For electromagnetism, the matrix form

$$\Bbb{F}^{\mu \nu}=\begin{pmatrix} 0 & E_x/c, & E_y/c & E_z /c \\ -E_x/c & 0 & B_z & -B_y \\ -E_y /c & -B_z & 0 & B_x \\ -E_z/c & B_y & -B_x &0 \end{pmatrix}$$

What is the equivalent form for QCD's $G$ tensor?

  • 2
    $\begingroup$ You can simply define the chromoelectric and chromomagnetic fields in the exact same way. (After all, formally speaking, the equation you wrote is how ordinary electric and magnetic fields are defined in the first place.) The only difference is that each field component now also needs a color index. $\endgroup$
    – knzhou
    Dec 25, 2019 at 21:09

1 Answer 1


You have to add a further index due to the gauge group. So, you will get a set of matrices like the one of the $U(1)$ case. The different components of these matrices are mixed up by the non-linear terms also present in the equations of motions. E.g., if your gauge group is $SU(N)$, you will get $N^2-1$ matrices. In QCD you will have 8 of these and so, this is the number of independent gluon fields.

  • $\begingroup$ Do you have a reference where the matrices are explicitly provided? $\endgroup$
    – Anon21
    Dec 24, 2019 at 18:10
  • $\begingroup$ The problem here is that the components of the fields have also non-linear terms depending on the potentials. So, it is quite natural to work with the components like $E^a_i$ and $B^a_i$, being $a$ the group index and $i$ the vector components rather than write them through matrices. You can find a good presentation in this form in the book by Faddeev and Slavnov, Gauge Fields: Introduction to Quantum Theory (Addisn-Wesley, 1991), Ch. 3. Just remember that here the fields are defined through $F_{\mu\nu}^a=\partial_\mu A_\nu^a-\partial_\nu A^a_\mu+gf^{abc}A_{\mu}^bA_{\nu}^c$. $\endgroup$
    – Jon
    Dec 25, 2019 at 18:57

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