Gauge choice.

The potential $\phi$ is determined only up to an arbitrary function of time; if we set $\tilde\phi(\vec x,t)=\phi(\vec x,t)-\int^t f(s)\,ds$ then $\partial_t\tilde\phi=\partial_t\phi-f(t)$ and $\nabla\tilde\phi=\nabla\phi$. Substituting into the Bernoulli relation removes $f(t)$. Thus one may (without loss of generality) choose the gauge so $f(t)\equiv 0$.