I think that you misunderstood slightly. When you say that S^{ab} u_{b}=0 for all u, it sounds like you're considering S to be fixed while u is varied. This is not correct. S itself depends on both the base point and u. The latter quantity effectively defines a preferred time direction. So there is only one constraint.
There is no lack of invariant quantities either. S^{a} S_{a} is a scalar invariant, for example. More generally, S^{ab} S_{ab} is always an invariant no matter whether we're using center-of-mass definitions or not. These quantities are related to the square of the angular momentum operator in quantum mechanics. In any case, here are some details of the classical theory:
As mentioned, the angular momentum depends on both a choice of base point and a preferred time direction. This would usually (though not always) be dealt with by defining a 'reference' worldline z^{a}(s). Of course, this has a 4-velocity u^{a}(s) = \dot{z}^{a}(s) (assume it is normalized, so s is a proper time for someone moving along z). Now, the direction of the 4-velocity is naturally interpreted as the time direction of an observer on z, so define a set of spacelike hyperplanes \Sigma(s) that intersect z^{a}(s), and are orthogonal to u^{a}(s). These can be taken as naturally-defined 'moments of time.' Note, however, that these surfaces will intersect each other if they are extended too far (unless \dot{u}^{a} =0).
Now, both the linear and angular momenta are naturally defined as objects on z. In terms of 'normal' variables, though, they are integrals over the hyperplanes. The most common definitions (again, these are not universal) look like
<br />
p^{a}(s) = \int_{\Sigma(s)} T^{ab}(x) d\Sigma_{b} ,<br />
<br />
S^{ab}(s) = \int_{\Sigma(s)} \left(x-z(s) \right)^{[a} T^{b]c}(x) d\Sigma_{c} .<br />
In the quantum theory, particles have intrinsic spins that can add extra terms here.
Anyway, you can see that the angular momentum depends on the base point z, but also on the surface of integration (defined here in terms of z and u). It is therefore instructive to rewrite these quantities more abstractly as p^{a}\left(\Sigma(u,z) \right) and S^{ab}\left(z, \Sigma(u,z) \right).
Let's now define a center-of-mass. This fixes the reference line, so let's forget about the one we've assumed to exist for now. Instead, choose a single point z(s_{0}). Then it can be shown that under reasonable conditions, there exists a unique vector u^{a} such that
<br />
p^{a} \left( \Sigma(z,u) \right) \propto u^{a}(z)<br />
From looking at the integral form of p, you can see that this is actually a highly implicit definition. In any case, given any z, it fixes a u.
We now need to find a way to pick out a unique z. This is done with
<br />
S^{ab} \left( z, \Sigma(z,u) \right) p_{b} \left( \Sigma(z,u) \right) = S^{ab} u_{b} =0<br />
Again, this is extremely implicit, but it works. Using the integral expression for the angular momentum, you can show that this is equivalent to the usual expression defining a center-of-mass expression:
<br />
\int_{\Sigma} (x-z)^{a} \left( u_{b} u_{c} T^{bc} \right) d\Sigma =0<br />
where the quantity in parentheses is usually interpreted as the mass density seen by on observer with velocity u.
Anyway, taking these two conditions together defines a unique reference line usually called the center of mass. Rewriting the integral for S^{ab} so that it gives S^{a} shows that this quantity is basically an integral over the cross products \bf{r} \times \bf{\mathcal{P}}, where \bf{\mathcal{P}} is a momentum density. This is just as it is in Newton's theory.
We also have (at least) two natural scalar invariants in this formalism:
m^{2} := -p^{a} p_{a}
|S|^{2} := S^{a} S_{a}
Neither of these quantities are necessarily constants (ie independent of s) unless there are no forces acting on the system.
Continuing in the next post...
