A good question about Noether's theorem that most textbooks do not address: what is the conservation law corresponding to Lorentzian boost?

We first start with some background material.

1. We are interested in working on Minkowski space, i.e. R¹⁺ⁿ with the Minkowski metric m^αβdx_αdx_β given by -c²dt²+∑_1≤i≤ndx_i². We typically normalize by setting c = 1.
2. The symmetries in Minkowski space is given by the Poincaré group, a.k.a. inhomogeneous Lorentz group, which consists of space-time translations, spacial rotations, and Lorentz boosts (rotations about the light-cone; what produces the time-dilation effect in special relativity). These are the isometries on Minkowski space. The Lorentz group has a corresponding Lie Algebra, the generators of which are the following:
   • T^μ = ∂/∂x^μ
   • L_μν = x_μ∂_ν - x_ν∂_μ
   (Notice that we follow the convention of Einstein summation and that lowering and raising of indices being done with respect to the metric.) The vector fields T corresponds to the various translations. The vector fields L corresponds to the Lorentz transformations: the spacial restrictions of which (L_ij,≠0) corresponds to the special orthogonal group (i.e. rotations) on Rⁿ.
3. The above generators of the symmetry groups are also called Killing vector fields. Recall that we can speak of the Lie derivative corresponding to a vector field X, which we denote by L_X. The deformation tensor of X is the two tensor given by L_Xg, where g is the metric of the manifold. We denote the deformation tensor by ^(X)Π. Killing's equation gives that a symmetry in space-time (i.e. one that doesn't change the metric) is generated by a vector field for which the deformation tensor is 0.
4. The energy-momentum tensor Q_μν is a symmetric, divergence free two-tensor that satisfies the positive energy condition (or in some cases, the dominant energy condition) of Q(X,Y) > 0 (≥ 0) for future-oriented time-like vector fields X and Y.

Noether's theorem says that symmetries corresponds to conservation laws. How do we arrive at that?

Consider the quantity P(X)_μ = Q_μνX^ν for an arbitrary vector field X. We look at its divergence.

∂^μP_μ = ∂^μQ_μνX^ν + Q_μν ∂^μX^ν

By the divergence free condition on Q, the first term on the right hand side is zero. By the symmetry condition, we can symmetrize the second term on the right hand side as following:

∂^μP_μ = ½ Q_μν ( ∂^μX^ν + ∂^νX^μ )

We raise the index on Q and lower it on the conjugate portions, and using the definition of the Lie derivative, we see

∂^μP_μ = ½ Q^μν (L_Xg)_μν = ½ Q^{μν (X)}Π_μν

So we see that in the case of X being a Killing v.f. the quantity P=P(X) is divergence free.

We then integrate over a space-time slab. We denote the space-like hyper-surface {s}×Rⁿ by Σ(s)

₀∫^t∫_Σ(s) ∂^μP_μdx ds = 0

by Stokes theorem, we can rewrite the left hand side as an integral over the boundary

∫_Σ(t)P₀ dx - ∫_Σ(0)P₀ dx = 0

which implies that the spacial integral of P₀ is a conserved quantity (doesn't change with time).

Putting in the T vectors for the Killing v.f. we see immediately that P(T^μ)₀ = Q_0μ. For μ = 0, this is the relationship that time translation symmetry equals conservation of energy. For μ ≠ 0, we get spacial translation symmetry equals conservation of momentum.

If we put in the rotational vectors (the spacial L vectors), we see rotational symmetry gives rise to conservation of angular momentum. The problem is, what about the Lorentzian boost?

Let's work it out: the Lorentzian boost is given by L_0i = t ∂_i+ x_i∂_t. Plugging it in to the P relation, we get that

∫Q₀₀x_i + Q_0it

is conserved. Noticing that t is constant on each time-slice Σ(t), we get the following equation

t2 ∫_t2Q_0i - t1 ∫_t1Q_0i = ∫_t1Q₀₀ x_i - ∫_t2Q₀₀ x_i

Well, we already know that Q_0i, the linear momentum, is a conserved quantity. But how do we interpret this equation? it says something about the change in time multiplied by the group momentum is equals to something else.

Here we need to recall our renormalization: c = 1. In fact, if we didn't renormalize, there would be a factor of c² floating around. In fact, we have

(t2-t1)∫_t2Q_0i = c^-2 (∫_t1Q₀₀ x_i - ∫_t2Q₀₀ x_i)

Well, Q₀₀ is the energy density. Its integral is the total energy. And we recall the famous equation of Einstein: E = mc². Morally speaking, then, c^-2Q₀₀ is a mass density on the space time. Which means that the integrals on the right hand side corresponds to mass moments, or otherwise known as center of mass.

So in reality, the boost symmetry produces a variant of the mass-transport equation. It states that the center of mass of the entire system will be transported linearly by the momentum over time.