- #1
yungman
- 5,718
- 241
This is copy from Griffiths Introduction to Electrodynamics page 424.
[tex]V = \frac 1 {4\pi\epsilon_0} \int \frac {\rho_{(\vec r',t_r)}}{ \eta } d \tau \hbox{'} \Rightarrow \; \nabla V = \frac 1 {4\pi\epsilon_0} \int \nabla \left ( \frac {\rho_{(\vec r',t_r)}}{\eta}\right ) d \tau \hbox{'} = \frac 1 {4\pi\epsilon_0} \int \left [ (\nabla \rho) \frac 1 {\eta} + \rho \nabla \left ( \frac 1 { \eta} \right ) \right ] d \tau \hbox{'} [/tex]
Where [tex]\eta = | \vec r - \vec r’| [/tex]
Please help me in deriving the following equation:
1) [tex] \nabla \rho = \dot{\rho} \nabla t_r = -\frac 1 c \dot {\rho} \nabla (\eta) [/tex]
Where [tex]\dot{\rho} = \frac {\partial \rho}{\partial t}[/tex]
2) Also why is [tex]\nabla (\eta) = \hat {(\eta) } [/tex]
Thanks
[tex]V = \frac 1 {4\pi\epsilon_0} \int \frac {\rho_{(\vec r',t_r)}}{ \eta } d \tau \hbox{'} \Rightarrow \; \nabla V = \frac 1 {4\pi\epsilon_0} \int \nabla \left ( \frac {\rho_{(\vec r',t_r)}}{\eta}\right ) d \tau \hbox{'} = \frac 1 {4\pi\epsilon_0} \int \left [ (\nabla \rho) \frac 1 {\eta} + \rho \nabla \left ( \frac 1 { \eta} \right ) \right ] d \tau \hbox{'} [/tex]
Where [tex]\eta = | \vec r - \vec r’| [/tex]
Please help me in deriving the following equation:
1) [tex] \nabla \rho = \dot{\rho} \nabla t_r = -\frac 1 c \dot {\rho} \nabla (\eta) [/tex]
Where [tex]\dot{\rho} = \frac {\partial \rho}{\partial t}[/tex]
2) Also why is [tex]\nabla (\eta) = \hat {(\eta) } [/tex]
Thanks