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I tried to proof that the Maxwell equations are invariant under parity transformations. Therefore I used the covariant formulation of the Maxwell equations

\begin{align} \partial_{\nu}F^{\nu\mu} &= \frac{4\pi}{c}j^{\mu}\\ \partial_{\nu}\tilde{F}^{\nu\mu} &= 0 \end{align} and the parity transformation given by

\begin{align} P = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \end{align}

Regarding only the first equation $\partial_{\nu}F^{\nu\mu} = \frac{4\pi}{c}j^{\mu}$ we have

\begin{align} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{c}\frac{\partial}{\partial \text{t}} \\ \vec{\nabla} \end{pmatrix} = \begin{pmatrix} \frac{1}{c}\frac{\partial}{\partial \text{t}} \\ -\vec{\nabla} \end{pmatrix} \end{align} as well as \begin{align} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} c\rho \\ \vec{j} \end{pmatrix} = \begin{pmatrix} c\rho \\ -\vec{j} \end{pmatrix} \end{align}

and

\begin{align} P \cdot F^{\nu\mu} &= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -E^1 & -E^2 & -E^3 \\ E^1 & 0 & -B^3 & B^2 \\ E^2 & B^3 & 0 & -B^1 \\ E^3 & -B^2 & B^1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -E^1 & -E^2 & -E^3 \\ -E^1 & 0 & B^3 & -B^2 \\ -E^2 & -B^3 & 0 & B^1 \\ -E^3 & B^2 & -B^1 & 0 \end{pmatrix} \end{align}

Based on these calculations, is there a way to see that Maxwell equations are invariant under parity transformations and if so how do I see it?

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You are not applying the transformations correctly. Your transformation, $P$, is linear map that changes a vector into another vector. Well, $F^{\mu\nu}$ is a rank (2,0) tensor, not a vector (rank (1,0) tensor). This all becomes much clearer if you use index notation, rather than writing matricies. I will work in Cartesian basis.

So, let us denote your $P$ with a rank (1,1) tensor, $P^\mu_\nu$, such that for a vector $V^\mu=\left(V^0, V^1, V^2, V^3\right)^\mu$, the effect of $P$ would be:

$V^\mu\to\bar{V}^\mu = P^\mu_\nu V^\nu = (V^0, -V^1, -V^2, -V^3)$

Now the transformation for the electromagnetic tensor would be then:

$F^{\mu\nu}\to\bar{F}^{\mu\nu}=P^\mu_\eta P^\nu_\sigma F^{\eta\sigma}$

In this case, because your transformation is simple, you can rewrite this as a matrix equation (in general, it may not work so flawlesly):

$\bar{F}=P\cdot F \cdot P = \left(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{matrix}\right) \left(\begin{matrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \\ \end{matrix}\right) \left(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{matrix}\right)=\left(\begin{matrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \\ \end{matrix}\right)$

So electric field transforms as a polar vector, whilst magnetic as axial. As is well known. You will also need to apply transform $J^\mu\to\bar{J}^\mu=P^\mu_\nu J^{\nu}$, and $\partial_\mu \to \bar{\partial}_\mu = \left(P^{-1}\right)^\nu_\mu \partial_\nu$ (i.e. you need to apply inverse transforms to co-variant tensors). When you do that, you will see that the equation for the new quantities will be the same as for the old ones, i.e. $\bar{\partial}_\mu \bar{F}^{\mu\nu}={4\pi}{c}\bar{J}^{\nu}$, so the equation does not change (vectors&tensors do)


Extra following the comment.

I really advise against matrix notation in these calculations, so I will stop using it. Here is how you work out $\bar{F}^{\mu\nu}$:

$F^{0\{1,2,3\}}=\{-E_x, -E_y, -E_z\}$

$F^{12}=-B_z,\, F^{13}=B_y,\, F^{23}=-B_x$

So, using $P^0_0=1, P^1_1=P^2_2=P^3_3=-1$, and zero otherwise:

$\bar{F}^{0i}=P^0_0 P^i_j F^{0j} = (1)(-1)F^{0i}$, so electric field gets a minus

$\bar{F}^{ij}=P^i_s P^j_k F^{sk}=(-1)^2 F^{ij}$, so magnetic field is unchanged

Regarding the equation, the term on LHS goes as $\partial_{\mu}F^{\mu\nu}\to\left(P^{-1}\right)^{\kappa}_\mu\partial_{\kappa}P^\mu_\sigma P^\nu_\rho F^{\sigma\rho}=\delta^\kappa_\sigma\partial_{\kappa} P^\nu_\rho F^{\sigma\rho}=P^\nu_\rho \partial_{\mu} F^{\mu\rho}$. You will find that the RHS transforms the same way, so the effect of transformation is to multiply the whole equation by an invertible operator $P$. Thus equation does not change.

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  • $\begingroup$ Why do you get $\bar{F} = P \cdot F \cdot P$ out of $F^{\mu\nu}\to\bar{F}^{\mu\nu}=P^\mu_\eta P^\nu_\sigma F^{\eta\sigma}$? Afterwards you say that I have to apply $J^{\mu} \to \bar{J}^{\mu} = P^\mu_\nu J^\nu$ and $\partial_\mu \to \bar{\partial}_\mu = (P^{-1})^\nu_\mu\partial_\nu$. As $P^{-1} = P$ that's what I did, when I asked the question, didn't I? So taking your result for $\bar{F}$ and my results for the other two transformations I still don't quite see why the Maxwell equations are invariant under parity transformation. Could you explain it a bit more detailed? $\endgroup$ – offline Feb 9 at 6:48
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As current is a vector, it is not invariant under parity. Therefore neither is Ampère's law.

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