In special relativity, the coordenates of a event are in general written using a 4-vector: $$x^{\mu} = \binom{ct}{\textbf{x}}$$ where $\textbf{x} = (x,y,z)$ are the spacial coordenates. This is a contravariant vector, and its covariant representation is written as $$x_{\mu} = \binom{ct}{-\textbf{x}}.$$ However, I know that the contravariant and covariant vectors are defined by the way that their components transforms. If $A$ is contravariant and $B$ is covariant: $$A'^{\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}} A^{\nu}, \ \ \ \ B' _{\mu} = \frac{\partial x^{\mu}}{\partial x'^{\nu}} B_{\nu}.$$ Using this definition, how can I show that $x_{\mu}$ is in fact covariant?

  • $\begingroup$ Note that $x^\mu$ only behaves like a vector if the coordinate transformations are Lorentz transforms (or more generally linear transformations). $\endgroup$ – jacob1729 May 24 at 14:30
  • $\begingroup$ If it eats covariant vectors (and spits out scalars),it's contravariant. If it eats contravariant vectors, it's covariant. $\endgroup$ – WillO May 24 at 14:30
  • $\begingroup$ The covariant vector should be a row vector. $\endgroup$ – Jon May 24 at 14:56

As a side issue, coordinates don't really act like vectors. Infinitesimal changes in the coordinates do. It makes a difference when spacetime isn't flat or when you use coordinates that aren't Minkowski. But this is not really relevant to your question.

Consider the coordinate transformation $(t,x)\mapsto(t',x')=(\alpha t,\alpha x)$, where $\alpha$ is a positive constant. This is a change of units. By comparing this with the rules for transforming vectors and covectors, you'll see that the pair of coordinates acts like a vector, not a covector.

If the metric was given by the line element $dt^2-dx^2$ in the original coordinates, then it's given by $\alpha^{-2}dt'^2-\alpha^{-2}dx'^2$ in the new coordinates. Then lowering an index gives the components $(\alpha^{-1}t',\alpha^{-1}x')$ for the covector in the primed coordinate system. You can now check that this behaves according to the rule for transforming covectors.


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