Quadrivectors in relativity

Question

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.

Answer 1

The answer is that it doesn't really matter as long as you are consistent !

To see this, let us define $X^\mu = [a\ b\ c\ d]^T$ and $\tilde{X}^\mu = [a\ -b\ -c\ -d]^T$ we will see that these are equivalent up to a change of basis. Which proves that it is indeed irrelevant as long as we are consistent!

Expanding both vectors onto the basis of vectors in 4D Minkowski space $\{ \partial_t, \partial_x, \partial_y, \partial_z \}$ gives us:

$$ X = a\partial_t+b\partial_x+c\partial_y+d\partial_z\\\text{and}\\ \tilde{X}=a\partial_t - b\partial_x - c\partial_y-d\partial_z$$

Those two vectors are indeed equivalent up to an inversion of the spatial part of the manifold.

Also, if I may give you a little tip. $X^\mu$ and $X_\mu$ are contractible to a scalar you can stress this when going to matrix notation by writing: $X^\mu = [...]$ and $X_\mu = [...]^T$.

I hope this helps. Feel free to ask questions if you are still confused