Finite Lorentz Transformation via Poisson Bracket

[andresB]

TL;DR

I've been unable to generate a Lorentz Transformation via finite canonical transformation

Let me define $L_{x;v}$ as the operator that produce a Lorentz boost in the $x$-direction with a speed of $v$. This operator acts on the components of the 4-position as follows
$$L_{x;v}(x) =\gamma_{v}(x-vt),$$
$$L_{x;v}(y) =y,$$
$$L_{x;v}(z) =z,$$
$$L_{x;v}(t) =\gamma_{v}(t-vx),$$
where $\gamma_{v}=\frac{1}{\sqrt{1-v^{2}}}$. Now, the infinitesimal generator of the Lorentz boost in the $x$-direction is $K_{x}=Hx-tP_{x}$, where $H=\sqrt{p^{2}+m^{2}}$. A finite Lorentz transformation should be given via the exponential operator
$$L_{x;v}=\exp\left[\left\{ \circ,K_{x}\right\} \right]=\sum_{n=0}^{\infty}\frac{1}{n!}\left\{ \circ,K_{x}\right\} ^{n}$$
But I can't see how this operator gives the correct Lorentz transformations.
For the $x$ coordinate we have that $\left\{ x,K_{x}\right\} =x\frac{p_{x}}{H}-t$, hence, the term $\left\{ \left\{ x,K_{x}\right\} ,K_{x}\right\}$ will be independent of time. So the term $\gamma_{v}vt$ will not appear.
Worse yet, $K_{x}$ generates a change in the $y$ coordinate $\left\{ y,K_{x}\right\} =x\frac{p_{y}}{H}\neq0$, in contradiction of what a Lorentz transformation should do.

My take is that the usual tools of Hamiltonian mechanics are unable to do the right answer since in this case we have a transformation that also change the time.

 

[vanhees71]

It's maybe difficult in the Hamiltonian formalism, because it involves the transformation of time. Within the Lagrangian formulation, the Noether theorem for the Poincare group leads to the conserved quantities $p^0=E/c$, $\vec{p}$ (temporal and spatial translations) and the angular-momentum-tensor $J^{\mu \nu}=x^{\mu} p^{\nu}-x^{\nu} p^{\mu}$. The space-space components are the usual angular-momentum and $J^{\mu 0}=E x^{\nu}-p^{\nu} c t$ is the said conserved quantity due to rotation-free boosts.

 

[andresB]

It seems, indeed, that the usual treatment is doomed to fail, and what is needed is a generalization of canonical transformation that includes time.

The concept of canonical transformation can be extended to include transformation in time once the phase space is extended, and there a generating function can be given that results in a Lorentz transformation.
https://aapt.scitation.org/doi/10.1119/1.16086
https://iopscience.iop.org/article/10.1088/0305-4470/38/6/006

However, I've been unable to find a treatment in terms of exponentiation of the infinitesimal generator of the Lorentz boost. It would be weird that all others space-time transformation can be obtained with an exponential operator but not Lorentz boosts.

 

[andresB]

Ok, the endeavour was doomed to fail since it was badly formulated from the the beginning. The reason is that the in special relativity the velocity is not additive, so it is not a good parameter to use in the exponential operator. Indeed, the exponential with the velocity do not form a one-parameter group
$$L_{x;v_{1}}L_{x;v_{2}}\neq L_{x;v_{1}+v_{2}}.$$

The right way to do it is to use the rapidity $s=\tanh v$, and make the definition
$$L_{x;s}=\exp\left[s\left\{ ,K_{x}\right\} \right]=\sum_{n=0}^{\infty}\frac{s^{n}}{n!}\left\{ \circ,K_{x}\right\} ^{n}.$$

With the above definition, the quantities $\left(H,\mathbf{p}\right)$ seem to transform as the components of a 4-vector
$$L_{x;s}(H)=H\cosh s-p_{x}\sinh s,$$
$$L_{x;s}(p_{x})=p_{x}\cosh s-H\sinh s,$$
$$L_{x;s}(p_{y})=p_{y},$$
$$L_{x;s}(p_{z})=p_{z}.$$
Moreover, the components of the velocity $V_{i}=\frac{p_{i}}{H}$ seems to transform as expected from Einstein addition formula.The position $\mathbf{r}$, however, still is not transforming as expected from a Loretnz boost.