[Physics] Why do people say that Hamilton’s principle is all of classical mechanics? How to get Newton’s third law

conservation-lawsforceslagrangian-formalismmomentumnewtonian-mechanics

From the principle of least (or stationary) action, we get that a classical system will evolve according to Euler-Lagrange equations:

$$\frac{d}{dt}\bigg (\frac{\partial L}{\partial \dot{q_i}}\bigg) = \frac{\partial L}{\partial q_i} .$$

I have often read and heard from physicists that this differential equation encapsulates all of classical mechanics. A glorious reformation of Newton's laws that are more general, compact and much more efficient.

I get that if you plug in the value of the Lagrangian, you re-obtain Newton's second law. But Newtonian mechanics is based on 3 laws, is it not? The law of inertia is a special consequence of the second law, so we don't need that, but what about the third law, namely that forces acts in pairs; action equals minus reaction?

My question is, can we obtain Newton's third law from this form of Euler-Lagrange equation? I understand that Newton's third law for an isolated $2$-body system follows from total momentum conservation, but what about a system with $N\geq 3$ particles?
If not why do people say that it's all of classical mechanics in a nutshell?

Best Answer

Newton's third law is that for every action, there is an equal and opposite reaction. This is a statement of momentum conversation.

In the Euler-Lagrange equation, the last term $$ \frac{\partial\mathcal{L}}{\partial q_i} $$ is a generalized force. Similarly, the generalized momentum is $$ \frac{\partial\mathcal{L}}{\partial \dot q_i}. $$ If the generalized force is zero, then $$ \frac{d}{dt} \frac{\partial\mathcal{L}}{\partial \dot q_i} = 0 $$ Mathematically, this means that the generalized momentum is constant over time, i.e. it is conserved, which is Newton's third law.

We don't even need the Lagrangian to summarize all of Newton's laws. As you likely know, Newton's second law $F=ma$ is a special case of $F = \frac{dp}{dt}$. This generally accounts for all of Newton's laws:

  • Newton's 1st Law - an object will persist in a state of uniform motion unless compelled by an external force: If $F = 0$, then $\frac{dp}{dt} = 0$, and thus $p$ is constant.
  • Newton's 2nd Law - $F = ma$: if m is constant, then $$ F = \frac{dp}{dt} = \frac{d(mv)}{dt} = m\frac{dv}{dt} = ma.$$
  • Newton's 3rd law - see the below derivation.

So, generally speaking, Newton's third laws are slightly redundant in the sense that they can all be described by $$ \vec F = \frac{d\vec p}{dt}. $$

(or by the Euler-Lagrange equation, as you argue.)


Edit: Derivation of N3L with conservation of momentum

Consider a system with total momentum $\vec p_{\rm tot}$ and two particles with momenta $\vec p_1$ and $\vec p_2$ such that $\vec p_{\rm tot} = \vec p_1 + \vec p_2$. If the system is closed, then the total momentum is conserved, so $$ \frac{d\vec p_{\rm tot}}{dt} = 0.$$ Differentiating both sides, you get $$ \frac{d}{dt}(\vec p_{\rm tot}) = \frac{d}{dt}(\vec p_1 + \vec p_2)$$ $$ 0 = \frac{d\vec p_1}{dt} + \frac{d\vec p_2}{dt} = \vec{F_1} + \vec{F_2}$$ $$ \vec F_1 = -\vec F_2 $$