Answer to Question #90877 in Field Theory for Vasiliy

Question

How does F^{mu nu} tensor is connected with the E and B vectors?

Answer 1

By definition 

$$F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$$

where (c=1)

$$\partial^\mu=(\frac{\partial}{\partial t},-\nabla)\,,\quad \nabla=(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})$$

Connection between the vectors $\vec{E}$ , $\vec{B}$ and 4-vector $A^\mu=(A^0,\vec{A})$ is

$$\vec{E}=-\nabla A^0-\partial^0\vec{A}\,,\quad \vec{B}=\operatorname{curl}\vec{A}=\nabla\times\vec{A}$$

So

$$F^{0i}=\partial^0 A^i-\partial^i A^0=\partial^0 A^i+\nabla_i A^0=-\vec{E}_i\,,\quad i=1,2,3$$

$$F^{12}=\partial^1 A^2-\partial^2A^1=-\partial_x A_y+\partial_y A_x=-[\nabla\times\vec{A}]_z=-{B}_z$$

$$F^{13}=\partial^1 A^3-\partial^3A^1=-\partial_x A_z+\partial_z A_x=[\nabla\times\vec{A}]_y={B}_y$$ $$F^{23}=\partial^2 A^3-\partial^3A^2=-\partial_y A_z+\partial_z A_y=-[\nabla\times\vec{A}]_x=-{B}_x$$

Taking into account that $F^{\mu\nu}=-F^{\nu\mu}$ and $F^{\mu\mu}=0$ we have

$$F^{\mu\nu}= \left( \begin{array}{cccc} 0&-E_x&-E_y&-E_y\\ E_x&0&-B_z&B_y\\ E_y&B_z&0&-B_x\\ E_z&-B_y&B_x&0\\ \end{array} \right)$$