Difference between the contravariant, covariant and mixed form of the electromagnetic tensor and Minkowski metric tensor?

Question

What is the difference between the contravariant, covariant and mixed form of the electromagnetic tensor and Minkowski metric tensor?

I know the difference in indices (superscript and subscript). Apart from that, what is the change in the matrix?

Please help me to understand the concept. Any help will be appreciated.

Answer 1

In special relativity there are two alternative sign conventions in use for the Minkowski metric: $(+---)$ and $(-+++)$.

Here I will use the $(+---)$ sign convention, which is the same as in Electromagnetic Tensor - Definition. That means, the covariant Minkowski metric is given by the matrix $$\eta_{\mu\nu}=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}. \tag{1}$$ And the contravariant Minkowski metric is the inverse of (1): $$\eta^{\mu\nu}=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}. \tag{2}$$

As usual, the first tensor index is identified with the row number, and the second tensor index with the column number of the matrix elements.

The electromagnetic tensor in its contravariant form (i.e. with 2 upper indices) is defined in terms of the electric and magnetic field ($\vec{E}$ and $\vec{B}$) as $$F^{\mu\nu}=\begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{pmatrix}. \tag{3}$$

For calculating tensor contractions like $C_{\mu\nu}=A_\mu{}^\alpha B_{\alpha\nu}$ or $C_\mu{}^\nu=A_{\mu\alpha}B^{\alpha\nu}$ or $C^\mu{}_\nu=A^{\mu\alpha}B_{\alpha\nu}$ the use of matrix algebra helps very much. Because of

• Einstein's summing convention,
• first index is row number of matrix element,
• second index is column number of matrix element

the above contractions are just equivalent to the matrix product $C=AB$.

Then the electromagnetic tensor in its mixed form (i.e. with one upper and one lower index) can be calculated from the contravariant form (3) by lowering one index with the $\eta$ metric:

$$\begin{align} F_\mu{}^\nu&=\eta_{\mu\alpha}F^{\alpha\nu} \\ &=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix} \begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{pmatrix} \\ &=\begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & B_z & -B_y \\ E_y/c & -B_z & 0 & B_x \\ E_z/c & B_y & -B_x & 0 \end{pmatrix} \end{align}$$ and $$\begin{align} F^\mu{}_\nu&=F^{\mu\beta}\eta_{\beta\nu} \\ &=\begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{pmatrix} \begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix} \\ &=\begin{pmatrix} 0 & E_x/c & E_y/c & E_z/c \\ -E_x/c & 0 & B_z & -B_y \\ -E_y/c & -B_z & 0 & B_x \\ -E_z/c & B_y & -B_x & 0 \end{pmatrix} \end{align}$$

Likewise, the electromagnetic tensor in its covariant form (with 2 lower indices) can be calculated by lowering both indices with the $\eta$ metric. $$\begin{align} F_{\mu\nu}&=\eta_{\mu\alpha}F^{\alpha\beta}\eta_{\beta\nu} \\ &=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix} \begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{pmatrix} \begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix} \\ &=\begin{pmatrix} 0 & E_x/c & E_y/c & E_z/c \\ -E_x/c & 0 & -B_z & B_y \\ -E_y/c & B_z & 0 & -B_x \\ -E_z/c & -B_y & B_x & 0 \end{pmatrix} \end{align}$$

You can also get the relations between the covariant and contravariant forms of the Minkowski metric. $$\begin{align} \eta_{\mu\alpha}\eta^{\alpha\nu} &=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix} \begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix} \\ &=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} =\delta_\mu^\nu \end{align}$$

Answer 2

Assume we are working on a Minkowski (i.e. flat) space.

The transformation from the contravariant electromagnetic two-tensor to the covariant electromagnetic tensor is:

\begin{equation} F_{\mu \nu} = \eta_{\mu \delta} \eta_{\nu \phi} F^{\delta \phi} = \eta_{\mu \delta} F^{\delta \phi}\eta_{\phi \nu} \; \; \; \textrm{(i)} \end{equation}

Likewise, the transformation from the covariant electromagnetic two-tensor to the contravariant electromagnetic tensor is:

\begin{equation} F^{\mu \nu} = \eta^{\mu \delta} \eta^{\nu \phi} F_{\delta \phi} = \eta^{\mu \delta} F_{\delta \phi}\eta^{\phi \nu}\; \; \; \textrm{(ii)} \end{equation}

We can also convert the contravariant electromagnetic two-tensor into a "mixed" electromagnetic tensor:

\begin{equation} F^{\mu}_ {\nu} = \eta_{\nu \delta} F^{\delta \mu} \; \; \; \textrm{(iii)} \end{equation}

Similarly, we can convert the covariant electromagnetic two-tensor to a "mixed" electromagnetic tensor:

\begin{equation} F^{\nu}_ {\mu} = \eta^{\nu \delta} F_{\delta \mu} \; \; \; \textrm{(iv)} \end{equation}

We also have the following relations between the covariant and contravariant forms of the Minkowski metric tensor:

\begin{equation} \eta ^{\mu \delta} \eta_{\delta \nu} = \delta^{\mu}_{\nu} \; \; \; \textrm{(v)} \end{equation}

\begin{equation} \eta _{\mu \delta} \eta^{\delta \nu} = \delta^{\nu}_{\mu} \; \; \; \textrm{(vi)} \end{equation}

That is, the covariant and contravariant metric tensors are inverses of each other.

Of course, all of these relations hold more generally (i.e. for any metric g and tensor T). For more information in this regard, I would encourage you to look at this Wikipedia link: https://en.wikipedia.org/wiki/Raising_and_lowering_indices