mbox{ (the components of $\mathbf{\vec{p}}$) }=\dfrac{\partial\mathcal{L}}{\partial\dot{q}_j}$. Here, $\dot{q}_j=\dfrac{\partial q_j}{\partial t}$, where $t$ is time. What is time? I will talk (Hamiltonian, or ADM) general relativistically. Let $M$ be the spacetime, and let $\Sigma$ be a three-dimensional Lorentzian manifold such that $M=\Sigma\times\mathbb{R}^1$. Therefore, an element $p\in M$ can be written as $p=(\sigma,t)$, with $\sigma\in\Sigma,t\in\mathbb{R}^1$. (Note that this disrupts the concept of the nonexistence of absolute time, but you can read about this online.) The $t$ "coordinate" of $p$ is the time coordinate.
2) Newton's second law states that a force on a body is the time rate of change of the momentum of that body, i.e., $\vec{\mathbf{F}}=k\dfrac{d\vec{\mathbf{p}}}{dt}$, where $k$ is a constant, which can chosen to be $1$ by choosing units correctly. When we write ($m$ is a scalar) $\vec{\mathbf{p}}=m\vec{\mathbf{v}}$,
$$\vec{\mathbf{F}}=\dfrac{d(m\vec{\mathbf{v}})}{dt}=\vec{\mathbf{v}}\dfrac{dm}{dt}+m\dfrac{d\vec{\mathbf{v}}}{dt}\\
\implies \vec{\mathbf{F}}=\vec{\mathbf{v}}\dfrac{dm}{dt}+m\vec{\mathbf{a}}$$
When, and only when, we take $\boxed{\dfrac{dm}{dt}=0}$, we get
$$\vec{\mathbf{F}}=\vec{\mathbf{v}}\dfrac{dm}{dt}+m\vec{\mathbf{a}}\\
\implies \boxed{\vec{\mathbf{F}}=m\vec{\mathbf{a}}}\mbox{, the famous Newton's second ''law''}$$
(When you take the magnitudes, you get $F=ma$.)
So? Is $\mathbf{\vec{F}}=m\mathbf{\vec{a}}$ a theorem or a law? It's a theorem if you define "theorem" as is defined in the first accepted answer in Difference between a theorem and a law. We used the fact that $\mathbf{\vec{F}}=\dfrac{d\mathbf{\vec{p}}}{dt}$ (not exactly an axiom, but instead a law), so $\mathbf{\vec{F}}=m\mathbf{\vec{a}}$ can be said to be a theorem.
