# EM vector potential

We can write the electromagnetic field tensor as $$\begin{bmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{bmatrix} = F^{\mu\nu}.$$ Erick J. Weinberg, Classical Solutions in Quantum Field Theory: Solitons and Instantons in High Energy Physics (p. 43), states:

For a static solution with vanishing electric fields $F_{0j}$, and hence $A_0 = 0$, [...]

How can this be proven?

• Ca you elaborate please? Is there any difference between $F^{\mu\nu}$ and $F_{0 j}$? Sep 8, 2013 at 8:13
• "An author" seems to refer to: Erick J. Weinberg, Classical Solutions in Quantum Field Theory: Solitons and Instantons in High Energy Physics, p. 43. Sep 8, 2013 at 8:32
• Comment to the question (v2): Weinberg's use of the word hence may be a bit misleading. All he means is that it is consistent to choose the temporal gauge $A_0=0$. But it is of course not the only consistent gauge-fixing choice possible. Sep 8, 2013 at 9:11
• @Qmechanic, Weinberg says that for stationary solution $A_0=0$ is it means that the magnetic properties in the tensor fields are zero? Can't we derive that from this equation which is come from the Tensor matrix $F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}.$? Sep 8, 2013 at 9:36
• and one more question, in the equation (3.23) Weinberg says equates the energy to zero why is that? Is it due to stationary solutions? Sep 8, 2013 at 9:42

• Can you please show this into the matrix notations with little math that how $A_0=0$ so the answer will be clear to me. Thanks. Sep 8, 2013 at 10:09