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 :

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


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.

  • $\begingroup$ In addition to the answer below: The concepts of covariance and contravariance are not restricted to relativity, they are much more fundamental (see for instance en.wikipedia.org/wiki/…). It may prove useful to take a look at what they mean in the familiar 3D case, and there is an excellent description in sec.2 of ita.uni-heidelberg.de/~dullemond/lectures/tensor/tensor.pdf. The generalization to Minkowski space becomes trivial. $\endgroup$
    – udrv
    Mar 26, 2017 at 4:38
  • $\begingroup$ Shouldn't your components of $X_\mu$ be $(t,-x,-y,-z)$? $\endgroup$
    – jim
    Apr 23, 2020 at 11:56

1 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

  • $\begingroup$ Hmmm but if I have a physical event $(t,x,y,z)$, it must transform using the contravariant $\Lambda$ physically. So if I say $X_\mu=(t,x,y,z)$ I will have my physical event that will transform with a $\Lambda$ covariant so I may end with a problem no ? Thank you for your answer $\endgroup$
    – StarBucK
    Mar 25, 2017 at 16:05
  • $\begingroup$ It is indeed somewhat more natural to define $X^\mu = (t,x,y,z)$ since we would get $X = t\partial_t + x\partial_x + ... $ according to our intuition. You should realize that measuring vectors is extremely difficult(direct measurement is impossible as far as I know) such that we can only measure scalars. It than doesn't matter weather which one has the - as long as one of the two has it ! You shouldn't really think about this to much to be honest. This is the kind of things that just works out ;) $\endgroup$
    – gertian
    Mar 25, 2017 at 16:10

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