$\omega_{\mu\nu}$ contains infinitesimal parameters and $J^{\mu\nu}$ contains generators of boost and rotation. Any 4-vector transforms as $p^\mu=\Lambda^\mu_\nu p^\nu$. Starting from given exponential form how shall I reach following form.

$$\begin{bmatrix} ct' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma & -\gamma \beta_x & -\gamma \beta_y & -\gamma \beta_z \\ -\gamma \beta_x & 1 + (\gamma - 1) \frac{\beta^2_x}{\beta^2} & (\gamma - 1) \frac{\beta_x \beta_y}{\beta^2} & (\gamma - 1) \frac{\beta_x \beta_z}{\beta^2} \\ -\gamma \beta_y & (\gamma - 1) \frac{\beta_y \beta_x}{\beta^2} & 1 + (\gamma - 1) \frac{\beta^2_y}{\beta^2} & (\gamma - 1) \frac{\beta_y \beta_z}{\beta^2} \\ -\gamma \beta_z & (\gamma - 1) \frac{\beta_z \beta_x}{\beta^2} & (\gamma - 1) \frac{\beta_z \beta_y}{\beta^2} & 1 + (\gamma - 1) \frac{\beta^2_z}{\beta^2} \end{bmatrix}\begin{bmatrix} ct \\ x \\ y \\ z \end{bmatrix}.$$ I don't know what's wrong with my concepts here.

  • $\begingroup$ 1) Start from the Lorentz transformations (see e.g. wikipedia and this Phys.SE answer); 2) make a linear approximation for $\beta \ll 1$ and write this infinitesimal transformation in matrix form; 3) exponentiate the matrix to obtain the desired result $\endgroup$ – glS Feb 7 '15 at 14:01
  • $\begingroup$ qr.ae/1LYmWg $\endgroup$ – bolbteppa Aug 7 '16 at 1:59

I. To obtain the matrix form you quote from the infinitesimal transformations you have to:

  1. Start from the Lorentz transformation as found e.g. in the wikipedia article.
  2. Write the infinitesimal form of these transformations. E.g. for a boost in the $x^1$ direction you have $$ \tag{1-1} x'^0 = x^0 + \beta x^1 + \mathcal{O}(\beta^2), $$ $$ \tag{1-2} x'^1 = x^1 + \beta x^0 + \mathcal{O}(\beta^2). $$ This corresponds to the approximation $$ \tag 2 \Lambda^\mu_{\,\,\nu} \approx \delta^\mu_{\,\,\nu} + \omega^\mu_{\,\,\nu},$$ where we have just used the $\delta$ to indicate the identity component of the transformation, and defined $\omega$ as the linearized part of the transfomation. In the example above we would have: $$ \tag 3 \omega^0_{\,\,1} = \omega^1_{\,\,0} = \beta, \quad \omega^\mu_{\,\,\nu}=0 \text{ otherwise}. $$
  3. Imagine the Lorentz transformation $\Lambda$ as composed of an infinite (say $N \gg 1$) succession of infinitesimal transformations of the form (2) with parameter $\omega/N$. Then you have $$ \tag 4\Lambda = \left( 1 + \omega/N \right)^N \rightarrow e^\omega, \,\, N \to \infty. $$ Finally, exponentiate the $\omega$ you found on step 2 to obtain your result.

II. Note that the formula you quoted in the title (V1), $$ \tag 5 \Lambda^a_{\,\,b} = \left[ \exp\left(-\frac{i}{2} \omega_{\mu\nu} J^{\mu\nu} \right) \right]^a_{\,\,b} \approx \delta^a_{\,\,b} - \frac{i}{2} \omega_{\mu\nu} (J^{\mu\nu})^a_{\,\,b}, $$ holds on more general grounds (for a generic, modulo mathematical subtleties, representation of the Lorentz group). To obtain the vector representation you are considering here you have to use the appropriate generators $J^{\mu\nu}$, which are in this case $$ \tag 6 (J^{\mu\nu})^{\rho\sigma} = i ( \eta^{\mu\rho} \eta^{\nu\sigma} - \eta^{\mu\sigma} \eta^{\nu\rho} ). $$ To see that this is consistent with (2) consider the following computation: $$ \tag 7 \Lambda^\rho_{\,\,\sigma} = \delta^\rho_{\,\,\sigma} - \frac{i}{2} \omega_{\mu\nu} (J^{\mu\nu})^\rho_{\,\,\sigma} = \delta^\rho_{\,\,\sigma} + \frac{1}{2} \omega_{\mu\nu} (\eta^{\mu\rho} \delta^\nu_{\,\,\sigma} - \eta^{\mu\sigma} \delta^\nu_{\,\,\rho}) = \delta^\rho_{\,\,\sigma} + \omega^\rho_{\,\,\sigma}, $$ where we have used (6) in (5) to reobtain (2).

See this Phys.SE post for further details on the various representations of the Lorenzt group.

  • $\begingroup$ Thanx dear i ll try to get explicit form this way. If i fail i ll come here again. $\endgroup$ – Zohaib Aarfi Feb 8 '15 at 3:40

The way to get from exponential form to the most general Lorentz transformation can be an extremely tedious computation to carry out. In principle there is nothing else to do but to explicitly compute the exponential of a given matrix. I would suggest you try to fix every generator and get an idea of the transformation it is generating. For example, fix the 1-2 component to be a certain parameter $\psi$ and let all the other components of $\omega$ be zero, i.e. $$\omega_{12} = \psi,\qquad\omega_{ij} = 0,\quad\forall 1<i<j\leq4,$$ and compute $$e^{-i\psi J^{12}}.$$


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