Previous discussions on this forum regarding the derivation of the law of conservation of angular momentum from Newton's Laws have pointed out that it supposes the strong form of Newton's Third Law. My question concerns why that law is reasonable for contact forces in rigid bodies and in fluids. To illustrate the form of the law I'm talking about, I reproduce the derivation of the law of conservation of angular momentum for a system of point particles: $$\vec{F_i} + \sum_j \vec{G_{ji}} = m_i \vec{a_i}$$ where $\vec{F_i}$ is the net external force on the $i$th particle, and $\vec{G_{ji}}$ is the force on the $i$th particle from the $j$th particle.

To go from change in linear momentum to angular momentum, we take a cross product with the position of this point particle: $$\vec{r_i} \times\vec{F_i} +\sum_j (\vec{r_i}\times\vec{G_{ji}}) = m_i \vec{r_i} \times \vec{a_i} = \frac{d}{dt}\vec{L_i}$$ In other words, the angular momentum of the $i$th particle changes in response to the external torque on that particle and to the torque owing to other particles. We would however like to show that the net angular momentum of an isolated system does not change, i.e, that the interaction of these particles cannot create a net torque internal to the system. In what follows we replace $\vec{r_i} \times \vec{F_i}$ with $\vec{\tau_i}$ for brevity.

Summing up over all particles $i$, we get: $$\frac{d}{dt}\vec{L} = \sum_i\frac{d}{dt}\vec{L_i} = \sum_i\vec{\tau_i} + \sum_i\sum_j(\vec{r_i}\times\vec{G_{ji}})$$ Recalling that action and reaction are equal and opposite, we can pair them up in the last term to obtain the equation: $$\frac{d}{dt}\vec{L} = \sum_i\vec{\tau_i} + \frac{1}{2}\sum_i\sum_j(\vec{r_i}\times\vec{G_{ji}} + \vec{r_j}\times\vec{G_{ij}}) = \sum_i\vec{\tau_i} + \frac{1}{2}\sum_i\sum_j[(\vec{r_i} - \vec{r_j})\times\vec{G_{ji}}]$$

This last term goes to zero assuming the strong form of Newton's third law, which makes an additional claim: that action and reaction are directed along the vector between the point particles. The answers I have seen then go on to talk about how this happens with magnetic forces and that it is resolved by considering the angular momentum carried by the electromagnetic fields. But what about contact forces in rigid bodies?

Consider a door being swung open about its hinge. I push the door, exerting a tangential force, at its extreme end away from the hinge. The door moves as a rigid body, with all points acquiring some tangential momentum. Thus, the internal interactions in the extended body between two adjacent points (in the plane of the door) were able to communicate a force in the direction normal to that plane. This is exactly the kind of force$-$satisfying the weak but not the strong form of the third law$-$that could potentially upset our conclusion that an isolated system ($\sum_i\vec{\tau_i}=0$) conserves its angular momentum.

In general, the approach I have seen in continuum mechanics is somewhat different. For example, in fluid mechanics, one decomposes the velocity gradient into a shear component, and a component owing to rigid body rotation. The argument then is that viscosity will only act when there is relative motion between different parts of the fluid. Thus, the viscous forces spring into action to dampen the shear and bring the fluid into a state resembling rigid body rotation or uniform translation. Once this is achieved, angular momentum is indeed conserved (because the offending force stops acting). The other contact force$-$hydrostatic pressure$-$appears to follow the strong form of the law as well and thus cannot create internal net torque.

I have little experience with solids, but I imagine a similar constraint is evoked? That is, do we resolve the existence of nonradial internal forces by saying they only come into the picture to ensure rigid body rotation? How do we then show that the action of these internal forces does not create a net torque? Further, in the solid case the internal forces are assumed to act instantaneously to produce rigid-body rotation, but the viscous fluid takes time to adjust; is the net angular momentum of the fluid conserved during this time? Finally, there seems to be some reason why everyone is focusing on magnetic forces$-$do nonradial contact forces result from microscopic magnetic interactions, if these are the only fundamental interactions to violate the strong form?

Thanks in advance!


1 Answer 1


It's common to assume non-polar media, whose regions don't exchange couples at microscopic level (neighbor material particles in a model where the Continuum hypothesis holds). For these materials, the balance equation of angular momentum is equivalent to the symmetry of the stress tensor: this is true since, in the limit of dimension of a volume going to zero, the leading order terms in the angular momentum equation reads $\sigma_{ij} - \sigma_{ji} = 0$. You can find a derivation here https://gitlab.com/davideMontagnani/fluidmechanics-ita/-/blob/master/qa/qa.pdf?ref_type=heads (download the doc if needed; you find it at page 13; only Italian, but you could follow the math formulas or use some translation service).

More than 99% of engineers and scientist study this model, since non-polar macroscopic media are the most common media in structural mechanics.

The interactions in fluid you're mentioning at the end of your answer don't occur at such a small scale, but on a "finite" scale: you can write the integral form of angular momentum balance equation to draw conclusion about the angular momentum in the region of your interest. In general, angular momentum is not conserved and the equation

\begin{equation} \dfrac{d}{dt} \int_{v_t} \rho \mathbf{r} \times \mathbf{u} + \oint_{\partial v_t} \rho \mathbf{r} \times \mathbf{u} ( \mathbf{u} - \mathbf{u}_{\partial v_t} ) \cdot \mathbf{\hat{n}} = \int_{v_t} \rho \mathbf{r} \times \mathbf{g} + \int_{\partial v_t} \mathbf{r} \times \mathbf{t_n} \ , \end{equation}

holds for an arbitrary volume $v_t$, whose boundary $\partial v_t$ moves with velocity $\mathbf{u}_{\partial v_t}$. Namely, the variation of the angular momentum in the volume equals the sum of the flux of angular momentum through its boundary $\partial v_t$ and the moment of external forces, both volume forces $\rho \mathbf{g}$ and surface stress $\mathbf{t_n}$.

As an example, you can use this equation in the analysis of rotating machines, like turbines or compressors, to link the entity and direction of the fluid flow with torque and power produced or required by the machine.

I'm not an expert about models of continuum media with magnetic forces, but I guess that when there is no contribution of magnetic couples, the influence of this interactions between neighboring material particles is negligible (goes to zero in the limit process as all the other terms of the angular momentum equation); anyway, at larger range, you can find models of continuum media with magnetic effects, as an example in magneto-hydrodynamics.

  • $\begingroup$ I'd like some clarification. You start by saying that the balance of angular momentum would imply symmetry of the stress tensor; do you mean that polar media are not subject to that balance? Specifically, if the stress tensor were asymmetric, would internal forces produce a net torque? Physically, why are such media rare compared to nonpolar media? $\endgroup$ Jan 10 at 11:32
  • $\begingroup$ Balance of angular momentum holds for polar (I'd should check the definition of these media used in continuum mechanics) media as well. Since neighboring material particles exchanges not only elementary forces but also elementary torques, the limit process for the dimension of a volume going to zero gives you a relation between these contributions and not only $\sigma_{ij} - \sigma_{ji} = 0$ (for polar media, there is not a zero but the contribution of torques, and thus you can easily realize that $\sigma_{ij}$ is not symmetric); $\endgroup$
    – basics
    Jan 10 at 12:05
  • $\begingroup$ so yes, forces exchanged between particles produces a net torque that is equilibrated by the exchanged torques. I don't know if such a kind of media are rare in terms of their quantity in the universe or on Earth, but I'm quite sure they're rare if you're studying some classical mechanics or structural mechanics, and thus most of the courses in engineering and physics mailny focuses on the mechanics of non-polar continuous media $\endgroup$
    – basics
    Jan 10 at 12:05
  • $\begingroup$ The proof I reproduce in my answer should ideally show that an isolated system conserves angular momentum. It is usually concluded at the step where I stopped, by assuming that all forces are central forces, and so there is no net internal torque. Can you complete the proof for continuous media, where the stresses show up as non-central forces (i.e, not aligned with the vector between material elements i and j), to show why the balance of angular momentum holds for this general case? $\endgroup$ Jan 10 at 12:51
  • $\begingroup$ if you're only interested for the proof of non-polar media, you can find it in the doc I shared with you, hosted on my github space. Page 13-14. All the conclusions come from the limit process as dimensions goes to zero: the leading terms that dominate gives you the symmetry of the stress tensor $\endgroup$
    – basics
    Jan 10 at 12:56

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