# I Relation between Poincare matrix and electromagnetic field t

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1. Sep 22, 2016

### Muratani

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. Sep 24, 2016

### robphy

What is the Poincare matrix? (Reference?)
What are K and J in this antisymmetric matrix?
(Can you write in $\LaTeX$?)

3. Sep 24, 2016

### Muratani

Poincare matrix is $M^{\mu\nu}$ in Poincare algebra which describes Lorentz transformations. You can look up wikipedia page https://en.wikipedia.org/wiki/PoincarĂ©_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?

4. Sep 24, 2016

### samalkhaiat

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

5. Sep 24, 2016

### vanhees71

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.,
$$x_{\mu}'=\Lambda_{\mu \nu} x^{\nu},$$
where
$$\Lambda_{\mu \nu} = \eta_{\mu \rho} {\Lambda^{\rho}}_{\nu}.$$
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.

6. Sep 25, 2016

### Muratani

Thank you very much! It explains a lot.