I'm doing something very wrong or It seems to me that I can't generate a finite Lorentz transformation using the exponential of the infinitesimal Lorentz boosts.

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 a Lorentz boost in the $x$-direction (in the Hamiltonian sense) 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 give the right answer because in this case we have a transformation that also change the time.


1 Answer 1


Now usually the generator of (proper orthochronous) LTs (generator of the Lie group; in the fundamental representation) is something like $$ L_{\mu \nu} = i(x_{\mu} \partial_{\nu} - x_{\nu} \partial_{\mu}) $$ and the corresponding finite transformation would be $\Lambda(\omega) = e^{i\omega^{\mu\nu}L_{\mu \nu}}$, where $\omega_{\mu \nu}$ are the parameters (which specify the group element). Acting e.g. on a position four vector: $$ x'^{\rho} \equiv e^{i\omega_{\mu\nu}L^{\mu \nu}} x^{\rho}. $$ To see what this is consider that $$ L^{\mu \nu} x^{\rho} = i(x^{\mu} g^{\nu \rho} - x^{\nu} g^{\mu \rho}) = \underbrace{i(g^{\mu \sigma} g^{\nu \rho} - g^{\nu \sigma} g^{\mu \rho})}_{\equiv (J^{\mu \nu})^{\rho \sigma}} x_{\sigma} $$ and therefore $$ x'^{\rho} = (e^{i\omega_{\mu \nu} J^{\mu \nu}})^{\rho \sigma} x_{\sigma}. $$ Note that $w_{\mu \nu} J^{\mu \nu}$ is a matrix and the LT is the corresponding matrix-exponential. E.g. $\omega_{10} = -\omega_{01} = \psi$ and all others components zero gives a boost in the $x$-direction ($\psi = \tanh^{-1}v/c$ is then the rapidity of the boost). Or $\omega_{12} = -\omega_{21} = \theta$, and all other components zero, corresponds to a rotation around the z-axis by an angle $\theta$. You can check all this by computing the matrix exponential. (See also Lorentz transformation boosts as matrices).

Also the momentum generator is $p_{\mu} = -i \partial_{\mu}$ (this generates space-time translations $e^{ia^{\mu} p_{\mu}} x^{\rho} = x^{\rho} + a^{\rho}$. Putting this into the LT generator gives $$ L_{\mu \nu} = x_{\nu} p_{\mu} - x_{\mu} p_{\nu} $$ and with the canonical Poisson bracket relation $\{ x^{\mu}, p^{\nu} \} = g^{\mu \nu}$ using your formula for the exponential of the Poisson-bracket is basically the same as what I have done. Note that $L_{\mu \nu}$ looks like angular momentum, which is of course expected, since LTs are nothing but rotations in 1+3 dimensional space-time.

Now in your formulation you made the identification $$ -i \partial_0 = p_0 = H = \sqrt{\vec{p}^2 + m^2}, $$ corresponding to a single, free, massive particle. But this messes up the poisson-bracket relation $\{ x^{\mu}, p^{\nu}\} = g^{\mu \nu}$, e.g. $\{ p_0, x^i \} = \frac{p^i}{\sqrt{\vec{p}^2 + m^2}} \neq 0$; $p^0$ and $\vec{p}$ are not independent anymore! This is precisely why your approach does not work. So one needs to set up a proper Hamiltonian formalism of relativistic mechanics and I am not very familiar with this. Maybe Hamiltonian mechanics and special relativity? can help you.


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