# Using the electromagnetic field tensor to derive Maxwell-Faraday equation

I'm currently going through Byron and Fuller, and problem 1.13 is a problem about deriving the electromagnetic field tensor. The field tensor we derive is the complex version:

$$F = \begin{bmatrix} 0 & H_z & -H_y & -iE_x \\ -H_z & 0 & H_x & -iE_y \\ H_y & -H_x & 0 & -iE_z \\ iE_x & iE_y & iE_z & 0 \\ \end{bmatrix}$$

Then Byron asks that we derive the Maxwell equations. However, I'm having a bit of a problem doing that.

We can using the Binachi identity to arrive at:

\begin{align*} \partial_l F_{ik} + \partial_i F_{kl} + \partial_k F_{li} &= 0 \\ l = 2 \quad i = 4 \quad &k = 3 \\ \partial_y (i E_z) + \partial_t (-H_x) + \partial_z(-i E_y) &= 0 \\ i(\nabla \times \vec{E})_x &= \partial_t H_x \\ \text{Similarly} \qquad& \\ i(\nabla \times \vec{E})_y &= \partial_t H_y \\ i(\nabla \times \vec{E})_z &= \partial_t H_z \\ \nabla \times \vec{E} = -i \frac{\partial \vec{H}}{\partial t} \\ \end{align*}

But the Maxwell-Faraday Equation is listed as:

$$\nabla \times \vec{E} = -\frac1{c} \frac{\partial \vec{B}}{\partial t}$$

How do I get to that point from here? Is it true that $$ic\vec{H} = \vec{B}$$?

• Why there are the imaginary units in front of the E field? This is "strange", try to compare with: en.wikipedia.org/wiki/Electromagnetic_tensor Oct 6 '20 at 17:07
• It's the form given in the book. I have also seen this form referenced in one other place while searching for ways to reconcile this with the example derivations I've seen. mathpages.com/home/kmath647/kmath647.htm (Though it ought to be noted that this page uses B in place of H, which is even more confusing.) Oct 6 '20 at 17:09
• The derivation of this form involves using $x_4 = ict$ and $A_4 = i\varphi$ specifically. Oct 6 '20 at 17:09
• I just realized that I answered my own question with that final comment. I took the derivative with respect to $t$ not $ict$ Oct 6 '20 at 17:11
• exactly.. it's because you are using "imaginary time", so no problems at all. Oct 6 '20 at 17:11