homework and exercises - Current Density and the Dipole Moment $int _v vec{J} dV = d vec{p}/dt$

Show that:

$$\int _v \vec{J} dV = \frac{d \vec{p}}{dt}$$

Here's my attempt:

$$\vec{p} = \int _v \rho \vec{r} dV \rightarrow \frac{d \vec{p}}{dt}= \frac{d}{dt} \int _v \rho \vec{r} dV = \int _v \frac{\partial \rho}{\partial t}\vec{r}$$

Because of the fact that $\nabla \cdot \vec{J} = - \frac{\partial \rho}{dt}$ we have

$$\frac{d \vec{p}}{dt}= -\int _v (\nabla \cdot \vec{J})\vec{r}dV$$

I'm not exactly sure how to turn this divergence of the current density per unit volume into the original $\int _v \vec{J} dV$

The last expression is very close to equaling the current, maybe we could use that?

Answer

You have the right idea. Let's use index notation for convenience. We have

$$\begin{align} p^i(t) = \int_{\mathbb R^3} d^3x \,x^i\rho(t,\mathbf x) \end{align}$$ and as you note, we have the continuity equation; $$\begin{align} \partial_jJ^j = -\frac{\partial\rho}{\partial t} \end{align}$$ so we get $$\begin{align} \dot p^i = -\int_{\mathbb R^3} d^3 x \,x^i\partial_j J^j \end{align}$$ where I'm using the summation convention for the dot product. This is the point to which you've basically gotten. The trick is now to essentialy perform an integration by parts. In other words, we use the product rule for differentiation to note that $$\begin{align} \partial_j(x^i J^j) &= (\partial_j x^i)J^j + x^i \partial_jJ^j \\ &= \delta^i_jJ^j +x^i\partial_jJ^j \\ &= J^i + x^i\partial_jJ^j \end{align}$$ and plugging this into the above expression for $\dot p^i$ gives $$\begin{align} \dot p^i = -\int_{\mathbb R^3} d^3 x \partial_j(x^iJ^j) + \int_{\mathbb R^3} d^3x J^i \end{align}$$ the first term is the volume integral of the divergence of $x^iJ^j$. Since we are integrating over all space, this turns into a boundary term at infinity which vanishes for any finite (or sufficiently rapidly decaying) charge density. The result you want then follows $$\begin{align} \dot p^i = \int_{\mathbb R^3} d^3 x J^i \end{align}$$