Help with a problem

1. Sep 20, 2006

Jimmy Snyder

This is from page 1 of "Photons and Atoms" by Cohen-Tannoudji et. al.

Given a set of particles of charge $q_{\alpha}$ and position $r_{\alpha}(t)$ the charge density $\rho$ and current j are given by
$$\rho(r,t) = \Sigma_{\alpha} q_{\alpha}\delta[r - r_{\alpha}(t)]$$
$$j(r,t) = \Sigma_{\alpha} q_{\alpha} v_{\alpha} \delta[r - r_{\alpha}(t)]$$

Show that:

$$\frac{\partial}{\partial t}\rho(r,t) + \nabla \cdot \j(r,t) = 0$$

I don't know how to get started. The only thing I know of that looks like it might help is:

$$x\delta'(x) = -\delta(x)$$

But I don't see how to apply it here.

Last edited: Sep 20, 2006
2. Sep 20, 2006

Galileo

Just use the following (product and chain) rules :

$$\vec \nabla \cdot (f\vec v)=(\vec \nabla f)\vec v+f(\vec \nabla \cdot \vec v)$$

$$\frac{\partial}{\partial t} f(\vec r(t))=(\vec \nabla f)\cdot\frac{\partial \vec r}{\partial t}$$
And it's be a one or two line proof.

Only thing is you get things like $$\vec \nabla \delta(\vec r-\vec r_\alpha(t))$$. I`m not really sure myself what that would be, but just leave it like that and you're fine.

3. Sep 20, 2006

Jimmy Snyder

Thanks for looking at this Galileo. I made a mistake when I posted it. The positions of the particles are a function of time $r_{\alpha}(t)$. I edited the original and I also repeat the corrected formulas here:

$$\rho(r,t) = \Sigma_{\alpha} q_{\alpha}\delta[r - r_{\alpha}(t)]$$
$$j(r,t) = \Sigma_{\alpha} q_{\alpha} v_{\alpha} \delta[r - r_{\alpha}(t)]$$

If I apply the partials to the formulas and use your suggestions, I get the following:

$$\frac{\partial}{\partial t}\rho(r,t) = \Sigma_{\alpha} q_{\alpha} \frac{\partial}{\partial t}\delta[r - r_{\alpha}(t)] = -\Sigma_{\alpha} q_{\alpha} (v_{\alpha} \nabla \delta(r - r_{\alpha}(t)))$$
$$\nabla j(r,t) = \Sigma_{\alpha} q_{\alpha} (\nabla v_{\alpha} \delta(r - r_{\alpha}(t)) + v_{\alpha} \nabla \delta(r - r_{\alpha}(t)))]$$

And so:
$$\frac{\partial}{\partial t}\rho(r,t) + \nabla \cdot \j(r,t) = \Sigma_{\alpha} q_{\alpha} \nabla v_{\alpha} \delta(r - r_{\alpha}(t))$$

I'm not there yet, but this is a great stride forward.

Last edited: Sep 20, 2006
4. Sep 20, 2006

George Jones

Staff Emeritus
Now, why is the right-hand side equal to zero?

5. Sep 20, 2006

Jimmy Snyder

I figured that out on my way home from work. Thanks to Galileo and to George. I remember your help with GR earlier.