# Derivations of Newton's laws?

I feel convinced that the mathematics behind newtons laws can be derived from simpler priciples. The fact that displacement s can be described by a cartesian coordinate system with a parameter t and that the laws are constant with both ds and dt, we could somehow (though I'm not certain how) arrive to the definition that dE = Fds and dP = Fdt. F is a function applied to a closed system and E and P are constants. It's tempting to come to the conclusion that dE/dP = ds/dt which is a constant, since that can lead to P = c*ds/dt (or m) and all of the other mechanics laws.

That last step seems like cheating, is it not possible to come to $F$ $\alpha$ $a$ using something else?

• Comment to the post (v1): Note that Noether's theorem assumes that there exists an action formulation. – Qmechanic Mar 8 '16 at 15:51
• Expanding on the previous comment: Noether's theorem explicitly assumes the existence of Lagrangian mechanics, and Lagrangian mechanics is equivalent to Newton's formulation of mechanics. Most text books are concerned with taking you from Newton to Lagrange; but you should be able to work it backwards as well. It takes a bit more work to show it in general, rather than just a special case. – Peter Diehr Mar 8 '16 at 17:08

My hint is too long for a comment, but maybe it is worth writing down. I'll not use symmetry arguments, but I'll try to get Newton's equations out of the non-degeneracy of a certain mathematical object.

Let $M$ be manifold and $\omega$ a 2-form on it. If $\omega$ is algebraically closed and non-degenerate, then the dimension of $M$ must be even (let's say $2n$) and, at least locally, $\omega = -d\alpha$. Now, since $\omega$ is non-degenerate, there's only one vector field $X_H$ associated to a differentiable function $H$ such that $\imath_{X_H} \omega = dH$, where $\imath$ stands for the inner product

\begin{align} \imath: \Omega^n(M) &\to \Omega^{n-1}(M) \\ \imath_X \omega &\mapsto \omega(X) \, . \end{align}

Consider now $n=1$ (generalization for any finite dimension is immediate), that is, the one degree of freedom case. Let's then express $X_H = a \partial_q + b \partial_p$, where $dq$ and $dp$ form a Darboux basis on $M$ (this means that, at least locally, $\omega = dq \wedge dp$). Then

\begin{align} \imath_{X_H} \omega &= dH \\ \Leftrightarrow [dq \wedge dp] (a \partial_q + b \partial_p) &= dH \\ \Leftrightarrow a dp - bdq &= dH \\ \Leftrightarrow a dp - bdq &= \frac{\partial H}{\partial q} dq + \frac{\partial H}{\partial p} dp \\ \end{align}

which means that,

\begin{align} \begin{cases} a = \partial_p H \\ b = -\partial_q H \end{cases} \, . \end{align}

If we consider $X_H$ to be the velocity field associated to $H$, that is, $a = \dot{q}$ and $b = \dot{p}$, then the non-degeneracy of $\omega$ leads unambiguously to Hamilton's equations. Starting from Hamilton's equations invariance under a certain group of transformations, called canonical, one can see that there is a certain integral that must be minimized along a classical allowed path (this can be done either in the Lagrangian or Hamiltonian formalisms). Using the inverse idea of d'Alemberts principle one can deduce Newton's second law and use it as a hint, together with some geometric considerations, to infere the first (if the acceleration, which is what changes momentum, is zero, then the momentum is constant in a certain class of reference frames, called inertial, which must exist if space is homogeneous). For a discussion about the third law, see P.S..

I believe what I've sketched is a hand-wavish way of getting to Newton's equations from pure mathematics. It is extremely artificial, since the area that gave birth to all Symplectic Geometry was Analytical Mechanics, which means each and every step I used in my "deduction" was historically done backwards.

P.S.: In an earlier version of this answer I said that it might have been possible to infer the third law from the absence of a law that could explain the rest of Nature (as Newton did). Since this answer is about deducing all laws from mathematics, which is pretty much impossible, I've said that the second law hints at the other two, but even though the first law is sort of necessary to provide 0 acceleration a meaning, the third law may not be discerned from the other two (this is not my opinion). For a discussion on that read "the battle of the third law", which was fought in the comments to this answer with Timaeus, whom I thank dearly for the exposure of his views and many tries to clarify my misleading interpretation of the Laws. Readers have definitely enough material to choose a stance.

• You're right in all your points. In every step of the way, as I myself pointed out when I wrote "from the second law and some geometric considerations", there will be one more assumption to deduce the first and third laws from the second. The second would only serve as a hint to the other two, but they would require original reasoning themselves. All I did was to point that it is possible to discover all laws using the second as a hint, using the historical accounts with reversed order. I'll clarify that in an edit. – QuantumBrick Jun 28 '16 at 10:50
• What is $n$? Dimension? Also, I'm not talking about a single body. I'm saying that if you apply a force to something Newton's second law says this something will respond back, and if there's no movement the force will be the same. I'm not stating what the third law, that is, I'm not talking about the whole system. And now you have to answer me how applying the second law in the situation I described does not facilitate the process of axiomatizing a third law, because as I see it's pretty good a hint. – QuantumBrick Jun 28 '16 at 14:29
• Ok, now I know what $n$ is. So you're saying that the fact that if I push a wall and it doesn't move, the conclusion obtained by the second law, which is that somehow the force that I apply does not move the damn wall, is NOT a hint that there must be another law that takes reaction into account. Is that it? – QuantumBrick Jun 28 '16 at 14:43
• Feel free to include friction in the wall-finger example. I'll insist that I'm not at all saying that the third law is not independent of the second. I'm talking about heuristic arguments that clarify why the third law is needed. If you push the wall with your finger what keeps you in place is the friction between you feet and the ground, but you can be absolutely sure that the wall is feeling that force you're applying and it is not moving. This means $a = 0$ even though $F \neq 0$. This suggests the second law is not enough and there's something missing. – QuantumBrick Jun 28 '16 at 15:50
• You are most welcome to check the excellent, most beautiful book on Mechanics ever written: Spivak's Physics for Mathematicians, Vol. 1: Mechanics. Check page 21 and read the full process Newton possibly used to arrive at his third law. It is a corollary of empirical observations of the conservation of total momentum, and his second law actually served as a hint. Check also page 276 for a fake proof given by Newton that uses a situation where the third law is not properly applied (in a way you assume I'm applying). – QuantumBrick Jun 28 '16 at 15:54