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I'm studying special relativity

I have read this:

We have $ x^u = (ct, x^1,x^2,x^3) $. If we apply Lorentz transformation we can write:

$x'^u = \Lambda^{u}_{\hspace{0,2 cm}\nu} x^{\nu} $

$x'_u = \Lambda_{u}^{\hspace{0,2 cm}\nu} x_{\nu} $

Where he have defined the Lorentz matrix:

$\Lambda^{u}_{\hspace{0,2 cm}\nu} (v) = \begin{bmatrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0& 1 \end{bmatrix} $ ; $\qquad\Lambda_{u}^{\hspace{0,2 cm}\nu} (v) = \begin{bmatrix} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0& 1 \end{bmatrix} $

Is the space between the $\Lambda$ and $ \nu$ (down in the former and up in the latter) written only to indicate that are two different matrix or is there something more?

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  • $\begingroup$ The order in which $u$ and $v$ appear is important as they are the tensor's indices. There is a space between $u$ and $v$ because $u$ corresponds to the row and $v$ to the column. $\endgroup$
    – gingras.ol
    May 1, 2019 at 16:27
  • $\begingroup$ I know that the index that is up is the row and the indes that is down the column $\endgroup$
    – MementoMori
    May 1, 2019 at 16:34
  • $\begingroup$ @MementoMori: gingras.ol's comment seems to me like an answer to your question, but it seems like it doesn't satisfy you as an answer. Could you clarify what your question is? $\endgroup$
    – user4552
    May 1, 2019 at 17:10
  • $\begingroup$ I don't understand why a letter is closer than the other letter and moreover why in the first matrix he put $u$ close to $\Lambda$ and in the second one he puts the low index close to $\Lambda$. $\endgroup$
    – MementoMori
    May 1, 2019 at 20:04
  • $\begingroup$ @Ben Crowell have you understood? $\endgroup$
    – MementoMori
    May 2, 2019 at 11:14

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The comment of gingras.ol is useful but this link is more useful to clarify my doubt

Difference between slanted indices on a tensor

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