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I Relation between Poincare matrix and electromagnetic field t

  1. Sep 22, 2016 #1
    We know that Poincare matrix which is 0 Kx Ky Kz
    ( -Kx 0 Jz -Jy ) describes the boost and rotation is very similar to
    -Ky -Jz 0 Jx
    -Kz -y -Jx 0

    to the electromagnetic field tensor 0 -Ex -Ey -Ez , in here E field like boost and B field like rotation.
    Ex 0 -Bz By
    Ey Bz 0 -Bx
    Ez -By Bx 0

    My question is how they are related theoretically? and how we can show that they stasfy the same algebra?
  2. jcsd
  3. Sep 24, 2016 #2


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    What is the Poincare matrix? (Reference?)
    What are K and J in this antisymmetric matrix?
    (Can you write in ##\LaTeX##?)
  4. Sep 24, 2016 #3
    Poincare matrix is $ M^{\mu\nu}$ in Poincare algebra which describes Lorentz transformations. You can look up wikipedia page https://en.wikipedia.org/wiki/Poincaré_group. If we write $ M^{\mu\nu}$ as matrix form, it looks like
    $$ M^{\mu\nu}=\begin{pmatrix} 0 & -K_x & -K_y & -K_z \\ K_x & 0 & J_z & -J_y \\ K_y & J_z &0 & J_x \\ K_z & J_y & -J_x &0 \end{pmatrix}$$.

    In parallel to this notation for uniform constant field the electromagnetic field tensor will have a similar effect on charge E field like boost and B field like rotation and field tensor $F^{\mu\nu} $ have similar structure to $ M^{\mu\nu}$

    $$ F^{\mu\nu}=\begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & B_z & -B_y \\ E_y & B_z &0 & B_x \\ E_z & B_y & -B_x &0 \end{pmatrix}$$.

    I was wondering what is theoretical connection between them?
  5. Sep 24, 2016 #4


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    [itex]M_{\mu\nu}[/itex] are the 6 abstract generators of the Lorentz group [itex]SO(1,3)[/itex]. In the vector representation, [itex]M_{\mu\nu}[/itex] are six [itex]4 \times 4[/itex] matrices whose matrix elements are given by [tex](M_{\mu\nu})^{\alpha}{}_{\beta} \sim \delta^{\alpha}_{\mu} \ \eta_{\nu\beta} - \delta^{\alpha}_{\nu} \ \eta_{\mu\beta} \ . [/tex]
    So, for example, the boost generators [itex]K_{i} = M_{i0}, \ i = x,y,z[/itex] are three [itex]4 \times 4[/itex] matrices with matrix elements given by [tex](K_{i})^{\alpha}{}_{\beta} \sim \delta^{\alpha}_{i} \ \eta_{0\beta} - \delta^{\alpha}_{0} \ \eta_{i\beta} .[/tex]
    So, in the vector representation, [itex]M_{\mu\nu}[/itex] is a collection of six [itex]4 \times 4[/itex] matrices. But, the field tensor [itex]F_{\mu\nu}[/itex] is a collection of 6 functions, i.e., six numbers which you can arrange them into one anti-symmetric [itex]4 \times 4[/itex] matrix. Under the Lorentz group, [itex]F_{\mu\nu}[/itex] transforms in the tensor representation of the generators [itex]M_{\mu\nu}[/itex]: [tex]\delta F_{\mu\nu} = \frac{i}{2} \left( \omega^{\rho\sigma}M_{\rho\sigma} \right)_{\mu\nu}{}^{\alpha\beta} \ F_{\alpha\beta} .[/tex]
  6. Sep 24, 2016 #5


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    The relation is that infinitesimal Lorentz transformations are given by antisymmetric ##4 \times 4## matrices. This is easy to see. For simplicity we work with the representation of Lorentz transformations with two lower indices, i.e.,
    $$x_{\mu}'=\Lambda_{\mu \nu} x^{\nu},$$
    $$\Lambda_{\mu \nu} = \eta_{\mu \rho} {\Lambda^{\rho}}_{\nu}.$$
    For a Lorentz transformation you have
    $$\eta^{\mu \nu} \Lambda_{\mu \rho} \Lambda_{\nu \sigma}=\eta_{\rho \sigma},$$
    and for an infinitesimal transformation, i.e., ##\Lambda_{\mu \rho}=\eta_{\mu \rho}+\delta \Omega_{\mu \rho}## yieds up to quantities of 2nd order in the ##\delta \Omega##
    $$\eta^{\mu \nu} (\eta_{\mu \rho}+\delta \Omega_{\mu \rho})(\eta_{\nu \sigma}+\delta \Omega_{\nu \sigma})=\eta_{\rho \sigma}+\delta \Omega_{\sigma \rho} + \delta \Omega_{\rho \sigma} \; \Rightarrow \; \delta \Omega_{\rho \sigma}=-\delta \Omega_{\sigma \rho}.$$
    For homogeneous static em. fields the ##F_{\mu \nu}## generate Lorentz transformations, i.e., the trajectories (world lines) of particles are given by Lorentz transformations of the initial momenta, because the equation of motion reads
    $$\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=\frac{q}{c} F^{\mu \nu} p_{\nu}.$$
    For ##F^{\mu \nu}=\text{const}## you get indeed Lorentz transformations. For ##\vec{B}=0## it's a rotation-free boost, for ##\vec{E}=0## it's spatial rotation.
  7. Sep 25, 2016 #6
    Thank you very much! It explains a lot.
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