- #1
yungman
- 5,718
- 241
This is regarding to derivative of retarded time t_r in static charge distribution vs moving charge distribution.
[tex]t_r=t-\frac{\eta}{c} \;\hbox { where } \;\eta = \vec r - \vec w(t_r) \;\hbox { where } \vec r \;\hbox { is the stationary point where the potential is measured and }[/tex]
[tex] \vec w(t_r) \;\hbox { is the vector point to the source point.}[/tex]
[tex] \vec r = \hat x x + \hat y y + \hat z z \;,\; \vec w(t_r) = \hat x w_x + \hat y w_y + \hat z w_z \;,\; \eta = \sqrt { (x-w_x)^2 + (y-w_y)^2 + (z-w_z)^2}[/tex]
1) In static case [itex] \eta[/itex] is a constant therefore [itex] d\;t_r = d\;t \;\hbox { and }\; \frac {d t_r}{dt}=1[/itex].
2) In moving charge case [itex] \eta [/itex] is not constant because [itex] \vec w(t_r)[/itex] change with time.
[tex]\frac {d\;t_r}{d\;t}= 1-\frac 1 c \frac {d\;\eta}{d\;t}[/tex]
[tex]t_r=t-\frac{\eta}{c} \;\hbox { where } \;\eta = \vec r - \vec w(t_r) \;\hbox { where } \vec r \;\hbox { is the stationary point where the potential is measured and }[/tex]
[tex] \vec w(t_r) \;\hbox { is the vector point to the source point.}[/tex]
[tex] \vec r = \hat x x + \hat y y + \hat z z \;,\; \vec w(t_r) = \hat x w_x + \hat y w_y + \hat z w_z \;,\; \eta = \sqrt { (x-w_x)^2 + (y-w_y)^2 + (z-w_z)^2}[/tex]
1) In static case [itex] \eta[/itex] is a constant therefore [itex] d\;t_r = d\;t \;\hbox { and }\; \frac {d t_r}{dt}=1[/itex].
2) In moving charge case [itex] \eta [/itex] is not constant because [itex] \vec w(t_r)[/itex] change with time.
[tex]\frac {d\;t_r}{d\;t}= 1-\frac 1 c \frac {d\;\eta}{d\;t}[/tex]