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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?

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    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – genneth
    Commented May 14, 2011 at 22:18
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    $\begingroup$ 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. $\endgroup$
    – Peter Morgan
    Commented May 14, 2011 at 22:42
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    $\begingroup$ 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" $\endgroup$
    – Helder Velez
    Commented May 16, 2011 at 10:45
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    $\begingroup$ @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. $\endgroup$
    – asmaier
    Commented May 16, 2011 at 20:56
  • $\begingroup$ good stuff. Lubos' answer is indeed right on the mark. $\endgroup$
    – genneth
    Commented May 17, 2011 at 10:34

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Zo the Relativist
    Commented May 15, 2011 at 18:50
  • $\begingroup$ 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? $\endgroup$
    – Gerard
    Commented Sep 11, 2017 at 10:53
  • $\begingroup$ 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"? $\endgroup$
    – Luboš Motl
    Commented Sep 11, 2017 at 16:23
  • $\begingroup$ $L$ 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. $\endgroup$
    – user76284
    Commented Oct 16, 2020 at 16:23

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