How Does Retarded Time Change Between Static and Moving Charge Distributions?

[yungman]

This is regarding to derivative of retarded time t_r in static charge distribution vs moving charge distribution.

$$t_r=t-\frac{\eta}{c} \;\hbox { where } \;\eta = \vec r - \vec w(t_r) \;\hbox { where } \vec r \;\hbox { is the stationary point where the potential is measured and }$$

$$\vec w(t_r) \;\hbox { is the vector point to the source point.}$$



$$\vec r = \hat x x + \hat y y + \hat z z \;,\; \vec w(t_r) = \hat x w_x + \hat y w_y + \hat z w_z \;,\; \eta = \sqrt { (x-w_x)^2 + (y-w_y)^2 + (z-w_z)^2}$$

1) In static case $\eta$ is a constant therefore $d\;t_r = d\;t \;\hbox { and }\; \frac {d t_r}{dt}=1$.

2) In moving charge case $\eta$ is not constant because $\vec w(t_r)$ change with time.

$$\frac {d\;t_r}{d\;t}= 1-\frac 1 c \frac {d\;\eta}{d\;t}$$

 

[henry_m]

Everything you have said is fine, and $$\eta$$ should not be constant. The physical interpretation of retarded time should make this clear.

Imagine that we are sitting at a point measuring the electromagnetic field. Each charged particle sends out little signals that move away at the speed of light, and we only know what its doing by these little signals (these little signals are called photons!), which tell us what the field should be. But if the charge is a long way away, we only see the signals from a long time ago as they have taken a finite time to reach us. So we are not measuring the field from what the particle is doing now, but as it was a time $$\eta$$ ago. But if the particle is moving towards us, $$\eta$$ will be getting smaller and smaller, since the signals have less far to travel.

If you know about using the retarded Green's function to solve for the field, this should become clearer. The field at a spacetime point depends only on the configuration of charges and currents on its past light cone, where the signals were sent at just the right time to reach that point.