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So I am studying special relativity and have been introduced to basic tensor calculus used in the theory. Recently, I came across a statement that is confusing me: $$\Lambda^\mu_{\,\,\nu} x^\nu = x^\nu \Lambda^\mu_{\,\,\nu}$$ where $\Lambda^{u} _{v}$ is the Lorentz transformation matrix and $x^u$ is a 4-vector. Now what I don't understand is why this is the case? More specifically why is it possible to swap the order of the 4-vector and Lorentz matrix, I thought that matrix multiplication was not commutative and so this should be wrong.

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  • $\begingroup$ Are you sure about the placement of the indices? Specifically, the $u$ index should appear in the bottom for the Lorentz transformation, not at the top. Otherwise, that's not a valid expression. $\endgroup$
    – enumaris
    Oct 24, 2018 at 20:05
  • $\begingroup$ @enumaris oh sorry yeah I edited the question $\endgroup$
    – user63248
    Oct 24, 2018 at 20:09
  • $\begingroup$ ... and also are you sure there is no difference between $\Lambda^\mu_{\,\,\nu}$, $\Lambda^{\,\,\mu}_\nu$ and $\Lambda^\mu_\nu$? $\endgroup$ Oct 24, 2018 at 20:13
  • $\begingroup$ @ZeroTheHero I improved the "notation". I wasn't aware of the appropriate Mathjax syntax for the notation $\endgroup$
    – user63248
    Oct 24, 2018 at 20:17

1 Answer 1

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You are not swapping the order of the 4-vector and the Lorentz matrix, this notation is contracted. What this equation is saying is that: $$\sum_u\Lambda^{v} _{u} x^u =\sum_u x^u \Lambda^{v} _{u}$$

So the symbols in the summation are actually components of the vector and the matrix. Being components, just numbers, they surely commute, and you can change their order without any problems.

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  • $\begingroup$ Ok so $$\Lambda^\mu_{\,\,\nu} x^\nu = \Lambda^\mu_{\,\,\nu} x^0 + \Lambda^\mu_{\,\,\nu} x^1 +\Lambda^\mu_{\,\,\nu} x^2 +\Lambda^\mu_{\,\,\nu} x^3$$ and NOT $$ \Lambda^\mu_{\,\,\nu} \begin{pmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \end{pmatrix}$$? Am I right? $\endgroup$
    – user63248
    Oct 25, 2018 at 12:31
  • $\begingroup$ Almost, the right answer is:$$ \Lambda^\mu_{\,\,\nu} x^\nu = \Lambda^\mu_{\,\,0} x^0 + \Lambda^\mu_{\,\,1} x^1 +\Lambda^\mu_{\,\,2} x^2 +\Lambda^\mu_{\,\,3} x^3$$ $\endgroup$
    – Hugo V
    Oct 25, 2018 at 12:38

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