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 3 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.

  • 2
    $\begingroup$ Search term: Noether’s theorem $\endgroup$
    – rob
    Feb 13, 2022 at 15:47

3 Answers 3


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).


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.

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    $\begingroup$ The strong form of Newton's third law (forces between a pair of particles are central and are equal but opposite) directly leads to conservation of angular momentum as well as conservation of linear momentum. The weak form (forces between a pair of particles are equal but opposite) only leads to conservation of linear momentum. $\endgroup$ Feb 14, 2022 at 13:57
  • $\begingroup$ What about the simple derivation from 2nd law $\vec{F}=m\vec{a}$ provided by @Eli in the comment to the question? Just take the vector product of $\vec{r}$ with each side of 2nd law, use the rule of the derivative of a vector product, and check that time dependence of angular momentum is zero if force is null or is a purely radial force. $\endgroup$
    – Arc
    Feb 14, 2022 at 14:09
  • $\begingroup$ @Arc That "derivation" is wrong, for multiple reasons. (1) $F=ma$ refers to net force. (2) The net force acting on a body is a vectorial sum of all of the external forces acting on a body. What allows that summation? It's a supposed corollary. Every derivation is circular. That forces are vectors is axiomatic in Newtonian mechanics. (3) The "proof" leaves $r\times \ddot r$ unresolved. This is indeed zero for individual forces if those individual forces obey the strong form of Newton's third law. But the "proof" does not reference that. $\endgroup$ Feb 14, 2022 at 14:27
  • $\begingroup$ @DavidHammen, $\vec{r}\times \ddot{\vec{r}}$ is not unresolved, it becomes the time derivative of $\vec{r}\times \vec{p}$, and thus the time derivative of angular momentum. The derivation is correct, the time derivative of angular momentum about a point for a particle is $\vec{r}\times \vec{F}$. I don't see the issue on the vectorial sum you talk about, just consider the case of a single force and you have a simple demostration that conservation of $L$ comes directly from 2nd law. $\endgroup$
    – Arc
    Feb 14, 2022 at 22:32
  • 1
    $\begingroup$ Obviously if the forces are opposite but NOT central then we have a torque and hence angular momentum won't be conserved $\endgroup$
    – obfuscated
    Feb 19, 2022 at 7:40

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.
  • $\begingroup$ Any case where Newton Third law is violated in real physics? $\endgroup$
    – obfuscated
    Feb 19, 2022 at 7:37
  • 1
    $\begingroup$ @obfuscated Absolutely! It's violated in classical electrodynamics. For example, see this old question of mine: physics.stackexchange.com/questions/114466/… $\endgroup$
    – user87745
    Feb 19, 2022 at 7:44

Look at this example

enter image description here

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

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


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

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

  • 1
    $\begingroup$ This is incorrect. You are implicitly assuming that forces between individual particles are radial. The basic form of Newton's third law only says that forces between individual particles are equal but opposite. It says nothing about directionality. $\endgroup$ Feb 15, 2022 at 16:57
  • $\begingroup$ Where you see that is radial ? Actio equal reactio eny why i used Newton law and got the conservation of the angular momentum , but we don’t Have to agree on it ! $\endgroup$
    – Eli
    Feb 15, 2022 at 19:53
  • $\begingroup$ Please read Sebastian Riese's answer. He gives a clear counterexample where angular momentum is not conserved in a situation that obeys the weak form of Newton's third law but does not obey the strong form of that law. Your proof is invalid. The strong form of Newton's third law is needed in order to derive conservation of angular momentum from Newtonian mechanics. On the other hand, the strong form of Newton's third law can be derived from conservation of linear and angular momentum if one assumes all interactions are pairwise and instantaneous. Noether's theorem is an even better start. $\endgroup$ Feb 15, 2022 at 20:16
  • $\begingroup$ Actually when there are magnetic field and we do not take into account angular momentum of the field, then yes angular momentum conservation may fail. That is the only way I know where angular momentum conservation may fail. In all other cases, strong newton third law applied $\endgroup$
    – obfuscated
    Feb 19, 2022 at 7:36
  • 1
    $\begingroup$ yes. You are correct. Just not the best way to explain it. But correct. So upvote $\endgroup$
    – obfuscated
    Feb 27, 2022 at 5:41

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