Is Newton's second law a consequence of the principle of conservation of energy? How can we arrive at
net force = rate of change of momentum
using only the law of conservation of energy?
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4 Answers
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Start with
$$\frac12 m \dot{x}^2 + V(x) = E$$
with $V$ your potential and $E$ a constant. Now take the time derivative:
$$m \dot{x} \ddot{x} + \frac{dV}{dx}\dot{x} = 0 \implies m\ddot{x} = - \frac{dV}{dx}$$
and there you have it, provided you identify the derivative of the potential with the force.
Do remember, though, that Newton's law is more general, because there exist nonconservative forces. This derivation only works for forces that come from a potential.
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1$\begingroup$ This shows that $(m\ddot{\mathbf{x}}+\boldsymbol{\nabla}V)\cdot\dot{\mathbf{x}}=0$, but $\mathbf{A}\cdot\mathbf{B}=0$ doesn't imply $\mathbf{A}=\mathbf{0}$. $\endgroup$– J.G.Commented Sep 14, 2017 at 18:02
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It is true that Newton's 2nd law with conservative force $$m_i{\bf a}_i ~=~-\frac{\partial V}{\partial {\bf r}_i}, \qquad i~\in~\{1, \ldots, N\},\tag{1}\label{eq:1}$$ implies conservation of the mechanical energy $$E=\sum_{i=1}^N \frac{m_i}{2}{\bf v}_i^2 + V({\bf r}_1, \ldots,{\bf r}_N).\tag{2}\label{eq:2}$$ But the opposite $\eqref{eq:2}\implies \eqref{eq:1}$ is not true in general. However, if there is only one particle $N=1$ and only one spatial dimension $d=1$, and the speed is non-zero, then the opposite $\eqref{eq:2}\implies \eqref{eq:1}$ is true.
answered Sep 14, 2017 at 18:00
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$\begingroup$ Can you direct me to some book or any known source regarding this? Basically I am trying to understand the redundancy of classical laws. From what I understand if Newton's laws imply conservation of energy and not vice versa then it is a very remarkable fact. As conservation laws seem universal and applicable across domains, newton's laws seem to me as a sore thumb sticking out. $\endgroup$ Commented Jul 28, 2018 at 5:25
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The connection is provided by the Hamiltonian formalism, if $F = -dV/dx$ is the net force acting on a particle of mass $m$, then the quantity
$$ H(x,p) = \frac{p^2}{2m} + V(x) = E $$
satisfies the expression
$$ \frac{dp}{dt} = -\frac{\partial H}{\partial x} = -\frac{dV}{dx} = F $$
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$\begingroup$ To add to this answer, you also obtain conservation of energy from this formalism: $\frac{dH}{dt} = \frac{\partial H}{\partial p}\frac{dp}{dt} + \frac{\partial H}{\partial x}\frac{dx}{dt} = \frac{dx}{dt}\frac{dp}{dt} - \frac{dp}{dt}\frac{dx}{dt} = 0$. $\endgroup$ Commented Aug 30, 2021 at 5:27
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Conservation of energy isn't strong enough to reproduce Newton's second law. As a counterexample, consider a situation with no potential energy, $U(\mathbf{x}) = 0$. There is no force, so according to Newton's laws, the particle should move with constant velocity, $\mathbf{x}(t) = \mathbf{v}t$. But conservation of energy tells us that the kinetic energy $\frac{1}{2}m\|\mathbf{v}\|^2$ remains constant, thus only constraining the magnitude of the velocity to be constant. This allows for nonphysical motions like $\mathbf{x}(t) = (\cos t, \sin t)$, whose velocity magnitude is constant but velocity is not.
