# Is it possible to have a physics where momentum conservation holds, but not angular momentum conservation?

In most basic physics classes, one learns that $F=ma$, then that in an isolated system total momentum is conserved because its time derivative is just the net force which is zero for an isolated system. And later you learn that $T=Ia$, so an isolated system conserves its angular momentum, because its time derivative is the total torque, which can be worked out to be zero when the force is radial (ie. not magnetic). Conservation of momentum and of angular momentum are two distinct postulates that hold in physical systems (when applied correctly), and if you never learned about angular momentum or about rotations, you would probably think that $F=ma$ would get you a full understanding of all of physics. Obviously this is not true in the real world.

If you know about Lagrangians, then you can see that doing this magically produces f=ma, or rather that $ma=-\mathrm{d}V/\mathrm{d}x$ etc. and that this agrees with 'all the laws of physics' (this is what Morin did in his textbook). But then if you move from cartesian to polar coordinates you find that theta is a free variable (forgetting the terminology here, sorry), then you discover that L (angular momentum) is conserved. So if we were trying to make a new way of saying that $ma=-\mathrm{d}V/\mathrm{d}x$ and came up with the Lagrangian idea, we'd accidentally get that angular momentum is conserved as well. We can't get linear momentum conservation without producing angular momentum conservation as well.

But is there any internally self-consistent system of physics in which linear momentum conservation is true (and/or derivable from some internal structure like $F=ma$), but angular momentum isn't? Such a physics would not describe the real world very well, but what kind of results would it produce? From my understanding, momentum conservation is a byproduct of invariance of space under linear translations, and angular momentum under rotations; what would a universe where the first works but the second doesn't, look like?

## 3 Answers

To answer your question in a word: yes! There are physical systems in which linear momentum is conserved but angular momentum is not conserved. Such systems aren't intuitive at the level of classical Lagrangian mechanics, but they are abound in the electromagnetic theory of materials.

First, let's get ourselves a deeper understanding of conservation laws. The true power of the Lagrangian formalism (and something that is even more clear in the Hamiltonian formalism) is the relationship between symmetries and conservation laws. You've most likely heard of this statement before, but I'll restate it for you here:

• Noether's Theorem: Continuous symmetries in a physical system generate conserved quantities.

This is just about one of the most important ideas in all of physics, so even if you've heard it before, it's worth hearing it again. I'm not going to dive into the details here (you can find them in an enormous number of references). For now, I will just list a few symmetries and their corresponding conservation laws:

• Time translation symmetry -- the symmetry $t\to t+t_0$ leads to the conservation of energy.

• Homogeneity -- the symmetry $\textbf{x}\to\textbf{x}+a$ leads to the conservation of momentum.

• Rotational symmetry -- the symmetry $\textbf{x}\to R_{\hat{\textbf{n}}}(\theta)\textbf{x}$, where $R$ is a rotation around the $\hat{\textbf{n}}$ axis for some unit vector $\hat{\textbf{n}}$, leads to the conservation of $L_{\hat{\textbf{n}}}\equiv\hat{\textbf{n}}\cdot\textbf{L}$, where $\textbf{L}$ is the angular momentum (that is, the angular momentum in the $\hat{\textbf{n}}$ direction is conserved).

• Isotropy -- the symmetry $\textbf{x}\to R\textbf{x}$, where $R$ is an arbitrary rotation implies the conservation of the angular momentum $\textbf{L}$.

Thus, Noether's theorem basically allows you to recast your question into the following form:

• Are there any homogeneous but anisotropic systems?

The answer is a resounding yes! As a toy model, simply consider the action

$$S=\int\mathrm{d}t\,A_{ij}\dot{x}^i\dot{x}^j,$$

where the summation over $i$ and $j$ is implied. Clearly, under translations this action is left invariant (there are no terms without derivatives in $\textbf{x}$). However, under rotations, it transforms as

$$S\to\int\mathrm{d}t\,A_{ij}R^{i}_{\,\,a}R^{j}_{\,\,b}\dot{x}^a\dot{x}^b,$$

which is not the same unless $A_{ij}R^{i}_{\,\,a}R^{j}_{\,\,b}=A_{ab}$ for all rotations $R$, which is not the case unless $A\propto\textbf{1}$ (by the definition of orthogonal matrices). Thus, we can't have that $\textbf{L}$ is conserved for general $A$. [Note: it is still perfectly possible (and I think inevitable, actually) that some projection of $\textbf{L}$ will be conserved in this model.]

Such a universe would be strange indeed if large objects obeyed this action principle. However, there are physical systems which aren't far off from this form. In optical physics, people work all the time with systems whose permittivity constant is an anisotropic tensor. Essentially this means that Ohm law, instead of taking the familiar form $\textbf{J}=\sigma\textbf{E}$ takes the modified form

$$J_i=\sigma_{ij}E_j,$$

with a summation over $j$ implied. In these situations, the total angular momentum of a single electromagnetic wave has no reason to be conserved. However, the system is still homogeneous, so that momentum is conserved!

Thank you for the great question. I hope this helped!

• I disagree a bit on the Lagrangian model if $A_{ij}$ is positively defined as expected. In this case you can diagonalize that quadratic form via Sylvester's theorem obtaining $\delta_{ij}$ replacing coordinates $x$ for coodinates $y$ linearly related. The procedure does not depend on the place as the system is supposed to be homogeneous. With respect to the new coordinates angular momentum is conserved. The point is the physical meaning of the used coordinates. – Valter Moretti Jun 20 '17 at 4:27
• Even adding a potential does not change the result because homogeneity requirement imolies that it is constant. – Valter Moretti Jun 20 '17 at 4:33
• The point is that the arising "angular momentum" has no the standard meaning because diagonalised coordinates are adapted to a particular physical system. For instance this angulsr momentum in general cannot be added to angular momenta of other systems. – Valter Moretti Jun 20 '17 at 4:41
• Very good points. I will edit accordingly in the morning. – Bob Knighton Jun 20 '17 at 4:47
• Thanks. Sorry, I am writing by my phone, thus inserting many typos. I hope the text is understandable... – Valter Moretti Jun 20 '17 at 4:52

Yes indeed. In Lagrangian mechanics, we see that momentum conservation is a consequence of translation invariance. The fact that every point is the same. Angular momentum conservation comes from rotation invariance. The fact that every direction is the same.

Can you have every point look the same without having every direction look the same. Yes, you can. Just draw in a little arrow at every point, all going in the x-direction. Space now has a special direction, but no special points.

Physically, we can realize this with a uniform electric field. What happens if we drop an electric dipole in there? Two charges $q$ and $-q$ at positions $r_1$ and $r_2$, constrained to be distance $d$ apart. The dipole feels a torque, but no net force. This is equivalent to saying that the system of a dipole with energy given by $Eq(x_1-x_2)$ has conservation of momentum but not angular momentum.

(Fun fact: this can't go the other way. Any system with conservation of $L$ must have conservation of $p$.)

• I like your examples, but it's definitely not true that angular momentum conservation implies momentum conservation. For instance, consider a gravitational point mass at the origin. The angular momentum around the origin is conserved but linear momentum is definitely not conserved. It is true, however, that if angular momentum calculated with respect to any point is conserved than linear momentum is conserved (at least to my knowledge). – Bob Knighton Jun 20 '17 at 4:09
• @BobKnighton yes, and I believe the argument is straightforward. If $\vec{r}\times m\vec{v}$ is conserved for every $\vec{r}$, then it must be true that $m\vec{v}$ itself is conserved. – user1936752 Jun 20 '17 at 7:17

It might be interesting to notice that a common case where momentum conservation holds, but not angular momentum conservation, is the world of all the numerical simulations using periodic boundary conditions.