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Relativistic center of mass

+ 6 like - 0 dislike
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Recently I realized the concept of center of mass makes sense in special relativity. Maybe it's explained in the textbooks, but I missed it. However, there's a puzzle regarding the zero mass case

Consider any (classical) relativistic system e.g. a relativistic field theory. Its state can be characterized by the conserved charges associated with Poincare symmetry. Namely, we have a covector $P$ associated with spacetime translation symmetry (the 4-momentum) and a 2-form $M$ associated with Lorentz symmetry. To define the center of mass of this state we seek a state of the free spinning relativistic point particle with the same values of conserved charges. This translates into the equations

$$x \wedge P + s = M$$ $${i_P}s = 0$$

Here $x$ is the spacetime coordinate of the particle and $s$ is a 2-form representing its spin (intrinsic angular momentum). I'm using the spacetime metric $\eta$ implicitely by identifying vectors and covectors

The system is invariant under the transformation

$$x'=x+{\tau}P$$

where $\tau$ is a real parameter

For $P^2 > 0$ and any $M$ these equations yields a unique timelike line in $x$-space, which can be identified with the worldline of the center of mass of the system. However, for $P^2=0$ the rank of the system is lower since it is invariant under the more general transformation

$$x'=x+y$$ $$s'=s-y \wedge P$$

where $y$ satisfies $y \cdot P = 0$

This has two consequences. First, if a solution exists it yields a null hyperplane rather than a line*. Second, a solution only exists if the following constaint holds:

$$i_P (M \wedge P) = 0$$

Are there natural situations in which this constaint is guaranteed to hold? In particular, does it hold for zero mass solutions of common relativistic field theories, for example Yang-Mills theory? I'm considering solutions with finite $P$ and $M$, of course

*For spacetime dimension $D = 3$ a canonical line can be chosen out of this hyperplane by imposing $s = 0$. For $D = 4$ this is in general impossible

This post has been migrated from (A51.SE)
asked Jan 21, 2012 in Theoretical Physics by Squark (1,700 points) [ no revision ]
Zero mass case is not obligatory a zero energy or zero energy density case. In relativistic case it is the energy that replaces the mass in the center of mass definition (center of inertia, if you like). If you have Landau-Lifshitz textbook on classical fields, then you may find there a definition of CI involving not only particle energies, but also the energy of their interactions and the field energy.

This post has been migrated from (A51.SE)
@Vladimir, I am referring to zero rest mass. My definition includes all these things since it involves the conserved charges of the entire system

This post has been migrated from (A51.SE)

1 Answer

+ 1 like - 0 dislike

This condition is due to the fact that for a free massless particle the Pauli-Lubanski vector $W=*(M\wedge P)$ must be proportional to the linear momentum (The proportionality factor being the helicity). Thus the condition must be valid to all free massless relativistic field theories.

This post has been migrated from (A51.SE)
answered Jan 22, 2012 by David Bar Moshe (3,505 points) [ no revision ]
In this case it probably holds as an identity in the universal enveloping algebra of the Poincare Lie algebra modulo the ideal generated by $P^2$. It would be cool to prove it directly. However, even this doesn't formally imply it holds in any *classical* field theory, although it would be very strange if it doesn't. Hence I suspect there must be a way to prove it for an arbitrary classical field theory in the Lagrangian formalism

This post has been migrated from (A51.SE)

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