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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?

a12345
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  • Possible duplicate of http://math.stackexchange.com/questions/465251/derivation-of-newtons-second-law-of-motion. On that thread I posted a passage from Terence Tao that addresses this issue. – littleO Mar 11 '14 at 00:57
  • Thank you very much for the link. It was helpful but it seems like that Tao's answer is vague – a12345 Mar 11 '14 at 15:32
  • No. It's a physical model which arise from several experiments. – Felix Marin Mar 11 '14 at 23:49
  • Are you familiar with the first chapter of Arnol'd's "Mathematical Methods of Classical Mechanics"? – Dominik Mar 19 '14 at 12:19
  • I will take a look at the book. I heard its a classic. – a12345 Mar 20 '14 at 21:58

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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.

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