Rena Cray said:
From my (rather renegade) point of view classical extended particles should be understood in terms of fields
There is one natural way to view classical extended objects in a space-time ##(M,g_{ab})## and it is microscopic in nature. By considering the constituent particles making up the object and working with a system in which the worldlines of the constituent particles don't intersect, we can note that the worldlines give rise to a vector field ##u^a## that encodes the local kinematical aspects of the extended object and physically represents the 4-velocity field of the constituent particles; for all intents and purposes, the congruence associated with ##u^a## is basically the space-time description of the extended object.
Letting ##\nabla_a## be the derivative operator associated with ##g_{ab}##, and letting ##h_{ab} = g_{ab}+ u_a u_b## be the spatial 3-metric relative to ##u^a##, we can define the aforementioned kinematical quantities as ##\theta_{ab} = h_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{(c}u_{d)}## and ##\omega_{ab} = h_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{[c}u_{d]}##.
##\theta_{ab}## is called the expansion tensor and it loosely speaking takes into account the relative radial velocities between constituent particles making up the extended object: if ##\xi^a## is an infinitesimal displacement vector between neighboring particles in the congruence then ##u^a \nabla_a (\xi^b\xi_b) = 2\theta_{ab}\xi^a \xi^b##.
##\omega_{ab}## on the other hand is called the rotation tensor and it takes into account the relative angular velocities between constituent particles. In fact if we define the twist ##\omega^{a} = \frac{1}{2}\epsilon^{abcd}u_{b}\omega_{cd}## then it is easy to show that ##\omega^{a} = \epsilon^{abcd}u_{b}\nabla_{c}u_{d}## and in the instantaneous rest frame of ##u^a## this is nothing more than ##\vec{\omega} = \vec{\nabla}\times \vec{u}##.
This is a kinematical decomposition because we can express the covariant derivative of the 4-velocity field as ##\nabla_a u_b = \theta_{ab} + \omega_{ab} + u_a a_b## where ##a^b = u^c\nabla_c u^b## is the 4-acceleration field. Using this essentially hydrodynamical formulation, the kinematics of extended objects in space-times can be worked out in principle.
Rena Cray said:
Any ideas on invariant mass of fields would be appreciated.
Continuing with what was said above, if we can represent an extended object in terms of, for simplicity, a perfect fluid with 4-velocity field ##u^a## then this has associated with it an energy-momentum tensor field ##T_{ab}## and ##\rho = T_{ab}u^a u^b## is the mass density in the instantaneous rest frame of ##u^a##; this is basically the analogue of ##m^2 = -p_a p^a## for a massive test particle.
Also, say we have a space-time ##(M,g_{ab})## with a time-translation symmetry; this time translation symmetry generates a time-like killing field ##\xi^a##. Note then that ##T_{ab}\xi^b## results in a conserved current because ##\nabla^{a}(T_{ab}\xi^{b}) = T_{(ab)}\nabla^{[a}\xi^{b]} + \xi^{b}\nabla^{a}T_{ab} = 0##.
If we now assume that ##T_{ab}## has compact support then ##E = \int_{\Sigma} T_{ab}n^{a}\xi^{b}## defines a globally conserved energy associated with ##T_{ab}## in the sense that ##E## is independent of the space-like hypersurface ##\Sigma## as a consequence of the divergence theorem.
Hope that helps somewhat xP