# How to integrate continuity equation to show that the probability density does not change with time?

$$\int_V\frac{\partial\rho}{\partial t}d^3x=\frac{\partial}{\partial t}\int_V\rho d^3x=-\int_v \nabla \cdot j d^3x=-\int_Fj.dF=0$$

What does the F means? How did the author convert $\nabla\cdot j$ to $j.dF$?

Edited: Why does it equate to zero?

• What do you mean by equate to zero? If you meant why $\dfrac{\mathrm d}{\mathrm dt}\displaystyle\int_V \rho~\mathrm dV + \displaystyle\int_F \mathbf J~\mathrm dF ~=~ 0,$ then it means that the probability is locally conserved.
– user36790
Dec 15, 2016 at 3:28
• I mean the $-\int_v \nabla \cdot j d^3x=-\int_Fj.dF=0$. I know why the overall continuity equation is zero. Dec 15, 2016 at 5:15
• That would simply mean $\rho$ is independent of time. $\boldsymbol\nabla\cdot \mathbf J~=~0$ means $\mathbf J$ is constant in time everywhere - the current density is a time-independent distribution.
– user36790
Dec 15, 2016 at 5:27
• So it is not obtain from the equation but rather is a given condition in order to ensure that $\rho$ is independent of time? Dec 15, 2016 at 5:45
• Indeed; if one imposes the condition of the steady state, then the continuity equation reduces to $\boldsymbol \nabla\cdot \mathbf J ~=~ 0\,.$
– user36790
Dec 15, 2016 at 5:52

$$\int_V \left(\nabla\cdot\vec{a}\right) \mathrm{d}V = \int_S \vec{a}\cdot \vec{\mathrm{d}S}$$
where $S$ is the bounding surface of the volume $V$.
$\mathrm{d}F$ is the infinitesimal vector, which is orthogonal to the surface and its magnitude is equal to the area of area element. For the derivation of last identity see any book on vector calculus or, say, this Wikipedia article.