Equivalence of two Distinct Definitions of the Current Density $\textbf{J}$

The current density is defined as:

$$\textbf{J} (\textbf{r},t) := \rho(\textbf{r},t) \cdot \textbf{v}(\textbf{r},t)$$

where $$\rho (\textbf{r},t)$$ and $$\textbf{v}(\textbf{r},t)$$ is the charge density and the particle velocity at point $$\textbf{r}$$ and at time $$t$$. Based on that definition, how could one derive the following relation?

$$I = \iint_{\mathcal{A}} \textbf{J}(\textbf{r},t)\cdot d\textbf{S}$$

where $$\mathcal{A}$$ is a surface. I am asking this because authors tend to define the current density $$\textbf{J}$$ through the latter equation (having the current $$I$$ predefined). I wish to understand how the two are equivalent.

• Isn't this the definition of current if $\mathbf{J}$ is defined like you say? Oct 4 '19 at 22:27
• Nice one. But let's assume that $I$ is separately defined as: $$I = \frac{dQ_{S}}{dt}$$ where $Q_S$ is the amount of charge passing through an area $S$. Oct 4 '19 at 23:17
• Thanks for clarifying that point by the way. It is really important. What I am asking here is for a proof that demonstrates the equivalence of the two definitions. Oct 4 '19 at 23:21

$$\textbf{J} (\textbf{r},t) = \rho(\textbf{r},t) \textbf{v}(\textbf{r},t)$$
Let $$\Delta Q(\textbf{r},t)$$ be the charge that crosses the infinitesimal surface $$\Delta S$$ in time $$\Delta t$$. You can clearly see that the amount of charge that crosses is the amount of charge present in the volume $$\Delta S \textbf{v}(\textbf{r},t)\Delta t \cdot \hat{\bf n}$$, $$\hat{\bf n}$$ being the normal to $$\Delta S$$. I can then write $$\frac{\Delta Q(\textbf{r},t)}{\Delta t} = \rho(\textbf{r},t) \textbf{v}(\textbf{r},t)\cdot \hat{ \bf n} \Delta S = \textbf{J}(\textbf{r},t)\cdot \hat{\bf n} \Delta S$$
$$I(t) = \frac{dQ_S(t)}{dt}= \int_{\mathcal{A}} \frac{dQ(\textbf{r},t)}{dt} = \iint_{\mathcal{A}} \textbf{J}(\textbf{r},t)\cdot d\textbf{S}$$ .