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transience
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Homework Statement
I am looking to demonstrate that the expressions for the charge and current density of point charges satisfy the equation of continuity of charge. Intuitively it makes sense to me but I run into trouble with the delta function when I try to prove it mathematically.
Homework Equations
[tex]\rho(\vec{x},t)=\sum_i q_i \delta (\vec{x}-\vec{x}_i(t))[/tex]
[tex]\vec{j}(\vec{x},t)=\sum_i q_i v_i(t) \delta (\vec{x}-\vec{x}_i(t))[/tex]
[tex]\frac{\partial}{\partial t}\rho(\vec{x},t)=- \nabla \cdot \vec{j}(\vec{x},t)[/tex]
[tex]\delta^{3}(\vec{x}-\vec{x}')=\frac{1}{r^2\sin\theta} \delta(r-r') \delta(\theta-\theta')\delta(\phi-\phi')[/tex]
The Attempt at a Solution
I start by constructing a sphere of radius [tex]a[/tex] centered at [tex]\vec{x}=0[/tex]
Using the integral form of the equation of continuity of charge in spherical coordinates I get
[tex]\frac{\partial}{\partial t}\int\int\int_V \rho(\vec{x},t) dV=-\int\int_S \vec{j}(\vec{x},t)dS [/tex]
Then
[tex]LHS=\frac{\partial}{\partial t}\int\int\int_V \rho(\vec{x},t) dV[/tex]
[tex]LHS=\frac{\partial}{\partial t}\int_0^{2 \pi} \int_0^\pi \int_0^a \sum_i q_i \frac{1}{r^2\sin\theta} \delta(r-r_i(t)) \delta(\theta-\theta_i(t))\delta(\phi-\phi_i(t)) r^2\sin{\theta} dr d \theta d \phi[/tex]
Working the maths through gives me
[tex]LHS=\sum_i q_i \frac{\partial}{\partial t}\int_0^a \delta(r-r_i(t)) dr[/tex]
Then I do basically the same thing on the right hand side
[tex]RHS=-\int_0^{2 \pi}\int_0^{\pi} \sum_i q_i v_{ir}(t) \frac{1}{r^2\sin\theta} \delta(r-r_i(t)) \delta(\theta-\theta_i(t))\delta(\phi-\phi_i(t)) a^2\sin{\theta} dr d \theta d \phi[/tex]
Which works out to be
[tex]RHS= \sum_i q_i v_{ir}(t) \frac{a^2}{r^2} \delta(r-r_i(t))[/tex]
I can't figure out where to go from here, any help would be greatly appreciated, thanks.