Given that a collection of classical particles can be modelled using Newton's three laws, it must be the case that both the conservation of momentum and the conservation of angular momentum are emergent features of Newton's three laws.

I can easily show the conservation of linear momentum from the third law, but how can the conservation of angular momentum be derived from Newton's three laws?

  • $\begingroup$ If you can show conservation of linear momentum from the Newton's 3rd law $F_{ab} =-F_{ba}$, then I'd believe you should be able to derive the conservation of angular momentum from the angular equivalent of Newton's 3rd law: $\tau_{ab} =-\tau_{ba}$ $\endgroup$
    – Steeven
    Jul 6, 2017 at 10:47
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    $\begingroup$ Angular momentum conservation assumes radial forces. See if you can take it from there. $\endgroup$
    – J.G.
    Jul 6, 2017 at 10:48
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    $\begingroup$ Related: physics.stackexchange.com/q/302487/50583, in particular CR Drost's answer that basically answers this question. $\endgroup$
    – ACuriousMind
    Jul 6, 2017 at 10:48
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    $\begingroup$ Possible duplicate of Why is conservation of angular momentum considered a law? $\endgroup$ Jul 6, 2017 at 11:50

2 Answers 2


Deriving the rotational form of $\mathbf{\vec{F} = m \vec{a}}$ (as per request in comments)

Begin with Newton's second law of motion

$$\vec{F} = m\vec{a}.$$

Multiply both sides by a vector cross product with position gives

$$ \underbrace{\vec{r} \times \vec{F}}_{\text{Definition of torque}} = \vec{r} \times m \vec{a}. $$

Using the definition of the cross product, the above equation can be equivalently expressed as

$$ |\vec{r} \times \vec{F}| = rma \; \sin{\theta}. $$

Notice that since the force and acceleration are parallel we may consider $a \sin \theta$ as the tangential acceleration $a_t$. Finally this can be cast into final form given by

$$\tau = mr^2 \left(a_t/r \right),$$

where moment of inertia and angular acceleration are given by $$I=mr^2, \;\; \alpha = a_t/r,$$ respectively.


You can consider the torque equation $$\vec{\tau} = I \vec{\alpha},$$ as the rotational equivalent to Newton's second law of motion where torque, moment of inertia and angular acceleration are given by $\vec{\tau}, \; I$ and $\vec{\alpha}$ respectively.

Now notice that angular acceleration is the time derivative of angular velocity $$ \vec{\tau} = \frac{d (I \vec{\omega})}{dt}.$$ which can be re-arranged into integral form $$\int\vec{\tau} dt= \int d(I \vec{\omega}).$$

Angular momentum $\vec{L}$ is given by the relation $\vec{L}=I\vec{\omega}$. Finally, in Newtonian mechanics, the conservation of angular momentum is used in analysing central force problems i.e. forces in the radial direction. With this is mind, I quote the definition of torque in terms of force

$$\vec{\tau} = \vec{r} \times \vec{F},$$ where the position and net force acting on a body are given by $\vec{r}$ and $\vec{F}$ respectively. If the force on some body is constantly pointed in the same direction then there is no torque on said body. Hence $$ C = I \omega = L, $$ and hence, $$ \frac{d L}{dt} \equiv 0. $$

  • $\begingroup$ -1: The rotational equivalent of Newton's second law is not a standard statement of Newton's laws of motion - from where the OP aims to obtain the conservation of angular momentum. Unless you show how the rotational equivalent stems from the standard statements of the Newtonian laws, in the light of what the OP wants to ask, this answer seems vacuous. $\endgroup$
    – user87745
    Jul 6, 2017 at 11:13
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    $\begingroup$ @Dvij Well... That this rotational form of the 2nd law is "not a standard statement" is not very relevant. Standard statement or not, it is correct. And well-known, so should not require derivation every single time. I do not find such harsh response fair on this very well setup answer. +1 from me to the answer. $\endgroup$
    – Steeven
    Jul 6, 2017 at 12:19
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    $\begingroup$ This is not the standard definition of the inertia tensor. The inertia tensor is typically defined in the frame. Moreover, this answer completely misses the importance of the strong form vs the weak form of Newton's third law. $\endgroup$ Jul 7, 2017 at 1:55
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    $\begingroup$ @Rumplestillskin -- This is standard fare for the undergraduate physics major class on classical mechanics (a course one takes after those first three introductory physics classes). The answer by CR Drost in the duplicate covers this completely. In that answer, CR also notes that angular momentum is not necessarily conserved if the strong form of Newton's 3rd law does not hold. Angular momentum is still conserved, of course. The resolution is that in the case where the strong form does not hold, the field that creates the force contains angular momentum. $\endgroup$ Jul 7, 2017 at 2:23
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    $\begingroup$ That's not good enough, @Rumplestillskin. Internal forces that do not obey the strong form of Newton's third law result in systems that do not conserve angular momentum. $\endgroup$ Jul 7, 2017 at 2:27

We repeatedly use the fact that a cross product of parallel vectors vanishes. Suppose body $i$ has position $r_i$ and momentum $p_i\parallel\dot{r}_i$, and angular momentum $L_I=r_i\times p_i$ so$$\dot{L}_I=\underbrace{\dot{r}_i\times p_i}_{0}+r_i\times \dot{p}_i=r_i\times \dot{p}_i.$$If body $j$ exerts a force $F_{ij}$ on body $i$, $\sum_i\dot{L}_i=\sum_{ij}r_i\times F_{ij}$.By Newton's third law, this is $\frac{1}{2}\sum_{ij}\left(r_i-r_j\right) \times F_{ij}$. The $ij$ term vanishes if $F_{ij}$ is radial.


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