# Transforming the field strength tensor - Why do we need to use $\Lambda^T$, instead of a row times row multiplication?

The transformation law is \begin{align} F'^{\mu\nu} = {\Lambda^{\mu}}_{\alpha} {\Lambda^{\nu}}_{\beta} F^{\alpha \beta} = {\Lambda^{\mu}}_{\alpha} F^{\alpha \beta} {\Lambda^{\nu}}_{\beta} \end{align} We are given $$F$$ as a matrix with respect to the basis $$e_\mu \otimes e_\nu$$, the entries of this matrix are then essentially also the coefficients $$F^{\mu \nu}$$. In the same way, we are given the Lorentz transformation $$\Lambda$$ as a matrix as well. We can then interpret $${\Lambda^{\mu}}_{\alpha} F^{\alpha \beta}$$ as the usual matrix multiplication: \begin{align} {\Lambda^{\mu}}_{\alpha} F^{\alpha \beta} = (\Lambda \times F)^{\mu \beta} = G^{\mu \beta} \end{align} However, for the 2nd tensor contraction \begin{align} G^{\mu \beta} {\Lambda^\nu }_\beta \end{align} the wrong order of indices doesn't permit such a way of calculating the coefficients at first glance. However, doesn't the Einstein summation convention (that is implicitly used in the tensor contraction operation) allow us to write: \begin{align} G^{\mu \beta} {\Lambda^\nu}_\beta = G^{\mu 0} {\Lambda^\nu}_0 + G^{\mu 1} {\Lambda^\nu}_1 + G^{\mu 2} {\Lambda^\nu}_2 + G^{\mu 3} {\Lambda^\nu}_3 \end{align}? I tried to calculate it this way, using the entries of $$\Lambda$$ as the coefficients $${\lambda^{\nu}}_0$$, but I don't get the right result, that I would get by calculating $$\Lambda F \Lambda^T$$. So where is my mistake? Is the Einstein summation convention not applicable here?

• My advice is to not get too enthusiastic about thinking about second-rank tensors as matrices. What will you do when you need to transform a higher-rank tensor? Commented Dec 13, 2022 at 1:41
• @Ghoster exactly what I try is not to use the matrix ( but only the coefficients), and calculate the tensor contraction in the way it is defined. But the result is wrong, and that's what I'd like to know about. Commented Dec 13, 2022 at 8:18

I'm not sure what you are confused about, but the tensor transformation law is $$F'_{\alpha\beta}=\Lambda^\mu_{\phantom\mu \alpha}\Lambda^\nu_{\phantom \nu\beta}F_{\mu\nu}\tag{1}$$

Then you notice that $$\Lambda^\mu_{\phantom\mu\alpha}=(\Lambda^T)_{\alpha}^{\phantom \alpha\mu}$$. As such this is

$$F'_{\alpha\beta}=(\Lambda^T)_\alpha^{\phantom \alpha \mu}F_{\mu\nu}\Lambda^\nu_{\phantom \nu\beta}\tag{2}$$

and this is exactly the formula for matrix multiplication so that $$F'=\Lambda^T F \Lambda$$. This is the tensor transformation law in matrix form, there is nothing wrong with it or with the summation convention. In fact it is just (1) written in a different notation.

• I know that can perform the calculation in the way you did. What I am confused about is that I must do it that way, and that it is apparently not possible to simply contract the tensors in the way it is defined (using summation convention). Commented Dec 13, 2022 at 8:23
• I don't understand what you mean by "it is apparently not possible to simply contract the tensors in the way it is defined ". What the calculation shows is that it is possible and the result expressed in terms of matrices is $\Lambda^T F\Lambda$. Maybe you mean that looking at the index version you have one intuition that it should be $\Lambda F\Lambda$, but then the intuition is just wrong.
– Gold
Commented Dec 13, 2022 at 9:37
• I mean that $G^{\mu \beta} {\Lambda^\nu}_\beta = G^{\mu 0} {\Lambda^\nu}_0 + G^{\mu 1} {\Lambda^\nu}_1 + G^{\mu 2} {\Lambda^\nu}_2 + G^{\mu 3} {\Lambda^\nu}_3$ (which is usual tensor contraction) will not yield the same result as $\Lambda F \Lambda^T$. And yes, expanding the tensor contraction amounts, number-wise, to the same as calculating $\Lambda F \Lambda$, because $\Lambda$ is symmetric. So yes, my intuition is wrong, but actually it's more than my intuition (because I argue mathematicaly) - the question remains: What goes wrong there? Commented Dec 13, 2022 at 9:47
• Sorry, but $G^{\mu\beta}\Lambda^\nu_{\phantom\nu\beta}$ will give exactly $\Lambda F \Lambda^T$ and not the other result. To recognize a matrix multiplication, you must flip the indices of the $\Lambda$ matrix giving rise to the transpose. Notice that the column index of $G$ is matched with the column index of $\Lambda$, which is the row index of $\Lambda^T$.
– Gold
Commented Dec 13, 2022 at 11:08
• If one wants to write this as a matrix multiplication, you are right. But I don't want to do that. What I have written down in my last comment ($G^{\mu \beta} {\Lambda^\nu}_\beta = G^{\mu 0} {\Lambda^\nu}_0 + G^{\mu 1} {\Lambda^\nu}_1 + G^{\mu 2} {\Lambda^\nu}_2 + G^{\mu 3} {\Lambda^\nu}_3$) could be completely oblivious to matrices. All I write out here is the Einstein summation convention. Why is that wrong? Commented Dec 13, 2022 at 11:16