It is a well-known result that angular momentum conservation is related to the invariance of the Lagrangian respect spatial rotations, here a demonstration of how infinitesimal rotations do not alter the Lagrangian is done. I will reproduce part of it. Namely an infinitesimal rotation $\delta \vec{\phi}$ causes a displacement of $\delta \vec{r}=\delta \vec{\phi} \times \vec{r}$ and a change in velocity of $\delta \vec{v}=\delta \vec{\phi} \times \vec{v}$.

These changes in position and velocity change the Lagrangian as:

$$\delta L=\frac{\partial L}{\partial \vec{r}}\cdot \delta \vec{r} + \frac{\partial L}{\partial \vec{v}}\cdot \delta \vec{v}.$$

Since $\frac{\partial L}{\partial \vec{v}}=\vec{p}$ and $\frac{\partial L}{\partial \vec{r}}=\dot{\vec{p}}$, and also $A\cdot(B\times C)=B\cdot(C\times A)$, the Lagrangian variation can be written as:

$$\delta L=\delta \vec{\phi}\cdot(\vec{r}\times\dot{\vec{p}} + \vec{v}\times\vec{p})$$

This is the same as $\delta \vec{\phi}\cdot\frac{d}{dt}(\vec{r}\times\vec{p})$, therefore, if the Lagrangian is conserved with infinitesimal rotations, the angular momentum must be conserved too.

However, when the linear momentum is locally conserved, it seems to me redundant to specify that the angular momentum is conserved since, for each point $\vec{r}$ is just a constant, therefore, if $\vec{p}$ is conserved, $\vec{r}\times\vec{p}$ must be conserved too.

I see the value of angular momentum conservation in non-local theories such as action at distance theories in which the force is exerted directly between separated particles and therefore if the force is not radial, it would cause a torque in the system breaking the angular momentum conservation.

I would like to know if angular momentum is important in theories in which the energy and momentum are conserved locally and if there is some conserved current for angular momentum as should be expected when a quantity is locally conserved.

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    $\begingroup$ for each point $\vec{r}$ is just a constant -- When a particle has non-zero momentum, the position is not a constant anymore. $\endgroup$ May 20 at 20:02

1 Answer 1


I) With the phrase the linear momentum is locally preserved OP presumably implies that there is no external force on each particle.

  1. If the only forces are internal forces that satisfy the weak Newton's 3rd law $$\vec{F}_{ij}+\vec{F}_{ji}~=~\vec{0},\tag{1}$$ then the total momentum $$\vec{p}_{\rm tot}~:=~ \sum_{i=1}^N \vec{p}_i\tag{2}$$ is conserved: $$ \dot{\vec{p}}_{\rm tot} \stackrel{(2)}{=} \sum_{i=1}^N \dot{\vec{p}}_i ~=~\sum_{i,j}^{i\neq j}\vec{F}_{ij} ~\stackrel{(1)}{=}~\vec{0}. \tag{3}$$

  2. If furthermore the internal forces are collinear, i.e. satisfy the strong Newton's 3rd law $$\vec{F}_{ij} ~\parallel ~\vec{r}_i-\vec{r}_j,\tag{4}$$ and if $$~\vec{p}_i \parallel ~\dot{\vec{r}}_i,\tag{5}$$ then the total angular momentum $$\vec{L}_{\rm tot} ~:=~ \sum_{i=1}^N \vec{r}_i\times\vec{p}_i\tag{6}$$ is conserved: $$\begin{align} \dot{\vec{L}}_{\rm tot} ~\stackrel{(5)+(6)}{=}&~ \sum_{i=1}^N \vec{r}_i\times\dot{\vec{p}}_i ~=~\sum_{i,j}^{i\neq j}\vec{r}_i\times\vec{F}_{ij}\cr ~\stackrel{(1)}{=}~&\sum_{i,j}^{i< j}(\vec{r}_i-\vec{r}_j)\times\vec{F}_{ij}~\stackrel{(4)}{=}~\vec{0}. \end{align} \tag{7}$$

II) More generally, in field theory the angular momentum current $$M^{\mu\alpha\beta}~=~x^{\alpha}T^{\mu\beta}-x^{\beta}T^{\mu\alpha}\tag{8}$$ satisfies a continuity equation $$ d_{\mu}M^{\mu\alpha\beta}~\stackrel{(8)+(10)+(11)}{=}~0\tag{9} $$ if the SEM tensor $T^{\mu\nu}$ is

  1. symmetric $$ T^{\mu\nu}~=~ T^{\nu\mu}\tag{10} $$
  2. and satisfies a continuity equation $$ d_{\mu}T^{\mu\nu}~=~0.\tag{11} $$

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