Friday, 28 October 2016

Quadrivectors in relativity


This is what I understood about 4-vectors in relativity.


We define the contravariant and covariant vectors like this : $$ A^\mu=\begin{bmatrix} A^0 \\ A^1 \\ A^2 \\ A^3 \end{bmatrix}$$


$$ A_\mu=\begin{bmatrix} A_0 \\ A_1 \\ A_2 \\ A_3 \end{bmatrix}$$


The relationship between them will be :


$$ A^\mu=\eta^{\mu \nu}A_\nu $$


In +--- convention it will lead to :


$$ A^\mu=\begin{bmatrix} A_0 \\ -A_1 \\ -A_2 \\ -A_3 \end{bmatrix}$$


Great.



But it doesn't give me information on the "absolute" sign of 4-vectors. For example if I take the 4-position.


I have an even at time $t$ at space coordinates $(x,y,z)$.


Will I have $$X^\mu=\begin{bmatrix} t \\ x \\ y \\ z \end{bmatrix}$$ Or


$$X_\mu=\begin{bmatrix} t \\ x \\ y \\ z \end{bmatrix}$$ I think it is the first answer because $A^\mu$ should transform the same way that the "real" coordinates $(t,x,y,z)$ transform, but I am not totally sure ?


Thank you.




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