# Questions about special relativity, index in the Lorentz matrix

I'm studying special relativity

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?

• 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. – gingras.ol May 1 '19 at 16:27
• I know that the index that is up is the row and the indes that is down the column – MementoMori May 1 '19 at 16:34
• @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? – user4552 May 1 '19 at 17:10
• 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 $\Lambda$. – MementoMori May 1 '19 at 20:04
• @Ben Crowell have you understood? – MementoMori May 2 '19 at 11:14