Why is it that $$\partial_\mu\partial^\mu=\partial_t^2-\nabla^2$$ (this I believe is called the D'Alembert operator.) but $$\partial_\mu j^\mu=\dot{j^0}+\nabla\cdot \vec j?$$
Why is there a minus on the first example but a plus in the second?
Thank you.
Hint: Vectors are naturally raised ($j^\mu$ for $\mu$ in $\{1,2,3\}$ is the $\mu$th component of the three-vector ${\bf j}$) while the derivative is naturally lowered ($\partial_\mu$ for $\mu$ in $\{1,2,3\}$ is the $\mu$th component of the three-vector $\nabla$). The minus sign in the d'Alembertian comes from the metric. The plus sign in the divergence comes from the metric and the convention that the derivative is naturally lowered.