1
$\begingroup$

The electromagnetic field tensor $F_{\mu\nu} $ in component form is:

$$ F_{\mu\nu} = \begin{pmatrix} 0 & \frac{E_x}{c} & \frac{E_y}{c} & \frac{E_z}{c} \\ -\frac{E_x}{c} & 0 & -B_z & B_y \\ -\frac{E_y}{c} & B_z & 0 & -B_x \\ -\frac{E_z}{c} & -B_y & B_x & 0 \end{pmatrix}$$

Where:

  • $E_x, E_y, E_z$ are the components of the electric field.
  • $ B_x, B_y, B_z $ are the components of the magnetic field.

The Faraday 2-form is defined as

\begin{align*} F &= \sum_{\mu} \sum_{\nu} \frac{1}{2} F_{\mu\nu} dx^{\mu} \wedge dx^{\nu} \\ & = \frac{1}{2} F_{11} dt \wedge dt + \frac{1}{2} F_{12} dt \wedge dx + \frac{1}{2} F_{13} dt \wedge dy + \frac{1}{2} F_{14} dt \wedge dz \\ &+ \frac{1}{2} F_{21} dx \wedge dt + \frac{1}{2} F_{22} dx \wedge dx + \frac{1}{2} F_{23} dx \wedge dy + \frac{1}{2} F_{24} dx \wedge dz\\ &+\frac{1}{2} F_{31} dy \wedge dt + \frac{1}{2} F_{32} dy \wedge dx + \frac{1}{2} F_{33} dy \wedge dy + \frac{1}{2} F_{34} dy \wedge dz \\ &+ \frac{1}{2} F_{41} dz \wedge dt + \frac{1}{2} F_{42} dz \wedge dx + \frac{1}{2} F_{43} dz \wedge dy + \frac{1}{2} F_{44} dz \wedge dz \\\\ % % & = \frac{1}{2} 0 \cdot 0 + \frac{1}{2} \frac{E_x}{c} dt \wedge dx + \frac{1}{2} \frac{E_y}{c} dt \wedge dy + \frac{1}{2} \frac{E_z}{c} dt \wedge dz \\ &+ \frac{1}{2} \left(-\frac{E_x}{c} \right) dx \wedge dt + \frac{1}{2} 0 \cdot 0 + \frac{1}{2} (-B_z) dx \wedge dy + \frac{1}{2} B_y dx \wedge dz\\ &+\frac{1}{2} \left(-\frac{E_y}{c}\right) dy \wedge dt + \frac{1}{2} B_z dy \wedge dx + \frac{1}{2} 0 \cdot 0 + \frac{1}{2} (-B_x) dy \wedge dz \\ &+ \frac{1}{2} \left(-\frac{E_z}{c}\right) dz \wedge dt + \frac{1}{2} (-B_y) dz \wedge dx + \frac{1}{2} -B_x dz \wedge dy + \frac{1}{2} 0 \cdot 0 \\\\ % % & = \left(\frac{1}{2} \frac{E_x}{c} dt \wedge dx +\frac{1}{2} \frac{E_x}{c} dt \wedge dx \right)+ \left(\frac{1}{2} \frac{E_y}{c} dt \wedge dy +\frac{1}{2} \frac{E_y}{c} dt \wedge dy \right)\\ &+ \left( \frac{1}{2} \frac{E_z}{c} dt \wedge dz +\frac{1}{2} \frac{E_z}{c} dt \wedge dz \right) + \left(\frac{1}{2} B_z dx \wedge dy + \frac{1}{2} B_z dx \wedge dy \right) \\ &+\left( \frac{1}{2} B_y dz \wedge dx + \frac{1}{2} B_y dz \wedge dx\right) + \left( \frac{1}{2} B_x dy \wedge dz + \frac{1}{2} B_x dy \wedge dz \right) \\\\ % % &= \frac{E_x}{c} dt \wedge dx + \frac{E_y}{c} dt \wedge dy + \frac{E_z}{c} dt \wedge dz + B_z dx \wedge dy + B_y dz \wedge dx + B_x dy \wedge dz \end{align*}

Is this right? It looks right to me. I felt ok with it. But both Wikipedia and Misner Wheeler and Thorne have

$$ F = \frac{E_x}{c}dx \wedge dt + \frac{E_y}{c} dy \wedge dt + \frac{E_z}{c} dz \wedge dt + B_z dx \wedge dy + B_y dz \wedge dx + B_x dy \wedge dz $$

See here for the wikipedia discussion.

$\endgroup$
3
  • $\begingroup$ I think $F_{0i}=-E_i$ rather than $E_i$...that will solve the ambiguity $\endgroup$
    – KP99
    Commented Oct 6 at 4:57
  • $\begingroup$ @KP99 agreed. but then is this definition on wikipedia inconsistent. i originally had that and then shifted it because i though i had the form of the tensor wrong en.wikipedia.org/wiki/Electromagnetic_tensor if that's the case, then yes that clears everything up $\endgroup$ Commented Oct 6 at 5:00
  • 1
    $\begingroup$ I haven't looked at MWT yet but I think you should pay attention to sign convention being used (if it's+--- or -+++). I think you are using+--- $\endgroup$
    – KP99
    Commented Oct 6 at 5:04

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.