div(grad f ) with the Special Relativity Metric

Question

I am looking at David Kay's "Tensor Calculus" 2011 Schaum's Outline; specifically equation 12.36 on page 172 and I am having difficulty making the leap from: $$ \text{div}(\text{grad}f) \equiv \Box f= \frac{\partial^2 f}{(\partial x^0)^2}- \frac{\partial^2 f}{(\partial x^1)^2}- \frac{\partial^2 f}{(\partial x^2)^2}- \frac{\partial^2 f}{(\partial x^3)^2} $$ to the final equation: $$ \text{div}(\text{grad}f) \equiv \Box f= \frac{1}{c^2}\frac{\partial^2f}{\partial t^2}- \nabla^2f. $$ I understand the $\nabla^2f$ term but I do not understand where the $1/c^2$ multiplier comes from on the first term. I believe $x^0$ translates directly to $t$ and is therefore not "proper time" so there is no proper time scaling taking place. Would someone please explain where this $1/c^2$ multiplier comes from? Thanks.

Answer 1

Differentiation operators with respect to coordinates $x^\mu$ should be understood as the covariant components of the four vector. A four gradient is then $\partial_\mu\equiv\displaystyle\frac{\partial}{\partial x^\mu}=\left(\frac{\partial_t}{c},\nabla\right)$ and div grad is given by $\partial^\mu\partial_\mu$ from which your desired formula follows.

The $c$ comes from the way the coordinates are defined, with $x^0=ct$, which is motivated by the physics: when $x^0=ct$ the magnitude of a differential displacement $c^2dt^2-dx^2-dy^2-dz^2$ is an invariant between reference frames.