Question 10.27: Check that the potentials of a point charge moving at consta...

Check that the potentials of a point charge moving at constant velocity (Eqs. 10.49 and 10.50) satisfy the Lorenz gauge condition (Eq. 10.12)

V( r , t)=\frac{1}{4 \pi \epsilon_{0}} \frac{q c}{\sqrt{\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)}}                               (10.49)

A ( r , t)=\frac{\mu_{0}}{4 \pi} \frac{q c v }{\sqrt{\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)}}                                 (10.50)

\nabla \cdot A =-\mu_{0} \epsilon_{0} \frac{\partial V}{\partial t}                                                            (10.12)

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Using Product Rule #5, Eq. 10.50 

\nabla \cdot A =\frac{\mu_{0}}{4 \pi} q c v \cdot \nabla\left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]^{-1 / 2}

 

=\frac{\mu_{0} q c}{4 \pi} v \cdot\left\{-\frac{1}{2}\left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]^{-3 / 2} \nabla \left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]\right\}

 

=-\frac{\mu_{0} q c}{8 \pi}\left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]^{-3 / 2} v \cdot\left\{-2\left(c^{2} t- r \cdot v \right) \nabla ( r \cdot v )+\left(c^{2}-v^{2}\right) \nabla \left(r^{2}\right)\right\}.

Product Rule #4 

\nabla ( r \cdot v )= v \times( \nabla \times r )+( v \cdot \nabla ) r , \text { but } \nabla \times r =0,

( v \cdot \nabla ) r =\left(v_{x} \frac{\partial}{\partial x}+v_{y} \frac{\partial}{\partial y}+v_{z} \frac{\partial}{\partial z}\right)(x \hat{ x }+y \hat{ y }+z \hat{ z })=v_{x} \hat{ x }+v_{y} \hat{ y }+v_{z} \hat{ z }= v , and

\nabla \left(r^{2}\right)= \nabla ( r \cdot r )=2 r \times( \nabla \times r )+2( r \cdot \nabla ) r =2 r . So

\nabla \cdot A =-\frac{\mu_{0} q c}{8 \pi}\left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]^{-3 / 2} v \cdot\left[-2\left(c^{2} t- r \cdot v \right) v +\left(c^{2}-v^{2}\right) 2 r \right]

 

=\frac{\mu_{0} q c}{4 \pi}\left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]^{-3 / 2}\left\{\left(c^{2} t- r \cdot v \right) v^{2}-\left(c^{2}-v^{2}\right)( r \cdot v )\right\}.

But the term in curly brackets is : c^{2} t v^{2}-v^{2}( r \cdot v )-c^{2}( r \cdot v )+v^{2}( r \cdot v )=c^{2}\left(v^{2} t- r \cdot v \right).

=\frac{\mu_{0} q c^{3}}{4 \pi} \frac{\left(v^{2} t- r \cdot v \right)}{\left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]^{3 / 2}}.

Meanwhile, from Eq. 10.49,

-\mu_{0} \epsilon_{0} \frac{\partial V}{\partial t}=-\mu_{0} \epsilon_{0} \frac{1}{4 \pi \epsilon_{0}} q c\left(-\frac{1}{2}\right)\left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]^{-3 / 2} \times

 

\frac{\partial}{\partial t}\left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]

 

=-\frac{\mu_{0} q c}{8 \pi}\left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]^{-3 / 2}\left[2\left(c^{2} t- r \cdot v \right) c^{2}+\left(c^{2}-v^{2}\right)\left(-2 c^{2} t\right)\right]

 

=-\frac{\mu_{0} q c^{3}}{4 \pi} \frac{\left(c^{2} t- r \cdot v -c^{2} t+v^{2} t\right)}{\left[\left(c^{2} t- r \cdot v \right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2} t^{2}\right)\right]^{3 / 2}}=\nabla \cdot A.

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