 Solution

Use $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and restore $\hslash$ temporarily. The free Dirac equation is $(i\hslash {\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$, with $c=1$. Its Hermitian conjugate is $-i\hslash ({\partial }_{\mu }{\psi }^{†})({\gamma }^{\mu }{)}^{†}-m{\psi }^{†}=0$. Multiplying on the right by ${\gamma }^0$ and using $({\gamma }^{\mu }{)}^{†}{\gamma }^0={\gamma }^0{\gamma }^{\mu }$ gives the adjoint equationThe Dirac adjoint and original equations give $({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }\psi =im\bar{\psi }\psi /\hslash$ and $\bar{\psi }{\gamma }^{\mu }{\partial }_{\mu }\psi =-im\bar{\psi }\psi /\hslash$. Adding gives the continuity equationThe Dirac current transforms as a four-vector; the positive density is the time component, not a Lorentz scalar. With ordinary units the number current is $c\bar{\psi }{\gamma }^{\mu }\psi$ and the electron charge current is $-e$ times it.