# Can I transform electromagnetic tensors by matrix multiplication?

I know that the eletromagnetic field tensor $F^{\mu\nu}$, can be transfomed to another reference frame by $$F^{\alpha\beta} = \varLambda^{\alpha}_{\mu}\varLambda^{\beta}_{\nu}F^{\mu\nu}$$

Since these tensors can be represented by matrices so I thought that I could represent the electromagnetic field tensor in another inertial reference frame by doing matrix multiplications, but I ended up with:

$$F^{\alpha\beta}= \begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & -\frac{1}{c}E_x & -\frac{1}{c}E_y & -\frac{1}{c}E_z \\ \frac{1}{c}E_x & 0 & -B_z & B_y\\ \frac{1}{c}E_y & B_z &0 & -B_x\\ \frac{1}{c}E_z & -B_y & B_x & 0 \end{bmatrix} =$$ $$=\begin{bmatrix} -2\frac{\gamma^2\beta}{c}E_x & -\frac{\gamma^2}{c}E_x(1+\beta^2)& -\frac{\gamma^2}{c}E_y(1+\beta^2)+2\gamma^2\beta^2B_z & -\frac{\gamma^2}{c}E_z(1+\beta^2)-2\gamma^2\beta B_y \\ \frac{1}{c}E_x & 2\frac{\gamma^2\beta}{c}E_x & 2\frac{\gamma^2\beta}{c}E_y-\gamma^2B_z(\beta^2+1) & 2\frac{\gamma^2\beta}{c}E_z+\gamma^2B_y(\beta^2+1)\\ \frac{1}{c}E_y & B_z &0 & -B_x\\ \frac{1}{c}E_z & -B_y & B_x & 0 \end{bmatrix}$$ But as it can easily be seen, this matix is not anti-symmetric as an eletromagnetic field tensor should be, by definition, so does this mean that I cannot use matrix multiplication on tensors or does it mean that I made a mistake somewhere in the calculations of the matrix products?

• This is an example of a similarity transformation. Nov 25, 2017 at 5:17
• @Bobak Hashemi : This is NOT a similarity transformation : $F'=\Lambda^{-1}F\Lambda$ . Here we have $F'=\Lambda F \Lambda$ and $\Lambda^{-1}\ne \Lambda$. Nov 25, 2017 at 7:12
• If they used $F_\mu^\nu$, then you would have the desired form (since F then acts like a matrix by taking a vector and covector to a number). In any case, this is clearly the same idea as similarity transformation, you are rotating the coordinates for each index. F is expressed in one coordinate system, and you can find how it looks in another coordinate system by sandwiching it between the coordinate transformations for vectors, $\Lambda$, and covectors, $\Lambda^{-1}$. Nov 25, 2017 at 7:48
• @Bobak Hashemi : Yes, precisely. But the OP using $F^{\mu\nu}$ is already confused, so it's not good timing to refer to similarity transformations without your explanation comment above. Nov 25, 2017 at 8:01

## 3 Answers

\begin{align} F^{\alpha\beta} & = \begin{bmatrix} \hphantom{-}\gamma & -\gamma\beta & \hphantom{-} 0 & \hphantom{-}0\hphantom{-} \vphantom{\dfrac12}\\ -\gamma\beta & \hphantom{-}\gamma & \hphantom{-}0 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-}0 & \hphantom{-}0 & \hphantom{-}1 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 1 \hphantom{-} \vphantom{\dfrac12} \end{bmatrix} \begin{bmatrix} 0 & -E_x & -E_y & -E_z \vphantom{\dfrac12}\\ E_x & \hphantom{-} 0 & -cB_z & \hphantom{-} cB_y \vphantom{\dfrac12} \\ E_y & \hphantom{-} cB_z & \hphantom{-} 0 & -cB_x \vphantom{\dfrac12} \\ E_z & -cB_y & \hphantom{-} cB_x & \hphantom{-} 0 \vphantom{\dfrac12} \end{bmatrix} \begin{bmatrix} \hphantom{-}\gamma & -\gamma\beta & \hphantom{-} 0 & \hphantom{-}0\hphantom{-} \vphantom{\dfrac12}\\ -\gamma\beta & \hphantom{-}\gamma & \hphantom{-}0 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-}0 & \hphantom{-}0 & \hphantom{-}1 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 1 \hphantom{-} \vphantom{\dfrac12} \end{bmatrix} \nonumber\\ & = \begin{bmatrix} \!\!-\gamma\beta E_x &\!\! -\gamma E_x & -\gamma(E_y-\beta cB_z) & -\gamma(E_z+\beta cB_y)\vphantom{\dfrac12} \hphantom{-} \\ \hphantom{-}\gamma E_x & \!\gamma\beta E_x & \hphantom{-}\gamma(\beta E_y- cB_z) & \hphantom{-}\gamma(\beta E_z+cB_y)\vphantom{\dfrac12}\hphantom{-}\\ \hphantom{\gamma\beta} E_y & \hphantom{-} cB_z &0 & -cB_x\vphantom{\dfrac12}\hphantom{-}\\ \hphantom{\gamma\beta} E_z & -cB_y & cB_x & \hphantom{-}0\vphantom{\dfrac12}\hphantom{-} \end{bmatrix} \begin{bmatrix} \hphantom{-}\gamma & -\gamma\beta & \hphantom{-} 0 & \hphantom{-}0\hphantom{-} \vphantom{\dfrac12}\\ -\gamma\beta & \hphantom{-}\gamma & \hphantom{-}0 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-}0 & \hphantom{-}0 & \hphantom{-}1 & \hphantom{-} 0 \hphantom{-} \vphantom{\dfrac12}\\ \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 0 & \hphantom{-} 1 \hphantom{-} \vphantom{\dfrac12} \end{bmatrix} \nonumber\\ & =\begin{bmatrix} \hphantom{-}0 & -E_x & -\gamma(E_y-\beta cB_z) &-\gamma(E_z+\beta cB_y)\hphantom{-}\vphantom{\dfrac12} \\ \hphantom{-} E_x & 0 & \hphantom{-}\gamma(\beta E_y- cB_z) & \hphantom{-}\gamma(\beta E_z+cB_y)\hphantom{-}\vphantom{\dfrac12}\\ \hphantom{-}\gamma(E_y-\beta cB_z) & -\gamma(\beta E_y- cB_z) &\hphantom{-} 0 & -cB_x\hphantom{-}\vphantom{\dfrac12}\\ \hphantom{-}\gamma(E_z+\beta cB_y) & -\gamma(\beta E_z+cB_y) & \hphantom{-} cB_x & \hphantom{-}0\hphantom{-}\vphantom{\dfrac12} \end{bmatrix} \nonumber\\ & =\begin{bmatrix} \hphantom{-}0 & -E'_x & -E'_y & -E'_z \hphantom{-}\vphantom{\dfrac12}\\ \hphantom{-}E'_x & \hphantom{-} 0 & -cB'_z & \hphantom{-} cB'_y \hphantom{-}\vphantom{\dfrac12} \\ \hphantom{-}E'_y & \hphantom{-} cB'_z & \hphantom{-} 0 & -cB'_x \hphantom{-}\vphantom{\dfrac12} \\ \hphantom{-}E'_z & -cB'_y & \hphantom{-} cB'_x & \hphantom{-} 0 \hphantom{-}\vphantom{\dfrac12} \end{bmatrix} \tag{01} \end{align} Since $\:\beta=\upsilon/c$ \begin{align} E'_x & = \hphantom{-}E_x \tag{02.x}\vphantom{\dfrac12}\\ E'_y & = \gamma(E_y-\upsilon B_z) \tag{02.y}\vphantom{\dfrac12}\\ E'_z & = \gamma(E_z+\upsilon B_y) \tag{02.z}\vphantom{\dfrac12}\\ B'_x & = \hphantom{-}B_x \tag{03.x}\vphantom{\dfrac12}\\ B'_y & = \gamma(B_y+\dfrac{\upsilon}{c^{2}} E_z) \tag{03.y}\\ B'_z & = \gamma(B_z-\dfrac{\upsilon}{c^{2}} E_y) \tag{03.z} \end{align}

For Your Information :

The equations of a more general Lorentz Transformation between two systems $\:\mathrm S(\mathbf{x},t)\:$ and $\:\mathrm S'(\mathbf{x}',t')\:$, the latter translating with constant velocity $\:\mathbf{v}\!=\!\upsilon\mathbf{n}\,,\Vert\mathbf{n}\Vert=1\,, \upsilon \in (-c,+c)$, with respect to the former, are : \begin{align} \mathbf{x}'& \!=\!\mathbf{x}\!\boldsymbol{+}\!(\gamma\!\boldsymbol{-}\!1)(\mathbf{n}\boldsymbol{\cdot}\mathbf{x})\mathbf{n}\!\boldsymbol{-}\!\gamma\mathbf{v}t \tag{ft-01a}\\ t' & \!=\! \gamma\left(t\!\boldsymbol{-}\!\dfrac{\mathbf{v}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right) \tag{ft-01b}\\ \gamma & \!=\!\left(1\!\boldsymbol{-}\!\dfrac{\upsilon^{2}}{c^{2}}\right)^{\boldsymbol{-}1/2} \tag{ft-01c} \end{align} see Figure.(1)

Under (ft-01) the vectors $\:\mathbf{E},\mathbf{B}\:$ of the electromagnetic field in empty space are transformed as follows :
\begin{align} \mathbf{E}'& \!=\!\gamma\mathbf{E}\!\boldsymbol{-}\!(\gamma\!\boldsymbol{-}\!1)(\mathbf{n}\boldsymbol{\cdot}\mathbf{E})\mathbf{n}\boldsymbol{+}\:\gamma\left(\mathbf{v}\boldsymbol{\times}\mathbf{B}\right) \tag{ft-02a}\\ \mathbf{B}'& \!=\!\gamma\mathbf{B}\!\boldsymbol{-}\!(\gamma\!\boldsymbol{-}\!1)(\mathbf{n}\boldsymbol{\cdot}\mathbf{B})\mathbf{n}\!\boldsymbol{-}\!\dfrac{\gamma}{c^{2}}\left(\mathbf{v}\boldsymbol{\times}\mathbf{E}\right) \tag{ft-02b} \end{align} Equations (02),(03) are a special case of (ft-02) for $\:\mathbf{n}=(1,0,0)$.

(1) See a 3D version of this Figure here : Figure 3D version

Check the definition of matrix multiplication. One of the two matrices of the Lorentz transform should be to the right of the matrix for electromagnetic field, but watch for other subtleties.

Matrix multiplication is like contracting the column-index of the first tensor with the file-index of the second tensor, but in this case you should contract the column-index in both tensors.