Help with Problem on Photons and Atoms | Cohen-Tannoudji et al.

[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.

 

[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.

 

[Jimmy Snyder]

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.

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.

 

[George Jones]

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))

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

 

[Jimmy Snyder]

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

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