After reading Wikipedia, I'm confused by the relativistic angular momentum definition. OK for the 4-angular momentum tensor. But does it mean that the following more intuitive "angular momentum" will not be exactly conserved at high speeds in a reference frame at rest? $${\bf M} = \sum_i r_i\times {\bf p_i},$$ where $\bf p_i$ is the relativistic momentum $\gamma m_iv_i$, for a system of masses $m_i$.
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Related : Vector product in a 4-dimensional Minkowski spacetime, $\color{blue}{\boldsymbol{\S 4-}\textbf{In the 4-dimensional Minskowski space-time}}$ – Frobenius May 11 '22 at 14:19
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For Minkowski or Schwartzschild spacetimes, the quantity $$m\left(X^i\frac{dX^j}{d\tau} - X^j\frac{dX^i}{d\tau}\right)$$ is conserved for masses following geodesic trajectories. It results from the existence of some Killing vectors.
In the Minkowski spacetime, the geodesics are straight lines, and it is the trivial fact that the relativistic angular momentum is just the distance to the line multiplied by the linear relativistic momentum (that is also conserved).
In the Schwartzschild spacetime, it means that the conservation of angular momentum of classical eliptical orbits is an approximation to the conservation of the relativistic angular momentum. Here it is supposed one big mass M, and only one small orbiting mass m, where M>>m.

Claudio Saspinski
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Thx. When you say " it is the trivial fact that the relativistic angular momentum is just the distance to the line multiplied by the linear relativistic momentum", multiplied means "cross product", that is ${\bf r}\times {\bf p}$ ? In other words, your answer is "yes, the quantity $\bf M$ in your question is conserved"? – MikeTeX May 13 '22 at 07:52
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1Yes, it is conserved. But for a particle in a geodesic, that is in this case (Minkowski) a straight line. – Claudio Saspinski May 13 '22 at 13:34