0
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ Ca you elaborate please? Is there any difference between $F^{\mu\nu}$ and $F_{0 j}$? $\endgroup$
    – Raisa
    Sep 8, 2013 at 8:13
  • 2
    $\begingroup$ "An author" seems to refer to: Erick J. Weinberg, Classical Solutions in Quantum Field Theory: Solitons and Instantons in High Energy Physics, p. 43. $\endgroup$
    – Řídící
    Sep 8, 2013 at 8:32
  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Sep 8, 2013 at 9:11
  • $\begingroup$ @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 $F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}.$? $\endgroup$
    – Raisa
    Sep 8, 2013 at 9:36
  • 1
    $\begingroup$ 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? $\endgroup$
    – Raisa
    Sep 8, 2013 at 9:42

1 Answer 1

1
$\begingroup$

EDIT: Answer was incorrect. So it was removed

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Raisa
    Sep 8, 2013 at 10:09
  • $\begingroup$ @dj_mummy There's a very easy MathJax quick reference guide here. $\endgroup$
    – Řídící
    Sep 8, 2013 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.