General Generator of Lorentz Transformation in Hamiltonian Formalism

[andresB]

In the Hamiltonian formalism, the space-time transformation are realized via canonical transformation, and the transformations are generated by Poisson brackets of certain functions of phase-space variables.
In Newtonian mechanics, Galilean boosts are generated by the sometimes called dynamic mass moment $$\overrightarrow{G}=m\overrightarrow{r}-t\overrightarrow{p}.$$
Now, in virtually every source I consult, the general generator of the Galilean boosts is not considered. Instead, people just use the $t=0$ generator
$$\overrightarrow{G}=m\overrightarrow{r}$$
The same happen for Lorentz transformations, people just use the $t=0$ generator
$$\overrightarrow{K}=H\overrightarrow{r}$$
where $H$ is the energy.
So, the question is, what is the most general form of $\overrightarrow{K}$?

 

[anuttarasammyak]

How about
$$t\mathbf{p}-\frac{E}{c^2}\mathbf{r}$$
which comes from $(M^{01},M^{02},M^{03})$ of angular momentum 4-tensor
$$M^{\mu\nu}=x^\mu p^\nu-x^\nu p^\mu$$ ?

 

[andresB]

How about
$$t\mathbf{p}-\frac{E}{c^2}\mathbf{r}$$
which comes from $(M^{01},M^{02},M^{03})$ of angular momentum 4-tensor
$$M^{\mu\nu}=x^\mu p^\nu-x^\nu p^\mu$$ ?

So it would have almost the same form of the Galilei boost generator with only $m$ replaced by $H$? I suppose it's a good Ansatz, I'll have to check the bracket relations of the Poincare group to see if they remain true.

 

[vanhees71]

Yes! The conclusion leads to the correct interpretation of the famous "energy-mass equivalence":

In Newtonian physics from invariance under Galilei boosts follows that the center of mass of a closed system moves with constant velocity. The measure for "inertia" is mass.

In relativistic physics invariance under Lorentz boosts implies that the center of energy of a closed system moves with constant velocity and thus the measure of inertia is energy.

Using the equivalence principle this implies that the source of a gravitational field must be energy, not mass, in relativistic physics.

 

[andresB]

So, yes, $\mathbf{K}=H\mathbf{r}-t\mathbf{p}$ has the same bracket relation than just $\mathbf{K}=H\mathbf{r}$. Thank you guys for the answer. I do wonder why the $t=0$ generator is preferred for the presentation of the Galilean and the Poincare algebra, it seems to hide some physical interpretation of the boost generator, and it is not much of a simplification in the math.