Is conservation of angular momentum derivable from Newtonian Mechanics?

Question

Why is conservation of angular momentum considered a law? This one says

Conservation of angular momentum really is a new phenomenon, one that does not follow from the Newtonian mechanics you already know; therefore it deserves its own place as a law. Specifically, you have proven that

I think it's wrong.

I think conservation of angular momentum does follow from the newtonian mechanics. However, I do not quite understand the derivation either.

So what's going on?

On the other hand, this question, shows how one can derive conservation of angular momentum from Newton 3rd laws.

Deriving Conservation of Angular Momentum from Newton's Laws

The only exception is when magnetic force create torque in moving charge. However, conservation of angular momentum actually hold because we need to take into account the angular momentum of the whole field.

Answer 1

It is not derivable from the Newton axioms as they are typically stated.

You need the extra assumption that your forces don't break rotation symmetry (as is pointed to by @rob's comment regarding the Noether theorem, which associates conserved quantities to symmetries).

Addendum

As an extended discussion about the derivation from Newtons axioms appeared under this post, let's just show what can be derived (and thereby see how one fails to derive conversation of total angular momentum without additional assumptions).

We have a system of $N$ particles at positions $\vec r_i(t)$ ($i = 1, \ldots, N$).

The equations of motion are (this is typically numberes as one of Newtons axioms): $$m_i \ddot{\vec r} = \vec F_i(\vec r_1, \ldots, \vec r_n). $$

Now, we can look at the time-derivate of the total angular momentum: \begin{align*} \dot{\vec L} &= \partial_t \sum_{i=1}^N m_i \vec r_i \times \dot{\vec r_i} = \sum_{i=1}^N m_i \underbrace{\dot{\vec r_i} \times \dot{\vec r_i}}_{=0} + \sum_{i=1}^N\vec r_i \times (m_i\ddot{\vec r_i}) \\ &= \sum_{i=1}^N \vec r_i \times \vec F_i(\vec r_1, \ldots, \vec r_N) \end{align*} That is, we can derive a formula for the change of the angular momentum. The result is, however, not zero in general even if the forces obey $\text{actio} = \text{reactio}$.

As a counterexample, take a two particle system with: \begin{align*} \vec F_1(\vec r_1, \vec r_2) &= -\vec F_2(\vec r_1, \vec r_2) = \vec F(\vec r_2 - \vec r_1) \\ \vec F(\vec d) &= \alpha \vec e_x (\vec d \cdot \vec e_y) \end{align*} These forces clearly fulfil $\text{actio} = \text{reactio}$.

But the angular momentum is not constant in general, take the following initial conditions: $$ \vec r_1(0) = \vec 0, \vec r_2(0) = l \vec e_y $$ Then (independently of the initial velocities), we have: $$ \dot{\vec L}(0) = m_1 \alpha l \vec 0 \times \vec e_y - m_2 \alpha l \vec e_y \times \vec e_x = m_2 \alpha l \vec e_y $$ As the derivative does not vanish, the angular momentum is obviously not conserved.

If you take the strong version of the second axiom${}^1$ (as noted in a comment by @DavidHammen) then conservation of angular momentum does hold (the proof is left as an exercise for the reader ;) ). More general forces are permissible, as long as they are invariant under rotations).

${}^1$ That all forces act as action-reaction pairs, and are central, that is, point along the connecting line.

Answer 2

I will give a very TLDR answer that clarifies the links between the various things that OP is asking about. However, I don't show why any of the statements I am making are true (in part, because they are very duckduckgoable once you know that this is what you have to search for).

• Newton's first two laws do not imply either the conservation of linear momentum or angular momentum.

• The weak form of Newton's third law implies the conservation of linear momentum but not that of the angular momentum. The strong form of Newton's third law implies the conservation of both linear and angular momentum.

• However, in nature, the law of conservation of linear momentum as well as that of angular momentum remain valid even in cases where Newton's third law is violated, both in its weak form and in its strong form.

  • In such cases, the conservation laws for linear momentum and angular momentum are obeyed because the laws of physics are still symmetric -- in particular, translationally symmetric and rotationally symmetric respectively. This relation between symmetries of the laws of nature and conservation law is established via Noether's theorems.

So, in conclusion:

• The laws of conservation of angular momentum and linear momentum are more universal than the third law of Newton.
• In the limited set of scenarios in which the third law of Newton is obeyed, the laws of conservation of angular momentum and linear momentum can be derived from the third law.
• But, they are empirically found to be obeyed even in cases where Newton's third law is violated. In such cases, they can be theoretically explained/derived from Noether's theorem which establishes the connection between symmetries of laws of physics and conservation laws.

Answer 3

Look at this example

Put the coordinate system at the center of mass and obtain the EOM's

$$m_1\ddot r_1=F_1+F_3\\ m_2\ddot r_2=F_2-F_1\\ m_3\ddot r_3=-F_2-F3$$

thus the

$$\sum m_i\ddot r_i=M\,\ddot r_{CM}=0$$

or

$$\frac {d}{dt}\underbrace{\,M\,(r_{CM}\times\dot r_{CM})}_{L}=0$$

so the angular momentum $~L~$ at the center of mass is conserved