The formal definition for Lorentz Transformation is a matrix $\Lambda$ such that $$\Lambda^\mu_{\ \ \alpha}\Lambda^\nu_{\ \ \beta}\eta_{\mu\nu}=\eta_{\alpha\beta.}$$ In some books I have found a definition that use the transposition: $$(\Lambda^T)\eta\Lambda=\eta.$$ My question is how to link them. My attempt, so far, is to multiply by the inverse but I get stuck very soon and I don't know how to reach the second equation. Probably the passagges are trivial. Thanks for any help.


2 Answers 2


Kontle's answer contains the main idea, but let me be a bit more specific about your example: all is about matrix notation.

Recall in general linear algebra that, given $n \times n$ matrices $X,A \in \mathbb{R}^{n \times n}$ one can write the $(i,j)$-coefficient of the matrix $B= X^TAX$ as follows: \begin{equation} (b_{ij})=\sum_{l=1}^n\left(\sum_{k=1}^n x_{ki}a_{kl}\right)x_{lj} = \sum_{k,l=1}^nx_{ki}a_{kl}x_{lj}, \end{equation}

To check this try to write explicitly the matrices and try to write in summation notation what you are multiplying (remembering that since at the beginning you have $X^T$, multiplying rows of $X$ is the same as multiplying columns of $X^T$).

More specifically, in your notation and for a $4 \times 4$ matrix, you have:

$B= \Lambda^T\eta \Lambda$ as follows: \begin{equation} (b_{\alpha\beta})=\sum_{\nu=0}^3\left(\sum_{\mu=0}^3 \Lambda_{\mu\alpha}\eta_{\mu\nu}\right)\Lambda_{\nu\beta} = \sum_{\mu,\nu=0}^3\Lambda_{\mu\alpha}\eta_{\mu\nu}\Lambda_{\nu\beta}, \end{equation}

and now you agree that by standard commutativity of product of real numbers you get: $$= \sum_{\mu,\nu=0}^3\Lambda_{\mu\alpha}\eta_{\mu\nu}\Lambda_{\nu\beta}= \sum_{\mu,\nu=0}^3\Lambda_{\mu\alpha}\Lambda_{\nu\beta}\eta_{\mu\nu}$$

Now, writing it with Einstein's convention you get exactly what you needed:

$$= \sum_{\mu,\nu=0}^3\Lambda_{\mu\alpha}\Lambda_{\nu\beta}\eta_{\mu\nu} = \Lambda^\mu_{\ \ \alpha}\Lambda^\nu_{\ \ \beta}\eta_{\mu\nu}$$


This has to simply to do with matrix multiplication. If you have a matrix $A$ that multiplies a vector $x$, this can be written as

$$ A_{ij}x_j = A x$$

where summation over double indices is assumed. Of course you can flip the expression around, as in

$$ A_{ij}x_j = x_jA_{ij}$$ Similarly, a vector can multiply a matrix, as in $$ x_i A_{ij} = xA $$

For the second step we have to recall that a matrix is transposed as

$$ A_{ij}^T = A_{ji}$$

Combining this, you directly obtain the expression you give. Also note that since we are in the context of relativity, each of the indices over which we sum is either up or low (not both up or both down).

  • 2
    $\begingroup$ I just elaborated a bit in his specific case, but I agree you basically put all the ingredients :) $\endgroup$
    – Son Gohan
    Mar 31, 2022 at 17:29

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