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For coordinates $x^{\mu }=(ct,x,y,z)$, the four-gradient of a scalar field is its covector of coordinate derivatives. Under a constant Lorentz transformation, ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. In metric signature $(+,-,-,-)$, raising its index gives ${\partial }^{\mu }=(c^{-1}{\partial }_t,-\nabla )$. Contracting it with a four-vector field produces the Lorentz scalar four-divergence. In particular ${\partial }_{\mu }{J}^{\mu }=0$ is charge continuity in every inertial frame.

Take $x^{\mu }=(ct,x,y,z)$ and metric signature $(+,-,-,-)$. The four-current, four-potential and four-gradient areThe first two transform as contravariant four-vectors under an inertial-frame Lorentz transformation, within a consistently transformed electromagnetic gauge. The lowered-index gradient is a covector. Their contraction is a Lorentz scalar:Indeed, $J^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{J}^{\nu }$ and ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\alpha }{}_{\mu }{\partial }_{\alpha }$, so the two matrices cancel in the contraction. Local charge conservation in one inertial frame implies local charge conservation in every inertial frame. Integrating over a region gives the charge change equal to minus its outward current flux; global conserved charge requires no flux through the system boundary.

Write the Lorentz transformation as $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$. Acting on a scalar field, the chain rule gives the four-gradientFor the specified Lorentz boost, the inverse is obtained by changing $\beta$ to $-\beta$, soThese are precisely the covector component rules for the lowered coordinates. Raising the index with the Minkowski metric ${\eta }^{\mu \nu }=\mathrm{diag}(-1,1,1,1)$ reverses the time component, yieldingIndeed ${\partial }^{\prime 0}=\gamma ({\partial }^0-\beta {\partial }^1)$ and ${\partial }^{\prime 1}=\gamma ({\partial }^1-\beta {\partial }^0)$, the contravariant four-vector rules. Thus the index on a partial derivative with respect to an upper coordinate is lower.

The four-gradient rules from a giveThe Lorentz transformation coefficients are constant, so the partial derivatives do not differentiate them. This proves the Lorentz invariance of the d'Alembert operator:For example, the two longitudinal terms transform as $-{\gamma }^2({\partial }_0+\beta {\partial }_1{)}^2+{\gamma }^2({\partial }_1+\beta {\partial }_0{)}^2=-{\partial }_0^2+{\partial }_1^2$, using ${\gamma }^2(1-{\beta }^2)=1$ and commuting partial derivatives. The transverse terms are unchanged. Applied to a scalar field, $\square$ therefore gives a Lorentz scalar.