What's the intuitive physical difference between the conservations of momentum and of the boost generator in SR? This is a soft question about classical special relativity (although a related question applies even to nonrelativistic classical mechanics).
The (connected) symmetry group of Minkowski space is the Poincare group, which is 10-dimensional and so has 10 generators corresponding to conserved quantities:

*

*energy (corresponding to time translational invariance)

*three components of spatial momentum (corresponding to space translational invariance)

*three components of angular momentum (corresponding to spatial rotational invariance)

*three generators which don't have a standard name (corresponding to boost invariance).

Spatial momentum and the boost generators correspond to completely different symmetries of spacetime, so there is no a priori relation between them. And on paper, the expressions look totally different: spatial momentum is given by
$$
P^i = \int d^3{\bf x}\ T^{0i}(x),
$$
where $T^{\mu \nu}$ is the stress-energy tensor, while the boost generators are given by
$$
M^{0i} = \int d^3{\bf x}\ \left[ x^0 T^{0i}(x) - x^i T^{00}(x) \right]
$$
or in more explicitly frame-dependent notation,
$$
{\bf N} = \int d^3{\bf x}\ \left[ t\ {\bf p}(x) - {\bf x}\ \epsilon(x) \right] \\
=  t\ {\bf P} - \int d^3{\bf x}\ \left[ {\bf x}\ \epsilon(x) \right]
$$
where ${\bf N} = M^{0i}$ is the boost Noether charge, ${\bf p}(x) := T^{0i}(x)$ is the momentum density, and $\epsilon(x) := T^{00}(x)$ is the energy density. The verbal description given to this Noether charge is usually that it reflects "the linear motion of the center of mass(/energy) of the system".
I understand all of the math fine, but I've never had any physical intuition for the difference between "momentum is conserved" and "the center of mass moves with constant linear motion". To me, these seem like such closely related statements that I can't really intuit the physical distinction. (I believe that the reason that the boost generators are rarely discussed is that they don't give you much physical intuition beyond conservation of momentum, in addition to the awkward fact of the explicit time-dependence). And yet, these two conservation laws have mathematically different forms and they correspond to completely different symmetries of Minkowski space, so it seems to me that they should have quite distinct physical content.
Can anyone help me out with how to intuit the difference between these two conservation laws? Or here's a more concrete way to pose the question: are there any semi-natural examples of Lagrangians that have one symmetry but not the other, so that one of these two quantities is conserved but the other isn't? If so, what do those systems look like?
There's a nonrelativistic analogue of this question where you consider Galilean invariance instead of Poincare invariance, but  I think that's a little less natural to think about.
 A: First, on a practical level, $M_{0i}$ does have information not contained in the momentum. This is clearest in the frame where the momentum is zero; the momentum vanishes, but the position of the center of mass (first moment of the energy density) does not. However, it is true that the time derivative of $M_{0i}$ vanishing implies the time derivative of $P_i$ vanishes.
Second, as far as I know, there is no requirement that Noether charges for different symmetries are independent of each other, in the sense that conservation of one does not imply conservation of another. While I can't think of any other examples like this offhand, there's no contradiction with Noether's theorem.
Third, I think it's useful to look at this from the perspective of the Poincaire algebra. The fact that the theory is Poincaire invariant means that (speaking quantum mechanically) the fields must be unitary representations of the Poincaire group. This implies there must be a set of operators that $P_\mu$, $M_{\mu\nu}$ that obey (using the conventions on wikipedia)
\begin{eqnarray}
[P_\mu, P_\nu] &=& 0\\
[M_{\mu\nu},P_\sigma] &=& i\left(\eta_{\mu\sigma} P_\nu - \eta_{\nu\sigma} P_\mu\right) \\
[M_{\mu\nu}, M_{\rho \sigma}] &=& i \left(\eta_{\mu\rho} M_{\nu\sigma} - \eta_{\mu\sigma} M_{\nu \rho} - \eta_{\nu\rho} M_{\mu\sigma} + \eta_{\nu\sigma} M_{\mu \rho}\right)
\end{eqnarray}
Of course, these operators are just the Noether charges associated with translations, rotations, and boosts.
In this sense, it's clear we need separate charges for $P_\mu$ and $M_{0i}$, since they play different roles in the algebra.
We can understand various properties of $M_{0i}$ from the algebra. In quantum mechanics, we normally say that an operator must commute with the Hamiltonian to be conserved. However, of course $M_{0i}$ does not commute with $P_0$. The loophole is that $M_{0i}$ is explicitly time dependent, and the explicit time dependence cancels the commutator with $P_0$ in the Heisenberg equation of motion.
From this algebra, we can determine that representations are determined entirely by their mass, spin, and any number of internal quantum numbers that are invariant under spacetime translations. These are the true "invariant" quantities of the system. Other quantities (in particular the momentum) label specific states of the system. The Noether currents must obey the above algebra and depend only on the mass, spin, and momenta.
For spin-$0$ particles, the only non-trivial invariant (ignoring internal quantum numbers) is the mass, so in fact the explicit expression for $M_{0i}$ cannot contain any more invariant information than what is already present in $P_\mu$. There is some "state dependent" information that appears the moment of the energy density; this term is needed to get the right commutation relations with $P_\mu$, and corresponds to the first point in the answer.
Therefore, from the point of view of the the algebra, we see that $M_{0i}$ needs to be a different operator from $P_\mu$ for things to hang together.
Finally, on a different note, I think that a non-maximally symmetric FLRW spacetime would satisfy your requirement of conserved spatial momentum but not a conserved boost.
