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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?

Qmechanic
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Kenshin
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2 Answers2

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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.

Answer:

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, $$ \frac{d L}{dt} \equiv 0. $$

Rumplestillskin
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  • -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. –  Jul 06 '17 at 11:13
  • @Kenshin I would have immediately followed the same route as the other answer. In fact this is what I normally do. I just thought the answer given above would show it from a different light! Cheers. – Rumplestillskin Jul 06 '17 at 11:58
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    @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. – Steeven Jul 06 '17 at 12:19
  • @Steeven I agree. I provided a derivation pre-answer. – Rumplestillskin Jul 06 '17 at 12:22
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    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. – David Hammen Jul 07 '17 at 01:55
  • @DavidHammen I'd be interested to see how you would go about it. Please enlighten us. – Rumplestillskin Jul 07 '17 at 02:09
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    @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. – David Hammen Jul 07 '17 at 02:23
  • @DavidHammen I too note that it is not necessarily conserved. It is conserved in the absence of external forces and, hence the torque is zero in an isolated system. – Rumplestillskin Jul 07 '17 at 02:26
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    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. – David Hammen Jul 07 '17 at 02:27
  • @DavidHammen I take your point. However, the question from the OP is how to show angular momentum is conserved in an isolated system starting with Newton's laws. I think my answer has sufficiently done so. – Rumplestillskin Jul 07 '17 at 02:31
  • Regardless of the issues discussed here, the argument is specious for a more basic reason: = ≡ × is the wrong formula! It's = + . No "definition" of the form = can be consistently maintained since it is not additive; i.e. × = ₀×₀ + ₁×₁ is generally false, when ₀ + ₁ = , m = m₀ + m₁ and m = m₀₀ + m₁₁. Instead, it's × + = ₀×₀ + ₀ + ₁×₁ + ₁ with = ₀ + ₁ + μ×, where = ₁ - ₀, μm = (m₀₁ - m₀₁), μm = m₀m₁. There is no law that entails d₀/dt = -d₁/dt. Therefore, d/dt = does not follow! – NinjaDarth Sep 24 '23 at 10:44
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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.

J.G.
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