Prove that a derivative with respect to a covariant 4-vector is a contravariant vector operator In special relativity, I know you can prove that the derivative with respect to a contravariant 4-vector component transforms like a covariant vector operator by using the chain rule, but I can't work out how to prove the inverse, that the derivative with respect to a covariant 4-vector component transforms like a contravariant vector operator.
 A: $$\partial'^{\mu}=\frac{\partial}{\partial x'_{\mu}}=\frac{\partial x'^{\lambda}}{\partial x'_{\mu}} \frac{\partial x^{\sigma}}{\partial x'^{\lambda}} \frac{\partial x_{\nu}}{\partial x^{\sigma}} \frac{\partial}{\partial x_{\nu}} = g^{\mu \lambda}\Lambda_{\lambda}^{\sigma}g_{\sigma \nu} \frac{\partial}{\partial x_{\nu}}=\Lambda^{\mu}_{\nu}\partial^{\nu}$$
So contravariant.
A: Let's have the inverse transformations for 4-vector components:
$$
\tag 1 \mathbf r = \mathbf r' + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r' )}{c^{2}} + \gamma \mathbf u t{'} , \quad t = \gamma \left(t' + \frac{(\mathbf u \cdot \mathbf r' )}{c^{2}}\right).
$$
Here $\Gamma = \frac{(\gamma - 1)}{\frac{u^{2}}{c^{2}}}$.
Then by using chain rule
$$
\frac{\partial }{\partial t' } = \frac{\partial t}{\partial t'}\frac{\partial }{\partial t} + \frac{\partial x_{j}}{\partial t{'}}\frac{\partial }{\partial x_{j}}, \quad \frac{\partial }{\partial x_{i}'} = \frac{\partial t}{\partial x_{i}'}\frac{\partial }{\partial t} + \frac{\partial x_{j}}{\partial x_{i}'}\frac{\partial }{\partial x_{j}}
$$
you may get by using $(1)$
$$ 
\frac{\partial t}{\partial t'} = \gamma, \quad \frac{\partial x_{j}}{\partial t{'}} = \gamma u_{j}, \quad \frac{\partial t}{\partial x_{i}'} = \frac{\gamma u_{i}}{c^{2}},
$$
$$\frac{\partial x_{j}}{\partial x_{i}{'}} = \delta_{ij} + \Gamma \frac{u_{j}u_{i}}{c^{2}} \Rightarrow 
$$
$$
\nabla_{i}{'} = \frac{\gamma u_{i}}{c^{2}}\frac{\partial }{\partial t} + \sum_{j}\left(\frac{\partial }{\partial x_{j}}\delta_{ij} + \frac{\Gamma u_{i}u_{j}}{c^{2}}\frac{\partial }{\partial x_{j}}\right) = \frac{\gamma u_{i}}{c^{2}}\frac{\partial }{\partial t} + \frac{\partial }{\partial x_{i}} + \frac{\Gamma}{c^{2}}u_{i}(\mathbf u \nabla).
$$
From these equations you may get
$$\frac{1}{c}\frac{\partial}{\partial t'} = \frac{1}{c}\gamma\frac{\partial }{\partial t} + \frac{1}{c}\gamma \sum_{j}u_{j}\frac{\partial }{\partial x_{j}} = \gamma \left( \frac{1}{c}\frac{\partial }{\partial t} + \frac{1}{c}(\mathbf u \nabla )\right),
$$
$$ 
\nabla ' = \nabla + \Gamma\frac{\mathbf u}{c^{2}}(\mathbf u \nabla) + \frac{\gamma \mathbf u}{c^{2}}\frac{\partial}{\partial t}.
$$
So $\partial_{\alpha}$ operator transforms as contravariant vector.
