How Does Charge Configuration Relate to Current Density and Dipole Moment?

[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 <br /> d\tau

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

Now, please help me with this calculus!

 

[George Jones]

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

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

Regards,
George

 

[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.

 

[George Jones]

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

 

[Reshma]

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

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]<br /> <br /> So,<br /> \int_V \vec J d\tau = \frac{d\vec p}{dt}

 

[George Jones]

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

 

[George Jones]

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

 

[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).

 

[George Jones]

George, You've made a mistake here.

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.

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).

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

 

[Reshma]

To make progress I suggest you write the integrand in component notation and then try to do an integration by parts (with the usual assumption that the current density vanishes sufficiently rapidly at infinity).

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:
\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?

 

[George Jones]

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?

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