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?
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$\begingroup$ Note that $x^\mu$ only behaves like a vector if the coordinate transformations are Lorentz transforms (or more generally linear transformations). $\endgroup$– jacob1729Commented May 24, 2019 at 14:30
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3$\begingroup$ If it eats covariant vectors (and spits out scalars),it's contravariant. If it eats contravariant vectors, it's covariant. $\endgroup$– WillOCommented May 24, 2019 at 14:30
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1$\begingroup$ The covariant vector should be a row vector. $\endgroup$– JonCommented May 24, 2019 at 14:56
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$\begingroup$ Strictly speaking, a vector $\bf X$ is neither contravariant nor covariant. It is a vector. It does not change when the coordinate system changes, but its components do. The components $x^\mu$ are contravariant components; the components $x_\mu$ are covariant components. $\endgroup$– Andrew SteaneCommented Mar 11, 2020 at 13:54
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2 Answers
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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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Simply by looking what it's a function of. If it's a multilinear function of vectors than it's covariant. If it's a function of Co vectors(linear functionals) it's contravariant.
