# 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?

• Remove the quantum field theory tag, it is not needed for this question. Look up the relationship between the 4-potential and E. Set all time derivatives to 0 and E to be 0. The gradient of A0 will thus be 0. – dj_mummy Sep 8 '13 at 8:11
• Ok I will type out the answer. – dj_mummy Sep 8 '13 at 8:15
• "An author" seems to refer to: Erick J. Weinberg, Classical Solutions in Quantum Field Theory: Solitons and Instantons in High Energy Physics, p. 43. – Keep these mind Sep 8 '13 at 8:32
• @Raisa If what aufkag says is true, that book maybe a little out of your reach yet. Start with J. D. Jackson's 'Classical Electrodynamics' to understand more about EM. – dj_mummy Sep 8 '13 at 8:40
• 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? – Raisa Sep 8 '13 at 9:42

The indices are important and essential when the problem involves the operation of the tensor on another vector or tensor. Depending on whether the indexes are up or down, the components of the tensor transform differently under co-ordinate transformations (like Lorentz boosts). If you choose upper indexes for a four-vector in your scheme, then a lower index must denote a dual vector object and vice versa. The indices determine how the tensor operates. Look up covariance and contra-variance of tensors on this link: "Ricci calculus".

I tried to type out the full steps but I am terrible at formatting, so I wrote it on a piece of paper and took a photo: Setting the constant to $0$ and by taking a stationary gauge one gets $A_0=0$.

I suggest you refer to a book on tensors to understand 'index gymnastics'. I recommend Murray Spiegel's book on vector analysis for a good intro to tensors.

• 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. – Raisa Sep 8 '13 at 10:09
• In fact you can absorb the constant into the time derivative of the gauge without any problem in this case. So the only condition required is that the gauge be stationary with time. – dj_mummy Sep 8 '13 at 10:48
• @dj_mummy There's a very easy MathJax quick reference guide here. – Keep these mind Sep 8 '13 at 11:33