Probability 4-current density confusion

Question

We know that the relativistic probability 4-current density can be written as $$ j^{\mu} = \frac{i\hbar}{2m}(\psi^*\partial^{\mu}\psi - \psi\partial^{\mu}\psi^*) $$

I am bit confused about the time part of this density. From various sources it seems that the time part, using (+---) metric, is $$ j^{0} = \frac{i\hbar}{2mc^2}(\psi^*\partial_{t}\psi - \psi\partial_{t}\psi^*) $$

But why is there a $\mathbf{c^2}$ in this equation when the 4-gradient is clearly written as $$ \partial^{\mu} = \left(\frac{\partial_{t}}{c},\nabla\right) $$ And therefore if we take the $1/c$ out from the parentheses, we should only get the power of one. Please correct me if I am making some horrible mathematical mistake or I am missing something from relativity. It's not my domain, I'm just interested.

Answer 1

It comes from $$\partial_\mu j^\mu=0=\partial_t\rho+\vec\nabla\cdot\vec j$$ Because the form with just one factor of $c$ is $$\frac1c\partial_t(c\rho)$$ then you will get $$c\rho=\frac{i\hslash}{2m}\left(\psi^*\frac1c\partial_t\psi-\psi\frac1c\partial_t\psi^*\right)$$