Demostrar que una derivada con respecto a un cuadrivector covariante es un operador vectorial contravariante

$$\partial'^{\mu}=\frac{\partial}{\partial x'_{\mu}}=\frac{\partial x'^{\lambda}}{\partial x'_{\mu}} \frac{\partial x^{\sigma}}{\partial x'^{\lambda}} \frac{\partial x_{\nu}}{\partial x^{\sigma}} \frac{\partial}{\partial x_{\nu}} = g^{\mu \lambda}\Lambda_{\lambda}^{\sigma}g_{\sigma \nu} \frac{\partial}{\partial x_{\nu}}=\Lambda^{\mu}_{\nu}\partial^{\nu}$$ Tan contravariante.

Tengamos las transformaciones inversas para componentes de 4 vectores:

$$\tag 1 \mathbf r = \mathbf r' + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r' )}{c^{2}} + \gamma \mathbf u t{'} , \quad t = \gamma \left(t' + \frac{(\mathbf u \cdot \mathbf r' )}{c^{2}}\right).$$

Aquí$\Gamma = \frac{(\gamma - 1)}{\frac{u^{2}}{c^{2}}}$.

Entonces usando la regla de la cadena

$$\frac{\partial }{\partial t' } = \frac{\partial t}{\partial t'}\frac{\partial }{\partial t} + \frac{\partial x_{j}}{\partial t{'}}\frac{\partial }{\partial x_{j}}, \quad \frac{\partial }{\partial x_{i}'} = \frac{\partial t}{\partial x_{i}'}\frac{\partial }{\partial t} + \frac{\partial x_{j}}{\partial x_{i}'}\frac{\partial }{\partial x_{j}}$$

puedes obtener usando$(1)$

$$\frac{\partial t}{\partial t'} = \gamma, \quad \frac{\partial x_{j}}{\partial t{'}} = \gamma u_{j}, \quad \frac{\partial t}{\partial x_{i}'} = \frac{\gamma u_{i}}{c^{2}},$$

$$\frac{\partial x_{j}}{\partial x_{i}{'}} = \delta_{ij} + \Gamma \frac{u_{j}u_{i}}{c^{2}} \Rightarrow$$ $$\nabla_{i}{'} = \frac{\gamma u_{i}}{c^{2}}\frac{\partial }{\partial t} + \sum_{j}\left(\frac{\partial }{\partial x_{j}}\delta_{ij} + \frac{\Gamma u_{i}u_{j}}{c^{2}}\frac{\partial }{\partial x_{j}}\right) = \frac{\gamma u_{i}}{c^{2}}\frac{\partial }{\partial t} + \frac{\partial }{\partial x_{i}} + \frac{\Gamma}{c^{2}}u_{i}(\mathbf u \nabla).$$

De estas ecuaciones se puede obtener

$$\frac{1}{c}\frac{\partial}{\partial t'} = \frac{1}{c}\gamma\frac{\partial }{\partial t} + \frac{1}{c}\gamma \sum_{j}u_{j}\frac{\partial }{\partial x_{j}} = \gamma \left( \frac{1}{c}\frac{\partial }{\partial t} + \frac{1}{c}(\mathbf u \nabla )\right),$$

$$\nabla ' = \nabla + \Gamma\frac{\mathbf u}{c^{2}}(\mathbf u \nabla) + \frac{\gamma \mathbf u}{c^{2}}\frac{\partial}{\partial t}.$$ Entonces $\partial_{\alpha}$el operador se transforma como vector contravariante.