# Current density problem

1. Feb 12, 2006

### Reshma

This one is from Griffiths' book on ED.
For a configuration of charges and currents confined within a volume V, show that,
$$\int_V \vec J d\tau = \frac{d\vec p}{dt}$$
where $\vec p$ is the total dipole moment.

Well, I tried it!
$$\frac{d\vec p}{dt} = \frac{d}{dt}\int_V \rho \vec r d\tau$$

$$\frac{d\vec p}{dt} = \int_V\frac{\partial \rho}{\partial t} \vec r d\tau$$

$$\frac{d\vec p}{dt} = -\int_V \left(\vec \nabla \cdot \vec J\right) \vec r d\tau$$

Last edited: Feb 12, 2006
2. Feb 12, 2006

### George Jones

Staff Emeritus
This line is a little wrong. The partial derivative acts on evrything under the integral. Use the product rule.

Regards,
George

3. Feb 12, 2006

### Reshma

Ok, thanks, I removed that 'd' from my equation.
For the Product rule(taking only x-component here):
$$\vec \nabla \cdot \left(x\vec J\right)= x(\vec \nabla \cdot \vec J) + \vec J \cdot (\vec \nabla x)$$
Now, I am not sure what to do with this equation.

4. Feb 12, 2006

### George Jones

Staff Emeritus
What you need is

$$\frac{\partial}{\partial t} \left( \rho \vec r \right) = \frac{\partial \rho}{\partial t} \vec r + \rho \frac{\partial \vec{r}}{\partial t}.$$

Use (you did) the continuity equation on the first term.

What is the second term?

What does this tell you about the first term?

Regards,
George

5. Feb 12, 2006

### Reshma

Terrific! Thanks(now, I don't have to slog through the calculus).

The second term gives:
$$\rho \frac{\partial \vec{r}}{\partial t} = \rho \vec v = \vec J$$

The first term will go to zero, since for steady currents; $\frac{\partial \rho}{\partial t} = 0[/tex] So, $$\int_V \vec J d\tau = \frac{d\vec p}{dt}$$ 6. Feb 12, 2006 ### George Jones Staff Emeritus Sorry - we're not done! The currents aren't necessarily steady. I'm in the middle of a post in another forum - I'll be back in a few minutes. Regards, George 7. Feb 12, 2006 ### George Jones Staff Emeritus The product rule gives $$\left( \vec{\nabla} \cdot \vec{J} \right) \vec{r} = \vec{\nabla } \left( \vec{J} \cdot {r} \right) - \vec{J} \vec{\nabla} \cdot \vec{r}$$ Each term on the right vanishes. Why? Regards, George 8. Feb 12, 2006 ### Physics Monkey George, You've made a mistake here. When taking the time derivative inside the integral you've got to remember that the $$\vec{r}$$ is just an integration variable. It doesn't have any time dependence and so that full time derivative becomes a partial time derivative acting on just the charge density precisely as Reshma originally wrote. Also, your identity in the last post looks wrong, just try taking $$\vec{J}$$ to be a constant for example. Reshma, You were right to take that time derivative inside where it acts only on the charge density. You also correctly used the conservation of charge equation, and you were left with an integral of the form $$- \int_V (\vec{\nabla}\cdot \vec{J}) \,\vec{r} \, dV$$. To make progress I suggest you write the integrand in component notation as $$(\partial_\alpha J_\alpha) r_\beta$$ and then try to do an integration by parts (with the usual assumption that the current density vanishes sufficiently rapidly at infinity). Last edited: Feb 12, 2006 9. Feb 12, 2006 ### George Jones Staff Emeritus Yikes!! In both cases, I was just "going with the flow" without really thinking - trying to get the answer by doing something I preach against, i.e., just by (in this case incorrect) manipulations. When typing the vector identity, alarm bells actually started sounding, but I hit the snooze button to silence them. Correct vector identities are, of course, $$\vec{\nabla} \left( \vec{a} \cdot \vec{b} \right) = \left( \vec{\nabla} \vec{a} \right) \cdot \vec{b} + \left( \vec{\nabla} \vec{b} \right) \cdot \vec{a}$$ and $$\vec{\nabla} \cdot \left( f \vec{a} \right) = \left( \vec{\nabla} f \right) \cdot \vec{a} + f \vec{\nabla} \cdot \vec{a},$$ with the second identity being applicable here. Students using Griffiths often have yet to see summation notation. Also, by the statement of the question, the current density vanishes outside of and on the boundary of a volume V. Regards, George 10. Feb 13, 2006 ### Reshma You mean taking one of the orthogonal components at a time? I tried it in post #3. $$\vec \nabla \cdot \left(x\vec J\right)= x(\vec \nabla \cdot \vec J) + \vec J \cdot (\vec \nabla x)$$ $$\vec \nabla \cdot \left(x\vec J\right) - J_x = x\left(\vec \nabla \cdot \vec J\right)$$ Integrating this over a volume V: $$\int_V \vec \nabla \cdot \left(x\vec J\right) d\tau - \int_V J_x d\tau = \int_V x\left(\vec \nabla \cdot \vec J\right) d\tau$$ The first term on the left by divergence theorem will be: [itex]\int_S x\vec J \cdot d\vec a$ which should go to zero as George said the density should vanish at the surface.

So, I'm left with:
$$\int_V \left(\vec \nabla \cdot \vec J\right)x d\tau = -\int_V J_x d\tau$$
Which resembles the result I need. Is my technique correct or am I going wrong somewhere?

Last edited: Feb 13, 2006
11. Feb 13, 2006

### George Jones

Staff Emeritus
This, together with similar expressions for y and z, is exactly the result that you need.

In this qestion, things are zero on the surface of V, but often a similar techique is used when the volume is all of space - integrate by parts, used the divergence theorem, and, as Physics Monkey said, assume things go to zero fast enough at large distances that the surface integral at infinity is zero.

I am very, very sorry that I misled you. I knew just by looking at the question the method that would be needed, but I didn't think carefully about what the "free" variable were, and what the "dummy" variables were.

Regards,
George