# How to define orbital angular momentum in other than three dimensions?

In classical mechanics with 3 space dimensions the orbital angular momentum is defined as

$$\mathbf{L} = \mathbf{r} \times \mathbf{p}.$$

In relativistic mechanics we have the 4-vectors $$x^{\mu}$$ and $$p^{\mu}$$, but the cross product in only defined for 3 dimensions. So how to define orbital angular momentum e.g. in special relativity in terms of 4-vectors? Or more generally in $$d$$ dimensions?

• en.wikipedia.org/wiki/… Commented May 14, 2011 at 22:18
• In classical relativistic field theory, there is an object called the Pauli-Lubanski vector which reduces to ordinary 3-dimensional angular momentum in the rest frame of the system (Google for this term unfortunately doesn't seem to find any elementary web page). There is also a generalized angular momentum tensor (of 3rd rank), which is constructed using the symmetric energy momentum tensor (which is of 2nd rank). Manifest Lorentz invariance is possible. Commented May 14, 2011 at 22:42
• very interesting: Relativistic Angular Momentum by Nick Menicucci, 2001 "Its relation to its 3-vector .. resulting consequence of uniform motion of the centroid. .. the most striking being the inability to compress a system of particles to inﬁnitesimal size, requiring new thoughts on just what “a point-particle with spin” really is. The spin vector and Pauli-Lubanski vector were discussed, The Thomas precession was explained and calculated, and two “paradoxes” involving torque and angular momentum were explored" Commented May 16, 2011 at 10:45
• @genneth I found the Wikipedia explanation "Angular momentum is the 2-form Noether charge associated with rotational invariance" not very helpful. So I added to the Wikipedia article the definition of the angular momentum as antisymmetric tensor of second order as explained by Lubos. Commented May 16, 2011 at 20:56
• good stuff. Lubos' answer is indeed right on the mark. Commented May 17, 2011 at 10:34

Dear asmaier, you shouldn't view $\vec L = \vec x \times \vec p$ as a primary "definition" of the quantity but rather as a nontrivial result of a calculation.

The angular momentum is defined as the quantity that is conserved because of the rotational symmetry - and this definition is completely general, whether the physical laws are quantum, relativistic, both, or nothing, and whether or not they're mechanics or field theory.

To derive a conserved charge, one may follow the Noether's procedure that holds for any pairs of a symmetry and a conservation law:

http://en.wikipedia.org/wiki/Noether_charge

In particular, the angular momentum has no problem to be evaluated in relativity - when the background is rotationally symmetric. The fact that you write $\vec L$ as a vector is just a bookkeeping device to remember the three components. More naturally, even outside relativity, you should imagine $$L_{ij} = x_i p_j - x_j p_i$$ i.e. $L_{ij}$ is an antisymmetric tensor with two indices. Such a tensor, or 2-form, may be mapped to a 3-vector via $L_{ij} = \epsilon_{ijk} L_k$ but it doesn't have to be. And in relativity, it shouldn't. So in relativity, one may derive the angular momentum $L_{\mu\nu}$ which contains the 3 usual components $yz,zx,xy$ (known as $x,y,z$ components of $\vec L$) as well as 3 extra components $tx,ty,tz$ associated with the Lorentz boosts that know something about the conservation of the velocity of the center-of-mass.

Incidentally, the general $x\times p$ Ansatz doesn't get any additional "gamma" or other corrections at high velocities. It's because you may imagine that it's the generator of rotations, and rotations are translations (generated by $\vec p$) that linearly depend on the position $x$. So the formula remains essentially unchanged. In typical curved backgrounds which still preserve the angular momentum, the other non-spatial components of the relativistic angular momentum tensor are usually not preserved because the background can't be Lorentz-boost-symmetric at the same moment.

• Also, all asymptotically flat spacetimes preserve a TOTAL angular momentum $L_{I}=\oint d^{2}x K_{ab}r^{a}e^{b}_{I}$, where $e^{b}_{I}$ is the dyad of the surface at infinity, and $K_{ab}$ is the extrinsic curvature of the 3-surface in the 4-spacetime, and the integral is over the intersection of the 3+1 slice and conformal spacelike infinity. There just wont' be any general, coordinate-invariant local angular momentum current in these spacetimes. Commented May 15, 2011 at 18:50
• An old page but .. Looking at $L_{ij} = x_i p_j - x_j p_i$ I notice that $x$ and $p$ are dual entities for which one would expect a contraction like in $P=F\cdot v$ ($F=\dot{p}$, $v=\dot{x}$) and not an exterior product like here? E.g. $v_i F_j - v_j F_i$ doesn't mean anything does it? Commented Sep 11, 2017 at 10:53
• Any two 3-vectors have some cross product. What's your problem with it? The inner product of x,p may also mean something - it's a generation of dilations - but the generators of rotations is the cross product. The cross product of the force and velocity could also have some importance in physics - at any rate, you can surely calculate it or define it, can't you? Why do you think or in what sense a well-defined cross product of two vectors "doesn't mean anything"? Commented Sep 11, 2017 at 16:23
• $L$ can be expressed concisely as the wedge product $x \wedge p$. The wedge product is a beautiful mathematical concept that should be taught more often. I wrote a related answer about it here. Commented Oct 16, 2020 at 16:23