# EM duality transformation

I have read that by defining a dual transformation as $$\begin{pmatrix} F'^{\mu \nu} \\ ^*F'^{\mu \nu} \end{pmatrix} = \begin{pmatrix} \cos (\alpha ) & \sin (\alpha ) \\ -\sin (\alpha ) & \cos (\alpha ) \end{pmatrix} \begin{pmatrix} F^{\mu \nu} \\ ^*F^{\mu \nu} \end{pmatrix}$$

where $$F^{\mu \nu} = \begin{pmatrix} 0 & -E^x & -E^y & -E^z \\ E^x & 0 & - B^z & B^y \\ E^y & B^z & 0 & - B^x \\ E^z & - B^y & B^x & 0 \end{pmatrix} \quad \text{and} \quad ^*F^{\mu \nu} = \begin{pmatrix} 0 & -B^x & -B^y & -B^z \\ B^x & 0 & E^z & -E^y \\ B^y & -E^z & 0 & E^x \\ B^z & E^y & -E^x & 0 \end{pmatrix},$$

it is equivalent to say that $E$ and $B$ transform as $$\begin{pmatrix} E' \\ B' \end{pmatrix} = \begin{pmatrix} \cos (\alpha ) & \sin (\alpha ) \\ -\sin (\alpha ) & \cos (\alpha ) \end{pmatrix} \begin{pmatrix} E \\ B \end{pmatrix}.$$

But how can I find it from the Faraday tensor? (replacing with $\mu$ and $\nu$ is a bad idea apparently)

• Have a look at this answer. It might be helpful. Note that in the complex formulation the duality transformation appears as a unit phase. – Diracology May 12 '17 at 18:44

## 1 Answer

This follows immediately from the first equation that you wrote. Just fix a $\mu \nu$ and obtain that the corresponding component of the Electric/Magnetic field is given by what you wrote in the last equation.