- #1
Jimmy Snyder
- 1,127
- 20
This is from page 1 of "Photons and Atoms" by Cohen-Tannoudji et. al.
Given a set of particles of charge [itex]q_{\alpha}[/itex] and position [itex]r_{\alpha}(t)[/itex] the charge density [itex]\rho[/itex] and current j are given by
[tex]\rho(r,t) = \Sigma_{\alpha} q_{\alpha}\delta[r - r_{\alpha}(t)][/tex]
[tex]j(r,t) = \Sigma_{\alpha} q_{\alpha} v_{\alpha} \delta[r - r_{\alpha}(t)][/tex]
Show that:
[tex]\frac{\partial}{\partial t}\rho(r,t) + \nabla \cdot \j(r,t) = 0[/tex]
I don't know how to get started. The only thing I know of that looks like it might help is:
[tex]x\delta'(x) = -\delta(x)[/tex]
But I don't see how to apply it here.
Given a set of particles of charge [itex]q_{\alpha}[/itex] and position [itex]r_{\alpha}(t)[/itex] the charge density [itex]\rho[/itex] and current j are given by
[tex]\rho(r,t) = \Sigma_{\alpha} q_{\alpha}\delta[r - r_{\alpha}(t)][/tex]
[tex]j(r,t) = \Sigma_{\alpha} q_{\alpha} v_{\alpha} \delta[r - r_{\alpha}(t)][/tex]
Show that:
[tex]\frac{\partial}{\partial t}\rho(r,t) + \nabla \cdot \j(r,t) = 0[/tex]
I don't know how to get started. The only thing I know of that looks like it might help is:
[tex]x\delta'(x) = -\delta(x)[/tex]
But I don't see how to apply it here.
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