Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In classical mechanics with 3 space dimension the 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 angular momentum e.g. in special relativity in terms of 4-vectors?

share|improve this question
en.wikipedia.org/wiki/… –  genneth May 14 '11 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. –  Peter Morgan May 14 '11 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" –  Helder Velez May 16 '11 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. –  asmaier May 16 '11 at 20:56
good stuff. Lubos' answer is indeed right on the mark. –  genneth May 17 '11 at 10:34
add comment

1 Answer

up vote 20 down vote accepted

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:


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.

share|improve this answer
This answer actually clears up a different question of mine +1 –  yayu May 15 '11 at 5:55
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. –  Jerry Schirmer May 15 '11 at 18:50
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.